TEST BANK for Nuclear Systems Volume 1: Thermal Hydraulic Fundamentals, 3rd Edition by Neil E. Todre by StudyGuide

Chapter 1 - Principal Characteristics of Power Reactors

PROBLEM 1.1 QUESTION World Utilization of Power Reactor Technology 1. For each position of Table 1.13 identify: (a) The principal technical reason for which this moderator-coolant combination cannot be exploited, and (b) The name of one (or more) terrestrial power reactor plants that have been built using this combination. TABLE 1.13 Worldwide Utilization of Power Reactor Technology - Thermal Reactor Types Coolant Light water

Heavy water

Organic HB-40 Santowax-OM

Pressurized Boiling Pressurized Boiling

Gas hydrogen, nitrogen, CO₂, helium

Liquid metal NaK Na

Vessel

Heavy water

Tube

Moderators

Light water

Graphite Beryllium

Organic References for thermal reactor types: 1. List of operational nuclear power plants. Nucl. News, August 1992. 2. Dietrich, J. R. and Zinn, W. H. Solid Fuel Reactors, Reading, MA: Addison-Wesley Pub. Co., 1958. 3. Directory of Nuclear Reactors, Vienna: International Atomic Energy Commission, published annually. 4. Kuljian, H. A. Nuclear Power Plant Design, Cranbury, NJ: A.S. Barnes & Co., 1968. 5. Meserve, R., Chairman, Safety Issues at the Defense Production Reactors: A Report to the U.S. Department of Energy, National Academy Press, Washington, DC, 1987. 6. Zinn, W. H., Pittman, F. K. and Hogerton, J. F. Nuclear Power, USA, McGraw-Hill, New York, NY, 1964.

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Chapter 1 - Principal Characteristics of Power Reactors

2. Which of these combinations would be best for submarine propulsion? Explain your choice. 3. State the single most important power cycle parameter affecting the cycle thermal efficiency. 4. Compare the primary side pressures of a PWR, a BWR and a MSR. Describe the reason for primary side pressure of the PWR being about twice the BWR and the advantage of MSR primary side pressure being close to atmospheric. 5. Comparing the core of a BWR with that of a PWR, explain the reason a BWR is considered a closed and a PWR an open core design. 6. Explain the reason that alloys of zirconium are used for fuel cladding. 7. Compare the number of coolant loops and fluids used in a BWR with those of a LMFR describe the reason for the differences, if any. 8. What is the key difference in the VVER design from other LWR designs? 9. How do you classify a LFR with respect to neutron spectrum and thermal efficiency as compared with the SFR? 10. Describe the key difference with respect to core coolant circulation between NuScale and a typical LWR.

PROBLEM 1.1 SOLUTION World Utilization of Power Reactor Technology Question 1 Note:

Descriptions are made with (coolant type) – (moderator). The following notes are keyed to Table 1.13

Pressurized Water Reactor (PWR). Used by: – USA – France – Germany – Former USSR

Boiling Water Reactor (BWR), Used by: – USA – Germany

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– Sweden C

Press. Heavy Water Heavy Water Vessel – AGESTA (Sweden) – R - 3/ ADAM (Sweden) – ATUCHA (Argentina) – MZFR (Germany) – BRUCE 1-8 (Canada) – CP - 5 (USA) – NPD -2 (USA)

Boiling Heavy Water - Heavy Water Vessel – MARVIKEN (Sweden) – HALDEN BWR (Norway) – ATUCHA 2 (Argentina)

Press. Light Water Heavy Water Tube – NRX (Canada) – GENKILLY 2 (Canada) – SGHRW – CIRUS/TROMBAY (India)

Boiling Light Water Heavy Water Tube – CIRENE (Italy) – BLW 250 (Canada) – FUGEN (Japan) – VENUS (Belgium)

Press. Heavy Water - Heavy Water Tube – CANDU (Canada) – PRTR (Canada) – CVTR (Canada) – NPD (Canada)

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Chapter 1 - Principal Characteristics of Power Reactors

– KARLSRUHE (Germany) – PICKERING (Canada) H

Boiling Heavy Water Heavy Water Tube – NPD CONVERSION (Canada)

Organic - Heavy Water Tube – WR-1 – ESSOR – ORGEL – DON – HWOCR

Gas - Heavy Water Tube – BOHUNIZE (Czechoslovakia) – KKN (Germany) – EL-2, EL-4 (France) – LUCENS (Switzerland)

Press. Light Water Graphite – APS OBNINSK (Former USSR) – CHERNOBYL (Ukraine) – HANFORD (USA) – RBMK (Former USSR)

Boiling Light Water - Graphite – SOSNOVY BORINSK (Former USSR) – BELOYARSK (Former USSR) – FIRST NUCLEAR REACTOR OF USSR – N REACTOR (USA)

Gas - Graphite – MAGNOX (Great Britain)

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Chapter 1 - Principal Characteristics of Power Reactors

– DRAGON (Great Britain) – HTGR (EGCR, PEACHBOTTOM, FT. ST. VRAIN, HINKLEY POINT B), (USA) – UTREX (USA) – AVR (Germany) – SIZEWELL NPS (Great Britain) – HECTOR – KIWI – NERVA – USA SPACE PROGRAM N

Liquid Metal - Graphite – HALLAM (USA) – MOLTEN SALT REACTOR EXPERIMENT (ORNL, USA)

Gas - Beryllium – EBOR (USA) – DANIEL’S PILE (ORNL, USA)

Organic - Organic – PIQUA (USA)

Press. Light Water -Beryllium – MIR (Russia) – GE TEST REACTOR (USA) – MR-2 – MIR (Russia)

Boiling Light Water Beryllium – BR 2 (Belgium)

Boiling Light Water Organic – AGN-211 (Switzerland)

Gas - Light Water – ESADA VESR (USA) – MOBILE LOW POWER PLANT (USA) – ML-1 (Idaho Falls, USA) – HEAT TRANSFER REACTOR EXPERIMENT (USAF)

Organic Heavy Water Vessel

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Chapter 1 - Principal Characteristics of Power Reactors

– ECO (Italy) V

Gas - Heavy Water Vessel – HWGCR (Former USSR)

Liquid Metal - Beryllium – SUBMARINE INTERMEDIATE REACTOR (USA)

Press, Light Water Organic – L-77 (Univ, of Nevada, USA)

Answer to Question 1 continued. Comments on moderator-coolants which have been utilized worldwide. 1. There is no advantage gained here by using a light water moderator with a different coolant, because light water is already being used to moderate, it is simpler and more economical to use it as a coolant as well. 2. Liquid sodium will react with water. Using sodium as a coolant would require extensive (and expensive) efforts to separate the two materials (i.e., a tertiary loop). 3. Separation of light and heavy water is not feasible in a vessel-type reactor. If they were both placed in the vessel, the advantages of heavy water would be lost by dilution with light water. 4. In a vessel type reactor you cannot separate the coolant from the moderator. 5. It is unnecessarily expensive to use heavy water as a coolant if it is not being used for its excellent moderating ability. Since graphite is the moderator here it is more economical to use light water as the coolant (even though it has a higher absorption cross section than heavy water.) 6. This is an uneconomical combination. If the organic is being used as the coolant it should be used as the moderator as well. 7. Beryllium is expensive, toxic, brittle, and hard to work with. For a power reactor there are plenty of better choices. 8. Because the organic is the moderator, it is more economical to sue it as the coolant as well. The decomposition of organics under radiation will pose a problem if the organic is not being used as the moderator AND the coolant. If it is not circulating as a coolant its residence time in the core will be too long, leading to decomposition.

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TABLE SM-1.1: Summary Coolant Pressurized Boiling light Pressurized Boiling (right), light water water heavy heavy Moderator water water (down)

Light Water A

Organic Gas Liquid (HB-40, (Hydrogen, Metal Santowax- Nitrogen, (NaK, Na) OM) Carbon dioxide, Helium) 1(USADA 1 2 VERS, USA) 44 (HWGCR, 2,4 USSR)

Graphite

K (APS) L

M (MSRE)

Beryllium

Q (MIR, Russia GE test reactor)

R (BR-2, Belgium)

S (AGN211, Basel, 8 Switzerland)

Heavy water 3 (vessel) Heavy E water (tube)

Organic

Question 2. Which of these combinations would be best for submarine propulsion? Explain your choice. Submarine propulsion reactors are generally designed for military ships. We may identify some characteristics which might be desirable for nuclear military submarines: – – – – – – – –

High reliability Low weight: high power density Low volume Very low toxic release, especially in cruise space Silent operation Tolerance to acceleration (in case of attack) and battle damage Power variation capability Ease of operation

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– Long fuel residence time – Need for both propulsion and electricity generation – Fuel cost is not a big concern In order to ensure reliability under acceleration, boiling water reactors should be avoided, because of the large density difference between the liquid and the vapor phase. Gas-cooled reactors should be avoided because of the large required volume, the risk of leaks and the high gas velocity, which may result into noisy operation. Since there is no limitation to fuel enrichment, there is no particular need for special moderating materials such as heavy water, graphite or organic. Therefore, pressurized light water-cooled and moderated reactors are good options. Light water has several advantages, among which are: – – – –

Small slowing down length, which allows for a compact core. Can be used in both the primary and secondary systems, simplifying the design. Neutron activation products have a short half-life. Can be easily distilled from seawater to provide makeup and safety function inventory.

Alternatively, liquid metal-cooled reactors may be used if corrosion is under control. If liquid metal coolants are used, electromagnetic pumps may be installed which allow more silent operation. However, sodium coolants have the drawback of reacting violently with water, producing explosive/flammable hydrogen. Question 3. State the single most important power cycle parameter affecting the cycle thermal efficiency. Coolant Outlet Temperature—it dictates reactor mission capability e.g. process heat and cycle thermal efficiency which affects capital cost. Also importantly it affects coolant corrosion performance. Question 4. Compare the primary side pressures of a PWR, a BWR and a MSR. Describe the reason for primary side pressure of the PWR being about twice the BWR and the advantage of MSR primary side pressure being close to atmospheric. PWR - 15.5MPa, BWR - 7.17 MPa, MSR - nominal 1 MPa PWR pressure is high to keep the primary coolant in the subcooled regime while still achieving high operating temperatures MSR nominal atmospheric primary operating pressure keeps stored energy in the salt coolant low which is a beneficial safety factor as well as allowing the primary coolant system pressure containment e.g. piping to be thin walled. Question 5. Comparing the core of a BWR with that of a PWR, explain the reason a BWR is considered a closed and a PWR an open core design. The BWR operates with a boiling mixture from about one third the distance from the core flow inlet. Particularly in the boiling region different axial pressure levels in adjacent coolant channels

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develop due to the core radial neutron flux shape which in turn cause radial pressure gradients to develop between coolant channels. These radial pressure gradients in turn cause radial flows to develop between coolant channels which are countered in BWRs by use of smaller radial sized bundles than in PWRs (a closed core design) . Also since the PWR does not operate in the boiling region, radial pressure gradients of BWR magnitude do not develop so that PWR fuel assemblies do not have bounding walls (an open core design). Question 6. Explain the reason that alloys of zirconium are used for fuel cladding. Zirconium has low neutron capture cross section and sufficient strength and corrosion resistance at water cooled reactor operating temperatures Question 7. Compare the number of coolant loops and fluids used in a BWR with those of a LMFR - describe the reason for the differences, if any. The BWR uses a direct cycle, hence a single water coolant loop while the LMFR uses a threecoolant loop system. Both use the steam cycle. The LMFR loops are the primary sodium loop, the intermediate loop between the intermediate heat exchange and the steam generator typically to date also using sodium coolant and finally a water coolant in which steam is produced which flows to the turbine in this third loop. Question 8. What is the key difference in the VVER design from other LWR designs? The VVER uses hexagonal shaped fuel assemblies and horizontal steam generators. Question 9. How do you classify a LFR with respect to neutron spectrum and thermal efficiency as compared with the SFR? Both are fast neutron spectrum reactors. The primary coolant outlet temperatures are respectively 550 degrees Centigrade with thermal efficiencies of 43-44 %. Question 10. Describe the key difference with respect to core coolant circulation between NuScale and a typical LWR. The Nuscale design is natural circulation using a helical coil steam generator operating in a oncethru producing superheated steam. A module produces 60 MWe with a plant composed of 12 modules. The typical LWR ( let’s take the PWR) is a forced circulation primary loop operating with a U tube pot type steam generator producing saturated steam (Westinghouse) or a steam generator of the once-thru design producing superheated steam (the Areva design).

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Chapter 2 Thermal Design Principles and Application Contents Problem 2.1 Relations among fuel element thermal properties in various power reactors ...... 16 Problem 2.2 Relationships between assemblies of different pin arrays ................................... 21 Problem 2.3 Minimum CHF ratio in a PWR for a flow coastdown transient ......................... 22 Problem 2.4 Minimum CPR in a BWR .................................................................................... 25 Problem 2.5 Primary cooling system pumping power for a PWR reactor .............................. 26 Problem 2.6 Relations among thermal design conditions in a PWR ....................................... 28

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PROBLEM 2.1 QUESTION Relations Among Fuel Element Thermal Parameters in Various Power Reactors (Section 2.3) Compute the core average values of the volumetric energy-generation rate in the fuel (q‴) and outside surface heat flux (q″) of cladding for the reactor types BWR, PWR, PHWR, HTGR, AGR and SFBR. Use the core average linear power levels derivable from data in Table 2.1 and the geometric parameters in Table 1.3. Answers (for BWR): –

q‴ = 243.2 MW/m3

–

″ 𝑞𝑞co = 500.2 kW/m2

TABLE 2.1 Typical Core Thermal Performance Characteristics for Six Reference Power Reactor Types Characteristics

BWR

PWR(W)

PHWRa

HTGR

AGR

SFBRb

Core Axis

Vertical

Horizontal

Vertical

Axial

Radial

764

193

380

493

332

364 (C) 233 (BR)

Assembly pitch (mm)

152

214

286

361

460

179

Active fuel height (m)

3.588

3.658

5.94

6.30

8.296

1.0 (C) 1.6 (C + BA)

Equivalent diameter (m)

4.75

3.37

6.29

8.41

9.458

3.66

Total fuel weight (MT)

160 UO2

101 UO2

89.3 UO2

1.56 U 34.0 Th

103.0 UO2

29 UO2

No. of assemblies

Reactor vessel Inside dimensions (m)

6.05D × 21.6H

4.83D × 13.4H 7.6D × 4L (5.94L for EC 6)

11.3D × 14.4H

20.25D × 21.87H 21D × 19.5H

Wall thickness (mm)

152

224

28.6

4720

5800

Materialc

SS-clad carbon steel

Stainless steel

Prestressed concrete

Concrete helical prestressed

Stainless steel

Pressure tubes

Steel liners

Steel lined

Pool type

Other features Power density core average (kW/L)

52.3

104.5

8.4

2.66

280

17.6

17.86

25.7

7.87

17.0

Linear heat rate Core average (kW/m)

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TABLE 2.1 Typical Core Thermal Performance Characteristics for Six Reference Power Reactor Types Characteristics

BWR

PWR(W)

PHWRa

Core maximum (kW/m)

47.24

44.62

~54d

Equilibrium burnup (MWd/MT)

50,000

Average assembly residence (full-power days)

2192

Sequence Outage time (days)

HTGR

AGR

SFBRb

23.0

29.8

8300

105,000

20,000

110,000

1644

470

1170

1320

640 (C) 320 (BR row 1) 640 (BR row 2)

1/4 per year

1/3 per year

Continuous online

1/2 per year

Continuous online

Variable

None for refueling

Unavailable

None for refueling

Performance

Refueling

Same for both CANDU 6 and Enhanced CANDU 6 (EC 6), except Reactor Vessel Inside Dimensions

SFBR: core (C), radial blanket (BR), axial blanket (BA)

SS = Stainless Steel

Assuming a 900 kW bundle, 37 elements and radial pin peaking factor of 1.12

Source: Principally Appendix K and (Knief, R. A. Nuclear Engineering: Theory and Technology of Commercial Nuclear Power, 2nd Ed. La Grange Park, IL: American Nuclear Society, 2008.) except for AGR data from M. A. H. G. Alderson (pers. comm., October 6, 1983) and A. A. Debenham (pers. comm., August 5, 1988) and PHWR data from CANDU 6 Technical Summary, CANDU 6 Program Team, Reactor Development Business Unit, May 2005.

Table 1.3 Typical Characteristics of the Fuel for Six Reference Power Reactor Types Characteristics

BWR

PWR

PHWR

HTGR

AGR

SFBR

General Atomic

National Nuclear Corp.

Novatome

(Fulton)

(Heysham 2)

(Superphenix 1)

D2O / D2O

Graphite

None

Reference Design Manufacturer

General Electric

Westinghouse

System (reactor station)

BWR/5 (NMP2)

(Seabrook)

Moderator

H2O

Neutron energy

Thermal

Thermal / Thermal

Thermal

Fast

Fuel production

Converter

Converter / Converter

Converter

Breeder

Atomic Energy of Canada, Ltd. CANDU 6

Enhanced CANDU-6 (EC 6)

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Table 1.3 Typical Characteristics of the Fuel for Six Reference Power Reactor Types BWR

Characteristics

PWR

PHWR

HTGR

AGR

SFBR

Fuelb Geometry

Cylindrical pellet

Cylindrical pellet / Cylindrical pellet

Microspheresc

Cylindrical pellet

Dimensions (mm)

9.60D × 10.0L

8.192D × 9.8L

12.2D × 16.4L / 12.2D × 16.4L

400–800 μm D

14.51D ×14.51L

7.14 D

Chemical form

UO2

UC/ ThO2

UO2

PuO2 / UO2

Fissile (first core avg. wt% unless designated as equilibrium core) Fertile

U (3.5 eq. core)

235

238

U (3.57 avg. eq. core)

235

238

UO2 / UO2

U (2 zones at 2.1 and 2.7)

235

U / 238U

238

U (93)

235

U (0.711) / 235U (0.711)

235

Pu (2 zones at 16 and 19.7)

239

Depleted U

Fuel Rodsb Geometry

Pellet stack Pellet stack in in clad tube clad tube

Pellet stack in clad tube

Cylindrical fuel compacts

Pellet stack in clad Pellet stack in clad tube tube

Dimensions

11.20 mm D 9.5 mm D × 3.588 m H ×3.658 m H (× 4.09 m L) (×3.876 m L)

13.1 mm D × 0.493 m L

15.7 mm D × ~0.742 m H (× 0.793 m L)

15.3 mm D × 0.987 8.5 mm D × 2.7 m m H (×1.04 m L) H(C)d 15.8 mm D × 1.94 m H(RB)d

Clad materiale

Zircaloy-2

Zirlo™

Zircaloy-4 / Zircaloy-4

No clad

Stainless steel

Clad thickness (mm)

0.71

0.572

0.42 / 0.40

Not applicable

0.37

0.56

Fuel Assembly Geometry

9 × 9 square rod arraya

17 × 17 square rod array

Rod pitch (mm)

14.37

12.60

14.6 / 14.6

289

No. rod locations 81

Concentric circles of rods

Cylindrical fuel Concentric circles compacts of rods within a hex. surrounded by graphite blockf graphite sleeve

Hexagonal rod array

25.7

9.8 (C)/17.0 (RB)

37 / 37

132 (SA)/76 (CA)g

271 (C)/91 (RB)

37 / 37

132 (SA)/76 (CA)g

271 (C)/91 (RB)

102D × 495L / 102D × 495L

359F × 793L

190.4 (sleeve inner 173F (inner) × 5.4 D) mL

No. fuel rods

74 (8 part length)

264

Outer dimensions (mm)

139 × 139

214 × 214

Channel

Yes

No / No

Yes

~640

~25 / 23.7

—

395

—

Total weight (kg) 273

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Table 1.3 Typical Characteristics of the Fuel for Six Reference Power Reactor Types Characteristics

BWR

PWR

PHWR

HTGR

AGR

SFBR

The most recent General Electric BWR fuel bundle designs, which have 9 × 9 and 10 × 10 rod arrays, have been introduced but data for many parameters are held proprietary. The table presents available data for a typical 9 × 9 GE11 design. Unique 11 × 11 and 12 × 12 rod arrays were employed at the early Big Rock Point BWR plant. b Fuel and fuel rod dimensions: diameter (D), heated length (H), total length (L), (across the) flats (F). c Blends of fuel microspheres are molded to form fuel cylinders each having a diameter of 15.7 mm and a length of 5–6 cm d SFBR-core (C), SFBR-radial blanket (RB). e Zircaloy and Zirlo are both trademarked alloys of zirconium. Zirlo stands for Zirconium low oxidation. f Early HTGRs (the German AVR and THTR) employed pebble fuel rather than prismatic graphite blocks. g HTGR-standard assembly (SA), HTGR-control assembly (CA). a

Source: BWR and PWR - Data from Appendix K. The BWR/5 is the Nine Mile Point 2 unit (NMP2) and the PWR is the four-loop Seabrook unit. PHWR - Adopted from Knief, R. A. Nuclear Engineering: Theory and Technology of Commercial Nuclear Power, pp. 707–717. American Nuclear Society, La Grange Park, IL, 2008. CANDU 6 Technical Summary, CANDU 6 Program Team, Reactor Development Business Unit, May 2005. HTGR - Breher, W., Neyland, A. and Shenoy, A. Modular High-Temperature Gas-Cooled Reactor (MHTGR) Status. GA Technologies, GAA18878, May 1987. AGR - Adopted from AEAT/R/PSEG/0405 Issue 3. Main Characteristics of Nuclear Power Plants in the European Union and Candidate Countries. Report for the European Commission, September 2001; Nuclear Engineering International. Supplement. August 1982. Alderson, M. A. H. G. (UKAEA, pers. comm., October 6, 1983 and December 6, 1983); Getting the most out of the AGRs, Nucl. Eng. Int., 28(358):8–11, August 1984. SFBR - Adopted from IAEA-TECDOC-1531. Fast Reactor Database 2006 Update. International Atomic Energy Agency. December 2006.

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PROBLEM 2.1 SOLUTION Relations Among Fuel Element Thermal Parameters in Various Power Reactors (Section 2.3) Compute the core average values of the volumetric energy-generation rate in the fuel (q‴) and outside surface heat flux (q″) of cladding for the reactor types BWR, PWR, PHWR, HTGR, AGR and SFBR. Use the core average linear power levels derivable from data in Table 2.1 and the geometric parameters in Table 1.3. The volumetric energy generation rate is calculated using 4𝑞𝑞 ′ 𝑞𝑞 = 2 𝜋𝜋𝐷𝐷fo ‴

(1)

and outside surface heat of the cladding is calculated with ″ 𝑞𝑞co =

𝑞𝑞 ′ 𝜋𝜋𝐷𝐷co

(2)

where Dfo = Diameter of fuel pellet, Dco = Diameter of fuel rod, q′ = Linear heat generation rate, ″ 𝑞𝑞co = Heat flux at cladding surface and q‴ = Volumetric heat generation rate.

Table SM-2.1 shows the data results for the different reactor types:

TABLE SM-2.1: Data and Results for Various Power Reactors BWR

PWR

PHWR

HTGR

AGR

SFBR

Dco [mm]

11.20

9.5

13.1

15.7

15.3

8.5

Dfo [mm]

9.6

8.192

12.2

15.7*

14.51

7.14

500.2

598.4

624.5

159.6

353.7

1086.0

243.2

338.9

219.8

40.7

102.8

724.3

″ 𝑞𝑞co �

𝑞𝑞 ‴ �

𝑘𝑘𝑘𝑘 � 𝑚𝑚2

𝑀𝑀𝑀𝑀 � 𝑚𝑚2

*Note: Fuel is coated with graphite and the coating thickness can be neglected.

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PROBLEM 2.2 QUESTION Relationships Between Assemblies of Different Pin Arrays (Section 2.3) A utility wishes to replace the fuel in its existing PWR from 15×15 fuel pin array assemblies to 17×17 fuel pin array assemblies. What is the ratio of the core average linear power, q′ in the new core to the old core, assuming reactor power, length, number of fuel assemblies and fuel mass are maintained constant. Repeat for the core average heat flux, q″. 𝑞𝑞

′

Answers: 17𝑥𝑥17 = 0.779 ′ 𝑞𝑞15𝑥𝑥15 ″

𝑞𝑞17𝑥𝑥17 ″

𝑞𝑞15𝑥𝑥15

= 0.882

PROBLEM 2.2 SOLUTION

Relationships Between Assemblies of Different Pin Arrays (Section 2.3) A utility wishes to replace the fuel in its existing PWR from 15×15 fuel pin array assemblies to 17×17 fuel pin array assemblies. What is the ratio of the core average linear power and core average heat flux, q′ and q″, in the new core to the old core, assuming reactor power, length, number of fuel assemblies and fuel mass are maintained constant. The core average power is defined as 𝑄𝑄 = 𝑁𝑁𝑁𝑁𝑞𝑞 ′

(1)

where Q is the core average power, N is the number of fuel pins, L is the length of the core and q′ is the core average linear power. If we keep the core power and length constant, we can determine the ratio of the linear power from new core to old core as 𝑄𝑄 𝑁𝑁17𝑥𝑥17 𝐿𝐿 𝑁𝑁15𝑥𝑥15 152 = = = = 0.779 ′ 𝑄𝑄 𝑁𝑁17𝑥𝑥17 172 𝑞𝑞15𝑥𝑥15 𝑁𝑁15𝑥𝑥15 𝐿𝐿 ′

𝑞𝑞17𝑥𝑥17

(2)

The heat flux is defined as

𝑞𝑞 ″ =

𝑞𝑞 ′ 𝜋𝜋𝜋𝜋

(3)

where q″ is the core average heat flux and D is the diameter of the fuel elements. We would like to not change the fuel mass in the core and therefore the diameter of the fuel elements must change when moving to the new core. To conserve mass or in this case just simply the area of fuel, we apply the following condition:

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Chapter 2 - Thermal Design Principles and Application

𝑁𝑁17𝑥𝑥17 =

𝜋𝜋 2 𝜋𝜋 2 𝐷𝐷17𝑥𝑥17 = 𝑁𝑁15𝑥𝑥15 = 𝐷𝐷15𝑥𝑥15 4 4

(4)

We can then determine the ratio of the old fuel element diameter to new element diameter as 𝐷𝐷15𝑥𝑥15 𝑁𝑁17𝑥𝑥17 =� 𝐷𝐷17𝑥𝑥17 𝑁𝑁15𝑥𝑥15

(5)

The ratio of the new to old core average heat flux can be determined by applying Equation (2) and the result from Equation (5): ′ 𝑞𝑞17𝑥𝑥17 ″ ′ 𝑞𝑞17𝑥𝑥17 𝜋𝜋𝐷𝐷17𝑥𝑥17 𝑞𝑞17𝑥𝑥17 𝐷𝐷15𝑥𝑥15 𝑁𝑁15𝑥𝑥15 𝑁𝑁17𝑥𝑥17 152 172 � � = ′ = ′ = = = 0.882 ″ 𝑞𝑞15𝑥𝑥15 𝑞𝑞15𝑥𝑥15 𝑞𝑞15𝑥𝑥15 𝐷𝐷17𝑥𝑥17 𝑁𝑁17𝑥𝑥17 𝑁𝑁15𝑥𝑥15 172 152

𝜋𝜋𝐷𝐷15𝑥𝑥15

(6)

PROBLEM 2.3 QUESTION

Minimum Critical Heat Flux Ratio in a PWR for a Flow Coastdown Transient (Section 2.4) Describe how you would determine the minimum critical heat flux ratio versus time for a flow coastdown transient by drawing the relevant channel operating curves and the CHF limit curves for several time values. Draw your sketches in relative proportion and be sure to state all assumptions.

PROBLEM 2.3 SOLUTION Minimum Critical Heat Flux Ratio (MCHFR) in a PWR for a Flow Coastdown Transient (Section 2.4) Describe how you would determine the minimum critical heat flux ratio versus time for a flow coastdown transient by drawing the relevant channel operating curves and the CHF limit curves for several time values. Draw your sketches in relative proportion and be sure to state all assumptions. For this transient, we can have two different situations depending on whether the reactor will be shut down or not. In a flow coastdown transient, the mass flow through the reactor follows the trends of Figure SM-2.1.

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Chapter 2 - Thermal Design Principles and Application

FIGURE SM-2.1: Mass flow rate behavior through reactor

As mentioned, there are two possibilities for the reactor power as stated below and shown in Figure SM-2.2: 1. Curve 1 - the reactor is shut down. 2. Curve 2 - the reactor is not shut down and the behavior of the power will be a function of reactivity feedback effects.

FIGURE SM-2.2: Reactor power behavior

There are two situations for change in operating heat flux with time. In both cases the critical heat flux decreases as the flow decreases. It is assumed that the inlet quality is constant. Condition 1 – operating heat flux decreases with time

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Chapter 2 - Thermal Design Principles and Application

Figure SM-2.3: Condition 1, Heat flux as a function of quality

At each instant we have a curve for the critical heat flux and for the maximum operating heat flux in the core. The CHF curve decreases with both decreasing heat flux and mass flux. The operating curve as shown in Figure SM-2.3 collapses primarily due to decrease in heat flux. The decrease in mass flux would cause the quality at any given position to increase if the heat flux were constant. Therefore, we can calculate the MCHFR at any time, t. The occurrence of CHF will depend on how the power and mass flux vary. Condition 2 – operating heat flux maximum is constant with time

Figure SM-2.4: Condition 2, Heat flux as a function of quality In this case, we can assume that the maximum operating heat flux does not vary with time.

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PROBLEM 2.4 QUESTION Minimum Critical Power Ratio in a RWR (Section 2.4) Calculate the minimum critical power ratio for a typical 1062 MWe BWR operating at 100% power using the data in Appendix K. Assume that: 1. The axial linear power shape can be expressed as ′ 𝑞𝑞 ′ (𝑧𝑧) = 𝑞𝑞ref exp �−

𝛼𝛼𝛼𝛼 𝜋𝜋𝜋𝜋 � sin � � 𝐿𝐿 𝐿𝐿

′ ′ where α = 1.96. Determine 𝑞𝑞ref such that 𝑞𝑞max = 47.24 kW⁄m.

2. The critical bundle power is 9319 kW. Answers: ′ 𝑞𝑞ref = 104.75 kW⁄m

MCPR = 1.28

PROBLEM 2.4 SOLUTION Minimum Critical Power Ratio in a BWR (Section 2.4) Calculate the minimum critical power ratio for a typical 1062 MWe BWR operating at 100% power using the data in Appendix K. The given parameters in the problem are listed below: –

α = 1.96

–

L = 3.588 m

–

′ 𝑞𝑞max = 47.24 m

–

Nrods = 74

–

qcr = 9319 kW

The linear heat rate profile is given as ′ 𝑞𝑞 ′ (𝑧𝑧) = 𝑞𝑞ref exp �−

𝛼𝛼𝛼𝛼 𝜋𝜋𝜋𝜋 � sin � � 𝐿𝐿 𝐿𝐿

(1)

′ ′ To begin, we must obtain the value of 𝑞𝑞ref such that the maximum linear heat rate is 𝑞𝑞max . Note ′ ′ that for a standard cosine profile, 𝑞𝑞max = 𝑞𝑞ref . To determine the axial location that yields the maximum linear heat rate, we can take the derivative of Equation (1) and solve for the roots,

𝑑𝑑𝑑𝑑 ′ (𝑧𝑧) = 0 ⟼ 𝑧𝑧max = 1.157 m 𝑑𝑑𝑑𝑑 25

(2)

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Chapter 2 - Thermal Design Principles and Application ′ can be solved for, noting that 𝑞𝑞 ′ (𝑧𝑧max ) = Substituting 𝑧𝑧max into Equation 1, the parameter 𝑞𝑞ref ′ 𝑞𝑞max . This results in ′ 𝑞𝑞ref = 104.75 kw⁄m

(3)

The linear heat rate (displayed in W/m) can be plotted as a function of axial location as shown in Figure SM-2.5, to verify the maximum linear heat rate condition of about 4.7×104 W/m.

FIGURE SM-2.5: Axial Profile of the Linear Heat Rate The power of the bundle, assuming all fuel rods have the same linear profile (no local peaking), 𝐿𝐿

(4)

𝑞𝑞 = 𝑁𝑁rods � 𝑞𝑞 ′(z) 𝑑𝑑𝑑𝑑 = 7270.4 kw 0

Then the Minimum Critical Power Ratio (MCPR) is MCPR =

𝑞𝑞cr = 1.28 𝑞𝑞

(5)

PROBLEM 2.5 QUESTION Pumping Power for a PWR Reactor Cooling System (Section 2.5) Calculate the pumping power under steady-state operating conditions for a typical PWR reactor coolant system using only the following operating conditions: –

Core thermal power, Q = 3411 MWth

–

Core temperature drop, ∆Tcore = 33.7 °C

–

Inlet temperature, Tin = 293 °C

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Chapter 2 - Thermal Design Principles and Application

–

Core system pressure, P = 15.5 MPa

–

Reactor coolant system pressure drop, ∆P = 778 kPa

–

Pump efficiency, ηp = 0.85

Answer: Pumping power = 21.8 MWe

PROBLEM 2.5 SOLUTION Pumping Power for a PWR Reactor Cooling System (Section 2.5) Calculate the pumping power under steady-state operating conditions for a typical PWR reactor coolant system using only the following operating conditions: –

Core thermal power, 𝑄𝑄̇ = 3411 MWth

–

Inlet temperature, Tin = 293 °C

–

Core system pressure, P = 15.5 MPa

–

Reactor coolant system pressure drop, ∆P = 778 kPa

–

Pump efficiency, ηp = 0.85

–

Core temperature drop, ∆Tcore = 33.7 °C

The pumping power can be calculated using Equation 2.9a in the text which applies under ideal conditions 𝑊𝑊̇𝑝𝑝 =

∆𝑃𝑃𝑚𝑚̇ = 𝛥𝛥𝛥𝛥𝛥𝛥𝑓𝑓 𝜐𝜐 𝜌𝜌

(1)

where Wp is the pumping power, Af is the flow area and υ is the velocity. We can perform a heat balance across the core to relate the core thermal power to the temperature drop with 𝑄𝑄̇ = 𝑚𝑚̇𝑐𝑐p Δ𝑇𝑇core = 𝜌𝜌𝜌𝜌f 𝜐𝜐𝜐𝜐p Δ𝑇𝑇core

(2)

To calculate the mass flow rate correctly and to relate the flow area and velocity correctly, we must determine the density of the fluid at inlet conditions, P = 15.5 MPa and T = 293 °C. This results in a density of ρ = 740.49.kg/m3. On the other hand the specific heat at constant pressure varies with temperature as the fluid is heated across the core, We may just take an average of this parameter, evaluating it at the core system pressure and the average coolant temperature, 𝑇𝑇core = 309.85 °𝐶𝐶. This result in an average specific heat of 𝑐𝑐p = 5736.03 J⁄kg − K. Combing Equations 1 and 2, solving for the ideal pumping power, we obtain the following expression and solution:

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Chapter 2 - Thermal Design Principles and Application

𝑊𝑊̇𝑝𝑝 = ∆𝑃𝑃

𝑄𝑄̇ = 18.5 MW 𝜌𝜌𝑐𝑐p ∆𝑇𝑇core

(3)

Taking into account the efficiency of the pump, we may divide by 0.85 to obtain the actual pumping power of 21.8 MWe.

PROBLEM 2.6 QUESTION Relations Among Thermal Design Conditions in a PWR (Section 2.5) Compute the margin to failure defined as the ratio of linear power rate at failure to the maximum ′ ′ linear power ratio 𝑞𝑞fail /𝑞𝑞max for a typical PWR having a core average linear power rate of 17.86 kW/m. Assume that the failure limit is established by centerline melting of the fuel at 70 kW/m. Use the following multiplication factors: –

Radial flux factor = 1.55

–

Axial and local flux factor = 1.70

–

Engineering uncertainty factor = 1.05

–

Overpower factor = 1.15

Answer: Margin = 1.23

PROBLEM 2.6 SOLUTION Relations Among Thermal Design Conditions in a PWR (Section 2.5) Compute the margin for a typical PWR having the following parameters and multiplication factors: –

Radial flux factor, Fr = 1.55

–

Axial and local flux factor, Fz = 1.70

–

Engineering uncertainty factor, Fe = 1.05

–

Overpower factor, Fop = 1.15

The average linear power rate and the failure limit, established by centerline melting of the fuel are: – –

′ Average linear power rate, 𝑞𝑞ave = 17.86 kW⁄m ′ = 70 kW⁄m Failure limit, 𝑞𝑞fail

The maximum linear power rate is

′ ′ 𝑞𝑞max = 𝑞𝑞ave 𝐹𝐹r 𝐹𝐹z 𝐹𝐹e 𝐹𝐹op = 56.83 kW⁄m

(1)

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The margin to failure is calculated as ′ 𝑞𝑞fail Margin = ′ = 1.23 𝑞𝑞max

(2)

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Chapter 3 Reactor Energy Distribution Contents Problem 3.1 Thermal Design Parameters For a Cylindrical Fuel Pin .....................................

Problem 3.2 Power Profile in a Homogeneous Reactor ..........................................................

Problem 3.3 Power Generation in Thermal Shield ..................................................................

Problem 3.4 Decay Heat Energy .............................................................................................

Problem 3.5 Decay Heat From a PWR Fuel Rod ....................................................................

Problem 3.6 Decay Power Calculations of a 3 Batch PWR Core ...........................................

Problem 3.7 Effect of Continuous Refueling on Decay Heat .................................................

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PROBLEM 3.1 QUESTION Thermal Design Parameters for a Cylindrical Fuel Pin (Section 3.4) Consider the PWR reactor of Example 3.1. 1. Evaluate the average neutron flux if the enrichment of the fuel is 3.25 wt %. 2. Evaluate the volumetric heat generation rate in the fuel in MW/m3. Assume the fuel density is 90% of theoretical density. 3. Calculate the average linear power of the fuel, assuming there are 264 fuel rods per assembly. Assume the fuel rod length to be 3658 mm. 4. Calculate the average heat flux at the cladding outer radius, when the cladding diameter is 9.5 mm. Answers: 1. 〈𝜙𝜙〉 = 3.7 × 1014

neutrons

2. 〈𝑞𝑞 ′′′ 〉 = 300.5 m3

cm2 s

3. 〈𝑞𝑞′〉 = 17.61 m

4. 〈𝑞𝑞″〉 = 590.0 m2

PROBLEM 3.1 SOLUTION

Thermal Design Parameters for a Cylindrical Fuel Pin (Section 3.4) Consider the PWR reactor of Example 3.1. 1. Evaluate the average neutron flux if the enrichment of the fuel is 3.25%. The molar mass of U can be calculated with −1

𝑟𝑟 1 − 𝑟𝑟 −1 0.0325 1 − 0.0325 � =� + + 𝑀𝑀𝑈𝑈 = � g g � 𝑀𝑀25 𝑀𝑀28 235.0439 238.0508 mol mol

= 237.952

g mol

The molar mass of UO2 and the weight percent of U in UO2 can be calculated as 𝑀𝑀𝑈𝑈𝑂𝑂2 𝑀𝑀𝑈𝑈 + 2𝑀𝑀𝑂𝑂 = 237.952

g g g + 15.9994 = 269.951 mol mol mol

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g 237.952 𝑀𝑀𝑈𝑈 mol 𝜔𝜔𝑈𝑈 = = = 0.881 𝑀𝑀𝑈𝑈𝑈𝑈2 269.951 g mol

The number of atoms of U-235 in the core can be calculated using parameters in Example 3.1 including the mass of UOX per assembly, 558.5 kg and the number of assemblies, 193. The total mass of UOX is the mass of U-235 is

𝑚𝑚𝑈𝑈𝑈𝑈2 = (193)(558.5 kg) = 1.078 × 108 g

𝑚𝑚25 = 𝜔𝜔𝜔𝜔𝜔𝜔𝑈𝑈𝑈𝑈2 = (0.881)(0.0325)(0.178 × 108 g) = 3.088 × 106 g

and the number of U-235 atoms

23 atoms

𝑚𝑚25 𝑁𝑁𝐴𝐴 (3.088 × 10 g) �6.022 × 10 𝑁𝑁25 = = g 𝑀𝑀25 235.0439 mol

mol

�

The average neutron flux can be calculated with Equation 3.15d, 〈𝜙𝜙〉 =

γ𝑄𝑄̇𝑡𝑡ℎ = 𝜒𝜒𝑓𝑓 𝜎𝜎 𝑓𝑓25 𝑁𝑁25 �3.2 × 10−11

= 7.912 × 1027 atoms

0.962(3411 MWt)

J � (35 × 10−24 cm2 )(7.912 × 1027 atoms) fission

Using 0.974 as the energy deposition fraction yields 3.75 × 1014

neutrons cm2 s

= 3.7 × 1014

neutrons cm2 s

. Here we use a

microscopic fission cross section of 35 b for U-235. This is a 1 group spectrum-averaged cross section. If a two-group approach is taken, the thermally-averaged microscopic fission cross section will be much higher. 2. Evaluate the volumetric heat generation rate in the fuel in MW/m3. Assume the fuel density is 90% of theoretical density (l0.97g/cm3). The fuel density and volume is calculated as 𝜌𝜌𝑈𝑈𝑈𝑈2 = 0.90(10.97 g⁄cm3 ) = 9.87 g⁄cm3

𝑉𝑉𝑈𝑈𝑈𝑈2 =

𝑚𝑚𝑈𝑈𝑈𝑈 2 1.078 × 108 g = = 1.092 × 107 cm3 9.87 g⁄cm3 𝜌𝜌𝑈𝑈𝑈𝑈2

The average power density is calculated with 〈𝑞𝑞 ′′′ 〉 =

γ𝑄𝑄̇𝑡𝑡ℎ 0.962(3411 MWt) MW = = 300.5 𝑉𝑉𝑈𝑈𝑈𝑈2 1.092 × 107 cm3 m3

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Chapter 3 - Reactor Energy Distribution MW

Using 0.974 as the energy deposition fraction yields 304.3 m3

3. Calculate the average linear power of the fuel, assuming there are 264 fuel rods per assembly. Assume the fuel rod length to be 3658 mm. The average linear power is 〈𝑞𝑞′〉 =

̇ γ𝑄𝑄𝑡𝑡ℎ 0.962(3411 MWt) kW = = 17.61 m 𝑛𝑛𝑎𝑎𝑎𝑎 𝑛𝑛𝑝𝑝𝑝𝑝𝑝𝑝 𝐿𝐿 (193)(264)(3.658 m) kW

Using 0.974 as the energy deposition fraction yields 17.83 m

4. Calculate the average heat flux at the cladding outer radius, when the cladding diameter is 9.5 mm. The average heat flux at the cladding outer radius is kW 16.95 m 〈𝑞𝑞′〉 kW 〈𝑞𝑞″〉 = = = 589.9 2 𝜋𝜋𝜋𝜋𝑐𝑐𝑐𝑐 𝜋𝜋(9.5 mm) m kW

Using 0.974 as the energy deposition fraction yields 597.3 m2

PROBLEM 3.2 QUESTION

Power Profile in a Homogeneous Reactor (Section 3.5) Consider an ideal core with the following characteristics: The U-235 enrichment is uniform throughout the core, and the flux distribution is characteristic of an unreflected, uniformly fueled cylindrical reactor, with extrapolation distances δz and δR of 10 cm. How closely do these assumptions allow prediction of the following characteristics of a PWR? 1. Ratio of peak to average power density and heat flux? 2. Maximum heat flux? 3. Maximum linear heat generation rate of the fuel rod? 4. Peak to average enthalpy rise ratio, assuming equal coolant mass flow rates in every fuel assembly? 5. Temperature of water leaving the central fuel assembly? Calculate the heat flux on the basis of the area formed by the cladding outside diameter and the active fuel length. Use as input only the following values from Tables 1.2, 1.3, 2.1 and Appendix K. —

Total power = 3411 MWt

—

Equivalent core diameter = 3.37 m

—

Active length = 3.658 m

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—

Fraction of energy released in fuel = 0.962

—

Total number of rods = 50,952

—

Rod outside diameter = 9.5 mm

—

Total flow rate = 17.476 × 103 kg/s

—

Inlet temperature = 293.1 °C

—

Core average pressure = 15.5 MPa

Answers: 1.

𝜙𝜙𝑜𝑜 ϕ

= 3.11

″ 2. 𝑞𝑞max = 1.88 MW⁄m2 ′ 3. 𝑞𝑞max = 56.1 kW⁄m

(Δℎ)max Δℎ

= 2.08

5. (Δ𝑇𝑇𝑜𝑜𝑜𝑜𝑜𝑜 )𝑚𝑚𝑚𝑚𝑚𝑚 = 344.8°C

PROBLEM 3.2 SOLUTION

Power Profile in a Homogeneous Reactor (Section 3.5) Consider an ideal core with the following characteristics: The U-235 enrichment is uniform throughout the core, and the flux distribution is characteristic of an unreflected, uniformly fueled cylindrical reactor, with extrapolation distances δz and δR of 10 cm. How closely do these assumptions allow prediction of the following characteristics of a PWR? 1. Ratio of peak to average power density and heat flux? We can define the radial and axial profile of the volumetric generation rate as 2.405𝑟𝑟 𝜋𝜋𝜋𝜋 ′′′ � 𝑐𝑐𝑐𝑐𝑐𝑐 � � 𝑞𝑞 ′′′ (𝑟𝑟, 𝑧𝑧) = 𝑞𝑞𝑚𝑚𝑚𝑚𝑚𝑚 𝐽𝐽0 � 𝑅𝑅𝑒𝑒 𝐿𝐿𝑒𝑒

The average volumetric heat generation rate can be calculated by averaging the volumetric generation rate, 𝑞𝑞 ′′′ =

𝐿𝐿 ⁄2

𝑅𝑅

∫−𝐿𝐿⁄2 ∫0 2𝜋𝜋𝜋𝜋𝑞𝑞 ′′′ (𝑟𝑟, 𝑧𝑧)𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝐿𝐿 ⁄2

𝑅𝑅

∫−𝐿𝐿⁄2 ∫0 2𝜋𝜋𝜋𝜋𝜋𝜋𝜋𝜋𝜋𝜋𝜋𝜋

The peak to average volumetric generation rate can be calculated as

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′′′ 𝑞𝑞𝑚𝑚𝑚𝑚𝑚𝑚 𝐹𝐹𝑟𝑟𝑟𝑟 = = 𝑞𝑞′′′

𝐿𝐿⁄2

𝑅𝑅

′′′ ∫−𝐿𝐿⁄2 ∫0 2𝜋𝜋𝜋𝜋𝜋𝜋𝜋𝜋𝜋𝜋𝜋𝜋 𝑞𝑞𝑚𝑚𝑚𝑚𝑚𝑚 = = 3.11 𝐿𝐿⁄2 𝑅𝑅 𝐿𝐿⁄2 𝑅𝑅 2.405𝑟𝑟 𝜋𝜋𝜋𝜋 2.405𝑟𝑟 𝜋𝜋𝜋𝜋 ′′′ J0 � 𝑅𝑅 � cos � 𝐿𝐿 � 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 ∫𝐿𝐿⁄2 ∫0 2𝜋𝜋𝜋𝜋 J0 � 𝑅𝑅 � cos � 𝐿𝐿 � 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 ∫−𝐿𝐿⁄2 ∫0 2𝜋𝜋𝜋𝜋𝑞𝑞𝑚𝑚𝑚𝑚𝑚𝑚 𝑒𝑒 𝑒𝑒 𝑒𝑒 𝑒𝑒 𝐿𝐿⁄2

𝑅𝑅

∫−𝐿𝐿⁄2 ∫0 2𝜋𝜋𝜋𝜋𝜋𝜋𝜋𝜋𝜋𝜋𝜋𝜋

where L = 3.658 m, Le = 3.858m, R = 1.658m and Re = 1.758m. 2. Maximum heat flux? The maximum heat flux is calculated with ″ = 𝐹𝐹𝑟𝑟𝑟𝑟 𝑞𝑞𝑚𝑚𝑚𝑚𝑚𝑚

γ𝑄𝑄̇𝑡𝑡ℎ 0.962(3411 MWt ) MW = 3.11 = 1.88 2 50952𝜋𝜋(9.5 mm)(3.568 m) m 𝑛𝑛𝑟𝑟𝑟𝑟𝑟𝑟 𝜋𝜋𝑑𝑑𝑐𝑐𝑐𝑐 𝐿𝐿

3. Maximum linear heat generation rate of the fuel rod?

The maximum linear heat generation rate is calculated as MW kW 𝜋𝜋(9.5 mm) = 54.8 m2 m

′ ″ 𝑞𝑞𝑚𝑚𝑚𝑚𝑚𝑚 = 𝑞𝑞𝑚𝑚𝑚𝑚𝑚𝑚 𝜋𝜋𝑑𝑑𝑐𝑐𝑐𝑐 = 1.77

4. Peak to average enthalpy rise ratio, assuming equal coolant mass flow rates in every fuel assembly? This ratio can be defined as 𝐹𝐹𝛥𝛥ℎ =

𝛥𝛥ℎ𝑚𝑚𝑚𝑚𝑚𝑚 𝛥𝛥ℎ

The average power can be calculated with 𝑞𝑞 =

𝑛𝑛𝑟𝑟𝑟𝑟𝑟𝑟

𝑞𝑞𝑚𝑚𝑚𝑚𝑚𝑚 𝑞𝑞𝑚𝑚𝑚𝑚𝑚𝑚 = 𝑚𝑚̇ = 𝑞𝑞 𝑞𝑞 𝑚𝑚̇

𝐿𝐿 ⁄2 𝑅𝑅

� � 2𝜋𝜋𝜋𝜋𝜋𝜋′′′(𝑟𝑟, 𝑧𝑧)𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

−𝐿𝐿⁄2 0

and the maximum power will be located at the center of the core and therefore the volumetric heat generation rate becomes 𝜋𝜋𝜋𝜋 ′′′ cos � � 𝑞𝑞 ′′′(0,𝑧𝑧) = 𝑞𝑞𝑚𝑚𝑚𝑚𝑚𝑚 𝐿𝐿𝑒𝑒

We can then calculate the maximum power as 𝑞𝑞𝑚𝑚𝑚𝑚𝑚𝑚 =

𝑛𝑛𝑟𝑟𝑟𝑟𝑟𝑟

𝐿𝐿 ⁄2

𝜋𝜋𝑅𝑅 2 � 𝑞𝑞′′′(0, 𝑧𝑧)𝑑𝑑𝑑𝑑 −𝐿𝐿⁄2

The peak to average enthalpy rise is determined to be

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Chapter 3 - Reactor Energy Distribution

1 1 𝜋𝜋𝜋𝜋 2 𝐿𝐿 ⁄2 2 𝐿𝐿 ⁄2 ′′′ 𝑛𝑛𝑟𝑟𝑟𝑟𝑟𝑟 𝜋𝜋𝑅𝑅 ∫−𝐿𝐿⁄2 𝑞𝑞′′′(0, 𝑧𝑧)𝑑𝑑𝑑𝑑 𝑛𝑛𝑟𝑟𝑟𝑟𝑟𝑟 𝜋𝜋𝑅𝑅 ∫−𝐿𝐿⁄2 𝑞𝑞𝑚𝑚𝑚𝑚𝑚𝑚 cos � 𝐿𝐿𝑒𝑒 � 𝑑𝑑𝑑𝑑 𝐹𝐹Δℎ = = 1 𝐿𝐿⁄2 𝑅𝑅 1 𝐿𝐿⁄2 𝑅𝑅 ′′′ J �2.405𝜏𝜏 � cos �𝜋𝜋𝜋𝜋� 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 2𝜋𝜋𝜋𝜋𝜋𝜋′′′(𝑟𝑟, 𝑧𝑧)𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 2𝜋𝜋𝜋𝜋𝑞𝑞𝑚𝑚𝑚𝑚𝑚𝑚 0 𝑛𝑛 ∫−𝐿𝐿⁄2 ∫0 𝑅𝑅 𝐿𝐿 𝑛𝑛 ∫−𝐿𝐿⁄2 ∫0 𝑟𝑟𝑟𝑟𝑟𝑟

𝑟𝑟𝑟𝑟𝑟𝑟

= 2.08

𝑒𝑒

5. Temperature of water leaving the central fuel assembly?

The core inlet enthalpy can be evaluated from the inlet temperature and the core pressure to be 1300 kJ/kg. Therefore the hot channel outlet enthalpy can be calculated as ℎ𝑜𝑜𝑜𝑜 ℎ𝑜𝑜𝑜𝑜𝑜𝑜 = ℎ𝑖𝑖𝑖𝑖 + 𝐹𝐹Δℎ

𝑄𝑄̇𝑡𝑡ℎ kJ 3411 MWt kJ = 1300 + 2.08 = 1706 𝑘𝑘𝑘𝑘 𝑚𝑚̇ kg kg 17476 𝑔𝑔

Using steam tables, the temperature can be determined to be in a saturated condition at the given ℎ𝑜𝑜𝑜𝑜 pressure, 𝑇𝑇𝑜𝑜𝑜𝑜𝑜𝑜 = 344.8°C.

PROBLEM 3.3 QUESTION

Power Generation in Thermal Shield (Section 3.8) Consider the heat-generation in the thermal shield discussed in Example 3.4. 1. Calculate the total power generation in the thermal shield if it is 4.0 m high. 2. How would this total power change if the thickness of the shield is increased from 12.7 cm to 15 cm? Assume uniform axial power profile. Answers: 1. 𝑄𝑄̇ = 23.21 MW 2. 𝑄𝑄̇ = 23.53 MW

PROBLEM 3.3 SOLUTION

Power Generation in Thermal Shield (Section 3.8) Consider the heat-generation in the thermal shield discussed in Example 3.4. 1. Calculate the total power generation in the thermal shield if it is 4.0 m high. General parameters needed for this calculation: — Height of thermal shield: L = 4.0 m — Inner radius of shield: ri = 1.206 m

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Chapter 3 - Reactor Energy Distribution — Outer radius of shield: ro = 1.333 m — Buildup factor in shield: B = 4.212 — Energy of gammas: Eo = 2 MeV — Linear attenuation and absorption coefficients: μ = 0.333 cm 1

Fast flux and gamma source: 𝜙𝜙𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓, 𝑆𝑆 = 1014 cm2

−1

and μa = 0.182 cm−1

The volumetric gamma heat generation rate is defined as 𝑞𝑞𝛾𝛾′′′ (𝑟𝑟) = 𝑆𝑆𝑆𝑆𝜇𝜇𝑎𝑎 𝐸𝐸𝑜𝑜 𝑒𝑒 −𝜇𝜇(𝑟𝑟−𝑟𝑟𝑖𝑖 )

Integrating this equation over the volume of the thermal shield, the gamma heat rate is 𝐿𝐿

𝑟𝑟𝑜𝑜

𝑞𝑞γ = � � 2𝜋𝜋𝜋𝜋𝑞𝑞γ′′′ (𝑟𝑟)𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 = 22.55 MW 0

𝜏𝜏𝑖𝑖

The neutron volumetric heat generation rate, made up of the sum of elastic and inelastic collisions, is determined from Example 3.4 to be ′′′ + 𝑞𝑞𝑖𝑖𝑖𝑖′′′ = 0.26 × 1012 𝑞𝑞𝑛𝑛′′′ = 𝑞𝑞𝑒𝑒𝑒𝑒

MeV MeV MeV 12 12 + 0.76 × 10 = 1.02 × 10 cm3 s cm3 s cm3 s

Integrating the neutron volumetric generation rate gives us the total neutron heat rate, 𝐿𝐿

𝑟𝑟𝑜𝑜

𝑞𝑞𝑛𝑛 = � � 2𝜋𝜋𝜋𝜋𝑞𝑞𝑛𝑛′′′ (𝜏𝜏)𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 = 0.66 𝑀𝑀𝑀𝑀 0

𝑟𝑟𝑖𝑖

Therefore the total heat generation rate due to gamma heating and neutron slo wing down is q = q γ + q n = 22.55 MW + 0.662 MW = 23.21 MW

2. How would this total change if the thickness of the shield is increased from 12.7 cm to 15 cm? We can apply the same process as part 1 and adjust the outer radius of the thermal shield from 1.333 m to 1.356 m. The gamma heat rate and neutron heat rate are then calculated to be: − qγ = 22.74 MW − qn = 0.79 MW The total power is then q = qγ + qn = 22.74 MW + 0.79 MW = 23.53 MW

PROBLEM 3.4 QUESTION Decay Heat Energy (Section 3.9) Using Equation 3.70c or 3.70d, evaluate the energy generated in a 3411 MWt PWR after the reactor shuts down. The reactor operated for 1 year at the equivalent of 75% of total power.

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Chapter 3 - Reactor Energy Distribution

1. Consider the following time periods after shutdown: (a) 1 hour (b) 1 day (c) 1 month 2. How would your answers be different if you had used Equation 3.71 (i.e., would higher or lower values be calculated)? Answers: 1a. 0.128 TJ (1 TJ = 1012 TJ) 1b. 1.42 TJ 1c. 14.8 TJ 2. Higher

PROBLEM 3.4 SOLUTION Decay Heat Energy (Section 3.9) Using Equation 3.70c or 3.70d, evaluate the energy generated in a 3411 MWt PWR after the reactor shuts down. The reactor operated for 1 year at the equivalent of 75% of total power. We may define the following parameters: — Steady state reactor power:

= 0.75(3411 MWt) = 2558 MWt

— Operating time: τs = 1 yr

1. Consider the following time periods after shutdown, (a) 1 hr, (b) 1 day, (c) 1 month. Equation 3.70d can be used to model the decay heat with

To determine the energy generated after the reactor shuts down, for example at a time ts = 1 hr, 1 day, 1 month, can be calculated by integrating Equation 3.70d, 𝑡𝑡𝑠𝑠

𝐸𝐸 = � 𝑄𝑄̇ (𝑡𝑡𝑠𝑠′ )𝑑𝑑𝑑𝑑𝑠𝑠′ 0

Evaluating parts (a), (b), and (c) gives 0.128 TJ, 1,42 TJ and 14,8 TJ, respectively, 2. How would your answers be different if you had used Equation 3.71 (i.e., would higher or lower values be calculated)?

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Chapter 3 - Reactor Energy Distribution

Equation 3.71 calculates the decay heat power as

We can integrate this equation to evaluate the energy generated to arrive at 0.188 TJ, 2.03 TJ and 19.1 TJ, respectively. Therefore a higher answer is calculated.

PROBLEM 3.5 QUESTION Decay Heat From a PWR Fuel Rod (Section 3.9) A decay heat cooling system is capable of removing 1 kW from the surface of a typical PWR (Seabrook) fuel rod (Appendix K). Assume the rod has operated for an essentially infinite period before shutdown. 1. At what time will the decay energy generation rate be matched by the cooling capability? 2. What is the maximum amount of decay heat that will be stored in the rod following shutdown? Answers: 1. t = 1490.6 s 2. Qmax = 3.73 × 105 J

PROBLEM 3.5 SOLUTION Decay Heat From a PWR Fuel Rod (Section 3.9) A decay heat cooling system is capable of removing 1 kW from the surface of a typical PWR (Seabrook) fuel rod (Appendix K). Assume the rod has operated for an essentially infinite period before shutdown. 1. At what time will the decay energy generation rate be matched by the cooling capability? For a rod that has been operated for an infinite period of time, Equation 3.70d becomes

The power generated by an average fuel pin can be determined from its average linear heat

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Chapter 3 - Reactor Energy Distribution kW

generation rate, 𝑞𝑞 ′ = 17.86 m and active length, L = 3.658 m. This power is calculated to be 𝑄𝑄̇𝑜𝑜 = 𝑞𝑞 ′ 𝐿𝐿 = �17.86

kW � (3.658 m) = 65.33 kW m

The time it takes for the decay power to be equivalent to the cooling power of 1 kW can be solved by setting Q(ts) = 1 kW. We can then determine ts to be 5 5 0.066𝑄𝑄̇𝑜𝑜 0.066(65.33kW) 𝑡𝑡𝑠𝑠 = � � =� � = 1490.6 s 1kW 𝑄𝑄̇ (𝑡𝑡𝑠𝑠 )

2. What is the maximum amount of decay heat energy that will be stored in the rod following a shutdown? One can write the conservation of energy equation after shutdown to be 𝑡𝑡

𝐸𝐸(𝑡𝑡) = � �0.066𝑄𝑄̇𝑜𝑜 𝑡𝑡 ′−0.2 − 1 kW�𝑑𝑑𝑡𝑡 ′ 0

To solve for this time we can take the derivative of the energy equation and set it equal to 0. This results in the expression,

0 0.066Q ot −0.2 − 1kW = This is the same equation that was solved for in part 1 and therefore, tmax = 1490.6 s. This can be substituted back in the energy formula to determine that the maximum energy is, 𝐸𝐸𝑚𝑚𝑚𝑚𝑚𝑚 = 3.73 × 105 J

PROBLEM 3.6 QUESTION Decay Power Calculations of a 3-Batch PWR Core (Section 3.9) A PWR core has been operated on a three-batch fuel management scheme on an 18 month refueling cycle, e.g., at every 18 months, one third the core loading is replaced with fresh fuel. A new batch is first loaded into the core in a distributed fashion such that it generates 43% of the core power. After 18 months of operation, it is shuffled to other core locations where it generates 33% of the core power. After another 18 months, it is moved to other core locations where is generates 24% of the core power. Question The plant rating is 3411 MWt. Assume it is shutdown after an 18-month operating cycle. What is the decay power of the plant one hour after shutdown if it has operated continuously at 100% power during each of the preceding three 18 month operating cycles and the shutdown periods for refueling were each of 35 days duration? Solve this problem two ways:

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Chapter 3 - Reactor Energy Distribution

1. Consider the explicit operating history of each of the three batches to the core decay power. 2. Assume the whole core had been operating for an infinite period before shutdown. Answers: 1. 𝑄𝑄̇ = 38.0 MWth 2. 𝑄𝑄̇ = 43.8 MWth

PROBLEM 3.6 SOLUTION

Decay Power Calculations of a 3-batch PWR Core (Section 3.9) A PWR core has been operated on a three-batch fuel management scheme on an 18 month refueling cycle, e.g., at every 18 months, one third the core loading is replaced with fresh fuel. A new batch is first loaded into the core in a distributed fashion such that it generates 43% of the core power. After 18 months of operation, it is shuffled to other core locations where it generates 33% of the core power. After another 18 months, it is moved to other core locations where is generates 24% of the core power. Question The plant rating is 3411 MWt. Assume it is shutdown after an 18-month operating cycle. What is the decay power of the plant one hour after shutdown if it has operated continuously at 100% power during each of the preceding three 18 month operating cycles and the shutdown periods for refueling were each of 35 days duration? Solve this problem two ways: 1. Consider the explicit operating history of each of the three batches to the core decay power. To describe the decay power, we will use Equation 3.70c. Let the fractional cycle power be described with the variable ω, such that 𝜔𝜔(𝜏𝜏s , 𝑡𝑡𝑠𝑠 ) = 0.066[𝑡𝑡𝑠𝑠−0.2 − (𝑡𝑡𝑠𝑠 + 𝜏𝜏𝑠𝑠 )−0.2 ]

To solve this question we will consider each cycle separately and determine the contributions of the batches in the cycle to the decay power one hour after shutdown. Cycle A The operating time for Cycle A is 18 months, τsA = 4.666 × 107s, and the time after the end of cycle to reach 1 hour after core final shutdown is tsA = 2 ∙ 18 months +2 ∙ 35 day + 1 hr = 1.007 × 108s. Cycle A contains only batch 1 operating at 43% of the power. The other two batches will not contribute the decay power after final shutdown of this core as they will be replaced by fresh batches. The fractional cycle power 1 hr after shutdown is therefore calculated as 𝜔𝜔𝐴𝐴 = (0.43)𝜔𝜔(𝜏𝜏𝑠𝑠𝑠𝑠 , 𝑡𝑡𝑠𝑠𝑠𝑠 ) = 5.218 × 10−5 41

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Chapter 3 - Reactor Energy Distribution

Cycle B The operating time for Cycle B is 18 months, τsB = 4.666 × 107s, and the time after the end of cycle to reach 1 hour after core final shutdown is tsB = 18 months + 35 day + 1 hr = 5.036 × 107s. Cycle B contains batch 1 operating now at 33% and batch 2 operating at 43% of the total power. The fractional cycle power 1 hr after shutdown is Cycle C

𝜔𝜔𝐵𝐵 = (0.43)𝜔𝜔(𝜏𝜏𝑠𝑠𝑠𝑠 , 𝑡𝑡𝑠𝑠𝑠𝑠 ) + (0.33)𝜔𝜔(𝜏𝜏𝑠𝑠𝑠𝑠 , 𝑡𝑡𝑠𝑠𝑠𝑠 ) = 1.776 × 10−4

The operating time for Cycle C is 18 months, τsC = 4.666 × 107s, and the time after the end of cycle to reach 1 hour after core final shutdown is tsC= 1 hr = 3600 s. Cycle C contains batch 1 operating now at 24% and batch 2 operating at 33% and batch 3 operating at 43% of the total power. The fractional cycle power 1 hr after shutdown is 𝜔𝜔𝐶𝐶 = (0.43)𝜔𝜔(𝜏𝜏𝑠𝑠𝑠𝑠 , 𝑡𝑡𝑠𝑠𝑠𝑠 ) + (0.33)𝜔𝜔(𝜏𝜏𝑠𝑠𝑠𝑠 , 𝑡𝑡𝑠𝑠𝑠𝑠 ) + (0.24)𝜔𝜔(𝜏𝜏𝑠𝑠𝑠𝑠 , 𝑡𝑡𝑠𝑠𝑠𝑠 ) = 1.09 × 10−2

The decay power after 1 hour of shutdown is determined to be

𝑄𝑄̇ = 𝑄𝑄̇𝑜𝑜 (𝜔𝜔𝐴𝐴 + 𝜔𝜔𝐵𝐵 + 𝜔𝜔𝐶𝐶 ) = 3411 MW(5.218 × 10−5 + 1.776 × 10−4 + 1.09 × 10−2 ) = 38.0MW

2. Assume the whole core had been operating for an infinite period before shutdown.

This easily be determined with the following expression derived from Equation 3.70d setting τs → ∞, 𝑄𝑄̇ = 0.066𝑄𝑄̇𝑜𝑜 𝑡𝑡𝑠𝑠−0.2 = 0.066(3411 MW)(3600 s)−0.2 = 43.8 MW

PROBLEM 3.7 QUESTION

Effect of Continuous Refueling on Decay Heat (Section 3.9) Using Equation 3.70c or 3.70d, estimate the decay heat rate in a 3000 MWt reactor in which 3.2% 235 U-enriched UO2 assemblies are being fed into the core. The burned-up fuel stays in the core for 3 years before being replaced. Consider two cases: 1. The core is replaced in two batches every 18 months. 2. The fuel replacement is so frequent that refueling can be considered a continuous process. (Note: The PHWR reactors and some of the water-cooled graphite-moderated reactors in the former Soviet Union are effectively continuously refueled.) Compare the two situations at 1 minute, 1 hour, 1 month and 1 year. Answers:

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Chapter 3 - Reactor Energy Distribution

Time

Case 1 (MWth)

Case 2 (MWth)

1 minute

81.9

81.0

1 hour

33.1

32.2

1 day

15.0

14.1

1 month

4.95

4.24

1 year

1.29

0.965

PROBLEM 3.7 SOLUTION Effect of Continuous Refueling on Decay Heat (Section 3.9) Using Equation 3.70c or 3.70d, estimate the decay heat rate in a 3000 MWt reactor in which 3.2% 235 U-enriched U02assemblies are being fed into the core. The burned-up fuel stays in the core for 3 years before being replaced. Consider two cases: 1. The core is replaced in two batches every 18 months. This core is divided into two zones. At the end of 3 years, 1 batch of the fuel will have been burned for 3 years and the other batch for 1.5 years. Each fuel bundle is assumed to generate half the total core power. Equation 3.70d is used to calculate the decay power at 1 minute, 1 hour, 1 day, 1 month and 1 year respectfully, 𝑄𝑄̇ = 0.5𝑄𝑄̇𝑜𝑜 (0.066){[𝑡𝑡𝑠𝑠−0.2 − (𝑡𝑡𝑠𝑠 + 𝜏𝜏𝑠𝑠1 )−0.2 ] + [𝑡𝑡𝑠𝑠−0.2 − (𝑡𝑡𝑠𝑠 + 𝜏𝜏𝑠𝑠2 )−0.2 ]}

In the above equation, 𝑄𝑄̇ is a vector of decay powers at time ts after shutdown, 𝑄𝑄𝑜𝑜̇ is the total core power, τs1 is the operating time for fuel bundle 1 which is 3 years and τs2 is the operating time for fuel bundle 2 which is 1.5 years. The results are shown below: 𝑸𝑸̇ in MWth 81.9 33.1 15.0 4.95 1.29

ts 1 minute 1 hour 1 day 1 month 1 year

2. The fuel replacement is so frequent that refueling can be considered a continuous process. For a continuously refueled core, the time between refuelings goes to 0 which implies that the number of batches goes to infinity. Therefore to get an average decay power we may integrate Equation 3.70d with respect to the core operating time and divide by the total time of 3 years (τs):

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Chapter 3 - Reactor Energy Distribution

𝑄𝑄̇ = 0.066

This results in the expression,

𝑄𝑄̇ = 0.066

𝑄𝑄̇o 𝜏𝜏𝑠𝑠 −0.2 � [𝑡𝑡 − (𝑡𝑡𝑠𝑠 + 𝜏𝜏𝑠𝑠′ )−0.2 ]𝑑𝑑𝜏𝜏𝑠𝑠′ 𝜏𝜏𝑠𝑠 0 𝑠𝑠

𝑄𝑄̇𝑜𝑜 −0.2 1 [(𝑡𝑡𝑠𝑠 + 𝜏𝜏𝑠𝑠 )0.8 − 𝑡𝑡𝑠𝑠0.8 ]� �𝑡𝑡𝑠𝑠 𝜏𝜏𝑠𝑠 − 0.8 𝜏𝜏𝑠𝑠

The resulting decay powers are shown below:

ts 1 minute 1 hour 1 day 1 month 1 year

𝑸𝑸̇ in MWth 81.0 32.2 14.1 4.24 0.965

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Chapter 4 Transport Equations for Single-Phase Flow Contents Problem 4.1 Various time derivatives ..................................................................................

Problem 4.2 Conservation of energy in a control volume ...................................................

Problem 4.3 Process-dependent heat addition to a control mass .........................................

Problem 4.4 Control mass energy balance ...........................................................................

Problem 4.5 Qualifying a claim against the first and second laws of thermodynamics ......

Problem 4.6 Determining rocket acceleration from an energy balance ...............................

Problem 4.7 Momentum balance for a control volume .......................................................

Problem 4.8 Internal conservation equations for an extensive property ..............................

Problem 4.9 Differential transport equations .......................................................................

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Chapter 4 - Transport Equations for Single-Phase Flow

PROBLEM 4.1 QUESTION Various Time Derivatives (Section 4.2) Air bubbles are being injected at a steady rate into the bottom of a vertical water channel of height L. The bubble rise velocity with respect to the water is υb . The water itself is flowing vertically at a velocity υ . The observer moves upward in the channel with a velocity Vo as defined below. What is the velocity of the bubbles in the channel as observed by the following: 1. A stationary observer 2. An observer who moves upward with velocity Vo = υ 3. An observer who moves upward with velocity Vo = 2υ

Answers: 1. υ + υb 2. υb

3. υb − υ

PROBLEM 4.1 SOLUTION Various Time Derivatives (Section 4.2) First, define the absolute bubble velocity, Vb, as the velocity of the bubble with respect to an (effectively) inertial reference frame. For this problem, the origin of this coordinate system can be at the entrance to the pipe. For simplicity this can be a 1-D coordinate system because we can assume that the bubbles are rising in the tube without changing their position in the cross section of the tube. For an observer at the origin of this absolute reference frame, the water is going past at υl. On top of this velocity, the bubbles are moving at speed υb with respect to the water due to the buoyancy force. So, the observer sees the bubble moving at an absolute velocity, 𝑉𝑉b = 𝜐𝜐 + 𝜐𝜐b

(1)

𝑉𝑉brel = 𝑉𝑉b − 𝑉𝑉0

(2)

The observer does not have to stay at the inertial origin—he is free to move in the tube at any velocity Vo where Vo is the absolute velocity of the observer with respect to the inertial frame. Using these two absolute velocities, it is possible to talk about the relative motion of the bubble with respect to any observer. The relative motion of the bubble is simply the difference between its absolute velocity and the absolute velocity of the observer, Substituting for Vb into Equation 2,

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Chapter 4 - Transport Equations for Single-Phase Flow

𝑉𝑉brel = 𝜐𝜐 + 𝜐𝜐b − 𝑉𝑉0

(3)

1. Stationary observer (Vo = 0). If Vo = 0, then Equation 3 reduces to, 𝑉𝑉brel = 𝜐𝜐 + 𝜐𝜐b

(4)

This is the same as assuming that the observer stays at a fixed reference point—such as the origin of the inertial system. 2. Observer moving upwards with velocity υ (Vo = υ). Now Equation 3 becomes, 𝑉𝑉brel = 𝜐𝜐 + 𝜐𝜐b − 𝜐𝜐1 = 𝜐𝜐b

(5)

This can be seen by assuming that the observer is moving with the liquid phase. The only component of the bubble velocity that he sees is the part due to buoyancy, υb. 3. Observer moving upwards with velocity 2υl (Vo = 2υl). Now Equation 3 becomes, 𝑉𝑉brel = 𝜐𝜐 + 𝜐𝜐b − 2𝜐𝜐1 = 𝜐𝜐𝑏𝑏 − 𝜐𝜐

(6)

Note that it is possible for this relative velocity to be negative if the buoyancy force is stronger than the force that is pumping the fluid through the tube. In that case, the observer would be “passing” the bubbles in the tube.

PROBLEM 4.2 QUESTION Conservation of Energy in a Control Volume (Section 4.3) A mass of 9 kg of gas with an internal energy of 1908 kJ is at rest in a rigid cylinder. A mass of 1.0 kg of the same gas with an internal energy of 95.4 kJ and a velocity of 30 m/s flows into the cylinder. In the absence of heat transfer to the surroundings and with negligible change in the center of gravity, find the internal energy of the 10 kg of gas finally at rest in the cylinder. The absolute pressure of the flowing gas crossing the control surface is 0.7 MPa, and the specific volume of the gas is 0.00125 m3/kg. Answer: Uf = 2004.7 kJ

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Chapter 4 - Transport Equations for Single-Phase Flow

PROBLEM 4.2 SOLUTION Conservation of Energy in a Control Volume (Section 4.3)

FIGURE SM-4.1 Collection cylinder of Problem 4.2 Operating Pressure = 0.7 MPa v = 0.00125 m3/kg First law of thermodynamics I

𝑑𝑑𝑑𝑑 𝑃𝑃i 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 � � = � 𝑚𝑚̇i �𝑢𝑢io + + 𝑔𝑔𝑔𝑔i � + � � � +� � −� −� −� 𝜌𝜌i 𝑑𝑑𝑑𝑑 cv 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 gen 𝑑𝑑𝑑𝑑 shaft 𝑑𝑑𝑑𝑑 normal 𝑑𝑑𝑑𝑑 shear i

For a Rigid body

Δ𝑉𝑉 = 0

No heat transfer

Δ𝑊𝑊 = 0 Δ𝑄𝑄 = 0

No gravitational change We can integrate Equation 1 from the initial state to the final state to get the change in energy 𝑣𝑣i2 𝑃𝑃i Δ𝐸𝐸 = 𝑚𝑚i �𝑢𝑢i + + � = 96.725 kJ 2 𝜌𝜌i where m i = 1kg, u i = 95.4kJ/kg, υi = 30 m/s, Pi = 0.7 MPa, ρi = (0.00125 m3/kg)− 1 . The final energy is 𝐸𝐸 = 𝐸𝐸𝑜𝑜 + Δ𝐸𝐸 = 2004.7 kJ where Eo = 1908 kJ.

PROBLEM 4.3 QUESTION Process-Dependent Heat Addition to a Control Mass (Section 4.3) Two tanks, A and B (Figure 4.10), each with a capacity of 20 ft3 (0.566 m3), are perfectly insulated from the surroundings. A diatomic perfect gas is initially confined in tank A at a pressure of 1.013 MPa and a temperature of 70 °F (21.2 °C). Valve C is initially closed, and tank B is completely evacuated.

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Chapter 4 - Transport Equations for Single-Phase Flow

1. If valve C is opened and the gas is allowed to reach the same temperature in both tanks, what is the final pressure and temperature? 2. If valve C is opened just until the pressure in the tanks is equalized and is then closed, what are the final temperature and pressure in tanks A and B? Assume no transfer of heat between tank A and tank B.

FIGURE 4.10 Initial state of perfectly insulated tanks for Problem 4.3. Answers: 1. P = 0.507 MPa; T = 530 R (21.1 ºC) 2. P = 0.507 MPa; TA = 434.5 R (-31.8 ºC), TB = 678.2 R (103.6 ºC)

PROBLEM 4.3 SOLUTION Process-Dependent Heat Addition to a Control Mass (Section 4.3) Two tanks, A and B (Figure 4.10), each with a capacity of 20 ft3 (0.566 m3), are perfectly insulated from the surroundings. A diatomic perfect gas is initially confined in tank A at a pressure of 1.013 MPa and a temperature of 70 °F (21.2 °C). Valve C is initially closed, and tank B is completely evacuated. Given Parameters: – TAi = 70 °F – PAi = 1.013 MPa – VA = VB = 20 ft3 1. Valve C is opened until thermal equilibrium is reached: We may begin by taking the control volume as the two tanks together. The first law of thermodynamics for a stationary closed system is: 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 = = 𝑄𝑄̇ − 𝑊𝑊̇ 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 49

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Chapter 4 - Transport Equations for Single-Phase Flow

Since we took the control volume over both tanks, there is no work. In addition, since the tanks are insulated, there is no heat transfer. Thus, Δ𝑈𝑈cv = 0

We may apply the energy constitutive relation of a perfect gas to determine that Δ𝑈𝑈cv = 𝑚𝑚cv 𝑐𝑐𝑐𝑐ΔTcv Δ𝑇𝑇cv = 0

Since this is an isothermal process, the final temperature of each tank is simply, 𝑇𝑇Af = 𝑇𝑇Bf = 𝑇𝑇Ai = 529.67R

We may use the equation of state for a perfect gas to determine the final pressure of the control volume, (1)

𝑃𝑃𝑃𝑃 = 𝑚𝑚𝑚𝑚𝑚𝑚

The right hand side of Equation (3) is the same at the initial and final states and therefore, The final pressure is 𝑃𝑃f =

𝑃𝑃𝑃𝑃 = const

𝑃𝑃Ai 𝑉𝑉A

𝑉𝑉A +𝑉𝑉B

= 0.507 MPa (5 atm)

2. Valve is opened until pressure equilibrium is attained: We may consider a control volume around tank A and a different control volume around tank B. The two systems taken together form a composite system, A+B, that is an isolated system. Therefore, the first law of thermodynamics for the composite system is (𝐸𝐸f − 𝐸𝐸i )𝐴𝐴+𝐵𝐵 = 0

(𝐸𝐸f − 𝐸𝐸i )𝐴𝐴+𝐵𝐵 = (𝐸𝐸f − 𝐸𝐸i )A + �𝐸𝐸j − 𝐸𝐸i �

Therefore, we can simply write that

Δ𝐸𝐸A + Δ𝐸𝐸B = 0

The constitutive relation for energy can be substituted in for each volume: 𝑚𝑚𝐴𝐴𝐴𝐴 𝑐𝑐𝑣𝑣 �𝑇𝑇𝐴𝐴𝐴𝐴 − 𝑇𝑇𝐴𝐴𝐴𝐴 � + 𝑚𝑚𝐵𝐵𝐵𝐵 𝑐𝑐𝑣𝑣 �𝑇𝑇𝐵𝐵𝐵𝐵 − 𝑇𝑇𝐴𝐴𝐴𝐴 � = 0

Simplifying the expression,

𝑚𝑚Af 𝑐𝑐v 𝑇𝑇Af + 𝑚𝑚Bf 𝑐𝑐𝑣𝑣 𝑇𝑇Bf = (𝑚𝑚Af 𝑐𝑐v + 𝑚𝑚B 𝑐𝑐v )𝑇𝑇Ai 𝑚𝑚Af 𝑇𝑇Af + 𝑚𝑚B 𝑇𝑇Bf = 𝑚𝑚Ai 𝑇𝑇Ai

(2)

In Equation (1), from conservation of mass, mAi = mAf + mBf. We can write the equation of state

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Chapter 4 - Transport Equations for Single-Phase Flow

for a perfect gas for each control volume, 𝑚𝑚Af 𝑅𝑅𝑅𝑅Af 𝑃𝑃Af = 𝑉𝑉A

𝑃𝑃Bf =

𝑚𝑚Bf 𝑅𝑅𝑅𝑅Bf 𝑉𝑉B

Since the final pressures are equivalent, PAf = PBf, the following relation is obtained: 𝑚𝑚Af 𝑇𝑇Af = 𝑚𝑚Bf 𝑇𝑇Bf

Substituting Equation (3) into Equation (1), 1 𝑚𝑚Af 𝑇𝑇Af = 𝑚𝑚Ai 𝑇𝑇Ai 2

(3) (4)

We can now write the equation of state for a perfect gas for control volume A at the initial and final state, 𝑃𝑃Ai 𝑉𝑉A 𝑃𝑃Af 𝑉𝑉A (5) = 𝑚𝑚Ai 𝑇𝑇Ai = 𝑚𝑚Af 𝑇𝑇Af 𝑅𝑅 𝑅𝑅 By inspection of Equations (4) and (5) we can see that 1 𝑃𝑃Af = 𝑃𝑃Ai = 5atm 2

(6)

Now that we have determined the final pressure, we can perform an energy balance on control volume A. Since this is an open system (mass will travel out of control volume A), the first law becomes 𝑑𝑑𝑑𝑑A = −𝑚𝑚̇out,A ℎA 𝑑𝑑𝑑𝑑

Since this is an ideal gas, the energy in the control volume is just the internal energy of the gas. We can replace the energy with the constitutive relation for an ideal gas. We can also replace the enthalpy with the constitutive relation for an ideal gas. The first law is rewritten as (for any instant in time), 𝑑𝑑(𝑚𝑚𝑚𝑚υ 𝑇𝑇)A = −𝑚𝑚̇out,A 𝑐𝑐p 𝑇𝑇A 𝑑𝑑𝑑𝑑

We can differentiate the left hand side to arrive at

𝑑𝑑𝑚𝑚A 𝑑𝑑𝑑𝑑A � � 𝐶𝐶υ 𝑇𝑇A + 𝑚𝑚A 𝑐𝑐υ = −𝑚𝑚̇out,A 𝑐𝑐p 𝑇𝑇A 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑

(7)

Applying continuity to control volume A we can determine that

This can be substituted into Equation (7), 𝑚𝑚A 𝑐𝑐𝜐𝜐

𝑑𝑑𝑑𝑑A = 𝑚𝑚̇out,A 𝑑𝑑𝑑𝑑

𝑑𝑑𝑑𝑑A = −𝑚𝑚̇out,A 𝑐𝑐p 𝑇𝑇A + −𝑚𝑚̇out,A 𝑐𝑐υ 𝑇𝑇A − 𝑚𝑚̇out,A �𝑐𝑐p − 𝑐𝑐υ �𝑇𝑇A 𝑑𝑑𝑑𝑑 51

(8)

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Chapter 4 - Transport Equations for Single-Phase Flow

The gas constant for a fluid can be expressed as This can be substituted into Equation (8), 𝑚𝑚A 𝑐𝑐υ

𝑅𝑅 = 𝑐𝑐p − 𝑐𝑐υ

𝑑𝑑𝑑𝑑A 𝑑𝑑𝑑𝑑A = 𝑚𝑚̇out,A 𝑅𝑅𝑅𝑅A = 𝑅𝑅𝑅𝑅A 𝑑𝑑𝑑𝑑 𝑑𝑑t

Separating variables, we arrive at

𝑐𝑐υ 1 𝑑𝑑𝑑𝑑A 𝑑𝑑𝑑𝑑A = 𝑅𝑅 𝑇𝑇A 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑

(9)

In order to eliminate the mass terms, we can apply the perfect gas law, 𝑃𝑃A 𝑉𝑉A = 𝑚𝑚A 𝑅𝑅𝑅𝑅A

Taking the logarithm of both sides we get,

ln(𝑃𝑃𝐴𝐴 ) + ln(𝑉𝑉A ) = ln(𝑚𝑚A ) + ln (𝑅𝑅) + ln(𝑇𝑇A )

Taking the derivative of both sides and noting that dVA is zero since the control volume is constant: 𝑑𝑑𝑑𝑑A 𝑑𝑑𝑑𝑑A 𝑑𝑑𝑑𝑑A = + , 𝑃𝑃A 𝑚𝑚A 𝑇𝑇A 1 𝑑𝑑𝑑𝑑A 1 𝑑𝑑𝑑𝑑A 1 𝑑𝑑𝑑𝑑𝐴𝐴 = − (10) 𝑚𝑚A 𝑑𝑑𝑑𝑑 𝑃𝑃A 𝑑𝑑𝑑𝑑 𝑇𝑇A 𝑑𝑑𝑑𝑑 Substituting Equation (10) into Equation (9) we obtain,

𝑐𝑐𝜐𝜐 1 𝑑𝑑𝑑𝑑A 1 𝑑𝑑𝑑𝑑A 1 𝑑𝑑𝑑𝑑A = − 𝑅𝑅 𝑇𝑇A 𝑑𝑑𝑑𝑑 𝑃𝑃A 𝑑𝑑𝑑𝑑 𝑇𝑇A 𝑑𝑑𝑑𝑑

1 𝑑𝑑𝑑𝑑A 𝑐𝑐𝜐𝜐 1 𝑑𝑑𝑑𝑑A 𝑐𝑐p 1 𝑑𝑑𝑑𝑑A = �1 + � − 𝑃𝑃A 𝑑𝑑𝑑𝑑 𝑅𝑅 𝑇𝑇A 𝑑𝑑𝑑𝑑 𝑅𝑅 𝑇𝑇A 𝑑𝑑𝑑𝑑

Taking the integral of both sides,

𝑇𝑇Af 𝑐𝑐 1 𝑑𝑑𝑑𝑑A p 1 𝑑𝑑𝑑𝑑A 𝑑𝑑𝑑𝑑 = � 𝑑𝑑𝑑𝑑 𝑃𝑃Ai 𝑃𝑃A 𝑑𝑑𝑑𝑑 𝑇𝑇Ai 𝑅𝑅 𝑇𝑇A 𝑑𝑑𝑑𝑑

�

results in:

𝑃𝑃Af

𝑐𝑐p 𝑃𝑃Af 𝑇𝑇Af ln � � = ln � � 𝑃𝑃Ai 𝑅𝑅 𝑇𝑇Ai 𝑅𝑅

𝑐𝑐𝑝𝑝

𝑃𝑃Af 𝑐𝑐𝑝𝑝 𝑇𝑇Af = 𝑇𝑇Ai � � 𝑃𝑃Ai

For a diatomic gas, 𝑅𝑅 = 2 and therefore the final temperature is 52

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𝑃𝑃Af 7 𝑇𝑇Af = 𝑇𝑇Ai � � = 434.507 R 𝑃𝑃Ai

Recalling from Equation (3) that mAfTAf = mBfTBf, we can eliminate mBf with mBf = mAi − mAf to get 𝑚𝑚A 𝑇𝑇Af = (𝑚𝑚Ai − 𝑚𝑚Af )𝑇𝑇Bf

We may divide both sides by mAi and solve for TBf to obtain, 𝑇𝑇Bf =

α𝑇𝑇Af 1 − 𝛼𝛼

(11)

where α is the ratio between final and initial mass of control volume A, α = mAf/mAi. Recalling the perfect gas law between the initial and final states of control volume A from Equation (5), 𝑃𝑃Ai 𝑉𝑉A = 𝑚𝑚Ai 𝑇𝑇Ai 𝑅𝑅

𝑃𝑃Af 𝑉𝑉A = 𝑚𝑚Af 𝑇𝑇Af 𝑅𝑅

Since the volume and gas constants are constants we can simplify to 𝑚𝑚Ai 𝑇𝑇Ai 𝑚𝑚Ai 𝑇𝑇Af = 𝑃𝑃Ai 𝑃𝑃𝐴𝐴𝐴𝐴

𝑚𝑚Af 𝑃𝑃Af 𝑇𝑇Ai = 𝛼𝛼 = 𝑚𝑚Ai 𝑃𝑃Ai 𝑇𝑇Af

The parameter α can be calculated to be

𝛼𝛼 =

𝑃𝑃Af 𝑇𝑇Ai = 0.6095 𝑃𝑃Ai 𝑇𝑇Af

Thus, the final temperature of control volume B, substituting α into Equation (6), is 𝑇𝑇Bf =

α𝑇𝑇Af = 678.2 R 1 − 𝛼𝛼

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PROBLEM 4.4 QUESTION Control Mass Energy Balance (Section 4.3) A perfectly insulated vessel contains 9.1 kg of water at an initial temperature of 277.8 K. An electric immersion heater with a mass of 0.454 kg is also at an initial temperature of 277.8 K. The water is slowly heated by passing an electric current through the heater until both water and heater attain a final temperature of 333.3 K. The specific heat of the water is 4.2 kJ/kg K and that of the heater is 0.504 kJ/kg K. Disregard volume changes and assume that the temperatures of water and heater are the same at all stages of the process. 𝒅𝒅𝒅𝒅

Calculate ∫ 𝑻𝑻 and ΔS for:

1. The water as a system

2. The heater as a system 3. The water and heater as a system Answers: 𝑑𝑑𝑑𝑑

1. Δ𝑆𝑆 = ∫ 𝑇𝑇 = 6.96 kJ⁄K 2. ΔS = 41.7 J/K

3. ΔS = 6.96 kJ/K

𝑑𝑑𝑑𝑑

∫ 𝑇𝑇 = − 6.96 kJ⁄K 𝑑𝑑𝑑𝑑

∫ 𝑇𝑇 = 0

PROBLEM 4.4 SOLUTION

Control Mass Energy Balance (Section 4.3) Perfectly insulated tank: ΔQ = 0. From the problem statement we are given: – mass of water, mw = 9.1kg – initial temperature of water, Tw = 277.8 K – mass of heater, = 0.454kg – initial temperature of heater, Th = 277.8 K – specific heat of water, Cpw — 4.2kJ/kg K – specific heat of heater, Cph = 0.504 kJ/kg K – final temperature, Tf = 333.3 K Water as a system: Since the temperature of water and heater are the same at all stages of the process, we can regard this process as reversible, 𝑑𝑑𝑑𝑑 = 𝑚𝑚w 𝑐𝑐pw 𝑑𝑑𝑑𝑑 54

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Chapter 4 - Transport Equations for Single-Phase Flow 𝑇𝑇f 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑇𝑇f 333.3 � = 𝑚𝑚w 𝑐𝑐pw � = 𝑚𝑚w 𝑐𝑐pw ln � � = 9.1(4.2) ln = 6.96 kJ⁄K 𝑇𝑇 𝑇𝑇w 277.8 𝑇𝑇w 𝑇𝑇

𝑑𝑑𝑑𝑑 =

∴ ∆𝑆𝑆 = � 𝑑𝑑𝑑𝑑 = �

𝑑𝑑𝑑𝑑 𝑇𝑇

𝑑𝑑𝑑𝑑 = 6.96 kJ⁄K 𝑇𝑇

Heater as a system: Considering the heater as the system we have �

𝑇𝑇f 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 = −𝑚𝑚w 𝑐𝑐pw � = 6.96 kJ⁄K 𝑇𝑇 𝑇𝑇𝑤𝑤 𝑇𝑇

∆𝑺𝑺 = � 𝒅𝒅𝒅𝒅 = �

𝑑𝑑𝑑𝑑 =

𝑑𝑑𝑑𝑑 𝑇𝑇

𝑻𝑻𝐰𝐰 𝒅𝒅𝒅𝒅 𝒅𝒅𝒅𝒅 = 𝒎𝒎𝐡𝐡 𝒄𝒄𝐩𝐩𝐩𝐩 � = 𝟒𝟒𝟒𝟒. 𝟕𝟕 𝐉𝐉⁄𝐤𝐤𝐤𝐤 𝑻𝑻 𝑻𝑻𝐡𝐡 𝑻𝑻

Water and heater as a system: For the water and heater as the system we have 𝑑𝑑𝑑𝑑 = 0

∆𝑆𝑆 = � 𝑑𝑑𝑑𝑑 = �

∴�

𝑑𝑑𝑑𝑑 =0 𝑇𝑇

𝑇𝑇f 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 = �𝑚𝑚w 𝑐𝑐pw + 𝑚𝑚h 𝑐𝑐ph � � = 6.92 + 0.042 = 6.96 kJ⁄K 𝑇𝑇 𝑇𝑇w 𝑇𝑇

PROBLEM 4.5 QUESTION

Qualifying a Claim Against the First and Second Laws of Thermodynamics (Section 4.3)

An engineer claims to have invented a new compressor that can be used with a small gas-cooled reactor. The CO2 used for cooling the reactor enters the compressor at 1.378 MPa and 48.9 °C and leaves the compressor at 2.067 MPa and −6.7°C. The compressor requires no input power but operates simply by transferring heat from the gas to a low-temperature reservoir surrounding the compressor. The inventor claims that the compressor can handle 0.908 kg of CO2 per second if the temperature of the reservoir is −95.6°C and the rate of heat transfer is -63.9 kJ/s. (This heat transfer is out of the control volume and hence taken negative.) Assume that the CO2 enters and leaves the device at very low velocities and that no significant elevation changes are involved. 1. Determine if the compressor violates the first or the second law of thermodynamics. 2. Draw a schematic of the process on a T-s diagram.

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3. Determine the change in the availability function (A) of the fluid flowing through the compressor. 4. Using the data from items 1 through 3, determine if the compressor is theoretically possible. For CO₂ supercritical properties of this problem, use the website: http://www.peacesoftware.de/einigewerte/co2_e.html Answers: 1. The compressor does not violate either the first or second law. 2. ΔA = 10.5 kJ/kg

PROBLEM 4.5 SOLUTION Qualifying a Claim Against the First and Second Laws of Thermodynamics (Section 4.3)

FIGURE SM-4.2 New compressor for small gas-cooled reactor From the problem statement we have: – entering pressure, Pi = 1.378 MPa – entering temperature, Ti = 48.9 °C – exiting pressure, Po = 2.067 MPa – exiting temperature, To = −6.7 °C – heat transfer rate, Q̇ = 63.9 kW – mass flow rate, ṁ = 0.908 kg/s – reservoir temperature, TR = −95 °C Thermodynamic properties evaluated at temperature and pressure for carbon dioxide (real gas) : ℎi = 515.9 kJ⁄kg

𝑠𝑠i = 2.3 kJ⁄kg K

ℎo = 452.0 kJ⁄kg 𝑠𝑠o = 2.0 kJ⁄kg K

The first law of thermodynamics for this case is

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Chapter 4 - Transport Equations for Single-Phase Flow

Q̇ + 𝑚𝑚̇i �ℎ𝑖𝑖 +

𝑣𝑣i2 𝑑𝑑𝑑𝑑 𝑣𝑣o2 + 𝑔𝑔𝑔𝑔i � = � � + 𝑚𝑚̇o �ℎo + + 𝑔𝑔𝑔𝑔o � + 𝑊𝑊 2 𝑑𝑑𝑑𝑑 CV 2

𝑑𝑑𝑑𝑑 No elevation change so zi = zo and we are at steady state so that 𝑚𝑚̇i = 𝑚𝑚̇̇ o = 𝑚𝑚̇, � 𝑑𝑑𝑑𝑑 �

Vi = Vo. There is also no work Ẇ = 0. The first law therefore reduces to

= 0 and

Q̇ = 𝑚𝑚̇(ℎo − ℎi ) = (452.0 − 515.9) = −63.9 kW

The first law of thermodynamics is satisfied.

For the second law analysis, use Equation 4.41:

 ∂S    cv=  ∂t 

∑ m s + S i i

gen

Q TS

which for our steady state case reduces to

Q 0 = m ( si − so ) + Sgen + TR Q Sgen= m ( so − si ) − TR

( −63.9 ) kW kg kJ Sgen = 0.908 (2.0 − 2.3) − s kgK 273.15 − 95.6 K kJ Q = −63.9 s since heat is being transferred out of the control volume. where kW Sgen = −0.272 + 0.360 K kW Sgen = 0.088 K ∆S = 𝑚𝑚̇(𝑠𝑠o − 𝑠𝑠i ) = −0.264 kW⁄K From the results, since

Sgen

𝑸𝑸̇ = −𝟎𝟎. 𝟑𝟑𝟑𝟑𝟑𝟑 𝐤𝐤𝐤𝐤⁄𝐊𝐊 𝑻𝑻𝐑𝐑

is positive, the second law of thermodynamics is satisfied.

2. Figure SM-4.3 is a schematic of the process.

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Chapter 4 - Transport Equations for Single-Phase Flow

FIGURE SM-4.3 Process diagram for compressor of Problem 4.5 3. The change in availability function is defined as follows (the Gibbs free energy): For this case E is equal to U.

𝐴𝐴 = 𝐸𝐸 + 𝑃𝑃𝑅𝑅 𝜐𝜐 − 𝑇𝑇𝑅𝑅 𝑠𝑠

Hence: ∆𝐴𝐴 = (𝐸𝐸𝑖𝑖 − 𝐸𝐸𝑜𝑜 ) − 𝑇𝑇𝑅𝑅 (𝑠𝑠𝑖𝑖 − 𝑠𝑠𝑜𝑜 ) = (ℎ𝑖𝑖 − ℎ𝑜𝑜 ) − 𝑇𝑇𝑅𝑅 (𝑠𝑠𝑖𝑖 − 𝑠𝑠𝑜𝑜 ) = 13.1 kJ⁄kg

∆Aper = = 10.46 {515.9 − 452.0} − ( −95.6 + 273.15){2.3 − 2.0= } 63.9 − (53.44) kg

kJ kg

4. The compressor is theoretically possible. However, since it needs a very low temperature heat sink, it is not practical.

PROBLEM 4.6 QUESTION Determining Rocket Acceleration from an Energy Balance (Section 4.3) A rocket ship is traveling in a straight line through outer space, beyond the range of all gravitational forces. At a certain instant, its velocity is V, its mass is M, the rate of consumption of propellant is P, the rate of energy liberation by the chemical reaction is Q̇R. From the first law of thermodynamics alone (i.e., without using a force balance on the rocket), derive a general expression for the rate of change of velocity with time as a function of the above parameters and the discharged gas velocity Vd and enthalpy hd. In solving this problem, place a control volume around the rocket ship within which the propellant is contained, neglect the mass of the rocket ship itself and take the discharge velocity Vd as that with respect to the rocket ship.

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Chapter 4 - Transport Equations for Single-Phase Flow

Answer: 𝑑𝑑𝑑𝑑 1 𝑑𝑑𝑑𝑑 𝑉𝑉 2 𝑉𝑉d2 = �𝑄𝑄̇R − 𝑀𝑀 + �𝑢𝑢 + − ℎ𝑑𝑑 − � 𝑃𝑃� 𝑑𝑑𝑑𝑑 𝑀𝑀𝑀𝑀 𝑑𝑑𝑑𝑑 2 2

PROBLEM 4.6 SOLUTION

Determining Rocket Acceleration from an Energy Balance (Section 4.3) Energy balance equation: 𝑑𝑑𝑑𝑑 𝑉𝑉d2 = 𝑚𝑚̇d �ℎd + � + 𝑄𝑄̇R 𝑑𝑑𝑑𝑑 2

(1)

where ṁd is the mass flow rate of discharged gas. The energy at any point in time is

From conservation of mass

𝐸𝐸 = 𝑀𝑀 �ℎd + 𝑚𝑚̇𝑑𝑑 =

𝑉𝑉2 2

�

𝑑𝑑𝑑𝑑 = −𝑃𝑃 𝑑𝑑𝑑𝑑

(2)

(3)

Substituting these equations into Equation 1:

�ℎ +

𝑑𝑑 𝑉𝑉𝟐𝟐 𝑉𝑉2d �𝑀𝑀 �ℎ + �� = −𝑃𝑃 �ℎ + � + 𝑄𝑄̇ R 𝑑𝑑𝑑𝑑 2 2

𝑉𝑉𝟐𝟐 𝑑𝑑𝑑𝑑 2

�

𝑑𝑑𝑑𝑑

+ 𝑀𝑀

𝑑𝑑ℎ 𝑑𝑑 𝑉𝑉2 𝑉𝑉2𝑑𝑑 + 𝑀𝑀 � � = −𝑃𝑃 �ℎd + � + 𝑄𝑄̇ R 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 2 2

𝑑𝑑𝑑𝑑 𝑑𝑑ℎ 𝑉𝑉2d 𝑉𝑉2 ̇ 𝑀𝑀𝑀𝑀 = −𝑃𝑃 �ℎd + � + 𝑄𝑄R + 𝑃𝑃 �ℎ + � − 𝑀𝑀 𝑑𝑑𝑑𝑑 2 2 𝑑𝑑𝑑𝑑

(4) (5) (6)

Neglecting the gravitational term, h = u and therefore

𝑑𝑑𝑑𝑑 1 𝑑𝑑𝑑𝑑 𝑉𝑉 2 𝑉𝑉d2 = �𝑄𝑄̇R − 𝑀𝑀 − 𝑃𝑃 �𝑢𝑢 + − ℎd − �� 𝑑𝑑𝑑𝑑 𝑀𝑀𝑀𝑀 𝑑𝑑𝑑𝑑 2 2

(7)

PROBLEM 4.7 QUESTION

Momentum Balance for a Control Volume (Section 4.3) A jet of water is directed at a vane (Figure 4.11) that could be a blade in a turbine. The water leaves the nozzle with a speed of 15 m/s and a mass flow of 250 kg/s; it enters the vane tangent to its surface (in the x direction). At the point the water leaves the vane, the angle to the positive

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Chapter 4 - Transport Equations for Single-Phase Flow

x direction of 120°. Compute the resultant force on the vane assuming there is no friction if: 1. The vane is held constant. 2. The vane moves with a velocity of 5 m/s in the x direction. Answers: 1. Fx = 5625 N, Fy = −3248 N 2. Fx = 2500 N, Fy = −1443 N

FIGURE 4.11 Jet of water directed at a vane.

PROBLEM 4.7 SOLUTION Momentum Balance for a Control Volume (Section 4.3)

FIGURE SM-4.4 Free body diagram of jet directed at vane of Problem 4.7 A jet of water is directed at a vane that could be a blade in a turbine. The water leaves the nozzle with a speed of 15 m/s and a mass flow of 250 kg/s; it enters the vane tangent to its surface (in the x direction). At the point the water leaves the vane, the angle to the positive x direction is

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Chapter 4 - Transport Equations for Single-Phase Flow

120°. 1. The vane is held constant: We can write conservation of momentum for a control volume surrounding the vane following Equation 4.32 of the text, � 𝐹𝐹⃗k = � k

𝑑𝑑𝑑𝑑𝜐𝜐⃗ 𝑑𝑑𝑑𝑑

�

𝑂𝑂

cυ

(1)

= � 𝑚𝑚̇i 𝜐𝜐⃗i o=1

where �𝑭𝑭⃗𝐤𝐤 is an external force acting on the control volume, ṁi is the mass inflow rate through a �⃗𝐢𝐢 is the velocity at which the fluid enters through that boundary of the control volume and 𝝊𝝊 boundary of the control volume. Similarly the subscript o is for outflows. For this problem we have two places where the fluid crosses the boundary of the control volume. Using continuity, we know that the mass flow rate entering and exiting the control volume is 𝑚𝑚̇in = 250

kg kg 𝑚𝑚̇out = 250 s s

Since the flow is assumed incompressible and the cross section flow are does not change, the speed of the fluid in and out of the control volume is the same, 𝜐𝜐𝑖𝑖𝑖𝑖 = 15

m m 𝜐𝜐𝑜𝑜𝑜𝑜𝑜𝑜 = 15 s s

We may break Equation 1 into vector components in the +x and +y directions: 𝐹𝐹x = 𝑚𝑚̇out 𝜐𝜐x,in − 𝑚𝑚̇𝑖𝑖𝑖𝑖 𝜐𝜐x,in 𝐹𝐹y = 𝑚𝑚̇out 𝜐𝜐y,out

We may define the components of velocity in the x direction as: 𝜐𝜐x,in = 𝜐𝜐in = 15

m s

𝜐𝜐x,out = 𝜐𝜐out [− cos(60°)] = −7.5

m s

We may define the components of velocity in the y direction as: 𝜐𝜐y,out = 𝜐𝜐out [sin(60°)] = 12.99

m s

We can now solve for the x and y components of the resultant force, 𝐹𝐹x = 250

m kg m �−7.5 − 15 � = −5625 N s s s

𝐹𝐹y = 250

kg m �12.99 � = 3248 N s s

Note these forces reflect the the force of the vane on the fluid. We may apply Newton’s 3rd Law

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Chapter 4 - Transport Equations for Single-Phase Flow

to determine that the force of the fluid on the vane is equal and opposite therefore: 𝐹𝐹xυ = 5625 N 𝐹𝐹yυ = −3248 N

2. The vane moves with a velocity of 5 m/s in the x direction: The same approach will be taken as in case 1, except a Lagrangian view will be used to solve this problem. If the control volume is formulated such that it is moving with the vane, the relative velocity of the entering fluid needs to be calculated. The velocity of the vane is given as

The relative velocity of the fluid entering is

𝜐𝜐o = 5

m s

𝜐𝜐r = 𝜐𝜐in − 𝜐𝜐o = 10

m s

Assuming that the flow is incompressible, the mass flow rate entering the control volume can be calculated by 𝑚𝑚̇ = 𝜌𝜌𝜌𝜌 = const υ

Therefore, we can calculate the mass flow rate of this new frame of reference as 𝜐𝜐r kg 𝑚𝑚̇r = 𝑚𝑚̇in � � = 166.667 𝜐𝜐in s

From conservation of mass, the mass flow rate entering the moving control volume will be equivalent to the mass flow rate exiting the control volume. Since the fluid is assumed incompressible and the flow area is not changing, the speed at the exit must be equivalent to the speed entering the control volume. From conservation of momentum we have 𝐹𝐹x,r = 𝑚𝑚̇r 𝜐𝜐x,r,out − 𝑚𝑚̇𝑟𝑟 𝜐𝜐x,r,in 𝐹𝐹y,r = 𝑚𝑚̇r 𝜐𝜐y,r,out

The velocity components in the x direction are

𝜐𝜐x,r,in = 𝜐𝜐r = 10

m s

𝜐𝜐x,r,in = 𝜐𝜐r [−cos(60°)] = −5

The velocity components in the y direction are

𝜐𝜐y,r,out = 𝜐𝜐r [sin(60°)] = 8.66

m s

We can now solve for the x and y components of the resultant force,

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Chapter 4 - Transport Equations for Single-Phase Flow

𝐹𝐹x,r = 166.667

kg m m �−5 − 10 � = −2500 N s s s

𝐹𝐹y,r = 166.667

kg m �8.66 � = 1443 N s s

Note these forces reflect the the force of the vane on the fluid. We may apply Newton’s 3rd Law to determine that the force of the fluid on the vane is equal and opposite, therefore: υ υ 𝐹𝐹x,r = 2500 N 𝐹𝐹y,r = −1443 N

PROBLEM 4.8 QUESTION Internal Conservation Equations for an Extensive Property (Section 4.4) Consider a coolant in laminar flow in a circular tube. The one-dimensional velocity is given by 𝑇𝑇 2 𝜐𝜐⃗ = 𝜐𝜐max �1 − � � � 𝚤𝚤⃗z 𝑅𝑅

where υmax = 2.0 m/s; R = radius of the tube = 0.05 m; 𝚤𝚤⃗z = 𝑎𝑎 unit vector in the axial direction. Assume the fluid density is uniform within the tube (ρ0 = 300 kg/m3). 1. What is the coolant flow rate in the rube?

2. What is the coolant average velocity (V) in the tube? 𝟏𝟏

3. What is the true kinetic head of this flow? Does the kinetic head equal 𝟐𝟐 𝝆𝝆𝟎𝟎 𝑽𝑽𝟐𝟐 ?

Answers:

 1. V = 7.854 × 10−3 m3/s 2. V = 1 m/s 𝝆𝝆 𝑽𝑽𝟐𝟐

3. 𝐊𝐊𝐊𝐊𝐊𝐊𝐊𝐊𝐊𝐊𝐊𝐊𝐊𝐊 𝐡𝐡𝐡𝐡𝐡𝐡𝐡𝐡 = 𝟏𝟏. 𝟑𝟑𝟑𝟑 𝐨𝐨𝟐𝟐

PROBLEM 4.8 SOLUTION

Internal Conservation Equations for an Extensive Property (Section 4.4) From the problem statement we are give: – velocity profile

𝑟𝑟 2 𝜐𝜐⃗(𝑟𝑟) = 𝜐𝜐max �1 − � � � 𝚤𝚤⃗z 𝑅𝑅

– maximum velocity, υmax = 2.0 m/s – radius, R = 0.05 m

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Chapter 4 - Transport Equations for Single-Phase Flow

– density, ρ0 = 300 kg/m3 Calculate coolant flow rate: The coolant flow rate can be determine by integrating the velocity profile over the cross sectional area R

V̇ = � 𝜐𝜐(𝑟𝑟)𝑑𝑑𝑑𝑑 = � 𝜐𝜐(𝑟𝑟)2𝜋𝜋𝜋𝜋𝜋𝜋𝜋𝜋 = 𝜋𝜋𝜐𝜐max 0

𝑅𝑅 2 = 7.854 × 10−3 m3 ⁄s 2

(1)

Average velocity: The average velocity can be determined by averaging over the cross sectional area 𝑹𝑹𝟐𝟐 ̇ 𝝅𝝅𝝅𝝅 𝝊𝝊𝐦𝐦𝐦𝐦𝐦𝐦 𝐦𝐦𝐦𝐦𝐦𝐦 𝟐𝟐 ∬ 𝝊𝝊(𝒓𝒓)𝒅𝒅𝒅𝒅 𝐕𝐕 𝑽𝑽 = = = = = 𝟏𝟏 𝐦𝐦⁄𝐬𝐬 𝟐𝟐 𝑨𝑨 𝝅𝝅𝑹𝑹 𝟐𝟐 ∬ 𝒅𝒅𝒅𝒅

(2)

Kinetic head: The kinetic head is determined with 𝑃𝑃k =

𝜌𝜌𝜌𝜌(𝑟𝑟)2 2

(3)

The average kinetic is therefore determined with 𝑃𝑃�k =

2 1 𝜌𝜌0 𝜐𝜐max 4𝜌𝜌0 𝑉𝑉 2 𝑉𝑉 2 2 � 𝜌𝜌0 𝜐𝜐(𝑟𝑟) 2𝜋𝜋𝜋𝜋𝜋𝜋𝜋𝜋 = = 1.333𝜌𝜌0 6 0 2 2𝐴𝐴

(4)

𝑉𝑉 2

Thus, the kinetic head does not equal 2 𝜌𝜌0 2

PROBLEM 4.9 QUESTION Differential Transport Equations (Section 4.5) Can the following sets of velocities belong to possible incompressible flow cases? Case 1: υx = x − y + z2 υy = x − y + z υz = 2xy + y2 Case 2: υx = xyzt υy = −xyzt2 z2 𝜐𝜐𝑧𝑧 = (xt 2 − yt) 2 64

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Chapter 4 - Transport Equations for Single-Phase Flow

Answers: 1. Yes 2. Yes

PROBLEM 4.9 SOLUTION Differential Transport Equations (Section 4.5) Case 1: For case one we have 𝜐𝜐x = 𝑥𝑥 − 𝑦𝑦 + 𝑧𝑧 2 𝜐𝜐y = 𝑥𝑥 − 𝑦𝑦 + 𝑧𝑧

Continuity equation:

For incompressible flow:

𝜐𝜐z = 2𝑥𝑥𝑥𝑥 + 𝑦𝑦 2

𝐷𝐷𝐷𝐷 + 𝜌𝜌(∇ · 𝜐𝜐⃗) = 0 𝐷𝐷𝐷𝐷 ∇ · 𝜐𝜐⃗ = 0

∇ · 𝜐𝜐⃗ =

𝜕𝜕𝜕𝜕x 𝜕𝜕𝜕𝜕y 𝜕𝜕𝜕𝜕z 𝜕𝜕 𝜕𝜕 𝜕𝜕 + + = = (𝑥𝑥 + 𝑦𝑦 − 𝑧𝑧 2 ) + (𝑥𝑥 − 𝑦𝑦 + 𝑧𝑧) + (2𝑥𝑥𝑥𝑥 + 𝑦𝑦 2 ) 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 ∇ · 𝜐𝜐⃗ = 1 − 1 + 0 = 0

Therefore this flow can belong to an incompressible flow. Case 2: For case two we have 𝜐𝜐x = 𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥

𝜐𝜐y = −𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥 2

Continuity equation: For incompressible flow:

𝑧𝑧 2 𝜐𝜐z = (𝑥𝑥𝑥𝑥 2 − 𝑦𝑦𝑦𝑦) 2 𝐷𝐷𝐷𝐷 + 𝜌𝜌(∇ · 𝜐𝜐⃗) = 0 𝐷𝐷𝐷𝐷 ∇ · 𝜐𝜐⃗ = 0

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Chapter 4 - Transport Equations for Single-Phase Flow

∇ · 𝜐𝜐⃗ =

𝜕𝜕𝜕𝜕x 𝜕𝜕𝜕𝜕y 𝜕𝜕𝜕𝜕z 𝜕𝜕 𝜕𝜕 𝜕𝜕 𝑧𝑧 2 (−𝑥𝑥𝑥𝑥𝑥𝑥𝑡𝑡 2 ) + � (𝑥𝑥𝑥𝑥 2 − 𝑦𝑦𝑦𝑦)� + + = = (𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥) + 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 2 ∇ · 𝜐𝜐⃗ = 𝑦𝑦𝑦𝑦𝑦𝑦 − 𝑥𝑥𝑥𝑥𝑡𝑡 2 + 𝑥𝑥𝑥𝑥𝑡𝑡 2 − 𝑦𝑦𝑦𝑦𝑦𝑦 = 0

Therefore this flow can belong to an incompressible flow.

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Chapter 5 Transport Equations for Two-Phase Flow Contents Problem 5.1 Area-averaged parameters ...............................................................................

Problem 5.2 Momentum balance for a two-phase jet load ..................................................

Problem 5.3 Estimating phase velocity differential ..............................................................

Problem 5.4 Torque on vessel due to jet from hot leg break ................................................

Problem 5.5 Interfacial term in the momentum equation .....................................................

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Chapter 5 - Transport Equations for Two-Phase Flow

PROBLEM 5.1 QUESTION Area Averaged Parameters (Section 5.4) In a BWR assembly it is estimated that the exit quality is 0.15 and the mass flow rate is 17.5 kg/s. If the pressure is 7.2 MPa, and the slip ratio can be given as S = 1.5, determine {α}, {β}, {jv}, Gv, and Gl. The flow area of the assembly is 1.2 × 102 m2. Answers: {α} = 0.6968 {β} = 0.7751 { j v } = 5.80 m/s Gv = 218.75 kg/m2 s Gl = 1239.6 kg/m2 s

PROBLEM 5.1 SOLUTION Area Averaged Parameters (Section 5.4) From the problem statement we are given: – exit quality, xex = 0.15 – mass flow rate, ṁ = 17.5 kg/s – slip ratio, S = 1.5 – cross sectional area, A = 1.2 × 10−2 m2 – operating pressure, P = 7.2 MPa At the operating pressure, the densities of the vapor and liquid phase are ρg = 37.71 kg/m3

ρf = 736.49kg/m3

Void Fraction To determine the void fraction we use {𝛼𝛼} =

1 = 0.6968 1 − 𝑥𝑥 𝜌𝜌𝑔𝑔 1 + 𝑥𝑥 𝑒𝑒𝑒𝑒 𝜌𝜌 𝑆𝑆 𝑒𝑒𝑒𝑒 𝑓𝑓

Volumetric Flow Fraction The volumetric flow fraction is {𝛽𝛽} =

1 = 0.7751 1 − 𝑥𝑥𝑒𝑒𝑒𝑒 𝜌𝜌𝑔𝑔 1 + 𝑥𝑥 𝜌𝜌𝑓𝑓 𝑒𝑒𝑒𝑒

Superficial Vapor Velocity The superficial vapor velocity is

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Chapter 5 - Transport Equations for Two-Phase Flow

{𝑗𝑗𝑣𝑣 } =

𝑚𝑚̇ 𝑣𝑣 𝑚𝑚̇𝑥𝑥 = = 5.80 m⁄s 𝜌𝜌𝑔𝑔 𝐴𝐴 𝜌𝜌𝑔𝑔 𝐴𝐴

Vapor Mass Flux The vapor mass flux is {𝐺𝐺𝑣𝑣 } =

𝑚𝑚̇𝑣𝑣 𝑚𝑚̇𝑥𝑥 = = 218.75kg/m2 s 𝐴𝐴 𝐴𝐴

Liquid Mass Flux The liquid mass flux is {𝐺𝐺𝑙𝑙 } =

𝑚𝑚̇𝑙𝑙 𝑚𝑚̇(1 − 𝑥𝑥) = = 1239.6kg⁄m2 s 𝐴𝐴 𝐴𝐴

PROBLEM 5.2 QUESTION

Momentum Balance for a Two-phase Jet Load (Section 5.5) Calculate the force on a wall subjected to a two-phase jet (Figure 5.10) that has the following parameters: Mass flux at exit, Gm = 10.75 × 103 kg/m2 s Exit diameter, D = 0.3 m Upstream pressure, P = 7.2 MPa Pressure at throat, Po = 3.96 MPa Quality at exit, xo = 0.68 Slip ratio, So = 1.5

FIGURE 5.10 Two-phase jet impacting a wall. Answer: Force = 0.416 MN

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Chapter 5 - Transport Equations for Two-Phase Flow

PROBLEM 5.2 SOLUTION Momentum Balance for a Two-phase Jet Load (Section 5.5)

G = 10.75 x 103 kg/m2 sec Po (Pressure at throat) = 3.96 MPa xo = 0.68 s = 1.5

FIGURE SM-5.1 Control volume for Problem 5.2 Momentum Equation: 𝜕𝜕 [𝑚𝑚 𝑣𝑣⃗ ] = �(𝑚𝑚̇𝐾𝐾 𝑣𝑣⃗𝐾𝐾 ) + 𝑚𝑚̇𝐾𝐾 𝑣𝑣⃗𝐾𝐾𝑠𝑠 + � 𝐹𝐹⃗𝑗𝑗𝑗𝑗 − ��𝑃𝑃𝑃𝑃𝐴𝐴⃗𝐾𝐾 � + 𝐹𝐹⃗𝑠𝑠𝑠𝑠 + 𝑚𝑚𝐾𝐾 𝑔𝑔⃗ 𝑗𝑗 𝜕𝜕𝜕𝜕 𝐾𝐾 𝐾𝐾

Assumptions:

𝑗𝑗

(1)

𝑗𝑗

– steady state – the cross section of jet area is constant – horizontal flow – thermal equilibrium (no phase transfer inside control volume) – no friction In the x direction: [𝑥𝑥𝑜𝑜 𝜐𝜐𝜐𝜐𝜐𝜐 + (1 − 𝑥𝑥𝑜𝑜 )𝜐𝜐𝑙𝑙𝑙𝑙 ]𝑚𝑚̇ + 𝑃𝑃𝑜𝑜 𝐴𝐴𝑜𝑜 − 𝑃𝑃𝑤𝑤 𝐴𝐴𝑤𝑤 = 0

(2)

𝐹𝐹 = [𝑥𝑥𝑜𝑜 𝜐𝜐𝜐𝜐𝜐𝜐 + (1 − 𝑥𝑥𝑜𝑜 )𝜐𝜐𝑙𝑙𝑙𝑙 ]𝑚𝑚̇ + 𝑃𝑃𝑜𝑜 𝐴𝐴𝑜𝑜

(3)

where Ao = Aw since the cross sectional area is constant and the force is given by F = PwAw. Therefore we can rewrite as The vapor velocity can be determined as

and the liquid velocity is

𝜐𝜐𝜐𝜐𝜐𝜐 =

𝐺𝐺𝑚𝑚 𝑥𝑥𝑜𝑜 𝜌𝜌𝜐𝜐 𝛼𝛼𝑜𝑜

(4)

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Chapter 5 - Transport Equations for Two-Phase Flow

𝜐𝜐𝑙𝑙𝑙𝑙 =

𝐺𝐺𝑚𝑚 (1 − 𝑥𝑥𝑜𝑜 ) 𝜌𝜌𝑙𝑙 (1 − 𝛼𝛼𝑜𝑜 )

(5)

The void fraction can be determined from the slip ratio and quality 𝛼𝛼𝑜𝑜 =

1 1~𝑥𝑥 𝜌𝜌 1 + 𝑥𝑥 𝑜𝑜 𝜌𝜌𝑣𝑣 𝑆𝑆 𝑜𝑜 𝑙𝑙

(6)

Substituting these equations into Equation (3) and substituting GmAo = ṁ, we get 2 𝐹𝐹 = 𝐴𝐴𝑜𝑜 𝐺𝐺𝑚𝑚 �

(1 − 𝑥𝑥𝑜𝑜 )2 𝑥𝑥𝑜𝑜2 + � + 𝑝𝑝𝑜𝑜 𝐴𝐴𝑜𝑜 𝜌𝜌𝜐𝜐 𝛼𝛼𝑜𝑜 𝜌𝜌𝑙𝑙 (1 − 𝛼𝛼𝑜𝑜 )

(7)

At the upstream pressure (7.2 MPa), the densities of vapor and liquid are 𝜌𝜌𝜐𝜐 = 𝜌𝜌𝑔𝑔 = 37.71kg/m3

The cross sectional area is

𝜌𝜌𝑙𝑙 = 𝜌𝜌𝑓𝑓 = 736.84kg/m3

𝜋𝜋𝐷𝐷2 = 7.0686 × 10−2 m2 𝐴𝐴𝑜𝑜 = 4

(8)

(9)

The void fraction can be determined with Equation (6) to be αo = 0.9651. Solving for the force in Equation (7) we get 𝐹𝐹 = 0.461 MN

(10)

PROBLEM 5.3 QUESTION Estimating phase velocity differential (Section 5.4) A vertical tube is operating at high pressure conditions as follows (Figure 5.11): Operating Conditions Pressure, P = 7.4 MPa Mass flux, G = 2000 kg/m2 s Quality at exit, xe = 0.0693 Geometry D = 10.0 mm L = 3.66 m Saturated water properties at 7.4 MPa hf = 1331.33 kJ/kg hg = 2759.60 kJ/kg hfg = 1448.27 kJ/kg vf = 0.001381 m3/kg vg = 0.02390 m3/kg

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Chapter 5 - Transport Equations for Two-Phase Flow

vfg = 0.02252 m3/kg

FIGURE 5.11 Cross-section of a high-pressure tube. Assuming that thermal equilibrium between steam and water has been attained at the tube exit, find: 1. The tube exit cross-sectional averaged true and superficial vapor velocities; that is, find {υυ}υ and {jυ}. 2. The difference between the tube exit cross sectional averaged vapor and liquid velocities; that is, find {υυ}υ − {υl}l at the exit. 3. The difference between the tube exit cross-sectional averaged vapor and liquid superficial velocities; that is, find {j υ} − {j l }. Answers: 1. {υυ}υ = 7.45 m/s, {j υ} = 3.31 m/s 2. 2.83 m/s 3. 0.74 m/s

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Chapter 5 - Transport Equations for Two-Phase Flow

PROBLEM 5.3 SOLUTION Estimating phase velocity differential (Section 5.4)

G = 2000 kg/m2 sec D = 10.0 mm = 1.0 x 10-2m L = 3.66 m P = 7. MPa Figure 5.11 Cross-section of a high-pressure tube.

xe = 0.0693

1. The tube exit cross-section averaged true and superficial vapor velocity 𝐺𝐺𝑥𝑥𝑒𝑒 = 𝐺𝐺𝑥𝑥𝑒𝑒 v𝑔𝑔 = 3.31 m⁄s 𝜌𝜌𝜐𝜐 𝜋𝜋 𝐷𝐷 2 void area 4 4 � 3 � 4 {𝛼𝛼}𝑒𝑒 = = 𝜋𝜋 = = 0.444 2 flow area 9 4 𝐷𝐷 {𝑗𝑗𝜐𝜐 } =

{𝜐𝜐𝜐𝜐 }𝜐𝜐 =

{𝑗𝑗𝜐𝜐 } = 7.45 m⁄s {𝛼𝛼}𝑒𝑒

2. The difference between the tube exit cross-sectional averaged vapor and liquid velocities {𝑗𝑗𝑙𝑙 } =

𝐺𝐺(1 − 𝑥𝑥𝑒𝑒 ) = 𝐺𝐺(1 − 𝑥𝑥𝑒𝑒 )v𝑓𝑓 = 2.57 m⁄s 𝜌𝜌𝑙𝑙 {𝑗𝑗𝑙𝑙 } {𝜐𝜐𝑙𝑙 }𝑙𝑙 = = 4.62 m⁄s {1 − 𝛼𝛼𝑒𝑒 } ∴ {𝜐𝜐𝜐𝜐 }𝜐𝜐 − {𝜐𝜐𝑙𝑙 }𝑙𝑙 = 2.83 m⁄s

3. The difference between the tube cross-sectional averaged vapor and liquid superficial velocities {𝑗𝑗𝜐𝜐 }𝜐𝜐 − {𝑗𝑗𝑙𝑙 } = 0.74 m⁄s

PROBLEM 5.4 QUESTION Torque on Vessel due to Jet from Hot Leg Break (Section 5.5) Calculate the torque on a pressure vessel when a break develops in the hot leg as shown in Figure 5.12. The conditions of the two phase emerging jet are: Ger = 10.75 × 103 kg/m2 s

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Chapter 5 - Transport Equations for Two-Phase Flow

xer = 0.68 L=5m Angle θ = 70° S = 1.5 Po = 3.96 MPa P = 7.2 Mpa Flow area at the rupture ≈ 0.08 m2

FIGURE 5.12 Two-phase jet at a pipe break. Answer: Torque ≈ 2.8 MN m

PROBLEM 5.4 SOLUTION Torque on Vessel due to Jet from Hot Leg Break (Section 5.5)

G = 10.75 x 103 kg/m2 sec x = 0.68 L=5m S = 1.5 Po = 3.96 MPa P = 7.2 MPa A = 0.08 m2 FIGURE 5.12

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Chapter 5 - Transport Equations for Two-Phase Flow

Momentum Equation: 𝜕𝜕 [𝑚𝑚 𝑣𝑣⃗ ] = �(𝑚𝑚̇𝐾𝐾 𝑣𝑣⃗𝐾𝐾 ) + 𝑚𝑚̇𝐾𝐾 𝑠𝑠 𝑣𝑣⃗𝐾𝐾𝑠𝑠 + � 𝐹𝐹⃗𝑗𝑗𝑗𝑗 − ��𝑃𝑃𝐾𝐾 𝐴𝐴⃗𝐾𝐾 � + 𝐹𝐹⃗𝑠𝑠𝑠𝑠 + 𝑚𝑚𝐾𝐾 𝑔𝑔⃗ 𝑗𝑗 𝜕𝜕𝜕𝜕 𝐾𝐾 𝐾𝐾

Assumptions: – steady state

𝑗𝑗

(1)

𝑗𝑗

– uniform pressure across the cross-section – no gravitational change 0 = [𝑥𝑥𝑖𝑖 𝜐𝜐𝜐𝜐𝜐𝜐 + (1 − 𝑥𝑥𝑖𝑖 )𝜐𝜐𝑙𝑙𝑙𝑙 ]𝑚𝑚̇𝚤𝚤⃗ − [𝑥𝑥𝑜𝑜 𝜐𝜐𝜐𝜐𝜐𝜐 + (1 − 𝑥𝑥𝑜𝑜 )𝜐𝜐𝑙𝑙𝑙𝑙 ]𝑚𝑚̇𝑛𝑛�⃗𝑜𝑜

(2)

0 = 𝑃𝑃𝐴𝐴𝑖𝑖 𝚤𝚤⃗ − (𝑃𝑃𝑜𝑜 − 𝑃𝑃𝑎𝑎𝑎𝑎𝑎𝑎 )𝐴𝐴𝑜𝑜 𝑛𝑛�⃗𝑜𝑜 + 𝐹𝐹⃗

(3)

𝐹𝐹𝑦𝑦 = [𝑥𝑥𝑜𝑜 𝜐𝜐𝜐𝜐𝜐𝜐 + (1 − 𝑥𝑥𝑜𝑜 )𝜐𝜐𝑙𝑙𝑙𝑙 ]𝑚𝑚̇ sin 𝜃𝜃 + (𝑃𝑃𝑜𝑜 − 𝑃𝑃𝑎𝑎𝑎𝑎𝑎𝑎 )𝐴𝐴𝑜𝑜 sin 𝜃𝜃

(4)

where 𝑛𝑛�⃗𝑜𝑜 is normal vector of rupture cross section. Only the y-direction component of 𝐹𝐹⃗ will cause a torque on the vessel. The y-direction balance: where

𝜐𝜐𝜐𝜐𝜐𝜐 =

and

𝜐𝜐𝑙𝑙𝑙𝑙 =

𝑗𝑗𝜐𝜐 𝐺𝐺𝑐𝑐𝑐𝑐 𝑥𝑥𝑐𝑐𝑐𝑐 = 𝛼𝛼𝑜𝑜 𝜌𝜌𝜐𝜐 𝛼𝛼𝑜𝑜

𝑗𝑗𝜐𝜐 𝐺𝐺𝑐𝑐𝑐𝑐 (1 − 𝑥𝑥𝑐𝑐𝑐𝑐 ) = 1 − 𝛼𝛼𝑜𝑜 𝜌𝜌𝑙𝑙 (1 − 𝛼𝛼𝑜𝑜 )

(5)

(6)

Substituting the velocities in the momentum equation we get and using mass flux instead of mass flow rate (ṁ= Gcr Ao), (1 − 𝑥𝑥𝑐𝑐𝑐𝑐 )2 𝑥𝑥𝑜𝑜2 (7) 𝐹𝐹𝑦𝑦 = � + � 𝐺𝐺 2 𝐴𝐴 sin 𝜃𝜃 + (𝑝𝑝𝑜𝑜 − 𝑝𝑝𝑎𝑎𝑎𝑎𝑎𝑎 )𝐴𝐴𝑜𝑜 sin 𝜃𝜃 𝜌𝜌𝜐𝜐 𝛼𝛼𝑜𝑜 𝜌𝜌𝑙𝑙 (1 − 𝛼𝛼𝑜𝑜 ) 𝑐𝑐𝑐𝑐 𝑜𝑜 The torque is then

(8)

𝜏𝜏 = 𝐿𝐿𝐿𝐿𝑦𝑦

Thus,

(1 − 𝑥𝑥𝑜𝑜 )2 𝑥𝑥𝑜𝑜2 𝜏𝜏 = �� + � 𝐺𝐺 2 𝐴𝐴 sin 𝜃𝜃 + (𝑃𝑃𝑜𝑜 − 𝑃𝑃𝑎𝑎𝑎𝑎𝑎𝑎 )𝐴𝐴𝑜𝑜 sin 𝜃𝜃� 𝐿𝐿 𝜌𝜌𝜐𝜐 𝛼𝛼𝑜𝑜 𝜌𝜌𝑙𝑙 (1 − 𝛼𝛼𝑜𝑜 ) 𝑐𝑐𝑐𝑐 𝑜𝑜

(9)

The densities of vapor and liquid at 7.2 MPa are The void fraction is

𝜌𝜌𝜐𝜐 = 37.31 kg⁄m3

𝜌𝜌𝑙𝑙 = 736.48 kg⁄m3

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Chapter 5 - Transport Equations for Two-Phase Flow

𝛼𝛼𝑜𝑜 =

and area

1 = 0.9651 1 − 𝑥𝑥𝑐𝑐𝑐𝑐 𝜌𝜌𝜐𝜐 1 + 𝑥𝑥 𝜌𝜌𝑙𝑙 𝑆𝑆 𝑐𝑐𝑐𝑐 𝐴𝐴𝑜𝑜 = 0.08 m2

Solving for the torque we get

𝜏𝜏 = 2.18 MN m

PROBLEM 5.5 QUESTION Interfacial Term in the Momentum Equation (Section 5.6) In a two-fluid model the momentum equation of the vapor in one-dimensional flow may be written from Equation 5.138 for flow in the z direction as 𝜕𝜕 𝜕𝜕 𝜕𝜕𝜕𝜕𝜕𝜕 (𝛼𝛼𝛼𝛼𝜐𝜐 𝑉𝑉𝜐𝜐 𝐴𝐴) = − (𝛼𝛼𝛼𝛼𝜐𝜐 𝑉𝑉𝜐𝜐 𝐴𝐴) + 𝛤𝛤𝐴𝐴𝐴𝐴𝜐𝜐𝜐𝜐 − 𝐴𝐴 − 𝐹𝐹𝑤𝑤𝑤𝑤 − 𝐹𝐹𝑠𝑠𝑠𝑠𝑠𝑠 − 𝛼𝛼𝛼𝛼𝜐𝜐 𝑔𝑔𝑔𝑔 cos 𝜃𝜃 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕

where Γ = rate of mass exchange between vapor and liquid per unit volume; Fwυ = rate of momentum loss at the wall due to friction; A = flow area; Vυs = vapor velocity at the liquidvapor interface; and Fsυ = rate of momentum exchange between vapor and liquid, Given the following: Steady-state flow condition in a tube Tube diameter is D Uniform axial heat flux q″ Annular flow conditions prevail 1. Write appropriate mass balance and energy balance equations for vapor to complete the model. State any assumption you made. 2. Provide an expression for Γ. Justify your answer. Answer: 4𝑔𝑔″

2. 𝛤𝛤 = 𝐷𝐷ℎ

𝑓𝑓𝑓𝑓

PROBLEM 5.5 SOLUTION Interfacial Term in the Momentum Equation (Section 5.6)

Momentum equation of vapor (from Equation 5.138 for flow in the z direction):

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Chapter 5 - Transport Equations for Two-Phase Flow

𝜕𝜕 𝜕𝜕 𝜕𝜕𝜕𝜕𝜕𝜕 (𝛼𝛼𝛼𝛼𝜐𝜐 𝑉𝑉𝜐𝜐 𝐴𝐴) = − (𝛼𝛼𝛼𝛼𝜐𝜐 𝑉𝑉𝜐𝜐 𝐴𝐴) + 𝛤𝛤𝐴𝐴𝐴𝐴𝜐𝜐𝜐𝜐 − 𝐴𝐴 − 𝐹𝐹𝑤𝑤𝑤𝑤 − 𝐹𝐹𝑠𝑠𝑠𝑠 − 𝛼𝛼𝛼𝛼𝜐𝜐 𝑔𝑔𝑔𝑔 cos 𝜃𝜃 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕

where Γ = rate of mass exchange between vapor and liquid per unit volume; Fwυ = rate of momentum loss at the wall due to friction; A = flow area; Vυs = vapor velocity at the liquid-vapor interface; and Fsυ = rate of momentum exchange between vapor and liquid. Mass balance equation for vapor (from Equation 5.132 for flow in the z direction): 𝜕𝜕 𝜕𝜕 (𝛼𝛼𝛼𝛼𝜐𝜐 𝐴𝐴) + [(𝛼𝛼𝛼𝛼𝜐𝜐 𝑉𝑉𝜐𝜐 )𝐴𝐴] = 𝛤𝛤𝛤𝛤 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕

Energy balance equation (from Equation 5.145 for flow in the z direction, neglecting shear effects): 𝜕𝜕 𝜕𝜕 𝜕𝜕𝜕𝜕 [(𝜌𝜌𝜐𝜐 𝑢𝑢𝜐𝜐𝑜𝑜 𝛼𝛼𝜐𝜐 )𝐴𝐴] + [(𝜌𝜌𝜐𝜐 ℎ𝜐𝜐𝑜𝑜 𝑉𝑉𝜐𝜐 𝛼𝛼𝜐𝜐 )𝐴𝐴] = 𝛤𝛤ℎ𝜐𝜐 𝐴𝐴 − �𝑃𝑃𝑘𝑘 � 𝐴𝐴 + (𝑞𝑞𝜐𝜐‴ 𝛼𝛼𝜐𝜐 )𝐴𝐴 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 ‴ ) (𝜌𝜌 𝛼𝛼 )𝐴𝐴 − (𝜌𝜌𝜐𝜐 𝑔𝑔𝑔𝑔𝜐𝜐 𝛼𝛼𝜐𝜐 ) 𝐴𝐴 −(𝑞𝑞𝜐𝜐‴ 𝛼𝛼𝑤𝑤𝑤𝑤 P𝑤𝑤 ) − (𝑞𝑞𝑠𝑠𝑠𝑠 P𝑠𝑠 − 𝜕𝜕𝜕𝜕 𝜐𝜐 𝜐𝜐

Where Pw is the wall perimeter in the Az plane and Ps is the interface perimeter in the Az plane. 2. Assume all heat is used to generate vapor in the steady state conditions. 𝛤𝛤 =

𝑚𝑚̇𝑘𝑘𝑘𝑘 V

(Equation 5.88)

𝑞𝑞 ″ . (π𝐷𝐷. 1)⁄ℎ𝑓𝑓𝑓𝑓 4𝑞𝑞 ″ 𝛤𝛤 = = π𝐷𝐷2 𝐷𝐷ℎ𝑓𝑓𝑓𝑓 . 1 4

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Chapter 6 Thermodynamics of Nuclear Energy Conversion Systems - Nonflow and Steady Flow: First- and SecondLaw Applications Contents Problem 6.1

Work output of a fuel-water interaction............................................................ 79

Problem 6.2

Evaluation of alternate ideal Rankine cycles .................................................... 80

Problem 6.3

Availability analysis of a simplified BWR ....................................................... 85

Problem 6.4

Thermodynamic analysis of a gas turbine ........................................................ 94

Problem 6.5

Analysis of a steam turbine ............................................................................... 95

Problem 6.6

Irreversibility problems involving the Rankine cycle ....................................... 96

Problem 6.7

Replacement of a steam generator in a PWR with a flash tank ....................... 100

Problem 6.8

Advantages of moisture separation and feedwater heating............................... 113

Problem 6.9

Ideal Brayton Cycle .......................................................................................... 114

Problem 6.10 Complex real Brayton cycle.............................................................................. 117 Problem 6.11 Cycle thermal efficiency problem involving a bottoming cycle ...................... 121 Problem 6.12 Nuclear cogeneration plant ............................................................................... 125

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Chapter 6 - Thermodynamics of Nuclear Energy Conversion Systems Nonflow and Steady Flow: First- and Second-Law Applications

PROBLEM 6.1 QUESTION Work Output of a Fuel-Water Interaction (Section 6.2) Compute the work done by a fuel-water interaction assuming that the 40,000 kg of mixed oxide fuel and 4000 kg of water expand independently and isentropically to 1 atmosphere. Assume that the initial fuel and water conditions are such that equilibrium mixture temperature (Te) achieved is 1945 K. Other water conditions are as follows: Tinitial = 400K; ρinitial = 945kg/m3; cυ = 4184J/kg–K. Caution: Equation 6.9 is inappropriate for these conditions, as the coolant at state e is supercritical. Answer: 1.67 × 1010 J

PROBLEM 6.1 SOLUTION Work Output of a Fuel-Water Interaction (Section 6.2) Compute the work done by a fuel-water interaction assuming that the 40,000 kg of mixed oxide fuel and 4000 kg of water (mw) expand independently and isentropically to 1 atmosphere. Assume that the initial fuel and water conditions are such that equilibrium mixture temperature (Te) achieved is 1945 K. Other water conditions are as follows: Tintial = 400 K; ρinitial = 945 kg/m3; cυ = 4184J/kg – K. Let us model the transformation in two steps. First, water and fuel undergo an isochore transformation to reach thermal equilibrium. This means that volume in conserved. This isochore transformation, also called isometric, does, by nature, not produce any work because W= 0 . This transformation turns state 1 to state e. Then the water expands ∫ − pdV =

isentropically and produces the work. This second transformation leads to the final state, state 2. As this expansion is very quick, let us postulate that the transformation is adiabatic. Obtain State 1 Properties To begin, the specific entropy and specific internal energy at state 1 can be determined, using steam tables, from the initial temperature and initial density. − s1 = 1.58787 kJ/kg – K − u1 = 527.081 kJ/kg Find specific entropy at state e For a pure substance the following relation exists between thermodynamic properties for two infinitesimally close equilibrium states (Equation 6.10a): 𝑑𝑑ℎ = 𝑇𝑇𝑇𝑇𝑇𝑇 + 𝜐𝜐𝜐𝜐𝜐𝜐

(1)

𝑑𝑑ℎ = 𝑑𝑑𝑑𝑑 + 𝑝𝑝𝑝𝑝𝑝𝑝 + 𝜐𝜐𝜐𝜐𝜐𝜐

(2)

Recalling that specific enthalpy is defined as h = u +pυi and we may differentiate this to get We may subtract Equation (2) from Equation (1) to get the relation

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Chapter 6 - Thermodynamics of Nuclear Energy Conversion Systems Nonflow and Steady Flow: First- and Second-Law Applications

(3)

0 = 𝑇𝑇𝑇𝑇𝑇𝑇 − 𝑑𝑑𝑑𝑑 − 𝑝𝑝𝑝𝑝𝑝𝑝

In the first process (state 1 to e) an equilibrium temperature is found for the fuel and coolant under the condition of no expansion and therefore, dυ = 0. Rearranging Equation (3): 𝑑𝑑𝑑𝑑 =

𝑑𝑑𝑑𝑑 𝑇𝑇

(4)

Using the state equation that expresses specific internal energy as a function of temperature: 𝑑𝑑𝑑𝑑 =

𝑐𝑐𝜐𝜐 𝑑𝑑𝑑𝑑 𝑇𝑇

(5)

Integrating Equation (5) from state 1 to e, we have 𝑇𝑇e 𝑠𝑠𝑒𝑒 − 𝑠𝑠1 = 𝑐𝑐𝜐𝜐 ln � � 𝑇𝑇1

(6)

We may now evaluate the specific entropy at state e to be: 𝑠𝑠𝑒𝑒 = 1.58787

kJ kJ 1945 K kJ � = 8.205 + 4.184 ln � kg K kg K 400 K kg K

(7)

Calculate specific internal energy at state e Using the state equation we can calculate the specific internal energy at state e: 𝑢𝑢𝑒𝑒 = 𝑢𝑢1 + 𝑐𝑐𝜐𝜐 (𝑇𝑇e − 𝑇𝑇1 ) = 6.99 × 106

J kg

(8)

Obtain specific internal energy at state 2 At state 2 we know that the pressure is P2 = 101.325 kPa and that the process from state 2 to e is isentropic and therefore, 𝑠𝑠2 = 𝑠𝑠𝑒𝑒 = J

8205.1 kg K . Since we have identified two independent state variables, we may use steam tables to determine specific internal energy:

𝑢𝑢2 = 2807.53

kJ kg

(9)

Calculate the work from state e to 2 The process form state e to state 2 is adiabatic and therefore using the first law we arrive at the following expression for the work. 𝑊𝑊𝑒𝑒→2 = 𝑚𝑚𝑤𝑤 (𝑢𝑢𝑒𝑒 − 𝑢𝑢2 ) = 1.67 × 1010 J

PROBLEM 6.2 QUESTION

Evaluation of Alternate Ideal Rankine Cycles (Section 6.3) Three alternative steam cycles illustrated in Figure 6.37 are proposed for a nuclear power station capable of producing either saturated steam or superheated steam at a temperature of 293 °C. The condensing steam temperature is 33 °C.

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Chapter 6 - Thermodynamics of Nuclear Energy Conversion Systems Nonflow and Steady Flow: First- and Second-Law Applications kg steam

1. Assuming ideal machinery, calculate the cycle thermal efficiency and steam rate � kWe hr � for each cycle using steam tables.

2. For each cycle, compare the amount of heat added per unit mass of working fluid in the legs 3′ → 4 and 4 → 1.

3. Briefly compare advantages and disadvantages of each of these cycles. Which would you use? Answers: kg steam

1. ηth= 38.2%, 45.9%, 36.8% and the steam rate = 3.60, 5.38, 3.54 � kWe hr � kJ

2. 𝑄𝑄3′ →4 = 1.163 × 103 , 0, 1.011 × 103 �kg� ;

𝑄𝑄4→1 = 1.456 × 103 , 1.456 × 103 , 1.748 × 103 �

kJ � kg

FIGURE 6.37 Alternate ideal Rankine cycles.

PROBLEM 6.2 SOLUTION Evaluation of Alternate Ideal Rankine Cycles (Section 6.3) Three alternative steam cycles illustrated in Figure 6.37 are proposed for a nuclear power station capable of producing either saturated steam or superheated steam at a temperature of 293 °C. The condensate steam temperature is 33 °C. 𝐤𝐤𝐤𝐤 𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬

1. Assuming ideal machinery, calculate the cycle thermal efficiency and steam rate � 𝐤𝐤𝐤𝐤𝐤𝐤 𝐡𝐡𝐡𝐡 � for each cycle using steam tables.

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Chapter 6 - Thermodynamics of Nuclear Energy Conversion Systems Nonflow and Steady Flow: First- and Second-Law Applications

Cycle 1 (Rankine Saturated Steam Cycle): We may begin at point 4. At this point we know two thermodynamic properties (T4 = 293 °C and x4 = 0) and therefore define all other properties of interest at this point, such as the pressure and the enthalpy. 𝑃𝑃4 = 7.77 MPa ℎ4 = 1306.3

kJ kg

Similarly we may look at point 1, where we know temperature and quality, T1= 293 °C and x =

𝑃𝑃1 = 7.77 MPa ℎ1 = 2762

kJ kg

𝑠𝑠1 = 5.7604

kJ kg K

We may know determine the properties at point 2 since this process is an isentropic expansion. kJ

This means that 𝑠𝑠2 = 𝑠𝑠1 = 5.760 kg K . Now, we have two properties defined at point 2, entropy

and temperature, T2 = 33 °C, and can calculate the quality. The saturated liquid entropy and kJ

saturated vapor entropy from steam tables are 𝑠𝑠f = 0.47792 kg K and 𝑠𝑠g = 8.3913 kg K . 𝑠𝑠2 − 𝑠𝑠𝑓𝑓 𝑥𝑥2 = = 0.668 𝑠𝑠𝑔𝑔 − 𝑠𝑠𝑓𝑓

We may now use this quality and temperature to look up the enthalpy in steam tables: ℎ2 = kJ

1756.62 kg .

Next, we can determine the enthalpy and the entropy at point 3 using the temperature, T3 = 33 °C, and the quality, x3 = 0. ℎ3 = 138.274

kJ kJ 𝑠𝑠3 = 0.47792 kg kg K

The process to get to point 3′ is an isentropic compression and therefore, 𝑠𝑠3′ = 𝑠𝑠3 = kJ

0.47792 kg K . We also know that the pressure at this point is the same at point 4, P3′ = P4 = 7.77 MPa. Therefore, the enthalpy can be determine from steam tables to be: ℎ3′ = 143.241

The efficiency of the cycle can be calculated: 𝜂𝜂1 =

kJ kg

𝑊𝑊̇𝑇𝑇 − 𝑊𝑊̇𝐶𝐶 (ℎ1 − ℎ2 ) − (ℎ3′ − ℎ3 ) = = 0.382 ℎ1 − ℎ3′ 𝑄𝑄̇𝑖𝑖𝑖𝑖

(1)

The steam rate is calculated to be: 𝑚𝑚̇1 =

𝑊𝑊̇𝑛𝑛𝑛𝑛𝑛𝑛

1 kg = 3.599 (ℎ1 − ℎ2 ) − (ℎ3′ − ℎ3 ) kWe hr

(2)

Cycle 2 (Carnot Cycle): We can do the same type of approach as cycle 1. We may begin at

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Chapter 6 - Thermodynamics of Nuclear Energy Conversion Systems Nonflow and Steady Flow: First- and Second-Law Applications

point 4 and use the temperature and quality to determine the enthalpy and entropy. ℎ4 = 1306.3

kJ kg

𝑠𝑠4 = 3.1892

kJ kg K

We may move to point 1 and determine the enthalpy and entropy from the temperature and quality. ℎ1 = 2762

kJ kJ 𝑠𝑠1 = 5.7604 kg kg K

For point 2 we may do the same analysis as cycle 1. Moving from point 1 to point 2 there is an kJ

isentropic expansion and therefore 𝑠𝑠2 = 𝑠𝑠1 = 5.7604 kg·K . Therefore we may determine the

quality and enthalpy as before to be:

𝑥𝑥2 = 0.668

ℎ2 = 1756.62

kJ kg

To determine the enthalpy at point 3, we must work backwards from point 4. The process from kJ

point 3 to point 4 is an isentropic compression and therefore, 𝑠𝑠3 = 𝑠𝑠4 = 3.1892 kg K . Using the kJ

saturated liquid and vapor entropies at this temperature, T3 = 33 °C , 𝑠𝑠𝑓𝑓 = 0.47792 kg k and 𝑠𝑠𝑔𝑔 = kJ

8.3913 kg K . The quality at point 3 is therefore 𝑥𝑥3 =

𝑠𝑠3 − 𝑠𝑠𝑓𝑓 = 0.343 𝑠𝑠𝑔𝑔 − 𝑠𝑠𝑓𝑓

We may then use steam tables to find that the enthalpy at this point is ℎ3 = 969.253 kg . The efficiency of the cycle is 𝜂𝜂2 =

𝑊𝑊̇𝑇𝑇 − 𝑊𝑊̇𝐶𝐶 (ℎ1 − ℎ2 ) − (ℎ4 − ℎ3 ) = = 0.459 ℎ1 − ℎ4 𝑄𝑄̇𝑖𝑖𝑖𝑖

(3)

The steam rate is calculated to be: 𝑚𝑚̇2 =

𝑊𝑊̇𝑛𝑛𝑛𝑛𝑛𝑛

1 kg = 5.387 (ℎ1 − ℎ2 ) − (ℎ4 − ℎ3 ) kWe hr

(4)

Cycle 3 (Superheated Rankine Cycle): Again, we may begin at point 4 and use the pressure, kJ

P4 = 5 MPa and quality, x4 = 0, to determine the enthalpy: ℎ4 = 1154.64 kg. Similarly the enthalpy kJ

at point 5 can be determine except using a quality of 𝑥𝑥5 = 1: ℎ5 = 2794.21 kg.

The enthalpy and entropy at point 1 can be determined using the pressure, P1 = 5 MPa, and temperature, T1 = 293 °C.

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Chapter 6 - Thermodynamics of Nuclear Energy Conversion Systems Nonflow and Steady Flow: First- and Second-Law Applications

ℎ1 = 2903.05

kJ kJ 𝑠𝑠1 = 6.17128 kg kg K

We can assume from the diagram that going from point 1 to 2 is an isentropic expansion and kJ

therefore, 𝑠𝑠2 = 𝑠𝑠1 = 6.17128 kg K . Using the saturated liquid and vapor entropies at this kJ

temperature, T2 = 33 °C, 𝑠𝑠𝑓𝑓 = 0.47792 kg K and 𝑠𝑠𝑔𝑔 = 8.3913 kg·K . The quality at point 2 is therefore

𝑥𝑥2 =

𝑠𝑠2 − 𝑠𝑠𝑓𝑓 = 0.719 𝑠𝑠𝑔𝑔 − 𝑠𝑠𝑓𝑓

We can use the temperature and quality to determine the enthalpy to be ℎ2 = 1880.18 kg

We can use the temperature and quality at point 3, T3 = 33 °C and x3 = 0, to determine the enthalpy and entropy to be ℎ3 = 138.274

kJ kJ 𝑠𝑠3 = 0.47792 kg kg K

At point 3′ we know the pressure, P3′ = 5 MPa and the entropy from the isentropic compression, kJ

𝑠𝑠3′ = 𝑠𝑠3 = 0.47792 kg K

The efficiency of cycle 3 is 𝑊𝑊̇𝑇𝑇 − 𝑊𝑊̇𝐶𝐶 (ℎ1 − ℎ2 ) − (ℎ3′ − ℎ3 ) 𝜂𝜂3 = = = 0.369 ℎ1 − ℎ3′ 𝑄𝑄̇𝑖𝑖𝑖𝑖

(5)

The steam rate is calculated to be: 𝑚𝑚̇3 =

𝑊𝑊̇𝑛𝑛𝑛𝑛𝑛𝑛

1 kg = 3.537 (ℎ1 − ℎ2 ) − (ℎ3′ − ℎ3 ) kWe hr

(6)

2. For each cycle, compare the amount of heat added per unit mass of working fluid in legs 3′ → 4 and 4 → 1: Cycle 1

𝑄𝑄3′ →4

ℎ4 − ℎ3′ = 1.163 × 103

kJ kg

ℎ4 − ℎ3′ = 1.011 × 103

kJ kg

__________

2 3

𝑄𝑄4→1

kJ kg kJ ℎ1 − ℎ4 = 1.456 × 103 kg kJ ℎ1 − ℎ4 = 1.748 × 103 kg ℎ1 − ℎ4 = 1.456 × 103

Comparing Case 2 against Case l, for the same amount of energy added per mass between 4 and 1, Case 2 supplies less energy/mass between 3′ and 4. Since the work out per mass will also be the same for these two cases, Case 2 adds less heat for this work out and therefore will have a higher efficiency. Comparing Case 3 to Case l the heat added in both cases is larger. This will tend to

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Chapter 6 - Thermodynamics of Nuclear Energy Conversion Systems Nonflow and Steady Flow: First- and Second-Law Applications

drive the efficiency down, however, since Case 3 adds more heat to create a superheated vapor, it extracts more work out per mass and this will raise the thermal efficiency. 3. Briefly compare advantages and disadvantages of each of these cycles. Which would you use? Case 2 has the highest efficiency and therefore would be the most advantageous to use. However, this case is ideal and uses the assumption that the fluid expands and compresses isentropically. In reality this does not happen and entropy is generated during these processes. In Case 2 the compression of the fluid occurs within the two phase dome. Pumps only like work on a liquid and compressors work on a single phase gas. Therefore to be more realistic this compression will have to take place outside of the two phase dome. In Case 1, the isentropic assumptions are still made except the compression is now occurring outside the two phase dome. This is probably the simplest Rankine cycle to create. However with the operating temperatures it does not maximize the thermal efficiency and the quality coming out of the turbine is wetter which may damage the turbine blades. Cycle 3 has the lowest steam rate, but requires more superheating equipment. Note that in the third cycle, the efficiency is actually lower than the first cycle even though we are superheating. This is because we have a fixed maximum temperature. If the temperature of the saturated vapor were kept constant, this efficiency would actually have increased.

PROBLEM 6.3 QUESTION Availability Analysis of a Simplified BWR (Section 6.4) A BWR system with a one-stage moisture separation is shown in Figure 6.38. The conditions in Table 6.20 may be used. TABLE 6.20 State Conditions for Problem 6.3 State 1 2 3 4 5 6 7 8 9

P (kPa) 6890 1380 1380 1380 6.89 6.89 1380 1380 6890

Condition Saturated vapor Saturated vapor Saturated liquid Saturated liquid

– Turbine isentropic efficiency, ηT = 0.90 – Pump isentropic efficiency, ηC = 0.85 – Environmental Temperature, To = 30 °C

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Chapter 6 - Thermodynamics of Nuclear Energy Conversion Systems Nonflow and Steady Flow: First- and Second-Law Applications

1. Calculate the cycle thermal efficiency 2. Recalculate the thermal efficiency of the cycle assuming that the pumps and turbines have isentropic efficiency of 100%. 3. Calculate the lost work due to the irreversibility of each component in the cycle and show numerically that the available work equals the sum of the lost work and the net work. Answers: 1. ηth = 34.2% 2. ηthi = 37.7% kJ 3. 𝑊𝑊̇𝑢𝑢1𝑚𝑚𝑚𝑚𝑚𝑚 = 2510.9 kg

kJ 𝑊𝑊̇𝑛𝑛𝑛𝑛𝑛𝑛 + 𝐼𝐼𝑇𝑇𝑇𝑇𝑇𝑇 = 2510.9 kg

FIGURE 6.38 BWR plant.

PROBLEM 6.3 SOLUTION Availability Analysis of a Simplified BWR (Section 6.4) 1. Calculate the cycle thermal efficiency. Point 1: We can begin the analysis at point 1. At this point we know the pressure and quality: 𝑃𝑃1 = 6890 kPa

𝑥𝑥1 = 1

We can then use steam tables to determine the enthalpy and entropy. ℎ1 = 2774.05

kJ kJ 𝑠𝑠1 = 5.82273 kg kg K

Point 2s: We can move to the analysis of point 2. First, we can use an isentropic assumption

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Chapter 6 - Thermodynamics of Nuclear Energy Conversion Systems Nonflow and Steady Flow: First- and Second-Law Applications

and correct with the turbine efficiency. Therefore at point 2s we know the pressure and entropy 𝑠𝑠2𝑠𝑠 = 𝑠𝑠1 = 5.82273

kJ kg K

𝑃𝑃2𝑠𝑠 = 1380 kPa

At this pressure, the saturated liquid/vapor entropies are: 𝑠𝑠2𝑓𝑓 = 2.27714

The quality at 2s can then be calculated:

𝑥𝑥2𝑠𝑠 =

kJ kg K

𝑠𝑠2𝑔𝑔 = 6.47256

𝑠𝑠2𝑠𝑠 − 𝑠𝑠2𝑓𝑓 = 0.845 𝑠𝑠2𝑔𝑔 − 𝑠𝑠2𝑓𝑓

kJ kg K

The enthalpy can now be determined with this quality and pressure from steam tables: ℎ2𝑠𝑠 = 2484.59

kJ kg

Point 2: The enthalpy at point 2 can be calculated using the isentropic efficiency of the high pressure turbine. The work per unit mass flow rate of a turbine is defined as 𝑊𝑊̇𝐻𝐻𝐻𝐻𝐻𝐻 = ℎ1 − ℎ2 = 𝜂𝜂𝑇𝑇 (ℎ1 − ℎ2𝑠𝑠 )

The isentropic assumption is therefore corrected with the isentropic turbine efficiency and the true enthalpy at point 2 is ℎ2 = ℎ1 − 𝜂𝜂𝑇𝑇 (ℎ1 − ℎ2𝑠𝑠 ) = 2513.54

kJ kg

The saturated liquid/vapor enthalpies are:

ℎ2𝑓𝑓 = 826.959

The quality at point 2 is therefore,

𝑥𝑥2 =

ℎ2𝑔𝑔 = 2788.39

ℎ2 − ℎ2𝑓𝑓 = 0.8599 ℎ2𝑔𝑔 − ℎ2𝑓𝑓

We may also determine the entropy here since it will be needed for part 3: 𝑠𝑠2 = (1 − 𝑥𝑥2 )𝑠𝑠2𝑓𝑓 + 𝑥𝑥2 𝑠𝑠2𝑔𝑔 = 5.885

Point 3: At point 3 we know the pressure and quality: P3 = 1380 kPa

kJ kg K

x3 = 1

Using steam tables we may determine the enthalpy and entropy:

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Chapter 6 - Thermodynamics of Nuclear Energy Conversion Systems Nonflow and Steady Flow: First- and Second-Law Applications

h3 = 2788.9

kJ kg

s3 = 6.47256

kJ kg K

In addition, we need to define another property at point 3 which is the fraction of the total mass flow rate. At the moisture separator, a portion of the liquid goes to the low pressure turbine (LPT) and portion goes to the OFWH. It can be understood that the fraction that travels to the LPT is simply just the quality at point 2 which is the fraction of steam mass flow rate to the total mass flow rate. We will denote this property with at y: y3 = x2 = 0.86

Point 4: At point 4 we know the pressure and quality: P4 = 1380 kPa

x4 = 0

Using steam tables we may determine the enthalpy and entropy: h4 = 826.959

kJ kg

s4 = 2.27714

kJ kg K

The fractional mass flow rate is just one minus the quality at point 2, y4 = 1 − x2 = 0.1401

Point 5s: To determine the properties at point 5, we must first analyze with an isentropic assumption. Therefore, we know the entropy and pressure at point 5s: 𝑠𝑠5𝑠𝑠 = 𝑠𝑠3 = 6.473

kJ kg K

𝑃𝑃5𝑠𝑠 = 6.89 kPa

The saturated liquid/vapor entropies at this pressure are: 𝑠𝑠5𝑓𝑓 = 0.555081

The quality at 5s can be calculated,

𝑥𝑥5𝑠𝑠 =

kJ kg K

𝑠𝑠5𝑔𝑔 = 8.28006

𝑠𝑠5𝑠𝑠 − 𝑠𝑠5𝑓𝑓 = 0.766 𝑠𝑠5𝑔𝑔 − 𝑠𝑠5𝑓𝑓

kJ kg K

The enthalpy can now be evaluated using steam tables,

ℎ5𝑠𝑠 = 2007.52

kJ kg

Point 5: The work of the LPT per unit mass flow rate can be defined as 𝑊𝑊̇𝐿𝐿𝐿𝐿𝐿𝐿 = ℎ3 − ℎ5 = 𝜂𝜂𝑇𝑇 (ℎ3 − ℎ5𝑠𝑠 )

Therefore, the true enthalpy at point 5 can be determined:

ℎ5 = ℎ3 − 𝜂𝜂𝑇𝑇 (ℎ3 − ℎ5𝑠𝑠 ) = 2085.61 88

kJ kg rev 112420

Chapter 6 - Thermodynamics of Nuclear Energy Conversion Systems Nonflow and Steady Flow: First- and Second-Law Applications kJ

The saturated liquid/vapor enthalpies at point 5 are ℎ5𝑓𝑓 = 162.12 kg and ℎ5𝑔𝑔 = 2571.19 kg. The quality is calculated to be

𝑥𝑥5 =

ℎ5 − ℎ5𝑓𝑓 = 0.7984 ℎ5𝑔𝑔 − ℎ5𝑓𝑓 kJ

We may also determine the entropy, where 𝑠𝑠5𝑓𝑓 = 0.555 kg K and 𝑠𝑠5𝑔𝑔 = 8.28 kg K 𝑠𝑠5 = (1 − 𝑥𝑥5 )𝑠𝑠5𝑓𝑓 + 𝑥𝑥5 𝑠𝑠5𝑔𝑔 = 6.723

kJ kg K

Finally, the fractional mass flow rate at point 5 is equivalent to the fractional mass flow rate at point 3, y5 = y3 = 0.86

Point 6: At point 6 we know the pressure and the quality: P6 = 6.89 kPa

x6 = 0

Using steam tables, the enthalpy and entropy can be determined: h6 = 162.12

kJ kg

s6 = 0.555081

kJ kg K

The fractional mass flow rate at this point is the same as point 5, y6 = y5 = 0.86

Point 7s To determine the properties at point 7, we must first use an isentropic assumption. Using this assumption, we know entropy and pressure: s7s = 𝑠𝑠6 = 0.555081

kJ kg K

P7s = 1380 kPa.

We can use these two properties to determine the enthalpy,

kJ kg Point 7 The pumping power of the main condensate pump per unit mass flow rate can be defined as: h7s = 163.487

𝑊𝑊̇𝑀𝑀𝑀𝑀𝑀𝑀 = ℎ7 − ℎ6 =

Therefore, the real enthalpy at point 7 is ℎ7 = ℎ6 +

1 (ℎ − ℎ6 ) 𝜂𝜂𝑃𝑃 7𝑠𝑠

1 kJ (ℎ7𝑠𝑠 − ℎ6 ) = 163.73 𝜂𝜂𝑃𝑃 kg

The fractional mass flow rate is equivalent to point 6:

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Chapter 6 - Thermodynamics of Nuclear Energy Conversion Systems Nonflow and Steady Flow: First- and Second-Law Applications

y7 = y6 = 0.86

We may also determine the entropy from the enthalpy and pressure at point 7 (1380 kPa) using kJ

steam tables 𝑠𝑠7 = 0.555834 kg K

Point 8 To evaluate the enthalpy at this point we must apply conservation of mass and energy: – Mass: ṁ8 = ṁ4 + ṁ7 – Energy: ṁ4h4 + ṁ7h7 = ṁ8h8

Here, the mass flow rate at point 8 is just the total mass flow rate of the system. We may then divide the energy equation by this flow and determine the enthalpy at point 8, ℎ8 = 𝑦𝑦4 ℎ4 + 𝑦𝑦7 ℎ7 = 256.666

kJ kg

Two independent properties are known at point 8, the enthalpy and pressure. We may use steam tables to determine entropy: 𝑠𝑠8 = 0.843481

kJ kg K

Point 9s In order to determine the properties at point 9, we must first use the isentropic condition. Therefore, we know the entropy and pressure at point 9s: 𝑠𝑠9𝑠𝑠 = 𝑠𝑠8 = 0.843

Therefore, we can determine the enthalpy,

kJ kg K

𝑃𝑃9𝑠𝑠 = 6890 kPa

ℎ9𝑠𝑠 = 262.158

kJ kg

Point 9 We can again define the work of the feed water pump per unit mass flow rate 𝑊𝑊̇𝐹𝐹𝐹𝐹 = ℎ9 − ℎ8 =

The real enthalpy at point 9 is therefore, ℎ9 = ℎ8 +

1 (ℎ − ℎ8 ) 𝜂𝜂𝑃𝑃 9𝑠𝑠

1 kJ (ℎ9𝑠𝑠 − ℎ8 ) = 263.13 𝜂𝜂𝑃𝑃 kg

Knowing the pressure and enthalpy, we may determine the entropy using steam tables 𝑠𝑠9 = 0.846389

kJ kg K

Cycle efficiency The thermal efficiency of the cycle is defined as

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Chapter 6 - Thermodynamics of Nuclear Energy Conversion Systems Nonflow and Steady Flow: First- and Second-Law Applications

𝜂𝜂𝑡𝑡ℎ =

𝑊𝑊̇𝐻𝐻𝐻𝐻𝐻𝐻 + 𝑊𝑊̇𝐿𝐿𝐿𝐿𝐿𝐿 − 𝑊𝑊̇𝑀𝑀𝑀𝑀𝑀𝑀 − 𝑊𝑊̇𝐹𝐹𝐹𝐹 𝑄𝑄̇𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟

𝑚𝑚̇1 (ℎ1 − ℎ2 ) + 𝑚𝑚̇3 (ℎ3 − ℎ5 ) − 𝑚𝑚̇7 (ℎ7 − ℎ6 ) − 𝑚𝑚̇8 (ℎ9 − ℎ8 ) 𝑚𝑚̇1 (ℎ1 − ℎ9 )

We may now divide the numerator and denominator my the total mass flow rate and the cycle efficiency may be calculated: 𝜂𝜂𝑡𝑡ℎ =

(ℎ1 − ℎ2 ) + 𝑦𝑦3 (ℎ3 − ℎ5 ) − 𝑦𝑦7 (ℎ7 − ℎ6 ) − (ℎ9 − ℎ8 ) = 0.3413 (ℎ1 − ℎ9 )

(1)

2. Recalculate the thermal efficiency of the cycle assuming that the pumps and turbines have isentropic efficiencies of 100%. We have already determined most of the properties needed for this part. We will briefly go through each point and relate them to previous quantities determined above. All properties here will be denoted with a subscript “i”. Point 1

Point 2

Point 3

Point 4

ℎ1𝑖𝑖 = ℎ1 = 2.774 × 103

kJ kg

ℎ2𝑖𝑖 = ℎ2𝑠𝑠 = 2.485 × 103

kJ kg

ℎ3𝑖𝑖 = ℎ3 = 2.788 × 103

kJ kg

𝑥𝑥2𝑖𝑖 = 𝑥𝑥2𝑠𝑠 = 0.845

𝑦𝑦3𝑖𝑖 = 𝑥𝑥2𝑖𝑖 = 0.845

ℎ4𝑖𝑖 = ℎ4 = 826.959 Point 5

kJ kg

𝑦𝑦4𝑖𝑖 = 1 − 𝑥𝑥2𝑖𝑖 = 0.155

ℎ5𝑖𝑖 = ℎ5𝑠𝑠 = 2.008 × 108 𝑥𝑥5𝑖𝑖 = 𝑥𝑥5𝑠𝑠 = 0.766

Point 6

kJ kg

𝑦𝑦5𝑖𝑖 = 𝑦𝑦3𝑖𝑖 = 0.845

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Chapter 6 - Thermodynamics of Nuclear Energy Conversion Systems Nonflow and Steady Flow: First- and Second-Law Applications

ℎ6𝑖𝑖 = ℎ6 = 162.12 Point 7

kJ kg

𝑦𝑦6𝑖𝑖 = 𝑦𝑦5𝑖𝑖 = 0.845

ℎ7𝑖𝑖 = ℎ7𝑠𝑠 = 163.487 𝑦𝑦7𝑖𝑖 = 𝑦𝑦6𝑖𝑖 = 0.845

kJ kg

Point 8 Conservation of mass and energy can be applied again, except some of the properties have changed due to the isentropic assumption. ℎ8𝑖𝑖 = 𝑦𝑦4𝑖𝑖 ℎ4𝑖𝑖 + 𝑦𝑦7𝑖𝑖 ℎ7𝑖𝑖 = 266.252

The new entropy at point 8 is now, using steam tables: 𝑠𝑠8𝑖𝑖 = 0.872172

kJ kg

kJ kg · K

Point 9 Keeping entropy constant we know the pressure and entropy at point 9, we may determine the enthalpy from steam tables: 𝑠𝑠9𝑖𝑖 = 𝑠𝑠8𝑖𝑖 = 0.872172

kJ kg K

ℎ9𝑖𝑖 = 271.952

𝑃𝑃9 = 6.89 MPa

kJ kg

Cycle efficiency We can now determine the thermal efficiency of the cycle, 𝜂𝜂𝑡𝑡ℎ𝑖𝑖 =

(ℎ1𝑖𝑖 − ℎ2𝑖𝑖 ) + 𝑦𝑦3𝑖𝑖 (ℎ3𝑖𝑖 − ℎ5𝑖𝑖 ) − 𝑦𝑦7𝑖𝑖 (ℎ7𝑖𝑖 − ℎ6𝑖𝑖 ) − (ℎ9𝑖𝑖 − ℎ8𝑖𝑖 ) = 0.3767 ℎ1𝑖𝑖 − ℎ9𝑖𝑖

3. Calculate the lost work due to the irreversibility of each component in the cycle and show numerically that the available work equals the sum of the lost work and the net work. To calculate the lost work due to irreversibility, we need to define the temperature where heat is rejected. For this problem it is the environment temperature, Te = 303.15 K. We will also be referencing variables from part 1 of this solution. For each of these components, we will be calculating quantities per unit mass flow rate. Therefore, one needs to multiply each of these quantities by the fraction of the total mass flow (y) passing through it. High Pressure Turbine (HPT) We may use Equation 6 65 to determine the lost work due to irreversibility. 𝐼𝐼𝐻𝐻𝐻𝐻𝐻𝐻 = 𝑇𝑇𝑒𝑒 (𝑠𝑠2 − 𝑠𝑠1 ) = 18.774 92

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Chapter 6 - Thermodynamics of Nuclear Energy Conversion Systems Nonflow and Steady Flow: First- and Second-Law Applications

Low Pressure Turbine (LPT) 𝐼𝐼𝐿𝐿𝐿𝐿𝐿𝐿 = 𝑇𝑇𝑒𝑒 𝑦𝑦3 (𝑠𝑠5 − 𝑠𝑠3 ) = 65.277

kJ kg

Condenser (CD) We may use Equation 6-72a for the condenser.

Pump 1 (p1)

−(ℎ6 − ℎ5 ) kJ + 𝑠𝑠6 − 𝑠𝑠5 � = 46.166 𝐼𝐼𝐶𝐶𝐶𝐶 = 𝑇𝑇𝑒𝑒 𝑦𝑦5 � 𝑇𝑇𝑒𝑒 kg

Pump 2

𝐼𝐼𝑝𝑝1 = 𝑇𝑇𝑒𝑒 𝑦𝑦6 (𝑠𝑠7 − 𝑠𝑠6 ) = 0.196

Reactor

𝐼𝐼𝑝𝑝2 = 𝑇𝑇𝑒𝑒 (𝑠𝑠9 − 𝑠𝑠8 ) = 0.882

Moisture Separator

kJ kg

𝐼𝐼𝑅𝑅 = 𝑇𝑇𝑒𝑒 (𝑠𝑠1 − 𝑠𝑠9 ) = 1.509 × 103 𝐼𝐼𝑀𝑀𝑀𝑀 = 𝑇𝑇𝑒𝑒 (𝑦𝑦4 𝑠𝑠4 + 𝑦𝑦3 𝑠𝑠3 − 𝑠𝑠2 ) = 0

The moisture separator is an adiabatic, reversible process.

kJ kg

OFWH 𝐼𝐼𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂 = 𝑇𝑇𝑒𝑒 (𝑠𝑠8 − 𝑦𝑦4 𝑠𝑠4 − 𝑦𝑦7 𝑠𝑠7 ) = 14.079

kJ kg

Total Irreversibility The total lost work due to irreversibility is simply the sum of all of the individual components. kJ 𝐼𝐼𝑇𝑇𝑇𝑇𝑇𝑇 = 𝐼𝐼𝐻𝐻𝐻𝐻𝐻𝐻 + 𝐼𝐼𝐿𝐿𝐿𝐿𝐿𝐿 + 𝐼𝐼𝐶𝐶𝐶𝐶 + 𝐼𝐼𝑝𝑝1 + 𝐼𝐼𝑝𝑝2 + 𝐼𝐼𝑅𝑅 + 𝐼𝐼𝑀𝑀𝑀𝑀 + 𝐼𝐼𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂 = 1.654 × 103 kg Available Work Using Equation 6-55, we may determine the amount of available work. 𝑊𝑊𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢 = ℎ1 − ℎ9 = 2510.9

kJ kg

Net Work The net work can be determined as the numerator of Equation (1). 𝑊𝑊𝑛𝑛𝑛𝑛𝑛𝑛 = (ℎ1 − ℎ2 ) + 𝑦𝑦3 (ℎ3 − ℎ5 ) − 𝑦𝑦7 (ℎ7 − ℎ6 ) − (ℎ9 − ℎ8 ) = 856.973

kJ kg

Answer We can now sum the net work and the work lost due to irreversibility and show that it

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Chapter 6 - Thermodynamics of Nuclear Energy Conversion Systems Nonflow and Steady Flow: First- and Second-Law Applications

is equivalent to the available work. 𝑊𝑊𝑛𝑛𝑛𝑛𝑛𝑛 + 𝐼𝐼𝑇𝑇𝑇𝑇𝑇𝑇 = 2510.9

kJ kg

PROBLEM 6.4 QUESTION Thermodynamic Analysis of a Gas Turbine (Section 6.4) Is transformation 1 → 2, shown in Figure 6.39, thermodynamically possible for a gas turbine? If so, under what conditions? If not, why? (Assume steady-state)

FIGURE 6.39 Temperature–entropy diagram for transformation in the turbine.

PROBLEM 6.4 SOLUTION Thermodynamic Analysis of a Gas Turbine (Section 6.4) Is transformation 1 → 2, shown in Figure 6.39, thermodynamically possible for a gas turbine? If so, under what conditions? If not, why? (Assume steady-state) Solution: Consider the entropy equation for transformation 1→ 2: 𝑠𝑠2 = 𝑠𝑠1 +

̇ 𝑆𝑆𝑔𝑔𝑔𝑔𝑔𝑔 𝑄𝑄̇ + 𝑚𝑚̇ 𝑚𝑚̇𝑇𝑇𝑠𝑠

(1)

̇ (> 0) is the entropy generation due to irreversibilities (e.g., where 𝑚𝑚̇ is the mass flow rate, 𝑆𝑆𝑔𝑔𝑔𝑔𝑔𝑔 friction), Q̇ is the heat rate exchanged between the turbine and the surroundings and Ts is the

temperature at which this exchange occurs. Normally, for a turbine it is assumed that Q̇ = 0 (the turbine is adiabatic). Then Equation (1) gives:

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Chapter 6 - Thermodynamics of Nuclear Energy Conversion Systems Nonflow and Steady Flow: First- and Second-Law Applications

𝑠𝑠2 = 𝑠𝑠1 +

̇ 𝑆𝑆𝑔𝑔𝑔𝑔𝑔𝑔 > 𝑠𝑠1 𝑚𝑚̇

(2)

and one has to conclude that transformation 1 → 2 shown in Figure 6.39 where 𝑠𝑠2 < 𝑠𝑠1 is not thermodynamically possible. However, the transformation becomes thermodynamically possible if one assumes that Q̇ < 0, i.e., the turbine is cooled.

PROBLEM 6.5 QUESTION

Analysis of a Steam Turbine (Section 6.4 ) In the test of a steam turbine the following data were observed: – – –

ℎ1 = 3000 kg 𝑃𝑃1 = 10 MPa kJ

ℎ2 = 2600 kg 𝑉𝑉2 negligible 𝑧𝑧2 = 𝑧𝑧1

kJ 𝑊𝑊̇1,2 = 384.45 kg

𝑉𝑉1 = 150 m⁄s

𝑃𝑃2 = 0.5 MPa

1. Assume steady flow, and determine the heat transferred to the surroundings per kilogram of steam. 2. What is the quality of the exit steam? Answers: 1. Q̇ = −26.8 kJ⁄kg 2. x = 92.9%

PROBLEM 6.5 SOLUTION Analysis of Steam Turbine (Section 6.4) In the test of a steam turbine the following data were observed: – – –

ℎ1 = 3000 kg 𝑃𝑃1 = 10 MPa kJ

ℎ2 = 2600 kg 𝑉𝑉2 negligible 𝑧𝑧2 = 𝑧𝑧1

kJ 𝑊𝑊̇1,2 = 384.45 kg

𝑉𝑉1 = 150 m⁄s

𝑃𝑃2 = 0.5 MPa

1. Assume steady flow, and determine the heat transferred to the surroundings per kilogram of steam: We can start with conservation of energy in the form of the steady flow energy equation: �ℎ2 +

𝑉𝑉22 𝑉𝑉12 + 𝑔𝑔𝑔𝑔2 � − �ℎ1 + + 𝑔𝑔𝑔𝑔1 � = 𝑄𝑄̇ − 𝑊𝑊̇ 2 2

where by convention heat added and work extracted from the system is positive.

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Chapter 6 - Thermodynamics of Nuclear Energy Conversion Systems Nonflow and Steady Flow: First- and Second-Law Applications

We may neglect the velocity at state 2 as given by the problem statement. We can solve for heat transferred to the surrounding per kilogram of steam, 𝑄𝑄̇𝑜𝑜𝑜𝑜𝑜𝑜 : 𝑉𝑉12 kJ 𝑄𝑄̇𝑜𝑜𝑜𝑜𝑜𝑜 = ℎ1 + − ℎ2 − 𝑊𝑊̇1,2 = −26.8 2 kg

This negative answer indicates that heat flow is out of the control volume. 2. What is the quality of the exit steam? The saturated liquid/vapor enthalpies at state 2 can be kJ

determined from steam tables using pressure, P2 = 0.5 MPa and enthalpy, ℎ2 = 2600 kg: ℎ2𝑓𝑓 = 640.085

The quality can therefore be calculated as 𝑥𝑥2 =

kJ kg

ℎ2𝑔𝑔 = 2748.11

kJ kg

ℎ2 − ℎ2𝑓𝑓 = 0.929 ℎ2𝑔𝑔 − ℎ2𝑓𝑓

PROBLEM 6.6 QUESTION Irreversibility Problems Involving the Rankine Cycle (Section 6.4) Consider the Rankine cycles given in the T-s diagram of Figure 6.40 and defined by operating conditions of Table 6.21. The cycles differ in the temperature and pressure of the condensation process. What are the differences in cycle irreversibilities between the two cases for irreversibilities defined as: ̇ 𝑚𝑚̇, and 1. Irreversibility per unit mass flowrate of working fluid, 𝐼𝐼 ⁄ ̇ 𝑄𝑄̇𝑖𝑖𝑖𝑖 𝑚𝑚̇𝑠𝑠 2. Irreversibility per unit mass flowrate of working fluid and energy output, i.e., 𝐼𝐼 ⁄ TABLE 6.21 Operating Conditions of Cycles of Problem 6.6 Cycle

State

Pressure (kPa)

Condition

7.0

Saturated Liquid

6800.0

7.0

1′

6.0

2′

6800.0

4′

6.0

1′

Saturated Vapor Saturated Liquid Saturated Vapor

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Chapter 6 - Thermodynamics of Nuclear Energy Conversion Systems Nonflow and Steady Flow: First- and Second-Law Applications

FIGURE 6.40 Temperature-entropy characteristics of two cycles. Answers: 1. I ≡ I / m s for cycle 1234 = 1645.1 kJ/kg  s for cycle 1′ 234′ = 1642.0 kJ/kg I ≡ I / m

2. I / Q in ≡ I / Q in m s for cycle 1234 = 0.632 I / Q in ≡ I / Q in m s for cycle 1'234' = 0.627

PROBLEM 6.6 SOLUTION Irreversibility Problems Involving the Rankine Cycle (Section 6.4) Consider the Rankine cycles given in the T-s diagram of Figure 6.40 and defined by operating conditions of Table 6.21. The cycles differ in the temperature and pressure of the condensation process. What are the differences in cycle irreversibilities between the two cases for irreversibilities defined as: 1. Irreversibility per unit mass flowrate of working fluid: This irreversibility can be determined using Equation 6.84. We will denote this quantity simply with I. 𝐼𝐼 = ℎ4 − ℎ1

Therefore, we must go around the cycle and determine the enthalpies at each point.

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Chapter 6 - Thermodynamics of Nuclear Energy Conversion Systems Nonflow and Steady Flow: First- and Second-Law Applications

State 1: At state 1, we know the pressure and quality: 𝑃𝑃1 = 7.0 kPa

𝑥𝑥1 = 0

We can therefore determine the enthalpy and entropy using steam tables: ℎ1 = 163.351

kJ kg

𝑠𝑠1 = 0.559028

State 1′: At state 1′, we know the pressure and quality: 𝑃𝑃1′ = 6.0 kPa

kJ kg K

𝑥𝑥1 = 0.

We can therefore determine the enthalpy and entropy using steam tables: ℎ1′ = 1551.478

kJ kg

𝑃𝑃2 = 6800 kpa

𝑠𝑠2 = 𝑠𝑠1 = 0.559028

𝑠𝑠1′ = 0.52082

kJ kg K

State 2: At state 2, we know the pressure and entropy, since the process from 1 to 2 is isentropic: kJ kg K

We may then use these properties to the determine the enthalpy with steam tables: kJ kg State 2′: At state 2′, we know the pressure and entropy, since the process from 1′ to 2′ is isentropic: kJ 𝑃𝑃2′ = 6800 kpa 𝑠𝑠2′ = 𝑠𝑠1′ = 0.52082 kg K ℎ2 = 170.211

We may then use these properties to the determine the enthalpy with steam tables: ℎ2′ = 158.306

kJ kg

State 3: At state 3, we know the pressure and quality: 𝑃𝑃3 = 6800 kPa

𝑥𝑥3 = 1

We may then determine the enthalpy and entropy using steam tables: kJ kJ 𝑠𝑠3 = 5.82931 kg kg K State 4: At state 4, we know the pressure and entropy, since the process from 3 to 4 is isentropic: kJ 𝑃𝑃4 = 7.0 kpa 𝑠𝑠4 = 𝑠𝑠3 = 5.82931 kg K ℎ3 = 2775.19

We can then look up the saturated liquid/vapor entropies from steam tables at this pressure:

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Chapter 6 - Thermodynamics of Nuclear Energy Conversion Systems Nonflow and Steady Flow: First- and Second-Law Applications

𝑠𝑠4𝑓𝑓 = 0.559028

kJ kg K

The quality at state 4 can then be determined: 𝑥𝑥4 =

𝑠𝑠4𝑔𝑔 = 8.27446

𝑠𝑠4 − 𝑠𝑠4𝑓𝑓 = 0.683 𝑠𝑠4𝑔𝑔 − 𝑠𝑠4𝑓𝑓

kJ kg K

Finally, we can use pressure and quality to determine enthalpy from steam tables: kJ kg State 4′: At state 4′, we know the pressure and entropy, since the process from 3 to 4′ is isentropic: ℎ4 = 1808.47

𝑃𝑃4′ = 6.0 kpa 𝑠𝑠4′ = 𝑠𝑠3 = 5.82931

kJ kg K

We can then look up the saturated liquid/vapor entropies from steam tables at this pressure: 𝑠𝑠4′ 𝑓𝑓 = 0.52082

𝑠𝑠4′ 𝑔𝑔 = 8.32904

The quality at state 4′ can then be determined: 𝑠𝑠4′ − 𝑠𝑠4′ 𝑓𝑓 𝑥𝑥4′ = = .680 𝑠𝑠4′ 𝑔𝑔 − 𝑠𝑠4′ 𝑓𝑓

Finally, we can use pressure and quality to determine enthalpy from steam tables: ℎ4′ = 1793.44

kJ kg

Irreversibility: The irreversibility per unit mass flow rate can now be determined for each cycle: kJ

𝐼𝐼 ≡ 𝐼𝐼 /̇ 𝑚𝑚̇𝑠𝑠 = ℎ4 − ℎ1 = 1645.1 kg for cycle 1

kJ 𝐼𝐼 ′ ≡ 𝐼𝐼 /̇ 𝑚𝑚̇𝑠𝑠 = ℎ4′ − ℎ1′ = 1642.0 kg for cycle 1′

2. Irreversibility per unit mass flow rate of working fluid and energy output: To determine this quantity, we must first define the heat rate added to the system per unit mass flow rate: kJ 𝑄𝑄̇𝑖𝑖𝑖𝑖 = ℎ3 − ℎ2 = 2605.0 kg for cycle 1

Hence

kJ ′ 𝑄𝑄̇𝑖𝑖𝑖𝑖 = ℎ3 − ℎ2′ = 2616.9 kg for cycle 1′

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Chapter 6 - Thermodynamics of Nuclear Energy Conversion Systems Nonflow and Steady Flow: First- and Second-Law Applications

1645.1 I / Q in ≡ I / Q in m s = =0.6315 for cycle 1 2605.0 1642.0 I / Q in ≡ I / Q in m s = =0.6275 for cycle 1′ 2616.9

PROBLEM 6.7 QUESTION Replacement of a Steam Generator in a PWR with a Flash Tank (Sections 6.4 and 6.5) Consider a “direct” cycle plant with a pressurized water-cooled reactor. This proposed design consists of using most of the typical PWR plant components except the steam generator. In place of the steam generator, a large “flash tank” is incorporated with the capability to take the primary coolant and reduce the pressure to the typical secondary side pressure. The resulting steam is separated, dried, and taken to the balance of the plant. The feedwater from the condenser returns to this flash tank. The primary water from the flash tank is repressurized and circulated back to the core. Make a schematic drawing of this direct cycle plant, and discuss the benefits and/or problems with this design. Also, compare a typical PWR plant design with this direct cycle design with respect to: — Plant thermal efficiencies (perform a numerical comparison and explain your results), and — Nuclear plant safety (perform a qualitative comparison).

Assume the reactor, steam generator and condensation conditions are those of Table 6.7 of Example 6.4.

PROBLEM 6.7 SOLUTION Replacement of a Steam Generator in a PWR with a Flash Tank (Sections 6.4 and 6.5) Purpose: The purpose of this problem is to compare the characteristics of a direct cycle nuclear power system to those of a conventional pressurized water reactor. A direct plant uses a “flash tank” in place of the steam generator found in PWRs. To evaluate feasibility of this process, the net work, heat supplied, mass flow ratios, and overall thermal efficiency will be calculated for several cases. Description of Cases Considered: In order to allow easy comparison, the cases considered will use the same design parameters. They are as follows: — Primary loop pressure: 15.5 MPa

100

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Chapter 6 - Thermodynamics of Nuclear Energy Conversion Systems Nonflow and Steady Flow: First- and Second-Law Applications — Secondary loop pressure (Single Turbine): 7.75 MPa — Sink Pressure: 0.00689 MPa — Pressure Losses: 0.0 MPa — Turbine efficiencies: 100% — Pump efficiencies: 100% — Mass flow rate through core: 1.0 units

Component efficiencies of 100% were chosen to make the comparison between the cycles clearer, and to make the mathematical solution easy to follow. Steam Generator Cases (Figure SM-6.1): For a conventional PWR with a steam generator, the analysis is identical to the solution of Example 6.4, except that all expansion and duct pressure losses are neglected for simplicity. Figure SM-6.1 shows the block diagram and associated T-s diagram for the single turbine cycle with a steam generator. Because there are no losses in the primary, no primary pump is included.

FIGURE SM-6.1: Single Turbine, Steam Generator Cycle Nonmixed Direct Cycle Cases (Figure SM-6.2): In these cases the steam generator from a conventional PWR has been replaced with a flash tank for generating steam. A porous stopper is placed in the pipe from the core to the flash tank to control the low rate of high pressure to the tank. Upon passing through the plug, a fraction (xf) of the water is “flashed” to steam at the secondary system pressure (assumed here to be a constant enthalpy-process). In the nonmixed cycle this steam is sent out of the flash tank and towards the turbine. The fraction that did not flash to steam (1 − xf) mixes with the stream of water returning from the condenser. These two streams mix completely and exit to pump P2 and go back to the reactor. One hundred percent efficiency in the steam separator and drying process are assumed in the flash tank.

101

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Chapter 6 - Thermodynamics of Nuclear Energy Conversion Systems Nonflow and Steady Flow: First- and Second-Law Applications

FIGURE SM-6.2: Single Turbine, Nonmixed Direct Cycle Fully-Mixed Direct Cycle Cases (Figure SM-6.3): The fully-mixed cycles represent an alternative way of modeling the process in the flash tank. In the analysis of the fully-mixed case, it is assumed that the entire mass of fluid entering the flash tank from the core mixes with the subcooled liquid from the condenser before any steam leaves. All of the enthalpy of the primary fluid is mixed with the enthalpy from the subcooled liquid. The result is a new mass of fluid with an average enthalpy including contributions from both streams. The high enthalpy fluid from the reactor gives the heat to the subcooled liquid, bringing it up to the saturation temperature. The excess enthalpy is given to saturated liquid to make saturated steam. The process is illustrated in the T-s diagram in Figure SM-6.3. It is apparent that this process will yield less steam than the nonmixed case, but the fluid going to the core will be slightly pre-heated. The correct method of modeling the flash tank would depend on the physical set up of the tank, but would surely be a hybrid of both models described here.

FIGURE SM-6.3: Single Turbine, Mixed Direct Cycle

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TABLE SM-6.1: Summary of Results Net Work (kJ/kg) Heat Added (kJ/kg) Thermal Efficiency (%) Steam Generator (1T)

968

2592

37.4

Nonmixed Direct (1T)

113

318

35.4

Mixed Direct (1T)

174

33.6

Mass Flow Ratio Steam Generator (1T)

13.18

Nonmixed Direct (1T)

7.9

Mixed Direct (1T)

14.0

As shown in Table SM-6.1, the most efficient cycle is the conventional steam generator cycle. The net work is several times greater in this case than for the direct cycles. The combination of the superior efficiency and much greater net work output makes the configuration the best choice. The nonmixed cycle, while achieving high efficiency, produces very little work compared to the indirect cycle. Because the flashing process is highly irreversible, it introduces inefficiency into the steam production process. In this example only 113% of the fluid flashes to steam, resulting in a low mass flowrate through the turbines. The fully mixed cycle is even worse than the nonmixed cycle. Its work output is low, and its efficiency is worse than the other 2 cases. Because it is fully mixed with subcooled liquid before producing steam, only 7.6% of the primary fluid is turned into steam. The calculations for each of these cases are included in Appendices A through F. Safety Considerations: The use of the direct cycle for the steam generator creates several safety issues that must be addressed. Ensuring the safety of this type of plant has implications in the areas of: 1. Accident Probability and Consequences 2. Plant Operations 3. Economics I. Accident Probability and Consequences Large Potential for Loss of Coolant Accident: The use of a flash tank requires new complexities not found in a conventional PWR. These complexities lead to more system piping and coupling between systems. Any increase in piping length has associated problems of corrosion, radiation damage, and fatigue that raise the probabilities of LOCA. Overpressurization of Secondary: A failure in the plug (and/or valve) between the primary side and the flash tank would lead to overpressurization of secondary as the steam inventory

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expanded. The secondary piping would rupture, releasing a potentially large volume of radioactive steam. Loss of Core Flow: A related problem is the loss of coolant flow through the core in the case of failure of the flash tarde valve. The flashing steam would severely reduce the heat transfer in the core. Some kind of circulation system would need to be engineered because normal forced circulation and natural circulation would both be unavailable, II. Plant Operations Large Volume of Radioactivity: Because the primary and secondary cycles are not physically separate in a direct cycle, there will be radioactive water and steam flowing throughout the system. This is similar to the problem encountered in a BWR but the coolant inventory here is larger (comparable to the inventory in a PWR). This amplifies the BWR problems of radiation damage to piping, turbines, pumps and other components. System Chemistry Control: The use of neutron poisons to control reactivity is common in PWRs and in BWR primary loops. Poisons used in the direct cycle would vaporize in the severe pressure drop of the flash tank. Some of the chemicals would also accumulate in the flash tank. These chemicals could potentially accumulate in the porous plug used to separate the primary from the flash tank. Furthermore, these chemicals (such as boron) would be circulated through the secondary, possibly accelerating corrosion damage to components. Flash Tank Level Control: The fluid level in the flash tank would have to be carefully monitored in order to prevent it from drying out. In this cycle, the fluid is pumped directly from the flash tank to the core. If the supply coolant disappears, the core will begin to dry out. The primary pump in the direct cycle is much stronger than in a conventional PWR (7.75 MPa vs. 0.5 MPa), so it would quickly begin to suck steam or air into the core. Startup Procedure: In a normal PWR, the primary pumps can be turned on to begin to heat the system. Because there is no way to run only a primary system in a direct cycle, this would not be possible. The flash tank would have to be bypassed until the system is ready to begin steam generation. This creates more tubing and more connections-adding to the complexity of the design. III. Economics Flash Tank: The flash tank and its associated valve system would be considerably more expensive than a conventional steam generator. If a predominantly nonmixing design is employed (as was proven to be more desirable), the internal structure of the flash tank will be more complex than a conventional steam generator. Steam flow must be directed one way, while merging liquid flows are sent another. The construction of the tank itself would have to be very strong to accommodate the high stresses associated with the flashing process. In addition, radioactive degradation of the material would be a concern for the entire structure (in a PWR only the piping carrying the primary fluid is radioactive). However, these costs could be offset by reducing the

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number of steam generation loops used in the design. A flash tank would not have the size limitation imposed on a conventional steam generator. All the primary loops could conceivable lead into one large flash tank. Primary Pump: The primary pump, as pointed earlier, must pump the fluid from secondary to primary pressure. This is an increase of 7.75 MPa. A normal PWR primary pump only has to cover the pressure drops due to friction in the loop. A 7.75 MPa pump would be very expensive and would require several stages (with associated leales and losses). In addition to its large size, it would have to withstand reactivity control chemicals, be exposed to radioactive water, and be desiged to high reliability (it is responsible for cooling the core). Secondary Piping: The threat of rapid pressurization of the secondary discussed earlier would require strong piping and precautions in its design. The associated costs could be quite high because of the length of piping and number of components that must be reinforced. Conclusion: The use of the direct cycle plant with a flash tank does not appear to be a desirable option for conventional power production. The efficiency is slightly less than that of a conventional PWR, and the net work is several times less. The safety, operational, and economic implications of the design make it appear even less favorable. The conventional PWR design with a steam generator is a better choice. Solution Manual Appendices: –

A. Calculations: Single-Turbine, Indirect Cycle

–

B. Calculations: Single-Turbine, Nonmixed direct Cycle

–

C. Calculations: Single-Turbine, Fully-Mixed direct Cycle

SM-Appendix A: Single Turbine, Indirect Cycle (Figure SM-6.1) Point 1: At point 1 we know the pressure and quality: 𝑃𝑃𝐴𝐴1 = 7.75 Mpa 𝑥𝑥𝐴𝐴1 = 1.

We can use steam tables to calculate enthalpy and entropy: ℎ𝐴𝐴1 = 2762.34

kJ kJ 𝑠𝑠𝐴𝐴1 = 5.76201 kg kg K

Point 2: At point 2 we know the entropy and pressure: 𝑠𝑠𝐴𝐴2 = 𝑠𝑠𝐴𝐴1 = 5.76201

kJ kg K

We can look up the saturated liquid/vapor entropies: 𝑠𝑠𝐴𝐴2𝑓𝑓 = 0.555081

kJ kg K 105

𝑃𝑃𝐴𝐴2 = 6.89 kPa

𝑠𝑠𝐴𝐴2𝑔𝑔 = 8.28006

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The quality is calculated to be: 𝑥𝑥𝐴𝐴2 =

𝑠𝑠𝐴𝐴2 − 𝑠𝑠𝐴𝐴2𝑓𝑓 = 0.674 𝑠𝑠𝐴𝐴2𝑔𝑔 − 𝑠𝑠𝐴𝐴2𝑓𝑓

We can use the quality and pressure in steam tables to determine the enthalpy: ℎ𝐴𝐴2 = 1785.93

kJ kg

Point 3: At point 3 we know the pressure and quality: 𝑃𝑃𝐴𝐴3 = 6.89 kPa

𝑥𝑥𝐴𝐴3 = 0

We may use steam tables to look up the enthalpy and entropy: ℎ𝐴𝐴3 = 162.12

kJ kg

𝑠𝑠𝐴𝐴3 = 0.555081

Point 4: At point 4 we know the entropy and pressure: 𝑠𝑠𝐴𝐴4 = 𝑠𝑠𝐴𝐴3 = 0.555081

kJ kg K

𝑃𝑃𝐴𝐴4 = 7.75 MPa

We can use steam tables to determine the enthalpy to be: ℎ𝐴𝐴4 = 170.22

Point 7: We the pressure and temperature to be:

kJ kg

𝑃𝑃𝐴𝐴7 = 15.5 MPa 𝑇𝑇𝐴𝐴7 = 599 K = 325.85°C

We can use steam tables to determine the enthalpy to be: ℎ𝐴𝐴7 = 1489.87

kJ kg

Thermal Efficiency: The work of the turbine per mass flow rate is 𝑊𝑊̇𝐴𝐴𝐴𝐴 = ℎ𝐴𝐴1 − ℎ𝐴𝐴2 = 976.41

The work of the pump per mass flow rate is

The net work is therefore,

𝑊𝑊̇𝐴𝐴𝐴𝐴 = ℎ𝐴𝐴4 − ℎ𝐴𝐴3 = 8.1

kJ kg

𝑊𝑊̇𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 = 𝑊𝑊̇𝐴𝐴𝐴𝐴 − 𝑊𝑊̇𝐴𝐴𝐴𝐴 = 968.31

The heat added to the system per unit mass is

106

kJ kg

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𝑄𝑄̇𝐴𝐴𝐴𝐴𝐴𝐴 = ℎ𝐴𝐴1 − ℎ𝐴𝐴4 = 2592.1

The thermal efficiency is therefore,

kJ kg

𝑊𝑊̇𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 = 0.374 𝑄𝑄̇𝐴𝐴𝐴𝐴𝐴𝐴

𝜂𝜂𝐴𝐴 =

The ratio of the primary to secondary mass flow rates is 13.18 as given in Table 6.7.

SM-Appendix B, Single Turbine, Direct, Nonmixed (Figure SM-6.2) We may follow much of the same procedure as Appendix A. Point 1: At point 1 we know the pressure and quality: 𝑃𝑃𝐵𝐵1 = 7.75 MPa 𝑥𝑥𝐵𝐵1 = 1

We can use steam tables to calculate enthalpy and entropy: ℎ𝐵𝐵1 = 2762.34

kJ kJ 𝑠𝑠𝐵𝐵1 = 5.76201 kg kg K

Point 2: At point 2 we know the entropy and pressure: 𝑠𝑠𝐵𝐵2 = 𝑠𝑠𝐵𝐵1 = 5.76201

kJ kg ∙ K

𝑃𝑃𝐵𝐵2 = 6.89 kPa

We can look up the saturated liquid/vapor entropies: 𝑠𝑠𝐵𝐵2𝑓𝑓 = 0.555081

The quality is calculated to be:

𝑥𝑥𝐵𝐵2 =

kJ kg K

𝑠𝑠𝐵𝐵2𝑔𝑔 = 8.28006

𝑠𝑠𝐵𝐵2 − 𝑠𝑠𝐵𝐵2𝑓𝑓 = 0.674 𝑠𝑠𝐵𝐵2𝑔𝑔 − 𝑠𝑠𝐵𝐵2𝑓𝑓

kJ kg K

We can use the quality and pressure in steam tables to determine the enthalpy: ℎ𝐵𝐵2 = 1785.93

kJ kg

Point 3: At point 3 we know the pressure and quality: 𝑃𝑃𝐵𝐵3 = 6.89 kPa

𝑥𝑥𝐴𝐴3 = 0.

We may use steam tables to look up the enthalpy and entropy: ℎ𝐵𝐵3 = 162.12

kJ kg

𝑠𝑠𝐵𝐵3 = 0.555081

Point 4: At point 4 we know the entropy and pressure:

107

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Chapter 6 - Thermodynamics of Nuclear Energy Conversion Systems Nonflow and Steady Flow: First- and Second-Law Applications

𝑠𝑠𝐵𝐵4 = 𝑠𝑠𝐵𝐵3 = 0.555081

kJ kg K

𝑃𝑃𝐵𝐵4 = 7.75 MPa

We can use steam tables to determine the enthalpy to be: ℎ𝐵𝐵4 = 170.22

Point 7: We the pressure and temperature to be:

kJ kg

𝑃𝑃𝐵𝐵7 = 15.5 MPa 𝑇𝑇𝐵𝐵7 = 599 𝐾𝐾 = 325.85 °C

We can use steam tables to determine the enthalpy to be: ℎ𝐵𝐵7 = 1489.87

kJ kg

Point F: At point F we know the pressure, and since this is an isenthalpic process, we also know the enthalpy: 𝑃𝑃𝐵𝐵𝐵𝐵 = 7.75 MPa ℎ𝐵𝐵𝐵𝐵 = ℎ𝐵𝐵7 = 1489.9

The saturated liquid and vapor enthalpies at this pressure are: ℎ𝐵𝐵𝐵𝐵𝐵𝐵 = 1305.22

The quality can therefore be calculated as, 𝑥𝑥𝐵𝐵𝐵𝐵 =

kJ kg

ℎ𝐵𝐵𝐵𝐵𝐵𝐵 = 2762.34

kJ kg

ℎ𝐵𝐵𝐵𝐵 − ℎ𝐵𝐵𝐵𝐵𝐵𝐵 = 0.127 ℎ𝐵𝐵𝐵𝐵𝐵𝐵 − ℎ𝐵𝐵𝐵𝐵𝐵𝐵

Point 8: At point 8 we know the pressure and quality of the fluid: 𝑃𝑃𝐵𝐵8 = 7.75 MPa 𝑥𝑥𝐵𝐵8 = 0

We can use this to determine the enthalpy from steam tables to be: ℎ𝐵𝐵8 = 1305.22

kJ kg

The fraction of the total mass flow rate at this point is, equivalent to the portion of liquid that is separated off from point F. We can relate this to the quality: 𝑦𝑦𝐵𝐵8 = 1 − 𝑥𝑥𝐵𝐵𝐵𝐵 = 0.873

Therefore, the rest of the fluid passes through points 1, 2, 3, 4: 𝑦𝑦𝐵𝐵1 = 𝑦𝑦𝐵𝐵2 = 𝑦𝑦𝐵𝐵3 = 𝑦𝑦𝐵𝐵4 = 0.127

Point 5: At point 5 we know the pressure to be:

𝑃𝑃𝐵𝐵5 = 7.75 MPa 108

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Chapter 6 - Thermodynamics of Nuclear Energy Conversion Systems Nonflow and Steady Flow: First- and Second-Law Applications

We can calculate the enthalpy using conservation of energy. Stream 4 and stream 8 combine to stream 5 and therefore: 𝑚𝑚̇𝐵𝐵5 ℎ𝐵𝐵5 = 𝑚𝑚̇𝐵𝐵4 ℎ𝐵𝐵4 + 𝑚𝑚̇𝐵𝐵8 ℎ𝐵𝐵8

If we divide by the total mass flow rate of the system, we may determine the enthalpy: ℎ𝐵𝐵5 = 𝑦𝑦𝐵𝐵4 ℎ𝐵𝐵4 + 𝑦𝑦𝐵𝐵8 ℎ𝐵𝐵8 = 1161.4

kJ kg

We can use the pressure and the enthalpy to determine the entropy from steam tables: 𝑠𝑠𝐵𝐵5 = 2.92689

kJ kg K

Point 6: At point we know the pressure and the entropy: 𝑃𝑃𝐵𝐵6 = 15.5 MPa

𝑠𝑠𝐵𝐵5 = 2.92689

From steam tables, we can determine the enthalpy to be: ℎ𝐵𝐵6 = 1171.44

kJ kg K

kJ kg

Thermal Efficiency: The work of the turbine per mass flow rate is 𝑊𝑊̇𝐵𝐵𝐵𝐵 = ℎ𝐵𝐵1 − ℎ𝐵𝐵2 = 976.41

The work of the pump 1 per mass flow rate is

𝑊𝑊̇𝐵𝐵𝐵𝐵1 = ℎ𝐵𝐵4 − ℎ𝐵𝐵3 = 8.1

The work of the pump 1 per mass flow rate is

kJ kg

𝑊𝑊̇𝐵𝐵𝐵𝐵2 = ℎ𝐵𝐵6 − ℎ𝐵𝐵5 = 10.05

kJ kg

The net work taking into account the fraction of the total flow rate going through each component is, 𝑊𝑊̇𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵 = 𝑦𝑦𝐵𝐵1 𝑊𝑊̇𝐵𝐵𝐵𝐵 − 𝑦𝑦𝐵𝐵4 𝑊𝑊̇𝐵𝐵𝐵𝐵1 − 𝑊𝑊̇𝐵𝐵𝐵𝐵2 = 112.657

The heat added to the system per unit mass is

𝑄𝑄̇𝐵𝐵𝐵𝐵𝐵𝐵 = ℎ𝐵𝐵7 − ℎ𝐵𝐵6 = 318.43

The thermal efficiency is therefore,

109

kJ kg

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Chapter 6 - Thermodynamics of Nuclear Energy Conversion Systems Nonflow and Steady Flow: First- and Second-Law Applications

𝜂𝜂𝐵𝐵 =

𝑊𝑊̇𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵 = 0.354 𝑄𝑄̇𝐵𝐵𝐵𝐵𝐵𝐵

The ratio of the mass flow rate of the “primary” side to “secondary” side is calculated using fractional flow rates. This primary side has the total mass flow rate and is therefore 1. The secondary side fractional mass flow rate, say through the turbine yB1. We can therefore calculate this ratio to be: 1 = 7.891 𝑦𝑦𝐵𝐵1

SM-Appendix C, Single Turbine, Direct, Mixed (Figure SM-6.3) We may follow much of the same procedure as Appendices A and B. Point 1: At point 1 we know the pressure and quality: 𝑃𝑃𝐶𝐶1 = 7.75 MPa 𝑥𝑥𝐶𝐶1 = 1

We can use steam tables to calculate enthalpy and entropy: ℎ𝐶𝐶1 = 2762.34

kJ kJ 𝑠𝑠𝐶𝐶1 = 5.76201 kg kg K

Point 2: At point 2 we know the entropy and pressure: 𝑠𝑠𝐶𝐶2 = 𝑠𝑠𝐶𝐶1 = 5.76201

kJ kg K

𝑃𝑃𝐶𝐶2 = 6.89 kPa

We can look up the saturated liquid/vapor entropies: 𝑠𝑠𝐶𝐶2𝑓𝑓 = 0.555081

The quality is calculated to be:

𝑥𝑥𝐶𝐶2 =

kJ kg K

𝑠𝑠𝐶𝐶2𝑔𝑔 = 8.28006

𝑠𝑠𝐶𝐶2 − 𝑠𝑠𝐶𝐶2𝑓𝑓 = 0.674 𝑠𝑠𝐶𝐶2𝑔𝑔 − 𝑠𝑠𝐶𝐶2𝑓𝑓

kJ kg K

We can use the quality and pressure in steam tables to determine the enthalpy: ℎ𝐶𝐶2 = 1785.93

kJ kg

Point 3: At point 3 we know the pressure and quality: 𝑃𝑃𝐵𝐵3 = 6.89 kPa

𝑥𝑥𝐴𝐴3 = 0

We may use steam tables to look up the enthalpy and entropy: ℎ𝐶𝐶3 = 162.12

kJ kg

𝑠𝑠𝐶𝐶3 = 0. 555081 110

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Chapter 6 - Thermodynamics of Nuclear Energy Conversion Systems Nonflow and Steady Flow: First- and Second-Law Applications

Point 4: At point 4 we know the entropy and pressure: 𝑠𝑠𝐶𝐶4 = 𝑠𝑠𝐶𝐶3 = 0.555081

kJ kg K

𝑃𝑃𝐶𝐶4 = 7. 75 MPa

We can use steam tables to determine the enthalpy to be: ℎ𝐶𝐶4 = 170.22

kJ kg

Point 5: At point 5 we know the pressure and quality:

𝑃𝑃𝐶𝐶5 = 7.75 MPa 𝑥𝑥𝐶𝐶5 = 0

We can then go to steam tables to look up the enthalpy and entropy: ℎ𝐶𝐶5 = 1305.22

kJ kJ 𝑠𝑠𝐶𝐶5 = 3.18736 kg kg K

Point 7: We the pressure and temperature to be:

𝑃𝑃𝐶𝐶7 = 15.5 MPa 𝑇𝑇𝐶𝐶7 = 599 K = 325.85 °C

We can use steam tables to determine the enthalpy to be: ℎ𝐶𝐶7 = 1489.87

kJ kg

Point F: At point F we know the pressure) and since this is an isenthalpic process, we also know the enthalpy: 𝑃𝑃𝐵𝐵𝐵𝐵 = 7.75 MPa ℎ𝐵𝐵𝐵𝐵 = ℎ𝐵𝐵7 = 1489.9

We can perform an energy balance on the fully mixed flash tank:

kJ kg

𝑚𝑚̇𝐶𝐶7 ℎ𝐶𝐶7 + 𝑚𝑚̇𝐶𝐶4 ℎ𝐶𝐶4 = 𝑚𝑚̇𝐶𝐶5 ℎ𝐶𝐶5 + 𝑚𝑚̇𝐶𝐶1 ℎ𝐶𝐶1

We can now divide by the total the mass flow rate. The fraction of steam mass flow rate that goes to the turbine will be denoted as xF: ℎ𝐶𝐶7 + 𝑥𝑥𝐹𝐹 ℎ𝐶𝐶4 = ℎ𝐶𝐶5 + 𝑥𝑥𝐹𝐹 ℎ𝐶𝐶1

Therefore, we can calculate this fraction of steam: 𝑥𝑥𝐹𝐹 =

ℎ𝐶𝐶7 − ℎ𝐶𝐶5 = 0.071 ℎ𝐶𝐶1 − ℎ𝐶𝐶4

This is parameter will also be denoted as the fraction of the total mass flow rate that travels through the turbine, condenser and pump 1: 𝑦𝑦𝐶𝐶1 = 𝑦𝑦𝐶𝐶2 = 𝑦𝑦𝐶𝐶3 = 𝑦𝑦𝐶𝐶4 = 0.071

Point 6: At point we know the pressure and the entropy:

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𝑃𝑃𝐶𝐶6 = 15.5 MPa 𝑠𝑠𝐶𝐶6 = 𝑠𝑠𝐶𝐶5 = 3.18736

From steam tables, we can determine the enthalpy to be: ℎ𝐶𝐶6 = 1315.68

kJ kg K

kJ kg

Thermal Efficiency: The work of the turbine per mass flow rate is 𝑊𝑊̇𝐶𝐶𝐶𝐶 = ℎ𝐶𝐶1 − ℎ𝐶𝐶2 = 976.41

The work of the pump 1 per mass flow rate is

𝑊𝑊̇𝐶𝐶𝐶𝐶1 = ℎ𝐶𝐶4 − ℎ𝐶𝐶3 = 8.1

The work of the pump 1 per mass flow rate is

kJ kg

𝑊𝑊̇𝐶𝐶𝐶𝐶2 = ℎ𝐶𝐶6 − ℎ𝐶𝐶5 = 10.46

kJ kg

The net work taking into account the fraction of the total flow rate going through each component is, 𝑊𝑊̇𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 = 𝑦𝑦𝐶𝐶1 𝑊𝑊̇𝐶𝐶𝐶𝐶 − 𝑦𝑦𝐶𝐶4 𝑊𝑊̇𝐶𝐶𝐶𝐶1 − 𝑊𝑊̇𝐶𝐶𝐶𝐶2 = 58.518

The heat added to the system per unit mass is

𝑄𝑄̇𝐶𝐶𝐶𝐶𝐶𝐶 = 𝑦𝑦𝐶𝐶7 − ℎ𝐶𝐶6 = 174.19

The thermal efficiency is therefore,

𝜂𝜂𝐶𝐶 =

kJ kg

𝑊𝑊̇𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 = 0.336 𝑄𝑄̇𝐶𝐶𝐶𝐶𝐶𝐶

112

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PROBLEM 6.8 QUESTION Advantages of Moisture Separation and Feedwater Heating (Section 6.5) A simplified BWR system with moisture separation and an open feedwater heater is described in Problem 6.3. Compute the improvement in thermal efficiency that results from the inclusion of these two components in the power cycle. Do you think the thermal efficiency improvement from these components is sufficient to justify the capital investment required? Are there other reasons for having moisture separation? Answer: ηth changes from 36.9% to 37.7%

PROBLEM 6.8 SOLUTION Advantages of Moisture Separation and Feedwater Heating (Section 6.5) To solve this problem, we will re-solve Problem 6.3 without the inclusion of the moisture separator and open feedwater heater. We will assume that the turbine and pump are isentropic components. Point 1 At this point we know the pressure and quality: 𝑃𝑃1 = 6890 kPa

𝑥𝑥1 = 1

We can use steam tables to determine the enthalpy and entropy: ℎ1 = 2774.05

kJ kJ 𝑠𝑠1 = 5.82273 kg kg K

Point 5 The two properties that are known at this state are the pressure and the entropy. The entropy is evaluated from point 3 using the isentropic expansion assumption, 𝑠𝑠5 = 𝑠𝑠1 = 5.82273

kJ 𝑃𝑃5 = 6.89 kPa kg K

The saturated liquid/vapor entropies at the pressure are:

The quality at 5s:

𝑠𝑠5𝑓𝑓 = 0.555081 𝑥𝑥5 =

kJ kg K

𝑠𝑠5𝑔𝑔 = 8.28006

𝑠𝑠5 − 𝑠𝑠5𝑓𝑓 = 0.682 𝑠𝑠5𝑞𝑞 − 𝑠𝑠5𝑓𝑓

kJ kg K

The enthalpy can now be evaluated using steam tables:

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ℎ5 = 1805.11

kJ kg

Point 6 At point 6 we know the pressure and quality: 𝑃𝑃6 = 6.89 kPa

𝑥𝑥6 = 0

We can use steam tables to determine the enthalpy and entropy: ℎ6 = 162.12

kJ kg

𝑠𝑠6 = 0.555081

kJ kg K

Point 9 At this state we know the pressure and entropy. The entropy is evaluated from point 6 using the isentropic assumption: 𝑠𝑠9 = 𝑠𝑠6 0.555081

kJ kg K

𝑃𝑃9 = 6890 kPa

We can use these properties to determine the enthalpy from steam tables: ℎ9 = 169.042

Cycle Efficiency 𝜂𝜂𝑡𝑡ℎ =

kJ kg

𝑊𝑊̇𝑇𝑇 − 𝑊𝑊̇𝑝𝑝 (ℎ1 − ℎ5 ) − (ℎ9 − ℎ5 ) = = 0.3693 ℎ1 − ℎ9 𝑄𝑄̇𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟

Is the 0.8% increase in efficiency worth it? Since it is such a low gain in thermal efficiency it would not be work the capital investment. There are other reasons for having the moisture separator. The less moisture that the turbine blades are exposed to, the less they will erode. This is very important for the low pressure turbine because the quality exiting the high pressure turbine is not close to 1. Therefore it is necessary to separate out the moisture and use this hot water for other applications such as a open feedwater heater to raise the average temperature of the water that is being heated in the reactor, This will inherently raise the thermal efficiency.

PROBLEM 6.9 QUESTION Ideal Brayton Cycle (Section 6.6) The Brayton cycle shown in Figure 6.41 operates using CO2 as a working fluid with compressor and turbine isentropic efficiencies of 1.0. Calculate the thermal efficiency of this cycle when the working fluid is modeled as: 1. A perfect gas of γ = 1.30. 2. A real fluid (see below for extracted values from Keenan and Kaye’s gas tables). 3. A real fluid and the compressor and turbine both have isentropic efficiencies of 0.95.

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FIGURE 6.41 Ideal Brayton cycle (State a: P = 137.9 kPa, T = 32.2°C, State c: P = 689.5 kPa, T = 537.8°C).

The parameters needed for a real fluid (from Keenan and Kaye’s gas tables) are shown in Table 6.22. Table 6.22 Parameters for Problem 6.9 Parameter T [°C]

a 32.2

b –

c 537.8

d –

P [psia]

137.9

689.5

137.9

Pr*

0.16108

0.8054

31.5

6.3

kJ � kg ∙ mole

9643.6

14513.8

33038.5

23439.1

ℎ�

*Pr is relative pressure. The ratio of pressures Pa and Pb corresponding to the temperatures Ta and Tb, respectively, for an isentropic process is equal to the ratio of the relative pressures Pra and Prb as tabulated for Ta and Tb, respectively. Thus:

Answers:

�

𝑃𝑃𝑎𝑎 𝑃𝑃𝑟𝑟𝑟𝑟 � = 𝑃𝑃𝑏𝑏 𝑠𝑠=𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑃𝑃𝑟𝑟𝑟𝑟

1. ηth = 31.0% 2. ηth = 25.5% 3. ηth = 21.9%

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Chapter 6 - Thermodynamics of Nuclear Energy Conversion Systems Nonflow and Steady Flow: First- and Second-Law Applications

PROBLEM 6.9 SOLUTION Ideal Brayton Cycle (Section 6.6) The Brayton cycle shown in Figure 6.41 operates using CO2 as a working fluid with compressor and turbine isentropic efficiencies of 1.0. Calculate the thermal efficiency of this cycle when the working fluid is modeled as: 1. A perfect gas of γ = 1.30 The thermodynamic efficiency of this cycle can be calculated as

𝜂𝜂 =

𝑊𝑊̇𝑇𝑇 − 𝑊𝑊̇𝐶𝐶 𝑄𝑄̇

� 𝑇𝑇 is the turbine work per mass flow rate, 𝑊𝑊 �𝐶𝐶 is the compressor work per mass flow rate where 𝑊𝑊 and 𝑄𝑄� is the heat rate added per mass flow rate. Each of these quantities can be determined using the following equations:

𝑊𝑊̇𝑇𝑇 = ℎ𝑐𝑐 − ℎ𝑑𝑑 = 𝑐𝑐𝑝𝑝 (𝑇𝑇𝑐𝑐 − 𝑇𝑇𝑑𝑑 )

𝑊𝑊̇𝐶𝐶 = ℎ𝑏𝑏 − ℎ𝑎𝑎 = 𝑐𝑐𝑝𝑝 (𝑇𝑇𝑏𝑏 − 𝑇𝑇𝑎𝑎 ) 𝑄𝑄̇ = ℎ𝑐𝑐 − ℎ𝑏𝑏 = 𝑐𝑐𝑝𝑝 (𝑇𝑇𝑐𝑐 − 𝑇𝑇𝑏𝑏 )

Temperatures at state points a and c are known, but b and d need to be determined. For an isentropic process, Pvk = const. Using this relationship and the equation of state, Pv = RT, the following relation can be derived relating pressure and temperature of two different state points: 1−𝛾𝛾

𝑇𝑇2 𝑃𝑃1 𝛾𝛾 =� � 𝑇𝑇1 𝑃𝑃2

Using this equation the temperature at points b and d can be determined, 1−1.30

1−𝛾𝛾

𝑃𝑃 𝑎𝑎 𝛾𝛾 1 1.30 = 305.35[K] � � = 442.69 K = 169.54 °C 𝑇𝑇𝑏𝑏 = 𝑇𝑇𝑎𝑎 � � 𝑃𝑃𝑏𝑏 5 1−𝛾𝛾

1−1.30 𝑃𝑃𝑐𝑐 𝛾𝛾 𝑇𝑇𝑑𝑑 = 𝑇𝑇𝑐𝑐 � � = 810.95[K](5) 1.30 = 559.36 K = 286.21 °C 𝑃𝑃𝑑𝑑

The efficiency can then be calculated: 𝜂𝜂1 =

(𝑇𝑇𝑐𝑐 − 𝑇𝑇𝑑𝑑 ) − (𝑇𝑇𝑏𝑏 − 𝑇𝑇𝑎𝑎 ) (537.8 − 286.21)[°C] − (19.54 − 32.2)[°C] = = 0.31 (537.8 − 169.54)[°C] 𝑇𝑇𝑐𝑐 − 𝑇𝑇𝑏𝑏

2. For a real fluid, using values given in the table

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To deterime the efficiency for a real fluid using the values in the table, we may just subtract the enthalpies per mole: kJ kJ (ℎ𝑐𝑐 − 𝑇𝑇𝑑𝑑 ) − (𝑇𝑇𝑏𝑏 − 𝑇𝑇𝑎𝑎 ) 33038.5 − 23439.1 �kg ∙ mole� − (14513.8 − 9643.6) �kg ∙ mole� = 𝜂𝜂2 = kJ 𝑇𝑇𝑐𝑐 − 𝑇𝑇𝑏𝑏 (33038.5 − 14513.8) � � kg ∙ mole = 0.255

3. For a real fluid with turbine and compressor isentropic efficiencies, ηT = ηC = 0.95

Since now we are working with isentropic efficiencies, we may redefine the work of the turbine and compressor: 𝑊𝑊̇𝑇𝑇 = ℎ𝑐𝑐 − ℎ𝑑𝑑′ = 𝜂𝜂𝑇𝑇 (ℎ𝑐𝑐 − ℎ𝑑𝑑 )

𝑊𝑊̇𝐶𝐶 = ℎ𝑏𝑏′ − ℎ𝑎𝑎 =

1 (ℎ − ℎ𝑎𝑎 ) 𝜂𝜂𝐶𝐶 𝑏𝑏

The real enthalpies at points b and d can then be calculated from the equations above, ℎ𝑏𝑏′ = ℎ𝑎𝑎 +

1 kJ (ℎ𝑏𝑏 − ℎ𝑎𝑎 ) = 14770.1 𝜂𝜂𝐶𝐶 kg ∙ mole

ℎ𝑑𝑑′ = ℎ𝑐𝑐 − 𝜂𝜂𝑇𝑇 (ℎ𝑐𝑐 − ℎ𝑑𝑑 ) = 23919.1

kJ kg ∙ mole

The thermodynamic efficiency can now be calculated using the real enthalpies, 𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 (ℎ𝑐𝑐 − ℎ𝑑𝑑′ ) − (ℎ𝑏𝑏′ − ℎ𝑎𝑎 ) (33038.5 − 23919.1) �𝑘𝑘𝑘𝑘 ∙ 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚� − (14770.1 − 9643.6) �𝑘𝑘𝑘𝑘 ∙ 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚� 𝜂𝜂3 = = 𝑘𝑘𝑘𝑘 ℎ𝑐𝑐 − ℎ𝑏𝑏′ (33038.5 − 14770.1) � � 𝑘𝑘𝑘𝑘 ∙ 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 = 0.219

PROBLEM 6.10 QUESTION

Complex Real Brayton Cycle (Section 6.7) A gas cooled reactor is designed to heat helium gas to a maximum temperature of 540 °C. The helium flow through a gas turbine, generating work to run the compressors and an electric generator, and then through a regenerative heat exchanger and two stages of compression with precooling to 40 °C before entering each compressor. Each compressor and the turbine have an isentropic efficiency of 85%, and the exchanger drop factor β is equal to 1.05. Each compression stage has a pressure ratio (rp) of 1.27. The heat exchanger effectiveness, ξ, is 0.90. Determine the cycle thermal efficiency. The Brayton cycle system is illustrated in Figure 6.42. The pressure drop factor β is defined as:

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Chapter 6 - Thermodynamics of Nuclear Energy Conversion Systems Nonflow and Steady Flow: First- and Second-Law Applications 𝛾𝛾−1

Answer: ηth = 13.9%

𝑃𝑃4 𝑃𝑃7 𝑃𝑃2 𝛾𝛾 𝛽𝛽 ≡ � ∙ ∙ � = 1.05 𝑃𝑃6 𝑃𝑃1 𝑃𝑃3

FIGURE 6.42 Complex Brayton cycle.

PROBLEM 6-10 SOLUTION Complex Real Brayton Cycle (Section 6.7) To begin a cycle analysis a T-s diagram should be created. Figure SM-6.4 shows the T-s diagram for this real Brayton cycle.

FIGURE SM-6.4: T-s diagram

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The given parameters in the problem statement are listed below: –

Temperatures: T6 = 540 °C T1 = T3 = 40 °C

–

Turbine and Compressor Efficiencies: ηT = ηC = 0.85

–

Compressor Pressure Ratio: rp = 1.27

–

Pressure drop factor: β = 1.05

–

Heat Exchanger Effectiveness: ξ = 0.90

–

Ratio of specific heats for helium (monatomic gas): 𝛾𝛾 = 𝑐𝑐 = 3⁄2𝑅𝑅 = 1.667

𝑐𝑐𝑝𝑝 𝜐𝜐

The cycle efficiency can be calculated with 𝜂𝜂𝑡𝑡ℎ =

5⁄2𝑅𝑅

𝑊𝑊̇𝑛𝑛𝑛𝑛𝑛𝑛 𝑊𝑊̇𝑇𝑇 − 𝑊𝑊̇𝐶𝐶1 − 𝑊𝑊̇𝐶𝐶2 = 𝑄𝑄̇𝑖𝑖𝑖𝑖 𝑄𝑄̇𝑖𝑖𝑖𝑖

(1)

Each of the components in Equation (1) are defined below: 𝑊𝑊̇𝑇𝑇 = 𝑚𝑚̇𝑐𝑐𝑝𝑝 (𝑇𝑇6 − 𝑇𝑇7 ) 𝑊𝑊̇𝐶𝐶1 = 𝑚𝑚̇𝑐𝑐𝑝𝑝 (𝑇𝑇2 − 𝑇𝑇1 )

(2) (3)

𝑊𝑊̇𝐶𝐶2 = 𝑚𝑚̇𝑐𝑐𝑝𝑝 (𝑇𝑇4 − 𝑇𝑇3 )

(4)

𝑄𝑄̇𝑖𝑖𝑖𝑖 = 𝑚𝑚̇𝑐𝑐𝑝𝑝 (𝑇𝑇6 − 𝑇𝑇5 )

(5)

Since the mass flow rate through each of these components is the same, and for this question we will also assume that the specific heat is constant, only the temperature drop need to be determined. Turbine Work The main goal here is to determine the temperature at point 7. First, we may solve for the ideal temperature assuming that the turbine is isentropic and then apply the turbine efficiency to get the real temperature. For an isentropic process, Pυγ = const. Using the equation of state, the following relation can be derived to relate the pressures and temperatures at points 6 and 7 (note a subscript “s” denotes a temperature derived from an isentropic calculation): 𝛾𝛾−1

𝑃𝑃7 𝛾𝛾 𝑇𝑇7𝑠𝑠 = 𝑇𝑇6 � � 𝑃𝑃6

(6)

The ratio of these pressures is unknown, however it may be determined using the pressure drop factor shown below: 𝛾𝛾−1

𝛾𝛾−1

𝑃𝑃2 𝑃𝑃4 𝛾𝛾 𝑃𝑃7 𝛾𝛾 𝑃𝑃4 𝑃𝑃7 𝑃𝑃2 𝛾𝛾 � � 𝛽𝛽 = � ∙ ∙ � =� ∙ � 𝑃𝑃6 𝑃𝑃1 𝑃𝑃3 𝑃𝑃1 𝑃𝑃3 𝑃𝑃6

(7)

The right most quantity in Equation (7) is the unknown quantity needed to solve for the temperature 𝑃𝑃

𝑃𝑃

in Equation (6). We can rearrange Equation (7) and substitute the pressure ratio, 𝑟𝑟𝑝𝑝 = 𝑃𝑃2 = 𝑃𝑃4 1

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Chapter 6 - Thermodynamics of Nuclear Energy Conversion Systems Nonflow and Steady Flow: First- and Second-Law Applications 𝛾𝛾−1

𝑃𝑃7 𝛾𝛾 � � = 𝑃𝑃6

𝛽𝛽

𝛾𝛾−1 = 0.867 �𝑟𝑟𝑝𝑝 ∙ 𝑟𝑟𝑝𝑝 � 𝛾𝛾

(8)

This result can be substituted into Equation (6) to determine the ideal temperature at point 7, 𝑇𝑇7𝑠𝑠 = (813.15 K) 0.867 = 706.27 K = 431.9 °C

(9)

𝑊𝑊̇𝑇𝑇 = 𝑚𝑚̇𝑐𝑐𝑝𝑝 (𝑇𝑇6 − 𝑇𝑇7 ) = 𝑚𝑚̇𝑐𝑐𝑝𝑝 𝜂𝜂𝑇𝑇 (𝑇𝑇6 − 𝑇𝑇7𝑠𝑠 ) 𝑇𝑇7 − 𝑇𝑇6 − 𝜂𝜂𝑇𝑇 (𝑇𝑇6 − 𝑇𝑇7𝑠𝑠 ) = 448.07 °C

(10) (11)

𝑊𝑊̇𝑇𝑇 − 𝑇𝑇6 − 𝑇𝑇7 = 91.93 °C

(12)

Using the turbine efficiency, we can rewrite Equation (2) and solve for the real temperature at point 7,

And therefore the work of the turbine (just need temperature drop is): Compressor 1 Work The main goal here is to determine the temperature at point 2. The ideal temperature at point 2 can be calculated similar to the temperature at point 7, 𝛾𝛾−1

𝑃𝑃2 𝛾𝛾 𝑇𝑇2𝑠𝑠 = 𝑇𝑇1 � � 𝑃𝑃1

(13)

However in this case, we know that the ratio of the pressures is just the given compressor pressure ratio, rp: 𝛾𝛾−1

𝑇𝑇2𝑠𝑠 = 𝑇𝑇1 �𝑟𝑟𝑝𝑝 � 𝛾𝛾 = 344.31 K = 71.43 °C

(14)

The real temperature can be calculated using the definition of the compressor work in Equation (3) and the compressor efficiency. 𝑊𝑊̇𝐶𝐶1 = 𝑚𝑚̇𝑐𝑐𝑝𝑝 (𝑇𝑇2 − 𝑇𝑇1 ) = 𝑚𝑚̇𝑐𝑐𝑝𝑝 𝑇𝑇2 − 𝑇𝑇1 +

1 (𝑇𝑇 − 𝑇𝑇1 ) 𝜂𝜂𝐶𝐶 2𝑠𝑠

1 (𝑇𝑇 − 𝑇𝑇1 ) = 76.97 °C 𝜂𝜂𝐶𝐶 2𝑠𝑠

(15) (16)

Therefore, the temperature drop over compressor 1 is

∆𝑇𝑇𝐶𝐶1 = 36.97 °C

(17)

𝑇𝑇4 = 𝑇𝑇2 = 76.97 °C

(18)

Compressor 2 Work Since the compressors are identical, they have the same efficiency, pressure ratio, and inlet temperature. Therefore the outlet temperature and temperature drop are equivalent,

∆𝑇𝑇𝐶𝐶2 = 𝑇𝑇4 − 𝑇𝑇3 = 36.97 °C

(19)

Reactor Heat Addition The main goal here is to determine the temperature at point 5. The heat

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exchanger effectiveness can be calculated with:

Solving for Temperature at point 5,

𝜉𝜉 =

𝑇𝑇5 − 𝑇𝑇4 𝑇𝑇7 − 𝑇𝑇4

(20)

𝑇𝑇5 − 𝑇𝑇4 + 𝜉𝜉(𝑇𝑇7 − 𝑇𝑇4 ) = 410.96 °C

(21)

𝑄𝑄̇𝑖𝑖𝑖𝑖 = 𝑇𝑇6 − 𝑇𝑇5 = 129.04 °C

(22)

The temperature drop across the reactor is then

Efficiency of Cycle: Recalling that for this problem, the mass flow rate is the same through all components and that we are neglecting any changes in the specific heat, the thermodynamic efficiency can just be calculated with temperature differences: 𝜂𝜂𝑡𝑡ℎ =

𝑊𝑊̇𝑇𝑇 − 𝑊𝑊̇𝐶𝐶1 − 𝑊𝑊̇𝐶𝐶2 = 0.138 = 13.8% 𝑄𝑄̇𝑖𝑖𝑖𝑖

(23)

PROBLEM 6.11 QUESTION

Cycle Thermal Efficiency Problem Involving a Bottoming Cycle (Section 6.7) In Example 6.10A it is shown that the cycle thermal efficiency of the simple Brayton cycle shown in Figure 6.24 can be increased by utilizing regeneration. Specifically, it was found that, with the addition of a regenerator effectiveness 0.75, the cycle thermal efficiency was increased from 42.3% to 48.1%. Another way of improving the efficiency of the simple Brayton cycle is to use a bottoming cycle. To this end, consider the system shown in Figure 6.43. It shows the simple Brayton cycle with a Brayton bottoming cycle. Parameters and assumptions for this Problem are given in Table 6.23.

TABLE 6.23 Parameters and Assumptions for Problem 6.11 T1 = 278 K T3 = 972 K T9 = T1 (ΔTp)1 = pinch point of heat exchanger #1= 15°C All turbine and compressors in both cycles are ideal

rp for the simple Brayton cycle = 4.0 cp for both cycles = 5230 J/kg K γ for both cycles = 1.658 Mass flow rate for the simple Brayton cycle = twice the mass flow rate for the Brayton bottoming cycle No duct pressure losses in either cycle

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FIGURE 6.43 A simple Brayton cycle with a Brayton bottoming cycle.

QUESTIONS 1. Draw the T-s diagram for the entire system. 2. What must the pressure ratio be of Turbine #2 and Compressor #2 such that the cycle thermal efficiency of the entire system is maximized? 3. What is the maximum cycle thermal efficiency? Answers: 2. rp = 2.34 3. ηth = 46.9%

PROBLEM 6.11 SOLUTION Cycle Thermal Efficiency Problem Involving a Bottoming Cycle (Section 6.7) Given Parameters: –

Temperature at point 1, T1 = 278 K

–

Temperature at point 3, T3 = 972 K

–

Temperature at point 9, T9 = T1 = 278 K

–

Pressure ratio of simple Brayton Cycle, rp1 = 4.0

–

Specific heat in both cycles, 𝑐𝑐𝑝𝑝 = 5230 kg K

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–

Gamma factor for both cycles, γ = 1.658

–

Pinch-point temperature of heat exchanger #1, (ΔTp)1 = 15°C

1. Draw the T-s diagram for the entire system (Figure SM-6.5).

FIGURE SM-6.5. T-s diagram for entire system.

2. What must the pressure ratio be of Turbine #2 and Compressor #2 such that the cycle thermal efficiency of the entire system is maximized? The thermal efficiency of the cycle can be defined as 𝜂𝜂𝑡𝑡ℎ =

𝑊𝑊̇𝑡𝑡1 + 𝑊𝑊̇𝑡𝑡2 − 𝑊𝑊̇𝑐𝑐1 − 𝑊𝑊̇𝑐𝑐2 𝑄𝑄̇𝑖𝑖𝑖𝑖

(1)

Work of Turbine 1: We may define the work of turbine 1 as 𝑊𝑊̇𝑡𝑡1 = 𝑚𝑚̇1 𝑐𝑐𝑝𝑝 (𝑇𝑇3 − 𝑇𝑇4 )

(2)

To determine the temperature at 4 we may use the isentropic condition that 1−𝛾𝛾

𝑇𝑇𝑇𝑇 𝛾𝛾 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐

The temperature at point 4 can now be calculated with,

1−𝛾𝛾 𝑇𝑇4 = 𝑇𝑇3 𝑟𝑟𝑝𝑝1𝛾𝛾 = 560.7 K

(3)

Work of Turbine 2: We may define the work of turbine 2 as 𝑊𝑊̇𝑡𝑡2 = 𝑚𝑚̇2 𝑐𝑐𝑝𝑝 (𝑇𝑇7 − 𝑇𝑇8 )

(4)

𝑇𝑇7 = 𝑇𝑇4 − (Δ𝑇𝑇𝑃𝑃 )1 = 545.7 K

(5)

Note that from the problem statement, ṁ1 = 2ṁ2. The temperature at point 7 may be calculate from the pinch point temperature of heat exchanger 1, Unlike with Turbine 1, we do not know the pressure ratio in this bottoming cycle. Therefore we

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will leave T8 in terms of the pressure ratio, rp2, 1−𝛾𝛾 𝛾𝛾 𝑇𝑇8 �𝑟𝑟𝑝𝑝2 � = 𝑇𝑇7 𝑟𝑟𝑝𝑝2

(6)

Work of Compressor 1: We may define the work of compressor 1 as 𝑊𝑊̇𝑐𝑐1 = 𝑚𝑚̇1 𝑐𝑐𝑝𝑝 (𝑇𝑇2 − 𝑇𝑇1 )

(7)

The temperature at point 2 can be determined from the isentropic condition, 𝛾𝛾−1 𝑇𝑇2 = 𝑇𝑇1 𝑟𝑟𝑝𝑝1𝛾𝛾 = 481.9 K

(8)

Note the change in the exponent for the compressor as compared to the turbine. Work of Compressor 2: We may define the work of compressor 2 as 𝑊𝑊̇𝑐𝑐2 = 𝑚𝑚̇2 𝑐𝑐𝑝𝑝 (𝑇𝑇6 − 𝑇𝑇9 )

(9)

Unlike with Compressor 1, we do not know the pressure ratio in this bottoming cycle. Therefore we will leave T6 in terms of the pressure ratio, rp2 𝛾𝛾−1 𝑇𝑇6 �𝑟𝑟𝑝𝑝2 � = 𝑇𝑇9 𝑟𝑟𝑝𝑝2𝛾𝛾

(10)

Heat added: The rate of addition of heat to the system can be defined as 𝑄𝑄̇𝑖𝑖𝑖𝑖 = 𝑚𝑚̇1 𝑐𝑐𝑝𝑝 (𝑇𝑇3 − 𝑇𝑇2 )

(11)

Thermal Efficiency: We can now rewrite Equation (1), 𝜂𝜂𝑡𝑡ℎ =

𝑚𝑚̇1 𝑐𝑐𝑝𝑝 (𝑇𝑇3 − 𝑇𝑇4 ) + 𝑚𝑚̇2 𝑐𝑐𝑝𝑝 �𝑇𝑇7 − 𝑇𝑇8 �𝑟𝑟𝑝𝑝2 �� − 𝑚𝑚̇1 𝑐𝑐𝑝𝑝 (𝑇𝑇2 − 𝑇𝑇1 ) − 𝑚𝑚̇2 𝑐𝑐𝑝𝑝 �𝑇𝑇6 �𝑟𝑟𝑝𝑝2 � − 𝑇𝑇9 � 𝑚𝑚̇1 𝑐𝑐𝑝𝑝 (𝑇𝑇3 − 𝑇𝑇2 )

(12)

Note that for unknown temperatures, we have written them as a function of the bottoming cycle pressure ratio, rp2. We may divide Equation (12) by the mass flow rate of the bottoming cycle, taking into account the relationships between the flow rates and getting rid of specific heat, 𝜂𝜂𝑡𝑡ℎ =

2(𝑇𝑇3 − 𝑇𝑇4 ) + �𝑇𝑇7 − 𝑇𝑇8 �𝑟𝑟𝑝𝑝2 �� − 2(𝑇𝑇2 − 𝑇𝑇1 ) − �𝑇𝑇6 �𝑟𝑟𝑝𝑝2 � − 𝑇𝑇9 � 2(𝑇𝑇3 − 𝑇𝑇2 )

(13)

We may take the derivative of Equation (13) using the relationships in Equations (6) and (10), and set it equal to zero to determine the pressure ratio such that the thermal efficiency is a max, 𝑑𝑑𝜂𝜂𝑡𝑡ℎ =0 𝑑𝑑𝑑𝑑𝑝𝑝2

(14)

Finding the roots of Equation (14), we determine that the pressure ratio is 𝑟𝑟𝑝𝑝2 = 2.34

(15)

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We can also graph Equation (13) in Figure SM-6.6 as a function of the pressure ratio to check that this is the maximum.

FIGURE SM-6.6 Cycle thermal efficiency as a function of pressure ratio (bottoming cycle).

3. What is the maximum cycle thermal efficiency? We can plug the resulting pressure ratio into Equation (13) to determine the that maximum cycle thermal efficiency is 𝜂𝜂𝑡𝑡ℎ,𝑚𝑚𝑚𝑚𝑚𝑚 = 0.469 = 46.9%

(16)

PROBLEM 6.12 QUESTION Nuclear Cogeneration Plant (Section 6.8) A High-Temperature Gas Reactor (HTGR) is being considered for cogeneration of electricity and heat for residential heating. This HTGR uses the direct Brayton cycle shown in Figure 6.44, which comprises a turbine, a regenerator, a cogeneration heat exchanger (3 → 4) and a compressor. The cogeneration heat exchanger is used to generate steam, which is then sent to the residential area served by the plant. The helium temperature and pressure at the turbine inlet are 1000 K and 9 MPa, respectively. The minimum temperature in the cycle is 373 K. The cycle operates with a compression ratio equal to 2. The isentropic efficiency for the turbo-machines (turbine and compressor) is 0.9. Assume negligible pressure losses throughout the cycle.

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FIGURE 6.44 Schematic of the nuclear cogeneration plant.

1. Sketch the T-s diagram for the cycle. 2. An important parameter to select is the cogeneration temperature T3. If T3 is too high, regeneration is minimal and the cycle thermal efficiency becomes too low. If T3 is too low, the amount of heat delivered to the residential heating system may be too low. Find the value of T3 that will give a cycle thermal efficiency equal to 30%. 3. What is the energy utilization factor (EUF) of this cycle? The EUF is defined as the ratio of the energy utilized (net work + cogeneration heat) to the heat input (reactor heat). 4. What is the reactor thermal power if the plant is to produce 100 kg/s of saturated steam at 0.5 MPa from the return water at 80°C? 5. Nuclear cogeneration for residential heating has been rarely done although the Agesta reactor performed this function in Stockholm for a decade starting in 1963. What are in your opinion the drawbacks of this approach? Useful Properties: –

–

Helium: Treat as an ideal gas with 𝑐𝑐𝑝𝑝 = 5193 kg K ; 𝑅𝑅𝑠𝑠𝑠𝑠 = 2077 kg K ; 𝛾𝛾 = 1.667

–

Water enthalpy of vaporization = 2109 kJ/kg

Water at 0.5 MPa (Tsat = 152°C);

specific heat = 4.24 kg K

Answers : 2. T3 = 572.5 K 3. EUF = 1 4. 𝑄𝑄̇𝑅𝑅 = 344.9 MW

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Chapter 6 - Thermodynamics of Nuclear Energy Conversion Systems Nonflow and Steady Flow: First- and Second-Law Applications

PROBLEM 6.12 SOLUTION Nuclear Cogeneration Plant (Section 6.8) Given Parameters: – Temperatures at points 1 and 4, T1 = 1000 K T4 = 373 K –

Pressure at point 1, P1 = 9 MPa

–

Compressor Pressure Ratio, rp = 2

–

Turbine and Compressor Isentropic Efficiencies, ηT = ηC = 0.9

–

Specific Heat, Gas Constant and gamma factor for He,

–

𝑐𝑐𝑝𝑝𝑝𝑝𝑝𝑝 = 5.193 kg∙K 𝑅𝑅𝐻𝐻𝐻𝐻 = 2077 kg∙K 𝛾𝛾𝐻𝐻𝐻𝐻 = 1.667

Pressure and specific heat of water, 𝑃𝑃𝑤𝑤 = 0.5 MPa 𝑐𝑐𝑝𝑝𝑝𝑝 = 4.24 kg∙K

1. Sketch the T-s diagram for the cycle (Figure SM-6.7).

FIGURE SM-6.7 T-s diagram for the Cogeneration Brayton Cycle.

2. Find the value of T3 that will give a cycle thermal efficiency equal to 30%. The efficiency of the cycle can be defined as 𝜂𝜂𝑡𝑡ℎ =

𝑊𝑊̇𝑇𝑇 − 𝑊𝑊̇𝐶𝐶 = 0.3 𝑄𝑄̇𝑖𝑖𝑖𝑖

(1)

Turbine Work: Considering the process from 1 to 2 to be isentropic, we can determine the ideal temperature at 2: 1−𝛾𝛾𝐻𝐻𝐻𝐻 𝛾𝛾 𝑇𝑇2𝑠𝑠 = 𝑇𝑇1 𝑟𝑟𝑝𝑝 𝐻𝐻𝐻𝐻 = 757.8 K

We can define the work of the turbine per unit mass flowrate as:

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Chapter 6 - Thermodynamics of Nuclear Energy Conversion Systems Nonflow and Steady Flow: First- and Second-Law Applications

𝑊𝑊̇𝑇𝑇 = 𝑐𝑐𝑝𝑝𝑝𝑝𝑝𝑝 (𝑇𝑇1 − 𝑇𝑇2 ) = 𝑐𝑐𝑝𝑝𝑝𝑝𝑝𝑝 𝜂𝜂𝑇𝑇 (𝑇𝑇1 − 𝑇𝑇2𝑠𝑠 )

Therefore, we can calculate the real temperature at state 2 to be,

𝑇𝑇2 = 𝑇𝑇1 − 𝜂𝜂𝑇𝑇 (𝑇𝑇1 − 𝑇𝑇2𝑠𝑠 ) = 782.0 K

The turbine work per unit mass flow rate is

𝑊𝑊̇𝑇𝑇 = 1132

kJ kg

(2)

Compressor Work: Considering the process from 4 to 5 to be isentropic, we can determine the ideal temperature at 5: 𝛾𝛾𝐻𝐻𝐻𝐻 −1 𝛾𝛾 𝑇𝑇5𝑠𝑠 = 𝑇𝑇4 𝑟𝑟𝑝𝑝 𝐻𝐻𝐻𝐻 = 492.2 K

We can define the work of the compressor per unit mass flowrate as: 𝑊𝑊̇𝐶𝐶 = 𝑐𝑐𝑝𝑝𝑝𝑝𝑝𝑝 (𝑇𝑇5 − 𝑇𝑇4 ) = 𝑐𝑐𝑝𝑝𝑝𝑝𝑝𝑝

1 (𝑇𝑇 − 𝑇𝑇4 ) 𝜂𝜂𝐶𝐶 5𝑠𝑠

Therefore, we can calculate the real temperature at state 5 to be, 𝑇𝑇5 = 𝑇𝑇4 +

1 (𝑇𝑇 − 𝑇𝑇4 ) = 505.5 K 𝜂𝜂𝐶𝐶 5𝑠𝑠

The compressor work per unit mass flow rate is

𝑊𝑊̇𝐶𝐶 = 687.91

kJ kg

(3)

Regenerator We can use conservation of energy to write the equation for this component as 𝑐𝑐𝑝𝑝𝑝𝑝𝑝𝑝 (𝑇𝑇6 − 𝑇𝑇5 ) = 𝑐𝑐𝑝𝑝𝑝𝑝𝑝𝑝 (𝑇𝑇2 − 𝑇𝑇3 )

Therefore, we can solve for the temperature at 6 in terms of 3: 𝑇𝑇6 = 𝑇𝑇5 + 𝑇𝑇2 − 𝑇𝑇3

(4)

𝑄𝑄̇𝑖𝑖𝑖𝑖 = 𝑐𝑐𝑝𝑝𝑝𝑝𝑝𝑝 (𝑇𝑇1 − 𝑇𝑇6 )

(5)

Reactor The heat added to the system per unit mass is: Temperature at 3 We can now substitute Equation (4) into Equation (5), substitute this result into Equation (1) and solve for T3: 𝑇𝑇3 =

𝑊𝑊̇𝑇𝑇 − 𝑊𝑊̇𝐶𝐶 + 𝑇𝑇2 + 𝑇𝑇5 − 𝑇𝑇1 = 572.5 K 𝜂𝜂𝑡𝑡ℎ 𝑐𝑐𝑝𝑝𝑝𝑝𝑝𝑝

And therefore the temperature at 6 is T6 = 714.9 K.

3. What is the energy utilization factor (EUF) of this cycle? The EUR can be determined as follows:

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𝐸𝐸𝐸𝐸𝐸𝐸 =

𝑊𝑊̇𝑇𝑇 − 𝑊𝑊̇𝐶𝐶 + 𝑄𝑄̇3→4 =1 𝑄𝑄̇𝑖𝑖𝑖𝑖

4. The power of the cogeneration heat exchanger. We must first define some additional parameters: –

𝑘𝑘𝑘𝑘

–

steam flow rate, 𝑚𝑚̇𝑤𝑤 = 100 𝑠𝑠

–

water inlet temperature, Tin = 80°C

water saturation temperature, Tsat = 152°C

To begin we must calculate the power of the cogneration heat exchanger from the water side: 𝑄𝑄̇𝐶𝐶𝐺𝐺𝐺𝐺𝐺𝐺 = 𝑚𝑚̇𝑤𝑤 �𝑐𝑐𝑝𝑝𝑝𝑝 (𝑇𝑇𝑠𝑠𝑠𝑠𝑠𝑠 − 𝑇𝑇𝑖𝑖𝑖𝑖 ) + ℎ𝑓𝑓𝑓𝑓 � = 241.4 MW

We can now move to the Helium side and determine the mass flow rate of this gas: 𝑚𝑚̇𝐻𝐻𝐻𝐻 =

Therefore, the reactor power is just

𝑄𝑄̇𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 kg = 233.0 𝑐𝑐𝑝𝑝𝑝𝑝𝑝𝑝 (𝑇𝑇3 − 𝑇𝑇4 ) s

𝑄𝑄̇𝑅𝑅 = 𝑚𝑚̇𝐻𝐻𝐻𝐻 𝑐𝑐𝑝𝑝𝑝𝑝𝑝𝑝 (𝑇𝑇1 − 𝑇𝑇6 ) = 344.9 MW

5. Nuclear cogeneration for residential heating has been rarely done. The main issues hindering development of nuclear cogeneration for residential heating on a large scale are as follows: –

For safety reasons, nuclear power plants are sited far from major residential areas. As such, heat transport losses and costs from the plant to the end users would be high.

–

The residential heating load varies greatly throughout the year and even during a single day. This may force frequent changes in the operating conditions of the plant, something nuclear plants are not particularly suitable for.

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Chapter 7 Thermodynamics of Nuclear Energy Conversion Systems - Nonsteady Flow First Law Analysis Contents Problem 7.1

Containment pressure analysis ....................................................................... 131

Problem 7.2

Ice condenser containment analysis ............................................................... 134

Problem 7.3

Containment pressure increase due to residual core heat ............................... 137

Problem 7.4

Loss of heat sink in a sodium-cooled reactor ................................................. 140

Problem 7.5

Response of a BWR suppression pool to safety/relief valve discharge ......... 143

Problem 7.6

Containment sizing for a gas cooled reactor with passive emergency cooling ......................................................................................................................... 146

Problem 7.7

Containment problem involving a LOCA ...................................................... 151

Problem 7.8

Analysis of a transient overpower in the PWR steam generator .................... 154

Problem 7.9

Drain tank pressurization problem ................................................................. 157

Problem 7.10 Containment pressurization following zircaloy-hydrogen reaction ............... 161 Problem 7.11 Effect of noncondensable gas on pressurizer response to an insurge ............. 164 Problem 7.12 Pressurizer sizing analysis .............................................................................. 168 Problem 7.13 Pressurizer insurge problem ........................................................................... 170 Problem 7.14 Behavior of a fully contained pressurized pool reactor under decay power conditions ....................................................................................................... 172 Problem 7.15 Depressurization of a primary system ............................................................ 175

Note on Nomenclature of Chapter 7: This Solution Manual uses italics for sub-scripts, whereas the Nuclear Systems I 3rd edition textbook uses non-italics.

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PROBLEM 7.1 QUESTION Containment Pressure Analysis (Section 7.2) For the plant analyzed in Example 7.1, Problem A, the peak containment pressure resulting from primary system blowdown is 0.523 MPa. Assume that the primary system failure analyzed in that accident sends an acoustic wave through the primary system that causes a massive failure of steam generator tubes. Although main and auxiliary feedwater to all steam generators are shut off promptly, the entire secondary system inventory of 89 m3 at 6.89 MPa is also now released to containment by blowdown through the primary system. Assume for this case that the secondary system inventory is all at saturated liquid conditions. 1. What is the new containment pressure? 2. Can it be less than that resulting from only primary system failure? Answers: 1. 0.6 MPa 2. Yes, if the secondary fluid properties are such that this fluid acts as an effective heat sink.

PROBLEM 7.1 SOLUTION Containment Pressure Analysis (Section 7.2) For the plant analyzed in Example 7.1, Problem A, the peak containment pressure resulting from primary system blowdown is 0.523 MPa. Assume that the primary system failure analyzed in that accident sends an acoustic wave through the primary system that causes a massive failure of steam generator tubes. Although main and auxiliary feedwater to all steam generators are shut off promptly, the entire secondary system inventory of 89 m3 at 6.89 MPa is also now released to containment by blowdown through the primary system. Assume for this case that the secondary system inventory is all at saturated liquid conditions. 1. What is the new containment pressure? The containment for this problem will be taken as the sum of the original containment volume plus the primary system. General parameters are listed below that were given or the result from the analysis performed in Example 7.1. —

Initial pressure before this transient, P1 = 0.523 MPa

—

Initial temperature of the containment, T1 = 415.6 K

—

Quality of water mixture in the containment, x1 − 0.505

—

Volume of containment and primary system, Vc = 51,324 m3

—

Volume of secondary system, Vss = 89 m3

—

Initial pressure of secondary system, Pss = 6.89 MPa

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Chapter 7 - Thermodynamics of Nuclear Energy Conversion Systems Nonsteady Flow First Law Analysis —

Mass of air, ma = 5.9 × 104kg

—

Mass of water initially in containment, mwa = 2.11 × 105 kg

We can begin by writing the conservation of energy for this system where we take a control volume of the fluid that is present after the primary system rupture and the volume of the secondary system, 𝜕𝜕𝜕𝜕𝐶𝐶𝐶𝐶 = � 𝑄𝑄𝑛𝑛 𝜕𝜕𝜕𝜕

(1)

𝑛𝑛

In Equation (9) ECV is the energy stored in the control volume and Qn are the heat sources and sinks. For this problem, we will neglect all heat transfer to the structures as well as the decay heat, heat from fission and chemical reaction. Thus Equation (9) becomes 𝜕𝜕𝜕𝜕𝐶𝐶𝐶𝐶 =0 𝜕𝜕𝑡𝑡

This can be integrated to give

(2)

𝐸𝐸2 − 𝐸𝐸1 = 0

(3)

𝑚𝑚𝑎𝑎 (𝑢𝑢𝑎𝑎2 − 𝑢𝑢𝑎𝑎1 ) + (𝑚𝑚𝑤𝑤𝑤𝑤𝑤𝑤 + 𝑚𝑚𝑤𝑤𝑤𝑤 )𝑢𝑢𝑤𝑤2 − (𝑚𝑚𝑤𝑤𝑤𝑤𝑤𝑤 𝑢𝑢𝑤𝑤𝑤𝑤𝑤𝑤1 + 𝑚𝑚𝑤𝑤𝑤𝑤 𝑢𝑢𝑤𝑤𝑤𝑤1 ) = 0

(4)

We can write the stored energy from air and water at the final (2) and initial (1) states,

In Equation (2) ua2 and ua1 are the final and initial internal energies of the air in containment, mwSS is the mass of water in the secondary system, uw2 is the final internal energy of water after the break, uwSS1 is the initial internal energy of the water in the secondary system and uwa1 is the initial internal energy of the water in the containment. Note that we have assume that the water from the secondary system and the initial water in the containment will be in thermal equilibrium at the final state. We will take Equation (2) apart looking at each component. We can assume that air is a perfect gas where (5) 𝑢𝑢𝑎𝑎2 − 𝑢𝑢𝑎𝑎1 = 𝑐𝑐𝜐𝜐𝜐𝜐 (𝑇𝑇2 − 𝑇𝑇1 ) However, we do not know the temperature T2 at the final state. The specific heat is taken to be J

719 kg K taken from Example 7.1. Next, the mass of water initially in the secondary system can be

determined from the volume and specific volume of the saturated liquid water. V𝑆𝑆𝑆𝑆 = 89 m3

v𝑆𝑆𝑆𝑆 = v�𝑃𝑃𝑆𝑆𝑆𝑆, 𝑥𝑥 = 0� = 0.00134827 𝑚𝑚𝑤𝑤𝑤𝑤𝑤𝑤 =

V𝑆𝑆𝑆𝑆 = 6.601 × 104 kg v𝑆𝑆𝑆𝑆 132

(6) m3 kg

(7) (8)

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Chapter 7 - Thermodynamics of Nuclear Energy Conversion Systems Nonsteady Flow First Law Analysis

Next, the specific internal energy at state 2 will be assumed to be saturated and therefore can be expressed as 𝑢𝑢𝑤𝑤2 = 𝑢𝑢𝑔𝑔𝑔𝑔2 (𝑇𝑇2 )𝑥𝑥𝑤𝑤2 + 𝑢𝑢𝑓𝑓𝑓𝑓2 (𝑇𝑇2 )(1 − 𝑥𝑥𝑤𝑤2 )

(9)

The internal energy of water at saturated liquid and vapor conditions are determined from T2 which is unknown still. The final specific volume of water in the system can be calculated by determining the total volume of the system and dividing by the total mass of water, v𝑤𝑤2 =

V𝑆𝑆𝑆𝑆 + V𝐶𝐶 m3 = 0.1856 𝑚𝑚𝑤𝑤𝑤𝑤𝑤𝑤 + 𝑚𝑚𝑤𝑤𝑤𝑤 kg

This can be used eventually to determine the quality at state 2 𝑥𝑥𝑤𝑤2 =

v𝑤𝑤2 − v𝑓𝑓𝑓𝑓2 (𝑇𝑇2 ) v𝑤𝑤2 − v𝑓𝑓𝑓𝑓2 (𝑇𝑇2 ) = v𝑔𝑔𝑔𝑔2 (𝑇𝑇2 ) − v𝑓𝑓𝑓𝑓2 (𝑇𝑇2 ) v𝑔𝑔𝑔𝑔2 (𝑇𝑇2 ) − v𝑓𝑓𝑓𝑓2 (𝑇𝑇2 )

(10)

So, we still have T2 as an unknown. The specific internal energy of the water initially in the secondary system can be determined as kJ 𝑢𝑢𝑤𝑤𝑤𝑤𝑤𝑤1 = 𝑢𝑢(𝑃𝑃𝑆𝑆𝑆𝑆 , 𝑥𝑥 = 0) = 1252.68 (11) kg The specific internal energy of the water in the containment initially, is determined from the temperature and the quality, kJ 𝑢𝑢𝑤𝑤𝑤𝑤1 = 𝑢𝑢(𝑇𝑇1 = 415.6 K, 𝑥𝑥1 = 0.505) = 1585.39 kg Looking at the equations that have been constructed above, we can formulate the following equations to be solved simultaneously, where we have substituted Equation (5) into Equation (2), 𝑚𝑚𝑎𝑎 𝑐𝑐𝜐𝜐𝜐𝜐 (𝑇𝑇2 − 𝑇𝑇1 ) + (𝑚𝑚𝑤𝑤𝑤𝑤𝑤𝑤 + 𝑚𝑚𝑤𝑤𝑤𝑤 )𝑢𝑢𝑤𝑤2 − (𝑚𝑚𝑤𝑤𝑤𝑤𝑤𝑤 𝑢𝑢𝑤𝑤𝑤𝑤𝑤𝑤1 + 𝑚𝑚𝑤𝑤𝑤𝑤 𝑢𝑢𝑤𝑤𝑤𝑤1 ) = 0 𝑢𝑢𝑤𝑤2 − 𝑢𝑢𝑔𝑔𝑔𝑔2 (𝑇𝑇2 )𝑥𝑥𝑤𝑤2 + 𝑢𝑢𝑓𝑓𝑓𝑓2 (𝑇𝑇2 )(1 − 𝑥𝑥𝑤𝑤2 )

𝑥𝑥𝑤𝑤2 =

v𝑤𝑤2 − v𝑓𝑓𝑓𝑓2 (𝑇𝑇2 ) 𝑢𝑢𝑤𝑤2 − 𝑢𝑢𝑓𝑓𝑓𝑓2 (𝑇𝑇2 ) = v𝑔𝑔𝑔𝑔2 (𝑇𝑇2 ) − v𝑓𝑓𝑓𝑓2 (𝑇𝑇2 ) 𝑢𝑢𝑔𝑔𝑔𝑔2 (𝑇𝑇2 ) − 𝑢𝑢𝑓𝑓𝑓𝑓2 (𝑇𝑇2 )

Here we have 3 unknowns, T2, uw2, xw2 and 3 equations. Note it is not trivial to solve these equations since there are thermal properties that depend on the unknowns. A MATLAB script was used to iterate on thefinal temperature determining properties on the fly until the equations converged. The important results from the script are m3 𝑇𝑇2 = 421.7 K 𝑥𝑥𝑤𝑤2 = 0.4547 v𝑓𝑓𝑓𝑓2 = 0.0011 kg Knowing the temperature the total pressure at state 2 can be calculated. The total pressure is made up of the partial pressure of water and the partial pressure of air. The pressure of the water is just the saturation pressure at the temperature above. The pressure of air will be calculated from the

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perfect gas law. The partial pressure of water at the final pressure is just 𝑃𝑃𝑤𝑤2 = 𝑃𝑃𝑠𝑠𝑠𝑠𝑠𝑠 (𝑇𝑇2 ) = 0.4594 MPa

The volume of air is the total volume subtracted by the volume of the liquid water droplets. The volume of liquid water is first calculated as V𝑓𝑓𝑓𝑓2 = (𝑚𝑚𝑤𝑤𝑤𝑤𝑤𝑤 + 𝑚𝑚𝑤𝑤𝑤𝑤 )(1 − 𝑥𝑥𝑤𝑤2 )v𝑓𝑓𝑓𝑓2 = 166.159 m3

The volume of air is therefore

V𝑎𝑎2 = (V𝑆𝑆𝑆𝑆 + V𝐶𝐶 ) − V𝑓𝑓𝑓𝑓2 = 5.125 × 104 m3 J

The partial pressure of air is determine from ideal gas law 𝑅𝑅𝑎𝑎 = 286 kg K taken from Example 7.1, 𝑃𝑃𝑎𝑎2 =

𝑚𝑚𝑎𝑎 𝑅𝑅𝑎𝑎 𝑇𝑇2 = 0.139 MPa V𝑎𝑎2

Thus, the total pressure is just the sum of the partial pressures,

𝑃𝑃2 = 𝑃𝑃𝑤𝑤2 + 𝑃𝑃𝑎𝑎2 = 0.598 MPa ≈ 0.6 MPa

2. Can the final containment pressure be less than that from only primary system failure? This situation could happen if the thermal properties of the fluid in the secondary system were such that energy is absorbed at failure of this secondary side. This could happen if the fluid is very subcooled.

PROBLEM 7.2 QUESTION Ice Condenser Containment Analysis (Section 7.2) Calculate the minimum mass of ice needed to keep the final containment pressure below 0.4 MPa, assuming that the total volume consists of a 5.05 × 104 m3 containment volume and a 500 m3 primary volume. Neglect the initial volume of the ice. Additionally, assume the following initial conditions: Containment pressure = 1.013 × 105 Pa (1 atm) Containment temperature = 300 K Ice temperature = 263 K (−10°C) Ice pressure = 1.013 × 105 Pa (1 atm) Primary pressure = 15.5 MPa (saturated liquid conditions) Relevant properties are: cυ, for air = 719 J/kg K Rsp,a for air = 268 J/kg K c for ice = 4.23 × 103 J/kg K Heat of fusion for water = 3.33 × 105 J/kg (at 1 atm)

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Answer: 1.78 × 105 kg of ice

PROBLEM 7.2 SOLUTION Ice Condenser Containment Analysis (Section 7.2) From the problem statement, we are given that: —

final containment pressure, Pf = 0.4 MPa

—

total volume of containment, Vc = 5.05 × 104 m3

—

primary system volume, Vp = 500 m3

—

initial containment pressure, Po = 1.013 × 105 Pa

—

initial containment temperature, To = 300 K

—

temperature of ice, Ti = 263 K

—

primary system pressure, Pp = 15.5 MPa

Thermodynamic properties are: —

specific heat of air, cυa = 719 J/kg K

—

gas constant for air, Rsp,a = 286 J/kg K

—

specific heat of ice, ci = 4.23 × 103 J/kg K

—

heat of fusion for ice, ufus = 3.33 × 105 J/kg

—

melting temperature, Tm = 273 K

The internal energy of primary system water initially (evaluated from steam tables) is 𝑢𝑢𝑤𝑤𝑤𝑤1 = 1603.8 kJ⁄kg

(1)

𝑢𝑢𝑖𝑖1 = −𝑢𝑢𝑓𝑓𝑓𝑓𝑓𝑓 − 𝑐𝑐𝑖𝑖 (𝑇𝑇𝑚𝑚 − 𝑇𝑇𝑖𝑖 ) = −375.3 kJ⁄kg

(2)

v𝑤𝑤𝑤𝑤1 = 0.00168243 m3 ⁄kg

(3)

The internal energy of ice initially is

From steam tables, the specific volume in the primary system initially is Thus, the mass of water in the primary system is 𝑚𝑚𝑤𝑤𝑤𝑤 =

V𝑝𝑝 = 2.972 × 105 kg v𝑤𝑤𝑤𝑤1

(4)

The mass of air in the containment can be determined from the ideal gas law to be

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𝑚𝑚𝑎𝑎 =

𝑃𝑃𝑜𝑜 V𝑜𝑜 = 5.962 × 104 kg 𝑅𝑅𝑠𝑠𝑠𝑠,𝑎𝑎 𝑇𝑇𝑜𝑜

(5)

For this situation, conservation of energy is

𝑚𝑚𝑤𝑤𝑤𝑤 �𝑢𝑢𝑤𝑤𝑤𝑤2 − 𝑢𝑢𝑤𝑤𝑤𝑤1 � + 𝑚𝑚𝑎𝑎 𝑐𝑐𝜐𝜐𝜐𝜐 (𝑇𝑇2 − 𝑇𝑇𝑜𝑜 ) + 𝑚𝑚𝑖𝑖 (𝑢𝑢𝑖𝑖2 − 𝑢𝑢𝑖𝑖1 ) = 0

(6)

𝑢𝑢𝑤𝑤𝑤𝑤2 = 𝑢𝑢𝑖𝑖2

(7)

𝑢𝑢𝑤𝑤𝑤𝑤2 = 𝑢𝑢(𝑇𝑇2 , 𝑥𝑥2 )

(8)

In this equation, the only unknowns are the final specific internal energy of water uwp2, final specific internal energy of ice ui2, the mass of ice mi and the final containment temperature T2. If we assume thermodynamic equilibrium conditions at the final state, then

To define this internal energy, we need two independent state properties. Here we state that the internal energy can be evaluated from the final containment temperature and final quality At the final state; the specific volume can be calculated with v2 =

V𝑐𝑐 + V𝑝𝑝 𝑚𝑚𝑤𝑤𝑤𝑤 + 𝑚𝑚𝑖𝑖

(9)

We can define this specific volume similar to the specific internal energy as v2 = v(𝑇𝑇2 , 𝑥𝑥2 )

(10)

𝑃𝑃2 = 𝑃𝑃𝑤𝑤2 + 𝑃𝑃𝑎𝑎2

(11)

𝑃𝑃𝑤𝑤2 = 𝑃𝑃𝑠𝑠𝑠𝑠𝑠𝑠 (𝑇𝑇2 )

(12)

The final pressure of the containment, which was given in the problem statement, is made up of the partial pressure of water and partial pressure of air, The partial pressure of water is the saturation pressure of the system at the final state, and the partial pressure of air can again be calculated with the ideal gas law 𝑃𝑃𝑎𝑎2 =

𝑚𝑚𝑎𝑎 𝑅𝑅𝑠𝑠𝑠𝑠,𝑎𝑎 𝑇𝑇2 V𝑐𝑐

(13)

assuming V𝑐𝑐 ⋙ V𝑝𝑝 and neglecting the volume that liquid water takes up. The number of unknowns, 8, in this formulation are T2 , uwp2 , ui2, mi, υ2, x2, Pw2 Pa2. We have 8 equations, (6)-(13) to solve for them. The final result is 𝑚𝑚𝑖𝑖 = 1.78 × 105 kg

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PROBLEM 7.3 QUESTION Containment Pressure Increase Due to Residual Core Heat (Section 7.2) Consider the containment system as shown in Figure 7.14 after a loss of coolant accident (LOCA). Assume that the containment is filled with saturated liquid, saturated vapor, and air and nitrogen gas released from rupture of one of the accumulators, all at thermal equilibrium. The containment spray system has begun its recirculating mode, during which the sump water is pumped by the residual heat removal (RHR) pump through the RHR heat exchanger and sprayed into the containment.

FIGURE 7.14 Containment.

Assume that after 1 h of operation the RHR pump fails and the containment is heated by core decay heat. The decay heat from the core is assumed constant at 1% of the rated power of 2441 MWth. Find the time when the containment pressure reaches its design limit, 0.827 MPa (121.6 psia). Additional necessary information is given below. Conditions after 1 h: Water mixture mass (mw) = 1.56 × 106 kg Water mixture quality (xst) = 0.0249 Air mass (ma) = 5.9 × 104 kg Nitrogen mass =(mN2) = 1.0×l03 kg Initial temperature (T) = 381.6 K

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Thermodynamic properties of gases: Rsp,a= 0.286 kJ/kg K cυa = 0.719 kJ/kg K Rsp,N = 0.296 kJ/kg K cυN = 0.742 kJ/kg K Answer: 7.08 h

PROBLEM 7.3 SOLUTION Containment Pressure Increase Due to Residual Core Heat (Section 7.2) From the problem statement, we are given: − −

reactor power, 𝑄𝑄̇𝑅𝑅 = 2441 MWth

containment pressure limit, Pc = 0.827 MPa

Conditions after 1 hr: −

mass of water mixture, mw = 1.56 × 106 kg

−

water mixture quality, xst = 0.0249

−

mass of air, ma = 5.9 × 104 kg

−

mass of nitrogen, mN = 1.0 × 103 kg

−

initial temperature, T1 = 381.6 K

Thermodynamic properties: −

gas constant for air, Rsp,a = 0.286 kJ/kg

−

specific heat for air, cυa = 0.719 kJ/kg

−

gas constant for nitrogen, Rsp,N = 0.296 kJ/kg

−

specific heat for nitrogen, cυN.= 0.742 kJ/kg

The decay heat is at 1% of the reactor power 𝑄𝑄̇𝑑𝑑 = 0.01𝑄𝑄𝑅𝑅 = 24.41 MWth

(1)

𝑃𝑃𝑤𝑤1 = 1.361 bar

(2)

The initial partial pressure of water in the containment corresponds to the saturation pressure at the initial temperature,

The corresponding saturated specific volumes for liquid and vapor are

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v𝑓𝑓1 = 0.00105028 m3 ⁄kg

(3)

v𝑔𝑔1 = 1.26993 m3 ⁄kg

The specific volume of the water mixture is therefore,

v𝑓𝑓1 = v𝑓𝑓1 + 𝑥𝑥𝑠𝑠𝑠𝑠 �v𝑔𝑔1 − v𝑓𝑓1 � = 0.032645 m3 ⁄kg

(4)

V𝑐𝑐 = v1 𝑚𝑚𝑤𝑤 = 5.093 × 104 m3

(5)

The volume of the containment is given by

Conservation of energy for this situation is

𝑚𝑚𝑤𝑤 (𝑢𝑢𝑤𝑤2 − 𝑢𝑢𝑤𝑤1 ) + 𝑚𝑚𝑎𝑎 𝑐𝑐𝜐𝜐𝜐𝜐 (𝑇𝑇2 − 𝑇𝑇1 ) + 𝑚𝑚𝑁𝑁 𝑐𝑐𝜐𝜐𝜐𝜐 (𝑇𝑇2 − 𝑇𝑇1 ) = 𝑄𝑄̇𝑑𝑑 𝑡𝑡

(6)

The initial specific internal energy of water is calculated from steam tables at temperature, T1, and quality xst, 𝑢𝑢1𝑤𝑤 = 5.06 × 105 J⁄kg

(7)

𝑢𝑢2𝑤𝑤 = 𝑢𝑢(𝑇𝑇2 , v1 )

(8)

𝑃𝑃2 = 𝑃𝑃2𝑤𝑤 + 𝑃𝑃2𝑎𝑎 + 𝑃𝑃2𝑁𝑁

(9)

𝑃𝑃2𝑤𝑤 = 𝑃𝑃𝑠𝑠𝑠𝑠𝑠𝑠 (𝑇𝑇2 )

(10)

The final specific internal energy of water is calculated from steam tables from the final temperature, T2 and final specific volume which is v1, The final pressure, given above, is made up of partial pressures (water, air, and nitrogen), The partial pressure of water is the saturation pressure at the final temperature,

The partial pressures of air and nitrogen are given by the ideal gas law,

and

𝑃𝑃2𝑎𝑎 =

𝑚𝑚𝑎𝑎 𝑅𝑅𝑠𝑠𝑠𝑠,𝑎𝑎 𝑇𝑇2 V𝑐𝑐

𝑃𝑃2𝑁𝑁 =

𝑚𝑚𝑁𝑁 𝑅𝑅𝑠𝑠𝑠𝑠,𝑁𝑁 𝑇𝑇2 𝑉𝑉𝑐𝑐

(11)

(12)

There are 6 unknowns (uw2 , T2 , t, P2w, P2a and P2N) and 6 corresponding equations (Equations (7), (12), (10), (11), (12) and (8)). Solving these equations simultaneously yields a time of 7.08 hours.

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PROBLEM 7.4 QUESTION Loss of Heat Sink in a Sodium-cooled Reactor (Section 7.2) A 1000 MWth SFR has three identical coolant loops. The reactor vessel is filled with sodium to a prescribed level, with the remainder of the vessel being occupied by an inert cover gas, as shown in Figure 7.15. Under the steady operating condition, the ratio of the volume of cover gas to that of sodium in the primary system is 0.1. At time t = 0, the primary system of this reactor suffers a complete loss of heat sink accident, and the reactor power instantly drops to 2% of full power.

FIGURE 7.15 Reactor vessel in a sodium-cooled reactor.

Using a lumped parameter approach, calculate the pressure of the primary system at t = 60 seconds. At this time, check if the sodium is boiling. Useful data: Initial average primary system temperature = 527°C Initial primary system pressure = 345 kPa Total primary system coolant mass = 8165 kg β (volumetric coefficient of thermal expansion of sodium) = 2.88 × 10−4 °C cp for sodium = 1256 J/kg K (at 527°C) ρ for sodium = 823.3 kg/m3 (527°C) Sodium saturated vapor pressure: 𝑃𝑃 = exp[18.832 − (13113⁄𝑇𝑇) − 1.0948 ln 𝑇𝑇 + 1.9777 × 10−4 𝑇𝑇]

where P is in atmospheres and T is in K. Answers: P = 607 kPa T = 644°C, thus no boiling occurs.

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PROBLEM 7.4 S OLUTION Loss of Heat Sink in a Sodium-cooled Reactor (Section 7.2) From the problem statement, we are given that: −

initial temperature, T1 = 527°C

−

initial pressure, P1 = 345 kPa

−

mass of sodium, ms = 8165 kg

−

volumetric thermal expansion coefficient of sodium, β = 2.88 × 10−4 1/K

−

specific heat of sodium, cps = 1256 J/kg K

−

density of sodium, ρs = 823.3 kg/m3

−

reactor power, 𝑄𝑄̇𝑅𝑅 = 1000 MWth

−

time, t = 60 s

The decay power of the reactor is 2% the operating power,

The volume of sodium is

𝑄𝑄̇𝑑𝑑 = 0.02𝑄𝑄̇𝑅𝑅 = 20 MWth V𝑠𝑠I =

𝑚𝑚𝑠𝑠 = 9.917m3 𝜌𝜌𝑠𝑠

(1) (2)

From the problem statement, the ratio of the nitrogen cover gas 1/10 that of the volume of sodium, therefore, V𝑁𝑁1 = 0.1 V𝑠𝑠1 = 0.9917 m3

(3)

V𝑐𝑐 = V𝑠𝑠1 + V𝑁𝑁1 = 10.91 m3

(4)

𝑃𝑃𝑠𝑠1 = exp[18.832 − (13113⁄𝑇𝑇1 ) − 1.0948 ln 𝑇𝑇1 + 1.9777 × 10−4 𝑇𝑇1 ] = 906.5 Pa

(5)

v𝑁𝑁1 = 0.6902 m3 ⁄kg

(6)

Thus, the total volume is

The initial partial pressure of sodium is given by the provided formula to be

The specific volume of nitrogen can be evaluated from thermodynamic tables (in EES, ideal nitrogen) at the initial temperature and pressure, Thus, the mass of nitrogen is

𝑚𝑚𝑁𝑁 =

𝑉𝑉𝑁𝑁1 = 1.437kg v𝑁𝑁1

(7)

The initial specific internal energy of nitrogen is evaluated from thermodynamic tables (in EES,

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ideal nitro gen) at the initial temperature, T1, and pressure partial pressure of nitrogen, PN1 = P1− Ps1 = 344 kPa, 𝑢𝑢1𝑁𝑁 = 3.003 × 105 J⁄kg

(8)

𝑚𝑚𝑠𝑠 𝑐𝑐𝑝𝑝𝑝𝑝 (𝑇𝑇2 − 𝑇𝑇1 ) + 𝑚𝑚𝑁𝑁 (𝑢𝑢2𝑁𝑁 − 𝑢𝑢1𝑁𝑁 ) = 𝑄𝑄̇𝑑𝑑 𝑡𝑡

(9)

𝑢𝑢2𝑁𝑁 = 𝑢𝑢(𝑇𝑇2 , 𝑃𝑃𝑁𝑁2 )

(10)

𝑃𝑃𝑠𝑠2 = exp[18.832 − (13113⁄𝑇𝑇2 ) − 1.0948 ln 𝑇𝑇2 + 1.9777 × 104 𝑇𝑇2 ]

(11)

V𝑁𝑁2 = V𝑁𝑁1 + V𝑠𝑠1 (1 − exp[𝛽𝛽(𝑇𝑇2 − 𝑇𝑇1 )])

(12)

Conservation of energy for this situation is

The final specific internal energy of nitrogen can be evaluated at the final temperature, T2 and nitrogen partial pressure, PN2,

The final partial pressure of sodium is

The final volume of nitrogen after expansion is

Therefore the final specific volume is

v𝑁𝑁2 =

V𝑁𝑁2 𝑚𝑚𝑁𝑁

(13)

The final partial pressure of nitrogen gas can be evaluated from thermodynamic tables with the final temperature T2 and final specific volume vN2 , (14) 𝑃𝑃𝑁𝑁2 = 𝑃𝑃�𝑇𝑇2, v𝑁𝑁2 �

The final pressure of the system is the summation of partial pressures, 𝑃𝑃2 = 𝑃𝑃𝑁𝑁2 + 𝑃𝑃𝑠𝑠2

(15)

(16)

and

𝑃𝑃2 = 607 kPa

(17)

Thus, no boiling occurred.

𝑇𝑇2 = 644°C

There are 7 number of unknowns (T2 , u2N, PN2, Ps2, VN2, vN2 P2) and 7 equations (Equations (7)(9)). After solving them simultaneously, the final pressure and temperature are

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PROBLEM 7.5 QUESTION Response of a BWR Suppression Pool to Safety/Relief Valve Discharge (Section 7.2) Compute the suppression pool temperature after 5 min for a case in which the reactor is scrammed and steam is discharged from the reactor pressure vessel (RPV) into the suppression pool such that the temperature of the RPV coolant is reduced at a specific cooldown rate. During this process, makeup water is supplied to the RPV. Heat input to the RPV is only from long-term decay energy generation. Numerical parameters applicable to this problem are given in Table 7.4. Answer: 34.6°C TABLE 7.4 Conditions for Suppression Pool Heat-Up Analysis Parameter

Value

Specified cooldown rate

38°C/h (68.4°F/h)

Reactor power level prior to scram

3434 MWth

RPV initial pressure

7 MPa (1015.3 psia)

Saturation properties at 7 MPa

T = 285.88°C νf = 1.3513 × 10−3 m3/kg νfg = 26.0187 × 10−3 m3/kg uf = 1257.55 kJ/kg ufg = 1323.0 kJ/kg sf = 3.1211 kJ/kg K sfg = 2.6922 kJ/kg K

Discharge period

5 min

Makeup water flow rate

32 kg/s (70.64 lbm/s)

Makeup water enthalpy

800 kJ/kg (350 Btu/lbm)

RPV free volume

656.3 m3 (2.3184 × 104 ft3)

RPV initial liquid mass

0.303 × 106 kg (0.668 × 106 lbm)

RPV initial steam mass

9.0264 × 103 kg (19.9 × 103 lbm)

Suppression pool initial temperature

32°C (90°F)

Suppression pool initial pressure

0.1 MPa (14.5 psia)

Suppression pool water mass

3.44 × 106 kg (7.6 × 106 lbm)

RHR heat exchanger is actuated at high suppression pool temperature: 43.4°C (110°F)

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PROBLEM 7.5 SOLUTION Response of a BWR Suppression Pool to Safety/Relief Valve Discharge (Section 7.2) From the problem statement, we are given: 𝑑𝑑𝑑𝑑

−

cooldown rate, 𝑑𝑑𝑑𝑑 = 38∆° C⁄hr

−

time, tf = 5 min

−

− − −

reactor power, 𝑄𝑄̇𝑅𝑅 = 3434 MW

inlet makeup water flow rate, 𝑚𝑚̇𝑖𝑖𝑖𝑖 = 32 kg⁄s

makeup water flow enthalpy, ℎ𝑖𝑖𝑖𝑖 = 800 kJ⁄kg volume of primary system, VPS = 656.3 m3

−

mass of saturated liquid initially in primary system, mf1 = 0.303 × 106 kg

−

mass of saturated vapor initially in primary system, mg1 = 9.0264 × 103 kg

−

initial suppression pool temperature, T1 = 32 °C

−

suppression pool initial pressure, P1 = 0.1 MPa

−

suppression pool water mass, mw1 = 3.44 × 106 kg

−

initial primary system pressure, Pi = 7 MPa

Saturation properties at 7 MPa: −

Ti = 285.88 °C

−

vf = 1.3513 × 10−3 m3/kg

− −

vfg = 26.0187 × 10−3 m3/kg uf = 1257.55 kJ/kg

−

ufg = 1323 kJ/kg

−

hg1 = 2773 kJ/kg

Reactor First, we look at the reactor system and determine the total amount of energy leaving the system to the suppression pool over 5 minutes. The final temperature in the reactor after cooldown is 𝑇𝑇𝑓𝑓 = 𝑇𝑇𝑖𝑖 −

𝑑𝑑𝑑𝑑 𝑡𝑡 = 282.713°C 𝑑𝑑𝑑𝑑 𝑓𝑓 144

(1)

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For conservation of energy we have 𝑡𝑡𝑓𝑓

𝑚𝑚𝑃𝑃𝑃𝑃 (𝑢𝑢𝑃𝑃𝑃𝑃2 − 𝑢𝑢𝑃𝑃𝑃𝑃1 ) = 𝑚𝑚̇𝑖𝑖𝑖𝑖 ℎ𝑖𝑖𝑖𝑖 𝑡𝑡𝑓𝑓 + � 𝑄𝑄̇𝑑𝑑 (𝑡𝑡)𝑑𝑑𝑑𝑑 − 𝑚𝑚̇𝑜𝑜𝑜𝑜𝑜𝑜 ℎ𝑜𝑜𝑜𝑜𝑜𝑜 𝑡𝑡𝑓𝑓

(2)

𝑚𝑚𝑃𝑃𝑃𝑃 = 𝑚𝑚𝑓𝑓1 + 𝑚𝑚𝑔𝑔1 = 3.12 × 105 kg

(3)

Here, we assume that the mass in the primary system will not change significantly in 5 minutes. Therefore we need to calculate all quantities and solve for 𝑚𝑚̇𝑜𝑜𝑜𝑜𝑜𝑜. Mass in the primary system is The initial quality in the primary system is 𝑥𝑥𝑃𝑃𝑃𝑃1 =

𝑚𝑚𝑔𝑔1 = 0.028928 𝑚𝑚𝑃𝑃𝑃𝑃

(4)

Since we know the pressure and quality, the initial specific internal energy is 𝑢𝑢𝑃𝑃𝑃𝑃1 = 𝑢𝑢�𝑃𝑃𝑖𝑖, 𝑥𝑥𝑃𝑃𝑃𝑃1 � = 1296 kJ⁄kg

(5)

The specific volume can be calculated to be v𝑖𝑖 =

V𝑃𝑃𝑃𝑃 = 2.103 × 10−3 m3 ⁄kg 𝑚𝑚𝑃𝑃𝑃𝑃

(6)

Since we know the final temperature from Equation (1), we can look up the saturation pressure to be 𝑃𝑃2𝑃𝑃𝑃𝑃 = 𝑃𝑃𝑠𝑠𝑠𝑠𝑠𝑠 �𝑇𝑇𝑓𝑓 � = 6.683 MPa

(7)

𝑢𝑢𝑃𝑃𝑃𝑃2 = 𝑢𝑢�𝑃𝑃2𝑃𝑃𝑃𝑃, v𝑖𝑖 � = 1279 kJ⁄kg

(8)

Since we are assuming that the mass of water in the primary system does not change much, the initial specific volume will be approximately equivalent to the final specific volume. Therefore, using pressure and specific volume, we can look up the final specific internal energy to be

We can evaluate the energy entering the system due to decay heat with 𝑡𝑡𝑓𝑓

𝑡𝑡𝑓𝑓

𝑄𝑄𝑑𝑑 = � 𝑄𝑄̇𝑑𝑑 (𝑡𝑡)𝑑𝑑𝑑𝑑 = � 0.066𝑄𝑄̇𝑅𝑅 𝑡𝑡 −0.2 𝑑𝑑𝑑𝑑 = 2.716 × 1010 J

(9)

ℎout = 0.5 �ℎ𝑔𝑔 (𝑃𝑃1𝑃𝑃𝑃𝑃 ) + ℎ𝑔𝑔 (𝑃𝑃2𝑃𝑃𝑃𝑃 )� = 2775 kJ⁄kg

(10)

Finally, the enthalpy leaving the system to the suppression pool will be at saturated vapor conditions. Since the pressure doesn’t change significantly, we will just take a simple average

where hg (P1PS) — 2773 kJ/kg and hg (P2PS) = 2777 kJ/kg. Thus, we

can solve Equation (2) for the flow rate going to the suppression pool, 𝑚𝑚̇𝑜𝑜𝑜𝑜𝑜𝑜 =

𝑄𝑄𝑑𝑑 + 𝑚𝑚̇𝑖𝑖𝑖𝑖 ℎ𝑖𝑖𝑖𝑖 𝑡𝑡𝑓𝑓 − 𝑚𝑚𝑃𝑃𝑃𝑃 (𝑢𝑢𝑃𝑃𝑃𝑃2 − 𝑢𝑢𝑃𝑃𝑃𝑃1 ) = 48.1 kg⁄s ℎ𝑜𝑜𝑜𝑜𝑜𝑜 𝑡𝑡𝑓𝑓

(11)

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Therefore, in the 5 minutes the change in primary system mass is (𝑚𝑚̇𝑜𝑜𝑜𝑜𝑜𝑜 − 𝑚𝑚̇𝑖𝑖𝑖𝑖 )𝑡𝑡𝑓𝑓 = 4.8 × 103 kg

(12)

which is small compared to 3.12 × 105 kg which is the total mass initially in the primary system. Suppression Pool We will now consider just the suppression where the outgoing mass flow rate from the reactor is now the incoming mass flow rate into the suppression pool, 𝑚𝑚̇𝑖𝑖𝑖𝑖 − 𝑚𝑚̇𝑜𝑜𝑜𝑜𝑜𝑜 = 48.1 kg⁄s

(13)

ℎ𝑖𝑖𝑖𝑖 − ℎ𝑜𝑜𝑜𝑜𝑜𝑜 = 2775 kJ⁄kg

(14)

𝑚𝑚𝑤𝑤2 = 𝑚𝑚𝑤𝑤1 + 𝑚𝑚̇𝑖𝑖𝑖𝑖 𝑡𝑡𝑓𝑓 = 3.454 × 106 kg

(15)

𝑚𝑚𝑤𝑤2 𝑢𝑢𝑤𝑤2 − 𝑚𝑚𝑤𝑤1 𝑢𝑢𝑤𝑤1 = 𝑚𝑚̇𝑖𝑖𝑖𝑖 ℎ𝑖𝑖𝑖𝑖 𝑡𝑡𝑓𝑓

(16)

𝑢𝑢𝑤𝑤1 = 𝑢𝑢(𝑇𝑇1 , 𝑃𝑃1 ) = 134.1 kJ⁄kg

(17)

𝑚𝑚̇𝑖𝑖𝑖𝑖 ℎ𝑖𝑖𝑖𝑖 𝑡𝑡𝑓𝑓 + 𝑚𝑚𝑤𝑤1 𝑢𝑢𝑤𝑤1 = 145.126 kJ⁄kg 𝑚𝑚𝑤𝑤2

(18)

Similarly the enthalpy of the flow going into the suppression pool is the same as the enthalpy leaving the reactor

Applying conservation of mass we can determine the final amount of mass in the suppression pool Conservation of energy for the suppression is

The initial specific internal energy can be determined from the temperature, T1 and pressure, P1, Solving for the final specific internal energy we get, 𝑢𝑢𝑤𝑤2 =

The final temperature can be evaluated from steam tables at a pressure P1 (assuming it doesn’t change much) and specific internal energy, uw1, 𝑇𝑇2 = (𝑃𝑃1 , 𝑢𝑢𝑤𝑤1 ) = 34.6°C

(19)

PROBLEM 7.6 QUESTION Containment Sizing for a Gas-cooled Reactor with Passive Emergency Cooling (Section 7.2) An advanced helium-cooled graphite-moderated reactor generates a nominal thermal power of 300 MW. To prevent air ingress in the core during a Loss Of Coolant Accident (LOCA), the reactor containment is filled with helium at atmospheric pressure and room temperature (Figure 7.16). The reactor also features an emergency cooling system to remove the decay heat from the containment

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during a LOCA. To function properly, this system, which is passive and based on natural circulation of helium inside the containment, requires a minimum containment pressure of 1.3 MPa.

FIGURE 7.16 Helium-cooled reactor with helium-filled containment: (a) normal operating conditions and (b) post-LOCA situation.

1. Find the containment volume, so that the pressure in the containment is 1.3 MPa immediately after a large-break LOCA occurs (Figure 7.16). (Assume that thermodynamic equilibrium within the containment is achieved instantaneously after the break) 2. Assuming that the emergency cooling system removes 2% of the nominal reactor thermal power, calculate at what time the pressure in the containment reaches its peak value after the LOCA as well as the peak temperature and pressure: (Calculate the decay heat rate assuming infinite operation time) 3. To reduce the peak pressure in the containment, a nuclear engineer suggests venting the containment gas to the atmosphere through a filter. What would be the advantages and disadvantages of this approach? Assumptions: − Treat helium as an ideal gas. − Neglect the heat contribution from fission and chemical reactions. − Neglect the thermal capacity of the structures. Data: −

Gas volume in the primary system: 200 m3

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Chapter 7 - Thermodynamics of Nuclear Energy Conversion Systems Nonsteady Flow First Law Analysis −

Initial primary system temperature and pressure: 673 K, 7.0 MPa

−

Initial containment temperature and pressure: 300 K, 0.1 MPa

−

Helium specific heat at constant volume: cυ=12.5 J/(mol-K)

−

Helium atomic weight: A= 0.004 kg/mol

−

Universal gas constant: R = 8.31 J/(mol-K)

Answers: 1. 950 m3 2. 787K and 1.6MPa at 391s

PROBLEM 7.6 SOLUTION Containment Sizing for a Gas-cooled Reactor with. Passive Emergency Cooling (Section 7.2) An advanced helium-cooled graphite-moderated reactor generates a nominal thermal power of 300 MW. To prevent air ingress in the core during a Loss Of Coolant Accident (LOCA), the reactor containment is filled with helium at atmospheric pressure and room temperature (Figure 7.16). The reactor also features an emergency cooling system to remove the decay heat from the containment during a LOCA. To function properly, this system, which is passive and based on natural circulation of helium inside the containment, requires a minimum containment pressure of 1.3 MPa. The following parameters are used in solving the problem: −

Gas volume of primary system, VPS = 200 m3

−

Temperature of primary system, TPS = 673 K

−

Pressure of primary system, PPS = 7.0 MPa

−

Initial temperature of containment, T1 = 300 K

−

Initial pressure of containment, P1 =0.1 MPa

−

Helium specific heat capacity at constant volume, 𝑐𝑐𝜐𝜐 = 12.5 mol K kg

−

Helium molar mass, 𝐴𝐴 = 0.004 mol

−

Final pressure of containment, p2 = 1.3 MPa

−

Universal Gas constant, 𝑅𝑅 = 8.31 mol K

Answers: 1. 950 m3 2. 787K and 1.64MPa at 391s

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(1)

𝑛𝑛𝑃𝑃𝑃𝑃 𝑐𝑐𝜐𝜐 (𝑇𝑇2 − 𝑇𝑇𝑃𝑃𝑃𝑃 ) + 𝑛𝑛𝐶𝐶 𝑐𝑐𝜐𝜐 (𝑇𝑇2 − 𝑇𝑇1 ) = 0

(2)

We can expand this equation by substituting in the constitutive relation for internal energy of an ideal gas. Here we will express it in a molar form,

We can rearrange Equation (2) to arrive at 𝑇𝑇2 =

𝑛𝑛𝑃𝑃𝑃𝑃 𝑇𝑇𝑃𝑃𝑃𝑃 + 𝑛𝑛𝐶𝐶 𝑇𝑇1 𝑛𝑛𝑃𝑃𝑃𝑃 + 𝑛𝑛𝐶𝐶

(3)

The amount of helium initially in the primary system is 𝑛𝑛𝑃𝑃𝑃𝑃 =

𝑃𝑃𝑃𝑃𝑃𝑃 V𝑃𝑃𝑃𝑃 = 2.5 × 105 mol 𝑅𝑅𝑅𝑅𝑃𝑃𝑃𝑃

The amount of helium initially in the containment is 𝑛𝑛𝐶𝐶 =

𝑃𝑃1 V𝐶𝐶 𝑅𝑅𝑅𝑅1

(4)

In order to calculate this we need the volume of the containment which is the parameter of interest. We can represent the final state as 𝑃𝑃2 =

(𝑛𝑛𝑃𝑃𝑃𝑃 + 𝑛𝑛𝐶𝐶 )(𝑅𝑅𝑅𝑅2 ) V𝐶𝐶 + V𝑃𝑃𝑃𝑃

(5)

We may substitute Equations (3) and (4) into Equation (5) to determine the volume of the containment to be

Therefore, T2 and nC are:

V𝐶𝐶 =

𝑛𝑛𝑃𝑃𝑃𝑃 𝑅𝑅𝑅𝑅𝑃𝑃𝑃𝑃 − 𝑃𝑃2 V𝑃𝑃𝑃𝑃 = 950m3 𝑃𝑃2 − 𝑃𝑃1

𝑛𝑛𝐶𝐶 =

𝑇𝑇2 =

𝑃𝑃1 V𝐶𝐶 = 3.811 × 104 mol 𝑅𝑅𝑅𝑅1

𝑛𝑛𝑃𝑃𝑃𝑃 𝑇𝑇𝑃𝑃𝑃𝑃 + 𝑛𝑛𝐶𝐶 𝑇𝑇1 = 623.7 K 𝑛𝑛𝑃𝑃𝑃𝑃 + 𝑛𝑛𝐶𝐶 149

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2. Assuming that the emergency cooling system removes 2% of the nominal reactor thermal power, calculate at what time the pressure in the containment reaches its peak value after the LOCA as well as the peak temperature and pressure. (Calculate the decay heat assuming infinite reactor time) The core power and heat removal are: 𝑄𝑄̇𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = 300 MW 𝑄𝑄̇𝑟𝑟𝑟𝑟𝑟𝑟 = 0.02𝑄𝑄̇𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = 6MW

The decay heat as a function of time is

𝑄𝑄̇𝑑𝑑𝑑𝑑𝑑𝑑 (𝑡𝑡) = 0.066 𝑄𝑄̇𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑡𝑡 −0.2

Conservation of energy can be written as 𝑡𝑡

𝐸𝐸𝑐𝑐𝑐𝑐 (𝑡𝑡) = � �𝑄𝑄̇𝑑𝑑𝑑𝑑𝑑𝑑 (𝑡𝑡′) − 𝑄𝑄̇𝑟𝑟𝑟𝑟𝑟𝑟 �𝑑𝑑𝑡𝑡 ′ 0

(6)

To find the peak time we may take the derivative and find the root 𝑑𝑑𝐸𝐸𝑐𝑐𝑐𝑐 (𝑡𝑡) = 𝑄𝑄̇𝑑𝑑𝑑𝑑𝑑𝑑 (𝑡𝑡) − 𝑄𝑄̇𝑟𝑟𝑟𝑟𝑟𝑟 = 0 𝑑𝑑𝑑𝑑

𝑄𝑄̇𝑑𝑑𝑑𝑑𝑑𝑑 (𝑡𝑡) = 0.066 𝑄𝑄̇𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑡𝑡 −0.2 = 𝑄𝑄̇𝑟𝑟𝑟𝑟𝑟𝑟

Solving for time,

𝑡𝑡𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 = 391.4 s

We can also write an equation for the temperature, relating it to the energy of the control volume, 𝐸𝐸𝑐𝑐𝑐𝑐 (𝑡𝑡) = (𝑛𝑛𝑃𝑃𝑃𝑃 + 𝑛𝑛𝐶𝐶 )𝑐𝑐𝜐𝜐 (𝑇𝑇(𝑡𝑡) − 𝑇𝑇2 )

Solving for T(t) and substituting into Equation (6) we can write 𝑡𝑡 1 𝑇𝑇(𝑡𝑡) = 𝑇𝑇2 + � �𝑄𝑄̇ (𝑡𝑡′) − 𝑄𝑄̇𝑟𝑟𝑟𝑟𝑟𝑟 �𝑑𝑑𝑑𝑑′ 𝑐𝑐𝜐𝜐 (𝑛𝑛𝑃𝑃𝑃𝑃 + 𝑛𝑛𝐶𝐶 ) 0 𝑑𝑑𝑑𝑑𝑑𝑑 Integrating to the peak time results in The peak pressure is therefore

𝑇𝑇𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 = 𝑇𝑇�𝑡𝑡𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 � = 786.5 K

𝑃𝑃𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 =

(𝑛𝑛𝑃𝑃𝑃𝑃 + 𝑛𝑛𝐶𝐶 )𝑅𝑅𝑅𝑅𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 = 1.639 MPa 𝑉𝑉𝑃𝑃𝑃𝑃 + 𝑉𝑉𝐶𝐶

3. To reduce the peak pressure in the containment, a nuclear engineer suggests venting the containment gas to the atmosphere through a filter. What would the advantages and disadvantages of this approach?

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The advantages are: —

Lower loads on the containment

—

Can reduce containment thickness, which results in lower capital costs

The disadvantages are: – Release of potentially radioactive gas to the environment, depending on the efficiency of the filter. – If the vent valve failed open, the containment would lose its function.

PROBLEM 7.7 QUESTION Containment problem involving a LOCA (Section 7.2) Upon a loss of primary coolant accident (LOCA) the primary system flashes as it discharges into the containment. At the resulting final equilibrium condition, the containment and primary system are filled with a mixture of steam and liquid. A containment is being designed as shown in Figure 7.17 which directs the liquid portion of this mixture to flood into a reactor cavity in which primary system is located. The condensate which passes back into the core through the break can satisfactorily cool the core if it can submerge it, that is, if the condensate level is high enough.

FIGURE 7.17 Containment with liquid fraction of discharge from LOCA directed to core cooling.

Find the containment volume which will yield a final equilibrium pressure following primary system rupture sufficient to create the 125 m3 of liquid required to fill the cavity and submerge the core. The pressure and volume of the primary system are 15.5 MPa and 354 m3, respectively.

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Chapter 7 - Thermodynamics of Nuclear Energy Conversion Systems Nonsteady Flow First Law Analysis −

Neglect the initial relative humidity

−

Neglect 𝑄𝑄̇𝑐𝑐−𝑠𝑠𝑠𝑠 and 𝑄𝑄̇𝑐𝑐−𝑎𝑎𝑎𝑎𝑎𝑎

Answers.

VT = 2.32 × 104 m3 VC = 2.27 × 104 m3 PT = 0.97 MPa

PROBLEM 7.7 S OLUTION Containment problem involving a LOCA (Section 7.2) From the problem statement, we are given: −

cavity volume, VL = 125 m3

−

primary system pressure, PPS = 15.5 MPa

−

primary system volume, VPS = 354 m3

−

initial containment pressure, P1 = 0.101 MPa

−

initial containment temperature, T1 = 300 K

Conservation of energy is 𝑚𝑚𝑃𝑃𝑃𝑃 (𝑢𝑢2𝑤𝑤 − 𝑢𝑢1𝑤𝑤 ) + 𝑚𝑚𝑎𝑎 (𝑢𝑢2𝑎𝑎 − 𝑢𝑢1𝑎𝑎 ) = 0

(1)

v𝑃𝑃𝑃𝑃 = v(𝑃𝑃𝑃𝑃𝑃𝑃 , 𝑥𝑥 = 0) = 0.001683 m3 ⁄kg

(2)

The specific volume of primary system fluid can be evaluated from pressure and quality (saturated liquid), The mass of primary system fluid is then 𝑚𝑚𝑃𝑃𝑃𝑃 =

V𝑃𝑃𝑃𝑃 = 210356 kg v𝑃𝑃𝑃𝑃

(3)

The initial specific internal energy of water in the primary system can be calculated with pressure and quality, 𝑢𝑢1𝑤𝑤 = 𝑢𝑢(𝑃𝑃𝑃𝑃𝑃𝑃 , 𝑥𝑥 = 0) = 1604 kJ⁄kg

(4)

𝑢𝑢1𝑎𝑎 = 𝑢𝑢(𝑇𝑇1 , 𝑃𝑃1 ) = 214.3 kJ⁄kg

(5)

The initial specific internal energy of air can be evaluated (in EES, ideal gas) with the initial containment temperature, The specific volume of air can be evaluated from the initial temperature and pressure of the

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containment, v𝑎𝑎 = v(𝑃𝑃1 , 𝑇𝑇1 ) = 0.8526 m3 ⁄kg

(6)

If the containment volume was known, the mass of air could be calculated with 𝑚𝑚𝑎𝑎 =

V𝐶𝐶 v𝑎𝑎

(7)

The final specific internal energy of air can be evaluated from the final temperature 𝑢𝑢2𝑎𝑎 = 𝑢𝑢(𝑇𝑇2 )

(8)

V𝑇𝑇 = V𝐿𝐿 + V𝑆𝑆 + V𝑃𝑃𝑃𝑃

(9)

V𝐶𝐶 = V𝐿𝐿 + V𝑆𝑆

(10)

𝑥𝑥2 = 𝑥𝑥(v2 , 𝑃𝑃2𝑤𝑤 )

(11)

The total volume (VT) of the system consists of the cavity where liquid water will fill (VL), a region for steam (VS) and the volume of the primary system where the containment volume (VC) is

The final quality of the system can be evaluated from the specific volume and partial pressure of water where the specific volume is defined as

v2 =

V𝑇𝑇 𝑚𝑚𝑃𝑃𝑃𝑃

(12)

The liquid cavity will just be filled with saturated liquid at the final equilibrium conditions. The specific volume of saturated liquid water can be evaluated from the final partial pressure of water v𝑓𝑓2 = v(𝑃𝑃2𝑤𝑤 , 𝑥𝑥 = 0)

(13)

V𝐿𝐿 = v𝑓𝑓2 (1 − 𝑥𝑥2 )𝑚𝑚𝑃𝑃𝑃𝑃

(14)

𝑢𝑢2𝑤𝑤 = 𝑢𝑢(𝑃𝑃2𝑤𝑤 , v2 )

(15)

𝑃𝑃2𝑤𝑤 = 𝑃𝑃𝑠𝑠𝑠𝑠𝑠𝑠 (𝑇𝑇2 )

(16)

Therefore, the volume of the cavity is

The final specific internal energy of water can be evaluated from the final partial pressure of water and specific volume, This partial pressure is the saturation pressure of water at the final temperature, The overall pressure of the system can be estimated with (assuming air as an ideal gas) 𝑃𝑃2 = 𝑃𝑃2𝑤𝑤 + 𝑃𝑃2𝑎𝑎 = 𝑃𝑃2𝑤𝑤 + 𝑃𝑃1

153

𝑇𝑇2 𝑇𝑇1

(17)

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The number of unknowns from these equations is 17 (mPS, u2w, u1w, ma, u2a, u1a vPS, va, VC, T2, VT VS, x 2 , v2, P2w , vf2, P2). The number of equations to solve is also 17. Solving them simultaneously yields V𝑇𝑇 = 2.32 × 104 m3

(18)

𝑃𝑃2 = 0.97 MPa

(20)

V𝐶𝐶 = 2.29 × 10 m

and

(19)

PROBLEM 7.8 QUESTION Analysis of a Transient Overpower in the PWR Steam Generator (Section 7.2) The steam generator of a large PWR delivers dry saturated steam at 5.7 MPa to the turbine. Consider the steam generator secondary side, which has a volume of 100 m3 and receives a thermal power 𝑄𝑄̇ from the primary coolant flowing in the U-tubes (Figure 7.18).

FIGURE 7.18 Schematic of the steam generator. At steady state the operating conditions for the secondary coolant are as follows: − Inlet mass flow rate 𝑚𝑚̇𝑖𝑖 = 456 kg/s

− Inlet temperature Ti = 267°C (hi= 1170 kJ/kg) − Mass of steam 880 kg − Mass of liquid 54,000 kg

Properties of saturated water at 5.7 MPa are given in Table 7.5.

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TABLE 7.5 Properties of Saturated Water at 5.7 MPa Parameter

Value

Tsat

272°C

1.3 × 10−3 m3/kg

0.034 m3/kg

1196 kJ/kg

2788 kJ/kg

cp,f

5.2 kJ/kg °C

cp,g

4.7 kJ/kg °C

1189 kJ/kg

2592 kJ/kg

1. Calculate 𝑄𝑄̇ , At one point in time the operator maneuvers the reactor so that the thermal power supplied to the secondary coolant increases to 1.2𝑄𝑄̇ . Assume that the secondary coolant pressure, inlet mass flow rate and inlet temperature do not change during the transient.

2. Write a complete set of equations that would allow you to find how the secondary coolant mass (Msc (t)) in the steam generator changes during the transient. Clearly identify all known and unknown parameters in the equations. You may neglect kinetic and gravitational terms. State all your assumptions. 3. Does the secondary coolant outlet mass flow rate increase, decrease or stay the same during the transient? 4. Now imagine that after 2 min both the secondary coolant inlet and outlet are suddenly and simultaneously closed shut, while the thermal power remains at 1.2𝑄𝑄̇ . Does the secondary coolant pressure increase or decrease during this transient? Write a complete set of equations that would allow you to find the pressure change in the secondary coolant during this transient. Answers: 1. 737.8 MWth 3. Increases 4. Increases

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P ROBLEM 7.8 S OLUTION Analysis of a Transient Overpower in the PWR Steam Generator (Section 7.2) 1. The control volume selected to analyze the problem is the volume occupied by the secondary coolant in the steam generator. The conservation of mass at steady state is: 0 = 𝑚𝑚̇𝑖𝑖 − 𝑚𝑚̇𝑜𝑜 → 𝑚𝑚̇𝑜𝑜 = 𝑚𝑚̇𝑖𝑖

(1)

0 = 𝑄𝑄̇ + 𝑚𝑚̇𝑖𝑖 ℎ𝑔𝑔 → 𝑚𝑚̇𝑜𝑜 ℎ𝑔𝑔 → 𝑄𝑄̇ = 𝑚𝑚̇𝑖𝑖 �ℎ𝑔𝑔 − ℎ𝑖𝑖 � = 737.8 MW

(2)

where 𝑚𝑚̇𝑜𝑜 is the secondary coolant outlet mass flow rate. The conservation of energy for steady state yields the following equation: where kinetic and gravitational terms were neglected and hg is the specific enthalpy of dry saturated steam at 5.7 MPa. 2. The conservation of mass equation is: 𝑑𝑑𝑑𝑑𝑆𝑆𝑆𝑆 = 𝑚𝑚̇𝑖𝑖 − 𝑚𝑚̇𝑜𝑜 𝑑𝑑𝑑𝑑

(3)

The conservation of energy equation is:

𝑑𝑑𝑑𝑑𝑆𝑆𝑆𝑆 = 1.2𝑄𝑄̇ + 𝑚𝑚̇𝑖𝑖 ℎ𝑖𝑖 − 𝑚𝑚̇𝑜𝑜 ℎ𝑔𝑔 𝑑𝑑𝑑𝑑

(4)

where the total energy of the secondary coolant is:

𝐸𝐸𝑆𝑆𝑆𝑆 = 𝑀𝑀𝑆𝑆𝑆𝑆 �𝑢𝑢𝑓𝑓 + 𝑢𝑢𝑓𝑓𝑓𝑓 𝑥𝑥�

(5)

V𝑆𝑆𝑆𝑆 = 𝑀𝑀𝑆𝑆𝑆𝑆 �v𝑓𝑓 + v𝑓𝑓𝑓𝑓 𝑥𝑥�

(6)

and x is the steam quality of the secondary coolant. The total volume of the secondary coolant in the steam generator is Vsc = 100 m3 and can be written as:

In Equations (7) through (11) 𝑚𝑚̇𝑜𝑜 , 𝑄𝑄̇ , hi hg, uf, Vsc, vf and vfg are all known. Therefore, these equations represent a system of four equations of the four unknowns Msc, 𝑚𝑚̇𝑜𝑜 and x, which can be solved to find the variation of Msc (t) during the transient. Note that for this particular problem it is possible to find

𝑑𝑑𝑑𝑑𝑆𝑆𝑆𝑆 𝑑𝑑𝑑𝑑

in close form, as follows. Solving Equation (11) for x, substituting into

Equation (10) and eliminating Esc from Equation (12) one gets: v𝑓𝑓𝑓𝑓 𝑑𝑑 �𝑀𝑀𝑆𝑆𝑆𝑆 𝑢𝑢𝑓𝑓 + �V − v𝑓𝑓 𝑀𝑀𝑆𝑆𝑆𝑆 �� = 1.2𝑄𝑄̇ + 𝑚𝑚̇𝑖𝑖 ℎ𝑖𝑖 − 𝑚𝑚̇𝑜𝑜 ℎ𝑔𝑔 𝑑𝑑𝑑𝑑 v𝑓𝑓𝑓𝑓 𝑆𝑆𝑆𝑆

(7)

The left hand side of Equation (12) can be simplified to give:

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v 𝑢 v

𝑢

𝑑𝑀 𝑑𝑡

1.2𝑄

𝑚ℎ

Eliminating 𝑚 from Equations (7) and (8), and solving for

Thus,

𝑑𝑀 𝑑𝑡

𝑚 ℎ ℎ

𝑢 𝑀

ℎ

1.2𝑄 v 𝑢 v

𝑡

𝑀

constant

𝑚 ℎ

(8)

, one gets:

89.2 kg⁄s

89.2𝑡

(9)

(10)

where Msc(0) is 54880 kg and t is in seconds. 3. Since the secondary coolant receives more heat, the rate at which steam is produced and delivered to the turbine increases, which means 𝑚 increases. 4. If the inlet and outlet are closed shut and heat is still being supplied to the secondary coolant, the pressure will increase. The equations are as follows. Mass: 𝑑𝑀 𝑑𝑡

0→𝑀

constant

(11)

That is, the secondary coolant mass in the steam generator does not change during the transient and can be treated as a constant, equal to 44176 kg from Equation (10), Energy: 𝑑𝐸 𝑑𝑡

where the stored energy is:

Volume:

(12)

1.2𝑄

𝐸

𝑀

𝑢 𝑃

𝑢

𝑃 𝑥

(13)

𝑀

v 𝑃

𝑃 𝑥

(14)

Equations (12), (13) and (14) are three equations of the three unknowns Esc, P and x, which can be solved to find P(t).

PROBLEM 7.9 QUESTION Drain Tank Pressurization Problem (Section 7.2) A drain tank is used to temporarily store water discharged from the pressurizer through the pressure-operated relief valve (PORV) (Figure 7.19). The drain tank has a burst disk on it which ruptures if the pressure inside the drain tank becomes too large. For this problem, assume that the PORV at the top of the pressurizer is stuck open, and saturated liquid water at 15.4 MPa leaves the

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pressurizer at a constant flow rate of 3 kg/s and enters a perfectly insulated drain tank of total volume 12 m3. In addition, assume that the initial conditions (before the water due to the stuckopen PORV has entered the drain tank) in the drain tank are No air present Initial vapor volume = 10 m3 Initial pressure = 3 MPa Initial liquid volume = 2 m3 Also assume that the liquid and the water vapor axe in thermal equilibrium at all times in the drain tank, and that the burst disk on the drain tank ruptures at 10 MPa.

FIGURE 7.19 Drain tank to store discharge from the PORV. Questions: a. Define the control mass or control volume you will use and the equation set you will develop. b. Solve for the elapsed time to burst disk rupture. c. Now assume 11.93 kg of air is present in the drain tank along with the liquid water and water vapor, Plw initial = 3 MPa and that the change in volume of the liquid water from the initial state to the final state is large. What is the new time to rupture? Answers: b. 1128 s c. 1044 s

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P ROBLEM 7.9 S OLUTION Drain Tank Pressurization Problem (Section 7.2) From the problem statement, we are given:  pressurizer pressure, PPZR = 15.4 MPa  mass flow rate out of the pressurizer, 𝑚 = 3 kg/s  drain tank total volume, VT = 12 m3

 the initial pressure of the drain tank, P1 = 3 MPa  initial vapor volume, V9 = 10 m3  initial liquid volume is therefore, Vf = 2 m3  final pressure, P2 = 10 MPa For this problem we will choose the control volume to be the drain tank only. The initial specific volumes of saturated liquid in vapor in the drain tank can be evaluated from the initial pressure, v

v P ,x

0.002117 m ⁄kg v

v P ,x

0.06667 m ⁄kg

(1)

The masses of each phase can then be determined with the volume, m

V v

1644 kg

(2)

V v

150 kg

(3)

and

Thus, the total initial mass of water in the drain tank is m

1794 kg

(4)

Therefore, the initial quality of the mixture is m m

0.07716

(5)

The specific enthalpy of saturated liquid flowing into the drain tank is determined from the pressure of the pressurizer, h

h P

1626 kJ⁄kg

(6)

The initial specific internal energy of the mixture in the drain tank can be evaluated from the pressure and quality u

u P ,x

1128 kJ⁄kg

159

(7)

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Chapter 7 - Thermodynamics of Nuclear Energy Conversion Systems Nonsteady Flow First Law Analysis

Conservation of energy can be written as The continuity equation is

mw2 uw2 − mw1 uw1 = ṁ hin t

(8)

mw2 = mw1 + ṁ t

(9)

u2w = u(P2 , x2 )

(10)

The final specific internal can be determined from the final pressure and quality The final specific volume can be calculated from the drain tank volume and final mass v2 =

VT mw2

(11)

The final quality can be evaluated from the specific volume and final pressure x2 = x(P2 , v2 )

(12)

t = 1128 s

(13)

mw2 uw2 − mw1 uw1 + ma (ua2 − ua1 ) = ṁ hin t

(14)

(15)

and

ua1 = u(T1 )

(16)

Again from continuity we have

ua2 = u(T2 ) mw2 = mw1 + ṁ t

(17)

Therefore, there are 5 unknowns (mw2, uw2, t, x2, v2) and 5 equations (Equations (7)-(12)). The time to disk rupture for this situation is In the last part, ma = 11.93 kg is present initially. All of the initial properties, Equations (l)-(7) are still valid. Conservation of energy for this situation is

The initial and final internal energy of air can be evaluated from the initial and final temperature,

The final specific volume can be calculated from the drain tank volume and final mass v2 =

VT mw2

(18)

The final quality can be evaluated from the specific volume and final pressure x2 = x(P2 , v2 )

(19)

uw2 = u(T2 , x2 )

(20)

The final specific internal energy can be evaluated from the final temperature and quality,

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The final pressure of the system is made up of the partial pressure of water and air, P2 = Pw2 + Pa2

(21)

Pw2 = Psat (T2 )

(22)

Va = VT − vf2 (1 − x2 )mw2

(23)

vf2 = υ(T2 , x = 0)

(24)

The partial pressure of water is the saturated pressure at the final temperature The final volume of air is the total volume minus the volume that the liquid water takes up. This can be represented with where υf2 is the specific volume of saturated liquid water at the final equilibrium conditions The specific volume of air is

va =

Va ma

(25)

The partial pressure of air can be evaluated from the final temperature and specific volume, Pa2 = P(T2 , va )

(26)

t = 1044 s

(27)

Therefore, there are 13 unknowns (mw2, uw2, t, x2, v2, P2, Pa2, ua2, ua2, T2, Va, vf2, va) and 13 equations (Equations (14)-(26)). Solving these equations simultaneously yields a time to disk rupture of

PROBLEM 7.10 QUESTION Containment Pressurization following Zircaloy-Hydrogen Reaction (Section 7.2) Consider a LOCA in a typical PWR in which the emergency cooling system is insufficient to prevent metal–water reaction of 75% of the Zircaloy clad and the hydrogen produced subsequently combusts. Using the results of Problem 3.6, this sequence of events yields the following material changes and energy releases relevant to the containment pressurization: Primary coolant released = 2,1 × 105 kg Zr reacted = 0.75 × 24,000 kg Energy released from Zr–H2O reaction = 1.18 × 1011 J H2 produced and reacted = 394.7 mols Energy released from H2 combustion = 9.54 × 1010 J O2 consumed = you must determine Net H2O change = you must determine

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Take the initial primary coolant and containment vessel geometry and conditions the same as Table 7.2. Also, assume that nitrogen has the same properties as air.

TABLE 7.2 Conditions for Containment Examples Heat Addition during Blowdown (Joules)

Fluid

Volume (m3) Pressure (MPa)

Temperature Quality (xst) or Relative (K) Humidity (φ)

Example 7.1: Saturated Water Mixture in Equilibrium with Air as Final State Primary coolant water (initial)

Vp = 354

15.5

617.9

Assumed saturated liquid

Containment vessel air (initial)

Vc = 50,970

0.101

300.0

φ = 80%

VT = 51,324

0.523

415.6

xst = 50.5%

Mixture (final)

Q=0

Example 7.2: Superheated Steam in Equilibrium with Air as Final State Secondary coolant water (initial)

Vs = 89

6.89

558

Assumed saturated liquid

Containment vessel air (initial)

Vc = 50,970

0.101

300

φ = 80%

VT = 51,059

0.446 (64.7 psia)

478

φ = 17%

Mixture (final)

Q = 1011

Question: For the sequence described (e.g., LOCA, 75% Zircaloy clad reaction and subsequent complete combustion of the hydrogen produced): Demonstrate that the final equilibrium temperature is 450 K, neglecting containment heat sinks using the initial conditions of Table 7.2. Find the final equilibrium pressure. Hint: Is the final state likely saturated water or superheated steam in equilibrium with the air? Consider the energy releases compared to those of Example 7.2. Answer: b. P2 (450 K) = 1.06 × 106 Pa

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P ROBLEM 7.10 S OLUTION Containment Pressurization following Zircaloy-Hydrogen Reaction (Section 7.2) From the problem statement, we are given that: − total heat released, Q = 1.180 × 1011 J + 9.54 × 1010 J = 2.134 × 1011 J − number of Zr mols reacted, Z = 394.7 mols − number of O2 mols consumed, O = 0.5Z = 197.4 mol − molar mass of oxygen, MMO = 32 kg/mol − mass of oxygen, mO = O ∙ MMO = 6315 kg Initial conditions taken from Table 7.2 and Example 7.1: − mass of air less the oxygen consumed, mai = 5.9 × 104 kg – mO = 52685 kg − mass of water in air, mwa1 = 1019 kg − mass of water in secondary system, mws1 = 2.10 × 105 kg − gas constant for air, Ra = 286 J/kg ∙ K − specific internal energy of water in primary system, uwp1 = 1.6 ×106 J/kg − specific internal energy of water in air, uwa1 = 2.41 × 106 J/kg − initial temperature, Ta1 = 300 K − specific heat of air, 719 J/kg ∙ K − total volume of containment, Vt = 51324 m3 The total energy of the initial state is 𝐸𝐸1 = 𝑚𝑚𝑤𝑤𝑤𝑤1 𝑢𝑢𝑤𝑤𝑤𝑤 + 𝑚𝑚𝑤𝑤𝑤𝑤1 𝑢𝑢𝑤𝑤𝑤𝑤1 + 𝑚𝑚𝑎𝑎1 𝑐𝑐𝑣𝑣𝑣𝑣 𝑇𝑇𝑎𝑎1

(1)

𝐸𝐸2 = 𝐸𝐸1 + 𝑄𝑄

(2)

𝐸𝐸2 = 𝑚𝑚𝑤𝑤𝑤𝑤1 𝑢𝑢𝑤𝑤2 + 𝑚𝑚𝑤𝑤𝑤𝑤1 𝑢𝑢𝑤𝑤2 + 𝑚𝑚𝑎𝑎1 𝑐𝑐𝑣𝑣𝑣𝑣 𝑇𝑇2

(3)

The total energy of the final state is the initial energy plus the total heat released This final energy is also

The specific volume of water is

v𝑤𝑤2 =

V𝑡𝑡 𝑚𝑚𝑤𝑤𝑤𝑤1 + 𝑚𝑚𝑤𝑤𝑤𝑤1 163

(4)

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The final specific internal energy can be evaluated from specific volume and temperature 𝑢𝑢𝑤𝑤2 = 𝑢𝑢(v𝑤𝑤2 , 𝑇𝑇2 )

(5)

𝑃𝑃2 = 𝑃𝑃2𝑎𝑎 + 𝑃𝑃2𝑤𝑤

(6)

The final pressure is made up of the partial pressure of air and water, The partial pressure of air can be determined from the ideal gas law, 𝑃𝑃2𝑎𝑎 =

𝑚𝑚𝑎𝑎1 𝑅𝑅𝑎𝑎 𝑇𝑇2 V𝑡𝑡

(7)

The partial pressure of water is the saturation pressure at the final temperature, (8)

𝑃𝑃2𝑤𝑤 = 𝑃𝑃𝑠𝑠𝑠𝑠𝑠𝑠 (𝑇𝑇2 )

The number of unknowns is 8 (E1, E2, uw2 , T 2 , v2 w , P 2 , P 2a, P2 w ) and the number of equations is 8. Solving the equations simultaneously yields a final pressure of 𝑃𝑃2 = 1.06 × 106 Pa

PROBLEM 7.11 QUESTION Effect of Noncondensable Gas on Pressurizer Response to an Insurge (Section 7.3) Compute the pressurizer and heater input resulting from an insurge of liquid from the primary system to a pressurizer containing a mass of air (ma). Use the following initial, final and operating conditions. Initial Conditions − Mass of liquid = mf1 − Mass of steam = mg1 − Mass of air (in steam space) = ma − Total pressure = P1 − Equilibrium temperature = T1 Operating Conditions − Mass of surge = msurge − Mass of spray = fmsurge − Enthalpy of surge = hsurge − Enthalpy of spray = hspray

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− Heater input = Qh Final Condition − Equilibrium temperature = T2 = T1 You may make the following assumptions for the solution: 1. Perfect phase separation 2. Thermal equilibrium throughout the pressurizer 3. Liquid water properties that are independent of pressure Answers: 𝑃𝑃2 = 𝑃𝑃𝑚𝑚 �𝑇𝑇1,𝑠𝑠𝑠𝑠𝑠𝑠 � + 𝑄𝑄ℎ =

𝑚𝑚𝑎𝑎 𝑅𝑅𝑠𝑠𝑠𝑠 𝑇𝑇2 v𝑔𝑔1 − v𝑓𝑓1 � � v𝑔𝑔1 𝑚𝑚𝑔𝑔1 v𝑔𝑔1 − 𝑚𝑚𝑔𝑔1 v𝑓𝑓1 − (1 + 𝑓𝑓)𝑚𝑚𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 v𝑓𝑓1

(1 + 𝑓𝑓)𝑚𝑚𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 �𝑢𝑢𝑓𝑓 v𝑔𝑔 − 𝑢𝑢𝑔𝑔 v𝑓𝑓 � − 𝑚𝑚𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 �ℎ𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 + 𝑓𝑓ℎ𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 � v𝑓𝑓𝑓𝑓

P ROBLEM 7.11 S OLUTION

Effect of Noncondensable Gas on Pressurizer Response to an Insurge (Section 7.3) Compute the pressurizer and heater input resulting from an insurge of liquid from the primary system to a pressurizer containing a mass of air (ma). Use the following initial, final and operating conditions. Initial Conditions − Mass of liquid = mf1 − Mass of steam = mg1 − Mass of air (in steam space) = ma − Total pressure = P1 − Equilibrium temperature = T1 Operating Conditions − Mass of surge = msurge − Mass of spray = fmsurge − Enthalpy of surge = hsurge − Enthalpy of spray = hspray

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− Heater input = Qh Final Condition − Equilibrium temperature = T2 = T1 The mass balance on the pressurizer is given as �𝑚𝑚𝑓𝑓2 + 𝑚𝑚𝑔𝑔2 + 𝑚𝑚𝑎𝑎 � − �𝑚𝑚𝑓𝑓1 + 𝑚𝑚𝑔𝑔1 + 𝑚𝑚𝑎𝑎 � = (1 + 𝑓𝑓)𝑚𝑚𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠

(1)

�𝑚𝑚𝑓𝑓2 + 𝑚𝑚𝑔𝑔2 � − �𝑚𝑚𝑓𝑓1 + 𝑚𝑚𝑔𝑔1 � = (1 + 𝑓𝑓)𝑚𝑚𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠

(2)

𝑚𝑚𝑓𝑓1 v𝑓𝑓1 + 𝑚𝑚𝑔𝑔1 v𝑔𝑔1 = 𝑚𝑚𝑓𝑓2 v𝑓𝑓2 + 𝑚𝑚𝑔𝑔2 v𝑔𝑔2

(3)

𝑃𝑃1 = 𝑃𝑃𝑤𝑤 �𝑇𝑇1,𝑠𝑠𝑠𝑠𝑠𝑠 � + 𝑃𝑃𝑎𝑎𝑎𝑎𝑎𝑎,1

(4)

𝑃𝑃1 ≠ 𝑃𝑃2

(6)

𝑚𝑚𝑓𝑓1 v𝑓𝑓1 + 𝑚𝑚𝑔𝑔1 v𝑔𝑔1 = 𝑚𝑚𝑓𝑓2 v𝑓𝑓1 + 𝑚𝑚𝑔𝑔2 v𝑔𝑔1

(7)

𝑃𝑃2 = 𝑃𝑃𝑤𝑤 �𝑇𝑇1,𝑠𝑠𝑠𝑠𝑠𝑠 � + 𝑃𝑃𝑎𝑎𝑎𝑎𝑎𝑎,2

(8)

Since the pressurizer is a rigid volume we can write

Since the temperature is the same at the initial and final state, the specific volume of the gas will be the same, vg1 = vg2. However, the pressure on the liquid will be the summation of the partial pressures of saturated water vapor and air. This pressure will be different from the initial state due to the noncondensable.

𝑃𝑃2 = 𝑃𝑃𝑤𝑤 �𝑇𝑇1,𝑠𝑠𝑠𝑠𝑠𝑠 � + 𝑃𝑃𝑎𝑎𝑎𝑎𝑎𝑎,2 We will however assume that vf1 ≃ vf2. Therefore, Equation (3) is

(5)

At the final state the pressure is

We can substitute in the ideal gas law for the air component, 𝑅𝑅𝑠𝑠𝑠𝑠 𝑇𝑇2 � 𝑃𝑃2 = 𝑃𝑃𝑤𝑤 �𝑇𝑇1,𝑠𝑠𝑠𝑠𝑠𝑠 � + � v𝑎𝑎2

(9)

We know that the final and initial temperatures are equivalent, T1 = T2. Also, the volume that the air takes up is the same volume that the saturated water vapor takes up and therefore we can say 𝑚𝑚𝑎𝑎 v𝑎𝑎2 = 𝑚𝑚𝑔𝑔2 v𝑔𝑔1

(10)

We may substitute Equation (10) into Equation (9) to obtain 𝑃𝑃2 = 𝑃𝑃𝑤𝑤 �𝑇𝑇1,𝑠𝑠𝑠𝑠𝑠𝑠 � +

𝑚𝑚𝑎𝑎 𝑅𝑅𝑠𝑠𝑠𝑠 𝑇𝑇2 𝑚𝑚𝑔𝑔2 v𝑔𝑔1

(11)

We may solve Equation (2) for the final saturated liquid water mass 𝑚𝑚𝑓𝑓2 = �𝑚𝑚𝑓𝑓1 + 𝑚𝑚𝑔𝑔1 − 𝑚𝑚𝑔𝑔2 � + (1 + 𝑓𝑓)𝑚𝑚𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 166

(12)

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Chapter 7 - Thermodynamics of Nuclear Energy Conversion Systems Nonsteady Flow First Law Analysis

We can multiply this equation by the initial specific volume of water 𝑚𝑚𝑓𝑓2 v𝑓𝑓1 = �𝑚𝑚𝑓𝑓1 v𝑓𝑓1 + 𝑚𝑚𝑔𝑔1 v𝑓𝑓1 − 𝑚𝑚𝑔𝑔2 v𝑓𝑓1 � + (1 + 𝑓𝑓)𝑚𝑚𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 v𝑓𝑓1

(13)

We may substitute Equation (13) into Equation (7) to get

𝑚𝑚𝑓𝑓1 v𝑓𝑓1 + 𝑚𝑚𝑔𝑔1 v𝑔𝑔1 = �𝑚𝑚𝑓𝑓1 v𝑓𝑓1 − 𝑚𝑚𝑔𝑔2 v𝑓𝑓1 � + (1 + 𝑓𝑓)𝑚𝑚𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 v𝑓𝑓1 + 𝑚𝑚𝑔𝑔2 v𝑔𝑔1

We can solve this for the final mass of saturated water vapor 𝑚𝑚𝑔𝑔2 =

𝑚𝑚𝑔𝑔1 v𝑔𝑔1 − 𝑚𝑚𝑔𝑔1 v𝑓𝑓1 − (1 + 𝑓𝑓)𝑚𝑚𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 v𝑓𝑓1 v𝑔𝑔1 − v𝑓𝑓1

(14)

The formulation for the final pressure can then be determined by substituting Equation (14) into Equation (11) 𝑃𝑃2 = 𝑃𝑃𝑤𝑤 �𝑇𝑇1,𝑠𝑠𝑠𝑠𝑠𝑠 � +

(15)

We can now write conservation of energy to determine the energy of the heater

𝑚𝑚𝑓𝑓2 𝑢𝑢𝑓𝑓2 − 𝑚𝑚𝑓𝑓1 − 𝑢𝑢𝑓𝑓1 + 𝑚𝑚𝑔𝑔2 𝑢𝑢𝑔𝑔2 − 𝑚𝑚𝑔𝑔1 𝑢𝑢𝑔𝑔1 − 𝑚𝑚𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 �ℎ𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 + 𝑓𝑓ℎ𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 � − 𝑄𝑄ℎ = 0

(16)

𝑄𝑄ℎ = 𝑚𝑚𝑓𝑓2 𝑢𝑢𝑓𝑓 − 𝑚𝑚𝑓𝑓1 − 𝑢𝑢𝑓𝑓 + 𝑚𝑚𝑔𝑔2 𝑢𝑢𝑔𝑔 − 𝑚𝑚𝑔𝑔1 𝑢𝑢𝑔𝑔 − 𝑚𝑚𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 �ℎ𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 + 𝑓𝑓ℎ𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 �

(17)

𝑚𝑚𝑓𝑓2 𝑢𝑢𝑓𝑓 = �𝑚𝑚𝑓𝑓1 𝑢𝑢𝑓𝑓 + 𝑚𝑚𝑔𝑔1 𝑢𝑢𝑓𝑓 − 𝑚𝑚𝑔𝑔2 𝑢𝑢𝑓𝑓 � + (1 + 𝑓𝑓)𝑚𝑚𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑢𝑢𝑓𝑓

(18)

We will assume again that uf2 = uf1 = uf and ug1 = ug2 = ug. Therefore we can rewrite Equation (16) as

We can multiply Equation (12) by the initial internal energy of water

We may substitute Equation (18) into Equation (17)

𝑄𝑄ℎ = �𝑚𝑚𝑓𝑓1 𝑢𝑢𝑓𝑓 + 𝑚𝑚𝑔𝑔1 𝑢𝑢𝑓𝑓 − 𝑚𝑚𝑔𝑔2 𝑢𝑢𝑓𝑓 � + (1 + 𝑓𝑓)𝑚𝑚𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑢𝑢𝑓𝑓 − 𝑚𝑚𝑓𝑓1 − 𝑢𝑢𝑓𝑓 + 𝑚𝑚𝑔𝑔2 𝑢𝑢2 − 𝑚𝑚𝑔𝑔1 𝑢𝑢𝑔𝑔 − 𝑚𝑚𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 �ℎ𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 + 𝑓𝑓ℎ𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 �

𝑄𝑄ℎ = 𝑚𝑚𝑔𝑔1 �𝑢𝑢𝑓𝑓 − 𝑢𝑢𝑔𝑔 � − 𝑚𝑚𝑔𝑔2 �𝑢𝑢𝑓𝑓 − 𝑢𝑢𝑔𝑔 � + (1 + 𝑓𝑓)𝑚𝑚𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑢𝑢𝑓𝑓 − 𝑚𝑚𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 �ℎ𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 + 𝑓𝑓ℎ𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 �

Equation (14) can be rewritten with vg1 = vg and vf1 = vf, We can rewrite this as

𝑚𝑚𝑔𝑔2 =

𝑚𝑚𝑔𝑔1 v𝑔𝑔 − 𝑚𝑚𝑔𝑔1 v𝑓𝑓 − (1 + 𝑓𝑓)𝑚𝑚𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 v𝑓𝑓 v𝑔𝑔 − v𝑓𝑓

𝑚𝑚𝑔𝑔1 − 𝑚𝑚𝑔𝑔2 =

(1 + 𝑓𝑓)𝑚𝑚𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 v𝑓𝑓 v𝑔𝑔 − v𝑓𝑓

(19) (20)

(21)

(22)

We may plug Equation (22) into Equation (20)

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Chapter 7 - Thermodynamics of Nuclear Energy Conversion Systems Nonsteady Flow First Law Analysis

𝑄𝑄ℎ =

(1 + 𝑓𝑓)𝑚𝑚𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 v𝑓𝑓 �𝑢𝑢𝑓𝑓 − 𝑢𝑢𝑔𝑔 � + (1 + 𝑓𝑓)𝑚𝑚𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑢𝑢𝑓𝑓 − 𝑚𝑚𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 �ℎ𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 + 𝑓𝑓ℎ𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 � v𝑔𝑔 − v𝑓𝑓

𝑄𝑄ℎ =

(1 + 𝑓𝑓)𝑚𝑚𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 �v𝑓𝑓 𝑢𝑢𝑓𝑓 − v𝑓𝑓 𝑢𝑢𝑔𝑔 + 𝑢𝑢𝑓𝑓 �v𝑔𝑔 − v𝑓𝑓 �� − 𝑚𝑚𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 �ℎ𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 + 𝑓𝑓ℎ𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 � v𝑔𝑔 − v𝑓𝑓 𝑄𝑄ℎ =

(23)

PROBLEM 7.12 QUESTION

Pressurizer Sizing Analysis (Section 7.3) The size of a pressurizer is determined by the criteria that the vapor volume must be capable of accommodating the largest insurge and the liquid volume must handle the outsurge. The important limitations of the design are that the pressurizer should not be totally liquid filled or the immersion heaters should not be uncovered after possible transients. To size the vapor volume, a maximum insurge is assumed to completely fill the pressurizer with liquid with some of the insurge being diverted to the spray to condense the vapor, Treating the entire pressurizer volume, Vt, as the control volume, find the vapor volume, Vg1, which will accommodate the insurge given below. Data: Initial pressurizer conditions Saturation at 15.51 MPa and 345ºC Initial liquid mass = 1827 kg Maximum insurge (includes spray) Mass = 2740 kg Enthalpy = 1.2 × 106 J/kg Final pressurizer condition Assume completely filled with liquid at 15.51 MPa Answer: Vg1 = 5.57 m3 Vt = 8.64 m3

P ROBLEM 7.12 S OLUTION Pressurizer sizing analysis (Section 7.3) From the problem statement, we are given: − pressurizer pressure, P = 15.5 MPa − mass of liquid, mf1 = 1827 kg

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Chapter 7 - Thermodynamics of Nuclear Energy Conversion Systems Nonsteady Flow First Law Analysis

− mass of insurge, mst = 2740 kg − enthalpy of vapor, hst = 1.2 × 106 J/kg Thermodynamic properties (from tables): − uf = 1604 kJ/kg − ug = 2444 k J/kg − ufg = ug – uf = 840 kJ/kg − vf = 0.001683 m3/kg

− vg = 0.009803 m3/kg

− vfg = vg – vf = 0.00812 m3/kg.

From conservation of mass

𝑚𝑚2 − 𝑚𝑚1 = 𝑚𝑚𝑠𝑠𝑠𝑠

(1) (2)

and

V𝑡𝑡 = 𝑚𝑚2 v2

(3)

From conservation of energy

V𝑡𝑡 = 𝑚𝑚1 v1 𝑚𝑚2 𝑢𝑢2 − 𝑚𝑚1 𝑢𝑢1 = 𝑚𝑚𝑠𝑠𝑠𝑠 ℎ𝑠𝑠𝑠𝑠

(4) (5)

where the specific volume is

𝑚𝑚𝑓𝑓1 = (1 − 𝑥𝑥1 )𝑚𝑚1

(6)

and specific internal energy

v1 = (1 − 𝑥𝑥1 )v𝑓𝑓 + 𝑥𝑥1 v𝑔𝑔 𝑢𝑢1 = (1 − 𝑥𝑥1 )𝑢𝑢𝑓𝑓 + 𝑥𝑥1 𝑢𝑢𝑔𝑔

(7)

v2 = v𝑓𝑓

(8)

V𝑔𝑔1 = 𝑉𝑉𝑡𝑡 − 𝑚𝑚𝑓𝑓1 v𝑓𝑓

(9)

The pressurizer is a rigid volume so therefore,

The mass of liquid can be calculated with

Since we are considering an insurge, the final specific volume is equal to the specific volume of saturated liquid only, The vapor volume to accommodate the insurge is therefore Therefore there are 9 unknowns, (m2, m1, Vt, v1, v2, u1, u2, x1, Vg1) and 9 equations. Solving these 169

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Chapter 7 - Thermodynamics of Nuclear Energy Conversion Systems Nonsteady Flow First Law Analysis

equations simultaneously yields and

V𝑔𝑔1 = 5.569 m3

(10)

V𝑡𝑡 = 8.644 m3

(11)

P ROBLEM 7.13 Q UESTION Pressurizer Insurge Problem (Section 7.3) For insurge case, why is latent heat of vaporization of vapor which is condensed insufficient to heat insurge mass to saturation?

P ROBLEM 7.13 S OLUTION Pressurizer Insurge Problem (Section 7.3) Sufficiency depends on the initial enthalpy of insurge assumed. The balance between these effects is close, but for the values taken (see Table 7.3), heat input is required. For example, see Figure SM-7.1, which shows a final pressurizer condition:

Figure SM-7.1

TABLE 7.3 Conditions for Pressurizer Design Problem Saturation pressure

15.5 MPa

170

2250 psia

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Chapter 7 - Thermodynamics of Nuclear Energy Conversion Systems Nonsteady Flow First Law Analysis Saturation temperature

618.3 K

652.9°F

1.60 × 106 J/kg

689.9 Btu/lbm

2.44 × 106 J/kg

1050.6 Btu/lbm

1.68 × 10−3 m3/kg

0.02698 ft3/lbm

9.81 × 10−3 m3/kg

0.15692 ft3/lbm

Mass of maximum outsurge

14,000 kg

30.86 × 103 lbm

Mass of maximum insurge

9500 kg

20.94 × 103 lbm

Hot leg insurge enthalpy

1.43 × 106 J/kg

612.8 Btu/lbm

Cold leg spray enthalpy

1.27 × 106 J/kg

546.8 Btu/lbm

Cold leg spray expressed as a fraction of hot leg insurge (f)

0.03

Outsurge enthalpy

1.63 × 106 J/kg

701.1 Btu/lbm

Mass of liquid water necessary to cover the heaters (requires an assumption about the pressurizer configuration)

1827 kg

4021.23 lbm

Saturation properties

Energy liberated by condensing vapor: 𝑚𝑚𝑔𝑔 𝑢𝑢𝑓𝑓𝑓𝑓 =

𝑚𝑚𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 (1 + 𝑓𝑓)v𝑓𝑓 𝑢𝑢𝑓𝑓𝑓𝑓 v𝑔𝑔 − v𝑓𝑓

since mg is given by Equation 7.51. The mass of surge and spray added: �𝑚𝑚𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 + 𝑚𝑚𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 �v𝑓𝑓 = 𝑚𝑚𝑔𝑔 vℎ𝑔𝑔 v𝑔𝑔 𝑚𝑚𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 (1 + 𝑓𝑓) = 𝑚𝑚𝑔𝑔 v𝑓𝑓

Energy needed to raise the mass added to saturation conditions:

𝑚𝑚𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 �𝑢𝑢𝑓𝑓 − ℎ𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 � + 𝑚𝑚𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑓𝑓�𝑢𝑢𝑓𝑓 − ℎ𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 � = 𝑚𝑚𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑢𝑢𝑓𝑓 (1 + 𝑓𝑓) − 𝑚𝑚𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 �ℎ𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 + 𝑓𝑓ℎ𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 �

since surge and spray masses are flow streams into the control volume. Summarizing: If 𝑄𝑄̇ℎ is extra energy needed, then 𝑄𝑄̇ℎ + 𝑚𝑚𝑔𝑔 𝑢𝑢𝑔𝑔 − 𝑚𝑚𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑢𝑢𝑓𝑓 (1 + 𝑓𝑓) + 𝑚𝑚𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 �ℎ𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 + 𝑓𝑓ℎ𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 � = 0

If the terms balance, 𝑄𝑄̇ℎ is zero. Depending on other values of parameters, 𝑄𝑄̇ℎ can be + or -.

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v𝑓𝑓 𝑄𝑄̇ℎ = −𝑚𝑚𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 (1 + 𝑓𝑓) � 𝑢𝑢 − 𝑢𝑢𝑓𝑓 � − 𝑚𝑚𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 �ℎ𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 + 𝑓𝑓ℎ𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 � v𝑔𝑔 − 𝜐𝜐𝑓𝑓 𝑓𝑓𝑓𝑓

v𝑓𝑓 𝑢𝑢𝑔𝑔 − v𝑓𝑓 𝑢𝑢𝑓𝑓 − v𝑔𝑔 𝑢𝑢𝑓𝑓 + v𝑓𝑓 𝑢𝑢𝑓𝑓 𝑄𝑄̇ℎ = −𝑚𝑚𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 (1 + 𝑓𝑓) � � − 𝑚𝑚𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 �ℎ𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 + 𝑓𝑓ℎ𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 � v𝑔𝑔 − v𝑓𝑓 v𝑔𝑔 𝑢𝑢𝑓𝑓 − v𝑓𝑓 𝑢𝑢𝑔𝑔 𝑄𝑄̇ℎ = 𝑚𝑚𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 (1 + 𝑓𝑓) � � − 𝑚𝑚𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 �ℎ𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 + 𝑓𝑓ℎ𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 � v𝑔𝑔 − v𝑓𝑓

Now take limiting case, i.e., let’s simplify equations making 𝑄𝑄̇ℎ as small as possible to see if it could be zero or negative. Take cold leg spray at hot leg surge conditions: hspray = hsurge

Then msurge �hsurge + fhspray � = msurge (1 + f)hsurge

vg uf − vfug Q̇ h = msurge (1 + f) � − hsurge � vfg

Further, take the surge enthalpy at saturation, then

ℎ𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 = 𝑢𝑢𝑓𝑓 + 𝑃𝑃v𝑓𝑓

Substituting

𝑄𝑄̇ℎ =

𝑚𝑚𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 (1 + 𝑓𝑓) �v𝑔𝑔 𝑢𝑢𝑓𝑓 − v𝑓𝑓 𝑢𝑢𝑔𝑔 − v𝑓𝑓𝑓𝑓 �𝑢𝑢𝑓𝑓 + 𝑃𝑃v𝑓𝑓 �� v𝑓𝑓𝑓𝑓

𝑚𝑚𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 (1 + 𝑓𝑓) �v𝑔𝑔 𝑢𝑢𝑓𝑓 − v𝑓𝑓 𝑢𝑢𝑔𝑔 − �𝑢𝑢𝑓𝑓 v𝑔𝑔 + 𝑃𝑃v𝑓𝑓 v𝑔𝑔 − 𝑢𝑢𝑓𝑓 v𝑓𝑓 − 𝑃𝑃v𝑓𝑓2 �� v𝑓𝑓𝑓𝑓 𝑄𝑄̇ℎ =

𝑚𝑚𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 (1 + 𝑓𝑓) �v𝑓𝑓 �v𝑓𝑓 − 𝑢𝑢𝑔𝑔 � − 𝑃𝑃v𝑓𝑓 �v𝑔𝑔 − v𝑓𝑓 �� v𝑓𝑓𝑓𝑓

The second term in brackets on the right hand side is negative. In this limiting case, as expected since the surge enthalpy is taken as saturation, 𝑄𝑄̇ℎ is negative.

PROBLEM 7.14 QUESTION

Behavior of a Fully Contained Pressurized Pool Reactor under Decay Power Conditions (Section 7.3) A 1600 MWth pressurized pool reactor has been proposed in which the entire primary coolant system is submerged in a large pressurized pool of cold water with a high boric acid content. The amount of water in the pool is sufficient to provide for core decay heat removal for at least 1 week following any incident, assuming no cooling systems are operating. In this mode, the pool water

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boils and is vented to the atmosphere. The vessel geometry is illustrated in Figure 7.20.

FIGURE 7.20 Pressurized pool reactor. The core volume can be neglected. Assume that at time t = 0 with an initial vessel pressure of 0.10135 MPa (1 atm), the venting to the atmosphere fails. Does water cover the core for all t? Plot the water level measured from the top of the vessel versus time. You may assume that the vessel volume can be subdivided into an upper saturated vapor and a lower saturated liquid volume. Also, assume that the decay heat rate is constant for the time interval of interest at 25 MWth. Answer: Water always covers the core.

P ROBLEM 7.14 S OLUTION Behavior of a Fully Contained Pressurized Pool Reactor under Decay Power Conditions (Section 7.3) From the problem statement, we are given that: − decay power, 𝑄𝑄̇𝑑𝑑 = 25 MWth

− initial pressure, P1 = 0.10135 MPa The initial volume of the pool reactor and phase volumes are calculated as follows from Figure 7.20:

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4 V = 𝜋𝜋(6.5)2 (14.5) + 0.5 � � 𝜋𝜋(6.5)3 = 2500 m3 3 V𝑔𝑔1 = 𝜋𝜋(6.5)2 (2.17) = 288 m3

(1) (2)

V𝑓𝑓1 = V − V𝑔𝑔1 = 2212 m3

(3)

v𝑓𝑓1 = v(𝑃𝑃1 , 𝑥𝑥 = 0) = 0.001043 m3 ⁄kg

(4)

The initial specific volumes of each phase are evaluated at the initial pressure,

The total mass of water is

v𝑔𝑔1 = v(𝑃𝑃1 , 𝑥𝑥 = 1) = 1.673 m3 ⁄kg 𝑚𝑚𝑤𝑤 =

V𝑓𝑓1 V𝑔𝑔1 + = 2.12 × 106 kg v𝑓𝑓1 v𝑔𝑔1

V = 0.001179 m3 ⁄kg 𝑚𝑚𝑤𝑤

(5)

(6)

The specific volume of the mixture (remains constant) is

(7)

The initial specific internal energy can be evaluated at the initial pressure and specific volume Conservation of energy is

𝑢𝑢1 = 𝑢𝑢(𝑃𝑃1 , v) = 419 kJ⁄kg

(8)

𝑚𝑚𝑤𝑤 (𝑢𝑢2 − 𝑢𝑢1 ) = 𝑄𝑄̇𝑑𝑑 𝑡𝑡

(9)

𝑃𝑃2 = 𝑃𝑃(𝑢𝑢2 , v)

(10)

𝑥𝑥2 = 𝑥𝑥(𝑢𝑢2 , v)

(11)

Therefore, for any given time, the final specific internal energy is the only unknown. This time will be varied in this problem to determine how the water level changes. Once the specific internal energy is known, the final pressure can be evaluated using the specific volume as the second independent state parameter, Similarly, the quality can be evaluated

Once the quality of the mixture is known, the mass of each phase can be calculated, 𝑚𝑚𝑔𝑔2 = 𝑚𝑚𝑤𝑤 𝑥𝑥2

(12)

𝑚𝑚𝑓𝑓2 = 𝑚𝑚𝑤𝑤 − 𝑚𝑚𝑔𝑔2

The final specific volumes of each phase can be determined from the final pressure, v𝑓𝑓2 = v(𝑃𝑃2 , 𝑥𝑥 = 0) 174

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Chapter 7 - Thermodynamics of Nuclear Energy Conversion Systems Nonsteady Flow First Law Analysis

The final volumes of each phase is

v𝑔𝑔2 = v(𝑃𝑃2 , 𝑥𝑥 = 1)

(14)

V𝑓𝑓2 = v𝑓𝑓2 𝑚𝑚𝑓𝑓2

(15)

V𝑔𝑔2 = v𝑔𝑔2 𝑚𝑚𝑔𝑔2

(16)

Since the cross sectional area is constant toward the top of the vessel, the height that the steam volume takes up is 𝐿𝐿 =

V𝑔𝑔2 𝜋𝜋(6.5)2

(17)

Finally, the water level height as measured from the top of the core is 𝐻𝐻 = 12.5 − 𝐿𝐿

(18)

As we vary the time an calculate the water level, the level increases as shown in Figure SM-7.2 below. Thus, the core is never uncovered.

Figure SM-7.2 Water level (measured from the vessel top) decrease with elapsed time.

PROBLEM 7.15 QUESTION Depressurization of a Primary System (Section 7.3) The pressure of the primary system of a PWR is controlled by the pressurizer via the heaters and spray, A simplified drawing of the primary system of a PWR is shown in Figure 7.21.

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FIGURE 7.21 Simplified drawing of the primary system of a PWR. If the spray valve were to fail in the open position, depressurization of the primary system would result. Calculate the time to depressurize from 15.5 MPa to 12.65 MPa if the spray rate is 30.6 kg/s with an enthalpy of 1252 kJ/kg (constant with time). Also calculate the final liquid and vapor volumes. You may assume that 1. Heaters do not operate. 2. Pressurizer wall is adiabatic. 3. Pressurizer vapor and liquid are in thermal equilibrium and occupy initial volumes of 20.39 and 30.58 m3, respectively. 4. Complete phase separation occurs in the pressurizer, 5. The subcooled liquid in the primary system external to the pressurizer is incompressible so that the spray mass flow rate is exactly balanced by the pressurizer outsurge. Answers: 225 s 22.13 m3 vapor 28.84 m3 liquid

P ROBLEM 7.15 S OLUTION Depressurization of a Primary System (Section 7.3) From the problem statement we are given: − initial pressure, P1 = 15.5 MPa

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− final pressure, P2 = 12.65 MPa − mass flow rate of spray, 𝑚𝑚̇𝑠𝑠𝑠𝑠 = 30.6 kg⁄s

− specific enthalpy of spray, hsp = 1252 kJ/kg − initial volume of vapor, Vg1 = 20.39 m3 − initial volume of liquid, Vf1 = 30.58 m3 Specific volume at initial pressure for saturated liquid and vapor, v𝑓𝑓1 = 0.001683 m3 ⁄kg

(1)

The specific volumes at the final pressure are

v𝑔𝑔1 = 0.009814 m3 ⁄kg

v𝑓𝑓2 = 0.001552 m3 ⁄kg

v𝑔𝑔2 = 0.01328 m3 ⁄kg

(2)

ℎ𝑓𝑓2 = 1517 kJ⁄kg

(3)

The specific enthalpy of saturated liquid at the initial and final pressure are ℎ𝑓𝑓1 = 1630 kJ⁄kg

Therefore, over the depressurization we will just use the average of these two quantities, ℎ𝑓𝑓 = 0.5�ℎ𝑓𝑓1 + ℎ𝑓𝑓2 � = 1574 kJ⁄kg

(4)

The total mass of water can be calculated from each phase 𝑚𝑚𝑤𝑤 =

V𝑓𝑓1 V𝑔𝑔1 + = 20249 kg v𝑓𝑓1 v𝑔𝑔1

(5)

Therefore, the quality at the initial state is

V𝑔𝑔1 𝑥𝑥1 = v = 0.1026 𝑔𝑔1

𝑚𝑚𝑤𝑤

(6)

Now that we have two independent thermodynamic properties, we can evaluate the specific internal energy 𝑢𝑢1 = 𝑢𝑢(𝑃𝑃1 , 𝑥𝑥1 ) = 1690 kJ⁄kg

(7)

The initial specific volume of the mixture is (constant throughout depressurization)

Conservation of energy is

V𝑓𝑓1 + V𝑔𝑔1 = 0.002517 m3 ⁄kg 𝑚𝑚𝑤𝑤

𝑚𝑚𝑤𝑤 (𝑢𝑢2 − 𝑢𝑢1 ) = 𝑚𝑚̇𝑠𝑠𝑠𝑠 �ℎ𝑠𝑠𝑠𝑠 − ℎ𝑓𝑓 �𝑡𝑡

(8)

(9)

where the final specific internal energy u2 and time t are unknown. Since the specific volume is constant, the final specific internal energy a the final pressure condition is

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𝑢𝑢2 = 𝑢𝑢(𝑃𝑃2 , v) = 1581 kJ⁄kg

(10)

𝑡𝑡 = 225 s

(11)

Solving the conservation of energy formula, the time is Similarly, the final quality can be evaluated to be

𝑥𝑥2 = 𝑥𝑥(𝑃𝑃2 , v) = 0.08233

Therefore, the mass of each phase is and

𝑚𝑚𝑓𝑓2 = (1 − 𝑥𝑥2 )𝑚𝑚𝑤𝑤 = 18582 kg

(12)

𝑚𝑚𝑔𝑔2 = 𝑥𝑥2 𝑚𝑚𝑤𝑤 = 1667 kg

(13)

V𝑓𝑓2 = v𝑓𝑓2 𝑚𝑚𝑓𝑓2 = 28.84 m3

(14)

V𝑔𝑔2 = v𝑔𝑔2 𝑚𝑚𝑔𝑔2 = 22.13 m3

(15)

Using the specific volumes that were evaluated at the final pressure conditions (listed above), the volume of each phase is and

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Chapter 8 Thermal Analysis of Fuel Elements Contents Problem 8.1

Application of Kirchhoff’s law to pellet temperature distribution ................. 180

Problem 8.2

Conductivity integral ...................................................................................... 183

Problem 8.3

Effect of cracking on UO2 conductivity ......................................................... 184

Problem 8.4

Temperature fields in fresh and irradiated fuel .............................................. 185

Problem 8. 5 Comparison of UO2 and UC fuel temperature fields ..................................... 188 Problem 8.6

Thermal conduction problem involving design of a BWR. core ................... 189

Problem 8.7

Comparison of thermal energy that can be extracted from a spherical hollow fuel pellet versus a cylindrical annular fuel pellet .......................................... 194

Problem 8.8

Fuel pin problem ............................................................................................ 197

Problem 8. 9 Radially averaged fuel temperature and stored energy in solid and annular pellet ......................................................................................................................... 198 Problem 8.10 Maximum linear power from a duplex fuel pellet .......................................... 202 Problem 8.11 Effect of internal cooling on fuel temperature ............................................... 204 Problem 8.12 Temperature field in a restructured fuel pin ................................................... 207 Problem 8.13 Eccentricity effects in a plate type fuel .......................................................... 208 Problem 8.14 Determining the linear power given a constraint on the fuel average temperature ......................................................................................................................... 213

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PROBLEM 8.1 QUESTION Application of Kirchoff’s Law to Pellet Temperature Distribution (Section 8.2) For a PWR cylindrical solid fuel pellet operating at a heat flux equal to 1.7 MW/m2 and a surface temperature of 400 °C, calculate the maximum temperature in the pellet for two assumed values of conductivities. 1. k = 3 W/m °C independent of temperature 2. 𝑘𝑘 = 1 + 3𝑒𝑒 −0.0005𝑇𝑇 where T is in °C –

UO2 pellet diameter = 8.192 mm

–

UO2 density, 95% of theoretical density

Answers: 1. Tmax = 1560.5 °C 2. Tmax: = 1627.8 °C

PROBLEM 8.1 SOLUTION Application of Kirchoff’s Law to Pellet Temperature Distribution (Section 8.2) For a PWR cylindrical solid fuel pellet operating at a heat flux equal to 1.7 MW/m2 and a surface temperature of 400 ° C, calculate the maximum temperature in the pellet for two assumed values of conductivities. Given parameters: –

heat flux, q″ = 1.7 MW/m2

–

surface temperature, Ts = 400 °C

–

fuel pellet diameter, df = 8.192 mm

1. Constant conductivity The heat conduction equation for a constant conductivity is 1 𝑑𝑑 𝑑𝑑𝑑𝑑 �𝑟𝑟𝑟𝑟𝑓𝑓 � = −𝑞𝑞 ″ 𝑟𝑟 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑

(1)

We may integrate this equation from the center of the pellet to an arbitrary radius r, 𝑇𝑇

𝑟𝑟 𝑑𝑑𝑑𝑑 � 𝑑𝑑 �𝑟𝑟𝑟𝑟𝑓𝑓 � = � −𝑞𝑞 ″ 𝑟𝑟𝑟𝑟𝑟𝑟 𝑑𝑑𝑑𝑑 0 0

𝑑𝑑𝑑𝑑 𝑟𝑟 2 𝑟𝑟𝑟𝑟𝑓𝑓 = −𝑞𝑞″ 𝑑𝑑𝑑𝑑 2 180

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𝑑𝑑𝑑𝑑 𝑟𝑟 = −𝑞𝑞″ 𝑑𝑑𝑑𝑑 2𝑘𝑘𝑓𝑓

(4)

We can now integrate this equation from the surface to the center of the fuel pellet, �

𝑇𝑇𝐶𝐶𝐶𝐶

𝑇𝑇𝑓𝑓

𝑑𝑑𝑑𝑑 = � −𝑞𝑞‴ 𝑅𝑅𝑅𝑅𝑜𝑜

𝑇𝑇𝐶𝐶𝐶𝐶 − 𝑇𝑇𝑓𝑓 = 𝑞𝑞‴

𝑟𝑟 𝑑𝑑𝑑𝑑 2𝑘𝑘𝑓𝑓

2 𝑅𝑅𝑓𝑓𝑓𝑓 4𝑘𝑘𝑓𝑓

(5) (6)

We can relate the volumetric heat generation to the heat flux 𝑞𝑞 ‴ =

2𝑞𝑞″ 𝑅𝑅𝑓𝑓𝑓𝑓

(7)

Substituting Equation (7) into Equation (6) we get

𝑞𝑞 ″𝐷𝐷𝑓𝑓𝑓𝑓 𝑇𝑇𝐶𝐶𝐶𝐶 − 𝑇𝑇𝑓𝑓 = 4𝑘𝑘𝑓𝑓

We can substitute the given parameters to calculate the centerline temperature to be 𝑇𝑇𝐶𝐶𝐶𝐶 = 𝑇𝑇𝑓𝑓 +

𝑞𝑞″𝐷𝐷𝑓𝑓𝑓𝑓 = 1560.5 °C 4𝑘𝑘𝑓𝑓

2. Temperature Dependent Thermal Conductivity Let us solve this part two ways: First, by a direct solution of Equation (1), Second, by the use of the Kirchoff transformation. Direct solution

Integrating between r = 0 and r:

The next step consists of separating the variable T and r and integrating between [TCL, Tfo] for the temperature and [0, Rfo] for the radius:

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R 2fo 3 −0.0005.T fo −0.0005TCL −e = −q′′′ (T fo − TCL ) − 0.0005 e 4

(

)

(8)

Substituting Equation (7) into Equation (8), we get: 3 (T − T ) − 0.0005 (e fo

−0.0005.T fo

)

− e −0.0005TCL = − q′′

R fo 2

(9)

= TCL 1627.9 °C A numerical solution gives Kirchoff transformation Using the Kirchoff transformation gives 1 𝑇𝑇 1 𝑇𝑇 𝜃𝜃 = � 𝑘𝑘(𝑇𝑇)𝑑𝑑𝑑𝑑 = � 𝑘𝑘 = 1 + 3e−0.0005𝑇𝑇 𝑑𝑑𝑑𝑑 𝑘𝑘0 𝑇𝑇𝑠𝑠 𝑘𝑘0 𝑇𝑇𝑠𝑠 1 𝜃𝜃 = [𝑇𝑇 − 6000𝑒𝑒𝑒𝑒𝑒𝑒(−0.0005𝑇𝑇)]� 𝑇𝑇𝑇𝑇𝑠𝑠 𝑘𝑘0 1 𝜃𝜃(𝑇𝑇) = [(𝑇𝑇 − 400) + 4912.38 − 6000exp(−0.0005𝑇𝑇)] 𝑘𝑘0

(10)

The heat conduction equation will therefore be with boundary conditions

𝑘𝑘0 ∇2 𝜃𝜃 + 𝑞𝑞 ‴ = 0 𝑟𝑟 = 0,

Solving Equation (9)

(11)

∇𝜃𝜃 = 0

𝑟𝑟 = 𝑅𝑅𝑓𝑓𝑓𝑓 , 𝑇𝑇�𝑅𝑅𝑓𝑓𝑓𝑓 � = 𝑇𝑇𝑠𝑠 → 𝜃𝜃�𝑅𝑅𝑓𝑓𝑓𝑓 � = 0 2 𝑞𝑞‴𝑅𝑅𝑓𝑓𝑓𝑓 𝑟𝑟 2 𝜃𝜃(𝑟𝑟) = �1 − 2 � 4𝑘𝑘𝑓𝑓 𝑅𝑅𝑓𝑓𝑓𝑓

2 𝑞𝑞‴𝑅𝑅𝑓𝑓𝑓𝑓 𝑞𝑞″𝑅𝑅𝑓𝑓𝑓𝑓 = 4𝑘𝑘0 2𝑘𝑘0 W 𝑘𝑘0 = 𝑘𝑘�𝑇𝑇𝑓𝑓𝑓𝑓 � = 3.46 mK

𝜃𝜃(0) = 𝜃𝜃𝐶𝐶𝐶𝐶 =

𝜃𝜃𝐶𝐶𝐶𝐶 = 1007°C

(12) (13) (14) (15)

Solving Equation (10) iteratively with Equation (15) we obtain that the maximum temperature is 𝑇𝑇𝐶𝐶𝐶𝐶 = 1627.8 °C 182

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PROBLEM 8.2 QUESTION Conductivity Integral (Section 8.2) Describe an experiment by which you would obtain the results of Figure 8.2, that is, the value of the conductivity integral. Be sure to explicitly state what measurements and observations are to be made and how the conductivity integral is to be determined from them.

FIGURE 8.2 Integral of thermal conductivity of UO2 and UO2 with 5% gadolinia—all cases at 95% of theoretical density. (Based on results of [20].) Note: 32.8 W/cm = 1kW/ft.

PROBLEM 8.2 SOLUTION Conductivity Integral (Section 8.2) The “Conductivity Integral” for a cylindrical fuel pellet is: 𝑻𝑻

� 𝑘𝑘𝜌𝜌 𝑑𝑑𝑑𝑑 = 𝑻𝑻𝒇𝒇𝒇𝒇

𝑞𝑞′(𝑧𝑧) 4𝜋𝜋

(1)

In order to evaluate the LHS of Equation (1) we must know the linear heat rate, q′(z) and temperature, T. 1. q′(z) (a) Irradiate a fuel pin in a reactor with a very well characterized axial flux distribution (b) measure the flow rate, inlet and exit temperature of the coolant flowing over the fuel pin (c) evaluate q′(z) using the know flux shape (hence power shape), coolant flow rate, inlet and exit

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temperature of the coolant and fluid properties. 𝑚𝑚̇𝑐𝑐𝑝𝑝 (𝑇𝑇𝑜𝑜𝑜𝑜𝑜𝑜 − 𝑇𝑇𝑖𝑖𝑖𝑖 ) = �

𝐿𝐿 ⁄2

𝑔𝑔0′ 𝑓𝑓(𝑧𝑧)𝑑𝑑𝑑𝑑

−𝐿𝐿⁄2

(2)

where f(z) is a given power shape. From Equation (2), we can calculate 𝑞𝑞0′ and q′ (z) using Equation (3). 𝑞𝑞 ′ (𝑧𝑧) = 𝑞𝑞0′ (𝑧𝑧)

2. Temperature and Density

(3)

(a) To get the temperature, it is very difficult, but not impossible to locate a high temperature thermocouple at the fuel centerline. This is the primary means to get the centerline fuel temperature. (b) Alternately we can destructively examine the fuel pin after irradiation looking at microstructure of the fuel centerline along its axis. We know the approximate relationship of certain fuel structure to temperature, i.e. onset of equiaxed grain structure occurs at about 1600 ºC to 1650 ºC and onset of columnar grain structure about 1800 ºC to 2150 ºC. These structures also are related to pellet densities. Ceramists also relate overall microstructural characteristics to temperature. Finally, we could insert a small amount of marker material (i.e., undergoes a phase transition at a given temperature) at various axial centerline positions. Hence we can establish several (but not a large number) of temperature values at different axial location from one pin or several pin irradiations. (c) The overall impression you should derive from this solution is that the curves of “Conductivity Integral” are drawn from a few experimentally determined points. The curves project more certainty that the information are derived from.

PROBLEM 8.3 QUESTION Effect of Cracking on UO2 Conductivity (Section 8.3) For the conditions given in Example 8.1, evaluate the effective conductivity of the UO2 pellet after cracking using the empirical relation of Equation 8.24. Assume the gas is helium at a temperature Tgas = 0.7Tcl + 0.3Tf. Compare the results to those obtained in Example 8.1. Answer: keff = 2.00 W/m K

PROBLEM 8.3 SOLUTION Effect of Cracking on UO2 Conductivity (Section 8.3) From Example 8.1 we are given that: –

outer clad diameter, Dco = 11.2 mm

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–

clad thickness, tc = 0.71mm

–

gap thickness, tg = 90 × 10−6 m

–

conductivity of UO2, kUO2 = 2.326 W/m K

–

hot gap thickness, δhot = 0.0516 mm

–

fuel temperature, Tf = 1000 °C

–

clad temperature, Tcl = 295 °C

–

thermal expansion coefficient of fuel, αf = 10.1 × 10−6 1/K

Therefore, the cold fuel pellet diameter is: (1)

𝐷𝐷𝑓𝑓,𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = 𝐷𝐷𝑐𝑐𝑐𝑐 − 2𝑡𝑡𝑐𝑐 − 2𝑡𝑡𝑔𝑔 = 9.6 mm

The parameters for Equation 8.24 are

𝐴𝐴 = 6.35 × 10−5 m 𝐵𝐵 = 0.077

𝐶𝐶 = 0.015

(2)

From the given formula in the problem statement, the temperature of the gas is (3)

𝑇𝑇𝑔𝑔𝑔𝑔𝑔𝑔 = 0.7𝑇𝑇𝑐𝑐𝑐𝑐 + 0.3𝑇𝑇𝑓𝑓 = 779.65 K

Using Equation 8.140, the conductivity of the helium gas is 𝑘𝑘𝑔𝑔𝑔𝑔𝑔𝑔 = 15.8 × 10−6 𝑇𝑇 0.79

The hot fuel pellet diameter is given as

W W = 3.043 × 10−3 cm K cm K

𝐷𝐷𝑓𝑓,ℎ𝑜𝑜𝑜𝑜 − 𝐷𝐷𝑓𝑓,𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 �1 + 𝛼𝛼𝑓𝑓 �𝑇𝑇𝑓𝑓 − 27°C�� = 9.694 mm

(4)

(5)

The effective conductivity is therefore, 𝑘𝑘𝑒𝑒𝑒𝑒𝑒𝑒 =

𝑘𝑘𝑈𝑈𝑈𝑈2

2𝛿𝛿ℎ𝑜𝑜𝑜𝑜 − 𝐴𝐴 � � 𝐵𝐵𝐵𝐵𝑔𝑔𝑔𝑔𝑔𝑔 𝐷𝐷𝑓𝑓,ℎ𝑜𝑜𝑜𝑜 � + 𝐶𝐶� + 1 𝑘𝑘𝑈𝑈𝑈𝑈2

= 2.00

W mK

(6)

PROBLEM 8.4 QUESTION

Temperature Fields in Fresh and Irradiated Fuel (Section 8.3) Consider two conditions for heat transfer in the pellet and the pellet-cladding gap of a BWR fuel pin: –

Initial uncracked pellet with no relocation

–

Cracked and relocated fuel

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1. For each combination, find the temperatures at the cladding inner surface, the pellet outer surface, and the pellet centerline. 2. Find for each case the volume weighted average temperature of the pellet. Geometry and material information: –

Cladding outside diameter = 11.20 mm

–

Cladding thickness = 0.71 mm

–

Fuel-cladding gap thickness = 90 μm

–

Initial solid pellet with density = 88%

Basis for heat transfer calculations: –

Cladding conductivity is constant at 17 W/m K

–

Gap conductance – Without fuel relocation, 4300 W/m2 K; – With fuel relocation, 31,000 W/m2 K;

–

Fuel conductivity (average) at 95% density – Uncracked, 2.7 W/m K, – Cracked, 2.4 W/m K;

–

Volumetric heat deposition rate: uniform in the fuel and zero in the cladding

–

Do not adjust the pellet conductivity for restructuring.

–

Use Biancharia’s porosity correction factor (Equation 8-21).

Operating conditions: –

Cladding outside temperature = 295 °C

–

Linear heat-generation rate = 44 kW/m

Answers: 1. Tmax (°C) Tfo (°C) Tci (°C) 2. Tave (°C)

Uncracked 2135 687 351 1411

Cracked 2026 398 351 1212

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PROBLEM 8.4 SOLUTION Temperature Fields in Fresh and Irradiated Fuel (Section 8.3) From the problem statement, we are given that: –

outer clad diameter, Dco = 11.20 mm

–

clad thickness, tc = 0.71 mm

–

gap thickness, tg = 90 × 10−6 m

–

fuel density, ρf = 0.88 × 10.97g/cm3 = 9.654 g/cm3

–

clad conductivity, kc = 17W/m K

–

gap conductance with no fuel relocation (case 1), hg1 = 4300W/m2 K

–

gap conductance with fuel relocation (case 2), hg2 = 31000 W/m2 K

–

fuel conductivity uncracked (case 1), kf1 = 2.7W/m K

–

fuel conductivity cracked (case 2), kf2 = 2.4 W/m K

–

clad outer temperature, Tco = 295 °C

–

linear heat rate, q′ = 44kW/m

First, the geometry of the fuel pin will be calculated. The radius to the outer clad surface is 𝑅𝑅𝑐𝑐𝑐𝑐 = 0.5𝐷𝐷𝑐𝑐𝑐𝑐 = 5.6 mm

(1)

𝑅𝑅𝑐𝑐𝑐𝑐 = 𝑅𝑅𝑐𝑐𝑐𝑐 − 𝑡𝑡𝑐𝑐 = 4.89 mm

(2)

𝑅𝑅𝑓𝑓𝑓𝑓 = 𝑅𝑅𝑐𝑐𝑐𝑐 − 𝑡𝑡𝑔𝑔 = 4.8 mm

(3)

The radius to the inner clad surface can be calculated with the clad thickness, The fuel pellet surface radius can be calculated by subtracting the gap thickness, Finally, the mean gap radius is computed by taking the average of the fuel surface radius and the inner clad radius, 𝑅𝑅𝑔𝑔 =

𝑅𝑅𝑐𝑐𝑐𝑐 + 𝑅𝑅𝑓𝑓𝑓𝑓 = 4.845 mm 2

(4)

Using Biancharia’s porosity correction factor, each of the fuel conductivities are adjusted as follows, 0.88 (1.5 1 + − 1)(1 − 0.88) ′ 𝑘𝑘𝑓𝑓1 = 𝑘𝑘𝑓𝑓1 = 2.418 W⁄m K 0.95 1 + (1.5 − 1)(1 − 0.95) 187

(5)

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0.88 (1.5 1 + − 1)(1 − 0.88) ′ 𝑘𝑘𝑓𝑓2 = 𝑘𝑘𝑓𝑓2 = 2.15 W⁄m K 0.95 1 + (1.5 − 1)(1 − 0.95)

(6)

The inner clad temperature can be calculated for each case to be (Equation 8.146) 𝑞𝑞′ 𝑅𝑅𝑐𝑐𝑐𝑐 ln � � = 350.8 °C 2𝜋𝜋𝜋𝜋𝑐𝑐 𝑅𝑅𝑐𝑐𝑐𝑐 𝑞𝑞′ 𝑅𝑅𝑐𝑐𝑐𝑐 𝑇𝑇𝑐𝑐𝑐𝑐2 = 𝑇𝑇𝑐𝑐𝑐𝑐 + ln � � = 350.8 °C 2𝜋𝜋𝜋𝜋𝑐𝑐 𝑅𝑅𝑐𝑐𝑐𝑐

(7)

𝑇𝑇𝑐𝑐𝑐𝑐1 = 𝑇𝑇𝑐𝑐𝑐𝑐 +

(8)

The fuel surface temperature can be computed using Equation 8.136

and

𝑇𝑇𝑓𝑓𝑓𝑓1 = 𝑇𝑇𝑐𝑐𝑐𝑐1 +

𝑞𝑞′ = 687 °C 2𝜋𝜋𝜋𝜋𝑔𝑔 ℎ𝑔𝑔1

𝑇𝑇𝑓𝑓𝑓𝑓2 = 𝑇𝑇𝑐𝑐𝑐𝑐2 +

𝑞𝑞′ = 397.5 °C 2𝜋𝜋𝜋𝜋𝑔𝑔 ℎ𝑔𝑔2

(9)

(10)

The fuel centerline temperature can be determined with Equation 8.145 for each case,

and

𝑇𝑇𝐶𝐶𝐶𝐶1 = 𝑇𝑇𝑓𝑓𝑓𝑓1 +

𝑞𝑞′ = 2135 °C 4𝜋𝜋𝜋𝜋𝑓𝑓1

𝑇𝑇𝐶𝐶𝐶𝐶2 = 𝑇𝑇𝑓𝑓𝑓𝑓2 +

𝑞𝑞′ = 2026 °C 4𝜋𝜋𝜋𝜋𝑓𝑓2

𝑇𝑇𝑎𝑎𝑎𝑎𝑎𝑎1 = 𝑇𝑇𝑓𝑓𝑓𝑓1 +

𝑞𝑞′ = 1411 °C 8𝜋𝜋𝜋𝜋𝑓𝑓1

(11)

(12)

Finally, the average fuel temperature is calculated with Equation 8.71a to give

and

𝑇𝑇𝑎𝑎𝑎𝑎𝑎𝑎2 = 𝑇𝑇𝑓𝑓𝑓𝑓2 +

(13)

𝑞𝑞′ = 1212 °C 8𝜋𝜋𝜋𝜋𝑓𝑓2

(14)

PROBLEM 8.5 QUESTION

Comparison of UO2 and UC Fuel Temperature Fields (Section 8.4) A fuel plate is of half width a = 10 mm and is clad in a zircaloy sheet of thickness δc = 2 mm. The heat is generated uniformly in the fuel. Compare the temperature drop across the fuel plate when the fuel is UO2 with that of a UC fuel

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for the same heat-generation rate (i.e., calculate the ratio of Tmax − Tco for the UC plate to that of the UO2 plate). Answer: ∆𝑇𝑇𝑈𝑈𝑈𝑈2 ⁄∆𝑇𝑇𝑈𝑈𝑈𝑈 = 4.16

PROBLEM 8.5 SOLUTION

Comparison of UO2 and UC Fuel Temperature Fields (Section 8.4) From the problem statement, we are given that: –

fuel plate half-width, a = 10 mm

–

Zircaloy cladding thickness, δc = 2 mm

From Table 8.1 we have: –

thermal conductivity of UO2, kUO2 = 3.6 W/m K

–

thermal conductivity of UC, kUC = 23 W/m K

From Table 8.2 we have: –

thermal conductivity of Zircaloy, kc = 13 W/m K

Substituting the volumetric heat generation rate from Equation 8.45 into Equation 8.47 we have 𝑎𝑎 𝛿𝛿𝑐𝑐 ∆𝑇𝑇 = 𝑞𝑞 ‴ 𝑎𝑎 � + � 2𝑘𝑘𝑓𝑓 𝑘𝑘𝑐𝑐

(1)

Taking the ratio of the temperature differences for the UO2 and the UC we have 𝑎𝑎 𝛿𝛿𝑐𝑐 𝑎𝑎 𝛿𝛿𝑐𝑐 ∆𝑇𝑇𝑈𝑈𝑈𝑈2 𝑞𝑞‴𝑎𝑎 �2𝑘𝑘𝑈𝑈𝑈𝑈2 + 𝑘𝑘𝑐𝑐 � 2𝑘𝑘𝑈𝑈𝑈𝑈𝑈𝑈2 + 𝑘𝑘𝑐𝑐 = = = 4.16 𝑎𝑎 𝛿𝛿𝑐𝑐 𝑎𝑎 𝛿𝛿𝑐𝑐 ∆𝑇𝑇𝑈𝑈𝑈𝑈 𝑞𝑞‴𝑎𝑎 � + � + 2𝑘𝑘𝑈𝑈𝑈𝑈 𝑘𝑘𝑐𝑐 2𝑘𝑘𝑈𝑈𝑈𝑈 𝑘𝑘𝑐𝑐

(2)

PROBLEM 8.6 QUESTION

Thermal Conduction Problem Involving Design of a BWR Core (Section 8.5) A core design is proposed in which BWR type UO2 pins are located in holes within graphite hexagonal blocks (Figure 8.29). These blocks then form a core of radius Ro. Constants and constraints are defined in Figure 8.30 and Table 8.10. Basically, these constraints exist under decay power conditions where the outside of the core radiates its energy to a passive air chimney. However, the outside of the core which is in touch with a vessel at the same

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temperature is limited to 500°C. The cladding outside temperature, Tco, which radiates to the graphite, Tgi, is also constrained, here to a temperature 649°C.

FIGURE 8.29 Unit cell dimensions.

FIGURE 8.30 Configuration of a solid core and variables.

TABLE 8.10 Constants and Constraints for Homogenized Core Power Analysis Constraints Tco < 649 °C Tgo < 500 °C

Constants Acell = 7.30 × 10−4 m2 d1 = 12.5 mm d2 = 19.8 mm ϵ1 = 0.6 ϵ2 = 0.7

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𝑘𝑘𝑔𝑔 = 60 m∙K

𝑊𝑊

𝜎𝜎 = 5.669 × 10−8 𝑚𝑚2 ∙𝐾𝐾4 Calculate the achievable linear thermal power of the core (MW/m) as a function of core radius, Ro(m) for constraints of Table 8.10. Present the result as a plot. Notes for Figure 8.29: – Unit Cell using BWR fuel pin in MHTGR prismatic block holding the ratio of fuel to graphite constant – Coolant channel size established by taking the area of water normally associated with a pin in a conventional BWR. Notes for Figure 8.30: – Tgi is the temperature at the inner surface of the matrix graphite surrounding the fuel pin –

Tgo is the temperature of the matrix graphite at the core outer surface

–

Tco is the temperature at the clad outer surface

–

dl and d2 are shown in Figure 8.29.

PROBLEM 8.6 SOLUTION Thermal Conduction Problem Involving Design of a BWR Core (Section 8.5) A core design is proposed in which BWR type UO2 pins are located in holes within graphite hexagonal blocks (Figure 8.29). These blocks then form a core of radius RO. The achievable linear heat power of the core (MW/m) is desired as a function of core radius, Ro(m) for constraints of Table 8.10. Present the result as a plot. Constants and terms are defined in Figure 8.30 and Table 8.10. Basically these constraints exist under decay power conditions where the outside of the core radiates its energy to a passive air chimney. However, the outside of the core which is in touch with a vessel at the same temperature is limited to 500 °C. The clad outside temperature, Tco, which radiates to the graphite, Tgi, is also constrained, here to a temperature 649 °C. In the problem there are two main heat transfer mechanisms: 1. From fuel rod to graphite: Radiation 2. Within the graphite: Conduction 1) Radiation Heat Transfer The net heat transfer by radiation between two surfaces (outer clad and inner graphite) is

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Chapter 8 - Thermal Analysis of Fuel Elements 4 4 𝜎𝜎�𝑇𝑇𝑐𝑐𝑐𝑐 − 𝑇𝑇𝑔𝑔𝑔𝑔 � (1) 𝑞𝑞̇ = 𝑅𝑅𝑡𝑡𝑡𝑡𝑡𝑡 In Equation (1), 𝑞𝑞̇ is the net heat transfer between the two surfaces, Tco is the temperature of the outer cladding surface, Tgi is the temperature of the inner surface of the graphite matrix and Rtot is the total thermal resistance between the two surfaces. For two-surface enclosures, the total thermal resistance is modeled with

𝑅𝑅𝑡𝑡𝑡𝑡𝑡𝑡 =

1 − 𝜖𝜖1 1 1 − 𝜖𝜖2 + + 𝐴𝐴1 𝜖𝜖1 𝐴𝐴1 𝐹𝐹1→2 𝐴𝐴2 𝜖𝜖2

(2)

where ϵ1 is the emissivity of surface 1 (outer clad), ϵ2 is the emissivity of surface 2 (inner graphite), A1 is the surface area of surface 1, A2 is the surface area of surface 2 and F1→2 is the view factor. For two long concentric cylinders, the view factor is unity and Equation (2) can be simplified into the form, 𝑅𝑅𝑡𝑡𝑡𝑡𝑡𝑡 = �

1 1 − 𝜖𝜖2 𝑑𝑑2 1 � + 𝜖𝜖1 𝜖𝜖2 𝑑𝑑2 𝐴𝐴1

(3)

Combining Equations (3) and (1) we arrive at an expression for the net heat transfer, 4 4 𝜎𝜎𝐴𝐴1 �𝑇𝑇𝑐𝑐𝑐𝑐 − 𝑇𝑇𝑔𝑔𝑔𝑔 � 𝑞𝑞̇ = 1 1 − 𝜖𝜖2 𝑑𝑑1 𝜖𝜖1 + 𝜖𝜖1 𝑑𝑑2

(4)

We may divide both sides of Equation (4) to get an expression for the linear heat rate, 4 4 𝜋𝜋𝑑𝑑1 𝜎𝜎�𝑇𝑇𝑐𝑐𝑐𝑐 − 𝑇𝑇𝑔𝑔𝑔𝑔 � 𝑞𝑞′ = 1 1 − 𝜖𝜖2 𝑑𝑑1 𝜖𝜖1 + 𝜖𝜖1 𝑑𝑑2

(5)

2) Conduction Heat Transfer within Homogeneous Core We can assume that the radius of the core, Ro, is much greater than the radius of the coolant 𝒓𝒓 channel, r2, i.e. 𝑅𝑅𝟐𝟐 ≪ 1. The core linear heat rate can then be described by the following 0

conduction equation:

𝑄𝑄 ′ = 4𝜋𝜋𝜋𝜋𝑔𝑔 �𝑇𝑇𝑔𝑔𝑔𝑔 − 𝑇𝑇𝑔𝑔𝑔𝑔 �

(6)

where Q′ is the core linear heat rate, kg is the thermal conductivity of graphite given in Table 8.10 and Tgo is the temperature of the outer surface of the graphite matrix. We can relate the core linear heat rate to the heat rate of one unit cell with 𝜋𝜋𝑅𝑅𝑜𝑜2 ′ 𝑄𝑄 = 𝑁𝑁𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑞𝑞′ = 𝑞𝑞 𝐴𝐴𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 ′

(7)

In Equation (7) Ncells is the number of unit cells which can be determined from the ratio of the core area, 𝜋𝜋𝑅𝑅𝑜𝑜2 and unit cell area, Acell. We first substitute Equation (6) into Equation (5) by eliminating Tgi, leaving us with 192

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𝑞𝑞 ′ =

4 𝑄𝑄′ 4 � � 𝜋𝜋𝑑𝑑1 𝜎𝜎 �𝑇𝑇𝑐𝑐𝑐𝑐 − �𝑇𝑇𝑔𝑔𝑔𝑔 + 4𝜋𝜋𝜋𝜋𝑔𝑔

(8)

1 1 − 𝜖𝜖2 𝑑𝑑1 + 𝜖𝜖 𝜖𝜖1 𝑑𝑑2 2

We can then covert the right hand side of Equation (8) from unit cell linear heat rate to core linear heat heat using Equation (7): 4

𝑄𝑄′ 4 ⎧ ⎫ 2 ⎪𝜋𝜋𝑑𝑑1 𝜎𝜎 �𝑇𝑇𝑐𝑐𝑐𝑐 − �𝑇𝑇𝑔𝑔𝑔𝑔 + 4𝜋𝜋𝜋𝜋 � �⎪ 𝜋𝜋𝑅𝑅𝑜𝑜 𝑔𝑔

(9) 1 1 − 𝜖𝜖2 𝑑𝑑1 𝐴𝐴𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 ⎨ ⎬ ⎪ ⎪ 𝜖𝜖1 + 𝜖𝜖2 𝑑𝑑2 ⎩ ⎭ The maximum achievable core linear power can be obtained when the temperature difference across the core is a maximum. This will occur when the inner surface temperature of the graphite matrix is equivalent to the outer cladding temperature and when the temperature are at their maximum allowed value, Tco = Tgi = 6490 °C and Tgo = 500 °C. Substituting these values into 𝑄𝑄 ′ =

𝑀𝑀𝑀𝑀

′ Equation (6), we find that the maximum core linear power is limited to 𝑄𝑄𝑚𝑚𝑚𝑚𝑚𝑚 = 0.112 𝑚𝑚 .

Looking at Eqution (9), we can see that it is easiest to solve this formula for the core radius as a function of core linear power,

𝑅𝑅𝑜𝑜 (𝑄𝑄′) =

⎧ ⎪ ⎪ ⎪

𝑄𝑄′𝐴𝐴𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐

⎫ ⎪ ⎪ ⎪

⎨ 2 4 − �𝑇𝑇 + 𝑄𝑄′ � � ⎬ 𝜋𝜋 𝑑𝑑1 𝜎𝜎 �𝑇𝑇𝑐𝑐𝑐𝑐 𝑔𝑔𝑔𝑔 4𝜋𝜋𝜋𝜋𝑔𝑔 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 1 1 − 𝜖𝜖2 𝑑𝑑1 + ⎩ ⎭ 𝜖𝜖1 𝜖𝜖2 𝑑𝑑2

1 2

(10)

The above function is plotted in Figure SM-8.1:

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Figure SM-8.1 BWR core thermal performance

PROBLEM 8.7 QUESTION Comparison of Thermal Energy that can be Extracted from a Spherical Hollow Fuel Pellet versus a Cylindrical Annular Fuel Pellet (Section 8.5) Consider a cylindrical annular fuel pellet of length L, inside radius R V , and outside radius R foc . It is operating at 𝑞𝑞c‴ , such that for a given outside surface temperature, Tfo, the inside surface temperature, TV, is just at the fuel melting limit Tmelt. A fellow engineer claims that if the same volume of fuel is arranged as a sphere with an inside voided region of radius R V and operated between the same two surface temperature limits, that is, T V and T fo , more power can be extracted from the spherical fuel volume then from a cylindrical fuel pellet. In both cases volumetric generation rate is radially constant. Is the claim correct? Prove or disprove it. Use the nomenclature of Figure 8.31. Assume no sintering occurs.

FIGURE 8.31 Fuel Pellet geometry. (a) cylindrical annular pellet. (b) spherical hollow pellet.

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Useful relations: The one dimensional heat conduction equation in the radial direction in spherical coordinates is 1 𝑑𝑑 𝑑𝑑𝑑𝑑 2 �𝑘𝑘𝑘𝑘 � + 𝑞𝑞 ‴ = 0 𝑟𝑟 2 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑

For a sphere: VS = 4/3πR3 and AS = πR2.

Cylindrical Annular Fuel Pellet

Spherical Hollow Fuel Pellet

𝑞𝑞C‴

RV = 0.25 mm

Rfoc = 1 cm

𝑞𝑞S‴

Rfos = ?

L = 1 cm

(to be determined from the problem statement)

Answer: The sphere does generate more power.

PROBLEM 8.7 SOLUTION Comparison of Thermal Energy that can be Extracted from a Spherical Hollow Fuel Pellet versus a Cylindrical Annular Fuel Pellet (Section 8.5) From the problem statement we are given that: –

length of annular pellet, L = 1 cm

–

annular pellet and spherical hollow pellet inside radius, RV = 0.25 mm

–

annular pellet outside radius, RfoC = 1cm

In this problem we assume that the volumes of the cylinder and sphere are the same. The heat rate can be related to the volumetric heat rate with 𝑄𝑄̇ = 𝑉𝑉𝑞𝑞 ‴

(1)

Therefore, we only need to compare the volumetric heat rate from the cylinder and sphere. From Equation 8.73 the volumetric heat generation rate for a annular pellet only cooled on the outside is 𝑞𝑞𝐶𝐶‴ =

𝑇𝑇

4 ∫𝑇𝑇 𝑚𝑚𝑚𝑚𝑚𝑚 𝑘𝑘𝑘𝑘𝑘𝑘 𝑓𝑓𝑓𝑓

= 4.0 × 10 𝑅𝑅𝑦𝑦 2 𝑅𝑅𝑓𝑓𝑓𝑓𝑓𝑓 2 𝑅𝑅V 2 2 𝑅𝑅𝑓𝑓𝑓𝑓𝑓𝑓 ��1 − �𝑅𝑅 � � − �𝑅𝑅 � 𝑙𝑙𝑙𝑙 � 𝑅𝑅 � � 𝑓𝑓𝑓𝑓𝑓𝑓 𝑓𝑓𝑓𝑓𝑓𝑓 V

�

𝑇𝑇𝑚𝑚𝑚𝑚𝑚𝑚

𝑇𝑇𝑓𝑓𝑓𝑓

𝑘𝑘𝑘𝑘𝑘𝑘

(2)

For the sphere we must first solve for the radius of the sphere given the volume (equivalent to annular cylinder),

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V𝑆𝑆 =

Therefore,

4 2 3 𝜋𝜋�𝑅𝑅𝑓𝑓𝑓𝑓𝑓𝑓 − 𝑅𝑅V3 � = V𝐶𝐶 = 𝜋𝜋𝐿𝐿�𝑅𝑅𝑓𝑓𝑓𝑓𝑓𝑓 − 𝑅𝑅V2 � 3

(3)

3 3 2 𝑅𝑅𝑓𝑓𝑓𝑓𝑓𝑓 = � 𝐿𝐿�𝑅𝑅𝑓𝑓𝑓𝑓𝑓𝑓 − 𝑅𝑅V2 � + 𝑅𝑅V2 = 0.908 cm 4

(4)

Starting with the one dimensional heat conduction equation in the radial direction in spherical coordinates: 1 𝜕𝜕 2 𝜕𝜕𝜕𝜕 �𝑟𝑟 𝑘𝑘 � + 𝑞𝑞𝑆𝑆‴ = 0 2 𝑟𝑟 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕

The boundary conditions are: – –

(5)

𝑑𝑑𝑑𝑑

at 𝑟𝑟 = 𝑅𝑅V , 𝑑𝑑𝑑𝑑 = 0

at 𝑟𝑟 = 𝑅𝑅𝑓𝑓𝑓𝑓𝑓𝑓 , 𝑇𝑇 = 𝑇𝑇𝑓𝑓𝑓𝑓𝑓𝑓

Integrating from RV to an arbitrary radius r: 𝑑𝑑 2 𝑑𝑑𝑑𝑑 �𝑟𝑟 𝑘𝑘 � = −𝑞𝑞𝑆𝑆‴ 𝑟𝑟 2 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑

𝑟𝑟

(6)

𝑇𝑇 𝑑𝑑 𝑑𝑑 ′2 ′ �𝑟𝑟 𝑘𝑘 � 𝑑𝑑𝑟𝑟 = � −𝑞𝑞𝑆𝑆‴ 𝑟𝑟 ′2 𝑑𝑑𝑑𝑑′ � 𝑑𝑑𝑑𝑑′ 𝑅𝑅V 𝑑𝑑𝑑𝑑′ 𝑅𝑅V

𝑟𝑟 2 𝑘𝑘

(7)

𝑑𝑑 𝑞𝑞𝑆𝑆‴ 3 (𝑟𝑟 − 𝑅𝑅V3 ) =− 𝑑𝑑𝑑𝑑 3

(8)

Integrating this formula again from Rfos to Rv we get: 𝑇𝑇𝑓𝑓𝑓𝑓𝑓𝑓

𝑇𝑇𝑚𝑚𝑚𝑚𝑚𝑚

𝑘𝑘𝑘𝑘𝑘𝑘 = �

𝑅𝑅𝑓𝑓𝑓𝑓𝑓𝑓

𝑅𝑅V

−

𝑞𝑞𝑆𝑆‴ 𝑅𝑅V3 �𝑟𝑟 − 2 � 𝑑𝑑𝑑𝑑 3 𝑟𝑟

(9)

2 − 𝑅𝑅V2 � 𝑞𝑞𝑆𝑆‴ �𝑅𝑅𝑓𝑓𝑓𝑓𝑓𝑓 1 1 � 𝑘𝑘𝑘𝑘𝑘𝑘 = − � + 𝑅𝑅V3 � − �� 3 2 𝑅𝑅𝑓𝑓𝑓𝑓𝑓𝑓 𝑅𝑅V 𝑇𝑇𝑚𝑚𝑚𝑚𝑚𝑚 𝑇𝑇𝑓𝑓𝑓𝑓𝑓𝑓

Thus,

�

𝑞𝑞𝑆𝑆‴ =

𝑇𝑇

− ∫𝑇𝑇 𝑓𝑓𝑓𝑓𝑓𝑓 𝑘𝑘𝑘𝑘𝑘𝑘 𝑚𝑚𝑚𝑚𝑚𝑚

2 �𝑅𝑅𝑓𝑓𝑓𝑓𝑓𝑓 − 𝑅𝑅V2 � 𝑅𝑅V3 1 1 � � + − 6 3 𝑅𝑅𝑓𝑓𝑓𝑓𝑓𝑓 𝑅𝑅V

= 7.29 × 104 �

𝑇𝑇𝑓𝑓𝑓𝑓𝑓𝑓

𝑇𝑇𝑚𝑚𝑚𝑚𝑚𝑚

(10)

𝑘𝑘𝑘𝑘𝑘𝑘

(11)

Comparing the total power generated in each of the pellets, we can see the fellow engineer was indeed correct.

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P ROBLEM 8.8 Q UESTION Fuel Pin Problem (Section 8.5) A fuel pin is operating with solid UO2 pellets of 88% theoretical density and outside radius 5 mm such that at the axial location of maximum fuel temperature, the fuel centerline temperature TCL, is 2500 °C and the fuel surface temperature Tfo, is 700 °C. It is desired to raise the pin linear power by 10% by employing one of the following alternative strategies (in each case all the other conditions except for the one cited are held constant): a. Raise the maximum allowable fuel temperature. b. Use an annular pellet with the center void of dimension Rv or c. Increase the pellet density. For each strategy find the new value of the cited parameter necessary to achieve the desired 10% increase of linear pin power. Sintering effects may be neglected. Answers: a. Tmax = 2600 °C b. RV = 0.75 mm c. ρ = 0.95ρTD

PROBLEM 8.8 S OLUTION Fuel Pin Problem (Section 8.5) a) for a solid fuel pellet: �

𝑇𝑇𝑚𝑚𝑚𝑚𝑚𝑚

𝑘𝑘𝑘𝑘𝑘𝑘 =

𝑇𝑇𝑓𝑓𝑓𝑓

𝑞𝑞𝑠𝑠′ 4𝜋𝜋

(1)

Using Figure 8.3, the conductivity integral can be calculated as 2500 2500 700 𝑞𝑞𝑠𝑠′ =� 𝑘𝑘0.88 𝑑𝑑𝑑𝑑 = � 𝑘𝑘0.88 𝑑𝑑𝑑𝑑 − � 𝑘𝑘0.88 𝑑𝑑𝑑𝑑 = 73 − 33 = 40 W⁄cm 4𝜋𝜋 700 100 100

(2)

Therefore for a 10% increase we need 𝑞𝑞𝑠𝑠′ = 44 W⁄cm. We can rewrite the conductivity integral now as �

𝑇𝑇𝑚𝑚𝑚𝑚𝑚𝑚

100

𝑘𝑘0.88 𝑑𝑑𝑑𝑑 = 44 + �

700

100

𝑘𝑘0.88 𝑑𝑑𝑑𝑑 = 44 + 33 = 77 W⁄cm

(3)

Using Figure 8.3, the maximum temperature is approximately, 𝑇𝑇𝑚𝑚𝑚𝑚𝑚𝑚 = 2600°C

(4)

b) for annular pellet without sintering:

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Chapter 8 - Thermal Analysis of Fuel Elements 2 𝑅𝑅𝑓𝑓𝑓𝑓 ⎡ ln � 𝑅𝑅 � ⎤ 𝑇𝑇𝑚𝑚𝑚𝑚𝑚𝑚 ′ 𝑅𝑅𝑓𝑓𝑓𝑓 𝑅𝑅𝑓𝑓𝑓𝑓 𝑞𝑞𝑎𝑎′ ⎢ 𝑞𝑞𝑎𝑎′ v ⎥ 𝑞𝑞𝑎𝑎 � � 𝑘𝑘𝑘𝑘𝑘𝑘 = 1 − = 𝐹𝐹v , 𝛽𝛽� == 𝐹𝐹v � , 1� ⎢ 2 ⎥ 4𝜋𝜋 4𝜋𝜋 𝑅𝑅V 4𝜋𝜋 𝑅𝑅V 𝑅𝑅 𝑇𝑇𝑓𝑓𝑓𝑓 𝑓𝑓𝑓𝑓 ⎢ � � − 1⎥ 𝑅𝑅V ⎣ ⎦

(5)

where β = 1 for fuel elements of uniform density. For a solid fuel pellet recall that:

Thus,

For a 10% increase in 𝑞𝑞𝑠𝑠′ , then:

�

𝑇𝑇𝑚𝑚𝑚𝑚𝑚𝑚

𝑇𝑇𝑓𝑓𝑓𝑓

𝑘𝑘𝑘𝑘𝑘𝑘 =

𝑞𝑞𝑠𝑠′ 4𝜋𝜋

(6)

𝑞𝑞𝑠𝑠′ 𝑞𝑞𝑎𝑎′ = 𝐹𝐹 4𝜋𝜋 4𝜋𝜋 𝑣𝑣

(7)

(8)

Hence by substitution

𝑞𝑞𝑠𝑠′ = 1.1𝑞𝑞𝑠𝑠′

(9)

and

𝑞𝑞𝑠𝑠′ = 1.1𝑞𝑞𝑠𝑠′ 𝐹𝐹𝑣𝑣 𝐹𝐹𝑣𝑣 =

1 = 0.909 1.1

(10)

Using Figure 8.14 for a void factor Fυ = 0.909 and β = 1, 1 = 0.15 𝛼𝛼

Therefore, 𝑅𝑅𝑉𝑉 =

𝑅𝑅𝑓𝑓𝑓𝑓 = 0.75 mm 𝛼𝛼

c) Recall from above that a 10% increase in

(11)

(12)

corresponds to 44 W/cm. From Figure 8.3 this

corresponds to about 95% theoretical density, keeping the maximum temperature 2500 °C.

PROBLEM 8.9 QUESTION Radially Averaged Fuel Temperature and Stored Energy in Solid and Annular Pellet (Section 8.5) Consider a solid pellet of radius b and an annular pellet of inner radius a, and outside radius b, each operating at the same linear power rate, q″.

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Define ∆𝑇𝑇(𝑟𝑟) ≡ 𝑇𝑇(𝑟𝑟) − 𝑇𝑇𝑏𝑏

and

�� ∆𝑇𝑇(𝑟𝑟) ≡ �� 𝑇𝑇(𝑟𝑟) − 𝑇𝑇𝑏𝑏

��⁄∆𝑇𝑇. Use the subscript “s” for solid and the subscript 1. Find across each pellet, the value of ∆𝑇𝑇 “a” for annular.

2. What is the ratio of the stored energy in the solid to the annular pellet? Answers: 1. Solid Pellet ��𝑠𝑠 ∆𝑇𝑇 = ∆𝑇𝑇𝑠𝑠

Annular Pellet -

2 �1 −

𝑟𝑟 2 � 𝑏𝑏 2

��𝑎𝑎 (2⁄𝐹𝐹 )(𝑏𝑏 4 ⁄4 − 𝑎𝑎2 𝑏𝑏 2 + 3⁄4𝑎𝑎4 + 𝑎𝑎4 ln 𝑏𝑏⁄𝑎𝑎 ) ∆𝑇𝑇 = (𝑏𝑏 2 − 𝑎𝑎2 ) ∆𝑇𝑇𝑎𝑎

where F is a geometric parameter equal

2. Ratio of stored energy −1 ��𝑠𝑠 𝑐𝑐𝑃𝑃𝑃𝑃 𝑀𝑀𝑠𝑠 ∆𝑇𝑇 𝑐𝑐𝑃𝑃𝑃𝑃 (∆𝑇𝑇𝑠𝑠 ) 𝜌𝜌𝑠𝑠 𝑏𝑏 4 𝑏𝑏 4 3 4 𝑏𝑏 2 2 4 = � − 𝑎𝑎 𝑏𝑏 + 𝑎𝑎 + 𝑎𝑎 ln � 𝑐𝑐𝑃𝑃𝑃𝑃 𝑀𝑀𝑎𝑎 ∆𝑇𝑇𝑎𝑎 𝑐𝑐𝑃𝑃𝑃𝑃 (∆𝑇𝑇𝑎𝑎 ) 𝜌𝜌𝑠𝑠 4 4 4 𝑎𝑎

PROBLEM 8.9 SOLUTION

Radially Averaged Fuel Temperature and Stored Energy in Solid and Annular Pellet (Section 8.5) Solid Fuel Pellet Starting from the one-dimensional heat conduction equation in the radial direction

The boundary conditions are:

1 𝑑𝑑 𝑑𝑑𝑑𝑑 �𝑘𝑘𝑘𝑘 + 𝑞𝑞 ‴ = 0� 𝑟𝑟 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 199

(1)

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Chapter 8 - Thermal Analysis of Fuel Elements 𝑑𝑑𝑑𝑑

– at 𝑟𝑟 = 0, 𝑑𝑑𝑑𝑑 = 0

– at r = b, T(r) = Tb After integrating from r = 0 to an arbitrary radius r and applying the first boundary condition, we get 𝑑𝑑𝑑𝑑 𝑞𝑞 ‴ 2 𝑘𝑘𝑘𝑘 = 𝑟𝑟 𝑑𝑑𝑑𝑑 2

(2)

After integrating from r = b to an arbitrary radius r and applying the second boundary condition we get ∆𝑇𝑇𝑠𝑠 = 𝑇𝑇(𝑟𝑟) − 𝑇𝑇𝑏𝑏 =

𝑞𝑞 ‴ 2 2 (𝑏𝑏 −𝑟𝑟 ) 4𝑘𝑘

(3)

The average temperature across a solid pellet is given as ��𝑠𝑠 = ∆𝑇𝑇

𝑏𝑏

∫0 2𝜋𝜋𝜋𝜋∆𝑇𝑇(𝑟𝑟)𝑑𝑑𝑑𝑑 𝑏𝑏 ∫0 2𝜋𝜋𝜋𝜋𝜋𝜋𝜋𝜋

Solving part 1 for the solid pellet gives

𝑏𝑏 𝑞𝑞 ‴ ∫0 4𝑘𝑘 (𝑏𝑏 2 𝑟𝑟 − 𝑟𝑟 3 ) 𝑏𝑏 ∫0 𝑟𝑟𝑟𝑟𝑟𝑟

𝑞𝑞 ‴ 𝑏𝑏 2 8𝑘𝑘

𝑞𝑞 ‴ 𝑏𝑏 2 �� 1 ∆𝑇𝑇𝑠𝑠 = ‴ 8𝑘𝑘 = 𝑟𝑟 2 ∆𝑇𝑇𝑠𝑠 𝑞𝑞 (𝑏𝑏 2 − 𝑟𝑟 2 ) �1 � 2 − 8𝑘𝑘 𝑏𝑏 2

(4)

(5)

Annular Pellet In this case, the first boundary condition changes to 𝑑𝑑𝑑𝑑

– at 𝑟𝑟 = 0, 𝑑𝑑𝑑𝑑 = 0

so that after the first integration we are left with 𝑑𝑑𝑑𝑑 𝑞𝑞 ‴ 2 2 (𝑟𝑟 −𝑎𝑎 ) = 𝑘𝑘𝑘𝑘 𝑑𝑑𝑑𝑑 2

(6)

After the second integration and boundary condition the temperature distribution is 𝑞𝑞 ‴ 2 𝑞𝑞 ‴ 𝑏𝑏 2 𝑏𝑏 𝑞𝑞 ‴ 𝑏𝑏 2) (𝑏𝑏 � � − 𝑟𝑟 = ln �(𝑏𝑏 2 − 𝑟𝑟 2 ) − 2𝑎𝑎2 ln � �� ∆𝑇𝑇𝑎𝑎 = 𝑇𝑇(𝑟𝑟) − 𝑇𝑇𝑏𝑏 = = 4𝑘𝑘 2𝑘𝑘 𝑟𝑟 4𝑘𝑘 𝑟𝑟

(7)

Now, we designate F such that

𝑏𝑏 (𝑏𝑏 2 −𝑟𝑟 2 ) − 2𝑎𝑎2 ln � � 𝑟𝑟 𝐹𝐹 = (𝑏𝑏 2 −𝑎𝑎2 )

and therefore ∆𝑇𝑇𝑎𝑎 =

𝑞𝑞 ‴ 𝑏𝑏 𝑞𝑞 ‴ �(𝑏𝑏 2 − 𝑟𝑟 2 ) − 2𝑎𝑎2 ln � �� = 𝐹𝐹(𝑏𝑏 2 − 𝑎𝑎2 ) 4𝑘𝑘 𝑟𝑟 4𝑘𝑘 200

(8)

(9)

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Again, we know that the definition of average change in temperature 𝑏𝑏 𝑞𝑞 ‴ 𝑏𝑏 𝑏𝑏 2 2 2 ∫0 2𝜋𝜋𝜋𝜋∆𝑇𝑇𝑎𝑎 (𝑟𝑟)𝑑𝑑𝑑𝑑 ∫𝑎𝑎 𝑟𝑟 4𝑘𝑘 (𝑏𝑏 − 𝑟𝑟 ) − 2𝑎𝑎 ln �𝑎𝑎� 𝑑𝑑𝑑𝑑 �� = ∆𝑇𝑇𝑠𝑠 = 𝑏𝑏 𝑏𝑏 ∫0 2𝜋𝜋𝜋𝜋𝜋𝜋𝜋𝜋 ∫0 𝑟𝑟𝑟𝑟𝑟𝑟 𝑞𝑞 ‴ 𝑏𝑏 𝑏𝑏 ∫𝑎𝑎 𝑟𝑟 �(𝑏𝑏 2 𝑟𝑟 − 𝑟𝑟 3 ) − 2𝑎𝑎2 𝑟𝑟 ln �𝑎𝑎�� 𝑑𝑑𝑑𝑑 4𝑘𝑘 = 𝑏𝑏 ∫0 𝑟𝑟𝑟𝑟𝑟𝑟

��𝑎𝑎 = ∆𝑇𝑇

(10)

𝑞𝑞 ‴ 𝑏𝑏 4 2 2 3 4 𝑏𝑏 � 𝑎𝑎 𝑏𝑏 + 𝑎𝑎 + 𝑎𝑎4 ln � 2 2 2𝑘𝑘(𝑏𝑏 −𝑎𝑎 ) 4 4 𝑎𝑎

(11)

Now solving for the annular pellet:

𝑞𝑞 ‴ 𝑏𝑏 4 3 4 𝑏𝑏 2 2 4 � ��𝑎𝑎 2𝑘𝑘(𝑏𝑏 2 −𝑎𝑎2 ) 4 − 𝑎𝑎 𝑏𝑏 + 4 𝑎𝑎 + 𝑎𝑎 ln 𝑎𝑎� ∆𝑇𝑇 = 𝑞𝑞 ‴ ∆𝑇𝑇𝑎𝑎 𝐹𝐹(𝑏𝑏 2 −𝑎𝑎2 ) 4𝑘𝑘 𝑏𝑏 4 3 𝑏𝑏 (2⁄𝐹𝐹 ) � − 𝑎𝑎2 𝑏𝑏 2 + 𝑎𝑎4 + 𝑎𝑎4 ln � 4 4 𝑎𝑎 = (𝑏𝑏 2 −𝑎𝑎2 )

(12)

Stored Energy is given by the equation

�� 𝐸𝐸 = 𝑀𝑀𝐶𝐶𝑝𝑝 ∆𝑇𝑇

(13)

𝑀𝑀𝑠𝑠 = 𝜌𝜌𝑠𝑠 𝜋𝜋𝑏𝑏 2 𝐿𝐿

(14)

The masses of the pellets are given by

(15)

𝑀𝑀𝑎𝑎 = 𝜌𝜌𝑠𝑠 𝜋𝜋(𝑏𝑏 2 − 𝑎𝑎2 )𝐿𝐿 ∴

𝑀𝑀𝑎𝑎 𝜌𝜌𝑎𝑎 𝑏𝑏 2 = 𝑀𝑀𝑎𝑎 𝜌𝜌𝑎𝑎 (𝑏𝑏 2 − 𝑎𝑎2 )

(16)

The average temperature ratio is given by ��𝑎𝑎 ∆𝑇𝑇 = ∆𝑇𝑇𝑎𝑎 Finally,

𝑞𝑞 ‴ 𝑏𝑏 2 8𝑘𝑘

𝑞𝑞 ‴ 𝑏𝑏 4 3 𝑏𝑏 � − 𝑎𝑎2 𝑏𝑏 2 + 𝑎𝑎4 + 𝑎𝑎4 ln � 2 2 4 𝑎𝑎 2𝑘𝑘(𝑏𝑏 −𝑎𝑎 ) 4 =�

= 𝑏𝑏 2

(17) −1

𝑏𝑏 − 𝑎𝑎 𝑏𝑏 3 𝑏𝑏 � � − 𝑎𝑎2 𝑏𝑏 2 + 𝑎𝑎4 + 𝑎𝑎4 ln � 4 4 4 𝑎𝑎

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Chapter 8 - Thermal Analysis of Fuel Elements −1 𝑐𝑐𝑝𝑝𝑝𝑝 𝑀𝑀𝑠𝑠 �� ∆𝑇𝑇𝑠𝑠 𝑐𝑐𝑝𝑝𝑝𝑝 𝜌𝜌𝑠𝑠 �� ∆𝑇𝑇𝑠𝑠 𝑏𝑏 4 3 4 𝑏𝑏 2 2 4 = � − 𝑎𝑎 𝑏𝑏 + 𝑎𝑎 + 𝑎𝑎 ln � ��𝑎𝑎 𝑐𝑐𝑝𝑝𝑝𝑝 𝜌𝜌𝑎𝑎 ∆𝑇𝑇 �� 𝑐𝑐𝑝𝑝𝑝𝑝 𝑀𝑀𝑠𝑠 ∆𝑇𝑇 4 𝑎𝑎 𝑎𝑎 4

(18)

PROBLEM 8.10 QUESTION

Maximum Linear Power from a Duplex Fuel Pellet (Section 8.5) To uprate the power of its Light Water Reactor fleet, an electric utility is considering the use of duplex fuel pellets. A duplex fuel pellet consists of two radial zones, one loaded with UO2 and one with PuO2 (Figure 8.32). The pellet outer temperature is fixed at 400 °C.

FIGURE 8.32 Cross-sectional view of a duplex fuel pellet. 1. Assuming that centerline fuel melting is the design limit for this pellet, which oxide would you put in Zone 1 (i.e., the inner zone)? 2. Using only the properties in Table 8.11 and assuming that the volumetric heat generation rate in PuO2 is 50% higher than in UO2, calculate the maximum linear power at which the pellet can be operated without melting the fuel? (Neglect the thermal conductivity dependence on temperature) 3. How does the maximum linear power in (2) compare with the maximum linear power for an all-UO2 pellet? TABLE 8.11 Properties of Oxide Fuels. Parameter UO2 PuO2 Density (g/cm3)

10.5

10.9

Thermal conductivity (W/m ∙° C)

3.0

2.5

Melting Point (°C)

2,800

2,300

Specific heat (J/kg∙°C)

410

380

Answers: 1. UO2 2. 95.6 kW/m 3. 90.5 kW/m

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PROBLEM 8.10 SOLUTION Maximum Linear Power from a Duplex Fuel Pellet (Section 8.5) From the problem statement, we are given that (uranium denoted with subscript “U” and plutonium with “Pu”: –

density, ρU = 10.5 g/cm3 and ρPU = 10.9 g/cm3

–

conductivity, kU = 3.0 W/m K and kPu = 2.5 W/m K

–

maximum temperature, TU = 2800 °C and Tpu = 2300 °C

–

specific heat, cpu = 410 J/kg K and Cppu = 380 J/kg K

–

fuel surface temperature, Tf0 = 400 °C

–

radius of zone 1, R1 = 4 mm

–

radius of fuel, Rf0 = 5 mm

1. Because the UO2 has a higher melting point, it should be placed in Zone 1. 2. The linear power for the duplex pellet can be expressed as: 2 𝑞𝑞 ′ = 𝜋𝜋𝑅𝑅12 𝑞𝑞1‴ + 𝜋𝜋�𝑅𝑅𝑓𝑓𝑓𝑓 − 𝑅𝑅12 �𝑞𝑞2‴ = �𝑅𝑅12 +

𝑞𝑞2‴ 2 �𝑅𝑅 − 𝑅𝑅12 �� 𝑞𝑞1‴ 𝑞𝑞1‴ 𝑓𝑓𝑓𝑓

(1)

where 𝑞𝑞1‴ and 𝑞𝑞2‴ are the volumetric heat generation rates in Zone 1 and 2, respectively, and 𝑞𝑞2‴ ⁄𝑞𝑞1‴ 1.5. To solve the problem, it is necessary to establish a relationship between 𝑞𝑞1‴ and the max temperature in the pellet. To do so, one needs to solve the heat conduction equation in both zones. Starting from Zone 2 (i.e., PuO2): 1 𝑑𝑑 𝑑𝑑𝑑𝑑 �𝑟𝑟𝑘𝑘2 � + 𝑞𝑞2‴ = 0 𝑟𝑟 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑

(2)

where k2 is the PuO2 thermal conductivity. Integrating twice and setting the boundary conditions 2𝜋𝜋𝑅𝑅1 𝑘𝑘2

and

𝑑𝑑𝑑𝑑 � = 𝑞𝑞1‴ 𝜋𝜋 𝑅𝑅12 𝑑𝑑𝑑𝑑 𝑅𝑅1

𝑇𝑇|𝑅𝑅𝑓𝑓𝑓𝑓 = 𝑇𝑇𝑅𝑅𝑓𝑓𝑓𝑓

one gets: 𝑇𝑇1 − 𝑇𝑇𝑅𝑅𝑓𝑓𝑓𝑓 = −

𝑅𝑅𝑓𝑓𝑓𝑓 𝑅𝑅12 ‴ 𝑞𝑞 ‴ 2 (𝑞𝑞2 − 𝑞𝑞1‴ ) ln � � + 2 �𝑅𝑅𝑓𝑓𝑓𝑓 − 𝑅𝑅12 � 2𝑘𝑘2 𝑅𝑅1 4𝑘𝑘2

(3)

(4)

(5)

where T1 is the temperature at R1. For Zone 1 the heat conduction equation yields:

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1 𝑑𝑑 𝑑𝑑𝑑𝑑 �𝑟𝑟𝑘𝑘1 � + 𝑞𝑞1‴ = 0 𝑟𝑟 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑

(6)

Integrating twice and setting the boundary conditions −𝑘𝑘1

and

𝑑𝑑𝑑𝑑 � =0 𝑑𝑑𝑑𝑑 0

𝑇𝑇|0 = 𝑇𝑇𝑐𝑐𝑐𝑐

one gets:

𝑇𝑇𝑐𝑐𝑐𝑐 = 𝑇𝑇1 =

From Equations (5) and (9) it follows that: 𝑇𝑇𝑐𝑐𝑐𝑐 = 𝑇𝑇𝑓𝑓𝑓𝑓 = 𝑞𝑞1‴ �

𝑞𝑞1‴ 𝑅𝑅12 4𝑘𝑘1

𝑅𝑅𝑓𝑓𝑓𝑓 𝑅𝑅12 𝑅𝑅12 𝑞𝑞2‴ 1 𝑞𝑞2‴ 2 − � − 1� ln � � + �𝑅𝑅 − 𝑅𝑅12 �� 4𝑘𝑘1 2𝑘𝑘2 𝑞𝑞1‴ 𝑅𝑅1 4𝑘𝑘2 𝑞𝑞1‴ 𝑓𝑓𝑓𝑓

(7)

(8)

(9)

(10)

For Tcl = 2800 °C, Equation (3) yields 𝑞𝑞1‴ and from Equation (1) it is easy to get the linear power, q′ = 95.6 kW/m. 3. For the traditional pellet the linear power is:

𝑞𝑞 ′ = 4𝜋𝜋𝑘𝑘1 �𝑇𝑇𝑐𝑐𝑐𝑐 −𝑇𝑇𝑓𝑓𝑓𝑓 � = 90.5k W⁄m which is somewhat lower than for the duplex pellet.

(11)

PROBLEM 8.11 QUESTION Effect of Internal Cooling on Fuel Temperature (Section 8.5) Consider the following three UO2 pellet configurations: – Solid pellet – Annular pellet with only external cooling – Annular pellet with simultaneous internal and external cooling The dimensions for all three pellets are in Table 8.12. Assume that the fuel thermal conductivity is kf =3 W/m K (independent of temperature), the pellet surface temperature is 700 °C and the linear power is q’ = 40 kW/m in all three cases.

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Table 8.12 Geometry Of The Three Uo2 Pellets ID (mm)

OD (mm)

Solid pellet

N/A

8.2

Annular pellet with only external cooling

2.0

8.44

Annular pellet with internal and external cooling

9.9

14.1

1. Calculate the maximum temperature for the solid pellet. 2. Calculate the maximum temperature for the annular pellet with only external cooling. 3. Calculate the maximum temperature for the annular pellet with simultaneous external and internal cooling, 4. For the annular pellet with simultaneous internal and external cooling calculate also the heat flux at the inner and outer surfaces. 5. What are the advantages and drawbacks of the annular fuel pellet with simultaneous internal and external cooling? Answers: 1. Tmax = 1761 °C 2. Tmax = 1579 °C 3. Tmax = 793 °C 4. Inner: q″ = − 567.9 kW/m2

Outer: q″ = 504.3 kW/m2

PROBLEM 8.11 S OLUTION Effect of Internal Cooling on Fuel Temperature (Section 8.5) From the problem statement we are given that: – conductivity, kf = 3 W/m K – fuel pellet surface temperature, Tf0 = 700 °C – linear heat rate, q′ = 40kW/m Dimensions from Table 8.12: – solid pellet diameter, D1 = 8.2 mm – externally cooled annular pellet, D2i = 2.0 mm and D2o = 8.44 mm – annular pellet with internal and external cooling, D3i = 9.9 mm and D3o = 14.1mm

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1. For a solid fuel pellet the maximum temperature can be calculated with Equation 8.71b 𝑞𝑞 ′ 𝑇𝑇𝑚𝑚𝑚𝑚𝑚𝑚1 = 𝑇𝑇𝑓𝑓𝑓𝑓 + = 1761°𝐶𝐶 4𝜋𝜋𝑘𝑘𝑓𝑓

(1)

2. For the annular pellet with external cooling the maximum temperature is given by Equation 8.75 𝑞𝑞 ′ 𝑙𝑙𝑙𝑙(𝐷𝐷20 ⁄𝐷𝐷2𝑖𝑖 )2 𝑇𝑇𝑚𝑚𝑚𝑚𝑚𝑚2 = 𝑇𝑇𝑓𝑓𝑓𝑓 + =� � = 1579°𝐶𝐶 (𝐷𝐷20 ⁄𝐷𝐷2𝑖𝑖 )2 − 1 4𝜋𝜋𝑘𝑘𝑓𝑓

(2)

3. For the annular pellet with both external and internal cooling, the location of the maximum temperature when both surfaces are held at the same temperature is given by Equation 8.83b

𝑅𝑅0 = �

𝐷𝐷 2 𝐷𝐷 2 � 230 � − � 23𝑖𝑖 � 𝐷𝐷 2 𝑙𝑙𝑙𝑙 � 𝐷𝐷30 � 3𝑖𝑖

(3)

= 5969 𝑚𝑚𝑚𝑚

From Equations 8.86 and 8.87, the maximum temperature is given by the following formulation: 𝑅𝑅 𝐷𝐷 2 𝑙𝑙𝑙𝑙 �2 𝐷𝐷 𝑜𝑜 � � 23𝑖𝑖 � − 𝑅𝑅𝑜𝑜2 𝑞𝑞 ′ 3𝑖𝑖 𝑇𝑇𝑚𝑚𝑚𝑚𝑚𝑚2 = 𝑇𝑇𝑓𝑓𝑓𝑓 + =� + � = 793 °𝐶𝐶 2 2 𝐷𝐷 4𝜋𝜋𝑘𝑘𝑓𝑓 3𝑜𝑜 𝐷𝐷30 𝐷𝐷3𝑖𝑖 � � 𝑙𝑙𝑙𝑙 � 2 � − � 2 � 𝐷𝐷3𝑖𝑖

(4)

The temperature distribution in the annular pellet can be derived from Eq, 8.81 to be 𝑟𝑟 � 𝑅𝑅𝑓𝑓𝑓𝑓 𝑞𝑞 2 2 2 2 � 𝑇𝑇(𝑟𝑟) = 𝑇𝑇𝑓𝑓𝑓𝑓 + = �𝑅𝑅𝑓𝑓𝑓𝑓 − 𝑟𝑟 + �𝑅𝑅𝑓𝑓𝑓𝑓 − 𝑅𝑅𝑓𝑓𝑓𝑓 � 2 2 𝐷𝐷3𝑜𝑜 4𝜋𝜋�𝑅𝑅𝑓𝑓𝑓𝑓 − 𝑅𝑅𝑓𝑓𝑓𝑓 � 𝑙𝑙𝑙𝑙 � 𝐷𝐷3𝑖𝑖 � 𝑙𝑙𝑙𝑙 �2

′

(5)

where Rf0 = 0.5D3o and Rfi = 0.51D3i. Talking the derivative of this expression, we get: 2 2 �𝑅𝑅𝑓𝑓𝑓𝑓 − 𝑅𝑅𝑓𝑓𝑓𝑓 � 𝑞𝑞 ′ 𝑑𝑑𝑑𝑑 2 2 ⎛𝑅𝑅𝑓𝑓𝑓𝑓 ⎞ = = − 𝑟𝑟 + 2 2 𝑅𝑅𝑓𝑓𝑓𝑓 𝑑𝑑𝑑𝑑 4𝜋𝜋�𝑅𝑅𝑓𝑓𝑓𝑓 − 𝑅𝑅𝑓𝑓𝑓𝑓 � 𝑙𝑙𝑙𝑙 � 𝑅𝑅 � ⎝ ⎠ 𝑓𝑓𝑓𝑓

(6)

The inner heat flux can therefore be determined with

and the outer heat flux as

𝑞𝑞𝑖𝑖″ = −𝑘𝑘𝑓𝑓

𝑑𝑑𝑑𝑑 � = −567.9 k W⁄m2 𝑑𝑑𝑑𝑑 𝑅𝑅𝑓𝑓𝑓𝑓

𝑞𝑞0″ = −𝑘𝑘𝑓𝑓

𝑑𝑑𝑑𝑑 � = 504.3 k W⁄m2 𝑑𝑑𝑑𝑑 𝑅𝑅𝑓𝑓𝑓𝑓

206

(7)

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PROBLEM 8.12 Q UESTION Temperature Field in a Restructured Fuel Pin (Section 8.6) Using the conditions of Problem 8.4 for the uncracked fuel, calculate the maximum fuel temperature for the given operating conditions. Assume two-zone sintering, with Tsintering = 1700 °C and ρsintered = 98% TD. Answer: Tv = 2278 K

PROBLEM 8.12 S OLUTION Temperature Field in a Restructured Fuel Pin (Section 8.6) From the problem statement and Problem 8.4 we are given that: – outer diameter, Dco = 11.20 mm – clad thickness, tc = 0.71 mm −6

– gap thickness, tg = 90 × 10

– clad conductivity, kc = 17 W/m K – gap conductance, hg = 4300 W/m

– fuel conductivity (95%), kf95 = 2.7W/m K – clad outer temperature, Tco = 295 °C – linear heat rate, q’ = 44kW/m – temperature at which sintering begins, Tsintering = 1700 °C – density of sintered region, ρsintered = 98% TD

First, using Biancharia’s porosity correction the fuel conductivity at 88% TD and 98% TD are

and

0.88 1 + (1.5 − 1)(1 − 0.88) 𝑘𝑘𝑓𝑓88 = 𝑘𝑘𝑓𝑓95 = 2.418 W⁄m K 0.95 1 + (1.5 − 1)(1 − 0.95) 0.98 1 + (1.5 − 1)(1 − 0.898) 𝑘𝑘𝑓𝑓98 = 𝑘𝑘𝑓𝑓95 = 2.827 W⁄m K 0.95 1 + (1.5 − 1)(1 − 0.95)

(1)

(2)

The results from the analysis in Problem 8.4 showed that:

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– outer fuel pellet temperature, Tf0 = 664 K – outer clad radius, Rco = 5.6mm – inner clad radius, Rci =4.89 mm – fuel pellet radius, Rf0 = 4.8mm – mean gap radius, Rg = 4.845 mm Using Equation 8.129, the radius at which sintering begins can be determined to be 2

𝑞𝑞 ′ 𝑅𝑅𝑠𝑠 𝑘𝑘𝑓𝑓88 �𝑇𝑇𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 − 𝑇𝑇𝑓𝑓𝑓𝑓 � = �1 − � � � 4𝜋𝜋 𝑅𝑅𝑓𝑓𝑓𝑓

(3)

Solving for the Rs we get

𝑅𝑅𝑠𝑠 = 2.56mm

(4)

Using Equation 8.128 the radius of the inner void is 𝑅𝑅𝜐𝜐 = �

𝜌𝜌98 − 𝜌𝜌88 2 𝑅𝑅𝑠𝑠 = 8.177 × 10−4 m 𝜌𝜌98

(5)

Finally using Equation 8.130 the maximum temperature can be determined to be 2

𝑞𝑞 ′ 𝑅𝑅𝜐𝜐 2 𝑅𝑅𝑠𝑠 2 𝜌𝜌98 𝑅𝑅𝑠𝑠 � � � � �1 − � � �1 + ln � � �� 𝑇𝑇𝜐𝜐 = 𝑇𝑇𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 + 4𝜋𝜋𝑘𝑘𝑓𝑓88 𝜌𝜌88 𝑅𝑅𝑓𝑓𝑓𝑓 𝑅𝑅𝑠𝑠 𝑅𝑅𝜐𝜐

(6)

2 2 2 44000  0.98  2.56    0.82    2.56    =+ 1973    1 −   1 + ln    4π ( 2.418 )  0.88  4.8    2.56    0.82   

= 1973 + 305 = 2278 K 𝜌𝜌

Note that the ratio, 𝜌𝜌98 is just 0.98/0.88 since the theoretical density will cancel out of the ratio. 88

PROBLEM 8.13 QUESTION

Eccentricity Effects in a Plate-type Fuel (Section 8.7) A nuclear fuel element is of plate geometry (Figure 8.33). It is desired to investigate the effects of fuel offset within the cladding. For simplicity, assume uniform heat generation in the fuel, temperature-independent fuel conductivity and heat conduction only in the gap.

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FIGURE 8.33 Effect of fuel offset. (a) Centered fuel pellet. (b) Effect of fuel pellet offset. Calculate: 1. The temperature difference between the offset fuel and the concentric fuel maximum temperatures: 𝑇𝑇𝑀𝑀𝑀𝑀 − 𝑇𝑇𝑀𝑀𝑀𝑀

2. The temperature difference between the cladding maximum temperatures: 3. The ratio of heat fluxes to the coolant. 𝑞𝑞″ 𝑞𝑞1″

𝑇𝑇7 − 𝑇𝑇4 𝑎𝑎𝑎𝑎𝑎𝑎

𝑞𝑞″ 𝑞𝑞2″

Tcoolant and heat transfer coefficient to coolant may be assumed constant on both sides and for both cases. Neglect interface contact resistance for fuel and cladding. kf = 3.011 W/m °C (PuO2 – UO2) kg= 0.289 W/m °C (He)

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kc= 21,63 W/m °C (SS) D = 6.352 mm Dg= 6.428 mm tc= 0.4054 mm q‴ = 9.313 × 105 kW/m3 hcoolant = 113.6 kW/m2 °C (Na) Answers: 1. TMO – TMC = −23.8 °C and equally -23.8 K 2. T7 −T4 = -8.83 °C and equally -8.83 K q″

q″

3. q″ = 0.902 and q″ = 1.121 1

PROBLEM 8.13 S OLUTION

Eccentricity Effects in a Plate-type Fuel (Section 8.7) From the problem statement we are given that: – fuel conductivity, kf = 3.011 W/m K – gap conductivity, kg = 0.289 W/m K – clad conductivity, kc = 21.63 W/m K – length of pellet , D = 6.352 mm – length between inner cladding, Dg = 6.428 mm – clad thickness, tc = 0.4054 mm – volumetric heat generation rate, q‴ = 9.313 × 105 kW/m3 – heat transfer coefficient to coolant, hc = 113.6 kW/m2 K Eccentric Case: For the eccentric case, the geometric parameters are as follows: – radius to inner clad: Rg = Dg/2 = 3.214mm – gap spacing, δg = Rg — D/2 = 0.038 mm – off radius of fuel, Rf = Rg – 2δg = 3.138 mm – fuel element radius, Rco = Rg +tc = 3.619 mm The temperature distribution in a one-dimensional geometry is 𝑇𝑇(𝑥𝑥) = −

𝑞𝑞 ‴ 2 𝑥𝑥 + 𝐶𝐶1 𝑥𝑥 + 𝐶𝐶2 2𝑘𝑘𝑓𝑓 210

(1)

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Chapter 8 - Thermal Analysis of Fuel Elements

The temperature drop through the gap is 𝑇𝑇1 − 𝑇𝑇2 = 𝑞𝑞2‴

𝑅𝑅𝑔𝑔 − 𝑅𝑅𝑓𝑓 𝑘𝑘𝑔𝑔

(2)

The temperature drop through the right hand side cladding is 𝑇𝑇2 − 𝑇𝑇3 = 𝑞𝑞2‴

𝑅𝑅𝑐𝑐𝑐𝑐 − 𝑅𝑅𝑔𝑔 𝑘𝑘𝑐𝑐

(3)

Finally, the temperature drop through the right hand side coolant is 𝑇𝑇3 − 𝑇𝑇𝑚𝑚 = 𝑞𝑞2‴

1 ℎ𝑐𝑐

(4)

Applying the two boundary conditions to Equation (1):

and

𝑇𝑇1 − 𝑇𝑇�𝑅𝑅𝑓𝑓 � = − 𝑞𝑞2‴ = −𝑘𝑘𝑓𝑓

Combining the last five equations

𝑞𝑞2‴ 2 𝑅𝑅 + 𝐶𝐶1 𝑅𝑅𝑓𝑓 + 𝐶𝐶2 2𝑘𝑘𝑓𝑓 𝑓𝑓

𝑑𝑑𝑑𝑑 � = 𝑞𝑞 ‴𝑅𝑅𝑓𝑓 − 𝑘𝑘𝑓𝑓 𝐶𝐶1 𝑑𝑑𝑑𝑑 𝑅𝑅𝑓𝑓

𝑅𝑅𝑔𝑔 − 𝑅𝑅𝑓𝑓 𝑅𝑅𝑐𝑐𝑐𝑐 − 𝑅𝑅𝑔𝑔 1 𝑞𝑞 ‴ 2 − 𝑅𝑅𝑓𝑓 + 𝐶𝐶1 𝑅𝑅𝑓𝑓 + 𝐶𝐶2 = �𝑞𝑞 ‴𝑅𝑅𝑓𝑓 − 𝑘𝑘𝑓𝑓 𝐶𝐶1 � � + + � + 𝑇𝑇𝑚𝑚 2𝑘𝑘𝑓𝑓 𝑘𝑘𝑔𝑔 𝑘𝑘𝑐𝑐 ℎ𝑐𝑐

(5)

(6)

(7)

For the left hand side heat flow:

The temperature through the left hand side cladding 𝑇𝑇4 − 𝑇𝑇5 = −𝑞𝑞1″

𝑅𝑅𝑐𝑐𝑐𝑐 − 𝑅𝑅𝑔𝑔 𝑘𝑘𝑐𝑐

(8)

and the temperature drop through the left hand side coolant is 𝑇𝑇5 − 𝑇𝑇𝑚𝑚 = −𝑞𝑞″

1 ℎ𝑐𝑐

(9)

Using Equation (1) we can apply the following boundary conditions:

and

𝑞𝑞 ‴ 2 𝑇𝑇4 = 𝑇𝑇�−𝑅𝑅𝑔𝑔 � = − 𝑅𝑅 − 𝐶𝐶1 𝑅𝑅𝑔𝑔 + 𝐶𝐶2 2𝑘𝑘𝑓𝑓 𝑔𝑔

211

(10)

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𝑞𝑞1″ = 𝑘𝑘𝑓𝑓

𝑑𝑑𝑑𝑑 � = −𝑞𝑞 ‴ 𝑅𝑅𝑔𝑔 − 𝑘𝑘𝑓𝑓 𝐶𝐶2 𝑑𝑑𝑑𝑑 −𝑅𝑅𝑔𝑔

(11)

Summing the last 4 equations we get −

𝑅𝑅𝑐𝑐𝑐𝑐 − 𝑅𝑅𝑔𝑔 1 𝑞𝑞 ‴ 2 𝑅𝑅𝑔𝑔 − 𝐶𝐶1 𝑅𝑅𝑔𝑔 + 𝐶𝐶2 = �−𝑞𝑞 ‴𝑅𝑅𝑔𝑔 − 𝑘𝑘𝑓𝑓 � � + � 2𝑘𝑘𝑓𝑓 𝑘𝑘𝑐𝑐 ℎ𝑐𝑐

(12)

Solving Equations (7) and (12) simultaneously we can obtain the integration constants C1 and C2 if we guess Tm = 500 K: 𝐶𝐶1 = 9.466 × 104

𝐶𝐶2 = 2.492 × 103

(13)

Now, we can take the derivative of Equation (1) with these integration constants and find the x location where the temperature is at a maximum: 𝑑𝑑𝑑𝑑 𝑞𝑞 ‴ =− 𝑥𝑥 + 𝐶𝐶1 = 0 𝑑𝑑𝑑𝑑 𝑘𝑘𝑓𝑓 𝑥𝑥0 = 3.06 × 10−4

(14) (15)

The temperature at this location, assuming Tm = 500 K is TMO = 2506.5 K. Concentric Case: For the concentric case we can just perform a simple resistance calculation 𝑅𝑅𝑓𝑓𝑓𝑓 𝑅𝑅𝑔𝑔 − 𝑅𝑅𝑓𝑓𝑓𝑓 𝑅𝑅𝑐𝑐𝑐𝑐 − 𝑅𝑅𝑔𝑔 1 𝑇𝑇𝑀𝑀𝑀𝑀 = 𝑞𝑞 ‴ 𝑅𝑅𝑓𝑓𝑓𝑓 � + + + � + 𝑇𝑇𝑚𝑚 = 2530.3 K 2𝑘𝑘𝑓𝑓 𝑘𝑘𝑔𝑔 𝑘𝑘𝑐𝑐 ℎ𝑐𝑐

(16)

where Rfo = 0.5D = 3.176 mm. Therefore,

𝑇𝑇𝑀𝑀𝑀𝑀 − 𝑇𝑇𝑀𝑀𝑀𝑀 = −23.8 K or equally − 23.8 ˚C

(17)

𝑇𝑇4 = 𝑇𝑇�−𝑅𝑅𝑔𝑔 � = 590.30 K

(18)

The temperature at point 4 for the eccentric can be determined by evaluating Equation (1) at x = −Rg so that

For the concentric case, point 7 is

Therefore,

𝑅𝑅𝑐𝑐𝑐𝑐 − 𝑅𝑅𝑔𝑔 1 𝑇𝑇7 = 𝑞𝑞‴𝑅𝑅𝑓𝑓𝑓𝑓 � + � + 𝑇𝑇𝑚𝑚 = 581.47 K 𝑘𝑘𝑐𝑐 ℎ𝑐𝑐

(19)

𝑇𝑇7 − 𝑇𝑇4 = −8.83 K or equally − 8.83 ˚C

(20)

𝑞𝑞1″ = �−𝑞𝑞‴𝑅𝑅𝑔𝑔 − 𝑘𝑘𝑓𝑓 𝐶𝐶1 � = 3.278 × 106 W⁄m2

(21)

𝑞𝑞2″ = 𝑞𝑞 ‴ 𝑅𝑅𝑓𝑓 − 𝑘𝑘𝑓𝑓 𝐶𝐶1 = 2.637 × 106 W⁄m2

(22)

Finally, the heat flux ratios for the eccentric case are (from boundary conditions)

and

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Note that an absolute value was applied to heat flux 1 since it is directed in the − x direction. The heat flux in the concentric case is Therefore the ratios are

𝑞𝑞 ″ = 𝑞𝑞 ‴ 𝑅𝑅𝑓𝑓𝑓𝑓 = 2.958 × W⁄m2 𝑞𝑞 ″ = 0.902 𝑞𝑞1″

𝑞𝑞 ″ = 1.121 𝑞𝑞2″

(23)

(24)

PROBLEM 8.14 QUESTION Determining the Linear Power Given a Constraint on the Fuel Average Temperature (Section 8.7) For a PWR fuel pin with pellet radius of 4.096 mm, cladding inner radius of 4.178 mm, and outer radius of 4.75 mm, calculate the maximum linear power that can be obtained from the pellet such that the mass average temperature in the fuel does not exceed 1204 °C (2200 °F). Take the bulk fluid temperature to be 307.5 °C and the coolant heat transfer coefficient to be 28,4 kW/m2 °C. Consider only conduction based on the effective gap width using the gap conductance model. Fuel conductivity kf = 3.011 W/m °C Cladding conductivity kc = 18.69 W/m °C Helium gas conductivity kg = 0.277 W/m °C Answer: q′ = 35.2 kW/m

PROBLEM 8.14 S OLUTION Determining the Linear Power Given a Constraint on the Fuel Average Temperature (Section 8.7) From the problem statement we are given that: –

fuel pellet radius, Rfo = 4.096mm

–

inner clad radius, Rci = 4.178 mm

–

outer clad radius, Rco = 4.75 mm

–

maximum mass average temperature, Tmax = 1204 °C

–

bulk fluid temperature, Tb = 307.5 °C

–

coolant heat transfer coefficient, hc = 28.4 kW/m2 K

–

fuel conductivity, kf = 3.011 W/m K

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–

clad conductivity, kc = 18.69 W/m K

–

gap conductivity, kg = 0.277 W/m K The jump distance for He at 1 atm is 𝛿𝛿𝑒𝑒𝑒𝑒𝑒𝑒 = 10 × 10−6 m

(1)

Therefore; the gap conductance (conduction only) can be estimated to be ℎ𝑔𝑔 =

The mean gap radius is

𝑘𝑘𝑔𝑔 = 3.842 × 103 W⁄m2 K 𝑅𝑅𝑐𝑐𝑐𝑐 − 𝑅𝑅𝑓𝑓 − 𝛿𝛿𝑒𝑒𝑒𝑒𝑒𝑒 𝑅𝑅𝑔𝑔 = 0.5�𝑅𝑅𝑓𝑓𝑓𝑓 + 𝑅𝑅𝑐𝑐𝑐𝑐 � = 4.137 mm

(2)

(3)

Therefore the linear heat rate is 𝑞𝑞 ′ =

𝑇𝑇𝑚𝑚𝑚𝑚𝑚𝑚 − 𝑇𝑇𝑏𝑏 = 35.2 kW⁄m 𝑅𝑅𝑐𝑐𝑐𝑐 𝑙𝑙𝑙𝑙 � 𝑅𝑅 � 1 1 1 𝑐𝑐𝑐𝑐 + + + 2𝜋𝜋𝑅𝑅𝑐𝑐𝑐𝑐 ℎ𝑐𝑐 2𝜋𝜋𝑘𝑘𝑐𝑐 2𝜋𝜋𝑅𝑅𝑔𝑔 ℎ𝑔𝑔 8𝜋𝜋𝜋𝜋𝑓𝑓

(4)

Note that the denominator represents the standard convective and conductive resistance networks. The last term represents the resistance in order to get the maximum mass average temperature.

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Chapter 9 Single-Phase Fluid Mechanics Contents Problem 9.1 Emptying of a liquid tank ................................................................................ 216 Problem 9.2 Laminar flow velocity distribution and pressure drop in parallel plate geometry ........................................................................................................................... 217 Problem 9.3 Velocity distribution in single-phase turbulent flow ....................................... 220 Problem 9.4 Flow split in downflow .................................................................................... 223 Problem 9.5 Sizing of an orificing device ............................................................................ 224 Problem 9.6 Pressure drop features of a PWR core ............................................................. 228 Problem 9.7 Comparison of laminar and turbulent friction factors of water and sodium heat exchangers ........................................................................................................ 231 Problem 9.8 Using the UCTD correlation to calculate the pressure drop for wire-wrapped and bare rod bundles ............................................................................................... 234

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PROBLEM 9.1 QUESTION Emptying of a Liquid Tank (Section 9.2) Consider an emergency water tank, shown in Figure 9.40, that is designed to deliver water to a reactor following a loss of coolant event. The tank is prepressurized by nitrogen at 1.0 MPa. The water is discharged through an 0.2 (inner diameter) pipe. What is the maximum flow rate delivered to the reactor if the flow is considered inviscid and the reactor pressure is: 1. 0.8 MPa 2. 0.2 Mpa

FIGURE 9.40 Emergency water tank. Answers: 1. m 1 = 827.3 kg / s 2. m 2 = 1366.6 kg / s

PROBLEM 9.1 SOLUTION Emptying of a Liquid Tank (Section 9.2) Consider an emergency water tank, shown in Figure 9.40, that is designed to deliver water to a reactor following a loss of coolant event. The tank is prepressurized by nitrogen at 1.0 MPa. Assume the water discharge through an 0.2 m (inner diameter) pipe is quasi-steady. What is the maximum flow rate delivered to the reactor if the flow is considered inviscid and the reactor pressure is:

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1. 0.8 MPa 2. 0.2 MPa The following parameters are given in the problem statement: – pressure of nitrogen tank, po = 1.0 MPa – pipe diameter, d = 0.2 m – height, H = 15 m – pressure 1, P1 = 0.8 MPa – pressure 2, P2 = 0.2 MPa The density of water is assumed to 1 g/cm3. The area of the discharge pipe is

The pressure at the pipe is

𝐴𝐴 =

𝜋𝜋 2 𝑑𝑑 = 0.0314 m2 4

𝑃𝑃𝑜𝑜𝑜𝑜 = 𝑃𝑃𝑜𝑜 + 𝜌𝜌𝜌𝜌𝜌𝜌 = 1.147MPa

(1)

(2)

The velocity can be calculated from Bernoulli's law as 𝜐𝜐𝑖𝑖 = �2

𝑃𝑃𝑜𝑜𝑜𝑜 − 𝑃𝑃𝑖𝑖 𝜌𝜌

(3)

Using pressure's 1 and 2, the corresponding velocities are 𝜐𝜐1 = 26.348 m⁄s

The mass flow rate can be calculated with

𝜐𝜐2 = 43.522 m⁄s

(4)

𝑚𝑚̇𝑖𝑖 = 𝜌𝜌𝜐𝜐𝑖𝑖 𝐴𝐴

(5)

𝑚𝑚̇1 = 827.3 kg⁄s 𝑚𝑚̇2 = 1366.6 kg⁄s

(6)

The corresponding mass flow rates are

PROBLEM 9.2 QUESTION

Laminar Flow Velocity Distribution and Pressure Drop in Parallel Plate Geometry (Section 9.4) For flow between parallel flat plates: 1. Show that the momentum equation for fully developed, steady-state, constant density and viscosity flow takes the form:

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𝑑𝑑𝑑𝑑 𝑑𝑑2 υ𝑥𝑥 = 𝜇𝜇 𝑑𝑑𝑑𝑑 𝑑𝑑𝑦𝑦 2

2. For plate separation of 2y0 show that the velocity profile is given by: 3 𝑦𝑦 2 𝜐𝜐𝑥𝑥 (𝑦𝑦) = 𝑉𝑉𝑚𝑚 �1 − � � � 2 𝑦𝑦𝑜𝑜

3. Show that:

−

𝑑𝑑𝑑𝑑 96 1 𝜌𝜌𝑉𝑉𝑚𝑚2 = 𝑑𝑑𝑑𝑑 𝑅𝑅𝑅𝑅 4𝑦𝑦𝑜𝑜 2

PROBLEM 9.2 SOLUTION Laminar Flow Velocity Distribution and Pressure Drop in Parallel Plate Geometry (Section 9.4) For fully developed laminar flow, there is no change in the velocity along the longitudinal flow direction x, (1) For fully developed laminar flow, the pressure gradients in the y and z direction are also zero, 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 = =0 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕

(2)

Therefore Equation 4.90a, assuming no gravity effects, becomes 𝜐𝜐𝑦𝑦

𝜕𝜕𝜐𝜐𝑥𝑥 𝜕𝜕𝜐𝜐𝑥𝑥 𝑑𝑑𝑑𝑑 𝜕𝜕 2 𝜐𝜐𝑥𝑥 𝜕𝜕 2 𝜐𝜐𝑥𝑥 𝜕𝜕 2 𝜐𝜐𝑥𝑥 + 𝜐𝜐𝑧𝑧 =− + 𝜇𝜇 � 2 + + � 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝑑𝑑𝑑𝑑 𝜕𝜕𝑥𝑥 𝜕𝜕𝑦𝑦 2 𝜕𝜕𝑧𝑧 2

(3)

Since the geometry is of infinite width in the z direction, 𝜕𝜕𝜐𝜐𝑥𝑥 =0 𝜕𝜕𝜕𝜕

From the continuity equation for incompressible flow, ∇ ∙ 𝜐𝜐⃗ = 0, 𝜕𝜕𝜐𝜐𝑦𝑦 =0 𝜕𝜕𝜕𝜕

(4)

(5)

Upon integration of Equation (5) and application of the boundary condition υy=0 at the wall, find that υy=0 throughout the flow region. Therefore, Equation (2) becomes 𝑑𝑑𝑑𝑑 𝜕𝜕 2 𝜐𝜐𝑥𝑥 = 𝜇𝜇 𝑑𝑑𝑑𝑑 𝜕𝜕𝑦𝑦 2 218

(6)

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This equation can be integrated twice to give 𝜐𝜐𝑥𝑥 (𝑦𝑦) =

Applying the boundary conditions that

1 𝑑𝑑𝑑𝑑 𝑦𝑦 2 = + 𝐶𝐶1 𝑦𝑦 + 𝐶𝐶2 𝜇𝜇 𝑑𝑑𝑑𝑑 2

𝜕𝜕𝜐𝜐

– at y = 0, 𝜕𝜕𝜕𝜕𝑥𝑥 = 0 which yields C1 = 0

– at y = yo, vx = 0 which yields 𝐶𝐶2 = −

yields the following expression

𝜐𝜐𝑥𝑥 (𝑦𝑦) =

(7)

1 𝑑𝑑𝑑𝑑 𝑦𝑦𝑜𝑜2

𝜇𝜇 𝑑𝑑𝑑𝑑 2

1 𝑑𝑑𝑑𝑑 𝑦𝑦 2 �− � 𝑦𝑦𝑜𝑜2 �1 − 2 � 2𝜇𝜇 𝑑𝑑𝑑𝑑 𝑦𝑦𝑜𝑜

(8)

The channel average flow velocity can be calculated as 𝑉𝑉𝑚𝑚 =

𝑦𝑦𝑜𝑜

∫−𝑦𝑦𝑜𝑜 𝜐𝜐𝑥𝑥 (𝑦𝑦)𝑑𝑑𝑑𝑑 1 1 𝑦𝑦𝑜𝑜 ∫−𝑦𝑦𝑜𝑜 𝑑𝑑𝑑𝑑

3𝜇𝜇

Therefore, the pressure gradient can be expressed as −

�−

𝑑𝑑𝑑𝑑 2 � 𝑦𝑦 𝑑𝑑𝑑𝑑 𝑜𝑜

𝑑𝑑𝑑𝑑 3𝜇𝜇 = 𝑉𝑉 𝑑𝑑𝑑𝑑 𝑦𝑦𝑜𝑜2 𝑚𝑚

(9)

(10)

The final expression can be obtained by substituting the pressure gradient into Equation (8) to get 𝜐𝜐𝑥𝑥 (𝑦𝑦) =

3 𝑦𝑦 2 𝑉𝑉 �1 − � � � 2 𝑚𝑚 𝑦𝑦𝑜𝑜

(11)

The pressure gradient can also be expressed as −

𝑑𝑑𝑑𝑑 𝑓𝑓 𝜌𝜌𝑉𝑉𝑚𝑚2 = 𝑑𝑑𝑑𝑑 𝐷𝐷𝑒𝑒 2

(12)

We can manipulate Equation (10) into the above form,

Therefore, by inspection,

−

𝑑𝑑𝑑𝑑 3𝜇𝜇 2 𝜌𝜌𝑉𝑉𝑚𝑚2 = 𝑑𝑑𝑑𝑑 𝑦𝑦𝑜𝑜2 𝜌𝜌𝜌𝜌𝑚𝑚 2 −

𝑓𝑓 3𝜇𝜇 2 = 2 𝐷𝐷𝑒𝑒 𝑦𝑦𝑜𝑜 𝜌𝜌𝜌𝜌𝑚𝑚

(13)

(14)

The hydraulic diameter can be calculated as 𝐷𝐷𝑒𝑒 =

4𝐴𝐴 4(2𝑦𝑦𝑜𝑜 ) = = 4𝑦𝑦𝑜𝑜 P𝑤𝑤 2 219

(15)

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Thus, 𝑓𝑓 =

The Reynolds number is given as 𝑅𝑅𝑒𝑒 =

24𝜇𝜇 𝜌𝜌𝑉𝑉𝑚𝑚 𝑦𝑦𝑜𝑜

(16)

𝜌𝜌𝑉𝑉𝑚𝑚 𝐷𝐷𝑒𝑒 𝜌𝜌𝑉𝑉𝑚𝑚 4𝑦𝑦𝑜𝑜 = 𝜇𝜇 𝜇𝜇

(17)

In order to use the Reynolds number in Equation (16), it must be multiplied and divided by 4 giving 𝑓𝑓 =

96𝜇𝜇 96 = 𝜌𝜌𝑉𝑉𝑚𝑚 4𝑦𝑦𝑜𝑜 𝑅𝑅𝑅𝑅

(18)

PROBLEM 9.3 QUESTION Velocity Distribution in Single-Phase Turbulent Flow (Section 9.5) Consider a smooth circular flow channel of diameter 13.5 mm (for a hydraulic simulation of flow through a PWR assembly). Operating Conditions: Assume two adiabatic, fully developed flow conditions: 1. High flow (mass flow rate = 0.5 kg/s) 2. Low flow (mass flow rate = 1 g/s) Properties (approximately those of pressurized water at 300 °C and 15.5 MPa) – Density = 720 kg/m3 – Viscosity = 91 μPa. s 1. For each flow condition draw a quantitative sketch to show the velocity distribution based on Martinelli's formalism (Equation 9.80a through 9.80c) 2. Find the positions of interfaces between the sublayer, the buffer zone, and the turbulent core. Answers: 2a. Laminar layer: 0 ≤ y ≤ 3.2 μm Buffer zone: 3.2 μm ≤ y ≤ 19.2 μm Turbulent core: 19.2 μm ≤ y ≤ 6.75 mm 2b. Laminar flow

PROBLEM 9.3 SOLUTION Velocity Distribution in Single-Phase Turbulent Flow (Section 9.5) From the problem statement, the following parameters are given:

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– High mass flow rate, ṁH = 0.5 kg/s – Low mass flow rate, ṀL = 10−3 kg/s – Channel diameter, d = 13.5 mm – Density, ρ = 720 kg/m3 – Viscosity, μ = 91 × 10−6 Pa s The flow area of the channel is 𝐴𝐴 =

𝜋𝜋 2 𝑑𝑑 = 1.431 × 10−4 m2 4

(1)

The high flow velocity and low flow velocity are respectively, 𝜐𝜐𝐻𝐻 =

𝑚𝑚̇𝐻𝐻 = 4.85 m⁄s 𝜌𝜌𝜌𝜌

𝜐𝜐𝐿𝐿 =

𝑚𝑚̇𝐿𝐿 = 0.0097 m⁄s 𝜌𝜌𝜌𝜌

(2)

The Reynolds number for the high and low flow are respectively, 𝑅𝑅𝑅𝑅𝐻𝐻 =

𝜌𝜌𝜌𝜌𝐻𝐻 𝑑𝑑 = 5.18 × 105 𝜇𝜇

𝑅𝑅𝑅𝑅𝐻𝐻 =

𝜌𝜌𝜌𝜌𝐿𝐿 𝑑𝑑 = 1036 𝜇𝜇

(3)

Therefore, the low flow case is laminar and will not be considered for the velocity boundary layer calculation. The friction factor for the high flow case can be calculated with the McAdam's correlation, 𝑓𝑓 = 0.184𝑅𝑅𝑅𝑅 −0.2 = 0.0132 From Equations 9.80a - 9.80c, the velocity profile is expressed as:

(4)

– laminar zone, 0 < y+ < 5 where v+ = y+

– buffer zone, 5 < y+ < 30 where v+ = −3.05 + 5.00 In (y+) – core zone, 30 > y+ where v+ = 5.5 + 2.5 In (y+) In this formulation, 𝑦𝑦 + =

𝑦𝑦𝑦𝑦 𝜏𝜏𝑤𝑤 𝑦𝑦𝑦𝑦𝑉𝑉𝑚𝑚 𝑓𝑓𝑤𝑤 � � = 𝜇𝜇 𝜌𝜌 2𝜇𝜇 2

(5)

where Vm represents the channel average velocity which was calculated in Equation (10). Solving for y+ and υ+, 𝑦𝑦 + =

𝑦𝑦𝑦𝑦𝑉𝑉𝑚𝑚 𝑓𝑓𝑤𝑤 � = 1.562 × 106 𝑦𝑦 2𝜇𝜇 2

(6)

There y can be expressed as a function of y+ with

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𝑦𝑦(𝑦𝑦

𝑦𝑦 + = 1.562 × 106

(7)

The boundaries of Equations 9.80a-9.80c can be used in the formula above to get the positions of interfaces between the sublayer, the buffer zone and the turbulent core, Laminar layer: Buffer zone:

0 ≤ y ≤ 3.2 μm

Turbulent core:

3.2 𝜇𝜇𝑚𝑚 ≤ 𝑦𝑦 ≤ 19.2 𝜇𝜇𝑚𝑚 19.2 μm ≤ y ≤ 6.75 mm

A sketch of each of these velocity profiles is shown in Figure SM-9.1.

Figure SM-9.1 Velocity profiles in turbulent flow

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PROBLEM 9.4 QUESTION Flow Split in Downflow (Section 9.5) High pressure water at mass flow rate ṁ enters a certain component having an inlet plenum and an outlet plenum, connected by two vertical tubes (Figure 9.41). These two tubes are identical except that one is heated and one is cooled. Is the mass flow rate in each tube the same? If not, which tube has the higher mass flow rate? Support your answer with an analytic proof. Assume the following: – The density of water decreases as temperature increases – Downflow exists in both tubes – The form and acceleration terms in the momentum equation are negligible – The friction factor is the same in both tubes – Single-phase and steady-state conditions apply

FIGURE 9.41

Parallel vertical heated channels.

Answer: The cooled channel has higher mass flow.

PROBLEM 9.4 SOLUTION Flow Split in Downflow (Section 9.5) High-pressure water at mass flow rate ṁ enters a certain component having an inlet plenum and an outlet plenum, connected by two vertical tubes (Figure 9.41). These two tubes are identical except that one is heated and one is cooled. Is the mass flow rate in each tube the same? Because the two channels are connected to the same inlet and outlet plena, the total pressure change in each channel, −ΔPtot is the same: −∆𝑃𝑃𝑡𝑡𝑡𝑡𝑡𝑡 = 𝑓𝑓

𝐿𝐿 𝐺𝐺12 − 𝜌𝜌1 𝑔𝑔𝑔𝑔 𝐷𝐷𝑒𝑒 2𝜌𝜌1 223

(1)

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𝐿𝐿 𝐺𝐺22 − 𝜌𝜌2 𝑔𝑔𝑔𝑔 −∆𝑃𝑃𝑡𝑡𝑡𝑡𝑡𝑡 = 𝑓𝑓 𝐷𝐷𝑒𝑒 2𝜌𝜌2

(2)

where the form and acceleration terms were neglected, f is the friction factor (assumed equal in both channels, as per the problem statement), De is the hydraulic diameter of the channels, G1 and G2 are the mass fluxes in channel 1 and 2, respectively, and ρ1 and ρ2 are the average water densities in channel 1 and 2, respectively. Eliminating −ΔPtot from Equation 1 and Equation 2 and recognizing that ρ1 > ρ2 (i.e. channel 1 is cooled, while channel 2 is heated), one gets: 𝐿𝐿 𝐺𝐺12 𝐿𝐿 𝐺𝐺22 − 𝜌𝜌1 𝑔𝑔𝑔𝑔 = 𝑓𝑓 − 𝜌𝜌1 𝑔𝑔𝑔𝑔 𝑓𝑓 𝐷𝐷𝑒𝑒 2𝜌𝜌1 𝐷𝐷𝑒𝑒 2𝜌𝜌2 𝐺𝐺12 𝐺𝐺22 𝐷𝐷𝑒𝑒 (𝜌𝜌 − 𝜌𝜌2 )𝑔𝑔𝑔𝑔 − = 2𝜌𝜌1 2𝜌𝜌2 𝑓𝑓𝑓𝑓 2

(3)

(4)

Therefore since the quantity on the right hand side must be positive, we can write that

Solving for the mass flux in channel,

𝐺𝐺12 𝐺𝐺22 − >0 2𝜌𝜌1 2𝜌𝜌2 𝐺𝐺1 > �

𝜌𝜌1 𝐺𝐺 𝜌𝜌2 2

(5)

(6)

Therefore, the mass flow rate in channel 1 is higher than the mass flow rate in channel 2 since the flow areas are equivalent. The result is also intuitive because cooling channel 1 and heating channel 2 creates a “chimney” effect that opposes downflow in channel 2.

PROBLEM 9.5 QUESTION Sizing of an Orificing Device (Section 9.6) In a hypothetical reactor an orificing scheme is sought such that the core is divided into two zones. Each zone produces one half of the total reactor power. However, zone 1 contains 100 assemblies, whereas zone 2 contains 80 assemblies. It is desired to obtain equal average temperature rises in the two zones. Therefore the flow in the assemblies of lower power (zone 1) is to be constricted by the use of orificing blocks, as shown in Figure 9.42.

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FIGURE 9.42 Comparison of unorificed and orificed fuel bundles. Determine the appropriate diameter (D) of the flow channels in the orificing block. Assume negligible pressure losses in all parts of the fuel assemblies other than the fuel rod bundle and the orifice block. The flow in all the assemblies may be assumed to be fully turbulent. All coolant channels have smooth surfaces. Data: – Pressure drop across assembly, ΔPA = 7.45 × 105 N/m2 – Total core flow rate, ṁT = 17.5 × 106 kg/h – Coolant viscosity, μ = 2 × 10−4 Ns/m2 – Coolant density, ρ = 0.8 g/cm3 – Contraction pressure loss coefficient, Kc = 0.5 – Expansion pressure loss coefficient, Ke = 1.0 Answer: D = 2.17 cm

PROBLEM 9.5 SOLUTION Sizing of an Orificing Device (Section 9.6) In a hypothetical reactor an orificing scheme is sought such that the core is divided into two zones. Each zone produces one half of the total reactor power. However, zone 1 contains 100 assemblies, whereas zone 2 contains 80 assemblies. It is desired to obtain equal average temperature rises in the two zones. Therefore the flow in the assemblies of lower power (zone 1) is to be constricted by the use of orificing blocks, as shown in Figure 9.42. Determine the the appropriate diameter (D) of the flow channels in the orificing block. Assume negligible pressure losses in all parts of the fuel assemblies other than the fuel rod bundle and the orifice block. The flow in all the assemblies may be assumed to be fully turbulent. All coolant

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channels have smooth surfaces. The following parameters are given in the problem statement and Figure 9.42: – Pressure drop across assembly, ΔPA = 7.45 × 105 N/m0 – Total core flow rate, ṁT = 17.5 × 106 kg/h – Coolant viscosity, μ = 2 × 10−4 Ns/m2 – Coolant density, ρ = 0.8 g/cm3 – Contraction pressure loss coefficient, Kc = 0.5 – Expansion pressure loss coefficient, Ke = 1.0 – Length of orifice region, Li = 40 cm – Length of fuel bundle, L1 = 360 cm – Total length of assembly, L = 400 cm The first step is to determine the flow rates in each of the assemblies. To get these quantities, an energy balance is performed where both assemblies produce the same amount of power and have equal temperature rises: 𝑁𝑁1 𝑚𝑚̇1 𝑐𝑐𝑝𝑝 ∆𝑇𝑇 = 𝑁𝑁2 𝑚𝑚̇2 𝑐𝑐𝑝𝑝 ∆𝑇𝑇

(1)

100𝑚𝑚̇1 = 80𝑚𝑚̇2

(2)

100𝑚𝑚̇1 + 80𝑚𝑚̇2 = 𝑚𝑚̇ 𝑇𝑇

(3)

where, N1 is the number of assemblies in zone 1, N2 is the number of assemblies in zone 2, cp is the specific heat and ΔT is the temperature rise. This expression reduces down to where from conservation mass it is also known that

Solving these two equations simultaneously yields a flow rate through and assembly in zone 1 and zone 2, respectively: 𝑚𝑚̇1 = 24.306 kg⁄s

𝑚𝑚̇2 = 30.382 kg⁄s

(4)

The next step that can be performed is to find a diameter of assembly 2. The flow area as a function of this diameter can be written as 𝜋𝜋 𝐴𝐴2 (𝐷𝐷2 ) = 𝐷𝐷22 (5) 4 The Reynolds number is then

𝑅𝑅𝑅𝑅2 𝐷𝐷2 =

The friction factor and then be represented as

𝑚𝑚̇2 𝐷𝐷2 𝐴𝐴2 (𝐷𝐷2 )𝜇𝜇

226

(6)

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𝑓𝑓2 (𝐷𝐷2 ) = 0.184𝑅𝑅𝑅𝑅2 (𝐷𝐷2 )−0.2

(7)

The pressure drop of assembly 2 is therefore

𝐿𝐿 𝑚𝑚̇22 + 𝜌𝜌𝜌𝜌𝜌𝜌 𝑃𝑃𝐴𝐴 = 𝑓𝑓2 (𝐷𝐷2 ) = 𝐷𝐷2 2𝐴𝐴2 (𝐷𝐷2 )2

(8)

The only unknown in the pressure drop formula is the diameter, which can be calculated to be 𝐷𝐷2 = 0.034 m

The same procedure can be written for the orifice region of assembly 1, 𝜋𝜋 𝐴𝐴𝑖𝑖 (𝐷𝐷𝑖𝑖 ) = 4 × 𝐷𝐷𝑖𝑖2 4 𝑅𝑅𝑅𝑅𝑖𝑖 𝐷𝐷𝑖𝑖 =

and the pressure drop is

𝑚𝑚̇1 𝐷𝐷1 𝐴𝐴𝑖𝑖 (𝐷𝐷𝑖𝑖 )𝜇𝜇

𝑓𝑓𝑖𝑖 (𝐷𝐷𝑖𝑖 ) = 0.184𝑅𝑅𝑅𝑅𝑖𝑖 (𝐷𝐷𝑖𝑖 )−0.2

∆𝑃𝑃𝑖𝑖 (𝐷𝐷𝑖𝑖 ) = �𝐾𝐾𝑒𝑒 + 𝐾𝐾𝑐𝑐 + 𝑓𝑓𝑖𝑖 (𝐷𝐷𝑖𝑖 )

𝐿𝐿𝑖𝑖 𝑚𝑚̇12 � + 𝜌𝜌𝜌𝜌𝐿𝐿𝑖𝑖 𝐷𝐷𝑖𝑖 2𝐴𝐴𝑖𝑖 (𝐷𝐷𝑖𝑖 )2

(9)

(10) (11) (12)

(13)

Note, that the there are 4 flow paths through the orifice. Unlike assembly 2, the pressure drop over the orifice sub-region is not known and so it is a function of the orifice diameter, Di. Since the fuel bundle region of assembly 1 is equivalent to assembly 2, they will have the same diameter, 𝐷𝐷1 = 𝐷𝐷2 = 0.0.34 m

(14)

Since this diameter is known now, the area, Reynolds number and friction factor can be calculated: 𝜋𝜋 𝐴𝐴1 = 𝐷𝐷12 = 8.872 × 10−4 m2 (15) 4 𝑅𝑅𝑅𝑅1 =

𝑚𝑚̇1 𝐷𝐷1 = 4.604 × 102 𝐴𝐴1 𝜇𝜇

𝑓𝑓1 = 0.184 𝑅𝑅𝑅𝑅1−0.2 = 0.009

(16)

(17)

The pressure drop just over the fuel bundle region to yield ∆𝑃𝑃1 = 𝑓𝑓1

𝐿𝐿1 𝑚𝑚̇12 + 𝜌𝜌𝜌𝜌𝐿𝐿1 = 458.047 kPa 𝐷𝐷1 2𝜌𝜌𝐴𝐴12

(18)

Therefore since the pressure drops are equivalent over the assemblies, the pressure drop over the orifice region is

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∆𝑃𝑃𝑖𝑖 = ∆𝑃𝑃𝐴𝐴 − ∆𝑃𝑃1 = 286.953 kPa

(19)

𝐷𝐷1 = 2.17 cm

(20)

The only unknown in Equation (13) is the orifice diameter, which is calculated to be

PROBLEM 9.6 QUESTION Pressure Drop Features of a PWR Core (Section 9.6) Consider a PWR core containing 50,952 fuel rods cooled with a total flow rate of 17.476 Mg/s. Each rod has a total length of 3.876 m and a smooth outside diameter of 9.5 mm. The rods are arranged in a square array with a pitch = 12.6 mm. The lower-end and upper end fittings are represented as a honeycomb grid spacer, with the thickness of individual grid elements being 1.5 mm. There are also five intermediate honeycomb grid spacers with thickness of 1 mm. Consider the upper and lower plenum regions to be entirely open. 1. What is the pressure drop for each of the following bundle regions: entrance, end fixtures, friction in the pin array between grids, gravity, grid spacers and exit? 2. What is the plenum-to-plenum pressure drop? Properties kg

– Water density = 720 𝑚𝑚3

– Water viscosity = 91 μPa s

Answers: – ΔPgravity = 27.37 kPa – ΔPfriction = 50.06 kPa – ΔPentrance + ΔPexit = 15.87 kPa – ΔPspacer = 26.07 kPa – ΔPfittings = 22.51 kPa – ΔPT = 141.88 kPa

PROBLEM 9.6 SOLUTION Pressure Drop Features of a PWR Core (Section 9.6) Consider a PWR core containing 50,952 fuel rods cooled with a total flow rate of 17.476 Mg/s. Each rod has a total length of 3.876 m and a smooth outside diameter of 9.5 mm. The rods are arranged in a square array with a pitch = 12.6 mm. The lower-end and upper end fittings are represented as a honeycomb grid spacer, with the thickness of individual grid elements being 1.5

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mm. There are also five intermediate honeycomb grid spacers with thickness of 1 mm. Consider the upper and lower plenum regions to be entirely open. Given Parameters: – Number of fuel rods: nrods = 50,952 – Total core flow rate: ṁcore = 17,476 kg/s – Fuel rod length: L = 3.876 m – Fuel rod outside diameter: dco = 9.5 mm – Fuel rod pitch: P = 12.6 mm – Spacer grid thickness: tspac = 1 mm – End fitting thickness: tfit = 1.5 mm Thermodynamic Properties: kg

– Water density: 𝜌𝜌 = 720 m3

– Water viscosity: μ = 91 μPa s Gravitational Pressure Drop The gravitational pressure drop can be calculated as

Frictional Pressure Drop

∆𝑃𝑃 = 𝜌𝜌𝜌𝜌𝜌𝜌 = 27.38 kPa

(1)

Before calculating the frictional pressure drop, some other parameters need to be determined. The mass flow rate through a unit cell (interior sub-channel), containing just 1 fuel rod is calculated to be, 𝑚𝑚̇𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 kg (2) 𝑚𝑚̇ = = 0.343 𝑛𝑛𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 s The flow area through this unit cell is 𝐴𝐴𝑓𝑓 = 𝑃𝑃2 −

𝜋𝜋 2 𝑑𝑑 = 8.788 × 10−5 m2 4 𝑐𝑐𝑐𝑐

(3)

The mass flux through the unit cell can now be determined as 𝐺𝐺 =

𝑚𝑚̇ kg = 3.903 × 103 2 𝐴𝐴𝑓𝑓 m s

(4)

The equivalent hydraulic diameter of the unit cell is 𝐷𝐷𝑒𝑒 =

4𝐴𝐴𝑓𝑓 𝜋𝜋𝑑𝑑𝑐𝑐𝑐𝑐 229

(5)

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The Reynolds number can now be determined to be 𝑅𝑅𝑒𝑒 =

𝐺𝐺𝐷𝐷𝑒𝑒 = 5.052 × 105 𝜇𝜇

(6)

Since we are dealing with a rod bundle geometry, we may use the Cheng and Todreas correlation (Equation 9.105) to determine the turbulent friction factor. First we must select the appropriate parameters in the correlation from Table 9.5b. In this problem we have square geometry, turbulent flow in an interior channel and a P/D ratio of 1.326. Therefore, the parameters are a = 0.1339, b1 = 0.09059, and b2 = −0.09926. We may now calculate the turbulent friction factor, and

′ 𝐶𝐶𝑓𝑓𝑓𝑓𝑓𝑓 = 𝑎𝑎 + 𝑏𝑏1 (𝑃𝑃⁄𝐷𝐷 − 1) + 𝑏𝑏2 (𝑃𝑃⁄𝐷𝐷 − 1)2 = 0.153

𝑓𝑓𝑖𝑖𝑖𝑖 =

′ 𝐶𝐶𝑓𝑓𝑓𝑓𝑓𝑓 ′ )0.18 = 0.01438 (𝑅𝑅𝑒𝑒𝑖𝑖𝑖𝑖

(7) (8)

The frictional pressure drop can now be determine to be

𝐿𝐿 𝐺𝐺 2 = 50.06 kPa ∆𝑃𝑃𝑓𝑓 = 𝑓𝑓𝑖𝑖𝑖𝑖 𝐷𝐷𝑒𝑒 2𝜌𝜌

(9)

Entrance and Exit Pressure Drop

The pressure losses at the entrance and exit of the fuel bundle develop from acceleration and form losses. Combining these losses we can derive that the entrance and exit pressure drops are

and

∆𝑃𝑃𝑖𝑖𝑖𝑖 = (𝐾𝐾𝑖𝑖 + 1)

𝐺𝐺 2 2𝜌𝜌

𝐺𝐺 2 ∆𝑃𝑃𝑒𝑒𝑒𝑒 = (𝐾𝐾𝑒𝑒 − 1) 2𝜌𝜌

(10)

(11)

One can look up the form loss coefficients for expansions and contractions in any fluid mechanics textbook. Assuming that the plena areas are much greater than the flow are in the rod bundles, we can determine that the inlet and exit form loss coefficients are Ki = 0.5 and Ke = 1.0. Therefore, the resulting pressure drops are (12) ∆𝑃𝑃𝑖𝑖𝑖𝑖 = 15.87 kPa and ∆Pex = 0 kPa Pressure Loss Due to Spacers

The pressure loss at spacers can be determined using the De Stordeur model modified by Rehme (Equation 9.111), ∆𝑃𝑃𝑠𝑠 = 𝑁𝑁𝑠𝑠 𝐶𝐶𝜐𝜐 �

𝜌𝜌𝜌𝜌 2 𝐴𝐴𝑠𝑠 2 �� 2 𝐴𝐴𝜐𝜐

(13)

For our case, the number of spacers is Ns = 5, the modified drag coefficient is Cυ = 6.5 and the average bundle fluid velocity can be derived from the mass flux to be υ = 5.421 m/s. The

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unrestricted flow area away from the spacer is equivalent to the already derived flow area, Aυ = Af = 8.788 × 10−5 m2. Finally, the projected frontal area of the spacer can be calculated from Example 9.9 to be 2

𝐴𝐴𝑠𝑠 = 𝑃𝑃2 − �𝑃𝑃 − 𝑡𝑡𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 � = 2.42 × 10−5 m2

(14)

∆𝑃𝑃𝑠𝑠 = 26.07 kPa

(15)

Therefore, the pressure loss due to the spacers can be calculated as Pressure Loss Due to End Fittings

This calculation follows the same procedure as the pressure loss due to spacers, 𝜌𝜌𝜌𝜌 2 𝐴𝐴𝑠𝑠 2 ∆𝑃𝑃𝑓𝑓𝑓𝑓𝑓𝑓 = 𝑁𝑁𝑓𝑓 𝐶𝐶𝜐𝜐 � �� 2 𝐴𝐴𝜐𝜐

(16)

However, in this calculation, the number of end fittings is Nf = 2, and the projected frontal area of the fittings is As = 3.555 × 10−5 m2. Therefore, the pressure drop over the fittings is (17)

∆𝑃𝑃𝑓𝑓𝑓𝑓𝑓𝑓 = 22.51 kPa

Total Pressure Drop

The total pressure drop is the summation of all of the individual pressure losses, ∆𝑃𝑃𝑇𝑇 = ∆𝑃𝑃𝑔𝑔 + ∆𝑃𝑃𝑓𝑓 + ∆𝑃𝑃𝑖𝑖𝑖𝑖 + ∆𝑃𝑃𝑒𝑒𝑒𝑒 + ∆𝑃𝑃𝑠𝑠 + ∆𝑃𝑃𝑓𝑓𝑓𝑓𝑓𝑓 = 141.9 kPa

(18)

PROBLEM 9.7 QUESTION

Comparison of Laminar and Turbulent Friction Factors of Water and Sodium Heat Exchangers (Sections 9.4 and 9.6) Consider square arrays of vertical tubes utilized in two applications: a recirculation PWR steam generator and an intermediate heat exchanger for SFR service. In each case primary system liquid flows through the tubes, and secondary system liquid flows outside the tubes within the shell side. The operating and geometric conditions of both units are given in Table 9.9. Assume the wall heat flux is axially constant. TABLE 9.9 Operating and Geometric Conditions Parameter

PWR Steam Generator

SFR Intermediate Heat Exchanger

System characteristics Primary fluid

Water

Sodium

Secondary fluid

Water

Sodium

P/D

1.5

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D (cm)

1.0

Pressure (MPa)

5.5

0.202

Temperature (°C)

270

480

Density (kg/m3)

767.9

837.1

Thermal conductivity (W/m °C)

0.581

Viscosity (kg/m s)

1.0 × 10

Heat capacity (J/kg K)

4990.0

Nominal shell-side properties

88.93 −4

2.92 × 10−4 1195.8

For the shell side of the tube array in each application answer the following questions: 1. Find the friction factor (f) for fully developed laminar flow at ReDe = 103. 2. Find the friction factor (f) for fully developed turbulent flow at ReDe = 105. 3. Can either of the above friction factors be found from the circular tube geometry using the equivalent diameter concept? Demonstrate and explain. 4. What length is needed to achieve fully developed laminar flow? 5. What length is needed to achieve fully developed turbulent flow? Answers: 1. f = 0.1198 2. f = 0.0194 3. Only for turbulent case 4. Cannot be determined 5. z = 0.466 m to 0.746 m

PROBLEM 9.7 SOLUTION Comparison of Laminar and Turbulent Friction Factors of Water and Sodium Heat Exchangers (Sections 9.4 and 9.6) Consider square arrays of vertical tubes utilized in two applications: a recirculation PWR steam generator and an intermediate heat exchanger for SFR service. In each case primary system liquid flows through the tubes, and secondary system liquid flows outside the tubes within the shell side. The following are given in Table 9.8, where a subscript ‘P’ denotes PWR and ‘S’ denotes SFR: – = diameter, DP 1.0 = cm and DS 1.0 cm

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–

pitch-to-diameter= ratio, P / DP 1.5 = and P / DS 1.5

– density, ρ P 767.9 kg/m3 and ρ S 837.1 kg/m3 = = –

viscosity, µ P = 1.0 ×10−4 Pa s and µ S = 2.92 ×10−4 Pa s

Find the friction factor for fully developed Laminar flow: The Reynolds number is (1)

Re1 = 103

The friction factor for an array of tubes on the shell side is given as 𝑓𝑓1 =

𝐶𝐶𝐿𝐿 Re1

(2)

For a square array with a pitch-to diameter ratio of 1.5, the coefficients are Therefore,

𝑎𝑎 = 35.55 𝑏𝑏1 = 263.7 𝑏𝑏2 = −190.2

(3)

𝐶𝐶𝐿𝐿 = 𝑎𝑎 + 𝑏𝑏1 (𝑃𝑃⁄𝐷𝐷 − 1) + 𝑏𝑏2 (𝑃𝑃⁄𝐷𝐷 − 1)2 = 119.85

(4)

For both heat exchangers, the friction factor is 𝑓𝑓1 =

𝐶𝐶𝐿𝐿 = 0.1198 Re1

(5)

Find the friction factor for fully developed turbulent flow: The Reynolds number is (6)

Re2 = 105

For a square array with a pitch-to diameter ratio of 1.5, the coefficients are Therefore,

𝑏𝑏2 = −0.09926

(7)

𝐶𝐶𝑇𝑇 = 𝑎𝑎 + 𝑏𝑏1 (𝑃𝑃⁄𝐷𝐷 − 1) + 𝑏𝑏2 (𝑃𝑃⁄𝐷𝐷 − 1)2 = 0.154

(8)

𝑎𝑎 = 0.1339

𝑏𝑏1 = 0.09059

For both heat exchangers, the friction factor is 𝑓𝑓2 =

𝐶𝐶𝑇𝑇 = 0.0194 Re0.18 2

(9)

Can either of the above friction factors be found from the circular tube geometry using the equivalent diameter concept? Only for the turbulent case. What is the length needed to achieve fully developed laminar flow? This cannot be determined. What length is needed to achieve fully developed turbulent flow? This hydrodynamic developing length ranges from 25 to 40 hydraulic diameters. Both heat

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exchangers have the same hydraulic diameters since there geometric characteristics are similar. The flow area is 𝜋𝜋 𝐴𝐴 = [(𝑃𝑃⁄𝐷𝐷 ) × 𝐷𝐷]2 − 𝐷𝐷2 = 1.465 × 10−4 m2 (10) 4 The hydraulic diameter is therefore

𝐷𝐷𝑒𝑒 =

4𝐴𝐴 = 0.019 m 𝜋𝜋𝜋𝜋

(11)

The developing region therefore ranges from to

𝑧𝑧 = 25𝐷𝐷𝑒𝑒 = 0.466 m

(12)

𝑧𝑧 = 40𝐷𝐷𝑒𝑒 = 0.746 m

(13)

PROBLEM 9.8 QUESTION Using the UCTD Correlation to Calculate the Pressure Drop for WireWrapped and Bare Rod Bundles (Sections 9.6.2.2) The preliminary design parameters for India’s PFBR 217-pin hexagonal wire-wrapped rod bundle are D = 6.6 mm, Dw = 1.65 mm, P/D =1.255, W/D= 1.255, and H/D = 30.30. Several experiments were performed using this set of geometrical data. The results obtained span laminar, transition and turbulent flow regimes. (Chun and Seo, 2001; Padmakumar et al., 2017) 1. Using the UCDT correlation calculate the pressure drop over one lead length for a Reynolds number of 10000, assuming the working fluid is water at room temperature. 2. What will be the result if no wire is used in the bundle, i.e., bare rod bundle? References Chun, M.H. and Seo, K.W., 2001. An experimental study and assessment of existing friction factor correlations for wire-wrapped fuel assemblies. Annals of Nuclear Energy, 28, 1683–1695. Padmakumar, G., Velusamy, K., Prasad, B.V.S.S., Rajan, K.K., 2017. Hydraulic characteristics of a fast reactor fuel subassembly: An experimental investigation. Annals of Nuclear Energy, 102, 255-267.

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PROBLEM 9.8 SOLUTION Using the UCTD Correlation to Calculate the Pressure Drop for WireWrapped and Bare Rod Bundles (Sections 9.6.2.2) Part 1 For the wire wrapped rod bundle a. For Re=10000, first determine the flow regime For this bundle, using Equations 9.118a and 9.118b respectively, RebL=320(100.255)=576 RebT=1000(100.7(0.255))=150834 Therefore the flow, at Re=10000, is in the transition regime. b. In the transition region, the formula for the bundle friction is fb = (CfbL/Reb)(1-Ψb )1/3(1-Ψb7)+( CfbT/Reb0.18) Ψb1/3

(9.117c)

Ψb = log(Reb/RebL)/log(RebT/RebL) for i=b CfbL = Deb(∑3i=1 (NiAi/Ab)(Dei/Deb)(Dei/CfiL))-1

(9.118c) (9.119a)

where

CfbT = Deb(∑3i=1 (NiAi/Ab)(Dei/Deb)0.0989(Dei/CfiT)0.54945)-1.82

(9.119b)

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Use the equations defined in UCTD given in Section 9.6.2.2 and the website in the next paragraph for calculation using the UCTD correlation as specified below, obtain the friction factor as 0.036 for Reb=10,000. d. The public website for calculating results using the UCTD correlation can be found by searching “upgraded Cheng and Todreas” or from the site http://wp.me/p61TQ-tI. Download the program directory, double click the file “project1.exe” the following window will show.

Chapter 9 - Single-Phase Fluid Mechanics

On the left side, input the required values. The first input is number of rings, and in this case it is 9 (for 217 pin). Input the rest values as specified. All parameters can be found in the output window. e. The pressure drop over one lead length H can be calculated from L

ρV2

∆𝑃𝑃 = f �De� 2 , where

f = 0.036

L = ***H = 30.3(6.6) =200 mm De = wire-wrapped rod bundle equivalent hydraulic diameter = 4Ab/Pwb = 4(5.583 ×10-3 m2 ) / 6.0566 m = 3.682 mm μ = water viscosity =101×10-5 Pa s V = 104μ/(ρDe)=104(101×10-5 Pa s) / (1000 kg/m3 )(3.683 × 10-3 m) = 2.74 m/s Therefore, L

ρV2

∆𝑃𝑃 = f(D ) 2 = 0.036(200/3.682)[1000(2.74)2 / 2] = 7.34 kPa 𝑒𝑒

Part 2 For the bare rod bundle f = 0.030

De = 4A b′ / Pwb′ = 4(6.051× 10−3 m 2 ) / 4.932 m = 4.908 mm

L = 200 mm V=104μ/(ρDe) =104(101×10-5 Pa s) / (1000 kg/m3 )(4.908×10-3 m) = 2.05 m/s L

ρV2

Yielding ∆𝑃𝑃 = f �D � 2 = 0.030(200/4.908)[1000(2.05)2 / 2] = 2.57 kPa 𝑒𝑒

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Chapter 10 Single-Phase Heat Transfer Contents Problem 10.1

Derivation of Nusselt number for laminar flow in rectangular geometry ..... 238

Problem 10.2

Derivation of Nusselt number for laminar flow in circular and flat plate geometry ........................................................................................................ 241

Problem 10.3

Derivation of Nusselt number for laminar flow in an equivalent annulus ..... 254

Problem 10.4

Estimating the effect of turbulence on heat transfer in SFBR fuel bundles ... 256

Problem 10.5

Reynolds analogy and equivalent diameter problem ..................................... 259

Problem 10.6

Determining the temperature of the primary side of a steam generator ........ 261

Problem 10.7

Comparison of heat transfer characteristics of water and helium .................. 265

Problem 10.8

Hydraulic and thermal analysis of the Emergency Core Spray System in a BWR ......................................................................................................................... 267

Problem 10.9

Coolant selection for an advanced high-temperature reactor ........................ 272

Problem 10.10 Effect of geometry on single phase heat transfer in straight tubes ................ 277

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PROBLEM 10.1 QUESTION Derivation of Nusselt Number for Laminar Flow in Rectangular Geometry Prove that the asymptotic Nusselt number for flow of a coolant with constant properties between two flat plates of infinite width, heated with uniform heat flux on both walls, is 8.235 for laminar flow (i.e., parabolic velocity distribution) and 12 for slug flow (i.e., uniform velocity distribution).

PROBLEM 10.1 SOLUTION Derivation of Nusselt Number for Laminar Flow in Rectangular Geometry Prove that the asymptotic Nusselt number for flow of a coolant with constant properties between two flat plates of infinite width, heated with uniform heat flux on both walls, is 8.235 for laminar flow (i.e., parabolic velocity distribution) and 12 for slug flow (i.e., uniform velocity distribution). For laminar flow in rectangular geometry, the momentum equations in the x and y directions are respectively,

The boundary conditions are

𝜕𝜕 2 𝜐𝜐𝑥𝑥 𝜕𝜕𝜕𝜕 + 𝜇𝜇 =0 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 2 𝜕𝜕𝜕𝜕 =0 𝜕𝜕𝜕𝜕

at y = y0 (wall location) and

𝜐𝜐𝑥𝑥 = 0

−

𝜕𝜕𝜕𝜕𝑥𝑥 =0 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕

(1) (2)

(3)

(4) 𝜕𝜕𝜕𝜕

at y = 0. Since P is not a function of y, then 𝜕𝜕𝜕𝜕 = 𝜕𝜕𝜕𝜕 , integrating Equation (1) we get 1 𝜕𝜕𝜕𝜕𝑥𝑥 𝑦𝑦 + 𝐶𝐶1 = 𝜇𝜇 𝜕𝜕𝜕𝜕

(5)

By the boundary condition in Equation (4), C1 = 0. Integrating Equation (5) again, 1 𝑑𝑑𝑑𝑑 2 𝑦𝑦 + 𝐶𝐶2 = 𝜐𝜐𝑥𝑥 2𝜇𝜇 𝑑𝑑𝑑𝑑

(6)

1 𝑑𝑑𝑑𝑑

By the boundary condition in Equation (3), 𝐶𝐶2 = − 2𝜇𝜇 𝑑𝑑𝑑𝑑 𝑦𝑦02 , 𝜐𝜐𝑥𝑥 =

1 𝑑𝑑𝑑𝑑 2 (𝑦𝑦 − 𝑦𝑦02 ) 2𝜇𝜇 𝑑𝑑𝑑𝑑 238

(7)

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Chapter 10 - Single-Phase Heat Transfer

The average velocity can be determined to be

and therefore)

𝜐𝜐𝑚𝑚 =

1 𝑦𝑦0 1 𝑑𝑑𝑑𝑑 2 𝑦𝑦 2 𝑑𝑑𝑑𝑑 (𝑦𝑦 − 𝑦𝑦02 )𝑑𝑑𝑑𝑑 = − 0 � 𝑦𝑦0 0 2𝜇𝜇 𝑑𝑑𝑑𝑑 3𝜇𝜇 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 3𝜇𝜇 = − 2 𝜐𝜐𝑚𝑚 𝑑𝑑𝑑𝑑 𝑦𝑦0

and

3 𝑦𝑦 2 𝜐𝜐𝑥𝑥 = 𝜐𝜐𝑚𝑚 �1 − � � � 2 𝑦𝑦0

(8)

(9)

(10)

The general form of the energy equation is 𝜌𝜌𝜌𝜌𝑝𝑝

𝐷𝐷𝐷𝐷 𝐷𝐷𝐷𝐷 = 𝑘𝑘∇2 𝑇𝑇 + 𝑞𝑞 ‴ + 𝛽𝛽𝛽𝛽 +Φ 𝐷𝐷𝐷𝐷 𝐷𝐷𝐷𝐷

(11)

We make following assumptions:

1. The flow is incompressible and fully developed 2. The flow is inviscid and steady state 3. Neglect axial heat condition and no heat generated in the fluid so that the energy equation becomes

For constant wall heat flux

𝜌𝜌𝜌𝜌𝑝𝑝 𝜐𝜐𝑥𝑥

𝜕𝜕𝜕𝜕 𝜕𝜕 2 𝑇𝑇 = 𝑘𝑘 2 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕

𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕𝑚𝑚 = 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕

(12)

(13)

is not a function of y. Substituting Equation (10) into Equation (12) and integrating twice, we get 3 𝜕𝜕𝜕𝜕 𝑦𝑦 2 𝑦𝑦 4 𝜌𝜌𝜌𝜌 𝜐𝜐 � − � = 𝑘𝑘𝑘𝑘 + 𝐶𝐶1 𝑦𝑦 + 𝐶𝐶2 2 𝑝𝑝 𝑚𝑚 𝜕𝜕𝜕𝜕 2 12𝑦𝑦02

(14)

The boundary conditions are at y = 0,

and at y = y0,

𝜕𝜕𝜕𝜕 =0 𝜕𝜕𝜕𝜕 𝑇𝑇 = 𝑇𝑇𝑤𝑤 (𝑥𝑥)

(15)

(16)

Substituting the boundary conditions into the energy equation, we find

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(17)

𝐶𝐶1 = 0

and 𝐶𝐶2 =

Therefore,

𝑇𝑇 − 𝑇𝑇𝑤𝑤 =

3 𝜕𝜕𝜕𝜕 5 2 � 𝑦𝑦 � − 𝑘𝑘𝑘𝑘𝑤𝑤 𝜌𝜌𝜌𝜌𝑝𝑝 𝜐𝜐𝑚𝑚 4 𝜕𝜕𝜕𝜕 6 0

3 𝜕𝜕𝜕𝜕 2 𝑦𝑦 4 5 𝜌𝜌𝜌𝜌𝑝𝑝 𝜐𝜐𝑚𝑚 �𝑦𝑦 − 2 − 𝑦𝑦02 � 4𝑘𝑘 𝜕𝜕𝜕𝜕 6𝑦𝑦0 6

(18)

(19)

The average temperature is obtained by

𝑦𝑦0 3 𝑦𝑦 2 3 𝜕𝜕𝜕𝜕 2 𝑦𝑦 4 5 2 𝑦𝑦0 𝜌𝜌 𝜐𝜐 �1 − � � � 𝜌𝜌𝜌𝜌 𝜐𝜐 �𝑦𝑦 − − 𝑦𝑦 � 𝑑𝑑𝑑𝑑 ∫ 𝑚𝑚 𝑝𝑝 𝑚𝑚 2 0 (𝑇𝑇 )𝑑𝑑𝑑𝑑 2 𝑦𝑦0 4𝑘𝑘 𝜕𝜕𝜕𝜕 ∫0 𝜌𝜌𝜌𝜌𝑥𝑥 − 𝑇𝑇𝑤𝑤 6𝑦𝑦0 6 0 𝑇𝑇𝑚𝑚 − 𝑇𝑇𝑤𝑤 = = 𝑦𝑦0 𝑦𝑦0 3 𝑦𝑦 2 ∫0 𝜌𝜌𝜌𝜌𝑥𝑥 𝑑𝑑𝑑𝑑 ∫0 𝜌𝜌 2 𝜐𝜐𝑚𝑚 �1 − �𝑦𝑦 � � 𝑑𝑑𝑑𝑑 0 𝜕𝜕𝜕𝜕

″ Since 𝑞𝑞𝑤𝑤 = −𝑘𝑘 𝜕𝜕𝜕𝜕 �

𝑦𝑦=𝑦𝑦0

temperature gradient is

=−

17 𝜕𝜕𝜕𝜕 2 𝜌𝜌𝜌𝜌𝑝𝑝 𝜐𝜐𝑚𝑚 𝑦𝑦 35 𝜕𝜕𝜕𝜕 𝐷𝐷

(20)

𝜕𝜕𝜕𝜕

= −𝜌𝜌𝜌𝜌𝑝𝑝 𝜐𝜐𝑚𝑚 𝜕𝜕𝜕𝜕 𝑦𝑦0 (from integrating energy equation once) the ″ 𝜕𝜕𝜕𝜕𝑚𝑚 𝑞𝑞𝑤𝑤 1 =− 𝜕𝜕𝜕𝜕 𝜌𝜌𝜌𝜌𝑝𝑝 𝜐𝜐𝑚𝑚 𝑦𝑦0

(21)

Combining Equations (20) and (21)

𝑇𝑇𝑚𝑚 − 𝑇𝑇𝑤𝑤 =

Since the heat transfer coefficient is

ℏ=

and the Nusselt number is

the Nusselt number is therefore

″ 17 𝑞𝑞𝑤𝑤 𝑦𝑦0 35 𝑘𝑘

″ 𝑞𝑞𝑤𝑤 𝑇𝑇𝑚𝑚 − 𝑇𝑇𝑤𝑤

Nu =

ℎ𝐷𝐷𝑒𝑒 𝑘𝑘

″ 𝑞𝑞𝑤𝑤 4𝑦𝑦0 4(35) Nu = = = 8.235 𝑇𝑇𝑚𝑚 − 𝑇𝑇𝑤𝑤 𝑘𝑘 17

(22)

(23)

(24)

(25)

Note that the hydraulic diameter is De = 4y0. For slug flow, υm = υx then Equation (12) becomes

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𝜕𝜕𝜕𝜕 𝜕𝜕 2 𝑇𝑇 𝜌𝜌𝜌𝜌𝑝𝑝 𝜐𝜐𝑚𝑚 = 𝑘𝑘 2 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕

After integrating twice,

𝜌𝜌𝜌𝜌𝑝𝑝 𝜐𝜐𝑚𝑚

𝜕𝜕𝜕𝜕 𝑦𝑦 2 = 𝑘𝑘𝑘𝑘 + 𝐶𝐶1 𝑦𝑦 + 𝐶𝐶2 𝜕𝜕𝜕𝜕 2

(26)

(27)

Using the same boundary conditions from Equations (15) and (16), 𝐶𝐶1 = 0

and

𝜕𝜕𝜕𝜕 𝑦𝑦02 𝐶𝐶2 = 𝜌𝜌𝜌𝜌𝑝𝑝 𝜐𝜐𝑚𝑚 − 𝑘𝑘𝑘𝑘𝑤𝑤 (𝑥𝑥) 𝜕𝜕𝜕𝜕 2

Thus,

𝑇𝑇 − 𝑇𝑇𝑤𝑤

The average temperature is 𝑇𝑇𝑚𝑚 − 𝑇𝑇𝑤𝑤 =

𝑦𝑦

1 𝜕𝜕𝜕𝜕 2 (𝑦𝑦 − 𝑦𝑦02 ) 𝜌𝜌𝜌𝜌𝑝𝑝 𝜐𝜐𝑚𝑚 2𝑘𝑘 𝜕𝜕𝜕𝜕

0 ∫0 𝜌𝜌𝜌𝜌𝑥𝑥 (𝑇𝑇 − 𝑇𝑇𝑤𝑤 )𝑑𝑑𝑑𝑑

𝑦𝑦

0 ∫0 𝜌𝜌𝜌𝜌𝑥𝑥 𝑑𝑑𝑑𝑑

𝑦𝑦0 1 𝜕𝜕𝜕𝜕 ∫0 𝜌𝜌𝜌𝜌𝑚𝑚 2𝑘𝑘 𝜌𝜌𝜌𝜌𝑝𝑝 𝜐𝜐𝑚𝑚 𝜕𝜕𝜕𝜕 (𝑦𝑦 2 − 𝑦𝑦02 )𝑑𝑑𝑑𝑑 = 𝑦𝑦0 ∫0 𝜌𝜌𝜌𝜌𝑚𝑚 𝑑𝑑𝑑𝑑

𝑇𝑇𝑚𝑚 − 𝑇𝑇𝑤𝑤 = −

Combining Equations (21) and (32)

so that,

1 𝜕𝜕𝜕𝜕 2 𝜌𝜌𝜌𝜌𝑝𝑝 𝜐𝜐𝑚𝑚 𝑦𝑦 3𝑘𝑘 𝜕𝜕𝜕𝜕 0

″ 𝑞𝑞𝑤𝑤 𝑦𝑦0 𝑇𝑇𝑚𝑚 − 𝑇𝑇𝑤𝑤 = 3𝑘𝑘 ″ (4𝑦𝑦 ) ℏ𝐷𝐷𝑒𝑒 𝑞𝑞𝑤𝑤 0 Nu = = ″ = 12 𝑞𝑞 𝑦𝑦 𝑘𝑘 𝑤𝑤 0 𝑘𝑘 3𝑘𝑘

(28)

(29)

(30)

(31)

(32)

(33)

(34)

PROBLEM 10.2 QUESTION

Derivation of Nusselt Number for Laminar Flow in Circular and Flat Plate Geometry (Section 10.2) Show that for a round tube, uniform wall temperature, and slug velocity conditions, the asymptotic Nusselt number for a round tube is 5.783, whereas for flow between two flat plates, it is 9.870.

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PROBLEM 10.2 SOLUTION Derivation of Nusselt Number for Laminar Flow in Circular and Flat Plate Geometry (Section 10.2) Show that for a round tube, uniform wall temperature, and slug velocity conditions, the asymptotic Nusselt number for a round tube is 5.783, whereas for flow between two flat plates, it is 9.87. Circular Tube To solve this problem, two methods will be considered. The first is an analytic iterative method and the second is a closed form approach to solve the differential equations directly. Iterative Method ITERATION 1 If the wall temperature is axially constant, Equation 10.16 states that 𝜕𝜕𝜕𝜕 𝑟𝑟 𝜕𝜕𝜕𝜕𝑚𝑚 = 𝑓𝑓 � � ; 𝑇𝑇𝑤𝑤 (𝑧𝑧) = constant 𝜕𝜕𝜕𝜕 𝑅𝑅 𝜕𝜕𝜕𝜕

where

𝑟𝑟 𝑇𝑇𝑤𝑤 − 𝑇𝑇 𝑓𝑓 � � = 𝑅𝑅 𝑇𝑇𝑤𝑤 − 𝑇𝑇𝑚𝑚

(1)

(2)

according to Equation 10.14c. The energy equation is: 𝜌𝜌𝜌𝜌𝑝𝑝 𝜐𝜐𝑧𝑧

𝜕𝜕𝜕𝜕 1 𝜕𝜕 𝜕𝜕𝜕𝜕 = �𝑟𝑟𝑟𝑟 � 𝜕𝜕𝜕𝜕 𝑟𝑟 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕

(3)

which follows from Equation 10.17 if axial heat conduction is neglected. For slug flow, υz = Vm, and thus 𝜌𝜌𝜌𝜌𝑝𝑝 𝑉𝑉𝑚𝑚

𝑇𝑇𝑤𝑤 − 𝑇𝑇(𝑟𝑟) 𝜕𝜕𝜕𝜕𝑚𝑚 1 𝜕𝜕 𝜕𝜕𝜕𝜕 = �𝑘𝑘𝑘𝑘 � 𝑇𝑇𝑤𝑤 − 𝑇𝑇𝑚𝑚 𝜕𝜕𝜕𝜕 𝑟𝑟 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕

(4)

We will solve this equation to obtain T(r) in an iterative method, First we will assume the same ″ , or (from Equation 10.25) temperature profiles as for the case of uniform 𝑞𝑞𝑤𝑤 2𝜌𝜌𝜌𝜌𝑝𝑝 𝜕𝜕𝜕𝜕 𝑟𝑟 2 𝑟𝑟 4 3𝑅𝑅 2 𝑇𝑇𝑤𝑤 − 𝑇𝑇(𝑟𝑟) = − 𝑉𝑉 � − − � 𝑘𝑘 𝑚𝑚 𝜕𝜕𝜕𝜕 4 16𝑅𝑅 2 16 𝜕𝜕𝜕𝜕

𝜕𝜕𝜕𝜕

where for a uniform heat flux, 𝜕𝜕𝜕𝜕 = 𝜕𝜕𝜕𝜕𝑚𝑚. Substituting this profile into Equation (4), 𝜌𝜌𝜌𝜌𝑝𝑝 𝑉𝑉𝑚𝑚

−

(5)

2𝜌𝜌𝜌𝜌𝑝𝑝 𝜕𝜕𝜕𝜕𝑚𝑚 𝑟𝑟 2 𝑟𝑟 4 3𝑅𝑅 2 � − � 𝑉𝑉𝑚𝑚 − 2 𝜕𝜕𝜕𝜕 16 𝜕𝜕𝜕𝜕𝑚𝑚 1 𝜕𝜕 𝑘𝑘 𝜕𝜕𝜕𝜕 4 16𝑅𝑅 = �𝑘𝑘𝑘𝑘 � 𝑇𝑇𝑤𝑤 − 𝑇𝑇𝑚𝑚 𝜕𝜕𝜕𝜕 𝑟𝑟 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕

(6)

We can rewrite the above equation so that

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1 𝜕𝜕 𝜕𝜕𝜕𝜕 𝑟𝑟 2 𝑟𝑟 4 3𝑅𝑅 2 �𝑟𝑟 � = 𝛼𝛼 � − − � 𝑟𝑟 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 4 16𝑅𝑅 2 16

where

(7)

2𝜌𝜌𝜌𝜌𝑝𝑝 𝜕𝜕𝜕𝜕𝑚𝑚 𝜌𝜌𝜌𝜌𝑝𝑝 V𝑚𝑚 𝑘𝑘 V𝑚𝑚 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕𝑚𝑚 ≠ 𝑓𝑓(𝑟𝑟) 𝛼𝛼 ≡ − 𝑘𝑘 𝑇𝑇𝑤𝑤 − 𝑇𝑇𝑚𝑚 𝜕𝜕𝜕𝜕

After multiplying each side by r and integration of Equation (7), we get 𝑟𝑟

𝜕𝜕𝜕𝜕 𝑟𝑟 4 3𝑅𝑅 2 𝑟𝑟 2 𝑟𝑟 6 = 𝛼𝛼 � − + � + 𝐶𝐶1 𝜕𝜕𝜕𝜕 16 32 96𝑅𝑅 2

After dividing both sides by r and a second integration, we get

𝑟𝑟 4 3𝑅𝑅 2 𝑟𝑟 2 𝑟𝑟 6 𝑇𝑇(𝑟𝑟) = 𝛼𝛼 � − − � + 𝐶𝐶1 ln(𝑟𝑟) + 𝐶𝐶2 64 64 576𝑅𝑅 2

We can apply the following boundary conditions: – –

𝜕𝜕𝜕𝜕

at 𝑟𝑟 = 0 ⟶ 𝜕𝜕𝜕𝜕 = 0 ⟹ 𝐶𝐶1 = 0

19𝛼𝛼𝑅𝑅 4

at 𝑟𝑟 = 𝑅𝑅 ⟶ 𝑇𝑇(𝑅𝑅) = 𝑇𝑇𝑤𝑤 ⟹ 𝐶𝐶2 = 𝑇𝑇𝑤𝑤 + 576

Therefore, the final form of the temperature profile is 𝑟𝑟 4 3𝑅𝑅 2 𝑟𝑟 2 𝑟𝑟 6 19𝑅𝑅 4 𝑇𝑇𝑤𝑤 − 𝑇𝑇(𝑟𝑟) = −𝛼𝛼 � − − + � 64 64 576𝑅𝑅 2 576

We can find the mean temperature with 𝑇𝑇𝑤𝑤 − 𝑇𝑇𝑚𝑚 =

𝑅𝑅

∫0 2𝜋𝜋𝜋𝜋𝑉𝑉𝑚𝑚 �𝑇𝑇𝑤𝑤 − 𝑇𝑇(𝑟𝑟)�𝑑𝑑𝑑𝑑

The wall heat flux is given by

𝑅𝑅

∫0 2𝜋𝜋𝜋𝜋𝑉𝑉𝑚𝑚 𝑑𝑑𝑑𝑑

″ 𝑞𝑞𝑤𝑤 = ℏ(𝑇𝑇𝑤𝑤 − 𝑇𝑇𝑚𝑚 ) = −𝑘𝑘

The partial derivative evaluated at the radius is

Solving for the heat transfer coefficient,

=−

11𝑅𝑅 4 𝛼𝛼 768

𝜕𝜕𝜕𝜕 � 𝜕𝜕𝜕𝜕 𝑅𝑅

𝜕𝜕𝜕𝜕 𝑅𝑅 3 𝛼𝛼 � = 𝜕𝜕𝜕𝜕 𝑅𝑅 24

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𝜕𝜕𝜕𝜕 𝑅𝑅 3 𝛼𝛼 � � � 32𝑘𝑘 −𝑘𝑘 𝜕𝜕𝜕𝜕 𝑅𝑅 24 ℏ= = = 11𝑅𝑅 4 𝛼𝛼 𝑇𝑇𝑤𝑤 − 𝑇𝑇𝑚𝑚 11𝑅𝑅 − 768 −𝑘𝑘

The Nusselt number is given by

32𝑘𝑘 ℎ𝐷𝐷 �11𝑅𝑅 � (2𝑅𝑅) 64 Nu = = = = 5.818 𝑘𝑘 𝑘𝑘 11

The temperature profile above will be used in the next iteration.

ITERATION 2 Starting from Equation (7), 1 𝜕𝜕 𝜕𝜕𝜕𝜕 𝑟𝑟 4 3𝑅𝑅 2 𝑟𝑟 2 𝑟𝑟 6 19𝑅𝑅 4 �𝑟𝑟 � = 𝛼𝛼 �− + + − � 𝑟𝑟 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 64 64 576𝑅𝑅 2 576

(8)

After multiplying each side by r and integration of Equation (8), we get

𝜕𝜕𝜕𝜕 𝑟𝑟 4 3𝑅𝑅 2 𝑟𝑟 4 19𝑅𝑅 4 𝑟𝑟 2 𝑟𝑟 8 𝑟𝑟 = 𝛼𝛼 �− − + + � + 𝐶𝐶1 𝜕𝜕𝜕𝜕 384 256 1152 4608𝑅𝑅 2

After dividing both sides by r and a second integration, we get

𝑟𝑟 6 3𝑅𝑅 2 𝑟𝑟 4 19𝑅𝑅 4 𝑟𝑟 2 𝑟𝑟 8 𝑇𝑇(𝑟𝑟) = 𝛼𝛼 � − + − � + 𝐶𝐶1 ln(𝑟𝑟) + 𝐶𝐶2 2304 1024 2304 36864𝑅𝑅 2

We can apply the following boundary conditions: 𝜕𝜕𝜕𝜕

− at 𝑟𝑟 = 0 → 𝜕𝜕𝜕𝜕 = 0 ⟹ 𝐶𝐶1 = 0

211𝑅𝑅 6 𝛼𝛼

− at 𝑟𝑟 = 𝑅𝑅 → 𝑇𝑇(𝑅𝑅) = 𝑇𝑇𝑤𝑤 ⟹ 𝐶𝐶2 = 𝑇𝑇𝑤𝑤 − 36864

Therefore, the final form of the temperature profile is 𝑟𝑟 6 3𝑅𝑅 2 𝑟𝑟 4 19𝑅𝑅 4 𝑟𝑟 2 𝑟𝑟 8 211𝑅𝑅 6 𝑇𝑇𝑤𝑤 − 𝑇𝑇(𝑟𝑟) = −𝛼𝛼 � − + − − � 2304 1024 2304 36864 36864

We can find the mean temperature with 𝑇𝑇𝑤𝑤 − 𝑇𝑇𝑚𝑚 =

The wall heat flux is given by

𝑅𝑅

∫0 2𝜋𝜋𝜋𝜋𝑉𝑉𝑚𝑚 �𝑇𝑇𝑤𝑤 − 𝑇𝑇(𝑟𝑟)�𝑑𝑑𝑑𝑑 𝑅𝑅

∫0 2𝜋𝜋𝜋𝜋 𝑉𝑉𝑚𝑚 𝑑𝑑𝑑𝑑

′′ 𝑞𝑞𝑤𝑤 = ℏ(𝑇𝑇𝑤𝑤 − 𝑇𝑇𝑚𝑚 ) = −𝑘𝑘

The partial derivative evaluated at the radius is

244

19𝑅𝑅 6 𝛼𝛼 7680

𝜕𝜕𝜕𝜕 � 𝜕𝜕𝜕𝜕 𝑅𝑅

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𝜕𝜕𝜕𝜕 11𝑅𝑅 5 𝛼𝛼 � =− 𝜕𝜕𝜕𝜕 𝑅𝑅 1536

Solving for the heat transfer coefficient,

𝜕𝜕𝜕𝜕 11𝑅𝑅 5 𝛼𝛼 � � � 55𝑘𝑘 −𝑘𝑘 𝜕𝜕𝜕𝜕 𝑅𝑅 1536 ℏ= = = 19𝑅𝑅 6 𝛼𝛼 𝑇𝑇𝑤𝑤 − 𝑇𝑇𝑚𝑚 19𝑅𝑅 7680 The Nusselt number is given by −𝑘𝑘

55𝑘𝑘 ℏ𝐷𝐷 �19𝑅𝑅 � (2𝑅𝑅) 110 Nu = = = = 5.789 𝑘𝑘 𝑘𝑘 19

ITERATION 3 Starting from Equation (7),

1 𝜕𝜕 𝜕𝜕𝜕𝜕 𝑟𝑟 6 3𝑅𝑅 2 𝑟𝑟 4 19𝑅𝑅 4 𝑟𝑟 2 𝑇𝑇 8 211𝑅𝑅 6 �𝑟𝑟 � = 𝛼𝛼 �− + − + + � 𝑟𝑟 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 2304 1024 2304 36864 36864

(9)

After multiplying each side by r and integration of Eq. (9), we get 𝑟𝑟

𝜕𝜕𝜕𝜕 𝑅𝑅 2 𝑟𝑟 6 𝑟𝑟 8 19𝑅𝑅 4 𝑟𝑟 4 211𝑅𝑅 6 𝑟𝑟 2 𝑟𝑟 10 = 𝛼𝛼 � − − − + � + 𝐶𝐶1 𝜕𝜕𝜕𝜕 2048 18432 9216 73728 368640𝑅𝑅 2

After dividing both sides by r and a second integration, we get

𝑅𝑅 2 𝑟𝑟 6 𝑟𝑟 8 19𝑅𝑅 4 𝑟𝑟 4 211𝑅𝑅 6 𝑟𝑟 2 𝑟𝑟 10 𝑇𝑇(𝑟𝑟) = 𝛼𝛼 � − − + + � + 𝐶𝐶1 ln(𝑟𝑟) + 𝐶𝐶2 12288 147456 36864 147456 3686400𝑅𝑅 2

We can apply the following boundary conditions: 𝜕𝜕𝜕𝜕

− at 𝑟𝑟 = 0 → 𝜕𝜕𝜕𝜕 = 0 ⟹ 𝐶𝐶1 = 0

1217𝑅𝑅 8 𝛼𝛼

− at 𝑟𝑟 = 𝑅𝑅 → 𝑇𝑇(𝑅𝑅) = 𝑇𝑇𝑤𝑤 ⟹ 𝐶𝐶2 = 𝑇𝑇𝑤𝑤 − 1228800

Therefore, the final form of the temperature profile is 𝑅𝑅 2 𝑟𝑟 6 𝑟𝑟 8 19𝑅𝑅 4 𝑟𝑟 4 211𝑅𝑅 6 𝑟𝑟 2 𝑟𝑟 10 1217𝑅𝑅 8 𝑇𝑇𝑤𝑤 − 𝑇𝑇(𝑟𝑟) = −𝛼𝛼 � − − + + − � 12288 147456 36864 147456 3686400𝑅𝑅 2 1228800

We can find the mean temperature with 𝑇𝑇𝑤𝑤 − 𝑇𝑇𝑚𝑚 =

The wall heat flux is given by

𝑅𝑅

∫0 2𝜋𝜋𝜋𝜋𝑉𝑉𝑚𝑚 �𝑇𝑇𝑚𝑚 − 𝑇𝑇(𝑟𝑟)�𝑑𝑑𝑑𝑑 𝑅𝑅

∫0 2𝜋𝜋𝜋𝜋𝑉𝑉𝑚𝑚 𝑑𝑑𝑑𝑑

245

473𝑅𝑅 8 𝛼𝛼 = 1105920

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′′ 𝑞𝑞𝑤𝑤 = ℏ(𝑇𝑇𝑤𝑤 − 𝑇𝑇𝑚𝑚 ) = −𝑘𝑘

The partial derivative evaluated at the radius is

Solving for the heat transfer coefficient,

𝜕𝜕𝜕𝜕 � 𝜕𝜕𝜕𝜕 𝑅𝑅

𝜕𝜕𝜕𝜕 19𝑅𝑅 7 𝛼𝛼 � = 𝜕𝜕𝜕𝜕 𝑅𝑅 15360

𝜕𝜕𝜕𝜕 19𝑅𝑅 7 𝛼𝛼 � 𝜕𝜕𝜕𝜕 𝑅𝑅 −𝑘𝑘 �− 15360 � 1368𝑘𝑘 = = ℏ= 473𝑅𝑅 8 𝛼𝛼 𝑇𝑇𝑤𝑤 − 𝑇𝑇𝑚𝑚 473𝑅𝑅 1105920 −𝑘𝑘

The Nusselt number is given by

1368𝑘𝑘 ℏ𝐷𝐷 � 473𝑅𝑅 � (2𝑅𝑅) 2736 Nu = = = = 5.784 𝑘𝑘 𝑘𝑘 473

ITERATION 4 Starting from Equation (7),

1 𝜕𝜕 𝜕𝜕𝜕𝜕 𝑅𝑅 2 𝑟𝑟 6 𝑟𝑟 8 19𝑅𝑅 4 𝑟𝑟 4 211𝑅𝑅 6 𝑟𝑟 2 𝑟𝑟 10 1217𝑅𝑅 8 𝛼𝛼 �𝑟𝑟 � = 𝛼𝛼 �− + + − − + � 𝑟𝑟 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 12288 147456 36864 147456 3686400𝑅𝑅 2 1228800

(10)

After multiplying each side by r and integration of Eq. (10), we get

𝜕𝜕𝜕𝜕 𝑟𝑟 10 𝑅𝑅 2 𝑟𝑟 8 19𝑅𝑅 4 𝑟𝑟 6 211𝑅𝑅 6 𝑟𝑟 4 1217𝑅𝑅 8 𝑟𝑟 2 𝑟𝑟 12 𝑟𝑟 = 𝛼𝛼 � − + − + − � + 𝐶𝐶1 𝜕𝜕𝜕𝜕 1474560 98304 221184 589824 2457600 44236800𝑅𝑅 2

After dividing both sides by r and a second integration, we get

𝑟𝑟 10 𝑅𝑅 2 𝑟𝑟 8 19𝑅𝑅 4 𝑟𝑟 6 211𝑅𝑅 6 𝑟𝑟 4 1217𝑅𝑅 8 𝑟𝑟 2 𝑟𝑟 12 𝑇𝑇(𝑟𝑟) = 𝛼𝛼 � − + − + − � 14745600 786432 1327104 2359296 4915200 53084𝑅𝑅 2 + 𝐶𝐶1 ln(𝑟𝑟) + 𝐶𝐶2

We can apply the following boundary conditions: 𝜕𝜕𝜕𝜕

− at 𝑟𝑟 = 0 → 𝜕𝜕𝜕𝜕 = 0 ⟹ 𝐶𝐶1 = 0

30307𝑅𝑅 10 𝛼𝛼

− at 𝑟𝑟 = 𝑅𝑅 → 𝑇𝑇(𝑅𝑅) = 𝑇𝑇𝑤𝑤 ⟹ 𝐶𝐶2 = 𝑇𝑇𝑤𝑤 − 176947200

Therefore, the final form of the temperature profile is

𝑟𝑟 10 𝑅𝑅 2 𝑟𝑟 8 19𝑅𝑅 4 𝑟𝑟 6 211𝑅𝑅 6 𝑟𝑟 4 1217𝑅𝑅 8 𝑟𝑟 2 𝑟𝑟 12 𝑇𝑇𝑤𝑤 − 𝑇𝑇(𝑟𝑟) = −𝛼𝛼 � − − + + + 14745600 786432 1327104 2359296 4915200 3530841600𝑅𝑅 2 30307𝑅𝑅10 − � 176947200

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We can find the mean temperature with 𝑇𝑇𝑤𝑤 − 𝑇𝑇𝑚𝑚 =

𝑅𝑅

∫0 2𝜋𝜋𝜋𝜋𝑉𝑉𝑚𝑚 �𝑇𝑇𝑚𝑚 − 𝑇𝑇(𝑟𝑟)�𝑑𝑑𝑑𝑑 𝑅𝑅

∫0 2𝜋𝜋𝜋𝜋𝑉𝑉𝑚𝑚 𝑑𝑑𝑑𝑑

The wall heat flux is given by

′′ 𝑞𝑞𝑤𝑤 = ℏ(𝑇𝑇𝑤𝑤 − 𝑇𝑇𝑚𝑚 ) = −𝑘𝑘

The partial derivative evaluated at the radius is

𝜕𝜕𝜕𝜕 473𝑅𝑅 9 𝛼𝛼 � = 𝜕𝜕𝜕𝜕 𝑅𝑅 2211840

299𝑅𝑅10 𝛼𝛼 3096576

𝜕𝜕𝜕𝜕 � 𝜕𝜕𝜕𝜕 𝑅𝑅

Solving for the heat transfer coefficient,

𝜕𝜕𝜕𝜕 473𝑅𝑅 9 𝛼𝛼 � �− � 3311𝑘𝑘 −𝑘𝑘 𝜕𝜕𝜕𝜕 𝑅𝑅 2211840 ℏ= = = 229𝑅𝑅10 𝛼𝛼 𝑇𝑇𝑤𝑤 − 𝑇𝑇𝑚𝑚 1145𝑅𝑅 3096576 The Nusselt number is given by −𝑘𝑘

3311𝑘𝑘 ℎ𝐷𝐷 �1145𝑅𝑅 � (2𝑅𝑅) 6622 Nu = = = = 5.783 𝑘𝑘 𝑘𝑘 1145

Since the Nusselt number did not change much from the previous iteration we will take this as the result. Therefore, for a circular tube with constant temperature and fully-developed laminar flow, the Nusselt number is about 5.783. Closed Form Method In the mathematical approach we again start at the energy equation,

We can define

Therefore, we can rewrite as

𝑃𝑃𝑃𝑃𝑝𝑝 𝜐𝜐𝑧𝑧

𝜕𝜕𝜕𝜕 1 𝜕𝜕 𝜕𝜕𝜕𝜕 = = �𝑟𝑟𝑟𝑟 � 𝜕𝜕𝜕𝜕 𝑟𝑟 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕

Θ(𝑟𝑟, 𝑧𝑧) = 𝑃𝑃𝑃𝑃𝑝𝑝 𝜐𝜐𝑧𝑧

𝑇𝑇𝑤𝑤 − 𝑇𝑇(𝑟𝑟, 𝑧𝑧) 𝑇𝑇𝑤𝑤 − 𝑇𝑇𝑖𝑖𝑖𝑖

𝜕𝜕Θ 1 𝜕𝜕 𝜕𝜕Θ = = �𝑟𝑟𝑟𝑟 � 𝜕𝜕𝜕𝜕 𝑟𝑟 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕

To solve this problem, we to separation of variables, where

Θ(𝑟𝑟, 𝑧𝑧) = Φ(𝑟𝑟)𝑍𝑍(𝑧𝑧)

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and

Φ(𝑟𝑟) =

𝑇𝑇𝑤𝑤 − 𝑇𝑇(𝑟𝑟) 𝑇𝑇𝑤𝑤 − 𝑇𝑇𝑚𝑚

𝑍𝑍(𝑧𝑧) =

𝑇𝑇𝑤𝑤 − 𝑇𝑇𝑚𝑚 𝑇𝑇𝑤𝑤 − 𝑇𝑇𝑖𝑖𝑖𝑖

Substituting this into the energy equation we get

𝑑𝑑𝑑𝑑 1 𝑑𝑑 𝑑𝑑Φ = 𝑘𝑘𝑘𝑘(𝑟𝑟) 𝑟𝑟 𝑑𝑑𝑑𝑑 𝑟𝑟 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑

We can collect terms as

𝜌𝜌𝑐𝑐𝑝𝑝 𝜐𝜐𝑧𝑧 Φ(𝑟𝑟)

Defining the following:

1 𝑑𝑑𝑑𝑑 𝑘𝑘 1 1 𝑑𝑑 𝑑𝑑Φ = 𝑟𝑟 𝑍𝑍(𝑧𝑧) 𝑑𝑑𝑑𝑑 𝜌𝜌𝑐𝑐𝑝𝑝 𝜐𝜐𝑧𝑧 Φ(𝑟𝑟) 𝑟𝑟 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑

and

we can rewrite as

𝛼𝛼 =

𝑘𝑘 𝜌𝜌𝑐𝑐𝑝𝑝

1 𝑑𝑑 𝑑𝑑Φ 𝑑𝑑2 Φ 1 𝑑𝑑Φ 𝑟𝑟 = + 𝑟𝑟 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 2 𝑟𝑟 𝑑𝑑𝑑𝑑

1 𝑑𝑑𝑑𝑑 𝛼𝛼 1 𝑑𝑑2 Φ 𝛼𝛼 1 𝑑𝑑Φ = + 𝑍𝑍(𝑧𝑧) 𝑑𝑑𝑑𝑑 𝜐𝜐𝑧𝑧 Φ(𝑟𝑟) 𝑑𝑑𝑑𝑑 2 𝜐𝜐𝑧𝑧 𝑟𝑟 𝑑𝑑𝑑𝑑

Since both sides are a function of a single variable we can set them equal to a constant, 1 𝑑𝑑𝑑𝑑 𝛼𝛼 1 𝑑𝑑2 Φ 𝛼𝛼 1 1 𝑑𝑑Φ = + = −λ 𝑍𝑍(𝑧𝑧) 𝑑𝑑𝑑𝑑 𝜐𝜐𝑧𝑧 Φ(𝑟𝑟) 𝑑𝑑𝑑𝑑 2 𝜐𝜐𝑧𝑧 Φ(𝑟𝑟) 𝑟𝑟 𝑑𝑑𝑑𝑑

Thus, we can rewrite the radial component as

where we define

so that

𝑑𝑑 2 Φ 1 𝑑𝑑Φ 𝜐𝜐𝑧𝑧 + + λΦ(𝑟𝑟) = 0 𝑑𝑑𝑑𝑑 2 𝑟𝑟 𝑑𝑑𝑑𝑑 𝛼𝛼 𝜐𝜐𝑧𝑧 λ 𝛼𝛼

𝛽𝛽 = �

𝑑𝑑 2 Φ 1 𝑑𝑑Φ + + 𝛽𝛽 2 Φ(𝑟𝑟) = 0 𝑑𝑑𝑑𝑑 2 𝑟𝑟 𝑑𝑑𝑑𝑑 248

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Multiplying both sides by r2 we get 𝑟𝑟 2

𝑑𝑑 2 Φ 𝑑𝑑Φ + 𝑟𝑟 + 𝛽𝛽 2 𝑟𝑟 2 Φ(𝑟𝑟) = 0 2 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑

This is the form of Bessel function and the solution can be written as Φ(𝑟𝑟) = 𝐶𝐶1 𝐽𝐽0 (𝛽𝛽𝛽𝛽) + 𝐶𝐶2 𝑌𝑌0 (𝛽𝛽𝛽𝛽)

As r → 0 the above expression will become unbounded. To have it bounded at 0, C2 = 0. Therefore, we are left with Φ(𝑟𝑟) = 𝐶𝐶1 𝐽𝐽0 (𝛽𝛽𝛽𝛽)

From the definition of Φ we know that when r = R, Φ(R) = 0, since T(R) = Tw. Therefore, 𝐽𝐽0 (𝛽𝛽𝛽𝛽) = 0

and thus

𝛽𝛽𝛽𝛽 = 2.4048

Now, we can also look at conservation of energy by only considering the axial variation of the bulk temperature. This is represented as 𝜌𝜌𝑐𝑐𝑝𝑝 𝑉𝑉𝑚𝑚 𝜋𝜋𝑅𝑅 2

We know from the definition of Z (z) that

Thus, after some rearranging we get

where

𝑑𝑑𝑑𝑑 1 𝑑𝑑𝑇𝑇𝑚𝑚 =− 𝑑𝑑𝑑𝑑 𝑇𝑇𝑤𝑤 − 𝑇𝑇𝑖𝑖𝑖𝑖 𝑑𝑑𝑑𝑑

−𝜌𝜌𝜌𝜌𝑝𝑝 𝑉𝑉𝑚𝑚 𝑅𝑅

𝑑𝑑𝑑𝑑 𝑇𝑇𝑤𝑤 − 𝑇𝑇𝑚𝑚 = 2ℏ 𝑑𝑑𝑑𝑑 𝑇𝑇𝑤𝑤 − 𝑇𝑇𝑖𝑖𝑖𝑖

𝑇𝑇𝑤𝑤 − 𝑇𝑇𝑚𝑚 = 𝑍𝑍 𝑇𝑇𝑤𝑤 − 𝑇𝑇𝑖𝑖𝑖𝑖

Also,

ℏ=

Therefore,

where

𝑑𝑑𝑇𝑇𝑚𝑚 = ℎ(𝑇𝑇𝑤𝑤 − 𝑇𝑇𝑚𝑚 )2π 𝑅𝑅 𝑑𝑑𝑑𝑑

−

Nu𝑅𝑅 2𝑅𝑅

1 𝑑𝑑𝑑𝑑 2 Nu𝑘𝑘 = 𝑍𝑍 𝑑𝑑𝑑𝑑 𝜌𝜌𝑐𝑐𝑝𝑝 V𝑚𝑚 𝑅𝑅 2𝑅𝑅 1 𝑑𝑑𝑑𝑑 = −λ 𝑍𝑍 𝑑𝑑𝑑𝑑 249

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and

Finally,

𝑘𝑘 = 𝛼𝛼 𝜌𝜌𝑐𝑐𝑝𝑝

From above we have

λ=

and

𝛽𝛽 = �

Combining the two we get

𝛽𝛽 =

Nu𝛼𝛼 V𝑚𝑚 𝑅𝑅 2 𝜐𝜐𝑧𝑧 λ 𝛼𝛼

2.4048 𝑅𝑅

2.4048 𝜐𝜐𝑧𝑧 =� λ 𝑅𝑅 𝛼𝛼 Plugging in the expression for λ that we just found, we get 2.4048 𝜐𝜐𝑧𝑧 Nu𝛼𝛼 =� 𝑅𝑅 𝛼𝛼 V𝑚𝑚 𝑅𝑅 2

Solving for the Nusselt number we get

Since this is slug flow we know that Thus,

Nu = 2.4048

V𝑚𝑚 𝜐𝜐𝑧𝑧

𝜐𝜐𝑧𝑧 = V𝑚𝑚

Nu = 2.40482 = 5.783

which is equivalent to the result from the iterative method.

Flat Plate: For the flat plate question, the mathematical approach will be used similar to above. In the mathematical approach we again start at the energy equation,

We can define

𝜌𝜌𝑐𝑐𝑝𝑝 𝜐𝜐𝑥𝑥 =

Therefore, we can rewrite as

Θ(𝑥𝑥, 𝑦𝑦) =

𝜕𝜕𝜕𝜕 𝜕𝜕 2 𝑇𝑇 = 𝑘𝑘 2 𝜕𝜕𝜕𝜕 𝜕𝜕𝑦𝑦

𝑇𝑇𝑤𝑤 − 𝑇𝑇(𝑥𝑥, 𝑦𝑦) 𝑇𝑇𝑤𝑤 − 𝑇𝑇𝑖𝑖𝑖𝑖

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𝜕𝜕Θ 𝜗𝜗 2 Θ 𝜌𝜌𝑐𝑐𝑝𝑝 𝜐𝜐𝑥𝑥 = 𝑘𝑘 2 𝜗𝜗𝜗𝜗 𝜗𝜗𝑦𝑦

To solve this problem, we to separation of variables, where

and

Θ = (𝑥𝑥, 𝑦𝑦) = 𝑋𝑋(𝑥𝑥)𝑌𝑌(𝑥𝑥) 𝑌𝑌(𝑦𝑦) =

𝑇𝑇𝑤𝑤 − 𝑇𝑇(𝑦𝑦) 𝑇𝑇𝑤𝑤 − 𝑇𝑇𝑖𝑖𝑖𝑖

𝑋𝑋(𝑥𝑥) =

Substituting this into the energy equation we get

𝑇𝑇𝑤𝑤 − 𝑇𝑇𝑚𝑚 𝑇𝑇𝑤𝑤 − 𝑇𝑇𝑖𝑖𝑖𝑖

𝑑𝑑𝑑𝑑 𝑑𝑑2 𝑌𝑌 = 𝑘𝑘𝑘𝑘(𝑥𝑥) 2 𝑑𝑑𝑑𝑑 𝑑𝑑𝑦𝑦

We can collect terms as

𝜌𝜌𝑐𝑐𝑝𝑝 𝜐𝜐𝑥𝑥 𝑌𝑌(𝑦𝑦)

Defining the following,

1 𝑑𝑑𝑑𝑑 𝑘𝑘 1 𝑑𝑑2 𝑌𝑌 = 𝑋𝑋(𝑥𝑥) 𝑑𝑑𝑑𝑑 𝜌𝜌𝑐𝑐𝑝𝑝 𝜐𝜐𝑥𝑥 𝑌𝑌(𝑦𝑦) 𝑑𝑑𝑦𝑦 2

we can rewrite as

𝛼𝛼 =

𝑘𝑘 𝜌𝜌𝑐𝑐𝑝𝑝

1 𝑑𝑑𝑑𝑑 𝛼𝛼 1 𝑑𝑑2 𝑌𝑌 = 𝑋𝑋(𝑥𝑥) 𝑑𝑑𝑑𝑑 𝜐𝜐𝑥𝑥 𝑌𝑌(𝑦𝑦) 𝑑𝑑𝑦𝑦 2 Since both sides are a function of a single variable we can set them equal to a constant, 1 𝑑𝑑𝑑𝑑 𝛼𝛼 1 𝑑𝑑 2 𝑌𝑌 = = −λ 𝑋𝑋(𝑥𝑥) 𝑑𝑑𝑑𝑑 𝜐𝜐𝑥𝑥 𝑌𝑌(𝑦𝑦) 𝑑𝑑𝑦𝑦 2 Thus, we can rewrite the radial component as where we define

so that

𝑑𝑑 2 𝑌𝑌 𝜐𝜐𝑥𝑥 + λ𝑌𝑌(𝑦𝑦) = 0 𝑑𝑑𝑦𝑦 2 𝛼𝛼 𝜐𝜐𝑥𝑥 λ 𝛼𝛼

𝛽𝛽 = �

𝑑𝑑2 𝑌𝑌 + 𝛽𝛽 2 𝑌𝑌(𝑦𝑦) = 0 𝑑𝑑𝑦𝑦 2 251

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The solution can be written as 𝑌𝑌(𝑦𝑦) = 𝐶𝐶1 cos(𝛽𝛽𝛽𝛽) + 𝐶𝐶2 sin(𝛽𝛽𝛽𝛽)

Since the two plate are identical and the flow is fully developed, we can apply a symmetry boundary condition at y = 0. Therefore 𝑑𝑑𝑑𝑑 � = −𝐶𝐶1 sin(0) + 𝐶𝐶2 cos(0) 𝑑𝑑𝑑𝑑 𝑦𝑦=0

Therefore, C2 = 0. From the definition of Y we know that when y = y0, Y (y0) = 0, since T(y0) = Tw. We define y0 is the distance from the center between the plates to the wall. Therefore, cos(𝛽𝛽𝑦𝑦0 ) = 0

and thus

𝛽𝛽𝑦𝑦0 =

π 2

Now, we can also look at conservation of energy by only considering the axial variation of the bulk temperature. This is represented as 𝑑𝑑𝑑𝑑𝑚𝑚 = ℏ(𝑇𝑇𝑤𝑤 − 𝑇𝑇𝑚𝑚 )2 𝑑𝑑𝑑𝑑 where we take the cross sectional area as 2y0 and the perimeter as just 2. We know from the definition of Χ(x) that 𝜌𝜌𝑐𝑐𝑝𝑝 𝑉𝑉𝑚𝑚 2𝑦𝑦0

𝑑𝑑𝑑𝑑 1 𝑑𝑑𝑑𝑑𝑚𝑚 = 𝑑𝑑𝑑𝑑 𝑇𝑇𝑤𝑤 − 𝑇𝑇𝑖𝑖𝑖𝑖 𝑑𝑑𝑑𝑑

Thus, after some rearranging we get

where

−𝜌𝜌𝑐𝑐𝑝𝑝 𝑉𝑉𝑚𝑚 2𝑦𝑦0

𝑑𝑑𝑑𝑑 𝑇𝑇𝑤𝑤 − 𝑇𝑇𝑚𝑚 = 2ℎ 𝑑𝑑𝑑𝑑 𝑇𝑇𝑤𝑤 − 𝑇𝑇𝑖𝑖𝑖𝑖

𝑇𝑇𝑤𝑤 − 𝑇𝑇𝑚𝑚 = 𝑋𝑋 𝑇𝑇𝑤𝑤 − 𝑇𝑇𝑖𝑖𝑖𝑖

Also,

ℏ=

where the hydraulic diameter is 4y0. Therefore,

where

−

Nu𝑘𝑘 4𝑦𝑦0

1 𝑑𝑑𝑑𝑑 1 Nu𝑘𝑘 = 𝑋𝑋 𝑑𝑑𝑑𝑑 𝜌𝜌𝑐𝑐𝑝𝑝 𝑉𝑉𝑚𝑚 𝑦𝑦0 4𝑦𝑦0

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and

1 𝑑𝑑𝑑𝑑 = −λ 𝑋𝑋 𝑑𝑑𝑑𝑑

Finally,

𝑘𝑘 = 𝛼𝛼 𝜌𝜌𝑐𝑐𝑝𝑝

From above we have

λ=

and

𝛽𝛽 = �

Combining the two we get

𝛽𝛽 =

Nu𝛼𝛼 4𝑉𝑉𝑚𝑚 𝑦𝑦02 𝜐𝜐𝑥𝑥 λ 𝛼𝛼

π 2𝑦𝑦0

π 𝜐𝜐𝑥𝑥 =� λ 2𝑦𝑦0 𝛼𝛼

Plugging in the expression for λ that we just found, we get

Solving for the Nusselt number we get

π 𝜐𝜐𝑥𝑥 Nu𝛼𝛼 =� 2𝑦𝑦0 𝛼𝛼 4V𝑚𝑚 𝑦𝑦02

Since this is slug flow we know that

Nu = π2

Thus,

V𝑚𝑚 𝜐𝜐𝑥𝑥

𝜐𝜐𝑥𝑥 = V𝑚𝑚

Nu = π2 = 9.870

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PROBLEM 10.3 QUESTION Derivation of Nusselt Number for Laminar Flow in an Equivalent Annulus (Section 10.2) Derive the Nusselt number for slug flow in the equivalent annulus of an infinite rod array by solving the differential energy equation subject to the appropriate boundary conditions, that is, 𝜕𝜕 2 𝑇𝑇 1 𝜕𝜕𝜕𝜕 𝜐𝜐𝜐𝜐𝑐𝑐𝑝𝑝 𝜕𝜕𝜕𝜕 + = 𝜕𝜕𝑟𝑟 2 𝑟𝑟 𝜕𝜕𝜕𝜕 𝑘𝑘 𝜕𝜕𝜕𝜕

Answer: Nu =

2(𝑟𝑟02 − 𝑟𝑟𝑖𝑖2 )3 𝑟𝑟𝑖𝑖2 [𝑟𝑟04 ln (𝑟𝑟0 ⁄𝑟𝑟𝑖𝑖 ) − (𝑟𝑟02 − 𝑟𝑟𝑖𝑖2 )(3𝑟𝑟02 − 𝑟𝑟𝑖𝑖2 )⁄4]

PROBLEM 10.3 SOLUTION

Derivation of Nusselt Number for Laminar Flow in an Equivalent Annulus (Section 10.2) The rod array and equivalent annulus are shown in Figure SM-10.1:

(Flow areas (shaded region) in both geometries are the same) Figure SM-10.1 Let's consider the case that heat is uniformly generated in a rod. Thus, we will use the boundary condition of the uniform heat flux at r = ri (This is the case in the reactor core, since q″ at the fuel rod surface is generally known). The energy equation is given as follows: 1 𝜕𝜕 𝜕𝜕𝜕𝜕 𝜌𝜌𝑐𝑐𝑝𝑝 𝜕𝜕𝜕𝜕 𝑟𝑟 = 𝜐𝜐 𝑟𝑟 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝑘𝑘 𝜕𝜕𝜕𝜕

(1)

where, υ = υm (mean velocity) for slug flow. Rearranging the equation,

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𝜕𝜕 𝜕𝜕𝜕𝜕 𝜌𝜌𝑐𝑐𝑝𝑝 𝜕𝜕𝜕𝜕 𝑟𝑟 = 𝜐𝜐𝑚𝑚 𝑟𝑟 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝑘𝑘 𝜕𝜕𝜕𝜕

(2)

For this problem, the boundary conditions are − at 𝑟𝑟 = 𝑟𝑟𝑖𝑖 , 𝑇𝑇 = 𝑇𝑇𝑤𝑤 (𝑧𝑧) 𝜕𝜕𝜕𝜕

− at 𝑟𝑟 = 𝑟𝑟0 , 𝜕𝜕𝜕𝜕 = 0.

Integrating Equation (2) from r0 to an arbitrary radius, r, and applying the second boundary condition we get 𝜕𝜕𝜕𝜕 𝜌𝜌𝑐𝑐𝑝𝑝 𝜐𝜐𝑚𝑚 𝜕𝜕𝜕𝜕 𝑟𝑟 2 𝑟𝑟02 = � − � 𝜕𝜕𝜕𝜕 𝑘𝑘 𝜕𝜕𝜕𝜕 2 2 After dividing by r, the equation becomes 𝑟𝑟

𝜕𝜕𝜕𝜕 𝜌𝜌𝑐𝑐𝑝𝑝 𝜐𝜐𝑚𝑚 𝜕𝜕𝜕𝜕 𝑟𝑟 𝑟𝑟02 = � − � 𝜕𝜕𝜕𝜕 𝑘𝑘 𝜕𝜕𝜕𝜕 2 2𝑟𝑟

(3)

(4)

We may integrate this equation from ri to an arbitrary radius r and apply the first boundary condition to get 𝑇𝑇(𝑟𝑟, 𝑧𝑧) =

𝜌𝜌𝑐𝑐𝑝𝑝 𝜐𝜐𝑚𝑚 𝜕𝜕𝜕𝜕 𝑟𝑟 2 𝑟𝑟𝑖𝑖2 𝑟𝑟02 𝑟𝑟 � − − ln � �� + 𝑇𝑇𝑤𝑤 (𝑧𝑧) 𝑘𝑘 𝜕𝜕𝜕𝜕 4 4 2 𝑟𝑟𝑖𝑖 𝜕𝜕𝜕𝜕

(5)

𝜕𝜕𝑇𝑇

In the case of a constant heat flux, 𝜕𝜕𝜕𝜕 = 𝜕𝜕𝜕𝜕𝑚𝑚 , where Tm is the mean temperature and is obtained from an energy balance on a cross section:

so that

𝜌𝜌𝜐𝜐𝑚𝑚 (𝑟𝑟02 − 𝑟𝑟𝑖𝑖2 )𝑐𝑐𝑝𝑝

𝜕𝜕𝑇𝑇𝑚𝑚 = −𝑞𝑞″2𝜋𝜋𝑟𝑟𝑖𝑖 𝜕𝜕𝜕𝜕

𝜕𝜕𝑇𝑇𝑚𝑚 2𝑞𝑞″ 𝑟𝑟𝑖𝑖 =− 𝜕𝜕𝜕𝜕 𝜌𝜌𝑐𝑐𝑝𝑝 𝜐𝜐𝑚𝑚 𝑟𝑟02 − 𝑟𝑟𝑖𝑖2

Therefore, 𝑇𝑇(𝑟𝑟, 𝑧𝑧) = −

2𝑞𝑞″ 𝑟𝑟𝑖𝑖 𝑟𝑟 2 𝑟𝑟𝑖𝑖2 𝑟𝑟02 𝑟𝑟 � − − ln � �� + 𝑇𝑇𝑤𝑤 (𝑧𝑧) 2 2 𝑘𝑘 𝑟𝑟0 − 𝑟𝑟𝑖𝑖 4 4 2 𝑟𝑟𝑖𝑖

(6)

(7)

(8)

The mean temperature can be calculated with 𝑇𝑇𝑚𝑚 (𝑧𝑧) − 𝑇𝑇𝑤𝑤 (𝑧𝑧) =

𝑟𝑟𝑜𝑜

∫𝑟𝑟 [𝑇𝑇(𝑟𝑟, 𝑧𝑧) − 𝑇𝑇𝑤𝑤 (𝑧𝑧)]𝜐𝜐𝑚𝑚 2𝜋𝜋𝜋𝜋𝜋𝜋𝜋𝜋

The resulting temperature difference is

𝑖𝑖

𝑟𝑟𝑜𝑜

∫𝑟𝑟 𝜐𝜐𝑚𝑚 2𝜋𝜋𝜋𝜋𝜋𝜋𝜋𝜋

(9)

𝑖𝑖

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𝑇𝑇𝑚𝑚 (𝑧𝑧) − 𝑇𝑇𝑤𝑤 (𝑧𝑧)

𝑇𝑇04 2 2𝑞𝑞″ 𝑟𝑟𝑖𝑖 1 2 𝑟𝑟0 2 2 (𝑟𝑟 ) � � = 2 �− � � − 𝑟𝑟 − ln 0 𝑖𝑖 𝑘𝑘 𝑟𝑟02 − 𝑟𝑟𝑖𝑖2 16 4 𝑟𝑟𝑖𝑖 𝑟𝑟0 − 𝑟𝑟𝑖𝑖2 𝑟𝑟02 + (𝑟𝑟02 − 𝑟𝑟𝑖𝑖2 )� 8

(10)

The heat transfer coefficient is given as

ℏ(𝑧𝑧) =

where the Nusselt number is

𝑞𝑞″ 𝑇𝑇𝑚𝑚 (𝑧𝑧) − 𝑇𝑇𝑤𝑤 (𝑧𝑧)

ℏ(𝑧𝑧)𝐷𝐷ℎ 𝑘𝑘 For the equivalent annulus, the hydraulic diameter is Nu(𝑧𝑧) =

4𝐴𝐴 4𝜋𝜋(𝑟𝑟02 − 𝑟𝑟𝑖𝑖2 ) = 𝐷𝐷ℎ = 𝑃𝑃ℎ 2𝜋𝜋𝑟𝑟𝑖𝑖 Therefore, the Nusselt number is

From Equation (10), 𝑞𝑞″

2(𝑟𝑟02 − 𝑟𝑟𝑖𝑖2 ) Nu(𝑧𝑧) = 𝑟𝑟𝑖𝑖 �𝑇𝑇𝑚𝑚 (𝑧𝑧) − 𝑇𝑇𝑤𝑤 (𝑧𝑧)�𝑘𝑘 𝑞𝑞″

�𝑇𝑇𝑚𝑚 (𝑧𝑧) − 𝑇𝑇𝑤𝑤 (𝑧𝑧)�𝑘𝑘

The resulting Nusselt number is Nu =

which is equivalent to

(𝑟𝑟02 − 𝑟𝑟𝑖𝑖2 )2

𝑟𝑟 2 1 𝑟𝑟 𝑇𝑇𝑖𝑖 �𝑟𝑟04 ln � 𝑟𝑟𝑜𝑜 � − 𝑟𝑟0 (𝑟𝑟02 − 𝑟𝑟𝑖𝑖2 ) − 4 (𝑟𝑟02 − 𝑟𝑟𝑖𝑖2 )2 � 𝑖𝑖

(11)

(12)

(13)

(14)

(15)

𝑖𝑖

2(𝑟𝑟02 − 𝑟𝑟𝑖𝑖2 )3 𝑟𝑟 2 1 𝑟𝑟 𝑟𝑟𝑖𝑖 �𝑟𝑟04 ln � 𝑟𝑟𝑜𝑜 � − 20 (𝑟𝑟02 − 𝑟𝑟𝑖𝑖2 ) − 4 (𝑟𝑟02 − 𝑟𝑟𝑖𝑖2 )2 � 𝑖𝑖

2(𝑟𝑟02 − 𝑟𝑟𝑖𝑖2 )3 Nu = 2 4 𝑟𝑟𝑖𝑖 [𝑟𝑟0 ln(𝑟𝑟0 ⁄𝑟𝑟𝑖𝑖 ) − (𝑟𝑟02 − 𝑟𝑟𝑖𝑖2 ) − (3𝑟𝑟02 − 𝑟𝑟𝑖𝑖2 )⁄4]

(16)

(17)

PROBLEM 10.4 QUESTION

Estimate the Effect of Turbulence on Heat Transfer in SFBR Fuel Bundles (Section 10.3) Consider the fuel bundle of a SFBR whose geometry is described in Table 1.3. Using Dwyer's recommendations for the values of ϵM/υ (Figure 10.7) and Equation 10.71, estimate the ratio of 256

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𝜖𝜖𝐻𝐻 ⁄𝜖𝜖𝑚𝑚 for the sodium velocities in the bundles: 1. For υ = 10 m/s 2. For υ = 1 m/s

FIGURE 10.7

Values of (εM/ν)max for fully developed turbulent flow of liquid metals through circular tubes, annuli, and rod bundles with equilateral triangular spacing [15].

Answers: 1. For υ = 10 m/s:

2. For υ = 1 m/s:

𝜖𝜖𝐻𝐻 𝜖𝜖𝑚𝑚 in core = 0.510; � � = 120 𝜖𝜖𝑚𝑚 V 𝑚𝑚𝑚𝑚𝑚𝑚 𝜖𝜖𝑚𝑚 𝜖𝜖𝐻𝐻 in blanket = 0.722; � � = 180 𝜖𝜖𝑚𝑚 V 𝑚𝑚𝑚𝑚𝑚𝑚 𝜖𝜖𝐻𝐻 in both core and blanket = 0 𝜖𝜖𝑚𝑚

PROBLEM 10.4 SOLUTION Estimate the Effect of Turbulence on Heat Transfer in SFBR Fuel Bundles (Section 10.3) Consider the fuel bundle of a SFBR whose geometry is described in Table 1.3. Using Dwyer's recommendations for the values of ϵm and ϵH (Equation 10.71), estimate the ratio of ϵH/ϵm for the sodium velocities in the bundles. Geometric parameters taken from Table 1.3 are: − pin diameter in core, DC = 8.5 mm

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− pin diameter in blanket, DB = 15.8 mm − pin pitch in core, PC = 9.8 mm − pin pitch in blanket, PB = 17.0 mm Sodium properties are: − density, ρ = 817.7 kg/m3 − specific heat, cp = 1254 J/kg K − conductivity, 62.6 W/m K − viscosity, μ = 2.276 × 10−4 Pa s 𝜇𝜇𝑐𝑐𝑝𝑝

− Prandtl number, 𝑃𝑃𝑃𝑃 = 𝑘𝑘 = 4.559 × 10−3

The dimensions (area and hydraulic diameter) of the core and blanket are: 4𝐴𝐴𝐶𝐶 √3 2 𝜋𝜋𝐷𝐷𝐶𝐶2 𝐴𝐴𝐶𝐶 = 𝑃𝑃𝐶𝐶 − = 1.321 × 10−5 m2 𝐷𝐷𝑒𝑒𝑒𝑒 = = 3.959 × 10−3 m 𝜋𝜋𝐷𝐷𝐶𝐶 4 8 2 2 𝜋𝜋𝐷𝐷 4𝐴𝐴 √3 2 𝐵𝐵 𝐵𝐵 𝑃𝑃𝐵𝐵 − = 2.711 × 10−5 m2 𝐷𝐷𝑒𝑒𝑒𝑒 = = 4.369 × 10−3 m 𝐴𝐴𝐵𝐵 = 𝜋𝜋𝐷𝐷 4 8 𝐵𝐵 2

For the 10 m/s case the Reynolds number for the core and blanket; respectfully, are: Re𝐶𝐶 =

𝜌𝜌𝜌𝜌10 𝐷𝐷𝑒𝑒𝑒𝑒 = 1.422 × 105 𝜇𝜇

Re𝐵𝐵 =

𝜌𝜌𝜌𝜌10 𝐷𝐷𝐷𝐷𝐵𝐵 = 1.57 × 105 𝜇𝜇

𝜌𝜌𝜌𝜌1 𝐷𝐷𝑒𝑒𝑒𝑒 = 1.422 × 104 𝜇𝜇

Re𝐵𝐵 =

𝜌𝜌𝜌𝜌1 𝐷𝐷𝐷𝐷𝐵𝐵 = 1.57 × 104 𝜇𝜇

For the 1 m/s case the Reynolds number for the core and blanket, respectfully, are: Re𝐶𝐶 =

From Figure 10.7, the maximum (ϵM/v)max is a function of Reynolds number and pitch-to-diameter ratio. For the 10 m/s case this parameter for the core and blanket are:

For the 1 m/s case they are

𝜖𝜖𝑀𝑀 � = 120 𝑣𝑣 𝑚𝑚𝑚𝑚𝑚𝑚,𝐶𝐶

�

𝜖𝜖𝑀𝑀 � = 180 𝑣𝑣 𝑚𝑚𝑚𝑚𝑚𝑚,𝐵𝐵

𝜖𝜖𝑀𝑀 � = 18 𝑣𝑣 𝑚𝑚𝑚𝑚𝑚𝑚,𝐶𝐶

�

𝜖𝜖𝑀𝑀 � = 22 𝑣𝑣 𝑚𝑚𝑚𝑚𝑚𝑚,𝐵𝐵

�

Using Equation 10.71, the ratio of ϵH/ϵm for the 10 m/s case is

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𝜖𝜖𝐻𝐻 � = 1− 𝜖𝜖𝑚𝑚 𝐶𝐶

1.82 = 0.510 𝜖𝜖𝑀𝑀 1.4 𝑃𝑃𝑃𝑃 � 𝑣𝑣 � 𝑚𝑚𝑚𝑚𝑚𝑚,𝐶𝐶

𝜖𝜖𝐻𝐻 � =1− 𝜖𝜖𝑚𝑚 𝐵𝐵

1.82 = −5.979 𝜖𝜖𝑀𝑀 1.4 𝑃𝑃𝑃𝑃 � 𝑣𝑣 � 𝑚𝑚𝑚𝑚𝑚𝑚,𝐶𝐶

𝜖𝜖𝐻𝐻 � =1− 𝜖𝜖𝑚𝑚 𝐵𝐵

Finally for the 1 m/s case, the results are 𝜖𝜖𝐻𝐻 � = 1− 𝜖𝜖𝑚𝑚 𝐶𝐶

1.82 = 0.722 𝜖𝜖𝑀𝑀 1.4 𝑃𝑃𝑃𝑃 � 𝑣𝑣 � 𝑚𝑚𝑚𝑚𝑚𝑚,𝐵𝐵

Both of these results are negative and therefore taken as 0.

1.82 = −4.27 𝜖𝜖𝑀𝑀 1.4 𝑃𝑃𝑃𝑃 � 𝑣𝑣 � 𝑚𝑚𝑚𝑚𝑚𝑚,𝐵𝐵

PROBLEM 10.5 QUESTION Reynolds Analogy and Equivalent Diameter Problem (Section 10.3) Consider a uniformly heated tube (constant heat flux) of diameter 0.025 m with fluid flowing at an average velocity of 0.5 m/s. Find the fully developed heat transfer coefficient for two different fluids (Fluid A and Fluid B, whose properties are given in Table 10.8) by the following two procedures: Procedure 1: Use only friction factor data. If you find this procedure not valid, state the reason. Procedure 2: Select the relevant heat transfer correlation. In summary, you are asked to provided four answers, that is, Procedure #1 Procedure #2

Fluid A

Fluid B

=? =?

TABLE 10.8 Fluid Properties for Problem 10.5 Fluid Properties

Fluid A

Fluid B

k (W/m°C)

0.5

ρ (kg/m3)

700

818

μ (kg/m s)

8.7 × 10−5

2.3 × 10−4

cp (J/kg°C)

6250

1250

Answers: Procedure #1:

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Fluid A:  = 5025 W/m2 K Fluid B:  = cannot be computed Procedure #2: Fluid A:  = 4780 W/m2 K Fluid B:  = 22,000 W/m2 K

PROBLEM 10.5 S OLUTION Reynolds Analogy and Equivalent Diameter Problem (Section 10.3) From the problem statement: − tube diameter, D = 0.025 m − fluid average velocity, υ = 0.5 m/s The following properties are given in Table 10.8; − conductivity, kA = 0.5 W/m K and kB = 63 W/m K − density, ρA = 700 kg/m3 and ρB = 818 kg/m3 − viscosity, μA = 8.7 × 10−5 Pa s and μB = 2.3 × 10−4 Pa s J

− specific heat, CPA = 6250 kg K and CPB = 1250 kg K

The Prandtl number for fluid A and fluid B are Pr𝐴𝐴 =

𝜇𝜇𝐴𝐴 𝑐𝑐𝑝𝑝 𝐴𝐴 𝜇𝜇𝐵𝐵 𝑐𝑐𝑝𝑝𝑝𝑝 = 1.087 Pr𝐵𝐵 = = 4.563 × 10−3 𝑘𝑘𝐴𝐴 𝑘𝑘𝐵𝐵

The Reynolds number for fluid A and fluid B are Re𝐴𝐴 =

𝜌𝜌𝐴𝐴 𝜐𝜐𝜐𝜐 = 1.006 × 105 𝜇𝜇𝐴𝐴

Re𝐵𝐵 =

𝜌𝜌𝐵𝐵 𝜐𝜐𝜐𝜐 = 4.446 × 104 𝜇𝜇𝐵𝐵

Procedure #1: Only fluid A is valid for the Reynolds analogy, fluid B has a low Pr number. The friction factor of fluid A using the McAdams correlation is 𝑓𝑓𝐴𝐴 = 0.184𝑅𝑅𝑅𝑅𝐴𝐴−0.2 = 0.018

The heat transfer coefficient from the Reynolds analogy is ℏ𝐴𝐴1 =

𝑓𝑓𝐴𝐴 𝜌𝜌𝐴𝐴 𝑐𝑐𝑝𝑝𝑝𝑝 𝜐𝜐 𝑊𝑊 = 5025 2 8 𝑚𝑚 𝐾𝐾

Procedure #2: For fluid A the Dittus Boelter correlation can be used to calculate the Nusselt number,

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Nu𝐴𝐴 = 0.023Re𝐴𝐴0.8 Pr𝐴𝐴0.4 = 238.941

The heat transfer coefficient is therefore

ℏ𝐴𝐴2 =

Nu𝐴𝐴 𝑘𝑘𝐴𝐴 𝑊𝑊 = 4779 2 𝐷𝐷 𝑚𝑚 𝐾𝐾

For fluid B, the Peclet number can be calculated with

Pe𝐵𝐵 = Re𝐵𝐵 Pr𝐵𝐵 = 202.877

Using the Lyon correlation for the Nusselt number (Equation 10.126a) yields Nu𝐵𝐵 = 7 + 0.025Pe0.8 = 8.753

The heat transfer coefficient is then

ℏ𝐵𝐵2 =

Nu𝐵𝐵 𝑘𝑘𝐵𝐵 W = 22057 2 𝐷𝐷 m K

PROBLEM 10.6 QUESTION Determining the Temperature of the Primary Side of a Steam Generator (Section 10.5) Consider the flow of high pressure water through the U-tubes of a PWR steam generator. There are 5700 tubes with outside diameter 19 mm, wall thickness 1.2 mm, and average length 16.0 m. The steady-state operating conditions are: − Total primary flow through the tubes = 5100 kg/s − Total heat transfer from primary to secondary = 820 MW − Secondary pressure = 5.6 MPa (272°C saturation) 1. What is the primary temperature at the tube inlet? 2. What is the primary temperature at the tube outlet? Use a Dittus–Boelter equation for the primary side heat transfer coefficient. Assume that the tube wall surface temperature on the secondary side is constant at 276°C. Properties For water at 300°C and 15 MPa: Density = 726 kg/m3 Specific heat = 5.7 kJ/kg K Viscosity = 92 μPa s Thermal conductivity = 0.56 W/m K For tube wall Thermal conductivity = 26 W/m K Hint: Consideration of the axial variation of the primary coolant bulk temperature is required.

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Answers: 1. Tp, in = 307.2°C 2. Tp, out = 279.0°C

PROBLEM 10.6 SOLUTION Determining the Temperature of the Primary Side of a Steam Generator (Section 10.5) From the problem statement: − the number of tubes, N = 5700 − tube outside diameter, Do = 19 mm − tube wall thickness, t = 1.2 mm − tube inside diameter, Di = Do – 2t = 16.6 mm − average tube length, L = 16 m − total primary flow through the tubes, 𝑚𝑚̇𝑡𝑡𝑡𝑡𝑡𝑡 = 5100 kg⁄s

− total heat transfer from primary to secondary, 𝑄𝑄̇ = 820 MW − secondary side pressure, Pss = 5.6 MPa

− secondary side wall surface temperature, Twss = 276 °C The following water properties are given for the primary side: − density, ρPS = 726 kg/m3 − specific heat, cpPS = 5.7 kJ/kg K − viscosity, μPS = 92 μPa s − thermal conductivity, kPS = 0.56 W/m K − Prandtl number, PrPS =

For tube wall

μPSC cpPS kPS

= 0.936

kw = 26W/m K The unknown primary temperatures at the tube inlet, T(0), and tube outlet, T(L), can be expressed in two equations. First, the following equation involving the total heat transfer from primary to the secondary side:

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(1) Second, following the hint in the problem statement develops the expression for the axial variation of the primary coolant temperature in a tube, i. For a differential length, conservation of energy for the primary side is (2) At an arbitrary location z, the heat transfer to the wall is = qi′′( z )  PS (T ( z ) − TwPS ( z ) )

(3)

Conduction through the wall can be represented by (4)

Eliminating the wall temperature from the primary side in Equation (4) by use of Equation (3), the heat flux is qi′′( z ) =

 PS (T ( z ) − TwSS ( z ) )

(5)

D   PS Di ln  o   Di  1+ 2k w

Substituting the heat flux from Equation (5) into Equation (2), the conservation of energy equation becomes  PS π Di dT = − − A (T ( z ) − TwSS ( z ) ) (T ( z ) − TwSS ( z ) ) = Do  dz    PS Di ln D  i  m i c pPS 1 + k 2   w  

To evaluate the parameter A, the heat transfer, the tube mass flow rate,

(6)

, and the heat transfer

coefficient, h, are needed and evaluated as follows: The mass flow rate for tube is

and the flow area per tube is

𝑚𝑚̇𝑖𝑖 =

𝑚𝑚̇𝑡𝑡𝑡𝑡𝑡𝑡 = 0.895 kg⁄s 𝑁𝑁

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𝐴𝐴𝑖𝑖 =

𝜋𝜋 2 𝐷𝐷 = 2.164 × 10−4 m2 4 𝑖𝑖

(8)

The primary side Reynolds number can be calculated with Re =

𝑚𝑚̇𝑖𝑖 𝐷𝐷𝑖𝑖 = 7.459 × 105 𝐴𝐴𝑖𝑖 𝜇𝜇𝑃𝑃𝑃𝑃

(9)

The friction factor can be calculated from the McAdams correlation 𝑓𝑓 = 0.184Re−0.2 = 0.012

(10)

Nu = 0.023Re0.8 Pr 0.4 = 1118.1

(11)

Using the Dittus-Boelter correlation the Nusselt number is and thus the heat transfer coefficient is ℏ𝑃𝑃𝑃𝑃 =

The mass flow rate for tube is = A

Nu𝑘𝑘𝑘𝑘𝑘𝑘 W = 3.772 × 104 2 𝐷𝐷𝑖𝑖 m K

 PS π Di 1 0.147 = m   Do     PS Di ln    Di   m i c pPS 1 +   2k w    

(12)

(13)

Solving the differential Equation (6) yields T (L) = TwSS + (T (0) – TwSS) exp (–AL)

(14)

In this equation neither the primary side inlet temperature, T(0), nor the primary side outlet temperature, T(L), are known. However, the primary side temperature difference is known and expressed in Equation (1). Therefore there are two equations, (1) and (14), and two unknowns, T(0) and T(L). Solving these equations simultaneously yields a primary inlet temperature of T(0) = 307.2ºC and a primary outlet temperature of T(L) = 279.0ºC

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PROBLEM 10.7 QUESTION Comparison of Heat Transfer Characteristics of Water and Helium (Section 10.5) Consider a new design of a thermal reactor that requires square arrays of fuel rods. Heat is being generated uniformly along the fuel rods. Water and helium are being considered as single-phase coolants. Relevant coolant properties are given in Table 10.9. The design condition is that the maximum cladding surface temperature should remain below 316°C. 1. Find the minimum mass flow rate of water to meet the design requirement. 2. Would the required mass flow rate of helium be higher or lower than that of water? Geometry of square array P = pitch = 1.4 cm D = fuel rod diameter = 1.09 cm H = fuel height = 3.66 m Operating conditions Heat flux = 78.9 W/cm2 Coolant inlet temperature = 260°C TABLE 10.9 Coolant Properties for Problem 10.7 Coolant ρ (kg/m3) cp (kJ/kg K)

μ (kg/m s)

k (W/m K)

Water

5.317

0.955 × 10−4

0.564

5.225

0.298 × 10−4

0.230

735.3

Helium

0.865

Answers: 1. 𝑚𝑚̇𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 = 0.451 kg⁄s 2. 𝑚𝑚̇𝐻𝐻𝐻𝐻 = 0.491 kg⁄s

PROBLEM 10.7 SOLUTION

Comparison of Heat Transfer Characteristics of Water and Helium (Section 10.5) From the problem statement: − pitch, P = 1.4 cm − fuel rod diameter, D = 1.09 cm

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− fuel height, H = 3.66 m − heat flux, q″ = 78.9 W/cm2 − coolant inlet temperature, Tin = 260 °C − max cladding surface temperature, Tco,max = 316 °C Thermodynamic properties from Table 10.9 (subscript w for water and h for helium): − density, ρw = 735.3 kg/m3 and ρh = 0.865 kg/m3 − specific heat, cpw = 5.317 kJ/kg K and cph = 5.225 kJ/kg K − viscosity, μw = 0.955 × 10−4 kg/m s and μh = 0.298 × 10−4 kg/m s − conductivity, kw = 0.564 W/m K and kh = 0.230 W/m K The flow area and hydraulic diameter of the geometry is

and

𝐴𝐴𝑓𝑓 = 𝑃𝑃2 −

𝜋𝜋 2 𝐷𝐷 = 1.027 × 10−4 m2 4

𝐷𝐷𝑒𝑒 =

4𝐴𝐴𝑓𝑓 = 0.012 m 𝜋𝜋𝜋𝜋

(1)

(2)

The Reynolds number is a function of the mass flow rate, which for each of the fluids is represented by

and

Re𝑤𝑤 (𝑚𝑚̇𝑤𝑤 ) =

𝑚𝑚̇𝑤𝑤 𝐷𝐷𝑒𝑒 𝐴𝐴𝑓𝑓 𝜇𝜇𝑤𝑤

Reℎ (𝑚𝑚̇ℎ ) =

𝑚𝑚̇ℎ 𝐷𝐷𝑒𝑒 𝐴𝐴𝑓𝑓 𝜇𝜇ℎ

(3)

(4)

The Prandtl number of each fluid is calculated with

and

Pr𝑤𝑤 = Prℎ =

𝜇𝜇𝑤𝑤 𝑐𝑐𝑝𝑝𝑝𝑝 = 0.9 𝑘𝑘𝑤𝑤

𝜇𝜇ℎ 𝑐𝑐𝑝𝑝ℎ = 0.677 𝑘𝑘ℎ

(5)

(6)

The Nusselt number for water can be calculated with Equation 10.104b and for helium with Equation 10.103b. Each of these formulations depend on the Reynolds number which is not known yet since we are trying to solve for the mass flow rate. Therefore we write this quantity as a function of mass flow rate for each fluid,

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and

𝑃𝑃 Nu𝑤𝑤 (𝑚𝑚̇𝑤𝑤 ) = 0.023Re𝑤𝑤 (𝑚𝑚̇𝑤𝑤 )0.8 Pr 0.333 �1.826 � � − 1.0430� 𝐷𝐷

(7)

𝑃𝑃 𝑃𝑃 Nuℎ (𝑚𝑚̇ℎ ) = 0.023Reℎ (𝑚𝑚̇ℎ )0.8 Pr 0.333 �0.9217 + 0.1478 � � − 0.1130 exp �−7 �� − 1�� 𝐷𝐷 𝐷𝐷

(8)

Thus, the heat transfer coefficients can be represented by

and

ℏ𝑤𝑤 (𝑚𝑚̇𝑤𝑤 ) = ℏℎ (𝑚𝑚̇ℎ ) =

Nu𝑤𝑤 (𝑚𝑚̇𝑤𝑤 )𝑘𝑘𝑤𝑤 𝐷𝐷𝑒𝑒

Nuℎ (𝑚𝑚̇ℎ )𝑘𝑘ℎ

𝐷𝐷𝑒𝑒

(9)

(10)

The maximum surface cladding temperature will occur at the exit first. Since we only have the bulk inlet temperature of the fluid, this temperature difference can be represented by 𝐻𝐻𝐻𝐻𝐻𝐻 1 𝑇𝑇𝑐𝑐𝑐𝑐,𝑚𝑚𝑚𝑚𝑚𝑚 − 𝑇𝑇𝑖𝑖𝑖𝑖 = 𝑞𝑞 ″ � + � 𝑚𝑚̇𝑐𝑐𝑝𝑝 ℏ(𝑚𝑚̇)

(11)

The first term in the parentheses represents the temperature difference to the bulk outlet and the second term represents the convective heat transfer from the surface of the cladding. Therefore, the only unknown in this formulation is the mass flow rate, which depends on the Reynolds number formulation, Nusselt number formulation and heat transfer coefficient. Solving these equations simultaneously yields a mass flow rate for water of and a mass flow rate for helium of

𝑚𝑚̇𝑤𝑤 = 0.451 kg⁄s

(12)

𝑚𝑚̇ℎ = 0.491 kg⁄s

(13)

PROBLEM 10.8 QUESTION Hydraulic and Thermal Analysis of the Emergency Core Spray System in a BWR (Chapters 9 and Section 10.5) The emergency spray system of a BWR delivers cold water to the core after a large-break loss of coolant accident has emptied the reactor vessel. The system comprises a large water pool, a pump, a spray nozzle, and connecting pipes (Figure 10.18). All pipes are smooth round tubes made of stainless steel with 10 cm internal diameter and 5 mm thickness. The pipe lengths are shown in Figure 10.18. Two sharp 90° elbows connect the vertical pipe to the horizontal pipe and the horizontal pipe to the spray nozzle. Each elbow has a form loss coefficient of 0.9. The spray nozzle

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has a total flow area of 26 cm2 and a form loss coefficient of 15. The suction pipe in the pool has a sharp-edged entrance with a form loss coefficient of 0.5. 1. Calculate the pumping power required to deliver 50 kg/s of cold water to the core. (Assume steady-state and constant water properties. Do not neglect the acceleration terms in the momentum equation. Neglect entry region effects in calculating the friction factor. To calculate the irreversible term of the spray nozzle form loss, use the value of the mass flux in the pipe. Neglect the vertical dimension of the pump. The isentropic efficiency of the pump is 80%). 2. The horizontal pipe leading to the spray nozzle is exposed to superheated steam at 200°C and 0.1 MPa. The length of the exposed section is 5 m. Estimate the heat transfer rate from the steam to the water inside the pipe. (Assume that the heat transfer coefficient on the outer surface of the pipe is 5000 W/m2 K. Neglect entry region effects in calculating the heat transfer coefficient within the pipe). 3. In light of the results in (2) judge the accuracy of the constant property assumption made in calculating the pumping power in (1). Relevant water, steam, and stainless steel properties are given in Table 10.10.

FIGURE 10.18 Emergency spray system.

TABLE 10.10 Property Values for Problem 10.8 Properties of Water at Room Temperature (25°C) Property

Value

Density

997 kg/m3

Viscosity

9 × 10−4 Pa s

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Thermal conductivity

0.61 W/m K

Specific heat

4.2 kJ/kg K

Properties of Steam at 200°C and 0.1 MPa Property

Value

Density

0.46 kg/m3

Viscosity

2 × 10−5 Pa s

Thermal conductivity

0.03 W/m K

Specific heat

2.0 kJ/kg K

Properties of Stainless Steel Property

Value

Density

8000 kg/m3

Thermal conductivity

14 W/m K

Specific heat

0.47 kJ/kg K

Answers: i. 𝑊𝑊̇𝑝𝑝 ≈ 48 kW ii. 𝑄𝑄̇ ≈ 463 kW

PROBLEM 10.8 S OLUTION

Hydraulic and Thermal Analysis of the Emergency Core Spray System in a BWR (Chapters 9 and Section 10.5) From the problem statement, we are given: − loss coefficient of elbow, Kelb = 0.9 − area of spray nozzle, As = 26 cm2 − exit loss coefficient, Kex = 15 − inlet loss coefficient, Kin = 0.5 − isentropic efficiency of pump, ηp = 0.8 − mass flow rate, ṁ = 50 kg/s − tube inner diameter, Di = 10 cm − tube wall thickness, tw = 5 mm

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− flow area of tube, A = 4 D2i = 7.854 × 10−3 m2 − steam temperature, Tυ = 200 °C

− length of exposed section, L = 5m − heat transfer coefficient from outer tube wall to steam, hυ = 5000 W/m2 K From Table 10.10, thermodynamic properties are (‘w’ for water, ‘v’ for steam and ‘s’ for stainless steel): − density, ρw = 997 kg/m3, ρυ = 0.46 kg/m3 and ρs = 8000 kg/m3 − viscosity, μw = 9 × 10−4 Pa s, μυ = 2 × 10−5 Pa s − thermal conductivity, kw = 0.61 W/m K, kυ = 0.03 W/m K and ks = 14 W/m K − specific heat, cpw = 4.2 kJ/kg K, Cpv = 2.0 kJ/kg K and cps = 0.47 kJ/kg K − primary side water temperature, T = 25 °C The Reynolds number of water is Re =

𝑚𝑚̇𝐷𝐷𝑖𝑖 = 7.074 × 105 𝐴𝐴𝜇𝜇𝑤𝑤

(1)

Since the flow is turbulent, the friction factor can be estimated using the McAdams correlation, 𝑓𝑓 = 0.184𝑅𝑅𝑅𝑅 −0.2 = 0.0124

(2)

∆𝑃𝑃𝑔𝑔 = 𝜌𝜌𝑤𝑤 𝑔𝑔𝑔𝑔 = 161.3 kPa

(3)

The overall pressure drop of the system is 0. The gravitational pressure drop can be calculated with where H is the net height of 16.5 m according to Figure 10.8. The frictional pressure loss including form losses can be calculated with ∆𝑃𝑃𝑓𝑓 = �𝐾𝐾𝑖𝑖𝑖𝑖 + 𝑓𝑓

𝐿𝐿𝑖𝑖 𝜌𝜌𝑤𝑤 𝜐𝜐 2 + 2𝐾𝐾𝑒𝑒𝑒𝑒𝑒𝑒 + 𝐾𝐾𝑒𝑒𝑒𝑒 � = 424.7 kPa 𝐷𝐷𝑖𝑖 2

(4)

In the frictional pressure loss, Li = 29 m, the length of the pipe and the velocity of water can be determined from continuity, 𝜐𝜐 =

𝑚𝑚̇ = 6.385 m/s 𝜌𝜌𝑤𝑤 𝐴𝐴

(5)

The inlet pressure form loss due to acceleration (reversible component) is ∆𝑃𝑃𝑎𝑎𝑎𝑎𝑎𝑎,𝑖𝑖 =

𝜌𝜌𝑣𝑣 2 2

(6)

while the exit pressure form loss due to acceleration is

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𝜌𝜌𝑣𝑣22 𝜌𝜌𝑣𝑣 2 ∆𝑃𝑃𝑎𝑎𝑎𝑎𝑎𝑎,𝑜𝑜 = − 2 2

(7)

The exit velocity can be determined from continuity since the mass flow rates must remain the same before and after the nozzle, 𝑣𝑣2 = 𝑣𝑣

𝐴𝐴 = 19.3 m/s 𝐴𝐴𝑠𝑠

(8)

Therefore, the overall acceleration pressure drop can be determined to be Δ𝑃𝑃𝑎𝑎𝑎𝑎𝑎𝑎 =

𝜌𝜌𝜐𝜐22 = 185.5 kPa 2

(9)

These pressure losses must be supplied by the pump,

Δ𝑃𝑃𝑝𝑝 = Δ𝑃𝑃𝑔𝑔 + Δ𝑃𝑃𝑓𝑓 + Δ𝑃𝑃𝑎𝑎𝑎𝑎𝑎𝑎 = 771.5 kPa

(10)

The pumping power is therefore,

The Prandtl number for water is

𝑊𝑊̇𝑝𝑝 =

Δ𝑃𝑃𝑝𝑝 𝑚𝑚̇ = 48.1 kW 𝜌𝜌𝑤𝑤 𝜂𝜂𝑝𝑝

Pr𝑤𝑤 =

𝜇𝜇𝑤𝑤 𝑐𝑐𝑝𝑝𝑝𝑝 = 6.197 𝑘𝑘𝑤𝑤

(11)

(12)

The Nusselt number can be determined from the Dittus-Boelter equation, Nu = 0.023Re0.8 Pr 0.4 = 2282

(13)

The primary side heat transfer coefficient is thus ℏ𝑝𝑝 =

Nu𝑘𝑘𝑤𝑤 = 1.392 × 104 W/m2 K 𝐷𝐷𝑖𝑖

(14)

The outer diameter of the pipe is

𝐷𝐷𝑜𝑜 = 𝐷𝐷𝑖𝑖 + 2𝑡𝑡𝑤𝑤 = 0.11 m

(15)

Therefore the linear heat rate is can be calculated from the convection of the primary side water to the wall, conduction through the wall and convection to the steam, 𝑞𝑞 ′ =

𝑇𝑇𝑠𝑠 − 𝑇𝑇𝑝𝑝 kW = 92.7 𝐷𝐷 m ln � 𝐷𝐷𝑜𝑜 � 1 1 𝑖𝑖 + + 𝜋𝜋𝐷𝐷𝑖𝑖 ℏ𝑝𝑝 2𝜋𝜋𝑘𝑘𝑠𝑠 𝜋𝜋𝐷𝐷𝑜𝑜 ℏ𝑠𝑠

(16)

The overall heat transfer rate is

𝑄𝑄̇ = 𝑞𝑞 ′ 𝐿𝐿 = 463.6 kW

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The water temperature rise due to heating from the steam is small, ∆𝑇𝑇𝑤𝑤 = 𝑚𝑚𝑚𝑚̇ assumption of constant properties used in part i is accurate.

𝑝𝑝𝑝𝑝

= 2.2°C, so the

PROBLEM 10.9 QUESTION Coolant Selection for an Advanced High-Temperature Reactor (Chapters 8, 9, 10 - Section 10.5) To improve the thermal–hydraulic performance of an advanced high temperature reactor (AHTR), a vendor wishes to compare two alternative coolants, that is, a liquid metal (Na) and a liquid salt (LiF – BeF2). In the AHTR core the coolant flows inside 10 m long round channels arranged in a hexagonal lattice and surrounded by a solid fuel matrix. Consider the unit cell of this core (Figure 10.19).

FIGURE 10.19 Cross-sectional view of the core unit cell. 1. The friction pressure drop in the coolant channel is to be limited to 200 kPa. Calculate the maximum allowable mass flow rate for the two candidate coolants. (Neglect surface roughness and entry effects). 2. Calculate the pumping power for the mass flow rates computed in (1). (Assume ηp = 100%). 3. The coolant temperature at the channel inlet is 600°C. Assuming that the temperature in the fuel cannot exceed 1000°C, calculate the maximum allowable linear power for each coolant. (Hint: approximate the geometry of the fuel around the coolant channel as an equivalent annulus that conserves the fuel volume. Then solve the heat conduction equation for this annulus with a zero heat flux boundary condition, Assume axially and radially uniform heat generation rate within the fuel). 4. In view of the above results, which coolant should the vendor select and why? The properties for all materials in the system are given in Table 10.11.

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TABLE 10.11 Properties (All Properties Constant with Temperature) for Problem 10.9 Material

ρ (kg/m3)

k (W/mK)

Liquid Na

780

1.7 × 10−4

1300

Liquid LiF-BeF2

1940

2.0 × 10−3

2410

Fuel matrix

8530

Not applicable

500

μ (Pa s)

cp (J/kg K)

Answers: Liquid Na (1) ṁ = 0.357 kg/s (2) Ẇp ≈ 91W (3) q′ = 9.4 kW/m

Liquid LiF – BeF2 ṁ = 0.443 kg/s Ẇp ≈ 46 W q′ = 12.5 kW/m

PROBLEM 10.9 SOLUTION Coolant Selection for an Advanced High-Temperature Reactor (Chapters 8, 9, 10 - Section 10.5) From the problem statement and Figure 10.19, the given geometric parameters are: − length of channel, L = 10 m − minimal diameter of hexagonal unit cell, Dhex = 3 cm − diameter of flow channel, Di = 1 cm 𝜋𝜋

− flow area, 𝐴𝐴𝑖𝑖 = 4 𝐷𝐷𝑖𝑖2 = 7.854 × 10−5 m2

Thermodynamic properties given in Table 10.11 (“Na” for liquid sodium, “Li” for liquid salt, “f” for fuel matrix): − density, ρNa = 780 kg/m3, ρLi = 1940 kg/m3 and ρf = 8530 kg/m3 − thermal conductivity, kNa = 60 W/m K, kLi = 1 W/m K and kf = 6 W/m K − viscosity, μNa = 1.7 × 10−4 Pa s and μLi = 2.0 × 10−3 Pa s − specific heat, cpNa = 1300 J/kg K, cpLi = 2410 J/kg K and cpf = 500 J/kg K Since the mass flow rate is not known yet, all of the equations presented will be functions of the mass flow rate. The Reynolds number for each fluid is

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Re𝑁𝑁𝑁𝑁 (𝑚𝑚̇𝑁𝑁𝑁𝑁 ) =

𝑚𝑚̇𝑁𝑁𝑁𝑁 𝐷𝐷𝑖𝑖 𝐴𝐴𝑖𝑖 𝜇𝜇𝑁𝑁𝑁𝑁

Re𝐿𝐿𝐿𝐿 (𝑚𝑚̇𝐿𝐿𝐿𝐿 ) =

𝑚𝑚̇𝐿𝐿𝐿𝐿 𝐷𝐷𝑖𝑖 𝐴𝐴𝑖𝑖 𝜇𝜇𝐿𝐿𝐿𝐿

(1)

Assuming turbulent flow, the friction factor can be calculated with the McAdams correlation for the liquid sodium and the Blasius relation for the liquid salt, 𝑓𝑓𝑁𝑁𝑁𝑁 (𝑚𝑚̇𝑁𝑁𝑁𝑁 ) = 0.184 Re𝑁𝑁𝑁𝑁 (𝑚𝑚̇𝑁𝑁𝑁𝑁 )−0.2 𝑓𝑓𝐿𝐿𝐿𝐿 (𝑚𝑚̇𝐿𝐿𝐿𝐿 ) = 0.316 Re𝐿𝐿𝐿𝐿 (𝑚𝑚̇𝐿𝐿𝐿𝐿 )−0.25

(2)

The frictional pressure drop, Δpf which is 200 kPa can be represented for each fluid with 2 𝐿𝐿 𝑚𝑚̇𝑁𝑁𝑁𝑁 ∆𝑝𝑝𝑝𝑝 = 𝑓𝑓𝑁𝑁𝑁𝑁 (𝑚𝑚̇𝑁𝑁𝑁𝑁 ) 𝐷𝐷𝑖𝑖 2𝜌𝜌𝑁𝑁𝑁𝑁 𝐴𝐴2𝑖𝑖

∆𝑝𝑝𝑝𝑝 = 𝑓𝑓𝐿𝐿𝐿𝐿 (𝑚𝑚̇𝐿𝐿𝐿𝐿 )

(3)

2 𝐿𝐿 𝑚𝑚̇𝐿𝐿𝐿𝐿 𝐷𝐷𝑖𝑖 2𝜌𝜌𝐿𝐿𝐿𝐿 𝐴𝐴2𝑖𝑖

(4)

Using the respective Reynolds number and friction factor relations for each fluid to solve for the mass flow rate in the pressure drop formula, the resulting flow rates are 𝑚𝑚̇𝑁𝑁𝑁𝑁 = 0.357 kg/s

𝑚𝑚̇𝐿𝐿𝐿𝐿 = 0.443 kg/s.

(5)

Re𝑁𝑁𝑁𝑁 = 2.672 × 105

Re𝐿𝐿𝐿𝐿 = 2.821 × 104

(6)

To verify that the correct friction factor relations were chosen, the Reynolds number were calculated: Therefore since the liquid salt Reynolds number is less than 30,000, it was correct to use the Blasius relation. Next, knowing the mass flow rate and associated pressure drop, the pumping power can be easily calculated for each fluid, 𝑊𝑊̇𝑝𝑝,𝑁𝑁𝑁𝑁 = 𝑚𝑚̇𝑁𝑁𝑁𝑁

∆𝑝𝑝𝑝𝑝 = 91 W 𝜌𝜌𝑁𝑁𝑁𝑁

𝑊𝑊̇𝑝𝑝,𝐿𝐿𝐿𝐿 = 𝑚𝑚̇𝐿𝐿𝐿𝐿

∆𝑝𝑝𝑝𝑝 = 46 W 𝜌𝜌𝐿𝐿𝐿𝐿

(7)

To calculated the linear heat rate, the fuel around the coolant channel must be converted into an equivalent annulus that conserves volume. The area of fuel is π √3 2 𝐷𝐷ℎ𝑒𝑒𝑒𝑒 − 𝐷𝐷𝑖𝑖2 = 7.009 × 10−4 m2 2 4

(8)

4 𝜋𝜋 𝐷𝐷𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 = � �𝐴𝐴𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 + 𝐷𝐷𝑖𝑖2 � = 0.031502 m 𝜋𝜋 4

(9)

𝐴𝐴𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 =

The outer diameter of the annulus can then be calculated by also including the area of the coolant channel so that

The Prandtl numbers of the fluid are Pr𝑁𝑁𝑁𝑁 =

𝜇𝜇𝑁𝑁𝑁𝑁 𝑐𝑐𝑝𝑝𝑝𝑝𝑝𝑝 = 3.683 × 10−3 𝑘𝑘𝑁𝑁𝑁𝑁 274

Pr𝐿𝐿𝐿𝐿 =

𝜇𝜇𝐿𝐿𝐿𝐿 𝑐𝑐𝑝𝑝𝑝𝑝𝑝𝑝 = 4.82 𝑘𝑘𝐿𝐿𝐿𝐿

(10)

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To calculate the Reynolds number, the Lyon correlation is used for the liquid sodium (metallic) and the Dittus-Boelter is used for the liquid salt, and

Nu𝑁𝑁𝑁𝑁 = 7 + 0.025(Re𝑁𝑁𝑁𝑁 Pr𝑁𝑁𝑁𝑁 )0.8 = 13.201

(11)

0.4 Nu𝐿𝐿𝐿𝐿 = 0.023Re0.8 𝐿𝐿𝐿𝐿 Pr𝐿𝐿𝐿𝐿 = 156.757

(12)

Thus, the heat transfer coefficients are

and

ℏ𝑁𝑁𝑁𝑁 =

Nu𝑁𝑁𝑁𝑁 𝑘𝑘𝑁𝑁𝑁𝑁 = 7.921 × 104 W/m2 K 𝐷𝐷𝑖𝑖

ℏ𝐿𝐿𝐿𝐿 =

Nu𝐿𝐿𝐿𝐿 𝑘𝑘𝐿𝐿𝐿𝐿 = 1.568 × 104 W/m2 K 𝐷𝐷𝑖𝑖

(13)

(14)

In general, the temperature difference between the bulk inlet temperature and the bulk outlet temperature is 𝑇𝑇𝑏𝑏,𝑜𝑜𝑜𝑜𝑜𝑜 − 𝑇𝑇𝑏𝑏,𝑖𝑖𝑖𝑖 =

𝑞𝑞 ′ 𝐿𝐿 𝑚𝑚̇𝑐𝑐𝑝𝑝

(15)

The temperature difference between the bulk fluid temperature at the outlet and inner surface temperature of the fuel matrix is 𝑇𝑇𝑠𝑠 − 𝑇𝑇𝑏𝑏,𝑜𝑜𝑜𝑜𝑜𝑜 =

𝑞𝑞 ′ 𝜋𝜋𝐷𝐷𝑖𝑖 ℏ

(16)

Finally, for an annulus with internal cooling, the temperature difference can be represented by 2 𝐷𝐷𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝐷𝐷𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝑞𝑞 ′ 1 �− � 𝑇𝑇𝑚𝑚𝑚𝑚𝑚𝑚 − 𝑇𝑇𝑠𝑠 = � 2 ln � 2 2𝜋𝜋𝑘𝑘𝑓𝑓 𝐷𝐷𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 − 𝐷𝐷𝑖𝑖 𝐷𝐷𝑖𝑖 2

(17)

This is very similar to Eq. 8.75 after some manipulation and assuming a constant thermal conductivity. Combining the three heat transfer equations above and solving for the linear heat rate, 𝑞𝑞 ′ =

2 𝐷𝐷𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓

𝑇𝑇𝑚𝑚𝑚𝑚𝑚𝑚 − 𝑇𝑇𝑏𝑏,𝑖𝑖𝑖𝑖

𝐷𝐷𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 1 1 1 1 � 2 2 ln � 𝐷𝐷 � − 2� + 𝜋𝜋𝐷𝐷 ℏ + 𝑚𝑚𝑐𝑐 2𝜋𝜋𝑘𝑘𝑓𝑓 𝐷𝐷𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 − 𝐷𝐷𝑖𝑖 𝑖𝑖 𝑝𝑝 𝑖𝑖

(18)

where Tmax = 1000 °C and Tb,in = 600 °C. Plugging in the appropriate parameters for each fluid, the resulting linear heat rates are and

′ 𝑞𝑞𝑁𝑁𝑁𝑁 = 9.40 kW/m

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Chapter 10 - Single-Phase Heat Transfer ′ 𝑞𝑞𝐿𝐿𝐿𝐿 = 12.51 kW/m

(20)

The thermal-hydraulic analysis indicates that a liquid salt coolant is the better choice, as it affords higher heat removal rates while requiring lower pumping power. The good thermalhydraulic performance of the liquid salt is due mainly to its high heat capacity.

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Chapter 10 - Single-Phase Heat Transfer

PROBLEM 10.10 QUESTION Effect of Geometry on Single-Phase Heat Transfer in Straight Tubes (Chapters 9, 10—Sections 10.2 (10.2.1, 10.2.2, 10.2.3) and 10.5 (10.5.4)) Consider a smooth round tube (2 cm in diameter) and a smooth square tube (2 cm × 2 cm), each 4 m in length (shown in Figure 10.20). Each tube has a fluid flowing through it at the conditions given in Table 10.12. 1. Determine if the flow is laminar or turbulent for both tubes. Justify your answer using calculations. 2. Given a uniform velocity profile at Point a (shown below in Figure 10.21), sketch the velocity profiles at Points b and c for the round tube. Explain changes you see, if any in the profiles at Points b and c. 3. Which tube has a higher heat transfer coefficient at Point c? Justify your answer using calculations. Give proper justification for the heat transfer correlation you choose. 4. Suppose that the flow rate triples. Which tube has a higher heat transfer coefficient at Point c?

FIGURE 10.20 Round and square tubes (figure not to scale).

FIGURE 10.21 Velocity profile at Point a.

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TABLE 10.12 Fluid Properties for Problem 10.10 Parameter

Value

Density (ρ)

914 kg/m3

Viscosity (μ)

9 × 10−5 Pa s

Specific heat (cp)

5.8 kJ/(kg‧°C)

Thermal conductivity (k) 0.54 W/(m‧°C) Mass flow rate ( m )

2.822 × 10−3 kg/s

Wall temperature

Constant throughout the length of the tube

Answers: 1. Laminar flow, Re = 1996 (round), Re = 1568 (square) 3. Round tube 4. Round tube

PROBLEM 10.10 S OLUTION Effect of Geometry on Single-Phase Heat Transfer in Straight Tubes From the problem statement and Table 10.12 the following parameters are given: − density, ρ = 914 kg/m3 − viscosity, μ = 9 × 10−5 Pa s − specific heat, cp = 5.8 kJ/kg K − thermal conductivity, k = 0.54 W/m K − mass flow rate, ṁ = 2.822 × 10−3 kg/s − round tube diameter, D = 2 cm 𝜋𝜋

− round tube flow area, 𝐴𝐴 = 4 𝐷𝐷2 = 3.142 × 10−4 m2

The Reynolds number for the round tube is

and for the square

Re1 =

𝑚𝑚̇𝐷𝐷 = 1996 𝐴𝐴𝐴𝐴

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Chapter 10 - Single-Phase Heat Transfer

Re2 =

𝑚𝑚̇𝐷𝐷 = 1568 𝐷𝐷2 𝜇𝜇

(2)

We must first calculate if Points b and c, shown in Figures A and B of SM-10.2, are in the entry or fully developed region. For laminar flow, the length of the entry region, ze, can be calculated as. Thus, ze = 1.996 m (round tube) and ze = 1.568 m (square tube), and it can be concluded that Points b and c are in the entry region and fully-developed region, respectively, for both tubes. The qualitative velocity profile at Point b and c is:

Fig. A - Point b

Fig. B - Point c Figure SM-10.2 Velocity Profiles in the Entry (Fig. A) and Fully Developed Regions (Fig. B) In order to select a heat transfer coefficient correlation, we note that: - Pr = 0.97, thus the fluid is non-metallic. − The flow regime is laminar. − The geometry is round tube and square tube. − The boundary condition is constant wall temperature, − Point c is in the fully-developed region (for both velocity and temperature profiles) Thus, the Nusselt number for laminar flow in a round tube with constant wall temperature is 3.66, while for a square tube is 2.98. The corresponding heat transfer coefficients are approx.

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99.8W/m2K and 80.5 W/m2K, respectively, i.e., the round tube has a higher heat transfer coefficient. If the flow rate triples, the Reynolds number triples (i.e., Re = 5998 for round tube and Re = 4704 for the square tube), taking the flow regime from laminar to turbulent. The situation is still one of fully-developed flow (ze=40 De=0.8m<zc). For turbulent flow in the fully developed region, the heat transfer is insensitive to geometry and boundary conditions. Since the round tube has a higher Reynolds number than the square tube, its heat transfer coefficient will also be higher. Note that the Dittus-Boelter correlation cannot be used because it is valid for Re>10,000.

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Chapter 11 Two-Phase Flow Dynamics Contents Problem 11.1

Methods of describing two phase flow ......................................................... 282

Problem 11.2

Flow regime transitions ................................................................................. 284

Problem 11.3

Regime map for vertical flow ....................................................................... 285

Problem 11.4

Comparison of correlations for flooding ....................................................... 287

Problem 11.5

Impact of slip model on the predicted void fraction ..................................... 288

Problem 11.6

Level swell in a vessel due to two-phase conditions .................................... 290

Problem 11.7

Void fraction evaluation using the EPRI correlation .................................... 293

Problem 11.8

Void, quality and pressure drop problem ...................................................... 296

Problem 11.9

Calculation of a pipe’s diameter for a specific pressure drop ....................... 299

Problem 11.10 Flow dynamics of nanofluids ........................................................................ 302 Problem 11.11 HEM pressure loss ........................................................................................ 305 Problem 11.12 Analysis of a liquid metal reactor vessel ...................................................... 307

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Chapter 11 - Two-Phase Flow Dynamics

PROBLEM 11.1 QUESTION Methods of Describing Two-Phase Flow (Section 11.2) Consider vertical flow through a subchannel formed by fuel rods (12.5 mm outer diameter) arranged in a square array (pitch 16.3 mm). Neglect effects of spacers and unheated walls and assume the following: 1. Saturated water at 7.2 MPa (288 °C); liquid density = 740kg/m3; vapor density = 38kg/m3. 2. For vapor fraction determination in the slug flow regime, use the drift flux representation (substituting subchannel hydraulic diameter for tube diameter). 3. Consider a simplified flow regime representation, including bubbly flow, slug flow, and annular flow. The slug-to-bubbly transition occurs at a vapor fraction of 0.15. The slug-to annular transition occurs at a vapor fraction of 0.75. Calculate and draw on a graph with axes the superficial vapor velocity, {jv} (abscissa, from 0 to 40 m/s); and the superficial liquid velocity, {jℓ} (ordinate, from 0 to 3 m/s). Answers: 1. Bubbly-to-slug {jℓ} = −0.10 + 4.55{jv} or {jv}= 0.0235 + 0.22 jv 2. Slug-to-annular {jℓ} = −0.10 + 0.11{jv} or {jv}= 0.966 + 9.0 jℓ

PROBLEM 11.1 SOLUTION Methods of Describing Two-Phase Flow (Section 11.2) The following parameters are given in the problem statement: – Fuel rod diameter, D = 12.5 mm – Fuel rod pitch, S = 16.3 mm – Liquid density, ρl = 740kg/m3 – Vapor density, ρυ = 38kg/m3 – Bubbly-to-slug transition void fraction, αsb = 0.15 – Slug-to-annular transition void fraction, αsa = 0.75 The hydraulic diameter can be calculated as 𝐷𝐷𝑒𝑒 =

The drift flux model is given as

𝜋𝜋 4 �𝑆𝑆 2 − 4 𝐷𝐷2 � 𝜋𝜋𝜋𝜋

= 0.01456 m

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Chapter 11 - Two-Phase Flow Dynamics

𝛼𝛼 =

𝑗𝑗𝜐𝜐 𝐶𝐶𝑜𝑜 𝑗𝑗 + 𝜐𝜐𝜐𝜐𝜐𝜐

(2)

where the drift velocity can be represented as

𝜐𝜐𝜐𝜐𝜐𝜐 = (1 − 𝛼𝛼)𝑛𝑛 𝜐𝜐∞

(3)

TABLE 11.3 Values of n and V∞ for Various Regimes Regime

V∞

Small bubbles (d < 0.5 cm)

g ( ρ − ρv ) d

Large bubbles (d < 2 cm)

1.5

18 µ 

σ g (ρ − ρ )  1.53   ρ

1/ 4

σ g (ρ − ρ )  1.53   ρ

1/ 4



Churn flow



Slug flow (in tube of diameter D)

0.35

ρ −ρ  D  ρ 

g 



For slug flow, Table 11.3 shows n = 0 and Co = 1.2. The drift velocity from Table 11.3 is therefore 𝜐𝜐𝜐𝜐𝜐𝜐 = 𝜐𝜐∞ = 0.35�𝑔𝑔 �

𝜌𝜌𝑙𝑙 − 𝜌𝜌𝜐𝜐 � 𝐷𝐷 = 0.119 m⁄s 𝜌𝜌𝑙𝑙

(4)

We can rearrange Equation (2) into the following form:

𝑗𝑗𝜐𝜐 = 𝛼𝛼𝐶𝐶𝑜𝑜 𝑗𝑗 + 𝛼𝛼𝛼𝛼𝜐𝜐𝜐𝜐 𝑗𝑗𝜐𝜐 = 𝛼𝛼𝐶𝐶𝑜𝑜 𝑗𝑗𝜐𝜐 + 𝛼𝛼𝛼𝛼𝑜𝑜 𝑗𝑗𝑙𝑙 + 𝛼𝛼𝛼𝛼𝜐𝜐𝜐𝜐 𝑗𝑗𝜐𝜐 =

𝛼𝛼𝜐𝜐𝜐𝜐𝜐𝜐 𝛼𝛼𝐶𝐶𝑜𝑜 𝑗𝑗𝑙𝑙 + 1 − 𝛼𝛼𝐶𝐶𝑜𝑜 1 − 𝛼𝛼𝛼𝛼𝑜𝑜

(5)

For bubbly-to-slug we can determine the coefficients as

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𝛼𝛼𝑠𝑠𝑠𝑠 𝜐𝜐𝜐𝜐𝜐𝜐 𝛼𝛼𝑠𝑠𝑠𝑠 𝐶𝐶𝑜𝑜 = 0.220 = 0.0218 m⁄s 1 − 𝛼𝛼𝑠𝑠𝑠𝑠 𝐶𝐶𝑜𝑜 1 − 𝛼𝛼𝑠𝑠𝑠𝑠 𝐶𝐶𝑜𝑜

For slug-to-annular we can determine the coefficients as

𝛼𝛼𝑠𝑠𝑠𝑠 𝜐𝜐𝜐𝜐𝜐𝜐 𝛼𝛼𝑠𝑠𝑠𝑠 𝐶𝐶𝑜𝑜 = 9.0 = 0.8952 m⁄s 1 − 𝛼𝛼𝑠𝑠𝑠𝑠 𝐶𝐶𝑜𝑜 1 − 𝛼𝛼𝑠𝑠𝑠𝑠 𝐶𝐶𝑜𝑜

After inverting Equation (5), we get the following transition equations: 1. Bubbly-to-slug {jℓ} = −0.10 + 4.55{jv} or {jv}= 0.0235 + 0.22 jv 2. Slug-to-annular {jℓ} = −0.10 + 0.11{jv} or {jv}= 0.966 + 9.0 jℓ

PROBLEM 11.2 QUESTION Flow Regime Transitions (Section 11.3) Line B of the flow regime map of Taitel et al. [70] is given as Equation 11.10. Derive this equation assuming that, because of turbulence, a Taylor bubble cannot exist with a diameter larger than:

where k = 1.14

𝜎𝜎 3⁄5 𝑑𝑑𝑚𝑚𝑚𝑚𝑚𝑚 = 𝑘𝑘 � � (𝜖𝜖)−2⁄5 𝑝𝑝𝑙𝑙 𝑑𝑑𝑑𝑑 𝑗𝑗 𝜖𝜖 = � � 𝑑𝑑𝑑𝑑 𝜌𝜌𝑚𝑚

Use the friction factor (f) = 0.046 (jD/υ)−0.2 and the fact that for small bubble the critical diameter that can be supported by the surface tension is given by: 0.5 0.4𝜎𝜎 𝑑𝑑𝑐𝑐𝑐𝑐 = � � (𝜌𝜌 − 𝜌𝜌𝑢𝑢 )𝑔𝑔

PROBLEM 11.2 SOLUTION Flow Regime Transitions (Section 11.3) From the problem statement we have

where k = 1.14

𝜎𝜎 3⁄5 𝑑𝑑𝑚𝑚𝑚𝑚𝑚𝑚 = 𝑘𝑘 � � (𝜖𝜖)−2⁄5 𝜌𝜌𝑙𝑙

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𝑑𝑑𝑑𝑑 𝑗𝑗 𝜖𝜖 = � � 𝑑𝑑𝑑𝑑 𝜌𝜌𝑚𝑚

(2)

The pressure gradient is related to the friction factor with 𝑑𝑑𝑑𝑑 2𝑓𝑓 = 𝜌𝜌 𝑗𝑗 2 𝑑𝑑𝑑𝑑 𝐷𝐷 𝑚𝑚

where

𝑓𝑓 = 0.046 �

(3)

𝑗𝑗𝑗𝑗 −0.2 � 𝜐𝜐𝑙𝑙

(4)

Substituting these equations, the max diameter is

𝜎𝜎 3⁄5 2 𝑗𝑗𝑗𝑗 −0.2 𝑗𝑗 𝑑𝑑𝑚𝑚𝑚𝑚𝑚𝑚 = 𝑘𝑘 � � � 0.046 � � 𝑃𝑃𝑚𝑚 𝑗𝑗 2 � 𝜌𝜌𝑙𝑙 𝐷𝐷 𝜐𝜐𝑙𝑙 𝜌𝜌𝑚𝑚

−2⁄5

(5)

The critical diameter is

𝑑𝑑𝑐𝑐𝑐𝑐 = �

0.5 0.4𝜎𝜎 � (𝜌𝜌𝑙𝑙 − 𝜌𝜌𝑢𝑢 )𝑔𝑔

(6)

If the critical diameter is set to the max diameter then

0.5 0.4𝜎𝜎 𝜎𝜎 3⁄5 2 𝑗𝑗𝑗𝑗 −0.2 𝑗𝑗 � � = 𝑘𝑘 � � � 0.046 � � 𝑃𝑃𝑚𝑚 𝑗𝑗 2 � (𝜌𝜌𝑙𝑙 − 𝜌𝜌𝜐𝜐 )𝑔𝑔 𝜐𝜐𝑙𝑙 𝜌𝜌𝑙𝑙 𝐷𝐷 𝜌𝜌𝑚𝑚

−2⁄5

(7)

Solving for the total superficial velocity

𝜎𝜎 0.0893 0.446 𝐷𝐷0.429 � � (𝜌𝜌𝑙𝑙 − 𝜌𝜌𝜐𝜐 )𝑔𝑔 𝜌𝜌𝑙𝑙 � � � 𝑗𝑗 = 𝑗𝑗𝑙𝑙 + 𝑗𝑗𝜐𝜐 = 4 � 𝜌𝜌𝑙𝑙 𝜐𝜐𝑙𝑙0.0714

(8)

PROBLEM 11.3 QUESTION Regime Map for Vertical Flow (Section 11.2) Construct a flow regime map based on coordinates {jυ} and {j} using the transition criteria of Taitel et al. [67] for the secondary side of the vertical steam generator. Assume that the length of the steam generator is 3.7 m and the characteristic hydraulic diameter is the volumetric hydraulic diameter:

Operation conditions Saturated water at 282°C

𝐷𝐷𝜐𝜐 =

4(𝑛𝑛𝑛𝑛𝑡𝑡 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣) 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠

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Liquid density = 747 kg/m3 Vapor density = 34 kg/m3 Surface tension = 17.6 × 10−3 N/m Answers: {j} = 3.0 {jv} - 0.14

Bubbly-to-slug:

Bubbly-to-dispersed bubbly: {j} + {jv} = 5.554 Disperse bubble-to-slug:

{j} = 0.923 {jv}

Slug to-churn:

{j} + {jv} = 0.527

Churn-to-annular:

{jv} = 1.78

PROBLEM 11.3 SOLUTION Regime Map for Vertical Flow (Section 11.2) From the problem statement, the operating conditions are – liquid density, ρ = 747 kg/m3 – vapor density, ρv = 34kg/m3 – surface tension, 17.6 N/m – D = 0.134 i) For bubbly-to-slug, the transition occurs at {α} = 0.25. Using Equation 11.5 1⁄4

{𝑗𝑗𝜐𝜐 } {𝑗𝑗 } 𝑔𝑔(𝜌𝜌 − 𝜌𝜌𝑣𝑣 ) = − 1.53 � � {1 − 𝛼𝛼} {𝛼𝛼} 𝜌𝜌2

(1)

Substituting in the appropriate conditions,

(2)

{𝑗𝑗 } = 3.0{𝑗𝑗𝜐𝜐 } − 0.14

ii) bubbly-to-dispersed bubbly, using Equation 11.10

0.446

𝐷𝐷0.429 (𝜎𝜎⁄𝜌𝜌 )0.089 𝑔𝑔(𝜌𝜌 − 𝜌𝜌𝑣𝑣 ) {𝑗𝑗 } + {𝑗𝑗𝜐𝜐 } = 4 � � � 𝜌𝜌 𝜐𝜐0.072

For saturated water, v = 0.13 × 10−6 m2/s

{𝑗𝑗 } + {𝑗𝑗𝜐𝜐 } = 5.554

�

(3)

(4)

iii) disperse bubbly-to-slug, using Equation 11.12

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0.52 =

Therefore,

{𝑗𝑗𝜐𝜐 } {𝑗𝑗 } + {𝑗𝑗𝜐𝜐 }

{𝑗𝑗 } = 0.923{𝑗𝑗𝜐𝜐 }

(5)

(6)

iv) slug-to-churn, using Equation 11.13

If we put lϵ = 3.7 m, Then

{𝑗𝑗} 𝑙𝑙𝜖𝜖 = 40.6 � + 0.22� 𝐷𝐷 �𝑔𝑔𝑔𝑔 {𝑗𝑗} = {𝑗𝑗 } + {𝑗𝑗𝜐𝜐 } = 0.527

(7)

(8)

v) churn-to-annular, using Equation 11.19

{𝑗𝑗𝜐𝜐 }𝜌𝜌𝜐𝜐1⁄2 = 3.1 [𝜎𝜎𝜎𝜎(𝜌𝜌 − 𝜌𝜌𝜐𝜐 )]1⁄4

Therefore,

{𝑗𝑗𝜐𝜐 } = 1.78

(9)

(10)

PROBLEM 11.4 QUESTION Comparison of Correlations for Flooding (Section 11.2) 1. It was experimentally verified that for tubes with a diameter larger than about 6.35 cm flooding is independent of diameter for air water mixtures at low pressure conditions. Test this assertion against Wallis correlation and the Pushkin and Sorokin correlation (Equation 11.28) by finding the diameter at which they are equal at atmospheric pressure. 2. Repeat question 1 for a saturated steam water mixture at 6.9 MPa. Note: For the j = 0 condition, the terms flooding and flow reversal are effectively synonymous. Answers: 1. D = 3.91 cm 2. D = 2.58 cm The Taylor bubble diameter should be less than the tube diameter if the tube diameter is not to influence flooding. At extremely low pressure situations, the limiting value of bubble diameter can reach 6.35 cm. Thus the assertion that for D > 6.35 cm, flooding is independent of diameter is applicable.

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PROBLEM 11.4 SOLUTION Comparison of Correlations for Flooding (Section 11.2) Properties at atmospheric pressure are — saturated liquid density, ρf = 958.37kg/m — saturated vapor density, ρg = 0.598 kg/m

— surface tension, σ = 0.05892 N/m — Kutateladze number, Kυ = 3.2 from Equation 11.28 — geometric constants, C = 0.9 and m = 1.0

Using the Wallis correlation (Equations 11.23 and 11.25), {𝑗𝑗𝜐𝜐 } �

0.5 𝜌𝜌𝜐𝜐 � = 𝐶𝐶 𝑔𝑔𝑔𝑔(𝜌𝜌𝑙𝑙 − 𝜌𝜌𝜐𝜐 )

(1)

The Kutateladze number is defined as

𝐾𝐾𝐾𝐾𝜐𝜐 = {𝑗𝑗𝜐𝜐 } �

0.5 𝜌𝜌𝜐𝜐 � [𝑔𝑔𝑔𝑔(𝜌𝜌𝑙𝑙 − 𝜌𝜌𝜐𝜐 )]0.5

(2)

Substituting Equation (1) into Equation (2) and solving for the diameter we get 𝐾𝐾𝐾𝐾𝜐𝜐 2 �𝑔𝑔𝑔𝑔�𝜌𝜌𝑓𝑓 − 𝜌𝜌𝑔𝑔 �� 𝐷𝐷 = � 2 � 𝐶𝐶 𝑔𝑔�𝜌𝜌𝑓𝑓 − 𝜌𝜌𝑔𝑔 �

0.5

(3)

Plugging in values the diameter is At 6.4 MPa, the properties are:

𝐷𝐷 = 3.91 cm

(4)

— saturated liquid density, ρf = 750.576 kg/m — saturated vapor density, ρg = 33.07kg/m

— surface tension, σ = 0.019033 N/m

Using Equation (3), and the new properties, the diameter is 𝐷𝐷 = 2.58 cm

(5)

PROBLEM 11.5 QUESTION Impact of Slip Model on the Predicted Void Fraction (Section 11.4) In a water channel, the flow conditions are such that: – Mass flow rate: ṁ = 0.29 kg/s

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– Flow area: A = 1.5 × 10−4 m2 – Flow quality: x = 0.15 – Operating pressure: P = 7.2 MPa Calculate the void fraction using (1) the HEM model, (2) Bankoff’s slip correlation, and (3) the drift flux model using Dix’s correlation for Co and Vυj calculated assuming churn flow conditions. Answers: 1. {α} = 0.775 2. {α} = 0.631 3. {α} = 0.703

PROBLEM 11.5 SOLUTION Impact of Slip Model on the Predicted Void Fraction (Section 11.4) In a water channel, the flow conditions are such that: – Mass flow rate: ṁ = 0.29 kg/s – Flow area: A = 1.5 × 10−4 m2 – Flow quality: x = 0.15 – Operating pressure: P = 7.2 MPa The thermodynamic properties for this operating condition are: – Saturated liquid water density, ρf = 736.2 kg/m3 – Saturated water vapor density, ρg = 37.7 kg/m3 HEM Model The void fraction for the HEM model is 𝛼𝛼𝐻𝐻𝐻𝐻𝐻𝐻 =

1 = 0.775 1 − 𝑥𝑥 𝜌𝜌𝑔𝑔 1 + 𝑥𝑥 𝜌𝜌 𝑓𝑓

(1)

Bankoff Correlation The K factor in the Bankoff correlation is given by 𝐾𝐾 = 0.71 + 0.0001𝑝𝑝 = 0.814

where p is in units of psi. The volumetric flow fraction is equivalent to the void fraction from the HEM model, 𝛽𝛽 = 𝛼𝛼𝐻𝐻𝐻𝐻𝐻𝐻 = 0.775

The void fraction from the Bankoff model is

𝛼𝛼𝐵𝐵𝐵𝐵 = 𝐾𝐾𝐾𝐾 = 0.631 289

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Drift Flux - Dix’s Correlation From Dix’s correlation, the b factor is 0.1

𝜌𝜌𝑔𝑔 𝑏𝑏 = � � 𝜌𝜌𝑓𝑓

The Co parameter is given as

= 0.743

𝑏𝑏 1 𝐶𝐶𝑜𝑜 = 𝛽𝛽 �1 + � − 1� � = 1.084 𝛽𝛽

For Churn flow, n = 0 and the drift velocity is given as

0.25

𝜎𝜎𝜎𝜎�𝜌𝜌𝑓𝑓 − 𝜌𝜌𝑔𝑔 � 𝑉𝑉𝜐𝜐𝜐𝜐 = 1.53 � � 𝜌𝜌𝑓𝑓2

The total superficial velocity is given as 𝑗𝑗 =

= 0.186 m⁄s

𝑚𝑚̇(1 − 𝑥𝑥) 𝑚𝑚̇𝑥𝑥 + = 9.925 m⁄s 𝐴𝐴𝜌𝜌𝑓𝑓 𝐴𝐴𝐴𝐴𝑔𝑔

The void fraction is calculated to be

𝛼𝛼𝐷𝐷𝐷𝐷 =

𝛽𝛽

𝑉𝑉𝜐𝜐𝜐𝜐 𝐶𝐶𝑜𝑜 + 𝑗𝑗

= 0.703

PROBLEM 11.6 QUESTION Level Swell in a Vessel due to Two-Phase Conditions (Section 11.4) Compute the level swell in a cylindrical vessel with volumetric heat generation under thermodynamic equilibrium and steady-state conditions. The vessel is filled with vertical fuel rods such that De = 0.0122 m. The collapsed water level is 2.13 m. Operating conditions No inlet flow to the vessel P = 5.516 MPa Q‴ (volumetric heat source) = 4.14 × 106 W/m3 Water properties σ = 0.07 N/m hfg = 1.6 × 106 J/kg ρg = 28.14kg/m3 ρf = 767.5 kg/m3 Assumptions Flow regime Selected values from Table 11.3 (n, V∞)

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Bubbly large bubbles TABLE 11.3 Values of n and V∞ for Various Regimes Regime

V∞

Small bubbles (d < 0.5

g ( ρ − ρv ) d

cm)

18 µ 

Large bubbles (d < 2 cm)

1.5

σ g (ρ − ρ )    ρ

1/ 4

σ g (ρ − ρ )  1.53   ρ

1/ 4

1.53



Churn flow



Slug flow (in tube of

0.35

diameter D)

ρ −ρ  D  ρ 

g 



Select appropriate transition for flow regimes so that there is a continuous shape for the α versus z curve. For these conditions, the following result for α versus z is known:

Answer: Hswell = 2.56 m

{𝛼𝛼}𝑉𝑉∞ (1 − {𝛼𝛼})𝑛𝑛−1 =

𝑄𝑄‴ 𝑧𝑧 ℎ𝑓𝑓𝑓𝑓 𝜌𝜌𝑔𝑔

PROBLEM 11.6 SOLUTION Level Swell in a Vessel due to Two-Phase Conditions (Section 11.4) From the problem statement, we are given: – pressure, P = 5.516 MPa — heat of vaporization, hfg = 1600 kJ/kg — equivalent diameter, De = 0.0122 m

– surface tension, σ = 0.07 N/m

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Chapter 11 - Two-Phase Flow Dynamics — saturated liquid density, ρf = 767.5 kg/m

– saturated vapor density, ρg = 28.14kg/m3 – collapsed height, Hc = 2.13 m ‴

— volumetric heat generation, Q

= 4.14 MW/m3

First, we need to determine the transition height between large bubbly and slug flow. From Table 11.3 the velocity of the bubbles for large bubbly (nb = 1.5) and slug (ns = 0) are 0.25

and

𝜎𝜎𝜎𝜎�𝜌𝜌𝑓𝑓 − 𝜌𝜌𝑔𝑔 � 𝜐𝜐𝑏𝑏 = 1.53 � � 𝜌𝜌𝑓𝑓2

= 0.262 m⁄s

(1)

= 0.119 m⁄s

(2)

𝛼𝛼𝑡𝑡𝑡𝑡 𝜐𝜐𝑏𝑏 (1 − 𝛼𝛼𝑡𝑡𝑡𝑡 )𝑛𝑛𝑏𝑏−1 = 𝛼𝛼𝑡𝑡𝑡𝑡 𝜐𝜐𝑠𝑠 (1 − 𝛼𝛼𝑡𝑡𝑡𝑡 )𝑛𝑛𝑜𝑜 −1

(3)

𝛼𝛼𝑡𝑡𝑡𝑡 = 0.41

(4)

0.5

𝜌𝜌𝑓𝑓 − 𝜌𝜌𝑔𝑔 � 𝐷𝐷𝑒𝑒 � 𝜐𝜐𝑠𝑠 = 0.35 �𝑔𝑔 � 𝜌𝜌𝑓𝑓

The transition void fraction can be determined with Solving for the transition void fraction, we get

The transition height can be determined from the given formula, {𝛼𝛼}𝑉𝑉∞ (1 − {𝛼𝛼})𝑛𝑛−1 =

This height is calculated to be

𝑄𝑄‴ 𝑧𝑧 ℎ𝑓𝑓𝑓𝑓 𝜌𝜌𝑔𝑔

(5)

𝐻𝐻𝑡𝑡𝑡𝑡 = 0.898 m

(6)

𝑚𝑚𝑓𝑓,𝑖𝑖 = 𝑚𝑚𝑓𝑓,𝑓𝑓 + 𝑚𝑚𝑔𝑔

(7)

To solve for the swelled level from the collapsed level, a mass balance approach can be used. All of the original liquid must either stay as liquid or turn to vapor. This can be represented mathematically as where mf,i is the original amount of liquid mass, mf, f is the final amount of liquid mass and mg is the final mass of vapor. Substituting in for density and volume (area is constant), these masses can be calculated as 𝐻𝐻𝑡𝑡

𝜌𝜌𝑓𝑓 𝐻𝐻𝑐𝑐 = � �𝜌𝜌𝑓𝑓 �1 − 𝛼𝛼(𝑧𝑧) + 𝜌𝜌𝑔𝑔 𝛼𝛼(𝑧𝑧)�� 𝑑𝑑𝑑𝑑 0

(8)

where Ht is the final swelled height. Therefore, we can break this calculation up for the bubbly and slug regions,

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𝜌𝜌𝑓𝑓 𝐻𝐻𝑐𝑐 = �

𝐻𝐻𝑡𝑡𝑡𝑡

�𝜌𝜌𝑓𝑓 �1 − 𝛼𝛼𝑏𝑏 (𝑧𝑧) + 𝜌𝜌𝑔𝑔 𝛼𝛼𝑏𝑏 (𝑧𝑧)�� 𝑑𝑑𝑑𝑑 +�

𝐻𝐻𝑡𝑡

𝐻𝐻𝐻𝐻𝐻𝐻

(9)

�𝜌𝜌𝑓𝑓 �1 − 𝛼𝛼𝑠𝑠 (𝑧𝑧) + 𝜌𝜌𝑔𝑔 𝛼𝛼𝑠𝑠 (𝑧𝑧)�� 𝑑𝑑𝑑𝑑

In the bubbly flow region, the void distribution is described by 𝛼𝛼𝑏𝑏 (𝑧𝑧)𝜐𝜐𝑏𝑏 �1 − 𝛼𝛼𝑏𝑏 (𝑧𝑧)�

𝑄𝑄‴ 𝑧𝑧 ℎ𝑓𝑓𝑓𝑓 𝜌𝜌𝑔𝑔

0.5

(10)

This formula cannot be explicitly solve for the void fraction. For the slug flow region, the void distribution is described by 𝛼𝛼𝑠𝑠 (𝑧𝑧)𝜐𝜐𝑠𝑠 �1 − 𝛼𝛼𝑠𝑠 (𝑧𝑧)�

−1

Solving for the void fraction, we get

𝑄𝑄‴ 𝑧𝑧 𝜐𝜐𝑠𝑠 ℎ𝑓𝑓𝑓𝑓 𝜌𝜌𝑔𝑔 𝛼𝛼𝑠𝑠 (𝑧𝑧) = 𝑄𝑄‴ 1+ 𝑧𝑧 𝜐𝜐𝑠𝑠 ℎ𝑓𝑓𝑓𝑓 𝜌𝜌𝑔𝑔

(11)

(12)

Solving Equations. (9), (10) and (12) numerically for the final swelled height, Ht, we get 𝐻𝐻𝑡𝑡 = 4.69 m

Therefore swelled level from the collapsed level is

𝐻𝐻𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 = 𝐻𝐻𝑡𝑡 − 𝐻𝐻𝑐𝑐 = 2.56 m

PROBLEM 11.7 QUESTION Void Fraction Evaluation Using the EPRI Correlation (Section 11.5.4) Using the EPRI correlation, evaluate the void fraction for a saturated water-steam mixture at a total mass flux, G, of 40 kg/m2s at 50% quality in a tube of diameter 20 mm at 7.45 MPa pressure. Use the fluid properties of Table 11.8. TABLE 11.8 Fluid Properties for Problem 11.7 ρf = 731.76 kg/m3 ρg = 39.18 kg/m3 μf = 89.63 μPa s μg = 19.16 μPa s σ = 0.0166 N/m

Answer: {α} = 0.62

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PROBLEM 11.7 SOLUTION Void Fraction Evaluation Using the EPRI Correlation (Section 11.5.4) Using the EPRI correlation, evaluate the void fraction for a saturated water-steam mixture at a total mass flux, G, of 40 kg/m2s at 50% quality in a tube of diameter 20 mm at 7.45 MPa pressure. Use the fluid properties of Table 11.8. Therefore for this problem we are given that: – total mass flux, G = 40 kg/m2s — quality, x = 0.5 — diameter, D = 20 mm — pressure, P = 7.45 MPa 3

— saturated liquid water density, ρf = 731.76 kg/m

– saturated water vapor density, ρg = 39.18 kg/m3 – saturated liquid water viscosity, μf =89.63 × 10−6 Pa · s −6

— saturated water vapor viscosity, μg = 19.16 × 10

Pa · s

– surface tension, σ = 0.0166 N/m – critical pressure, Pc = 22.1 MPa Total Superficial Velocity and Volumetric Flow Fraction: The phasic superficial velocities can be calculated as follows: 𝑥𝑥𝑥𝑥 = 0.51 m⁄s 𝜌𝜌𝑔𝑔 (1 − 𝑥𝑥)𝐺𝐺 = 0.027 m⁄s 𝑗𝑗𝑓𝑓 = 𝜌𝜌𝑓𝑓 𝑗𝑗𝑔𝑔 =

(1) (2)

The total superficial velocity is jus the sum of the individual phasic superficial velocities, 𝑗𝑗 = 𝑗𝑗𝑔𝑔 + 𝑗𝑗𝑓𝑓 = 0.538 m⁄s

(3)

The volumetric flow fraction can be expressed as the ratio of the gas phasic superficial velocity to the total superficial velocity, 𝛽𝛽 =

𝑗𝑗𝑔𝑔 = 0.949 𝑗𝑗

(4)

Evaluation of Co: To evaluate the parameter Co we can first calculated the parameter C1, 4𝑃𝑃𝑐𝑐2 𝐶𝐶1 = � � = 17.9 𝑃𝑃(𝑃𝑃𝑐𝑐 − 𝑃𝑃) 294

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To calculate the parameter A1, the Reynolds number of each phase is computed, 𝜌𝜌𝑓𝑓 𝑗𝑗𝑓𝑓 𝐷𝐷 = 4.463 × 103 𝜇𝜇𝑓𝑓 𝜌𝜌𝑔𝑔 𝑗𝑗𝑔𝑔 𝐷𝐷 = 2.088 × 104 𝑅𝑅𝑅𝑅𝑔𝑔 = 𝜇𝜇𝑔𝑔

𝑅𝑅𝑅𝑅𝑓𝑓 =

(6) (7)

Since the gas phase Reynolds number is greater than the liquid phase Reynolds number, the gas phase is chosen as the reference value to calculate the parameter A1, 𝐴𝐴1 =

−𝑅𝑅𝑅𝑅 = 0.586 1 + 𝑒𝑒 00000

(8)

Since A1 is less than 0.8, the parameter B1 is defined as 𝐵𝐵1 = 𝐴𝐴1 = 0.586

(9)

Knowing this parameter, the parameter K0 can be determined to be 0.25

𝜌𝜌𝑔𝑔 𝐾𝐾0 = 𝐵𝐵1 + (1 − 𝐵𝐵1 ) � � 𝜌𝜌𝑓𝑓

The parameter r is given as

𝑟𝑟 =

𝜌𝜌𝑔𝑔 0.25 1 + 1.57 �𝜌𝜌 � 𝑓𝑓

1 − 𝐵𝐵1

= 0.785

= 2.619

(10)

(11)

Finally, the parameter Co can be expressed as a function of the void fraction. Since we are ultimately trying to solve for the void fraction, having this be unknown at this step is acceptable: 1 − 𝑒𝑒 −𝐶𝐶1𝛼𝛼 � 1 − 𝑒𝑒 −𝐶𝐶1 𝐶𝐶𝑜𝑜 (𝛼𝛼) = 𝐾𝐾0 + (1 − 𝐾𝐾0 )𝛼𝛼 𝑟𝑟 �

(12)

Evaluation of Drift Velocity: To begin this evaluation, the parameter C5 can be evaluated as 1

𝜌𝜌𝑔𝑔 2 𝐶𝐶5 = �150 � �� = 2.834 𝜌𝜌𝑓𝑓

(13)

𝐶𝐶2 = 1

(14)

Since C5 is greater than unity and the ratio of the phasic densities is greater than 18, The parameter C3 is given as

−4462

𝐶𝐶3 = max �0.5, 𝑒𝑒 −300000 � = 1.97

(15)

The parameter C7 is given as

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0.09144 0.6 � = 2.489 𝐶𝐶7 = � 𝐷𝐷

(16)

Since C7 is greater than unity,

𝐶𝐶4 = 1

(17)

The drift velocity is also a function of the void fraction given as 1 �𝜌𝜌𝑓𝑓 − 𝜌𝜌𝑔𝑔 �𝑔𝑔𝑔𝑔 4 𝑉𝑉𝜐𝜐𝜐𝜐 (𝛼𝛼) = 1.41 � � (1 − 𝛼𝛼)𝐵𝐵1 𝐶𝐶2 𝐶𝐶3 𝐶𝐶4 𝜌𝜌𝑓𝑓2

(18)

We can define the void fraction in the drift flux form as 𝛼𝛼 =

𝛽𝛽

𝐶𝐶𝑜𝑜 (𝛼𝛼) +

𝑉𝑉𝜐𝜐𝜐𝜐 (𝛼𝛼) 𝑗𝑗

(19)

Here, the parameter Co and the drift velocity are functions of the void fraction. This expression can be solved iterative along with Equations. (3), (4) and (12) to yield a void fraction of 𝛼𝛼 = 0.62

(20)

PROBLEM 11.8 QUESTION Void, Quality and Pressure Drop Problem (Sections 11.4 and 11.5) Consider an adiabatic water channel 3 m long and 1 cm in diameter operating in homogeneous flow at 7.4 MPa pressure with a void (steam) distribution as shown in Fig. 11.31. The total flow rate is 0.3 kg/s. Talee the liquid viscosity at the operating conditions as 8.7 × 10−5 kg/m s. 1. Find the values of {α}, {β}, and x for the channel, i.e., volume averaged values. 2. Which values of Part 1 would change if the local flow velocities of the two phases remain equal but the flow rate was reduced sufficiently to yield a laminar flow distribution while the void distribution of Figure 11.31 remains unchanged? 3. Compare the pressure loss within the 3 m length

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FIGURE 11.31 Void distribution. Answers: 1. {α} = {β} = 0.041, x = 0.0023 2. {β} and x would change 3. ΔPtotal = 63.3 kPa

PROBLEM 11.8 SOLUTION Void, Quality and Pressure Drop Problem (Sections 11.4 and 11.5) Consider an adiabatic water channel 3 m long and 1 cm in diameter operating in homogeneous flow at 7.4 MPa pressure with a void (steam) distribution as shown in Fig. 11.31. The total flow rate is 0.3 kg/s. Take the liquid viscosity at the operating conditions as 8.7 × 10−5 kg/m · s. In this problem, we are given the following: — Channel length, L = 3 m — Channel diameter, D = 1cm

– Pressure, P = 7.4 MPa – Mass flow rate, ṁ = 0.3 kg/s – Liquid viscosity, μf = 8.7 × 10−5Pa·s — Bubble diameter, db = 0.25 cm

Part 1 Find the values of {α}, {β}, and x for the channel, i.e., volume averaged values. The cross section area of the channel is 𝜋𝜋 𝐴𝐴 = 𝐷𝐷2 = 7.854 × 10−5 m2 (1) 4

The liquid only Reynolds number is calculated as

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Re =

𝑚𝑚̇𝐷𝐷 = 4.39 × 105 𝜇𝜇𝑓𝑓 𝐴𝐴

(2)

Therefore turbulent flow and the HEM model can be assumed since this is a high mass flux. The volume of spherical bubbles (4 of them) is collectively 4 𝑑𝑑𝑏𝑏 3 𝑉𝑉𝑏𝑏 = 4 � 𝜋𝜋 � � � = 3.272 × 10−8 m3 3 2

The total volume of the region is 𝜋𝜋 𝑉𝑉 = 𝐷𝐷2 (1 cm) = 7.854 × 10−7 m3 4

(3)

(4)

The void fraction is calculated as the ratio of the volumes 𝛼𝛼 =

𝑉𝑉𝑏𝑏 = 0.041 𝑉𝑉

(5)

Since we me assuming HEM model, the volumetric flow fraction is equivalent to the void fraction, 𝛽𝛽 = 𝛼𝛼 = 0.041

(6)

The quality can then be determined from the void fraction assuming slip ratio is 1.0, 𝑥𝑥 =

1 = 0.0023 1 − 𝛼𝛼 𝜌𝜌𝑓𝑓 1 + 𝛼𝛼 𝜌𝜌 𝑔𝑔

(7)

where ρf = 732.64kg/m3 and ρg = 38.88kg/m3.

Part 2 Which values of Part 1 would change if the local flow velocities of the two phases remain equal but the flow rate was reduced sufficiently to yield a laminar flow distribution condition but the void distribution of Figure 11.31 remained unchanged? β and x would change. Explanation: In part A, because the Re was high, a turbulent flow exists, the HEM model was assumed applicable and a slip ratio of unity followed. In part B the problem statement indicates laminar flow with its attendant velocity distribution exists. Equation 11.45 illustrates that the slip ratio departs from unity due to non-uniform void distribution but there is no local slip per the problem statement because the problem statement explicitly states to assume the local flow velocities of the two phases remain equal. However the velocity distribution becomes laminar, with a central peak, which is quite different from the effectively uniform turbulent distribution. Hence more vapor volumetric flow than liquid flow occurs because the void distribution shown in Figure 11.31 places vapor preferentially in the center tube region where the velocity is higher than the average velocity. Thus, both the slip ratio exceeds unity and β (Equation 5.51) increases.

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The void fraction, a, determined solely by the areas occupied the phases, Equation 5.28, remains constant since the instantaneous void distribution of Figure 11.31 remains unchanged. However the quality, x, is a function involving S, and since S is higher than unity but alpha is constant, the value of x will increase as Equation 5.55 indicates. Part 3 Compare the pressure loss within the 3 m length. The photographic density is calculated as 𝜌𝜌𝑚𝑚 = (1 − 𝛼𝛼)𝜌𝜌𝑓𝑓 + 𝛼𝛼𝛼𝛼𝑔𝑔 = 703.733 kg⁄m

The pressure loss due to gravity is

(8)

∆𝑃𝑃𝑔𝑔 = 𝜌𝜌𝑚𝑚 𝑔𝑔𝑔𝑔 = 20.704 kPa

(9)

𝑓𝑓 = 0.184𝑅𝑅𝑅𝑅 −0.2 = 0.014

(10)

The friction factor can be calculated from the Reynolds number using the McAdams correlation The pressure loss due to friction is given as

The total pressure loss is

∆𝑃𝑃𝑓𝑓 = 𝑓𝑓

𝐿𝐿 𝑚𝑚̇2 = 42.566 kPa 𝐷𝐷 2𝜌𝜌𝑚𝑚 𝐴𝐴2

∆𝑃𝑃𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 = ∆𝑃𝑃𝑔𝑔 + ∆𝑃𝑃𝑓𝑓 = 63.3 kPa

(11)

(12)

Note there is no acceleration pressure drop since the void fraction is not changing in the channel.

PROBLEM 11.9 QUESTION Calculation of a Pipe’s Diameter for a Specific Pressure Drop (Section 11.5) For the vertical plate riser shown in Figure 11.32, calculate the steam-generation rate and the riser diameter necessary for operation at the flow conditions given below: Geometry: Downcomer height = riser height = 4.6 m Flow conditions: T (feed) = 226.7°C Steam pressure = 6.7 MPa Thermal power = 856 MWth (from heater primary side to fluid in riser)

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FIGURE 11.32 Vertical user schematic for Problem 11.9. Closed heater primary side not shown. Assumptions: Homogenous flow model is applicable. Friction losses in the downcomer, upper plenum, lower plenum and heater are negligible. Riser and downcomer are adiabatic. Quality at heater exit (xout) is 0.10. Answers: Steam flow rate = 475 kg/s D = 0.583 m

PROBLEM 11.9 SOLUTION Calculation of a Pipe’s Diameter for a Specific Pressure Drop (Section 11.5) For the vertical plate riser shown in Figure 11.32, calculate the steam-generation rate and the riser diameter necessary for operation at the flow conditions given below: — Downcomer and riser height, H = 4.6 m — Feedwater temperature, Tfw = 226.7 °C — Steam pressure, P = 6.7 MPa — Thermal power, 𝑄𝑄̇ = 856 MW — Exit quality, xout = 0.10

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Thermodynamic properties evaluated at the system pressure are: — Feedwater enthalpy, hfw = 975.7 kJ/kg — Saturated vapor enthalpy, hg = 2776 kJ/kg — Saturated liquid enthalpy, hf = 1252 kJ/kg 3

— Saturated liquid density, ρf = 745.1kg/m — Saturated vapor density, ρg = 34.8 kg/m

— Saturated liquid viscosity, μf = 92.386 μPa s

The steam mass flow rate can be determined from an energy balance, 𝑚𝑚̇𝑠𝑠𝑠𝑠 =

𝑄𝑄̇ = 475 kg⁄s ℎ𝑔𝑔 − ℎ𝑓𝑓𝑓𝑓

To determine the total mass flow rate 𝑚𝑚̇𝑡𝑡𝑡𝑡𝑡𝑡 observe that the feed water input flow rate 𝑚𝑚̇𝑓𝑓𝑓𝑓 makes up for the steam flow rate which is discharged which is 0.1 𝑚𝑚̇𝑡𝑡𝑡𝑡𝑡𝑡 . Hence 𝑚𝑚̇𝑓𝑓𝑓𝑓 = 0.1 𝑚𝑚̇𝑡𝑡𝑡𝑡𝑡𝑡

Applying the first law to the closed heater control volume obtain 𝑄𝑄̇ = ℎ𝑔𝑔 0.1𝑚𝑚̇𝑡𝑡 − ℎ𝑓𝑓𝑓𝑓 0.1𝑚𝑚̇𝑡𝑡

Hence 𝑚𝑚̇𝑡𝑡 =

𝑄𝑄̇ 856 × 106 W = 0.1(ℎ𝑔𝑔 − ℎ𝑓𝑓𝑓𝑓) 0.1(2.776 × 106 ) − (0.975 × 106 )J/kg

The total mass flow rate is thus

𝑚𝑚̇𝑡𝑡 = 4.75 × 103 kg⁄s

From above the feedwater input flow, 𝑚𝑚̇𝑓𝑓𝑓𝑓 which equals the steam generation rate, is 0.1 𝑚𝑚̇𝑡𝑡𝑡𝑡𝑡𝑡 Hence 𝑚𝑚̇𝑓𝑓𝑓𝑓 = 0.1(4.75 × 103 ) kg⁄s 𝑚𝑚̇𝑠𝑠𝑠𝑠 = 𝑚𝑚̇𝑓𝑓𝑓𝑓 = 475 kg⁄s

Pressure loss in the downcomer: The pressure loss in the downcomer is just due to gravity. Therefore the density of the liquid in the downcomer can be calculated from the enthalpy of the fluid in the downcomer. Using an energy in the mixing region, this enthalpy can be determined as ℎ𝐷𝐷𝐷𝐷 = ℎ𝑓𝑓𝑓𝑓 𝑥𝑥 + ℎ𝑓𝑓 (1 − 𝑥𝑥) = 1.224 × 103 kJ⁄kg

The density can be evaluated from this enthalpy and operating pressure, 𝜌𝜌𝐷𝐷𝐷𝐷 = 755.2 kg⁄m3 301

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The pressure loss can be calculated as ∆𝑃𝑃𝑔𝑔,𝐷𝐷𝐷𝐷 = −𝜌𝜌𝐷𝐷𝐷𝐷 𝑔𝑔𝑔𝑔 = −34.069 kPa Pressure loss in the riser: The pressure loss in the riser is due to both gravity and friction. The void fraction can be calculated as (assuming HEM) 𝛼𝛼 = The static mixture density is therefore

1 = 0.704 1 − 𝑥𝑥 𝜌𝜌𝑔𝑔 1 + 𝑥𝑥 𝜌𝜌 𝑓𝑓

𝜌𝜌𝑚𝑚 = (1 − 𝛼𝛼)𝜌𝜌𝑓𝑓 + 𝛼𝛼𝜌𝜌𝑔𝑔 = 244.936 kg⁄m3

The gravitational pressure drop is

∆𝑃𝑃𝑔𝑔,𝑇𝑇 = 𝜌𝜌𝑚𝑚 𝑔𝑔𝑔𝑔 = 11.049 kPa

Since the diameter is unknown, the frictional pressure drop will be left in terms of the diameter. The cross sectional area is 𝐴𝐴(𝐷𝐷) =

𝜋𝜋 2 𝐷𝐷 4

The liquid only Reynolds number is given as function of this diameter, Re(𝐷𝐷) =

𝑚𝑚̇𝑡𝑡𝑡𝑡𝑡𝑡 𝐷𝐷 𝐴𝐴(𝐷𝐷)𝜇𝜇𝑓𝑓

The friction factor, assuming McAdams correlation is

𝑓𝑓(𝐷𝐷) = 0.184𝑅𝑅𝑅𝑅(𝐷𝐷)−0.2

The frictional pressure drop is therefore,

2 𝐿𝐿 𝑚𝑚̇𝑡𝑡𝑡𝑡𝑡𝑡 ∆𝑃𝑃𝑓𝑓,𝑟𝑟 (𝐷𝐷) = 𝑓𝑓(𝐷𝐷) 𝐷𝐷 2𝜌𝜌𝑚𝑚 𝐴𝐴(𝐷𝐷)2

Overall Pressure Drop: Combining all of the pressure losses, we get an equation that is only a function of the unknown diameter, ∆𝑃𝑃𝑔𝑔,𝐷𝐷𝐷𝐷 + ∆𝑃𝑃𝑔𝑔,𝑟𝑟 + ∆𝑃𝑃(𝐷𝐷) = 0

Since the diameter is the only unknown, this equation can be solved yielding 𝐷𝐷 = 0.583 m

PROBLEM 11.10 QUESTION Flow Dynamics of Nanofluids (Section 11.5) Nanofluids are suspensions of solid nanoparticles in a base fluid (e.g., water), and were

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investigated at MIT for their potential as coolants in nuclear systems. The flow behavior of nanofluids can be analyzed as a two-phase liquid/solid mixture. Since the particles are so small, they can be assumed to move homogeneously with the base fluid, i.e., the slip ratio is equal to one. Consider a water-based nanofluid with alumina nanoparticles. It flows at steady state through a tube of 2.5 cm diameter. The nanofluid volumetric flowrate is 400 cm3/s and the nanoparticle volumetric fraction is 0.05. 1. Find the mass flow rate of the nanoparticles and the total mass flow rate of the nanofluid. 2. Find the pressure gradient within the tube if the flow direction is vertically downward. To calculate the friction pressure gradient, assume fully developed flow, zero surface roughness and fTP − flo. 3. If the nanoparticles were made of a material denser than alumina, how would the components of the pressure gradient (e.g. friction, gravity) change? Assume that the nanofluid volumetric flow rate and the nanoparticle volumetric fraction are held at 400 cm3/s and 0.05, respectively. The slip ratio is still equal to one. Provide a qualitative answer. Properties: Water: Density 1 g/cm3, viscosity 10−3 Pa s Alumina: Density 4 g/cm3, specific heat 780 J/kg K Answers: 1. ṁnp = 80g/s 𝑑𝑑𝑑𝑑

2. � 𝑑𝑑𝑑𝑑 �

𝑡𝑡𝑡𝑡𝑡𝑡

ṁtot = 460 g/s

= −10.9 𝑘𝑘𝑘𝑘𝑘𝑘⁄𝑚𝑚

PROBLEM 11.10 SOLUTION

Flow Dynamics of Nanofluids (Section 11.5) In the problem statement we are given: — Tube diameter, D = 2.5 cm 3

— Volumetric flow rate of nanofluid, Q = 400 cm /s — Volumetric flow fraction, β = 0.05 — Density of water, ρw = 1 g/cm −3

— Viscosity of water, μw = 10

Pa s 3

— Density of nanoparticle, ρnp = 4 g/cm

— Specific heat of nanoparticle, cnp = 780 J/kg K

Part 1: Find the mass flow rate of the nanoparticles and the total mass flow rate of the

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nanofluid. The volumetric flow rate of the nanoparticles is ̇ = 𝑄𝑄𝑄𝑄 = 2 × 10−5 m3 ⁄s 𝑉𝑉𝑛𝑛𝑛𝑛

The volumetric flow rate of the water is

𝑉𝑉𝑤𝑤̇ = 𝑉𝑉̇ (1 − 𝛽𝛽) = 3.8 × 10−5 m3 ⁄s

The mass flow rate of nanoparticles is

̇ 𝜌𝜌𝑛𝑛𝑛𝑛 = 80 g⁄s 𝑚𝑚̇𝑛𝑛𝑛𝑛 = 𝑉𝑉𝑛𝑛𝑛𝑛

The mass flow rate of water is

𝑚𝑚̇𝑤𝑤 = 380 g⁄s

The total mass flow rate is therefore

𝑚𝑚̇𝑡𝑡𝑡𝑡𝑡𝑡 = 𝑚𝑚̇𝑛𝑛𝑛𝑛 + 𝑚𝑚̇𝑤𝑤 = 460 g⁄s

Part 2: Find the pressure gradient within the tube if the flow direction is vertically downward. Assuming HEM model, the void fraction is equivalent to the volumetric flow fraction, 𝛼𝛼 = 𝛽𝛽 = 0.05

The mean density is therefore

𝜌𝜌𝑚𝑚 = (1 − 𝛼𝛼)𝜌𝜌𝑤𝑤 + 𝛼𝛼𝜌𝜌𝑛𝑛𝑛𝑛 = 1.15 g⁄cm3

The pressure change gradient due to gravity is �

𝑑𝑑𝑑𝑑 � = −𝜌𝜌𝑚𝑚 𝑔𝑔 = −11.278 kPa⁄m 𝑑𝑑𝑑𝑑 𝑔𝑔

The cross sectional area of the tube is

𝐴𝐴 =

The liquid only Reynolds number is

𝜋𝜋 2 𝐷𝐷 = 4.909 × 10−4 m2 4

𝑅𝑅𝑅𝑅 =

The friction factor is

𝑚𝑚̇𝑡𝑡𝑡𝑡𝑡𝑡 𝐷𝐷 = 2.343 × 104 𝐴𝐴𝐴𝐴𝑤𝑤

𝑓𝑓 = 0.316𝑅𝑅𝑅𝑅 −0.25 = 0.026

The pressure change gradient due to friction is

2 𝑑𝑑𝑑𝑑 1 𝑚𝑚̇𝑡𝑡𝑡𝑡𝑡𝑡 � � = 𝑓𝑓 = 0.39 kPa⁄m 𝑑𝑑𝑑𝑑 𝑓𝑓 𝐷𝐷 2𝜌𝜌𝑚𝑚 𝐴𝐴2

The total pressure loss gradient is

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𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 � � = � � + � � = −10.9 kPa⁄m 𝑑𝑑𝑑𝑑 𝑡𝑡𝑡𝑡𝑡𝑡 𝑑𝑑𝑑𝑑 𝑔𝑔 𝑑𝑑𝑑𝑑 𝑓𝑓

PROBLEM 11.11 QUESTION

HEM Pressure Loss (Section 11.5) Consider a 3 meter long water channel of circular cross sectional area 1.5 × 10−4 m2 operating in upflow at the following conditions: ṁ = 0.29 kg/s P = 7.2 MPa Compute the pressure loss under homogeneous equilibrium assumptions for the following additional conditions: 1. Adiabatic channel with inlet flow quality of 0.15. 2. Uniform axial heat flux of sufficient magnitude to heat the entering saturated coolant (xin = 0) to an exit quality of 0.15. Answers: 1. ΔPtot = 36.6 kPa 2. ΔPtot = 44.0 kPa

PROBLEM 11.11 SOLUTION HEM Pressure Loss (Section 11.5) In the problem statement we are given the following parameters: – mass flow rate, ṁ = 0.29 kg/s – pressure, P = 7.2 MPa – tube area, A = 1.5 × 10−4 m2 – channel length, L = 3 m Thermodynamic properties: – saturated liquid density, ρf = 736.2 kg/m3 – saturated vapor density, ρg = 37.7 kg/m3 – saturated liquid viscosity, μf = 90.5 × 10−6 Pa s Part 1: Adiabatic channel with inlet flow quality of 0.15. For this part the quality is constant at

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𝑥𝑥 = 0.15

The diameter of the channel is

𝐷𝐷 = �

4𝐴𝐴 = 0.014 m 𝜋𝜋

Assuming HEM model, the void fraction can be calculated as 𝛼𝛼 = The mean density is

1 = 0.775 1 − 𝑥𝑥 𝜌𝜌𝑔𝑔 1 + 𝑥𝑥 𝜌𝜌 𝑓𝑓

𝜌𝜌𝑚𝑚 = (1 − 𝛼𝛼)𝜌𝜌𝑓𝑓 + 𝛼𝛼𝜌𝜌𝑔𝑔 = 194.8 kg⁄m3

The gravitational pressure drop is therefore The Reynolds number is given as

∆𝑃𝑃𝑔𝑔 = 𝜌𝜌𝑚𝑚 𝑔𝑔𝑔𝑔 = 5.731 kPa 𝑅𝑅𝑅𝑅 =

𝑚𝑚̇𝐷𝐷 = 2.952 × 105 𝐴𝐴𝜇𝜇𝑓𝑓

The friction factor can be calculated with the McAdams correlation The frictional pressure drop is

The total pressure drop is

𝑓𝑓 = 0.184𝑅𝑅𝑅𝑅 −0.2 = 0.015 ∆𝑃𝑃𝑓𝑓 = 𝑓𝑓

𝐿𝐿 𝑚𝑚̇2 = 30.863 kPa 𝐷𝐷 2𝜌𝜌𝑚𝑚 𝐴𝐴2

∆𝑃𝑃𝑡𝑡𝑡𝑡𝑡𝑡 = ∆𝑃𝑃𝑔𝑔 + ∆𝑃𝑃𝑓𝑓 = 36.6 kPa

Part 2: Uniform axial heat flux. The inlet quality is while the exit quality is

𝑥𝑥𝑖𝑖𝑖𝑖 = 0 𝑥𝑥𝑜𝑜𝑜𝑜𝑜𝑜 = 0.15

Therefore, the quality can be expressed as a function of axial position 𝑥𝑥(𝑧𝑧) =

𝑥𝑥𝑜𝑜𝑜𝑜𝑜𝑜 𝑧𝑧 𝐿𝐿

The void fraction at any location can be computed using the HEM model

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𝛼𝛼(𝑧𝑧) = And the mean density is given as

1 1 − 𝑥𝑥(𝑧𝑧) 𝜌𝜌𝑔𝑔 1+ 𝑥𝑥(𝑧𝑧) 𝜌𝜌𝑓𝑓

𝜌𝜌𝑚𝑚 (𝑧𝑧) = �1 − 𝛼𝛼(𝑧𝑧)�𝜌𝜌𝑓𝑓 + 𝛼𝛼(𝑧𝑧)𝜌𝜌𝑔𝑔

The acceleration pressure drop is given as

𝑚𝑚̇ 2 1 1 ∆𝑃𝑃𝑎𝑎 = � � � − � = 14.112 kPa 𝐴𝐴 𝜌𝜌𝑚𝑚 (𝐿𝐿) 𝜌𝜌𝑓𝑓

The gravitational pressure drop is given as 𝐿𝐿

∆𝑃𝑃𝑔𝑔 = � 𝜌𝜌𝑚𝑚 (𝑧𝑧)𝑔𝑔𝑔𝑔𝑔𝑔 = 10.361 kPa 0

The frictional pressure drop (friction factor taken as liquid only friction factor computed above) is 𝐿𝐿

The total pressure drop is

∆𝑃𝑃𝑓𝑓 = � 𝑓𝑓 0

1 𝑚𝑚̇2 𝑑𝑑𝑑𝑑 = 19.515 kPa 𝐷𝐷 2𝜌𝜌𝑚𝑚 (𝑧𝑧)𝐴𝐴2

∆𝑃𝑃𝑡𝑡𝑡𝑡𝑡𝑡 = ∆𝑃𝑃𝑎𝑎 + ∆𝑃𝑃𝑔𝑔 + ∆𝑃𝑃𝑓𝑓 = 44.0 kPa

PROBLEM 11.12 QUESTION

Analysis of a Liquid-Metal Reactor Vessel (Section 11.7) Fast reactors have attracted renewed attention within the nuclear community because of their ability to consume the actinides from the LWR spent fuel. Consider the vessel of a liquid-leadcooled fast reactor, which is made of stainless steel with the dimensions shown in Figure 11.33. The top of the vessel is filled with a cover gas (nitrogen) whose opening temperature and pressure are 400 °C and 0.5 MPa, respectively. The pressure outside the vessel is 0.1 MPa. Relevant material and fluid properties are given in Table 11.9. 1. A high magnitude earthquake causes a 10 cm2 crack in the vessel. Calculate the mass flow rate through the crack, immediately after the earthquake, for the following two cases: (a) The crack occurs at the very bottom of the vessel (b) The crack occurs at the very top of the vessel, i.e., in the cover gas region (The flow through the crack can be treated as steady-state, adiabatic and inviscid in both cases. The pressure outside the vessel can be assumed constant at 0.1 MPa. Liquid lead can be treated as an incompressible liquid and nitrogen as a perfect gas.) 2. Which of the two cases considered in Part 1 would you judge more dangerous from a safety

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viewpoint?

FIGURE 11.33 Schematic and dimensions of the vessel. TABLE 11.9 Properties for Problem 11.12 Stainless

Maximum allowable stress intensity factor, (Sm) = 138 MPa at 400 °C,

steel

ρ = 7500 kg/m3

Lead

ρ = 10,500 kg/m3, μ = 1.6 × 10−3 Pa s, boiling point 1750°C

Nitrogen

cp = 1039 J/kg K, Rsp = 297 J/kg K, γ = 1.4

Answers: 1. ṁ = 215 kg/s 2. ṁ = 0.766 kg/s

PROBLEM 11.12 SOLUTION Analysis of a Liquid-Metal Reactor Vessel (Section 11.7) In the problem statement we are given: – pressure inside the vessel, P = 0.5 MPa – pressure outside the vessel, Pb = 0.1 MPa – break area, Ab = 10 cm2 – height of lead, H = 17.5 m – density of lead, ρl = 10500 kg/m3

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– viscosity of lead, μl = 1.6 × 10−3 Pa s – specific heat at constant pressure for nitrogen, cpN = 1039 J/kg K – gas constant for nitrogen, RN = 297 J/kg K – gamma factor for nitrogen, γ = 1.4 Part 1(a): Crack at bottom of vessel. The pressure at the bottom of the vessel is 𝑃𝑃𝑎𝑎 = 𝑃𝑃 + 𝜌𝜌𝑙𝑙 𝑔𝑔𝑔𝑔 = 2.302 MPa

Since the flow is assumed inviscid in both cases, the Bernoulli equation can be applied: 𝑃𝑃𝑎𝑎 𝑃𝑃𝑏𝑏 𝜐𝜐𝑏𝑏2 = + 𝑃𝑃𝑙𝑙 𝑃𝑃𝑙𝑙 2

Solving for the break velocity and converting it to a mass flow rate, 𝑃𝑃𝑎𝑎 − 𝑃𝑃𝑏𝑏 � = 215 kg⁄s 𝑚𝑚̇ = 𝜌𝜌𝑙𝑙 𝐴𝐴𝑏𝑏 �2 � 𝜌𝜌𝑙𝑙

Part 1(b): Crack at the top of vessel. Since the flow is compressible, the critical pressure for critical flow is calculated for a perfect gas 𝛾𝛾

2 𝛾𝛾−1 � 𝑃𝑃𝑐𝑐𝑐𝑐 = 𝑃𝑃 � = 0.264 MPa 𝛾𝛾 + 1

Since the back pressure is lower than the critical pressure, critical flow condition exists. The density of the nitrogen gas is calculated using the ideal gas law 𝜌𝜌𝑁𝑁 =

𝑃𝑃 = 2.501 kg⁄m3 𝑅𝑅𝑁𝑁 𝑇𝑇

The choked mass flow rate can then be calculated with the critical pressure as 2

𝛾𝛾+1

𝑃𝑃𝑐𝑐𝑐𝑐 𝛾𝛾 𝑃𝑃𝑐𝑐𝑐𝑐 𝛾𝛾 𝑚𝑚̇ = 𝐴𝐴𝑏𝑏 𝜌𝜌𝑁𝑁 �2𝑐𝑐𝑝𝑝𝑝𝑝 𝑇𝑇 �� − � � � = 0.766 kg⁄s 𝑃𝑃 𝑃𝑃

Part 2: The first case (i.e., break at the bottom of the vessel) is far more dangerous from a safety viewpoint, because it has the potential to empty the vessel and uncover the core. The second case merely results in a depressurization of the reactor without any loss of coolant (the boiling point of lead is very high, so it does not flash upon depressurization).

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Chapter 12 Pool Boiling Contents Problem 12.1 Comparison of liquid superheat required for nucleation in water and sodium ........................................................................................................... 311 Problem 12.2 Nucleate boiling initiation and termination on a heat exchanger tube ........... 313 Problem 12.3 Evaluation of pool boiling conditions at high pressure .................................. 316 Problem 12.4 Shell and tube horizontal evaporator .............................................................. 319 Problem 12.5 Comparison of stable film boiling conditions in water and sodium ............... 322 Problem 12.6 Analysis of decay heat removal during a severe accident .............................. 324 Problem 12.7 Void fraction and pressure drop in an isolation condenser ............................ 326

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PROBLEM 12.1 QUESTION Comparison of Liquid Superheat Required for Nucleation in Water and Sodium (Section 12.2) 1. Calculate the relation between superheat and equilibrium bubble radius for sodium at 1 atmosphere using Equation 12.7. Repeat for water at 1 atmosphere. Does sodium require higher or lower superheat than water? 2. Consider bubbles of 10μm radius. Evaluate the vapor superheat and the bubble pressure difference for the bubbles of question 1. Answer: 1

1. For sodium: 𝑇𝑇ℓ − 𝑇𝑇𝑠𝑠𝑠𝑠𝑠𝑠 = 2.517 × 10−4 𝑟𝑟 1

For water: 𝑇𝑇ℓ − 𝑇𝑇𝑠𝑠𝑠𝑠𝑠𝑠 = 3.250 × 10−5 𝑟𝑟

𝑒𝑒

2. (T −Tsat)sodium = 25.2°C; (T -Tsat)water = 3.25ºC (Pv - P)sodium = 22.6 kN/m2; (Pv - P)water = 11.7 kN/m2

PROBLEM 12.1 SOLUTION Comparison of Liquid Superheat Required for Nucleation in Water and Sodium (Section 12.2) 1. Calculate the relation between superheat and equilibrium bubble radius for sodium at 1 atmosphere using Equation 12.7. Repeat for water at 1 atmosphere. Does sodium require higher or lower superheat than water? The relation between superheated and equilibrium bubble radius from Equation 12.7, is: (12.7)

(a) For sodium at 1 atm (= 1.0135 × 105 Pa) Tsat = 881ºC = 1154 K

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R= sp

R = 361.5 J/ kgK M

hfg = 3.78 MJ/kg σ = 0.113 N/m vf = 1.3477 × 10-3 m3/kg vg = 3.6496 m3/kg Note: These properties must be searched on line. They fall just beyond Table E.8, and interpolation from Table E.7 would be too inaccurate. (b) For water at 1 atm Tsat = 373 K

R= sp

R = 461.9 J/ kgK M

hfg = 2.256 MJ/kg σ = 0.05878 N/m vf = 1.0435 × 10-3 m3/kg vg = 1.6736 m3/kg Thus, for sodium, Tb − Tsat=

2(0.113)(1154)(3.6496 − 1.3477 ×10−3 )  1  −4 1  = 2.517 ×10 6 3.78 ×10 r*  r*

(1)

2(0.05878)(373)(1.673 − 1.0435 ×10−3 )  1  −5 1  = 3.250 ×10 6 2.2569 ×10 r*  r*

(2)

and for water Tb − Tsat=

where r* is in meters From the two equations above, we see that sodium requires a much larger superheat. This is primarily due to the fact that σsodium > σwater.

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2. Vapor superheat and the bubble pressure difference for the bubble of question 1. For r* = 10μm Equation 12.6 is (12.6) where ΔTsat for sodium and water are available from Equations (1) and (2). Hence for sodium 2.517 ×10−4 ∆Tsat = = 25.2°C 10 ×10−6

While for water ∆Tsat =

3.250 ×10−5 = 3.25°C 10 × 10−6

PROBLEM 12.2 QUESTION Nucleate Boiling Initiation and Termination on a Heat Exchanger Tube (Chapter 8, 10, and 12 - Section 12.2) A heat exchanger tube is immersed in a water cooling tank at 290 K, as illustrated in Figure 12.13. Hot water (single phase, 550 K) enters the tube inlet and is cooled as it flows at 2 kg/s through the 316 grade stainless steel tube (19 mm outside diameter and 15.8 mm inside diameter). Neglect entrance effects. 1. Compute the length along the horizontal inlet length of the tube where nucleate boiling on the tube outside diameter is initiated. 2. Compute the length where nucleate boiling on the tube outside diameter is terminated. The heat transfer coefficient between the outer tube wall and the water cooling tank is 500 W/m2K for single phase conditions and 5000W/m2K for nucleate boiling conditions. The wall superheat for incipient nucleation is 15°C for this configuration. Estimate and justify any additional

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information you need to execute the solution. Fluid properties of inlet water (assume they stay constant) k, thermal conductivity = 0.5 W/mºC

ρ, density = 704 kg/m3 µ, viscosity = 8.69

10-5 kg/m s

cp, heat capacity =6270 J/kgºC

Answer:

FIGURE 12.13

Cooling tank—heat exchange diagram.

1. x = 0 m 2. L = 35.16 m

PROBLEM 12.2 SOLUTION Nucleate Boiling Initiation and Termination on a Heat Exchanger Tube (Chapter 8, 10, and 12 - Section 12.2) Neglecting entrance effects, the maximum wall temperature will occur at the entrance since the water in the tube begins cooling at the entrance continuously along the length of the tube. 1. Applying the concept of thermal resistance:

Hence,

𝑟𝑟 ln � 𝑟𝑟𝑜𝑜 � 1 1 𝑖𝑖 + + � 𝑇𝑇𝐵𝐵𝐵𝐵 − 𝑇𝑇𝐵𝐵𝐵𝐵 = 𝑞𝑞 ′ � 2𝜋𝜋𝑟𝑟𝑖𝑖 ℎ𝑖𝑖 2𝜋𝜋𝜋𝜋 2𝜋𝜋𝑟𝑟𝑜𝑜 ℎ𝑜𝑜

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𝑇𝑇𝐻𝐻𝐵𝐵 − 𝑇𝑇𝐶𝐶𝐵𝐵 𝑞𝑞 = 𝑟𝑟 ln � 𝑜𝑜 � 𝑟𝑟𝑖𝑖 1 1 + + 2𝜋𝜋𝑟𝑟𝑖𝑖 ℎ𝑖𝑖 2𝜋𝜋𝑟𝑟𝑜𝑜 ℎ𝑜𝑜 2𝜋𝜋𝜋𝜋 ′

ℎ𝑜𝑜 = 500W/mK

To calculate hi, u = constant ⇒ u (TBH) = u (Ti)

For u = uw we make use of the Dittus-Boelter correlation: 𝑚𝑚̇ = 𝜌𝜌𝜌𝜌𝜌𝜌 = 14.49m/s

𝜌𝜌𝜌𝜌𝜌𝜌 = 1.85 × 106 𝜇𝜇

Re =

Pr =

𝐶𝐶𝑃𝑃 𝜇𝜇 = 1.09 𝑘𝑘

For fully developed flow, all conditions are satisfied so hi can be found:

Thus, and since: then

ℎ𝑖𝑖 =

𝐾𝐾 (0.023)Re0.8 𝑃𝑃𝑃𝑃 0.3 = 77.24kW/km2 𝐷𝐷𝐻𝐻 𝑞𝑞 ′ = 7450.69W/m 𝑞𝑞 ′ = 2𝜋𝜋𝑟𝑟𝑜𝑜 ℎ𝑜𝑜 (𝑇𝑇0 − 𝑇𝑇𝐵𝐵𝐵𝐵 ) 𝑇𝑇0 = 539.65K

Pressure at the pipe is calculated by:

𝑃𝑃𝑎𝑎𝑎𝑎𝑎𝑎 + 𝜌𝜌𝜌𝜌𝜌𝜌 = 1.15 × 105Pa ⟹ 𝑇𝑇𝑠𝑠𝑠𝑠𝑠𝑠 = 103.58℃

since T0 > Tsat + 15 boiling will first occur at x = 0 m.

2. Boiling terminates at z > Z(T0 = Tsat + 15°C) where Tsat + 15 = 118.58°C = 391.73 K q′ at the location where boiling terminates is: 𝑞𝑞 ′ = 2𝜋𝜋𝑟𝑟𝑜𝑜 ℎ𝑜𝑜 (𝑇𝑇0 − 𝑇𝑇𝐵𝐵𝐵𝐵 ) = 30.36kW/m

𝑟𝑟 ln � 𝑜𝑜 � 1 1 𝑟𝑟𝑖𝑖 𝑇𝑇𝐵𝐵𝐵𝐵 = 𝑞𝑞 ′ � + + � + 𝑇𝑇𝐵𝐵𝐵𝐵 = 433.92K 2𝜋𝜋𝑟𝑟𝑖𝑖 ℎ𝑖𝑖 2𝜋𝜋𝜋𝜋 2𝜋𝜋𝑟𝑟𝑜𝑜 ℎ𝑜𝑜 315

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so q' simplifies to: 𝑞𝑞 ′ =

𝑇𝑇𝐵𝐵𝐵𝐵 − 290 4.7404 × 10−3

𝑞𝑞 ′ ∆𝑧𝑧 = 𝑚𝑚̇𝑐𝑐𝑝𝑝 (𝑇𝑇𝐵𝐵𝐵𝐵 − 𝑇𝑇𝐸𝐸 ) ⟹ 𝑇𝑇𝐸𝐸 = 𝑇𝑇𝐵𝐵𝐵𝐵 −

𝑞𝑞 ′ ∆𝑧𝑧 𝑚𝑚̇𝑐𝑐𝑝𝑝

Thus, to determine the location where boiling on the outside surface terminates, the following procedure must be executed: – Set TBH = 550 K 𝑇𝑇

−290

𝐵𝐵𝐵𝐵 – Solve for q′ using the equation, 𝑞𝑞 ′ = 4.7404×10 −3

– Chose a small ∆z – Calculate q′ ∆z

– Calculate TBH at the end of the interval and set TE equal to TBH – Set TBH = TE and repeat steps 1 through 6 until TE = 433.92 K – Sum the ∆z's Executing this procedure yields: −𝑚𝑚̇𝑐𝑐𝑝𝑝

𝑑𝑑𝑑𝑑𝐵𝐵𝐵𝐵 𝑇𝑇𝐵𝐵𝐵𝐵 (𝑧𝑧) − 290 = 𝑑𝑑𝑑𝑑 4.7404 × 10−3

Integrating for TBH from 550 to 433.92 from z = 0 to z =zb produces:

and finally:

[− ln(𝑇𝑇𝐵𝐵𝐵𝐵 (𝑧𝑧) − 290)]433.92 = 550

(𝑧𝑧𝑏𝑏 − 0) 𝑚𝑚𝑐𝑐𝑝𝑝 4.7404 × 10−3

𝑧𝑧𝑏𝑏 = 35.16m

PROBLEM 12.3 QUESTION Evaluation of Pool Boiling Conditions at High Pressure (Section 12.4) 1. A manufacturer has free access on the weekends to a supply of 8000 amps of 440 V electric power. If his boiler operates at 3.35 MPa, how many 2.5 cm diameter, 2 m-long electric emersion heaters would be required to utilize the entire available electric power? He desires to operate at 80% of critical heat flux. 2. At what heat flux would incipient boiling occur? Assume that the water at the saturation temperature corresponds to 3.35 MPa. The natural convection heat flux is given by:

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P =3.35 MPa Tsat = 240ºC hfg = 1766 kJ/kg ρf = 813 kg/m3 ρg = 16.8 kg/m3 σ = 0.0286 N/m kf = 0.63185 W/mºC cpf = 4.7719 kJ/kgºC µf = 1.109 10-4 kg/m s

Assume that the maximum cavity radius is very large. You may use the Rohsenow correlation for nucleate pool boiling taking Csf = 0.0130 in Equation 12.18c. Answer: 1. N = 6 (based on Cℓ = 0.18 in Equation 12.31) 2. 𝑞𝑞𝑖𝑖″ = 3.76 kW/m2

PROBLEM 12.3 SOLUTION

Evaluation of Pool Boiling Conditions at High Pressure (Section 12.4) 1. Total available power: Q = 8000(400) = 3.52 MW N

number of heaters

Heated Surface Area, As = NπDL = Nπ(0.025 m)(2 m) = 0.157 N m2 Then calculate the heat flux: The critical heat flux for saturated conditions in pool boiling is given by Equation 12.31 as:

= qcr′′ h fg ρ = g jg

jg 2.97 ×104 jg kJ m3 (1766 kJ kg ) (16.8 kg m3 )=

where

and C1 = 0.18 (Rohsenow), thus: which if run at 80% power, solving for N:

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2. At the incipient condition, where

N = 5.6 ≅ 6

= incipient heat flux = natural convection heat flux

Per the problem statement:

, hence

For the pool condition, the incipient heat flux and ∆Tsat are related by the Rohsenow correlation, Equation 12.18c: (12.18c) where, for water, Csf = 0.013, n = 0.33, and since conditions are saturated, = ρf and ρv = ρg . Now inserting the expression above for into the Rohsenow correlation obtain the following explicit equation for :

Inserting the properties of saturated water at 3.35 MPa, obtain

= 0.013 (370.08)(13.4)0.33 (3.67 × 10-6)0.167 (0.8376) = 0.013 (370.08)(2.355)(0.124)(0.8376)

Therefore,

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PROBLEM 12.4 QUESTION Shell and Tube Horizontal Evaporator (Chapters 9, 10, 12 - Section 12.4) A shell-and-tube horizontal evaporator is to be designed with 30 tubes 1 cm diameter. Inside the tubes water at 100 psia (690 kN/m2) enters at one end at 130°C and leaves at the other end at 120°C. The water velocity v in the tubes is 3 m/s. In the shell side, atmospheric pressure steam is generated at 100°C. Calculate: 1. The length of the tubes 2. Rate of evaporation, kg/s 3. Rate of flow of the water on the tube side, kg/s 4. Pressure drop in the tubes on the water side, (Assume fully developed flow and neglect entrance and exit losses). – For the boiling side, use the Rohsenow correlation (Equation 12.18c and Csf = 0.013). – Make your calculations for heat flux at mid-point where the liquid is at 125°C. Neglect thermal resistance of the thin tube wall. – Assume properties of liquid inside the tubes at 690 kPa in the range 120-130°C are the same as those at 1 atm and 100°C given in Table 12.4. TABLE 12.4 Water Properties for Problem 12.4 Liquid Vapor ρ (kg/m3)

960

0.60

cp (kJ/kg°C)

4.2

μ (kg/m s)

3 × 10

1.3 × 10−5

k (W/m°C)

0.68

0.025

2 −4

σ (N/m)

0.06

hfg (kJ/kg)

2280

1.75

0.97

Properties for water at P = 1 atm. Tsat = 100°C

Answer: 1. L = 1.31 m 2. Rate of evaporation = 0.125 kg/s 3. Mass flow rate = 6.79 kg/s 4. ∆P = 10.2 kPa

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PROBLEM 12.4 SOLUTION Shell and Tube Horizontal Evaporator (Chapters 9, 10, 12 - Section 12.4) Changes/corrections made to text problem statement in this solution: • •

For boiling use the Rohsenow correlation Equation 12.18c and take Csf as 0.013. Table 12.4 – specific heat for vapor = 2 kJ/kg-K Pr liquid at 100ºC = 1.75

Solution: First, let’s compute parameters needed later in the solution. D 2 (π 4 ) (10−2 )= m 2 7.85 ×10−5 m 2 (π 4 ) = = q m W cp (TIN − T= Total energy transfer: ( 6.7858 kg s )( 4.2 kJ kg°C )(130 − 120°C ) OUT )

Flow area per tube: = AF

= 2.85 ×105 W

Prandtl number:

µ

( 960 kg m ) ( 3m s ) (10 m=) 9.6 ×10 3

GD ρVD Reynolds number: Re= = = 

−2

−4

3 ×10 kg m s

(see above)

Dittus-Boelter heat transfer coefficient:  = 0.023 k D Re

0.8

Pr 0.3

 0.68 W m°C  0.3 4 0.8 2.3 ×10−2  9.6 10 = × (1.75) ( )  −2  10 m  = 1.79 ×104 W m 2 °C Question 1: Length of the tubes, L(m) Total surface area of tubes, = AF π= DLN

qT q′′

Now per problem statement, take q′′ as q′′ at mid-point of tube length where the liquid is 125ºC and neglect thermal resistance of the thin tube wall. Hence,

qT π DN qat′′ mid length

To find q′′ take a radial traverse through the evaporator from the mixed (bulk) liquid temperature of 125ºC (= TB) to the shell side steam at 100ºC (= TS) at the tube mid-length. Hence,

TB − TW = q′′ within the tube

(1)

and Equation 12.18c from Rohsenow

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12   µc   hfg ′′ q σ     p  TW − TS = Csf       cp  µ hfg  ( ρ − ρg ) g    k 

(2)

Taking TW from Equation 1 into Equation 2 yields

TB −

q′′ − TS = RHS of Equation 2  = RHS of Equation 2

Now

hfg cp

= C sf

(3)

2280 kJ kg = ( 0.013 ) 7.06°C 4.2 kJ kg°C

( 3 ×10 kg m s ) 4.2 ×10 J kg°C = = 1.85 −4

µ cp

0.68 W m°C

k

  0.06 N m or kg s 2   2.53 ×10−3 m = 3 2  ( 960 − 0.6 ) kg m ( 9.81 m s )   q′′ ( W m 2 )  1 q′′ ( W m 2 ) q′′   = = µ hfg ( 3 ×10−4 kg m s ) ( 2280 kJ kg )  684  m     σ =    ( ρ − ρg ) g 

Then Equation 3 becomes 1/3

 q′′ W m 2  q′′ W m 2 25 − = 7.06 2.53 ×10−3 )  1.85(°C) (  4 1.79 ×10  684  2 13 q′′ W m 2   ′′ = 25 − 0.202 q W m ( )   1.79 ×104

′′ 2.3 ×105 W m 2 Solving, obtain q= Now solving for L,

= L

qT 2.85 ×105 W = = π DNq′′ π ( 0.01 m ) 30 ( 2.3 ×105 W m 2 )

1.31 m

Question 2: Rate of evaporation (kg/s) q 2.85 ×105 W m= = = 0.125 kg s V hfg 2.28 ×106 W s kg Question 3: Rate of water flow on the tube side (kg/s) m W = N ρ vAF =30 ( 960 kg m3 ) ( 3 m s ) ( 7.85 ×10−5 m 2 ) = 6.788 kg s

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Question 4: Pressure drop in tubes on the water side L ρV 2 ∆P = f D 2 f per Figure 9.19 (Moody) ≅ 0.018

3 2 2 2  1.31  960 kg m ( 3 m s ) = = ∆ P 0.018  1.019 ×104 Pa  2  0.01 

PROBLEM 12.5 QUESTION Comparison of Stable Film Boiling Conditions in Water and Sodium (Section 12.5) Compare the value of the wall superheat required to sustain film boiling on a horizontal steel wall as predicted by Berenson’s correlation to that predicted by Henry’s correlation (Table 12.3). 1. Consider the cases of saturated water at (a) atmospheric pressure and (b) P = 7.0 MPa., 2. Consider the case of saturated sodium at atmospheric pressure. Answers: Berenson

Henry

1. H 2 O at 0.1 MPa: M

TB − Tsat = 68.9 °C

TH − Tsat = 187 °C

H 2 O at 7.0 MPa: M

TB − Tsat = 1042 °C

TH − Tsat = 1443 °C

(too high in physical sense) (too high in physical sense) 2. Na at 0.1 MPa: M

TB − Tsat = 30.6 °C

TH − Tsat = 411°C

(too low in physical sense)

PROBLEM 12.5 SOLUTION Comparison, of Stable Film Boiling Conditions in Water and Sodium (Section 12.5) Berenson’s correlation:

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𝑇𝑇𝐵𝐵𝑀𝑀 − 𝑇𝑇𝑠𝑠𝑠𝑠𝑠𝑠 = 0.127 Henry’s correlation:

𝜌𝜌𝑣𝑣𝑣𝑣 ℎ𝑓𝑓𝑓𝑓 𝑔𝑔�𝜌𝜌𝑓𝑓 − 𝜌𝜌𝑔𝑔 � � 𝜌𝜌 + 𝜌𝜌 � 𝑘𝑘𝑣𝑣𝑣𝑣 𝑓𝑓 𝑔𝑔 1⁄2

𝑔𝑔�𝜌𝜌𝑓𝑓 − 𝜌𝜌𝑔𝑔 � � � 𝑔𝑔𝑐𝑐 𝜎𝜎

1⁄3

𝑔𝑔�𝜌𝜌𝑓𝑓 − 𝜌𝜌𝑢𝑢𝑢𝑢 � � � 𝜇𝜇 𝑣𝑣𝑣𝑣

�𝜌𝜌𝜌𝜌𝑐𝑐𝑝𝑝 �ℓ ℎ𝑓𝑓𝑓𝑓 𝑇𝑇𝐻𝐻𝑀𝑀 − 𝑇𝑇𝐵𝐵𝑀𝑀 � = 0.42 � � 𝑀𝑀 𝑇𝑇𝐵𝐵 − 𝑇𝑇ℓ �𝜌𝜌𝜌𝜌𝑐𝑐𝑝𝑝 � 𝑐𝑐𝑝𝑝,𝑤𝑤 (𝑇𝑇𝐵𝐵𝑀𝑀 − 𝑇𝑇𝑠𝑠𝑠𝑠𝑠𝑠 )

0.6

𝑤𝑤

1. a) Applying Berenson’s correlation for water at 0.1 MPa: 𝑇𝑇𝐵𝐵𝑀𝑀 − 𝑇𝑇𝑠𝑠𝑠𝑠𝑠𝑠 = 68.9℃

which appears to be too high in a physical sense.

Applying Henry’s correlation for water at 0.1 MPa: 𝑇𝑇𝐻𝐻𝑀𝑀 − 𝑇𝑇𝑠𝑠𝑠𝑠𝑠𝑠 = 187℃

which appears to be too low in a physical sense,

b) Applying Berenson’s correlation for water at 7.0 MPa: 𝑇𝑇𝐵𝐵𝑀𝑀 − 𝑇𝑇𝑠𝑠𝑠𝑠𝑠𝑠 = 1042℃

which appears to be too high in a physical sense.

Applying Henry’s correlation for water at 7.0 MPa: 𝑇𝑇𝐻𝐻𝑀𝑀 − 𝑇𝑇𝑠𝑠𝑠𝑠𝑠𝑠 = 1443℃

which appears to be too low in a physical sense

2. a) Applying Berenson’s correlation for sodium at 0.1 MPa: 𝑇𝑇𝐵𝐵𝑀𝑀 − 𝑇𝑇𝑠𝑠𝑠𝑠𝑠𝑠 = 30.6℃

Applying Henry’s correlation for sodium at 0.1 MPa:

𝑇𝑇𝐻𝐻𝑀𝑀 − 𝑇𝑇𝑠𝑠𝑠𝑠𝑠𝑠 = 411℃

Both of which appears to be too low in a physical sense.

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PROBLEM 12.6 QUESTION Analysis of Decay Heat Removal During a Severe Accident (Chapter 3, 4, 12 Section 12.5) Extreme events in which the reactor core melts partially or completely are designated by nuclear engineers as ‘severe accidents’. Consider a severe accident during which the core has completely melted, thus falling to the bottom of the pressure vessel. The situation is illustrated in Figure 12.14. The molten mixture of fuel (UO2), fission products, clad (Zr), control rod material (B4C) and coresupporting structures (steel) is known in severe accident analysis as ‘corium’. In the situation considered here, the corium melt fills the bottom of the vessel up to the junction of the hemispherical lower head with the cylindrical beltline region. There is water above the corium and water outside the vessel. The fuel decay heat is removed by boiling above the corium, and by conduction through the vessel walk The whole system is at atmospheric pressure. Figure 12.14 The lower head of the reactor vessel during a severe accident with complete melting of the core.

FIGURE 12.14

The lower head of the reactor vessel during a severe accident with complete melting of the core.

1. At normal operating conditions the thermal power of this reactor is 3400 MWth. Three hours after reactor shutdown, the corium melt is at 2000ºC, the temperature on the outer surface of the vessel is 112ºC and the temperature of the water above the corium is 100ºC. At this time is the corium heating up, cooling down or staying at steady temperature? (Hint: assume that the temperature distribution within the corium melt is uniform). 2. At a certain time during the accident, the heat flux at the upper surface of the corium melt is 200 kW/m2. Calculate the average void fraction in the (stagnant) water above the corium melt. (Use the drift-flux model with Co = l and the expression of Vvj for churn flow. Assume water at saturated conditions). 3. With regard to Question 2, would an HEM approach be acceptable? Why? In solving Question 1 use the following film boiling heat transfer correlation with convection (by Berenson) and radiation components (see Equation 12.30b). Table 12.5 lists materials properties which should be used in the solution of this problem.

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TABLE 12.5 Materials Properties for Problem 12.6 Parameter Tsat (°C/K) ρf (kg/m3) ρg (kg/m3) hf (kJ/kg) hg (kJ/kg) cpf [kJ/(kg°C)] cpg [kJ/(kg°C)] μf (Pa s) μg (Pa s) kf [W/(m°C)] kg [W/(m°C)] σ (N/m)

Value 100 (373) 960 0.6 419 2675 4.2 2.1 2.8 × 10−4 1.2 × 10−5 0.68 0.02 0.06

Note: Corium—density: 8000 kg/m3; specific heat: 530 J/kg°C; Emissivity: 0.5. Vessel steel—density: 7500 kg/m3; thermal conductivity: 30 W/m°C. Water surrounding vessel—saturated at atmospheric pressure.

Answers: 1. Heating up. 2. α ≈ 0.381 3. HEM is a poor approach since the liquid is nearly stagnant while the vapor is flowing upward.

PROBLEM 12.6 SOLUTION Analysis of Decay Heat Removal During a Severe Accident (Chapter 3, 4, 12 Section 12.5) 1. The energy balance equation for the corium melt is: 𝑚𝑚𝑐𝑐 𝐶𝐶𝑐𝑐

𝑑𝑑𝑑𝑑𝑐𝑐 = 𝑄𝑄̇𝑑𝑑𝑑𝑑𝑑𝑑 − 𝑄𝑄̇𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 − 𝑄𝑄̇𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑑𝑑𝑑𝑑

Assuming the reactor was operating for a long time before shutdown: 𝑄𝑄̇𝑑𝑑𝑑𝑑𝑑𝑑 − 0.066𝑄𝑄̇0 𝑡𝑡 −0.2 ≈ 35.0 MW

Since the thickness-diameter ratio for the lower vessel head is small (<<1), it is approximated as a flat wall with little loss of accuracy: 𝑇𝑇𝑐𝑐 − 𝑇𝑇𝑣𝑣,𝑜𝑜 𝜋𝜋 2 𝑄𝑄̇𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = 𝑘𝑘𝑣𝑣 𝐷𝐷 ≈ 9.3 MW 𝛿𝛿 2

The film boiling coefficient, hFB, is calculated using the Berenson correlation to yield:

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𝜋𝜋 𝑄𝑄̇𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 = ℎ𝐹𝐹𝐹𝐹 (𝑇𝑇𝑐𝑐 − 𝑇𝑇𝑠𝑠𝑠𝑠𝑠𝑠 ) 𝐷𝐷2 ≈ 13.8 MW 4

Calculating the value of the right hand side of the energy balance equation yields: 𝑄𝑄̇𝑑𝑑𝑑𝑑𝑑𝑑 − 𝑄𝑄̇𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 − 𝑄𝑄̇𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = 11.9 MW

which is greater than 0 so the corium is heating up.

2. The void fraction is calculated with the drift-flux model: 𝛼𝛼 =

𝑗𝑗𝑣𝑣 𝐶𝐶0𝑗𝑗 + 𝑉𝑉𝑣𝑣𝑣𝑣

where for churn flow,𝐶𝐶0 = 1 and V𝑣𝑣𝑣𝑣 = 1.53 �

𝜎𝜎𝜎𝜎�𝜌𝜌𝑓𝑓 −𝜌𝜌𝑔𝑔 � 𝜌𝜌𝑓𝑓2

(11.38) �

0.25

≈ 0.24 m/s

Now, in the pool of water above the corium, the liquid is stagnant, while the vapor flows upward (due to buoyancy). Therefore, one has jl = 0 and also j = jv. The vapor superficial velocity, jv, in general can be calculated as xG/ ρg. In this case, x = l (only vapor is flowing so the flow quality is one) and G is equal to the vapor generation rate per unit area of the corium surface. Thus

From the drift-flux model: 𝑉𝑉

𝐺𝐺 =

𝑞𝑞 ″ = 0.09 kg/m2 s ℎ𝑓𝑓𝑓𝑓 α ≈ 0.381

3. HEM �𝑆𝑆 = 𝑉𝑉𝑣𝑣 � = 1 would have clearly been a bad choice because the velocity of the liquid is 𝑙𝑙

zero, while the velocity of the vapor is greater than zero. In fact, in this case the slip ratio S is infinite.

PROBLEM 12.7 QUESTION Void Fraction and Pressure Drop in an Isolation Condenser (Chapters 6, 9, 10, 11 - Section 12.7) A modern BWR uses an isolation condenser to remove the decay heat from the core following a feedwater pump trip. The isolation condenser receives 50 kg/s of saturated dry steam at 280ºC and condenses it completely. The isolation condenser consists of 200 horizontal round tubes of 3-cm inner diameter and 12-m length. The condensing steam flows inside the tubes (see Figure 12.15). The tubes sit in a pool of water at atmospheric pressure. 1. Calculate the isolation condenser heat removal rate. (The properties of saturated water at 280ºC are presented in Table 12.6) 2. Using the simplified Chato correlation (Eq. 12.46), estimate the temperature on the inner surface of the tubes. (Assume an axially uniform heat flux in the tubes)

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3. Sketch the axial profile of the void fraction in the tubes. (Assume linear variation of the quality in the tubes. Use the HEM to calculate the void fraction) 4. Calculate the acceleration, friction, gravity and total pressure drops within the tubes. (Use the HEM approach with fTP = fℓo to calculate the friction pressure drop) 5. How would the acceleration, friction and gravity pressure drops within the tubes change, if the tubes were vertical and the steam flow were downward? (A qualitative answer is acceptable)

FIGURE 12.15

Isolation condenser tubes.

TABLE 12.6 Properties of Saturated Water at 280°C for Use in Problem 12.7 Parameter vf (m3/kg) vg (m3/kg) hf (kJ/kg) hg (kJ/kg) μf (Pa s) μg (Pa s) kf [W/(m°C)] kg [W/(m°C)]

Value 0.0013 0.03 1237 2780 9.8 × 10−5 1.9 × 10−5 0.574 0.061

Answer: 1. Q̇ = 77.15 MWth 2. Tw ≈ 216ºC

4. ∆Pacc ≈ −3594 Pa ∆Pfric ≈ 7060 Pa ∆Ptot ≈ 3466 Pa

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PROBLEM 12.7 SOLUTION Void Fraction and Pressure Drop in an Isolation Condenser (Chapters 6, 9, 10, 11 - Section 12.7) 1) Saturated steam is completely condensed at a rate of 50 kg/s. Thus, the heat removal rate can be calculated from the energy balance as: =77.15 MWth where the subscripts “o” and “i” refer to the tube outlet and inlet, respectively. 2) The average heat flux at the inner surface of the tubes, q″, is: ≈ 341 kW/m2 The inner wall temperature can be readily found from Newton’s law of cooling:

where hD is from the simplified Chato correlation. Solving for Tw, one gets Tw≈216°C. 3) For HEM the void fraction can be calculated as:

where x is the flow quality, assumed to vary linearly with the axial location, z, from 1 (inlet) to 0 (outlet), i.e., x(z)=1-z/L. This equation can be used to sketch α vs. z, as shown in Figure SM-12.1. Void fraction 1

L Figure SM-12.1

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4) The gravity pressure drop is zero, because the tubes are horizontal. Acceleration:  1 1  ∆Pacc = G 2  + − +  ≈ -3,594 Pa  ρo ρi 

where the subscripts “o” and “i” refer to the tube outlet and inlet, respectively. Note that in this case, ρ i+ = ρ g =33.3 kg/m3, ρ o+ = ρ f =769.2 kg/m3, and G=50/200/(π/4×0.032) ≈ 354 kg/m2s. Friction:

1 G2  dP  = f ⋅   TP D 2ρTP  dz  fric

where D=3cm, f TP = f o = 0.184 ≈0.018 ( Re o = 0.2 Re o

(5)

GD ≈108,400). Also, for HEM, µf

1 x 1− x . Thus, Equation 5 can be integrated: = + ρTP ρ g ρf 1

1 L G2  x 1− x  L G 2  x 2 / 2 x − x 2 / 2  1 G2 dz = f o dx f ∆Pfric = ∫ f TP + = + ≈ 7,060 Pa   o ∫   ρ D D D 2 2 2 ρ ρ ρ ρ   TP f  f 0 0 g 0  g L

where again a linear variation of x along the tubes has been assumed (i.e., x(z)=1-z/L). Total: ≈ -3,594+7,060 = 3,466 Pa where ∆Ptot = Pin - Pout 5) The acceleration and friction pressure drops would not change. The gravity pressure drop would be negative because the flow direction is downward.

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Chapter 13 Flow Boiling Contents Problem 13.1 Nucleation in pool and flow boiling ............................................................. 331 Problem 13.2 Factors affecting incipient superheat in a flowing system ........................... 334 Problem 13.3 Heat transfer problems for a BWR channel ................................................. 337 Problem 13.4 Thermal parameters in a heated channel in two-phase flow ........................ 342 Problem 13.5 CHF calculation with the Bowring correlation ............................................ 344 Problem 13.6 Boiling crisis on the vessel outer surface during a severe accident ............. 346 Problem 13.7 Calculation of CPR for a BWR hot channel ................................................. 349 Problem 13.8 Reduction in a PHWR CHF margin upon a local channel power ................ 353

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PROBLEM 13.1 QUESTION Nucleation in Pool and Flow Boiling (Section 13.2) A platinum heat surface has conical cavities of uniform size, R, of 10 microns. 1. If the surface is used to heat water at 1 atm in pool boiling, what is the value of the wall superheat and surface heat flux required to initiate nucleation? 2. If the same surface is now used to heat water at 1 atm in forced circulation, what is the value of the wall superheat required to initiate nucleate boiling? What is the surface heat flux required to initiate nucleation? Answers: 1. ∆Tsat = 3.3 K Rohsenow method: q″ = 4.68 kW/m2 using Csf = 0.013 Stephan and Abdelsalam method: q″ = 2.85 kW/m2 Cooper method: q″ = 2.67 kW/m2 2. ∆Tsat = 6.5 K q″ = 221 kW/m2

PROBLEM 13.1 SOLUTION Nucleation in Pool and Flow Boiling (Section 13.2) All the cavities on the heat surface are assumed to be conical and of uniform size. 𝑟𝑟 = 10μm

The following properties of saturated water at 1 atm pressure (101.325 kPa) may be found using tables or computer programs. – saturation temperature, Tsat = 373 K – surface tension, σ = 0.05892 N/m – density, ρf = 958.4 kg/m3 and ρg = 0.5977 kg/m3 1

– v𝑔𝑔 = 𝑝𝑝 = 0.5977 = 1.673 m3 /kg 𝑔𝑔

– specific enthalpy, h f = 419 kJ/kg and h g = 2676 kJ/kg – h fg = h g = h f = 2257 kJ/kg – specific heat, cPf = 4.216 kJ/kg K – viscosity, μf = 281.8 × 10−6 Pa s

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– conductivity, kf = 0.6791 W/m K 1. If the surface is used to heat water at 1 atmosphere in pool boiling, what is the value of the wall superheat and surface heat flux required to initiate nucleation? The wall superheat required to initiate nucleation in pool boiling may be calculated rearranging Equation 12.7. ∆𝑇𝑇𝑠𝑠𝑠𝑠𝑠𝑠 = 𝑇𝑇𝑤𝑤 − 𝑇𝑇𝑠𝑠𝑠𝑠𝑠𝑠 =

2𝜎𝜎𝑇𝑇𝑠𝑠𝑠𝑠𝑠𝑠 𝑣𝑣𝑔𝑔 2(0)(05892)(373)(1.673) = = 3.259 K = 3.3 K ℎ𝑓𝑓𝑓𝑓 𝑟𝑟 (2257 × 103 )(10−6 )

(1)

Several correlations are available for calculating the surface heat flux required to initiate nucleation. Three different correlations have been used in this solution: Rohsenow, Stephan and Abdelsalam, and Cooper. Rohsenow method Equation 12.18c may be used applying a coefficient Csf = 0.013, which is suitable for a platinum heat surface in water. 1/2 0.33

𝑐𝑐𝑝𝑝𝑝𝑝 ∆𝑇𝑇𝑠𝑠𝑠𝑠𝑠𝑠 𝑞𝑞″ 𝜎𝜎 = 𝐶𝐶𝑠𝑠𝑠𝑠 � � � ℎ𝑓𝑓𝑓𝑓 𝜇𝜇𝑓𝑓 ℎ𝑓𝑓𝑓𝑓 �𝜌𝜌𝑓𝑓 − 𝜌𝜌𝑔𝑔 �𝑔𝑔

�

𝜇𝜇𝑓𝑓 𝑐𝑐𝑝𝑝𝑝𝑝 � � 𝑘𝑘𝑓𝑓

(2)

0.33

1/2 𝑞𝑞″ 4.216(3.259) 0.05892 � � � = 0.013 � 2257 (281.8 × 10−6 )2257 (958.4 − 0.5977)9.81

Solving for q″:

281.8 × 10−6 )(4.216 × 103 ) ×� � 0.6791 (3)

𝑞𝑞 ″ = 4.68 kW⁄m2

Stephan and Abdelsalam method Equation 12.22a may be used: 0.674

𝑞𝑞″𝑑𝑑𝑑𝑑 𝑞𝑞″𝑑𝑑𝑑𝑑 = 0.23 � � ∆𝑇𝑇𝑠𝑠𝑠𝑠𝑠𝑠 𝑘𝑘𝑓𝑓 𝑘𝑘𝑓𝑓 𝑇𝑇𝑠𝑠𝑠𝑠𝑠𝑠

0.297

𝜌𝜌𝑔𝑔 � � 𝜌𝜌𝑓𝑓

ℎ𝑓𝑓𝑓𝑓 𝑑𝑑𝑑𝑑2 � 2 � 𝛼𝛼𝑓𝑓

0.371

𝜌𝜌𝑓𝑓 − 𝜌𝜌𝑔𝑔 � � 𝜌𝜌𝑓𝑓

−1.73

0.35

𝛼𝛼𝑓𝑓2 𝑃𝑃𝑓𝑓 � � 𝜎𝜎. 𝑑𝑑𝑑𝑑

(4)

Where dd is given by Equation 12.22b, assuming a contact angle of θ = 25°: 2𝜎𝜎

1⁄2

𝑑𝑑𝑑𝑑 = (0.146)𝜃𝜃 � � 𝑔𝑔�𝜌𝜌𝑓𝑓 − 𝜌𝜌𝑔𝑔 � = 0.01293 m

⁄

1 2 2(0.05892) = 0.146(25) � � 9.81(958.4 − 0.5977)

(5)

And the thermal diffusivity of the liquid is: 𝛼𝛼𝑓𝑓 =

𝑘𝑘𝑓𝑓 0.6791 m2 −7 = = 1.681 × 10 𝜌𝜌𝑓𝑓 𝑐𝑐𝑝𝑝𝑝𝑝 958.4(4.216 × 103 ) s

(6)

The main equation can now be solved:

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3.259(0.6791) 𝑞𝑞 ″ (0.01293) 𝑞𝑞 = 0.23 � � 0.01293 0.6791(373) ″

0.371

(2257 × 103 )0.012932 � � (1.681 × 10−7 )2

�

0.5977 0.297 � 958.4

0.35

958.4 − 0.5977 −1.73 (1.681 × 10−7 )2 (958.4) � � � � 958.4 0.05892(0.01293) 𝑞𝑞 ″ = 2.85 kW/m2

(7)

Cooper method Applying Equation 12.23: (0.12−0.211 log10 𝜖𝜖)

ℎ = 𝐴𝐴 �𝑃𝑃𝑅𝑅

� [−log10 𝑃𝑃𝑅𝑅 ]−0.55 𝑀𝑀−0.5 𝑞𝑞″2⁄3

(8)

Where M = 18 is the molar mass of water, while the heat transfer coefficient is given by the general relation: ℏ=

We will consider 𝜖𝜖 = 1𝜇𝜇𝜇𝜇 and A = 55. The pressure ratio is given by: 𝑃𝑃𝑅𝑅 =

𝑞𝑞″ ∆𝑇𝑇𝑠𝑠𝑠𝑠𝑠𝑠

(9)

𝑃𝑃 101.325 = = 4.59 × 10−3 𝑃𝑃𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 22060

(10)

Substituting into the previous equation yields: (0.12−0.211log10 ϵ)

𝑞𝑞 ″ = �𝐴𝐴 �𝑃𝑃𝑅𝑅

� [−10log10 𝑃𝑃𝑅𝑅 ]−0.55 𝑀𝑀−0.5 ∆𝑇𝑇𝑠𝑠𝑠𝑠𝑠𝑠 �

𝑞𝑞 ″ = �55[4.59 × 10−3 ](0.12−0.211log101) [−log10 4.59 × 10−3 ]−0.55 18−0.5 3.259� = 2.67 kW⁄m2

(11)

2. If the same surface is now used to heat water at 1 atm in forced circulation, what is the value of the wall superheat required to initiate nucleate boiling? What is the surface heat flux required to initiate nucleation? The condition required for bubble nucleation in flow boiling is given by Equation 13.2: 𝑟𝑟 ∗ = �

2𝜎𝜎𝑇𝑇𝑠𝑠𝑠𝑠𝑠𝑠 𝑣𝑣𝑔𝑔 𝑘𝑘𝑓𝑓 ℎ𝑓𝑓𝑓𝑓 𝑞𝑞″

(12)

We may solve this equation to obtain the desired heat flux value. 10 × 10−6 = �

2(0.05892)(373)(1.673)(0.6791) 2257 × 103 𝑞𝑞″

⇒ 𝑞𝑞 ″ = 221 kW⁄m2

(13)

(14)

In forced convection, the minimum wall superheat required to initiate nucleation can be expressed by Equation 13.5.

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8𝜎𝜎𝑇𝑇𝑠𝑠𝑠𝑠𝑠𝑠 𝑣𝑣𝑔𝑔 𝑞𝑞 ″ ∆𝑇𝑇𝑠𝑠𝑠𝑠𝑠𝑠 = 𝑇𝑇𝑤𝑤 − 𝑇𝑇𝑠𝑠𝑠𝑠𝑠𝑠 = � � ℎ𝑓𝑓𝑓𝑓 𝑘𝑘𝑓𝑓 = 6.5 K

⁄

8(0.05892)(373)(1.673)(221000) 1 2 � =� (2257 × 103 )(0.6791)

(15)

PROBLEM 13.2 QUESTION

Factors Affecting Incipient Superheat in a Flowing System (Section 13.2) 1. Saturated liquid water at atmospheric pressure flows inside a 20 mm diameter tube. The mass velocity is adjusted to produce a single phase heat transfer coefficient equal to 10 kW/m2K, What is the incipient boiling heat flux? What is the corresponding wall superheat? 2. Provide answers to the same questions for saturated liquid water at 290°C, flowing through a tube of the same diameter, and with a mass velocity adjusted to produce the same singlephase heat transfer coefficient. 3. Provide answers to the same questions if the flow rate in the 290°C case is doubled. Answers: 1. q″ = 1.92×104 W/m2 TW – Tsat =1.92 °C 2. q″ = 230 W/m2 TW – Tsat = 0.023 °C 3. q" = 698.5 W/m2 TW – Tsat = 0.04 °C

PROBLEM 13.2 SOLUTION Factors Affecting Incipient Superheat in a Flowing System (Section 13.2) The tube diameter is: 𝐷𝐷 = 0.0220 m The following properties of saturated water at 1 atm pressure (101.325 kPa) may be found using tables or computer programs. – saturation temperature, Tsat = 373 K – surface tension, σ = 0.05892 N/m – specific volume, vg = 1.673 m3/kg

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– specific enthalpy, hf = 419 kJ/kg and hg = 2676 kJ/kg – hfg = hg − hf = 2256 kJ/kg – viscosity, μf = 0.0002818 Pa s – conductivity, kf = 0.6791 W/m K The mass velocity is adjusted to produce a single-phase heat transfer coefficient equal to: ℏ = 10 kW⁄m2 K

1. What is the incipient boiling heat flux? What the corresponding wall superheat? In forced convection, the minimum wall superheat required to initiate nucleation is given by Equation 13.5. 1⁄2

8𝜎𝜎𝑇𝑇𝑠𝑠𝑠𝑠𝑠𝑠 v𝑔𝑔 𝑞𝑞″ (1) ∆𝑇𝑇𝑠𝑠𝑠𝑠𝑠𝑠 = 𝑇𝑇𝑤𝑤 − 𝑇𝑇𝑠𝑠𝑠𝑠𝑠𝑠 = � � ℎ𝑓𝑓𝑓𝑓 𝑘𝑘𝑓𝑓 At the point of boiling inception, the condition of minimum wall superheat has to satisfy also the single-phase heat transfer equation: 𝑞𝑞 ″ = ℏ∆𝑇𝑇𝑠𝑠𝑠𝑠𝑠𝑠

(2)

Notice that per the problem statement at boiling inception the coolant bulk temperature is Tsat. Combining those two equations and solving: 1⁄2

8𝜎𝜎𝑇𝑇𝑠𝑠𝑠𝑠𝑠𝑠 v𝑔𝑔 𝑞𝑞″ 𝑞𝑞″ =� � ℏ ℎ𝑓𝑓𝑓𝑓 𝑘𝑘𝑓𝑓

1⁄2

8(0.05892)(373)(1.673)𝑞𝑞″ 𝑞𝑞″ =� � 3 10 × 10 (2256 × 103 )(0.6791) ⇒ 𝑞𝑞 ″ = 1.92 × 104 W⁄m2

⇒ ∆𝑇𝑇𝑠𝑠𝑠𝑠𝑠𝑠 =

𝑞𝑞″ 1.92 × 104 = = 1.92 °C ℏ 10 × 103

(3)

(4) (5)

2. Provide answers to the same questions for saturated liquid water at 290°C, flowing through a tube of the same diameter, and with a mass velocity adjusted to produce the same single-phase heat transfer coefficient. The following properties of saturated water at 290°C (563.1 K) temperature may be found using tables or computer programs. 𝜎𝜎 = 0.0167 N⁄m

v𝑔𝑔 = 0.02556 m3 ⁄kg ℎ𝑓𝑓 = 1290 kJ⁄kg

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ℎ𝑔𝑔 = 2677 kJ⁄kg

ℎ𝑓𝑓𝑓𝑓 = ℎ𝑔𝑔 − ℎ𝑓𝑓 = 2677 − 1290 = 1477 kJ⁄kg 𝜇𝜇𝑓𝑓 = 8.967 × 10−5 Pa s 𝑘𝑘𝑓𝑓 = 0.565 W⁄m K

Solving the equations already seen in Part 1, and keeping the same heat transfer coefficient: 1⁄2

8𝜎𝜎𝑇𝑇𝑠𝑠𝑠𝑠𝑠𝑠 v𝑔𝑔 𝑞𝑞″ 𝑞𝑞″ =� � ℎ𝑓𝑓𝑓𝑓 𝑘𝑘𝑓𝑓 ℏ

1⁄2

𝑞𝑞″ 8(0.0167)(563.1)(0.02556)𝑞𝑞″ =� � 3 10 × 10 (1477 × 103 )(0.565)

(7)

⇒ 𝑞𝑞 ″ = 230 W⁄m2

⇒ ∆𝑇𝑇𝑠𝑠𝑠𝑠𝑠𝑠 =

(6)

𝑞𝑞″ 230 = = 0.023 °C ℏ 10 × 103

(8)

3. Provide answers to the same questions if the flow rate in the 290°C case is doubled. The single-phase heat transfer coefficient may be expressed using the Dittus-Boelter/Mc Adams correlation, Equation 10.91, as: Nu = 0.023Re0.8 Pr 0.4 0.8

𝐺𝐺𝐺𝐺 ℏ𝐷𝐷 = 0.023 � � 𝑘𝑘𝑓𝑓 𝜇𝜇𝑓𝑓

(9) 0.4

𝑐𝑐𝑝𝑝𝑝𝑝 𝜇𝜇𝑓𝑓 � � 𝑘𝑘𝑓𝑓

(10)

Since the mass flow rate is doubled, G is doubled with respect to part 2 of this problem. All the terms in the equation above are constant except h and G. The new heat transfer coefficient may be calculated as: ℏ = ℏ𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝2 (20.8 ) = (10 × 103 )(20.8 ) = 17.41 × 103 W⁄m2 K

(11)

Updating the equations already seen in part 2 with the new heat transfer coefficient and solving: 1⁄2

8𝜎𝜎𝑇𝑇𝑠𝑠𝑠𝑠𝑠𝑠 𝑣𝑣𝑔𝑔 𝑞𝑞″ 𝑞𝑞″ =� � ℏ ℎ𝑓𝑓𝑓𝑓 𝑘𝑘𝑓𝑓

1⁄2

𝑞𝑞″ 8(0.0167)(563.1)(0.02556)𝑞𝑞″ = � � 17.41 × 103 (1477 × 103 )(0.565) 336

(12)

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⇒ 𝑞𝑞 ″ = 698.5 W⁄m2

⇒ ∆𝑇𝑇𝑠𝑠𝑠𝑠𝑠𝑠 =

𝑞𝑞″ 698.5 = = 0.040 °C ℏ 17.41 × 103

(13) (14)

PROBLEM 13.3 QUESTION Heat Transfer Problems for a BWR Channel (Section 13.3) Consider a heated tube operating at BWR pressure conditions with a cosine heat flux distribution. Relevant conditions are as follows: Channel geometry D = 11.20 mm L = 3.588 m Operating conditions P = 7.14 MPa Tin = 278.3 C G = 1625 kg/m2s q′max= 47.24 kW/m 1. Find the axial position where the equilibrium quality, xe, is zero. 2. What is the axial extent of the channel where the actual quality is zero? That is, this requires finding the axial location of boiling incipience, commonly called ONB, onset of nuclear boiling. (It is sufficient to provide a final equation with all parameters expressed numerically to determine this answer without solving for the final result). 3. Find the axial location of maximum wall temperature assuming the heat transfer coefficient given by the Thom et al. correlation for nuclear boiling heat transfer (Equation 13.23b). Hint! Example 14.5 treats a similar question. 4. Find the axial location of maximum wall temperature assuming the heat transfer coefficient is not constant but varies as is calculated by relevant correlations. Here, you are not asked for the exact location, but whether the location is upstream or downstream from the value from Part 3. Answers: 1. Z OSB = 0.61 m (from the bottom of the channel) 2. Z ONB = 0.21 m 3. 1.79 m 4. Upstream

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PROBLEM 13.3 SOLUTION Heat Transfer Problems for a BWR Channel (Section 13.3) The following parameters are known: – tube diameter, D = 11.2 ×10−3 m – tube length, L = 3.588 m – inlet temperature, Tin = 278.3 °C – pressure, P = 7.14 MPa – mass flux, G = 1625 kg/m2s – peak linear heat generation rate, q′max = 47.24 kW/m The following parameters may be taken from tables or software with water properties. – enthalpy: ℎ𝑓𝑓 = ℎ(𝑃𝑃 = 7.14 MPa, 𝑥𝑥 = 0) = 1275 kJ⁄kg

ℎ𝑔𝑔 = ℎ(𝑃𝑃 = 7.14 MPa, 𝑥𝑥 = 1) = 1771 kJ⁄kg

ℎ𝑓𝑓𝑓𝑓 = ℎ𝑔𝑔 − ℎ𝑓𝑓 = 2771 − 1275 = 2771 − 1275 = 1496 kJ⁄kg ℎ𝑖𝑖𝑖𝑖 = ℎ(𝑇𝑇 = 278.3 °C, 𝑝𝑝 = 7.14 MPa) = 1228 kJ⁄kg

– viscosity, μf = μ(P = 7.14MPa, x = 0) = 90.73 × 10−6 Pa s

– specific heat, cpf = cp (P = 7.14 MPa, x = 0) = 5.431 kJ/kg K – conductivity, kf = k (P = 7.14 MPa, x = 0) = 0.570 W/m K – saturation temperature, Tsat = Tsat (P = 7.14 MPa) = 287.2 °C For a cosine heat flux distribution, the axial peaking factor is equal to π/2. The total channel power is hence: 𝑞𝑞̇ = 𝑞𝑞� ′ 𝐿𝐿 =

𝑞𝑞′𝑚𝑚𝑚𝑚𝑚𝑚 47.24 𝐿𝐿 = 3.588 = 107.9 kW 𝜋𝜋⁄2 𝜋𝜋⁄2

(1)

The inlet equilibrium quality is: 𝑥𝑥𝑒𝑒,𝑖𝑖𝑖𝑖 =

ℎ𝑖𝑖𝑖𝑖 − ℎ𝑓𝑓 1228 − 1275 = = −0.0314 ℎ𝑔𝑔 − ℎ𝑓𝑓 2771 − 1275

(2)

The area of the heated tube is:

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𝜋𝜋𝜋𝜋2 𝜋𝜋(11.2 × 10−3 )2 𝐴𝐴 = = = 9.852 × 10−5 m2 4 4

(3)

The mass flow rate is:

𝑚𝑚̇ = 𝐺𝐺𝐺𝐺 = (1625)(9.852 × 10−5 ) = 0.160 kg⁄s

(4)

The heat flux distribution is:

𝜋𝜋𝜋𝜋 𝑞𝑞′𝑚𝑚𝑚𝑚𝑚𝑚 cos � 𝐿𝐿 � 𝑞𝑞′(𝑧𝑧) (5) 𝑞𝑞 ″(𝑧𝑧) = = 𝜋𝜋𝜋𝜋 𝜋𝜋𝜋𝜋 1. Find the axial position where the equilibrium quality, xe, is zero. Applying the conservation of energy to the heated tube with a cosine heat flux distribution, Equation 14.33b is obtained, which yields the equilibrium quality as a function of z. The equilibrium quality is set to be equal to zero to obtain zOSB. 𝑥𝑥𝑒𝑒 (𝑧𝑧) = 𝑥𝑥𝑒𝑒,𝑖𝑖𝑖𝑖 +

𝑞𝑞̇ 𝜋𝜋𝜋𝜋 �sin � � + 1� = 0 ̇ 𝑓𝑓𝑓𝑓 𝐿𝐿 2𝑚𝑚ℎ

107.9 𝜋𝜋𝜋𝜋𝑂𝑂𝑂𝑂𝑂𝑂 −0.0314 + �sin � � + 1� = 0 2(0.160)(1496) 3.588

(6)

Solving numerically:

𝑧𝑧𝑂𝑂𝑂𝑂𝑂𝑂 = −1.184 m

(7)

Measuring this location from the channel inlet instead of the midplane: 𝑍𝑍𝑂𝑂𝑂𝑂𝑂𝑂 = 𝑧𝑧𝑂𝑂𝑂𝑂𝑂𝑂 +

𝐿𝐿 3.588 = −1.184 + = 0.61 m 2 2

(8)

2. What is the axial extent of the channel where the actual quality is zero? That is, this requires finding the axial location of boiling incipience commonly called ONB, onset of nucleate boiling. (It is sufficient to provide a final equation with all parameters expressed numerically to determine this answer without solving for the final result). The location of axial boiling incipience may be determined as the location where the single-phase heat transfer curve intersects the nucleate boiling curve described by the Thom correlation. The Prandtl number is: Pr =

𝑐𝑐𝑝𝑝𝑝𝑝 𝜇𝜇𝑓𝑓 (5431)(90.73 × 10−6 ) = = 0.864 𝑘𝑘𝑓𝑓 0.570

Re𝑒𝑒𝑒𝑒 =

𝐺𝐺𝐺𝐺 (1625)(11.2 × 10−3 ) = = 2.01 × 105 𝜇𝜇𝑓𝑓 90.73 × 10−6

(9)

The liquid only Reynolds number is:

(10)

The flow is turbulent. The Dittus-Boelter/McAdams relation may be applied (Equation 10.91) to obtain the heat transfer coefficient hDB:

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Nu = 0.023Re0.8 Pr 0.4

Nu = 0.023(2.01 × 105 )0.8 (0.864)0.4 = 378.7 ℏ𝐷𝐷𝐷𝐷 =

𝑘𝑘𝑓𝑓 𝑛𝑛𝑢𝑢 0.570 × 378.7 = = 19.3 kW⁄m2 K 𝐷𝐷 11.2 × 10−3

(11)

(12)

The coolant bulk temperature is given by Equation 14.14: 𝑇𝑇𝑏𝑏 (𝑧𝑧) = 𝑇𝑇𝑖𝑖𝑖𝑖 +

𝑞𝑞′𝑚𝑚𝑚𝑚𝑚𝑚 𝐿𝐿 𝜋𝜋𝜋𝜋 �sin � � + 1� 𝑚𝑚̇𝑐𝑐𝑝𝑝 𝜋𝜋 𝐿𝐿

(13)

The wall temperature according to the Dittus-Boelter/McAdams relation is: 𝑇𝑇𝑤𝑤,𝐷𝐷𝐷𝐷 (𝑧𝑧) = 𝑇𝑇𝑏𝑏 (𝑧𝑧) +

𝑞𝑞″(𝑧𝑧) ℏ𝐷𝐷𝐷𝐷

(14)

𝜋𝜋𝜋𝜋 𝑞𝑞′𝑚𝑚𝑚𝑚𝑚𝑚 cos � 𝐿𝐿 � 𝑞𝑞′𝑚𝑚𝑚𝑚𝑚𝑚 𝐿𝐿 𝜋𝜋𝜋𝜋 𝜋𝜋𝜋𝜋 = 𝑇𝑇𝑖𝑖𝑖𝑖 + �sin � � + 1� + 𝑚𝑚̇𝑐𝑐𝑝𝑝 𝜋𝜋 𝐿𝐿 ℏ𝐷𝐷𝐷𝐷

𝜋𝜋𝜋𝜋 47.24 cos � � 47.24(3.588) 𝜋𝜋𝜋𝜋 3.588 = 278.3 + �sin � � + 1� + 3.588 0.160(5.431)𝜋𝜋 𝜋𝜋(11.2 × 10−3 )(19.3) = 278.3 + 62.1 sin(0.876𝑧𝑧) + 69.6 cos(0.876𝑧𝑧)

The wall temperature according to the Thom correlation is given rearranging Equation 13.23b: 𝑇𝑇𝑤𝑤,𝑇𝑇ℎ𝑜𝑜𝑜𝑜 (𝑧𝑧) = 𝑇𝑇𝑠𝑠𝑠𝑠𝑠𝑠 + 22.7�

𝑞𝑞″𝑀𝑀𝑀𝑀⁄𝑚𝑚2 (𝑧𝑧) exp(2𝑃𝑃⁄8.7)

(15)

� 𝜋𝜋𝜋𝜋 ⃓ (47.24)cos �3.688� ⃓ ⃓ � � ⃓ 1000𝜋𝜋(11.2 × 10−3 ) ⃓ ⃓ = 287.2 + 227⃓ ⃓ ⃓ 7.14 ⃓ exp �2 � 8.7 �� ⎷

The final expression is:

= 287.2 + 11.6�cos(0.876𝑧𝑧)

𝑇𝑇𝑤𝑤,𝐷𝐷𝐷𝐷 (𝑧𝑧𝑂𝑂𝑂𝑂𝑂𝑂 ) = 𝑇𝑇𝑤𝑤,𝑇𝑇ℎ𝑜𝑜𝑜𝑜 (𝑧𝑧𝑂𝑂𝑂𝑂𝑂𝑂 )

(16)

278.3 + 62.1 sin (0.876𝑧𝑧𝑂𝑂𝑂𝑂𝑂𝑂 ) + 69.6 cos(0.876𝑧𝑧𝑂𝑂𝑂𝑂𝑂𝑂 ) = 287.2 + 11.6�cos(0.876𝑧𝑧𝑂𝑂𝑂𝑂𝑂𝑂 ) 340

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Solving numerically: 𝑧𝑧𝑂𝑂𝑂𝑂𝑂𝑂 = −1.58 𝑚𝑚

(17)

Measuring this location from the channel inlet instead of the midplane: 𝐿𝐿 3.588 (18) = −1.58 + = 0.21 m 2 2 3. Find the axial location of maximum wall temperature assuming the heat transfer coefficient given by the Thom et al. correlation for nuclear boiling heat transfer (Equation 13.28b). Equation 13.28b provides the heat flux as: 𝑍𝑍𝑂𝑂𝑂𝑂𝑂𝑂 = 𝑧𝑧𝑂𝑂𝑂𝑂𝑂𝑂 +

𝑞𝑞″𝑀𝑀𝑀𝑀⁄𝑚𝑚2 =

𝑒𝑒𝑒𝑒𝑒𝑒(2𝑃𝑃MPa /8.7) (𝑇𝑇𝑤𝑤 − 𝑇𝑇sat )2 (22.7)2

(19)

The parameters in bold are constant in our model. Therefore, observing the Thom et al. correlation we can notice that the maximum wall temperature occurs at the location of maximum heat flux, which is the core midplane: (20)

𝑧𝑧 = 0.00 m

Measuring this location from the channel inlet instead of the midplane: 𝐿𝐿 3.588 (21) = 0+ = 1.79 m 2 2 4. Find the axial location of maximum wall temperature assuming the heat transfer coefficient is not constant but varies as is calculated by relevant correlations. Here, you are not asked for the exact location, but whether the location is upstream or downstream from the value from Part 3. The solution to this question may be given qualitatively. The general heat transfer equation is: 𝑍𝑍 = 𝑧𝑧 +

𝑞𝑞 ″ = ℏ(𝑇𝑇𝑤𝑤 − 𝑇𝑇𝑏𝑏 )

(22)

In the location of interest the water bulk has an enthalpy comprised between hf and hg. Therefore, its temperature is fixed as Tsat. The wall temperature is hence: 𝑇𝑇𝑤𝑤 = 𝑇𝑇𝑠𝑠𝑠𝑠𝑠𝑠 +

𝑞𝑞″ ℏ

(23)

Observing this equation we notice that the location of maximum wall temperature should have a high heat flux q" and a low heat transfer coefficient ℏ. The heat flux has a cosine axial shape and is maximum at the channel midplane, while the heat transfer coefficient increases as z increases. This occurs because the boiling process and the turbulence caused by boiling enhance the heat transfer. Therefore, if the location of maximum wall temperature is not at z = 0, it must be necessarily where the heat transfer coefficient is lower: upstream from that point.

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PROBLEM 13.4 QUESTION Thermal Parameters in a Heated Channel in Two-Phase Flow (Sections 13.3 and 13.4) Consider a 3 m long water channel of circular cross-sectional area 1.5 × 10−4 m2 operating at the following conditions: 𝑚𝑚̇ = 0.29 kg⁄s 𝑃𝑃 = 7.2 MPa

ℎ𝑖𝑖𝑖𝑖 = saturated liquid 𝑞𝑞 ″ = axially uniform 𝑥𝑥𝑜𝑜𝑜𝑜𝑜𝑜 = 0.15

Compute the following: 1. Fluid temperature

2. Wall temperature using Jens and Lottes correlation 3. Determine CPR using the Groeneveld lookup table Answers: 1. 287.7 °C 2. 294.3 °C 3. CPR= 2.75

PROBLEM 13.4 SOLUTION Thermal Parameters in a Heated Channel in Two-Phase Flow (Sections 13.3 and 13.4) The following parameters are known: – flow area, A = 1.5×10−4 m2 – mass flow rate, ṁ = 0.29 kg/s – channel length, L = 3m – pressure, P = 7.2 MPa – outlet equilibrium quality, xe,out = 0.15 The following parameters may be taken from water properties:

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– enthalpy: ℎ𝑖𝑖𝑖𝑖 = ℎ𝑓𝑓 = ℎ(𝑃𝑃 = 7.2 MPa, 𝑥𝑥 = 0) = 1278 kJ⁄kg ℎ𝑔𝑔 = ℎ(𝑃𝑃 = 7.2 MPa, 𝑥𝑥 = 1) = 2770 kJ⁄kg

– saturation temperature, Tsat = Tsat (P = 7.2 MPa) = 287.7 °C The following parameters are calculated:

– tube diameter, 𝐷𝐷 = 2�𝐴𝐴⁄𝜋𝜋 = 2�1.5 × 10−4⁄𝜋𝜋 = 0.01382 m – mass flux, G = ṁ/A = 0.29/1.5×10−4 = 1933 kg/m2s

The heat flux, which is axially constant, is given by solving the following energy balance (Equation 13.44b): 𝑥𝑥𝑒𝑒,𝑜𝑜𝑜𝑜𝑜𝑜 = 𝑥𝑥𝑒𝑒,𝑖𝑖𝑖𝑖 +

1 4𝐿𝐿𝐿𝐿″ ℎ𝑓𝑓𝑓𝑓 𝐺𝐺𝐺𝐺

1 4(3𝑞𝑞 ″ ) 0.15 + 0 + 2770 − 1278 1933(0.01382)

(1)

⇒ 𝑞𝑞 ″ = 498 kW⁄m2

(2)

𝑇𝑇𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 = 𝑇𝑇𝑠𝑠𝑠𝑠𝑠𝑠 = 287.7 °C

(3)

1. Compute the fluid temperature The fluid is at saturation, so it has the same temperature throughout the channel, which is: 2. Compute the wall temperature using Jens and Lottes correlation A numerical solution of the Jens and Lottes correlation, Equation 13.23a, yields the wall temperature: 𝑞𝑞″ 𝑀𝑀𝑀𝑀 �𝑚𝑚2 =

exp(4𝑃𝑃𝑀𝑀𝑀𝑀𝑀𝑀 /6.2) (𝑇𝑇𝑤𝑤 − 𝑇𝑇𝑠𝑠𝑠𝑠𝑠𝑠 )4 (25)4

498⁄1000 =

exp(4(7.2))⁄6.2) (𝑇𝑇𝑤𝑤 − 287.7)4 (25)4

(4)

⇒ 𝑇𝑇𝑤𝑤 = 294.3 °C

(5)

𝐾𝐾1 = �8⁄𝐷𝐷𝑚𝑚𝑚𝑚 = �0.008⁄0.01382 = 0.761

(6)

3. Determine MCPR using the Groeneveld Lookup Table First of all, we determine the diameter correction factor as given in Table 13.5:

The Groeneveld CHF lookup table provides the critical heat flux requiring three input parameters: quality, mass flux and pressure. In our case, mass flux and pressure are known and constant, while the quality varies with the channel power.

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The critical heat flux decreases with increasing quality. In this problem we are dealing with a constant heat flux distribution, so the location where we expect dryout to occur is the channel outlet. The Critical Power Ratio, which is the ratio between the channel power that leads to critical condition and the operating power, is calculated iteratively, increasing the power until the heat flux matches the CHF. The results of these iterations can be seen in Table SM-13.1, where the nominal power is multiplied by a tentative coefficient called “power ratio”. The exit quality is calculated using Equation 13.44b (see above). The uncorrected CHF is determined using the lookup table at the given exit quality, and then multiplied by K1 to obtain the corrected CHF. Note: See www.CRCPRESS.com/PRODUCT/ISBN/9781433980887 for the interpolation procedure of the Groeneveld lookup table. TABLE SM-13.1 Iterations for CHF calculation Power ratio Heat flux Exit quality Uncorrected CHF Corrected CHF 1 2 2.5 2.8 2.75

498 996 1245 1394 1370

0.15 0.30 0.37 0.42 0.41

3544 2545 2147 1712 1804

2697 1937 1634 1303 1373

The Critical Power Ratio is hence equal to 2.75.

PROBLEM 13.5 QUESTION CHF Calculation with the Bowring Correlation (Section 13.4) Using the data of Example 13.3, calculate the minimum critical heat flux ratio using Bowring correlation. Consider an axially-uniform linear heat generation rate: q′ = 35.86 kW/m. Answer: MCHFR = 2.13

PROBLEM 13.5 SOLUTION CHF Calculation with the Bowring Correlation (Section 13.4) The following parameters are known: – tube diameter, D = 10.0 × 10−3 m

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– pressure, P = 6.89 MPa – tube length, L = 3.66 m – inlet temperature, Tin = 204 °C – mass flux, G = 2000 kg/m2s – linear heat generation rate, q′ = 35.86 kW/m The following specific enthalpy values may be taken from tables or software with water properties: ℎ𝑓𝑓 = ℎ(𝑃𝑃 = 6.89 MPa, 𝑥𝑥 = 0) = 1262 kJ⁄kg ℎ𝑔𝑔 = ℎ(𝑃𝑃 = 6.89 MPa, 𝑥𝑥 = 1) = 2774 kJ⁄kg

ℎ𝑓𝑓𝑓𝑓 = ℎ𝑔𝑔 − ℎ𝑓𝑓 = 2774 − 1262 = 1512 kJ⁄kg

ℎ𝑖𝑖𝑖𝑖 = ℎ(𝑇𝑇 = 204 °C, 𝑃𝑃 = 6.89 MPa) = 872 kJ⁄kg

The area of the heated tube is: 𝐴𝐴 =

The mass flow rate is:

𝜋𝜋𝐷𝐷2 𝜋𝜋(10.0 × 10−3 )2 = = 7.854 × 10−5 m2 4 4

𝑚𝑚̇ = 𝐺𝐺𝐺𝐺 = 2000(7.854 × 10−5 ) = 0.157 kg⁄s

(1)

(2)

The exit enthalpy is:

ℎ𝑜𝑜𝑜𝑜𝑜𝑜 = ℎ𝑖𝑖𝑖𝑖 +

The exit quality is:

𝑥𝑥𝑒𝑒,𝑜𝑜𝑜𝑜𝑜𝑜 =

The operating heat flux is:

𝑞𝑞 ″ =

𝑞𝑞′𝐿𝐿 35.86(3.66) = 872 + = 1708 kJ⁄kg 𝑚𝑚̇ 0.157 ℎ𝑜𝑜𝑜𝑜𝑜𝑜 − ℎ𝑓𝑓 1708 − 1262 = = 0.295 ℎ𝑔𝑔 − ℎ𝑓𝑓 2774 − 1262

𝑞𝑞′ 35.86 = = 1141 kW⁄m2 𝜋𝜋𝜋𝜋 𝜋𝜋10.0 × 10−3

(3)

(4)

(5)

Calculate the minimum critical heat flux ratio using Bowring correlation. The Bowring correlation is given by Equation 13.49a ″ 𝑞𝑞𝑐𝑐𝑐𝑐 =

𝐴𝐴 − 𝐵𝐵ℎ𝑓𝑓𝑓𝑓 𝑥𝑥 𝐶𝐶

(6)

The parameters A, B and C are calculated as follows, using the equations from 13.49b to 13.49h. 𝑃𝑃𝑟𝑟 = 0.145𝑃𝑃 𝑀𝑀𝑀𝑀𝑀𝑀 = 0.145(6.89) = 1.00

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𝑛𝑛 = 2.0 − 0.5𝑃𝑃𝑟𝑟 = 2.0 − 0.5(1.00) = 1.5

(8)

𝐹𝐹1 = 𝐹𝐹2 = 𝐹𝐹3 = 𝐹𝐹4 = 1.00

(9)

For the particular case of Pr = 1, it may be easily verified (Equations 13.49g and 13.49h) that the coefficients F1, F2, F3, F4 are all equal to unity.

𝐴𝐴 =

2.317�ℎ𝑓𝑓𝑓𝑓 𝐷𝐷𝐷𝐷 ⁄4�𝐹𝐹1 2.317(1512(0.01) 2000⁄4) = = 4.54 × 103 ⁄ 1 2 1 + 0.0143𝐹𝐹2 𝐷𝐷 𝐺𝐺 1 + 0.0143√0.01(2000)

𝐶𝐶 =

𝐵𝐵 =

(10) (11)

𝐷𝐷𝐷𝐷 0.01(2000) = = 5.00 4 4

0.077𝐹𝐹3 𝐷𝐷𝐷𝐷 0.077(0.01)2000 𝑛𝑛 = 1.5 = 0.950 𝐺𝐺 2000 1 + 0.347𝐹𝐹4 � � 1 + 0.347 � � 1356 1356 ″ 𝑞𝑞𝑐𝑐𝑐𝑐 =

(12)

𝐴𝐴 − 𝐵𝐵ℎ𝑓𝑓𝑓𝑓 𝑥𝑥 4.54 × 103 − 5.00(1512)𝑥𝑥 = 𝐶𝐶 0.950

(13)

The critical heat flux decreases when quality increases. In this problem we are dealing with a constant heat flux distribution, so the location where we expect the minimum CHFR is the channel outlet. ″ 𝑞𝑞𝑐𝑐𝑐𝑐,𝑜𝑜𝑜𝑜𝑜𝑜 =

𝐴𝐴 − 𝐵𝐵ℎ𝑓𝑓𝑓𝑓 𝑥𝑥𝑒𝑒,𝑜𝑜𝑜𝑜𝑜𝑜 4.54 × 103 − 5.00(1512)0.295 = = 2433 kW⁄m2 𝐶𝐶 0.950

(14)

The MCHFR is given, applying Equation 13.64a, by: MCHFR =

″ 𝑞𝑞𝑐𝑐𝑐𝑐,𝑜𝑜𝑜𝑜𝑜𝑜 2433 = = 2.13 𝑞𝑞″ 1141

(15)

PROBLEM 13.6 QUESTION

Boiling Crisis on the Vessel Outer Surface during a Severe Accident (Section 13.4) Consider water boiling on the outer surface of the vessel where water flows in a hemispherical gap between the surface of the vessel and the vessel insulation (Figure 13.31). The gap thickness is 20 cm. The system is at atmospheric pressure. 1. The water inlet temperature is 80 °C and the flow rate in the gap is 300 kg/s. The heat flux on the outer surface of the vessel is a uniform 350 kW/m2 in the hemispherical region (0 ≤ θ ≤ 90°) and zero in the beltline region (θ > 90°). If a boiling crisis occurred in this system, what type of boiling crisis would it be (DNB or Dryout)? 2. At what angle θ within the channel would you expect the boiling crisis to occur first and why?

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Answers: 1. DNB 2. θ = 90°

FIGURE 13.31 Two-phase flow in the gap between the outer surface of the vessel and the vessel insulation.

PROBLEM 13.6 SOLUTION Boiling Crisis on the Vessel Outer Surface during a Severe Accident (Section 13.4) The system shown in Figure 13.31 is at atmospheric pressure. Some of the known parameters of the problem are written in the figure. The heat flux on the outer surface of the vessel, the mass flow rate and the pressure are, respectively: 𝑞𝑞 ″ = 350 kW/m2 𝑚𝑚̇ = 300 kg/s

𝑃𝑃 = 1 atm = 101.325 kPa

The heated surface can be calculated as follows, considering a hemisphere of radius r = 5.2 m: 𝐴𝐴 =

1 1 (4𝜋𝜋𝑟𝑟 2 ) = (4𝜋𝜋)(5.2⁄2)2 = 42.5 m2 2 2

(1)

The total heat produced by the hemispherical surface is:

𝑞𝑞̇ = 𝑞𝑞 ″ 𝐴𝐴 = 350(42.5) = 149 MW

(2)

1. If a boiling crisis occurred in this system, what type of boiling crisis would it be (DNB or Dryout)? Let us determine the water enthalpy at the inlet and at the outlet of the control volume. The inlet enthalpy can be determined from water properties as: ℎ𝑖𝑖𝑖𝑖 = ℎ(𝑇𝑇 = 80℃, 𝑃𝑃 = 101.325 kPa) = 335 kJ/kg

The outlet enthalpy is given by a simple energy balance:

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𝑞𝑞̇ 14.9 × 103 ℎ𝑜𝑜𝑜𝑜𝑜𝑜 = ℎ𝑖𝑖𝑖𝑖 + = 335 + = 384.7 kJ/kg 𝑚𝑚̇ 300

While the saturated liquid and vapor enthalpies are:

ℎ𝑓𝑓 = ℎ(𝑥𝑥𝑒𝑒 = 0, 𝑃𝑃 = 101.325 kPa) = 419 kJ/kg

ℎ𝑔𝑔 = ℎ(𝑥𝑥𝑒𝑒 = 1, 𝑃𝑃 = 101.325 kPa) = 2506 kJ/kg

The exit equilibrium quality is: 𝑥𝑥𝑜𝑜𝑜𝑜𝑜𝑜 =

ℎ𝑜𝑜𝑜𝑜𝑜𝑜 − ℎ𝑓𝑓 385 − 419 = = −0.016 (i. e. below saturation condition) ℎ𝑔𝑔 − ℎ𝑓𝑓 2506 − 419

(3)

The DNB is a boiling crisis observed at low-quality or even subcooled conditions, while film Dryout is observed at moderately high quality. Since our system has a maximum equilibrium quality of just -1.6% (and an outle temperature of 91.8ºC), the boiling crisis type which is expected to occur is DNB. To find the coolant temperature of 91.8 degrees C where cP,f = 4.2 kJ/ºC, solve the following equation

Tout =+ Tin

q 14.9 × 103 =+ 80 = 91.8º C  P, f mc 300 ( 4.2 )

2. At what angle θ within the channel would you expect the boiling crisis to occur first and why? Let us analyze the trends of the coolant and heat transfer parameters in the control volume. – Pressure: Decreases from inlet to outlet due to friction, gravity and acceleration. – Equilibrium quality: increases from inlet to outlet. – Mass flux: Decreases from inlet to outlet, as the flow area increases. – Heat flux: Constant. Therefore, at the exit of the control volume the coolant has the lowest pressure and mass flux and the highest equilibrium quality. We may notice observing the Groeneveld LUTs (or using other critical condition prediction methods) that at, a pressure around 1 atm and equilibrium quality lower than 0.1, the critical heat flux: – Decreases when equilibrium quality increases. – Decreases when mass flux decreases. – Decreases when pressure decreases. According to the trend of those three parameters, the lowest critical heat flux occurs at the outlet. Since the heat flux is constant on the whole surface, the location with the lowest CHFR is the outlet. Therefore, we expect that DNB could occur first at an angle: 𝜃𝜃 = 90∗ 348

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PROBLEM 13.7 QUESTION Calculation of CPR for a BWR Hot Channel (Section 13.4) Consider a BWR channel operating at 100% power at the conditions noted below. Using the Hench–Gillis correlation (Equation 13.57a) determine the CPR at 100% power. Operating conditions

𝑧𝑧 1 𝜋𝜋𝜋𝜋 ′ 𝑞𝑞 ′ (𝑧𝑧) = 𝑞𝑞𝑟𝑟𝑟𝑟𝑟𝑟 exp �−𝛼𝛼 � + �� cos � � 𝐿𝐿 2 𝐿𝐿

where z = 0 defines the channel midplane.

′ 𝑞𝑞𝑟𝑟𝑟𝑟𝑟𝑟 = 104.75 kW/m

𝛼𝛼 = 1.96

𝐺𝐺 = 1569.5 kg/m2 s 𝑃𝑃 = 7.14 MPa 𝑇𝑇𝑖𝑖𝑖𝑖 = 278.3℃

Channel conditions

𝐿𝐿 = 3.588 m

𝐷𝐷 = 0.0112 m

𝐴𝐴𝑓𝑓𝑓𝑓ℎ = 1.42 × 10−4 m2 𝐴𝐴𝑓𝑓 = 9.718 × 10−3 m2

Hench–Gillis correlation for simplicity assume, from Example 13.5:

𝐽𝐽 = 1.032

𝐹𝐹𝑝𝑝 = −1.66 × 10−3

Answer: CPR = 1.17

PROBLEM 13.7 SOLUTION Calculation of CPR for a BWR Hot Channel (Section 13.4) The following parameters are given: – reference linear heat generation rate, q′ref = 104.75 kW/m

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– axial heat distribution coefficient, α = 1.96 – inlet temperature, Tin = 278.3 °C – pressure, P = 7.14 MPa – mass flux, G = 1569.5 kg/m2s – heated length, L = 3.588 m – rod diameter, D = 0.0112 m – subchannel flow area, Afch = 1.42 × 10−4 m2 – bundle flow area, Af = 9.718 × 10−3 m2 The linear heat distribution is given by the following equation: 𝑧𝑧 1 𝜋𝜋𝜋𝜋 ′ exp �−𝛼𝛼 � + �� cos � � 𝑞𝑞 ′(𝑧𝑧) = 𝑞𝑞𝑟𝑟𝑟𝑟𝑟𝑟 𝐿𝐿 2 𝐿𝐿

(1)

The following parameters may be taken from tables or software with water properties: ℎ𝑓𝑓 = ℎ(𝑃𝑃 = 7.14 MPa, 𝑥𝑥 = 0) = 1275 kJ/kg

ℎ𝑔𝑔 = ℎ(𝑃𝑃 = 7.14 MPa, 𝑥𝑥 = 1) = 2771 kJ/kg

ℎ𝑓𝑓𝑓𝑓 = ℎ𝑔𝑔 − ℎ𝑓𝑓 = 2771 − 1275 = 2771 − 1275 = 1496 kJ/kg The inlet quality is:

ℎ𝑖𝑖𝑖𝑖 = ℎ(𝑇𝑇 = 278.3℃, 𝑃𝑃 = 7.14 MPa) = 1228 kJ/kg

𝑥𝑥𝑒𝑒,𝑖𝑖𝑖𝑖 =

ℎ𝑖𝑖𝑖𝑖 − ℎ𝑓𝑓 1228 − 1275 = = −0.0314 ℎ𝑔𝑔 − ℎ𝑓𝑓 2771 − 1275

(2)

The mass flow rate in the subchannel of interest is:

𝑚𝑚̇ = 𝐺𝐺(𝐴𝐴𝑓𝑓𝑓𝑓ℎ ) = 1569.5(1.42 × 10−4 ) = 0.223 kg/s

(3)

Calculation of zOSB The equilibrium quality distribution is given by Equation 14.33a as: 𝐿𝐿

1 𝑥𝑥𝑒𝑒 (𝑧𝑧) = 𝑥𝑥𝑒𝑒,𝑖𝑖𝑖𝑖 + � 𝑞𝑞′(𝑧𝑧)𝑑𝑑𝑑𝑑 𝑚𝑚̇ℎ𝑓𝑓𝑓𝑓

(4)

−𝐿𝐿⁄2

Setting the equilibrium quality equal to zero and integrating numerically, the OSB location zOSB can be found. 𝑥𝑥𝑒𝑒 (𝑧𝑧𝑂𝑂𝑂𝑂𝑂𝑂 ) = 0

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1 = 0.0314 + 0.223(1496)

𝑧𝑧𝑂𝑂𝑂𝑂𝑂𝑂

�

−3.588⁄2

104.75 exp �−1.96 � ⇒ 𝑧𝑧𝑂𝑂𝑂𝑂𝑂𝑂 = −1.26 m

𝑧𝑧 1 𝜋𝜋𝜋𝜋 + �� cos � � 𝑑𝑑𝑑𝑑 = 0 3.588 2 3.588

(6)

Equilibrium quality profile Applying the same equation, the equilibrium quality profile is: 1 𝑥𝑥𝑒𝑒 (𝑧𝑧) = −0.0314 + 0.223(1496)

𝑧𝑧

�

−3.588⁄2

104.75

𝑧𝑧 1 𝜋𝜋𝜋𝜋 × exp �−1.96 � + �� cos � � 𝑑𝑑𝑑𝑑 3.588 2 3.588

(7)

Hench-Gillis correlation The Hench-Gillis correlation is given by Equation 13.57a: 𝑥𝑥𝑐𝑐 =

𝐴𝐴𝐴𝐴 (2 − 𝐽𝐽) + 𝐹𝐹𝑝𝑝 𝐵𝐵 + 𝑍𝑍

(8)

To use the Hench Gillis correlation, we have to convert the mass flux into British units.

Where:

𝐺𝐺 = 1569.5 kg⁄m2 s ⇒ GBrit = 1.157 Mlbm ⁄hft 2

(9)

−0.43 𝐴𝐴 = 0.50𝐺𝐺𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵 = 0.5(1.157−0.43 ) = 0.470

(10)

𝐽𝐽 = 1.032

(12)

2.3 𝐵𝐵 = 165 + 115𝐺𝐺𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵 = 165 + 115(1.1572.3 ) = 326

(11)

𝐹𝐹𝑝𝑝 = −1.66 × 10−3

(13)

𝑍𝑍 =

𝜋𝜋𝜋𝜋𝜋𝜋𝐿𝐿𝐵𝐵 𝜋𝜋(0.0012)74𝐿𝐿𝐵𝐵 = = 267.9𝐿𝐿𝐵𝐵 𝐴𝐴𝑓𝑓 9.718 × 10−3

(14)

Substituting all those values in Equation 13.57a:

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𝑥𝑥𝑐𝑐 =

0.470(267.9𝐿𝐿𝐵𝐵 ) (2 − 1.032) − 1.66 × 10−3 326 + 267.9𝐿𝐿𝐵𝐵 =

(15)

121.9𝐿𝐿𝐵𝐵 − 1.66 × 10−3 326 + 267.9𝐿𝐿𝐵𝐵

The boiling length LB is defined as the distance from the OSB location: 𝐿𝐿𝐵𝐵 = 𝑧𝑧 − 𝑧𝑧𝑂𝑂𝑂𝑂𝑂𝑂 = 𝑧𝑧 − (−1.26 m) = z + 1.26m

(16)

𝐿𝐿𝐵𝐵 = 𝑧𝑧 − 𝑧𝑧𝑂𝑂𝑂𝑂𝑂𝑂

(17)

Determination of the CPR The critical power ratio can be determined graphically, by plotting the distributions of the Hench-Gillis critical quality x c (z) and of the equilibrium quality x e (z) while increasing the channel power. We will plot equilibrium quality and critical quality as a function of the boiling length L B , which is defined as the distance from the OSB location:

Notice that the OSB location is equal to -1.26 m, as calculated above, only for nominal power. When the power increases, the OSB location must be re-calculated. As the following figure shows, the CPR is 1.17.

FIGURE SM-13.1 Axial distribution of equilibrium quality and critical quality as a function of the boiling length

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PROBLEM 13.8 QUESTION Reduction in a PHWR CHF Margin upon a Local Channel Power Increase due to Increased Two Phase Pressure Drop and Corresponding Reduction in Channel Mass Flux Consider a horizontal flow channel of typical PHWR geometry operating at typical PHWR full power conditions of 2700 𝑀𝑀𝑊𝑊𝑡𝑡ℎ with 480 channels. The relevant geometric and steady state channel operating conditions are given in Table 13.7 below. In this example we will postulate a local increase in power and assess the reduction in CHF margin. Since there are more than hundred channels connected to the same inlet and outlet manifolds/headers (boundary conditions), a heat up in one or a few channels would not affect the boundary conditions (i.e. the inlet and outlet pressure conditions) noticeably. Local heat up / power increase could be due to movement of local reactivity devices, or (abnormal) online fueling of a channel. (Normally the on line fueling causes about 10% increase for central channels and as much as 15% for peripheral channels). Using the Groeneveld 2006 CHF lookup table [34] determine the following quantities: a) The margin to CHF at steady state full power channel conditions of 6 MW b) The changed margin to CHF upon an increase in channel power by 15% Assume: a) The heat flux is uniform along the channel b) Treat the friction factor in the subcooled region of the channel as single phase coolant c) Use heavy water properties d) Homogeneous Equilibrium Model (HEM) TABLE 13.7 PHWR Flow Channel Geometry and Operating Conditions at Full Power Parameter Inlet pressure Outlet pressure Inlet temperature Channel power Channel internal diameter Channel length Number of bundle pins in the channel Pin outside diameter

Symbol 𝑃𝑃𝑖𝑖𝑖𝑖 𝑃𝑃𝑜𝑜𝑜𝑜𝑜𝑜 𝑇𝑇𝑖𝑖𝑖𝑖 𝑞𝑞̇ 𝐷𝐷1 𝐿𝐿 𝑛𝑛 𝐷𝐷2

353

Value 11.0 MPa 10.0 MPa 267 ℃ 6 MW 0.1 m 6m 37 0.013 m

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Chapter 13 - Flow Boiling

Pressure loss coefficients Inlet feeder and end-fitting Outlet feeder and end-fitting Fuel channel minor losses Fuel channel skin friction

𝐾𝐾𝑖𝑖𝑖𝑖 𝐾𝐾𝑜𝑜𝑜𝑜𝑜𝑜 𝐾𝐾𝑚𝑚𝑚𝑚𝑚𝑚 𝐾𝐾𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓

4.609 3.770 6.396 7.755

Answer: a) Steady State: MCHFR = 5.82 b) Transient (channel power increase by 15%): MCHFR = 3.17

PROBLEM 13.8 SOLUTION Reduction in a PHWR CHF Margin upon a Local Channel Power Increase due to Increased Two Phase Pressure Drop and Corresponding Reduction in Channel Mass Flux 1. Steady state In steady state, the mass conservation equation (Equation 5.63) applied to the heated channel reduces to: 𝑚𝑚̇𝑖𝑖𝑖𝑖 = 𝑚𝑚̇𝑜𝑜𝑜𝑜𝑜𝑜 = 𝑚𝑚̇𝑠𝑠𝑠𝑠

(1)

𝑞𝑞̇ 𝑠𝑠𝑠𝑠 = 𝑚𝑚̇𝑠𝑠𝑠𝑠 (ℎ𝑠𝑠𝑠𝑠,𝑜𝑜𝑜𝑜𝑜𝑜 − ℎ𝑠𝑠𝑠𝑠,𝑖𝑖𝑖𝑖 )

(2)

where “ss” indicates the steady state. In the energy conservation equation (Equation 5.159), the time-dependent terms are zero because it is a steady state problem, the potential energy terms cancel out because the channel is horizontal, we neglect the kinetic terms, we neglect the shear term and the heat source term is zero. After integration we obtain:

where 𝑞𝑞̇ 𝑠𝑠𝑠𝑠 is given in Table 13.7, ℎ𝑠𝑠𝑠𝑠,𝑖𝑖𝑖𝑖 can be calculated from 𝑃𝑃𝑠𝑠𝑠𝑠,𝑖𝑖𝑖𝑖 and 𝑇𝑇𝑠𝑠𝑠𝑠,𝑖𝑖𝑖𝑖 , and ℎ𝑠𝑠𝑠𝑠,𝑜𝑜𝑜𝑜𝑜𝑜 can be expressed as a function of outlet quality 𝑥𝑥𝑠𝑠𝑠𝑠,𝑜𝑜𝑜𝑜𝑜𝑜 and saturation enthalpies at the outlet conditions, using heavy water properties: 𝑘𝑘𝑘𝑘

ℎ𝑠𝑠𝑠𝑠,𝑖𝑖𝑖𝑖 = ℎ�𝑃𝑃𝑠𝑠𝑠𝑠,𝑖𝑖𝑖𝑖 , 𝑇𝑇𝑠𝑠𝑠𝑠,𝑖𝑖𝑖𝑖 � = ℎ(11 𝑀𝑀𝑀𝑀𝑀𝑀, 267 ℃) = 1127.25 𝑘𝑘𝑘𝑘 ℎ𝑠𝑠𝑠𝑠,𝑜𝑜𝑜𝑜𝑜𝑜 = 𝑥𝑥𝑠𝑠𝑠𝑠,𝑜𝑜𝑜𝑜𝑜𝑜 ℎ𝑔𝑔,𝑠𝑠𝑠𝑠,𝑜𝑜𝑜𝑜𝑜𝑜 + �1 − 𝑥𝑥𝑠𝑠𝑠𝑠,𝑜𝑜𝑜𝑜𝑜𝑜 �ℎ𝑓𝑓,s𝑠𝑠,𝑜𝑜𝑜𝑜𝑜𝑜 𝑘𝑘𝑘𝑘

(3)

𝑘𝑘𝑘𝑘

ℎ𝑔𝑔,𝑠𝑠𝑠𝑠,𝑜𝑜𝑜𝑜𝑜𝑜 = 2529 𝑘𝑘𝑘𝑘 and ℎ𝑓𝑓,𝑠𝑠𝑠𝑠,𝑜𝑜𝑜𝑜𝑜𝑜 = 1350.14 𝑘𝑘𝑘𝑘

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Chapter 13 - Flow Boiling

The inlet quality can be calculated from the saturation enthalpies at the inlet conditions. At the inlet the fluid is sub-cooled since at the inlet pressure 11 𝑀𝑀𝑀𝑀𝑀𝑀 the saturation temperature is 317.06 ℃. kJ

Since ℎ𝑔𝑔,𝑠𝑠𝑠𝑠,𝑖𝑖𝑖𝑖 = 2511.8 kg and ℎ𝑓𝑓,𝑠𝑠𝑠𝑠,𝑖𝑖𝑖𝑖 = 1389.95 kg , then: 𝑥𝑥𝑠𝑠𝑠𝑠,𝑖𝑖𝑖𝑖 =

ℎ𝑠𝑠𝑠𝑠,𝑖𝑖𝑖𝑖 − ℎ𝑓𝑓,𝑠𝑠𝑠𝑠,𝑖𝑖𝑖𝑖 1127.25 − 1389.95 = = −0.234 ℎ𝑔𝑔,𝑠𝑠𝑠𝑠,𝑖𝑖𝑖𝑖 − ℎ𝑓𝑓,𝑠𝑠𝑠𝑠,𝑖𝑖𝑖𝑖 2511.8 − 1389.95

To calculate the coordinate at which the fluid transitions to 2-phase conditions, we start by guessing a value of 𝑚𝑚̇𝑠𝑠𝑠𝑠 , then we calculate the corresponding ℎ𝑠𝑠𝑠𝑠,𝑜𝑜𝑜𝑜𝑜𝑜 and 𝑥𝑥𝑠𝑠𝑠𝑠,𝑜𝑜𝑜𝑜𝑜𝑜 from Equation (2) and Equation (3). 𝑘𝑘𝑘𝑘

For 𝑚𝑚̇𝑠𝑠𝑠𝑠 = 24 𝑠𝑠 (𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 𝑡𝑡𝑡𝑡 𝑏𝑏𝑏𝑏 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑤𝑤𝑤𝑤𝑤𝑤ℎ 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠), obtain: 𝑘𝑘𝑘𝑘

ℎ𝑠𝑠𝑠𝑠,𝑜𝑜𝑜𝑜𝑜𝑜 = 1377.25 𝑘𝑘𝑘𝑘 and 𝑥𝑥𝑠𝑠𝑠𝑠,𝑜𝑜𝑜𝑜𝑜𝑜 = 0.023

Denoting 𝐿𝐿 the channel length and 𝑎𝑎 the coordinate at which the fluid transitions from 1-phase to 2-phase conditions, and assuming that the fluid’s quality grows linearly in the channel (which is reasonable, as the heat flux is constant), then: 𝑎𝑎 = 𝐿𝐿 �𝑥𝑥

−𝑥𝑥𝑠𝑠𝑠𝑠,𝑖𝑖𝑖𝑖

𝑠𝑠𝑠𝑠,𝑜𝑜𝑜𝑜𝑜𝑜 −𝑥𝑥𝑠𝑠𝑠𝑠,𝑖𝑖𝑖𝑖

� = 6�

−(−0.234)

(4)

� = 5.463 m

0.023−(−0.234)

Hydraulic loss coefficients provided in Table 13.7 are derived as follows: 𝐾𝐾𝑚𝑚𝑚𝑚𝑚𝑚 – There are 12 fuel bundles, each with two separate minor losses. One at mid-length where the spacers are located, and one at bundle beginning/end where the fuel elements are attached to the bundle end plates. The 𝐾𝐾𝑚𝑚𝑚𝑚𝑚𝑚 provided in Table 13.7 is the sum of all minor losses in the 6 𝑚𝑚 long channel.

𝐾𝐾𝑖𝑖𝑖𝑖 and 𝐾𝐾𝑜𝑜𝑜𝑜𝑜𝑜 − Provided to capture all hydraulic losses (friction and minor losses) in the inlet feeder pipe (between inlet header and beginning of heated section), and 𝐾𝐾𝑜𝑜𝑜𝑜𝑜𝑜 in the outlet feeder (between the exit of heated channel and the outlet header). 𝐾𝐾𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 – The fuel channel skin friction loss is given as 𝐾𝐾𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 in Table 13.7.

In the heated channel, to calculate the pressure drop due to friction and minor losses, minor losses (𝐾𝐾𝑚𝑚𝑚𝑚𝑚𝑚 ) in the heated length of the channel can be treated continuously for simplicity. Then, treating the single and two-phase regions separately: ∆𝑃𝑃𝑐𝑐ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 = ∆𝑃𝑃𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 + ∆𝑃𝑃𝑚𝑚𝑚𝑚𝑚𝑚 = (𝐾𝐾𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 + 𝐾𝐾𝑚𝑚𝑚𝑚𝑚𝑚 ) 2𝜌𝜌

355

2 𝑚𝑚̇𝑠𝑠𝑠𝑠

𝑓𝑓,𝑎𝑎𝑎𝑎𝑎𝑎

𝐴𝐴2

𝑎𝑎

𝐿𝐿−𝑎𝑎

2 �𝐿𝐿 + � 𝐿𝐿 � 𝜙𝜙𝐿𝐿𝐿𝐿 �

(5)

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Chapter 13 - Flow Boiling

where we used the liquid saturation density at an intermediate pressure (10.5 MPa): and:

𝜌𝜌𝑓𝑓,𝑎𝑎𝑎𝑎𝑎𝑎 = 750

𝜋𝜋

kg m3

𝐴𝐴 = 4 (𝐷𝐷12 − 37𝐷𝐷22 ) = 2.94 × 10−3 m2

(6)

For the HEM model, the 2-phase flow multiplier is expressed as: 𝑓𝑓2𝑝𝑝 𝜌𝜌𝑓𝑓

2 𝜙𝜙𝐿𝐿𝐿𝐿 = 𝑓𝑓

We make the following simplifying assumptions: • •

𝑓𝑓2𝑝𝑝 = 𝑓𝑓𝐿𝐿𝐿𝐿 𝜌𝜌𝑓𝑓

𝜌𝜌𝑚𝑚

(7)

𝐿𝐿𝐿𝐿 𝜌𝜌𝑚𝑚

𝜌𝜌

= 1 + 𝑥𝑥𝑎𝑎𝑎𝑎𝑎𝑎 �𝜌𝜌𝑓𝑓 − 1�

(11.82)

𝑔𝑔

In Equation 8 we assume that 𝑥𝑥𝑎𝑎𝑎𝑎𝑎𝑎 =

𝑥𝑥𝑠𝑠𝑠𝑠,𝑜𝑜𝑜𝑜𝑜𝑜 2

is an average quality value in the 2-phase zone, and

𝜌𝜌𝑓𝑓 and 𝜌𝜌𝑔𝑔 are calculated at the outlet conditions. Finally, Equation 5 becomes:

∆𝑃𝑃𝑐𝑐ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 = ∆𝑃𝑃𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 + ∆𝑃𝑃𝑚𝑚𝑚𝑚𝑚𝑚 = (𝐾𝐾𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 + 𝐾𝐾𝑚𝑚𝑚𝑚𝑚𝑚 ) 2𝜌𝜌

2 𝑚𝑚̇𝑠𝑠𝑠𝑠

𝑎𝑎

𝑥𝑥

𝐿𝐿−𝑎𝑎

𝜌𝜌

𝐴𝐴2

� 𝑓𝑓,𝑜𝑜𝑜𝑜𝑜𝑜 − 1�� (8) �𝐿𝐿 + � 𝐿𝐿 � �1 + 𝑠𝑠𝑠𝑠,𝑜𝑜𝑜𝑜𝑜𝑜 2 𝜌𝜌

𝑚𝑚̇2

(9)

𝑓𝑓,𝑎𝑎𝑎𝑎𝑎𝑎

The expressions of inlet and outlet pressure drop terms are: ∆𝑃𝑃𝑖𝑖𝑖𝑖 = 𝐾𝐾𝑖𝑖𝑖𝑖 2𝜌𝜌 𝑠𝑠𝑠𝑠𝐴𝐴2

𝑔𝑔,𝑜𝑜𝑜𝑜𝑜𝑜

𝑓𝑓,𝑖𝑖𝑖𝑖

∆𝑃𝑃𝑜𝑜𝑜𝑜𝑜𝑜 = 𝐾𝐾𝑖𝑖𝑖𝑖 2𝜌𝜌

2 𝑚𝑚̇𝑠𝑠𝑠𝑠

𝑓𝑓,𝑜𝑜𝑜𝑜𝑜𝑜

𝐴𝐴2

2 𝜙𝜙𝐿𝐿𝐿𝐿 = 𝐾𝐾𝑖𝑖𝑖𝑖 2𝜌𝜌

The acceleration pressure drop term is:

2 𝑚𝑚̇𝑠𝑠𝑠𝑠

𝑓𝑓,𝑜𝑜𝑜𝑜𝑜𝑜

𝑚𝑚̇2

∆𝑃𝑃𝑎𝑎𝑎𝑎𝑎𝑎 = � 𝐴𝐴𝑠𝑠𝑠𝑠 � �𝜌𝜌+

𝑚𝑚,𝑜𝑜𝑜𝑜𝑜𝑜

We make the following assumptions: • •

+ 𝜌𝜌𝑚𝑚,𝑖𝑖𝑖𝑖 = 𝜌𝜌𝑓𝑓,𝑖𝑖𝑖𝑖

−1 𝑥𝑥𝑠𝑠𝑠𝑠,𝑜𝑜𝑜𝑜𝑜𝑜 1−𝑥𝑥𝑠𝑠𝑠𝑠,𝑜𝑜𝑜𝑜𝑜𝑜 + � In HEM: 𝜌𝜌𝑚𝑚,𝑜𝑜𝑜𝑜𝑜𝑜 = � 𝜌𝜌 + 𝜌𝜌 𝑔𝑔,𝑜𝑜𝑜𝑜𝑜𝑜

𝐴𝐴2

𝑓𝑓,𝑜𝑜𝑜𝑜𝑜𝑜

𝜌𝜌

�1 + 𝑥𝑥𝑠𝑠𝑠𝑠,𝑜𝑜𝑜𝑜𝑜𝑜 �𝜌𝜌𝑓𝑓,𝑜𝑜𝑜𝑜𝑜𝑜 − 1�� 𝑔𝑔,𝑜𝑜𝑜𝑜𝑜𝑜

− 𝜌𝜌+ � 𝑚𝑚,𝑖𝑖𝑖𝑖

(10)

(11.59a)

(11.73)

Finally, Equation (11.59a) becomes:

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Chapter 13 - Flow Boiling 𝑚𝑚̇2

The total pressure loss is:

𝑥𝑥

∆𝑃𝑃𝑎𝑎𝑎𝑎𝑎𝑎 = � 𝐴𝐴𝑠𝑠𝑠𝑠 � �� 𝜌𝜌𝑠𝑠𝑠𝑠,𝑜𝑜𝑜𝑜𝑜𝑜 + 𝑔𝑔,𝑜𝑜𝑜𝑜𝑜𝑜

1−𝑥𝑥𝑠𝑠𝑠𝑠,𝑜𝑜𝑜𝑜𝑜𝑜 𝜌𝜌𝑓𝑓,𝑜𝑜𝑜𝑜𝑜𝑜

�−

𝜌𝜌𝑓𝑓,𝑖𝑖𝑖𝑖

(11)

�

(12)

∆𝑃𝑃𝑡𝑡𝑡𝑡𝑡𝑡 = 1𝑀𝑀𝑀𝑀𝑀𝑀 = ∆𝑃𝑃𝑖𝑖𝑖𝑖 + ∆𝑃𝑃𝑐𝑐ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 + ∆𝑃𝑃𝑜𝑜𝑜𝑜𝑜𝑜 + ∆𝑃𝑃𝑎𝑎𝑎𝑎𝑎𝑎

The only unknowns in Equation (11) and Equation (12) are 𝑚𝑚̇𝑠𝑠𝑠𝑠 and 𝑥𝑥𝑠𝑠𝑠𝑠,𝑜𝑜𝑜𝑜𝑜𝑜 . We can solve the system of equations iteratively, guessing a mass flow rate 𝑚𝑚̇𝑠𝑠𝑠𝑠 , then calculating the corresponding 𝑥𝑥𝑠𝑠𝑠𝑠,𝑜𝑜𝑜𝑜𝑜𝑜 until these values are consistent. The solution is: 𝑚𝑚̇𝑠𝑠𝑠𝑠 = 22.8

kg s

𝑥𝑥𝑠𝑠𝑠𝑠,𝑜𝑜𝑜𝑜𝑜𝑜 = 0.03

The corresponding steady state mass flux is: 𝑚𝑚̇

(13)

𝐺𝐺𝑠𝑠𝑠𝑠 = 𝐴𝐴𝑠𝑠𝑠𝑠 = 7755 m2 s

The Groeneveld lookup table [34] provide the values of Critical Heat Flux (CHF), as function of mass flux 𝐺𝐺, fluid pressure 𝑃𝑃 and thermodynamic quality 𝑥𝑥. In case of constant heat rate profile, the minimum departure from nucleate boiling ratio is at the channel exit. Thus, we use the tables to find CHF𝑠𝑠𝑠𝑠,𝑜𝑜𝑜𝑜𝑜𝑜 : CHF𝑠𝑠𝑠𝑠,𝑜𝑜𝑜𝑜𝑜𝑜 = CHF�𝐺𝐺𝑠𝑠𝑠𝑠 , 𝑃𝑃𝑜𝑜𝑜𝑜𝑜𝑜 , 𝑥𝑥𝑠𝑠𝑠𝑠,𝑜𝑜𝑜𝑜𝑜𝑜 � = CHF �7755

The steady state heat flux is:

′′ 𝑞𝑞𝑠𝑠𝑠𝑠 = 𝐴𝐴

𝑞𝑞̇ 𝑠𝑠𝑠𝑠

𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙

𝑞𝑞̇

kg kW , 10.0 MPa , 0.03� = 3849.4 2 2 m s m

6 MW

𝑠𝑠𝑠𝑠 = 37𝜋𝜋𝐷𝐷 = 37𝜋𝜋(0.013m×6m) = 661.77 m2 𝐿𝐿 2

(14)

The Minimum CHF Ratio (MCHFR) is then:

2. Transient

MCHFR = min(CHFR) =

CHF𝑠𝑠𝑠𝑠,𝑜𝑜𝑜𝑜𝑜𝑜 ′′ 𝑞𝑞𝑠𝑠𝑠𝑠

3849.4

= 661.77 = 5.82

(15)

The transient consists of a local heat rate increase of 15%, due potentially to at-power fueling of the fuel channel, which brings the system to a new equilibrium state. 𝑞𝑞̇ 𝑡𝑡 = 1.15𝑞𝑞̇ 𝑠𝑠𝑠𝑠 = 6.9 MW

(16)

The boundary conditions are pressure and temperature at the inlet plenum and pressure at the outlet plenum; they are assumed not to change with respect to the steady state analysis since

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Chapter 13 - Flow Boiling

increasing the power in a single channel does not change the boundary conditions since there are 90 to 120 channels connected to the same plena. On the other hand, the fluid quality at the outlet is higher in the new post-transient equilibrium state due to the larger heat rate. Saturation is reached earlier in the channel, so the 2-phase portion of the channel is longer. This results in a decreased post transient mass flow rate 𝑚𝑚̇𝑡𝑡 because of the increased hydraulic flow resistance within the channel. We follow a procedure analogous to the steady state case, with an initial guess of 𝑚𝑚̇𝑡𝑡 lower than in the steady state: 𝑚𝑚̇𝑡𝑡 = 21

kg s

Then we use the same equations as in the steady state to calculate: ℎ𝑡𝑡,𝑜𝑜𝑜𝑜𝑜𝑜 = 1455.821 𝑥𝑥𝑡𝑡,𝑜𝑜𝑜𝑜𝑜𝑜 = 0.09

kJ kg

The new coordinate 𝑎𝑎𝑡𝑡 (e.g., the coordinate where the fluid transitions from 1 to 2-phase conditions) is: 𝑎𝑎𝑡𝑡 = 4.338 m

We assume that form, minor and friction loss coefficients do not change in the transient. After a few iterations we find that the post-transient mass flow rate and exit quality are: 𝑚𝑚̇𝑡𝑡 = 20.6

kg s

𝑥𝑥𝑡𝑡,𝑜𝑜𝑜𝑜𝑜𝑜 = 0.095 kg

The post-transient mass flux is 𝐺𝐺𝑠𝑠𝑠𝑠 = 7000 m2 s. Using once again the Groeneveld lookup tables, we find that:

CHF𝑡𝑡,𝑜𝑜𝑜𝑜𝑜𝑜 = CHF𝑡𝑡 �𝐺𝐺𝑡𝑡 , 𝑃𝑃𝑜𝑜𝑜𝑜𝑜𝑜 , 𝑥𝑥𝑡𝑡,𝑜𝑜𝑜𝑜𝑜𝑜 � = CHF𝑡𝑡 �7000

kg kW , 10.0 MPa , 0.095� = 2413.3 2 2 m s m

The post-transient steady state heat flux is 15% higher than the steady state value: 𝑞𝑞𝑡𝑡′′ = 661.77

𝑘𝑘𝑘𝑘 kW (1.15) = 761.04 𝑚𝑚2 m2

The post-transient Minimum Departure from Nucleate Boiling Ratio (MDNBR) is then: MCHFR = min(CHFR) =

CHF𝑡𝑡,𝑜𝑜𝑢𝑢𝑢𝑢 2413.3 = = 3.17 𝑞𝑞𝑡𝑡′′ 761.04

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Chapter 13 - Flow Boiling

A 15% increase of the heat rate results in the MCHFR decreasing by almost half! 3. Transient*: result due to power increase only If we only accounted for the power increase (e.g. if we neglected the associated reduction in flow rate), the results would be different. We indicate the quantities related to this case with an asterisk (*). To show this, we assume that 𝑞𝑞 ∗̇ = 1.15𝑞𝑞̇ 𝑠𝑠𝑠𝑠 = 6.9 MW

but the mass flow rate doesn’t change with respect to the steady state: 𝑚𝑚̇ ∗ = 22.8

kg s

∗ ∗ At this point, the exit quality 𝑥𝑥𝑜𝑜𝑜𝑜𝑜𝑜 is the only unknown in the following equation., where ∆𝑃𝑃𝑡𝑡𝑡𝑡𝑡𝑡 is from Table 13.7. ∗ ∗ ∗ ∗ ∗ ∆𝑃𝑃𝑡𝑡𝑡𝑡𝑡𝑡 = 1𝑀𝑀𝑀𝑀𝑀𝑀 = ∆𝑃𝑃𝑖𝑖𝑖𝑖 + ∆𝑃𝑃𝑐𝑐ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 + ∆𝑃𝑃𝑜𝑜𝑜𝑜𝑜𝑜 + ∆𝑃𝑃𝑎𝑎𝑎𝑎𝑎𝑎

∗ Solving for 𝑥𝑥𝑜𝑜𝑜𝑜𝑜𝑜 we get:

∗ 𝑥𝑥𝑜𝑜𝑜𝑜𝑜𝑜 = 0.044

The mass flux is the same as in the steady state: 𝐺𝐺 ∗ =

𝑚𝑚̇𝑠𝑠𝑠𝑠 kg = 7755 2 𝐴𝐴 m s

The heat flux is the same as in the transient:

𝑞𝑞𝑡𝑡′′ = 770.8

kW m2

Using once again the Groeneveld lookup tables, we find that: ∗ ∗ ) CHF𝑜𝑜𝑜𝑜𝑜𝑜 = CHF ∗ (𝐺𝐺 ∗ , 𝑃𝑃𝑜𝑜𝑜𝑜𝑜𝑜 , 𝑥𝑥𝑜𝑜𝑜𝑜𝑜𝑜 = CHF𝑡𝑡 �7755

The Minimum CHF Ratio (MCHFR) is then:

MCHFR∗ = min(CHFR∗ ) =

359

kg kW , 10.0 MPa , 0.044� = 3529.6 m2 s m2

∗ CHF𝑜𝑜𝑜𝑜𝑜𝑜 3529.6 = = 4.6 ′′ 𝑞𝑞𝑡𝑡 761.04

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Chapter 13 - Flow Boiling

In Summary: Description MCHFR Comparison with steady state value

Steady State Normal operations

Transient* Power increase only

5.82

Transient Power increase and mass flow rate decrease 3.17

5.82/5.82 = 1

3.17/5.82 = 0.54

4.60/5.82 = 0.79

360

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Chapter 14 Single Heated Channel: Steady-State Analysis Contents Problem 14.1 Heated channel power limits ....................................................................... 362 Problem 14.2 Specification of power profile for a given clad temperature ....................... 366 Problem 14.3 Pressure drop-flow rate characteristic for a fuel channel ............................ 367 Problem 14.4 Pressure drop in a two-phase flow channel ................................................. 368 Problem 14.5 Critical heat flux for PWR channel ............................................................. 373 Problem 14.6 Nonuniform linear heat rate of a BWR ....................................................... 376 Problem 14.7 Thermal hydraulic analysis of a pressure tube reactor ................................ 380 Problem 14.8 Maximum clad temperature for SFBR ........................................................ 389 Problem 14.9 Flow-levitated control rod in a PWR .......................................................... 390 Problem 14.10 Thermal behavior of a plate fuel element following a loss of coolant ........ 395

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Chapter 14 - Single Heated Channel: Steady-State Analysis

PROBLEM 14.1 QUESTION Heated Channel Power Limits (Section 14.5) How much power can be extracted from a PWR with the geometry and operating conditions of Examples 14.1 and 14.2 and a Bessel Function radial power distribution if: 1. The coolant exit temperature is to remain subcooled. 2. The maximum clad temperature is to remain below saturation conditions, 3. The fuel maximum temperature is to remain below the melting temperature of 2400°C (ignore sintering effects). Answers: 1. Q̇ = 2421 MW 2. Q̇ = 1501 MW 3. Q̇ = 2312 MW

PROBLEM 14.1 SOLUTION Heated Channel Power Limits (Section 14.5) The following parameters are known: —

pressure, P = 15.51 MPa

—

inlet temperature, Tin = 293.1 °C

—

fuel rod outer diameter, D = 9.5 × 10−3 m

—

fuel pellet diameter, Df = 8.192 × 10−3 m

—

heated length, L = 3.658 m

—

peak linear heat generation rate, 𝑞𝑞0′ = 44.6 kW/m

—

mass flow rate, ṁ = 0.335 kg/s

—

number of fuel rods in the core, NrodsCore = Na Nrods = 193(264) = 50952

—

cladding thickness, tclad = 0.572 × 10−3 m

—

gap thickness, tgap = 0.0826 × 10−3 m

—

clad conductivity, kc = 13.85 W/m K

—

fuel conductivity, kf = 2.163 W/m K

362

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Chapter 14 - Single Heated Channel: Steady-State Analysis —

gap conductance, hgap = 5.7 kW/m2 K

The following specific enthalpy values may be taken from tables or software with water properties: ℎ𝑓𝑓 = ℎ(𝑃𝑃 = 15.51 MPa, 𝑥𝑥 = 0) = 1630 kJ/kg ℎ𝑔𝑔 = ℎ(𝑃𝑃 = 1551 MPa, 𝑥𝑥 = 1) = 2596 kJ/kg ℎ𝑓𝑓𝑓𝑓 = ℎ𝑔𝑔 − ℎ𝑓𝑓 = 2596 − 1630 = 966 kJ/kg

ℎ𝑖𝑖𝑖𝑖 = ℎ(𝑇𝑇 = 293.1 °C, 𝑃𝑃 = 15.51 MPa) = 1301 kJ/kg

The following geometrical parameters are calculated: — — — —

𝐷𝐷

0.0095

pellet outer radius, 𝑅𝑅𝑓𝑓 = 2𝑓𝑓 =

0.008192

cladding outer radius, 𝑅𝑅𝑐𝑐𝑐𝑐 = 2 =

= 0.00475 m

cladding inner radius, 𝑅𝑅𝑐𝑐𝑐𝑐 = 𝑅𝑅𝑐𝑐𝑐𝑐 − 𝑡𝑡𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = 0.00475 − 0.000572 = 0.00418 m gap radius, 𝑅𝑅𝑔𝑔 =

𝑅𝑅𝑓𝑓 +𝑅𝑅𝑐𝑐𝑐𝑐 2

𝐷𝐷

= 0.004096 m

0.004096+0.00418 2

= 0.00414 m

1. How much power can be extracted from a PWR, with the geometry and operating conditions of Examples 14.1 and 14.2 and a Bessel function radial power distribution, if the hot channel exit enthalpy is to remain subcooled. From a simple energy balance, the hot channel exit enthalpy is: ℎ𝑜𝑜𝑜𝑜𝑜𝑜,𝐻𝐻𝐻𝐻 = ℎ𝑖𝑖𝑖𝑖 +

𝑞𝑞̇ 𝐻𝐻𝐻𝐻 𝑚𝑚̇

(1)

If the hot channel exit enthalpy is to remain subcooled, hout,HC must be at most equal to hf . This condition may be used to obtain the hot channel power. ℎ𝑓𝑓 = ℎ𝑖𝑖𝑖𝑖 +

𝑞𝑞̇ 𝐻𝐻𝐻𝐻 𝑚𝑚̇

1630 = 1301 +

𝑞𝑞̇ 𝐻𝐻𝐻𝐻 0.335

⇒ 𝑞𝑞̇ 𝐻𝐻𝐻𝐻 = 110.2kW

(2)

(3)

The radial peaking factor is given in Table 3.6 and may be verified by integrating the Bessel function. (4) 𝑃𝑃𝑃𝑃𝑅𝑅 = 2.32 The total power that can be extracted from a PWR under those conditions is: 𝑄𝑄̇ =

𝑁𝑁𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑞𝑞̇ 𝐻𝐻𝐻𝐻 50952(110.2) = = 2421 MW 𝑃𝑃𝑃𝑃𝑅𝑅 2.32 363

(5)

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Chapter 14 - Single Heated Channel: Steady-State Analysis

2. How much power can be extracted from a PWR, with the geometry and operating conditions of Examples 14.1 and 14.2 and a Bessel function radial power distribution, if the maximum cladding temperature is to remain below saturation conditions. The saturation temperature at the operating pressure can be taken from tables with water properties: 𝑇𝑇𝑠𝑠𝑠𝑠𝑠𝑠 = 𝑇𝑇𝑠𝑠𝑠𝑠𝑠𝑠 (𝑃𝑃 = 15.51 MPa) = 344.8°C

(6)

ℏ = 34 kW/m2 K

(7)

For simplicity, we may take from Table K.2 a typical PWR single-phase heat transfer coefficient and the specific heat calculated at average PWR conditions:

(8)

𝑐𝑐𝑝𝑝 = 5.742 kJ/kg K

The location of maximum cladding temperature is given by Equation 14.22b: 𝑧𝑧𝑐𝑐 =

𝐿𝐿 𝐷𝐷𝐷𝐷ℏ tan−1 � � 𝜋𝜋 𝑚𝑚̇𝑐𝑐𝑝𝑝

(9)

(9.5 × 10−3 )(3.658)(34) 3.658 tan−1 � � = 0.641 m 𝜋𝜋 0.335(5.742)

Equation 14.19 provides the cladding temperature as a function of z. 𝑇𝑇𝑐𝑐𝑐𝑐 (𝑧𝑧) = 𝑇𝑇𝑖𝑖𝑖𝑖 + 𝑞𝑞0′ �

𝐿𝐿 𝜋𝜋𝜋𝜋 1 𝜋𝜋𝜋𝜋 �sin � � + 1� + cos � �� 𝜋𝜋𝑚𝑚̇𝑐𝑐𝑝𝑝 𝐿𝐿 2𝜋𝜋𝑅𝑅𝐶𝐶𝐶𝐶 ℏ 𝐿𝐿

(10)

We may impose the condition that, at the location of maximum cladding temperature, the cladding temperature is equal to the saturation temperature: 𝑇𝑇𝑐𝑐𝑐𝑐 (𝑧𝑧𝑐𝑐 ) = 𝑇𝑇𝑠𝑠𝑠𝑠𝑠𝑠 = 344.8°C. Thus: Solving for 𝑞𝑞0′ :

𝑇𝑇𝑖𝑖𝑖𝑖 + 𝑞𝑞0′ �

𝐿𝐿 𝜋𝜋𝑧𝑧𝑐𝑐 1 𝜋𝜋𝑧𝑧𝑐𝑐 �sin � � + 1� + cos � �� = 𝑇𝑇𝑠𝑠𝑠𝑠𝑠𝑠 𝜋𝜋𝑚𝑚̇𝑐𝑐𝑝𝑝 𝐿𝐿 2𝜋𝜋𝑅𝑅𝐶𝐶𝐶𝐶 ℏ 𝐿𝐿

𝑞𝑞0′ =

(11)

𝑇𝑇𝑠𝑠𝑠𝑠𝑠𝑠 − 𝑇𝑇𝑖𝑖𝑖𝑖

(12)

𝐿𝐿 𝜋𝜋𝑧𝑧 1 𝜋𝜋𝑧𝑧 �sin � 𝐿𝐿 𝑐𝑐 � + 1� + cos � 𝐿𝐿 𝑐𝑐 � 𝜋𝜋𝑚𝑚̇𝑐𝑐𝑝𝑝 2𝜋𝜋𝑅𝑅𝐶𝐶𝐶𝐶 ℏ

344.8 − 293.1 3.658 𝜋𝜋0.641 1 𝜋𝜋0.641 �sin � � + 1� + cos � � 3.658 3.658 2𝜋𝜋(0.00475)34 𝜋𝜋(0.335)(5.742) = 29.3 kW/m

The total power that can be extracted from a PWR core under the aforesaid conditions is: 𝑄𝑄̇ =

𝑁𝑁𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑞𝑞0′ 𝐿𝐿 50952(29.3)(3.658) = = 1501 MW 𝜋𝜋 𝑃𝑃𝑃𝑃𝑅𝑅 𝑃𝑃𝑃𝑃𝐴𝐴 2.32 2 364

(13)

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Chapter 14 - Single Heated Channel: Steady-State Analysis

Where the axial peaking factor for a cosine power distribution is PFA = π/2. 3. How much power can be extracted from a PWR with the geometry and operating conditions of Examples 14.1 and 14.2 and a Bessel function radial power distribution if the fuel maximum temperature is to remain below the melting temperature of 2400°C (ignore sintering effects). First of all, let us evaluate the sum of the thermal resistance terms between the fuel centerline and the coolant, neglecting for simplicity the oxide film: 𝑆𝑆 =

1 1 𝑅𝑅𝑐𝑐𝑐𝑐 1 1 + ln � � + + 4𝜋𝜋𝑘𝑘𝑓𝑓 2𝜋𝜋𝑘𝑘𝑐𝑐 𝑅𝑅𝑐𝑐𝑐𝑐 2𝜋𝜋𝑅𝑅𝑔𝑔 ℎ𝑔𝑔𝑔𝑔𝑔𝑔 2𝜋𝜋𝑅𝑅𝑐𝑐𝑐𝑐 ℏ

(14)

1 1 0.00475 1 �+ + ln � 4𝜋𝜋2.163 2𝜋𝜋13.85 0.00418 2𝜋𝜋(0.00414)(5.7 × 103 ) 1 + 2𝜋𝜋(0.00475)(34 × 103 )

= 0.0368 + 0.00147 + 0.00674 + 0.00085 = 0.0460 mK/W

The axial location where the fuel reaches the maximum temperature is given by Equation 14.25. 𝑧𝑧𝑓𝑓 =

𝐿𝐿 𝐿𝐿 tan−1 � � 𝜋𝜋 𝑆𝑆𝑆𝑆𝑚𝑚̇𝑐𝑐𝑝𝑝

(15)

3.658 3.658 � tan−1 � 𝜋𝜋 0.0460𝜋𝜋(0.335)(5.742 × 103 ) = 0.015 m

Equation 14.24 provides the fuel centerline temperature. 𝑇𝑇𝑐𝑐𝑐𝑐 (𝑧𝑧) = 𝑇𝑇𝑖𝑖𝑖𝑖 + 𝑞𝑞0′ �

𝐿𝐿 𝜋𝜋𝜋𝜋 𝜋𝜋𝜋𝜋 �sin � � + 1� + 𝑆𝑆 cos � �� 𝜋𝜋𝑚𝑚̇𝑐𝑐𝑝𝑝 𝐿𝐿 𝐿𝐿

(16)

We can impose the condition that, at the location of maximum fuel temperature, the fuel centerline temperature is equal to the melting temperature: Tcl(zf) = Tmelt = 2400°C.

Solving for 𝑞𝑞0′ :

𝑇𝑇𝑖𝑖𝑖𝑖 + 𝑞𝑞0′ �

𝜋𝜋𝑧𝑧𝑓𝑓 𝜋𝜋𝑧𝑧𝑓𝑓 𝐿𝐿 �sin � � + 1� + 𝑆𝑆 cos � �� = 𝑇𝑇𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝜋𝜋𝑚𝑚̇𝑐𝑐𝑝𝑝 𝐿𝐿 𝐿𝐿

𝑞𝑞0′ =

𝑇𝑇𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 − 𝑇𝑇𝑖𝑖𝑖𝑖 𝜋𝜋𝑧𝑧𝑓𝑓 𝜋𝜋𝑧𝑧𝑓𝑓 𝐿𝐿 �sin � � + 1� + 𝑆𝑆 cos � 𝜋𝜋𝑚𝑚̇𝑐𝑐𝑝𝑝 𝐿𝐿 𝐿𝐿 �

365

(17)

(18)

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Chapter 14 - Single Heated Channel: Steady-State Analysis

2400 − 293.1 3.658 𝜋𝜋0.015 𝜋𝜋0.015 �sin � � + 1� + (0.0460 × 103 ) cos � � 𝜋𝜋0.335.5.742 3.658 3.658 = 45.2kW/m

The total power that can be extracted from a PWR core under the aforesaid conditions is: 𝑄𝑄̇ =

𝑁𝑁𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑞𝑞0′ 𝐿𝐿 50952(45.2)(3.658) = = 2312 MW 𝜋𝜋 𝑃𝑃𝑃𝑃𝑅𝑅 𝑃𝑃𝑃𝑃𝐴𝐴 2.35 2

(19)

PROBLEM 14.2 QUESTION

Specification of Power Profile for a Given Cladding Temperature (Section 14.6) Consider a nuclear fuel rod whose cladding outer radius is a. Heat is transferred from the fuel rod to coolant with constant heat transfer coefficient ℏ. The coolant mass flow rate along the rod is ṁ. Coolant specific heat cp is independent of temperature. It is desired that the temperature of the outer surface of the fuel rod t (at radius a) be constant, independent of distance Z from the coolant inlet to the end of the fuel rod. Derive a formula showing how the linear power of the fuel rod q' should vary with Z if the temperature at the outer surface of the fuel rod is to be constant. Answer: 𝑞𝑞 ′ (𝑍𝑍) = 𝑞𝑞′(0)𝑒𝑒

−

ℏ(2π𝑎𝑎)𝑍𝑍 ṁcp

PROBLEM 14.2 SOLUTION

Specification of Power Profile for a Given Cladding Temperature (Section 14.6) The heat transfer equation that controls the cladding-to-coolant heat exchange along a portion of the channel of length dZ at the axial position Z is: 𝑞𝑞 ′ (𝑍𝑍)𝑑𝑑𝑑𝑑 = ℎ 2𝜋𝜋𝜋𝜋�𝑡𝑡 − 𝑇𝑇(𝑍𝑍)�𝑑𝑑𝑑𝑑

(1)

The coolant temperature as a function of the axial position T(Z) is given by an energy balance between the channel inlet and Z: 𝑍𝑍

� 𝑞𝑞′(𝑍𝑍 ′ )𝑑𝑑𝑑𝑑′ = 𝑚𝑚̇𝑐𝑐𝑝𝑝 (𝑇𝑇(𝑍𝑍) − 𝑇𝑇𝑖𝑖𝑖𝑖 )

(2)

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Chapter 14 - Single Heated Channel: Steady-State Analysis

We can now isolate T(Z) in the first equation and then substitute it in the second equation: 𝑇𝑇(𝑍𝑍) = 𝑡𝑡 −

𝑍𝑍

𝑞𝑞′(𝑍𝑍) ℏ2𝜋𝜋𝜋𝜋

� 𝑞𝑞′(𝑍𝑍 ′ )𝑑𝑑𝑑𝑑′ = 𝑚𝑚̇𝑐𝑐𝑝𝑝 �𝑡𝑡 − 0

� 𝑞𝑞′(𝑍𝑍 ′ )𝑑𝑑𝑑𝑑′ = 0

𝑞𝑞 ′(𝑍𝑍) − 𝑇𝑇𝑖𝑖𝑖𝑖 � ℏ2𝜋𝜋𝜋𝜋

𝑚𝑚̇𝑐𝑐𝑝𝑝 [(𝑡𝑡 − 𝑇𝑇𝑖𝑖𝑖𝑖 )ℏ2𝜋𝜋𝜋𝜋 − 𝑞𝑞′(𝑍𝑍)] ℏ2𝜋𝜋𝜋𝜋

(3) (4)

(5)

Where (t − Tin)ℏ2𝜋𝜋𝜋𝜋 is the linear heat generation rate at the channel inlet, q'(0) Hence: 𝑍𝑍

� 𝑞𝑞′(𝑍𝑍 ′ )𝑑𝑑𝑑𝑑′ = 0

𝑚𝑚̇𝑐𝑐𝑝𝑝 [−𝑞𝑞′(𝑍𝑍′)]0𝑍𝑍 ℏ2𝜋𝜋𝜋𝜋

(6)

The function q'(Z), which is solution to this equation, is: 𝑞𝑞

′ (𝑍𝑍)

= 𝑞𝑞′(0)𝑒𝑒

−

ℏ(2π𝑎𝑎)𝑍𝑍 ṁcp

(7)

PROBLEM 14.3 QUESTION Pressure Drop-Flow Rate Characteristic for a Fuel Channel (Section 14.6) A designer is interested in determining the effect of a 50% channel blockage on the downstream-clad temperatures in a BWR core. The engineering department proposes to assess this effect experimentally by inserting a prototypical BWR channel containing the requisite blockage in the test loop sketched in Figure 14.15 and running the centrifugal pump to deliver prototypic BWR pressure and single-channel flow conditions. Is the test plan acceptable to you? If not, what changes would you propose and why?

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Chapter 14 - Single Heated Channel: Steady-State Analysis

FIGURE 14.15 Test loop (a) and pump characteristic curve (b). Answer: It is not possible to specify both flow and pressure conditions to be identical to BWR conditions. A suitable modification to the test loop must be accomplished.

PROBLEM 14.3 SOLUTION Pressure Drop-Flow Rate Characteristic for a Fuel Channel (Section 14.6) The proposed test plan is not acceptable. By controlling the centrifugal pump it is not possible to specify both mass flow rate and pressure to be identical to BWR conditions. The test plan may be modified by connecting to the loop a pressurizer, which can be used to set a desired operating pressure. The mass flow rate and the inlet enthalpy at the test section may be adjusted by regulating the pump speed and the heat exchanger operating conditions.

PROBLEM 14.4 QUESTION Pressure Drop in a Two-Phase Flow Channel (Section 14.6) For the BWR conditions given below, using the HEM model determine the acceleration, frictional, gravitational, and total pressure drop (ignore the spacer pressure drop). Compare the frictional pressure drop obtained using the HEM model to that obtained from the Thom approach (Equation 11.97).

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Chapter 14 - Single Heated Channel: Steady-State Analysis

Geometry: – Consider a vertical tube of 17 mm I.D. and 3.8 m length.

Operating Conditions:

– Steady state – Operating pressure = 7550 kPa – Inlet temperature = 275 °C – Mass flux = 1700 kg/m2s – Heat flux (axially constant) = 670 kW/m2

Answers:

The pressure drops (HEM model) are as follows: ∆Pacc = 12.5 kPa ∆Pgrav = 16.0 kPa ∆Pfric = 14.1 kPa ∆Ptot = 42.6 kPa ∆Pfric,Thom = l5.1 kPa

PROBLEM 14.4 SOLUTION Pressure Drop in a Two-Phase Flow Channel (Section 14.6) The given conditions are: – – – – – –

tube inner diameter, D = 0.017 m tube length, L = 3.8m pressure, P = 7550 kPa inlet temperature, Tin = 275 °C mass flux, G = 1700 kg/m2s heat flux, q″ = 670 kW/m2

The following water properties (specific enthalpy, density, viscosity) are taken from tables or specific computer programs. ℎ𝑓𝑓 = ℎ(𝑃𝑃 = 7.55 MPa, 𝑥𝑥 = 0) = 1295 kJ/kg

ℎ𝑔𝑔 = ℎ(𝑃𝑃 = 7.55 MPa, 𝑥𝑥 = 1) = 1295 kJ/kg

ℎ𝑓𝑓𝑓𝑓 = ℎ𝑔𝑔 − ℎ𝑓𝑓 = 2765 − 1295 = 1470 kJ/kg

ℎ𝑖𝑖𝑖𝑖 = ℎ(𝑇𝑇 = 275°C, 𝑃𝑃 = 7.55 MPa) = 1210 kJ/kg 𝜌𝜌𝑓𝑓 = 𝜌𝜌(𝑃𝑃 = 7.55 MPa, 𝑥𝑥 = 0) = 730 kg/m3

𝜌𝜌𝑔𝑔 = 𝜌𝜌(𝑃𝑃 = 7.55 MPa, 𝑥𝑥 = 1) = 39.8 kg/m3 369

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Chapter 14 - Single Heated Channel: Steady-State Analysis

𝜌𝜌𝑖𝑖𝑖𝑖 = 𝜌𝜌(𝑇𝑇 = 275°C, 𝑃𝑃 = 7.55 MPa) = 761 kg/m3

𝜇𝜇𝑓𝑓 = 𝜇𝜇(𝑃𝑃 = 7.55 MPa, 𝑥𝑥 = 0) = 89.28 × 10−6 Pa 𝑠𝑠 𝜇𝜇𝑔𝑔 = 𝜇𝜇(𝑃𝑃 = 7.55 MPa, 𝑥𝑥 = 1) = 19.2 × 10−6 Pa 𝑠𝑠

The flow area, mass flow rate, and linear heat generation rate are calculated as follows, 𝐴𝐴 =

𝜋𝜋𝐷𝐷2 𝜋𝜋0.0172 = = 2.270 × 10−4 m2 4 4

𝑚𝑚̇ = 𝐺𝐺𝐺𝐺 = 1700(2.270 × 10−4 ) = 0.3859 𝑞𝑞 ′ = 𝜋𝜋𝜋𝜋𝑞𝑞 ″ = 𝜋𝜋0.017(670) = 35.78

kW m

(1) kg s

(2) (3)

The OSB location may be calculated applying a simple energy balance (Equation 14.29): 𝑍𝑍𝑂𝑂𝑂𝑂𝑂𝑂

𝑍𝑍𝑂𝑂𝑂𝑂𝑂𝑂

35.78 = 1295 − 1210 0.3859 ⇒ 𝑍𝑍𝑂𝑂𝑂𝑂𝑂𝑂 = 0.917 m

The inlet equilibrium quality is: 𝑥𝑥𝑒𝑒,𝑖𝑖𝑖𝑖 =

𝑞𝑞′ = ℎ𝑓𝑓 − ℎ𝑖𝑖𝑖𝑖 ṁ

ℎ𝑖𝑖𝑖𝑖 − ℎ𝑓𝑓 1210 − 1295 = = −0.0578 ℎ𝑔𝑔 − ℎ𝑓𝑓 2765 − 1295

(4)

(5)

(6)

The equilibrium quality profile is given by Equation = 14.33a: 𝑥𝑥𝑒𝑒 (𝑧𝑧) = 𝑥𝑥𝑒𝑒,𝑖𝑖𝑖𝑖 +

𝑞𝑞 ′ 35.78 𝑍𝑍 = −0.0578 + 𝑍𝑍 𝑚𝑚̇ℎ𝑓𝑓𝑓𝑓 0.3859(1470)

(7)

= −0.578 + 0.0631𝑍𝑍

In the region above ZOSB the void fraction profile is determined, considering HEM, as: 𝛼𝛼(𝑍𝑍) =

1 1 = (1 + 0.0578 − 0.0631𝑍𝑍) 39.8 1 − 𝑥𝑥𝑒𝑒 (𝑧𝑧)𝜌𝜌𝑔𝑔 1+ 1+ (−0.0578 + 0.0631𝑍𝑍) 730 𝑥𝑥𝑒𝑒 (𝑧𝑧)𝜌𝜌𝑓𝑓 370

(8)

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Chapter 14 - Single Heated Channel: Steady-State Analysis

Simplifying: 𝛼𝛼(𝑍𝑍) =

0.0578 − 0.0631𝑍𝑍 0.0001282 − 0.0597𝑍𝑍

(9)

In the region above ZOSB the mixture density is:

= 730 �1 −

(10)

𝜌𝜌𝑚𝑚 (𝑍𝑍) = 𝜌𝜌𝑓𝑓 �1 − 𝛼𝛼(𝑍𝑍)� + 𝜌𝜌𝑔𝑔 𝛼𝛼(𝑍𝑍)

0.0578 − 0.0631𝑍𝑍 0.0578 − 0.0631𝑍𝑍 � + 39.8 kg/m3 0.0001282 − 0.0597𝑍𝑍 0.0001282 − 0.0597𝑍𝑍

Simplifying and approximating:

At the channel outlet:

𝜌𝜌𝑚𝑚 (𝑍𝑍) =

−39.8 kg⁄m3 0.0001282 − 0.05966𝑍𝑍

𝜌𝜌𝑚𝑚,𝑜𝑜𝑜𝑜𝑜𝑜 = 𝜌𝜌𝑚𝑚 (𝐿𝐿) =

(11)

−39.8 175.7kg = 0.0001282 − 0.05966(3.8) m3

(12)

Acceleration pressure drop The acceleration pressure drop is given by Equation 14.38a: 1 1 1 1 � = 12.5 kPa ∆𝑃𝑃𝑎𝑎𝑎𝑎𝑎𝑎 = 𝐺𝐺 2 � − � = 17002 � − 175.7 730 𝜌𝜌𝑚𝑚,𝑜𝑜𝑜𝑜𝑜𝑜 𝜌𝜌𝑓𝑓

(13)

Gravity pressure drop The average density in the region below ZOSB is approximately: 𝜌𝜌̅ℓ =

𝜌𝜌𝑖𝑖𝑖𝑖 + 𝜌𝜌𝑓𝑓 761 + 730 = = 746 kg/m3 2 2

(14)

The gravity pressure drop is given by Equation 14.38b: 𝐿𝐿

∆𝑃𝑃𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 = 𝜌𝜌̅ℓ 𝑔𝑔 𝑍𝑍𝑂𝑂𝑂𝑂𝑂𝑂 + � 𝜌𝜌𝑚𝑚 (𝑍𝑍) 𝑔𝑔 𝑑𝑑𝑑𝑑

(15)

𝑍𝑍𝑂𝑂𝑂𝑂𝑂𝑂

3.8

= 746(8.91)(0.917) + �

0.917

−39.8 9.81𝑑𝑑𝑑𝑑 0.0001282 − 0.05966𝑍𝑍

Integrating numerically or analytically and solving:

∆𝑃𝑃𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 = 16.0 kPa 371

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Chapter 14 - Single Heated Channel: Steady-State Analysis

Friction pressure drop The liquid only Reynolds number is: Reℓ𝑜𝑜 =

𝐺𝐺𝐺𝐺 1700(0.017) = = 3.24 × 105 𝜇𝜇𝑓𝑓 89.28 × 10−6

(17)

The flow is turbulent, so the McAdams relation may be applied to calculate the liquid only friction factor: = 0.184(3.24 × 105 )−0.20 = 0.0145 𝑓𝑓ℓ𝑜𝑜 = 0.184Re−0.20 ℓ𝑜𝑜

(18)

The HEM two-phase friction multiplier can be expressed using Equation 11.82. 2 (𝑍𝑍) = 1 + 𝑥𝑥𝑒𝑒 (𝑍𝑍) � Φℓ𝑜𝑜

𝜌𝜌𝑓𝑓 730 − 1� = 1 + 𝑥𝑥𝑒𝑒 (𝑍𝑍) � − 1� = 1 + 17.37𝑥𝑥𝑒𝑒 (𝑍𝑍) 𝜌𝜌𝑔𝑔 39.8

(19)

= 1 + 17.34(−0.0578 + 0.0631𝑍𝑍) = −0.002252 + 1.094𝑍𝑍

We may express the friction pressure drop re-arranging Equation 14.47. 𝐿𝐿

𝑓𝑓ℓ𝑜𝑜 𝐺𝐺 2 𝑓𝑓ℓ𝑜𝑜 𝐺𝐺 2 2 (𝑍𝑍)𝑑𝑑𝑑𝑑 ∆𝑃𝑃𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 = 𝑍𝑍𝑂𝑂𝑂𝑂𝑂𝑂 + � Φℓ𝑜𝑜 2𝐷𝐷𝜌𝜌̅ℓ 2𝐷𝐷𝜌𝜌̅𝑓𝑓

(20)

𝑍𝑍𝑂𝑂𝑂𝑂𝑂𝑂 3.8

0.0145(17002 ) 0.0145(17002 ) = 0.917 + � (−0.002252 + 1.094𝑍𝑍)𝑑𝑑𝑑𝑑 2(0.017)(746) 2(0.017)(730) 0.917

Integrating and solving:

∆𝑃𝑃𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 = 14.1 kPa

(21)

∆𝑃𝑃𝑡𝑡𝑡𝑡𝑡𝑡 = ∆𝑃𝑃𝑎𝑎𝑎𝑎𝑎𝑎 + ∆𝑃𝑃𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 + ∆𝑃𝑃𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 = 12.5 + 16.0 + 14.1 = 42.6 kPa

(22)

Total pressure drop

Friction pressure drop using the Thom approach The Thom approach applies to saturated inlet conditions. The friction pressure drop using the Thom approach may be calculated for a boiling region of length LB using Equation 11.97: ∆𝑃𝑃𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓,𝑇𝑇ℎ𝑜𝑜𝑜𝑜 =

𝑓𝑓ℓ𝑜𝑜 𝐺𝐺 2 𝐿𝐿𝐵𝐵 𝑇𝑇 2𝐷𝐷𝜌𝜌ℓ 3

(23)

We may read the r3 coefficient from Figure 11.15 knowing the outlet quality and the operating pressure in psi.

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Chapter 14 - Single Heated Channel: Steady-State Analysis

𝑥𝑥𝑒𝑒,𝑜𝑜𝑜𝑜𝑜𝑜 = 𝑥𝑥𝑒𝑒,𝑖𝑖𝑖𝑖 +

𝑞𝑞 ′ 35.78 𝐿𝐿 = −0.0578 + 3.8 = 0.182 𝑚𝑚̇ℎ𝑓𝑓𝑓𝑓 0.3859(1470)

Reading from the table yields:

(24)

𝑃𝑃 = 7550 kPa ⇒ 1095 psi

(25)

⇒ 𝑟𝑟3 = 2.8

(26)

The friction pressure drop is composed of two parts: one for the liquid-only region and one for the boiling region. Hence: ∆𝑃𝑃𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓,𝑇𝑇ℎ𝑜𝑜𝑜𝑜 =

𝑓𝑓ℓ𝑜𝑜 𝐺𝐺 2 𝑍𝑍𝑂𝑂𝑂𝑂𝑂𝑂 (𝐿𝐿 − 𝑍𝑍𝑂𝑂𝑂𝑂𝑂𝑂 ) � + � 2𝐷𝐷 𝜌𝜌̅ℓ 𝜌𝜌𝑓𝑓 0.0145(17002 ) 0.917 2.8(3.8 − 0.917) = � + � 2(0.017) 746 730

(27)

= 15.1 kPa

PROBLEM 14.5 QUESTION Critical Heat Flux for PWR Channel (Section 14.6) Using the conditions of Examples 14.1 and 14.2, determine the minimum critical heat flux ratio (MCHFR) using the W-3 correlation. Answer: MCHFR = 2.41

PROBLEM 14.5 SOLUTION Critical Heat Flux for PWR Channel (Section 14.6) The following parameters are known: –

pressure, P = 15.51 MPa

–

heated length, L = 3.658 m

–

inlet temperature, Tin = 293.1 °C

–

peak linear heat generation rate, q0′ = 44.6 kW/m

–

fuel rod outer diameter, D = 9.5 ×10−3 m

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Chapter 14 - Single Heated Channel: Steady-State Analysis

–

mass flux, G = 3807 kg/m2s subchannel area, Ach = 8.79 ×10−5 m2 mass flow rate, ṁ = 0.335 kg/s

The following specific enthalpy values may be taken from tables or software with water properties: ℎ𝑓𝑓 = ℎ(𝑃𝑃 = 15.51 MPa, 𝑥𝑥 = 0) = 1630 kJ/kg

ℎ𝑔𝑔 = ℎ(𝑃𝑃 = 15.51 MPa, 𝑥𝑥 = 1) = 2596 kJ⁄kg ℎ𝑓𝑓𝑓𝑓 = ℎ𝑔𝑔 − ℎ𝑓𝑓 = 2596 − 1630 = 966 kJ⁄kg

ℎ𝑖𝑖𝑖𝑖 = ℎ(𝑇𝑇 = 293.1°C, 𝑃𝑃 = 15.51 MPa) = 1301 kJ⁄kg

The inlet equilibrium quality is:

The rod power is:

𝑥𝑥𝑒𝑒,𝑖𝑖𝑖𝑖 =

ℎ𝑖𝑖𝑖𝑖 − ℎ𝑓𝑓 1301 − 1630 = = −0.341 ℎ𝑔𝑔 − ℎ𝑓𝑓 2596 − 1630

𝑞𝑞̇ =

𝑞𝑞0′ 𝐿𝐿 44.6(3.658) = = 103.9 kW 𝜋𝜋⁄2 𝜋𝜋⁄2

The peak heat flux is:

𝑞𝑞0′′ =

The cosine heat flux distribution is:

𝑞𝑞0′ 44.6 kW = = 1494 2 𝜋𝜋𝜋𝜋 𝜋𝜋0.0095 m

𝑞𝑞 ′′ (𝑍𝑍) = 𝑞𝑞0′′ cos �

𝜋𝜋𝜋𝜋 𝜋𝜋𝜋𝜋 � = 1494 cos � � [kW⁄m2 ] 𝐿𝐿 3.658

(1)

(2)

(3)

(4)

Equation 14.33b describes the equilibrium quality axial distribution.

𝜋𝜋𝜋𝜋 � + 1� 𝐿𝐿 2𝑚𝑚ℎ̇ 𝑓𝑓𝑓𝑓 103.9 𝜋𝜋𝜋𝜋 = −0.341 + �sin � � + 1� 2(0.335)966 3.658

𝑥𝑥𝑒𝑒 (𝑍𝑍) = 𝑥𝑥𝑒𝑒,𝑠𝑠𝑠𝑠𝑠𝑠 + The equivalent diameter is:

𝐷𝐷𝑒𝑒 =

𝑞𝑞̇

�sin �

4𝐴𝐴𝑐𝑐ℎ 4(8.79 × 10−5 ) = = 0.01178 m 𝜋𝜋𝜋𝜋 𝜋𝜋0.0095

(5)

(6)

W-3 correlation The W-3 correlation is described, for uniform axial heat flux, by Equation

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13.51: ′′ (𝑍𝑍) = 𝐾𝐾1(𝑧𝑧) 𝐾𝐾2(𝑧𝑧) 𝐾𝐾3(𝑧𝑧) 𝐾𝐾4 𝑞𝑞𝑐𝑐𝑐𝑐,𝑢𝑢

(7)

Where the four coefficients are calculated below. To avoid confusion, the equilibrium quality axial distribution is indicated as xe(z) instead of substituting the expression above. 𝐾𝐾1(𝑧𝑧) = (2.022 − 0.06238𝑃𝑃) + (0.1722 − 0.01427𝑃𝑃) exp [(18.177 − 0.5987𝑃𝑃)𝑥𝑥𝑒𝑒 (𝑧𝑧)]

(8)

= [2.022 − 0.06238(15.51)] + [0.1722 − 0.01427(15.51)] exp[(18.177 − 0.5987(15.51))𝑥𝑥𝑒𝑒 (𝑧𝑧)] = 1.054 − 0.04913 exp(8.891𝑥𝑥𝑒𝑒 (𝑧𝑧))

(9)

(10)

𝐾𝐾3(𝑧𝑧) = �1.157 − 0.869𝑥𝑥𝑒𝑒 (𝑧𝑧)�[0.2664 + 0.8357 exp (−124.1𝐷𝐷𝑒𝑒 )]

= �1.157 − 0.869𝑥𝑥𝑒𝑒 (𝑧𝑧)�[0.2664 + 0.8357exp (−124.1)(0.01178)] = 0.4601(1.157 − 0.869𝑥𝑥𝑒𝑒 (𝑧𝑧))

(11)

𝐾𝐾4 = 0.8258 + 0.0003413�ℎ𝑓𝑓 − ℎ𝑖𝑖𝑖𝑖 �

= 0.8258 + 0.0003413(1630 − 1301) = 0.9381

Equations 13.51d and13.51c allow calculating the nonuniform axial heat flux factor, F(z). [1 − 𝑥𝑥𝑒𝑒 (𝑧𝑧)]4.31 𝐶𝐶(𝑧𝑧) = 185.6 𝐺𝐺 0.478

= 185.6 𝐹𝐹(𝑧𝑧) =

(12)

[1 − 𝑥𝑥𝑒𝑒 (𝑧𝑧)]4.31 − 3.606[1 − 𝑥𝑥𝑒𝑒 (𝑧𝑧)]4.31 38070.478 𝑧𝑧

𝐶𝐶(𝑧𝑧) ∫−𝐿𝐿⁄2 𝑞𝑞 ′′ (𝑧𝑧 ′ ) exp{−𝐶𝐶(𝑧𝑧)[𝑧𝑧 − 𝑧𝑧 ′ ]}𝑑𝑑𝑑𝑑′ 𝐿𝐿 𝑞𝑞′′(𝑧𝑧) �1 − exp �−𝐶𝐶(𝑧𝑧) �𝑧𝑧 + 2�� 375

(13)

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Chapter 14 - Single Heated Channel: Steady-State Analysis

𝑧𝑧 𝜋𝜋𝑧𝑧 ′ � exp{−[3.606[1 − 𝑥𝑥𝑒𝑒 (𝑧𝑧)]4.31 ][𝑧𝑧 − 𝑧𝑧 ′ ]}𝑑𝑑𝑧𝑧 ′ 3.606[1 − 𝑥𝑥𝑒𝑒 (𝑧𝑧)]4.31 ∫−𝐿𝐿⁄2 1494 cos � 3.658 = 𝜋𝜋𝜋𝜋 3.658 1494 cos � � �1 − exp �−[3.606[1 − 𝑥𝑥𝑒𝑒 (𝑧𝑧)]4.31 ] �𝑧𝑧 + 2 �� 3.658

Equation 13.51b gives the nonuniform DNB heat flux as the ratio between q″cr,u (z) and F(z): ′′ (𝑧𝑧) = 𝑞𝑞𝑐𝑐𝑐𝑐,𝑛𝑛

′′ 𝑞𝑞𝑐𝑐𝑐𝑐,𝑢𝑢 (𝑧𝑧) 𝐹𝐹(𝑧𝑧)

(14)

The equations above may be implemented in an electronic calculation tool, which may be used to compute the MCHFR:

The result is:

′′ 𝑞𝑞𝑐𝑐𝑐𝑐,𝑛𝑛 (𝑧𝑧) MCHFR = 𝑚𝑚𝑚𝑚𝑚𝑚 � � , −𝐿𝐿/2 ≤ 𝑧𝑧 ≤ 𝐿𝐿/2 𝑞𝑞′′(𝑧𝑧)

MCHFR = 2.41

(15)

(16)

The CHFR (or DNBR) is plotted in Figure SM-14.1 as a function of the distance from the core inlet.

FIGURE SM-14.1 DNBR as a function of the distance from the core inlet

PROBLEM 14.6 QUESTION Nonuniform Linear Heat Rate of a BWR (Section 14.6) For the hot subchannel of a BWR-5, calculate the axial location where bulk boiling starts to occur. Assuming the outlet quality is 29.2%, determine the axial profiles of the hot channel

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enthalpy, hm(z), and thermal equilibrium quality, xe(z). Assume thermodynamic equilibrium and the following axial heat generation profile:

Parameters –

𝜋𝜋𝜋𝜋 𝑧𝑧 1 ′ 𝑞𝑞 ′ (𝑧𝑧) = 𝑞𝑞𝑟𝑟𝑟𝑟𝑟𝑟 exp �−𝛼𝛼 � + �� cos � � 𝐿𝐿 2 𝐿𝐿

−

𝐿𝐿 𝐿𝐿 ≤ 𝑧𝑧 ≤ 2 2

peak linear heat generation rate, q′max = 47.24 kW/m

–

inlet temperature, Tin = 278.3 °C

–

heated length, L = 3.588 m

–

outlet equilibrium quality, xe, out = 0.292

–

pressure, P = 7.14 MPa

–

axial heat distribution constant, α = 1.96

Answer: 𝑧𝑧𝑂𝑂𝑂𝑂𝑂𝑂 = −1.29 𝑚𝑚

PROBLEM 14.6 SOLUTION Nonuniform Linear Heat Rate of a BWR (Section 14.6) The following parameters are given by the problem statement: –

peak linear heat generation rate, q′max = 47.24 kW/m

–

inlet temperature, Tin = 278.3 °C

–

heated length, L = 3.588 m

–

outlet equilibrium quality, xe, out = 0.292

–

pressure, P = 7.14 MPa

–

axial heat distribution constant, α = 1.96

The axial linear heat generation profile is: 𝑧𝑧 1 𝜋𝜋𝜋𝜋 ′ 𝑞𝑞 ′ (𝑧𝑧) = 𝑞𝑞𝑟𝑟𝑟𝑟𝑟𝑟 exp �−𝛼𝛼 � + �� cos � � 𝐿𝐿 2 𝐿𝐿

(1)

This equation represents a bottom shaped axial distribution. Therefore, we expect the location of maximum heat generation zmax to be below the core midplane. To find this point, we look for the location where the first derivative of the axial heat distribution is equal to zero.

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�

𝜕𝜕𝜕𝜕′(𝑧𝑧) � 𝜕𝜕𝜕𝜕 𝑧𝑧=𝑧𝑧

(2)

𝑧𝑧𝑚𝑚𝑚𝑚𝑚𝑚 = −0.637 m

(3)

′ 𝑞𝑞𝑚𝑚𝑚𝑚𝑚𝑚 = 𝑞𝑞′(𝑧𝑧𝑚𝑚𝑚𝑚𝑚𝑚 )

(4)

′ ⇒ 𝑞𝑞𝑟𝑟𝑟𝑟𝑟𝑟 = 104.75kW/m

(5)

𝑚𝑚𝑚𝑚𝑚𝑚

The solution may be found either numerically or analytically as:

Applying the maximum linear heat generation rate q′max to the location zmax we obtain q′ref.

−0.637 1 𝜋𝜋(−0.637) ′ � 47.24 = 𝑞𝑞𝑟𝑟𝑟𝑟𝑟𝑟 exp �−1.96 � + �� cos � 3.588 2 3.588 The following specific enthalpy values may be taken from water properties: ℎ𝑓𝑓 = ℎ(𝑃𝑃 = 7.14 MPa, 𝑥𝑥 = 0) = 1275 kJ⁄kg

ℎ𝑔𝑔 = ℎ(𝑃𝑃 = 7.14 MPa, 𝑥𝑥 = 1) = 2771 kJ⁄kg

ℎ𝑓𝑓𝑓𝑓 = ℎ𝑔𝑔 − ℎ𝑓𝑓 = 2771 − 1275 = 2771 − 1275 = 1496 kJ⁄kg ℎ𝑖𝑖𝑖𝑖 = ℎ(𝑇𝑇 = 278.3°C, 𝑃𝑃 = 7.14 MPa) = 1228 kJ⁄kg

The inlet equilibrium quality is:

𝑥𝑥𝑒𝑒,𝑖𝑖𝑖𝑖 =

ℎ𝑖𝑖𝑖𝑖 − ℎ𝑓𝑓 1228 − 1275 = = −0.0314 ℎ𝑔𝑔 − ℎ𝑓𝑓 2771 − 1275

(6)

1. Calculate the axial location where bulk boiling starts to occur. The equilibrium quality distribution is given by Equation 14.33a as: 𝑧𝑧

1 𝑥𝑥𝑒𝑒 (𝑧𝑧) = 𝑥𝑥𝑒𝑒,𝑖𝑖𝑖𝑖 + � 𝑞𝑞′(𝑧𝑧)𝑑𝑑𝑑𝑑 𝑚𝑚̇ℎ𝑓𝑓𝑓𝑓

(7)

−𝐿𝐿⁄2

First of all, we set the outlet equilibrium quality condition xe(zout) = 0.292 to obtain the mass flow rate. 0.292 = −0.0314

1 + 𝑚𝑚̇1496

3.588⁄2

�

−3.588⁄2

104.75 exp �−1.96 �

378

𝑧𝑧 1 𝜋𝜋𝜋𝜋 + �� cos � � 𝑑𝑑𝑑𝑑 3.588 2 3.588

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Solving this equation numerically yields: 𝑚𝑚̇ = 0.203 kg⁄s

Now that the mass flow rate is known, we may set the equilibrium quality equal to zero to find the OSB location ZOSB. 1 0 = −0.0314 + 0.203(1496)

𝑧𝑧𝑂𝑂𝑂𝑂𝑂𝑂

�

−3.588⁄2

1 𝜋𝜋𝜋𝜋 + �� cos � � 𝑑𝑑𝑑𝑑 2 3.588

𝑧𝑧 104.75 exp �−1.96 � 3.588

(9)

Solving this equation numerically yields:

(10)

𝑧𝑧𝑂𝑂𝑂𝑂𝑂𝑂 = −1.29 m

2. Determine the axial profile of the hot channel enthalpy and equilibrium quality. The axial enthalpy and equilibrium quality distributions are given by Equations 14.29 and 14.33a as: 𝑧𝑧

1 ℎ(𝑧𝑧) = ℎ𝑖𝑖𝑖𝑖 + � 𝑞𝑞′(𝑧𝑧)𝑑𝑑𝑑𝑑 𝑚𝑚̇ −𝐿𝐿⁄2

(11)

𝑧𝑧

1 𝑥𝑥𝑒𝑒 (𝑧𝑧) = 𝑥𝑥𝑒𝑒,𝑖𝑖𝑖𝑖 + � 𝑞𝑞 ′ (𝑧𝑧)𝑑𝑑𝑑𝑑 𝑚𝑚̇ℎ𝑓𝑓𝑓𝑓

(12)

−𝐿𝐿⁄2

Substituting all the parameters, the axial profiles of enthalpy and equilibrium quality are: 1 ℎ(𝑧𝑧) = 128 + 0.203 𝑥𝑥𝑒𝑒 (𝑧𝑧) = −0.314

𝑧𝑧

�

−3.588⁄2

1 + 0.203(1496)

𝑧𝑧 1 𝜋𝜋𝜋𝜋 104.75 exp �−1.96 � + �� cos � � 𝑑𝑑𝑑𝑑 3.588 2 3.588

𝑧𝑧

�

−3.588⁄2

𝑧𝑧 1 𝜋𝜋𝜋𝜋 104.75 exp �−1.96 � + �� cos � � 𝑑𝑑𝑑𝑑 3.588 2 3.588

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(13)

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Those profiles have been plotted in Figures SM-14.2 and SM-14.3.

FIGURE SM-14.2 Enthalpy profile

FIGURE SM-14.3 Equilibrium quality profile

PROBLEM 14.7 QUESTION Thermal Hydraulic Analysis of a Pressure Tube Reactor (Chapters 3, 8, 9, 10, 11, and Section 14.6) Consider the light water-cooled and moderated pressure tube reactor shown in Figure 14.16. The fuel and coolant in the pressure tube are within a graphite matrix. Each pressure tube consists of a graphite matrix that has 24 fuel holes and 12 coolant holes. Part of the graphite matrix and

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a unit cell are also shown in Figure 14.16. An equivalent annuli model for thermal analysis is shown in Figure 14.17. Consider the fuel (although composed of fuel particles in each fuel hole) as operating at a uniform volumetric heat generation rate in the r, ϑ plane. Operating conditions and property data are in Tables 14.5 and 14.6. Assumptions –

HEM (Homogeneous Equilibrium Model) for two phase flow analysis is valid.

–

Saturated coolant data at 6.89 MPa

–

Enthalpy: hf = 1261.6 kJ/kg, hfg = 1511.9 kJ/kg

–

Cosine axial heat flux (neglect extrapolation)

–

Density: ρf = 742.0 kg/m3, ρg = 35.94 kg/m3

FIGURE 14.16 Calandria with pressure tubes and unit cell in the pressure tube

FIGURE 14.17 Equivalent annuli model (not to scale)

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TABLE 14.5 Operating Data for Problem 14.7 Reactor system Core thermal power Number of pressure tubes Core radius Core length Primary system Pressure Inlet coolant temperature

Units

Data

MWth m m

2000 740 8.5 6

MPa °C

6.89 245

Units

Data

mm kg/hole

12.8 2.3

Coolant Hole Coolant hole diameter Coolant flow rate

mm kg/s

14.8 1.4

Unit cell pitch

27.5

Fuel Hole Fuel hole diameter Mass of UC

TABLE 14.6 Property Data for Problem 14.7 Parameters Thermal conductivity, k (W/m‧K) Dynamic viscosity, μ (Pa s) Specific heat, cp (J/kg‧K) Coolant inlet sp. enthalpy, hin (kJ/kg) Single phase density, ρ (kg/m3)

Fuel 23

Graphite 50

Coolant 0.59 101 × 10−6 5.0 × 103 1062.3 776.3

Questions 1. What is the radial peaking factor assuming an axial cosine and radial Bessel function flux shape (neglect extrapolation length)? For the following questions, assume that the total power of the fuel hole is 260 (kW/fuel hole) in the hot channel. 2. What is the coolant outlet temperature in the hot channel? 3. What is the coolant outlet enthalpy in the hot channel? 4. What is the outlet void fraction in the hot channel? 5. What is the nonboiling length in the hot channel? 6. What is the fuel centerline temperature at the position where bulk boiling starts in the hot channel 7. What is the pressure drop in the hot channel? To simplify you calculation, assume a uniform heat flux value that provides total power equivalent to the cosine shape heat flux distribution. Answers: 1. PR = 2.32 2. Tm,out = 284.8°C 3. hout =1433.7 kJ/kg 4. αout = 0.726

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5. ZOSB = 3.14 m 6. T CL (Z OSB ) = 599.0°C 7. ΔP = 542 kPa

PROBLEM 14.7 SOLUTION Thermal Hydraulic Analysis of a Pressure Tube Reactor (Chapters 3, 8, 9, 10, 11, and Section 14.6) The operating conditions and the thermodynamic data for the saturated water coolant given by the problem statement are: • pressure, P = 6.89 MPa • length, L = 6 m • density, ρf = 742.0 kg/m3, ρg = 35.94 kg/m3, ρl = 776.3 kg/m3 • specific enthalpy, hf = 1261.6 kJ/kg, hfg = 1511.9 kJ/kg, hin = 1062.3 kJ/kg • inlet temperature, Tin = 245ºC • liquid specific heat, cp,l = 5.0 × 103 J/kg K • fuel hole diameter, Df = 0.0128 m • coolant hole diameter, Dc = 0.0148 m • unit cell pitch, P = 0.0275 m • fuel conductivity, kf = 23 W/m K • graphite conductivity, kg = 50 W/m K • liquid conductivity, kl = 0.59 W/m K • liquid viscosity, µl = 101×10−6 Pa s 1. What is the radial peaking factor assuming an axial cosine and radial Bessel function flux shape (neglect extrapolation length)? The radial peaking factor is given in Table 3.6 and may be verified by integrating the Bessel function. PR = 2.32

(1)

2. What is the coolant outlet temperature in the hot channel? The fuel hole power in the hot channel is given by the problem statement: q˙ = 260 kW

(2)

The mass flow rate per coolant hole is: ṁCH = 1.4 kg/s

(3)

The Equivalent Annuli Model is applied using the triangle-shaped unit cell shown in Figure 14.16. It is geometrically evident that each unit cell covers one fuel hole and half a coolant hole. Therefore, the mass flow rate of the unit cell is half of the mass flow rate of the coolant hole.

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This is also justified by the fact that a pressure tube contains 24 fuel holes and 12 coolant holes, which means 0.5 coolant holes for each fuel hole. Therefore, in the Equivalent Annuli Model illustrated in Figure 14.17, the mass flow rate is:

(4) To find the coolant exit temperature it is convenient to first find the coolant exit enthalpy, which then together with the coolant pressure determines the exit temperature. Hence, (5) hout together with P = 6.89 MPa yields (6) Note that Tout is at the saturation condition indicating that the exit coolant is in an equilibrium two-phase state. 3. What is the coolant outlet enthalpy in the hot channel? (5) 4. What is the coolant outlet void fraction in the hot channel? The exit equilibrium quality is: ℎ𝑜𝑜𝑜𝑜𝑜𝑜 − ℎ𝑓𝑓 1433.7 − 1261.6 = = 0.1138 𝑥𝑥𝑒𝑒,𝑜𝑜𝑜𝑜𝑜𝑜 = (7) ℎ𝑓𝑓𝑓𝑓 1511.9 Assuming HEM, the exit void fraction is: 𝛼𝛼𝑜𝑜𝑜𝑜𝑜𝑜 =

1 1 = = 0.726 𝜌𝜌 1 − 𝑥𝑥𝑐𝑐,𝑜𝑜𝑜𝑜𝑜𝑜 1 − 0.1138 35.94 𝑔𝑔 1 + � � � � �� 1 + � 𝑥𝑥 0.1138 742 𝜌𝜌𝑓𝑓 𝑐𝑐,𝑜𝑜𝑜𝑜𝑜𝑜

(8)

5. What is the nonboiling length in the hot channel? The OSB location may be calculated using Equation 14.31b: ℎ𝑓𝑓 − ℎ𝑖𝑖𝑖𝑖 𝐿𝐿 −1 6 1261.6 − 1062.3 �� sin �−1 + 2 � �� = sin−1 �−1 + 2 � 𝜋𝜋 ℎ𝑜𝑜𝑜𝑜𝑜𝑜 − ℎ𝑖𝑖𝑖𝑖 𝜋𝜋 1433.7 − 1062.3 (9) = 0.140 m

𝑧𝑧𝑂𝑂𝑂𝑂𝑂𝑂 =

Calculating the OSB location from the inlet instead of the midplane yields the nonboiling length in the hot channel: 𝐿𝐿 6 (10) 𝑍𝑍𝑂𝑂𝑂𝑂𝑂𝑂 = 𝑧𝑧𝑂𝑂𝑂𝑂𝑂𝑂 + = 0.140 + = 3.14 m 2 2

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6. What is the fuel centerline temperature at the position where bulk boiling starts in the hot channel? The power generated by one unit cell in the Equivalent Annuli Model represented in Figure 14.17 is: (11) 𝑞𝑞̇ = 260 kW This power follows a cosine axial distribution. The peak linear heat generation rate is: 𝑞𝑞0′ =

𝑞𝑞̇ 𝜋𝜋 260 𝜋𝜋 � �= � � = 68.07 kW/m 𝐿𝐿 2 6 2

(12)

Now, let us evaluate the area occupied by each region in the unit cell. 𝜋𝜋𝐷𝐷𝑓𝑓2 𝜋𝜋0.01282 = 1.287 × 10−4 m2 𝐴𝐴𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 = 4 4

𝜋𝜋𝐷𝐷𝑐𝑐2 𝜋𝜋0.01482 𝐴𝐴𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = 0.5 = = 8.602 × 10−5 m2 4 4

𝑃𝑃2 √3 𝐴𝐴𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔ℎ𝑖𝑖𝑖𝑖𝑖𝑖 = − 𝐴𝐴𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 − 𝐴𝐴𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 4

0.02752 √3 = − 1.29 × 10−4 − 8.60 × 10−5 4 = 1.127 × 10−4 m2

(13) (14)

(15)

In the Equivalent Annuli Model of Figure 14.17, the central fuel pin is surrounded by an annulus of graphite, which is surrounded by the coolant. The inner and outer radii of the equivalent graphite annulus are respectively: 𝑅𝑅𝑒𝑒,𝑓𝑓 =

𝑅𝑅𝑒𝑒,𝑔𝑔 = �

𝐷𝐷𝑓𝑓 0.0128 = = 0.00640 m 2 2

(1.287 × 10−4 + 1.127 × 10−4 ) �𝐴𝐴𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 + 𝐴𝐴𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔ℎ𝑖𝑖𝑖𝑖𝑖𝑖 � =� 𝜋𝜋 𝜋𝜋

(16)

(17)

= 0.008766 m

The fuel centerline temperature is to be evaluated by considering the sum of the thermal resistance terms between the fuel centerline and the coolant. For the coolant resistance one could consider either that of the Equivalent Annuli Model or that of the simple circular coolant channel. Below we do both to determine the difference. First adopting the full equivalent annulus geometry: The equivalent diameter for the equivalent annular coolant region is:

The mass flux is:

𝐷𝐷𝑒𝑒 =

4𝐴𝐴𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 4(8.602 × 10−5 ) = = 0.006247 m 2𝜋𝜋𝑅𝑅𝑒𝑒,𝑔𝑔 2𝜋𝜋0.008766

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𝐺𝐺 =

The Reynolds number is:

The Prandtl number is:

𝑚𝑚̇

𝐴𝐴𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐

0.7 = 8138 kg⁄𝑚𝑚2 𝑠𝑠 8.602 × 10−5

𝐺𝐺𝐷𝐷𝑒𝑒 8138(0.006247) = = 5.033 × 105 𝜇𝜇𝑙𝑙 101 × 10−6

Re =

𝑐𝑐𝑝𝑝,𝑙𝑙 𝜇𝜇𝑙𝑙 (5.0 × 103 )(101 × 10−6 ) Pr = = = 0.8559 𝑘𝑘𝑙𝑙 0.59

(19)

(20)

(21)

The flow is turbulent, and the Dittus-Boelter/McAdams relation, Equation 10.91, may apply. Nu = 0.023Re0.8 Pr 0.4 = 0.023(5.033 × 105 )0.8 (0.8559)0.4 = 787.3 ℏ=

𝑘𝑘𝑙𝑙 Nu 0.59 × 787.3 kW = = 74.36 2 𝐷𝐷𝑒𝑒 0.006247 m K

𝑆𝑆 =

𝑅𝑅𝑒𝑒,𝑔𝑔 1 1 1 + In � �+ 4𝜋𝜋𝑘𝑘𝑓𝑓 2𝜋𝜋𝑘𝑘𝑔𝑔 𝑅𝑅𝑒𝑒,𝑓𝑓 2𝜋𝜋𝑅𝑅𝑒𝑒,𝑔𝑔 ℎ

(22) (23)

Let us evaluate the sum of the thermal resistance terms between the fuel centerline and the coolant in the Equivalent Annuli Model.

(24)

1 1 0.008766 1 �+ + In � 4𝜋𝜋7 2𝜋𝜋23 0.00640 2𝜋𝜋0.008766 × 74360

= 0.01137 + 0.002177 + 0.0002442 = 0.01379 m K/W

Equation 14.24 provides the fuel centerline temperature, which we calculate at the OSB position: (25) 𝐿𝐿 𝜋𝜋𝑧𝑧𝑂𝑂𝑂𝑂𝑂𝑂 𝜋𝜋𝑧𝑧𝑂𝑂𝑂𝑂𝑂𝑂 �sin � � + 1� + 𝑆𝑆 cos � �� 𝜋𝜋𝑚𝑚𝑐𝑐̇ 𝑝𝑝 𝐿𝐿 𝐿𝐿 6 𝜋𝜋0.140 𝜋𝜋0.140 � + 1� + 0.01379 cos � �� = 1221℃ = 245 + 68070 � �sin � 𝜋𝜋0.7(5000) 6 6 𝑇𝑇𝑐𝑐𝑐𝑐 (𝑧𝑧𝑂𝑂𝑂𝑂𝑂𝑂 ) = 𝑇𝑇𝑖𝑖𝑖𝑖 + 𝑞𝑞0′ �

= 245 + 68070 {0.0005 + 0.0047} = 599ºC

Second, considering the cylindrical coolant channel geometry: Re = �

0.0148 � = 5.033 × 105 = 1.19 × 106 0.006247

1.19 × 106 � 787.3 = 1566.7 5.033 × 105 0.59(1566.7) ℏ= = 62.46 kW⁄m2 K 0.0148

𝑁𝑁𝑢𝑢 = �

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74360 � 0.0002 = 0.0047 𝑆𝑆 = 0.0035 + 0.0010 + � 63460

Hence, since the value of S is unchanged for this geometry from the full equivalent annulus geometry, the value of the fuel centerline temperature per Equation (25) remains the same as above for the full equivalent annulus geometry at 599ºC. 7. What is the pressure drop in the hot channel? To simplify your calculation assume a uniform heat flux value that provides total power equivalent to the cosine shape heat flux distribution. The constant linear heat generation rate is: 𝑞𝑞̇ 260 = = 43.3 kW/m 𝐿𝐿 6

𝑞𝑞 ′ =

(26)

The OSB location may be calculated applying a simple energy balance (Equation 14.29), where Z, written in capital letter, is measured from the core inlet. 𝑍𝑍𝑂𝑂𝑂𝑂𝑂𝑂

𝑍𝑍𝑂𝑂𝑂𝑂𝑂𝑂 = The enthalpy distribution is: ℎ(𝑍𝑍) = ℎ𝑖𝑖𝑖𝑖 +

𝑞𝑞 ′ = ℎ𝑓𝑓 − ℎ𝑖𝑖𝑖𝑖 𝑚𝑚̇

(27)

43.3 = 1261.6 − 1062.3 0.7

(28)

⟹ 𝑍𝑍𝑂𝑂𝑂𝑂𝑂𝑂 = 3.22 m

𝑞𝑞 ′ 43.3 𝑍𝑍 = 1062.3 + 𝑍𝑍 = 1062.3 + 61.86𝑍𝑍 kJ/kg 𝑚𝑚̇ 0.7

(29)

The equilibrium quality distribution is: 𝑥𝑥𝑒𝑒 (𝑍𝑍)

ℎ(𝑍𝑍) − ℎ𝑓𝑓 1062.3 + 61.86𝑍𝑍 − 1261.6 = ℎ𝑓𝑓𝑓𝑓 1511.9 = −0.1318 + 0.04091𝑍𝑍

(30)

In the region above ZOSB the void fraction profile is determined, considering HEM, as: 𝛼𝛼(𝑍𝑍) =

Simplifying:

1 1 − 𝑥𝑥𝑒𝑒 (𝑍𝑍) 𝜌𝜌𝑔𝑔 1+ 𝑥𝑥𝑒𝑒 (𝑍𝑍) 𝜌𝜌𝑓𝑓 = �1 +

[1 − (−0.1318 + 0.04091𝑍𝑍)] 35.94 � −0.1318 + 0.04091𝑍𝑍 742.0

𝛼𝛼(𝑍𝑍) =

199.3 − 61.86𝑍𝑍 116.4 − 58.86𝑍𝑍 387

−1

(31)

(32)

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In the region above ZOSB the mixture density is: 𝜌𝜌𝑚𝑚 (𝑍𝑍) = 𝜌𝜌𝑓𝑓 �1 − 𝛼𝛼(𝑍𝑍)� + 𝜌𝜌𝑔𝑔 𝛼𝛼(𝑍𝑍)

(33)

199.3 − 61.86𝑍𝑍 199.3 − 61.86𝑍𝑍 � + 35.94 = 742.0 �1 − kg/m3 116.4 − 58.86𝑍𝑍 116.4 − 58.86𝑍𝑍

Simplifying and approximating:

𝜌𝜌𝑚𝑚 (𝑍𝑍) =

At the channel outlet:

𝜌𝜌𝑚𝑚,𝑜𝑜𝑜𝑜𝑜𝑜 = 𝜌𝜌𝑚𝑚 (𝐿𝐿) =

−54349 kg/m3 116.4 − 58.86𝑍𝑍

(34)

−54349 = 229.3 kg/m3 116.4 − 58.86(6)

(35)

Acceleration pressure drop The acceleration pressure drop is given by Equation 14.38a: ∆𝑃𝑃𝑎𝑎𝑎𝑎𝑎𝑎 = 𝐺𝐺 2 �

𝜌𝜌𝑚𝑚,𝑜𝑜𝑜𝑜𝑜𝑜

−

1 1 1 � = 203.6 kPa � = 81382 � − 𝜌𝜌𝑙𝑙 229.3 776.0

(36)

Gravity pressure drop The gravity pressure drop is given by Equation 14.38b: 𝐿𝐿

𝑍𝑍𝑂𝑂𝑂𝑂𝑂𝑂

3.22

∆𝑃𝑃𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 = 𝜌𝜌𝑙𝑙 𝑔𝑔𝑍𝑍𝑂𝑂𝑂𝑂𝑂𝑂 + � 𝜌𝜌𝑚𝑚 (𝑍𝑍)𝑔𝑔𝑔𝑔𝑔𝑔 = 776.3(9.81)3.22 + �

Integrating numerically or analytically and solving:

−54349 9.81𝑑𝑑𝑑𝑑 116.4 − 58.86𝑍𝑍

∆𝑃𝑃𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 = 35.2 kPa

(37)

(38)

Friction pressure drop For calculating the friction pressure drop we have to abandon the Equivalent Annuli Model and refer to the original coolant hole, whose diameter is Dc. The liquid only Reynolds number is: Reℓ𝑜𝑜 =

𝐺𝐺𝐷𝐷c 8138(0.0148) = = 1.192 × 106 𝜇𝜇ℓ 101 × 10−6

(39)

The flow is turbulent, so the McAdams relation, Equation 9.87, may be applied to calculate the liquid-only friction factor: 𝑓𝑓ℓ𝑜𝑜 = 0.184Re−0.20 = 0.184(1.192 × 106 )−0.20 = 0.0112 ℓ𝑜𝑜

(40)

The HEM two-phase friction multiplier can be expressed using Equation 11.82. 𝜌𝜌𝑓𝑓 742.0 2 (𝑍𝑍) = 1 + 𝑥𝑥𝑒𝑒 (𝑍𝑍) � − 1� = 1 + 𝑥𝑥𝑒𝑒 (𝑍𝑍) � Φℓ𝑜𝑜 − 1� = 1 + 19.65𝑥𝑥𝑒𝑒 (𝑍𝑍) 𝜌𝜌𝑔𝑔 35.94

(41)

= 1 + 19.65(−0.1318 + 0.04091𝑍𝑍) = −1.1590 + 0.504𝑍𝑍

We may express the friction pressure drop re-arranging Equation 14.47.

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𝑓𝑓ℓ𝑜𝑜 𝐺𝐺 2 𝑓𝑓ℓ𝑜𝑜 𝐺𝐺 2 2 (𝑍𝑍)𝑑𝑑𝑑𝑑 ∆𝑃𝑃𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 = 𝑍𝑍𝑂𝑂𝑂𝑂𝑂𝑂 + � Φ𝑙𝑙𝑙𝑙 2𝐷𝐷c 𝜌𝜌𝑙𝑙 2𝐷𝐷c 𝜌𝜌𝑓𝑓

(42)

𝑍𝑍𝑂𝑂𝑂𝑂𝑂𝑂 6

0.0112(81382 ) 0.0112(81382 ) = 3.22 + � (−1.590 + 0.804𝑍𝑍)𝑑𝑑𝑑𝑑 2(0.0148)776.3 2(0.0148)742.0 3.22

Integrating numerically or analytically and solving:

∆𝑃𝑃𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 = 302.8 kPa

(43)

∆𝑃𝑃𝑡𝑡𝑡𝑡𝑡𝑡 = ∆𝑃𝑃𝑎𝑎𝑎𝑎𝑎𝑎 + ∆𝑃𝑃𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 + ∆𝑃𝑃𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 = 203.6 + 35.2 + 302.8 = 542 kPa

(44)

Total pressure drop

PROBLEM 14.8 QUESTION Maximum Cladding Temperature for a SFBR (Chapters 3, 10, and Section 14.6) Derive the relationship between the physical and extrapolated axial lengths for a SFBR core such that the maximum clad temperature occurs at the core outlet during steady-state operating conditions. This relationship describes the truncation of the assumed sinusoidal thermal flux variation along the core axis. Ignore the reactor blankets and assume the following remain constant along the axial length of the core: –

Heated perimeter of channels

– Mass flux of coolant – Coolant specific heat –

Film heat transfer coefficient

Answer: 𝐿𝐿 ≤

𝐿𝐿𝑒𝑒 ℎ𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝑃𝑃ℎ 2𝐿𝐿𝑒𝑒 tan−1 � � 𝜋𝜋 𝜋𝜋𝑚𝑚̇𝑐𝑐𝑝𝑝

PROBLEM 14.8 SOLUTION

Maximum Cladding Temperature for a SFBR (Chapters 3, 10, and Section 14.6) As described in Section 14.5.1.2, the location of maximum cladding surface temperature may be determined by taking the derivative of Equation 14.19, and finding the maximum of that

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function. This leads to Equation 14.22b: 𝑧𝑧𝑐𝑐 =

Where the heated perimeter is:

2𝜋𝜋𝑅𝑅𝑐𝑐𝑐𝑐 𝐿𝐿𝑒𝑒 ℎ𝑓𝑓𝑓𝑓𝑙𝑙𝑙𝑙 𝐿𝐿𝑒𝑒 tan−1 � � 𝜋𝜋 𝜋𝜋𝑚𝑚̇𝑐𝑐𝑝𝑝 𝑃𝑃ℎ = 2𝜋𝜋𝑅𝑅𝑐𝑐𝑐𝑐

Hence: 𝑧𝑧𝑐𝑐 =

𝐿𝐿𝑒𝑒 ℎ𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝑃𝑃ℎ 𝐿𝐿𝑒𝑒 tan−1 � � 𝜋𝜋 𝜋𝜋𝑚𝑚̇𝑐𝑐𝑝𝑝

(1)

(2)

(3)

The maximum cladding temperature occurs at the core outlet when: 𝑧𝑧𝑐𝑐 ≥

Merging those two equations and re-arranging:

𝐿𝐿 2

𝐿𝐿𝑒𝑒 ℎ𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝑃𝑃ℎ 𝐿𝐿𝑒𝑒 𝐿𝐿 tan−1 � �≥ 𝜋𝜋 𝜋𝜋𝑚𝑚̇𝑐𝑐𝑝𝑝 2 𝐿𝐿 ≤

𝐿𝐿𝑒𝑒 ℎ𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝑃𝑃ℎ 2𝐿𝐿𝑒𝑒 tan−1 � � 𝜋𝜋 𝜋𝜋𝑚𝑚̇𝑐𝑐𝑝𝑝

(4)

(5)

(6)

PROBLEM 14.9 QUESTION Flow-Levitated Control Rod in a PWR (Chapters 8, 9, 10, and Section 14.5) An engineer has recently proposed a novel control rod design based on flow levitation, to be used in PWRs. The control rod consists of a slug of absorbing material (boron carbide, B4C) that can slide within a guide tube (see Figure 14.18). During normal operating conditions, the control rod is held out of the core by the coolant flow. If the coolant flow suddenly decreases, the control rod falls back into the core. The idea is to create a scram system that would automatically (passively) shut down the reactor in case of loss of flow and loss of coolant accidents. The control rod slug diameter is D = 2 cm and its length is L = 3.7 m. The guide tube diameter is Do = 2.6 cm. The properties of B4C and coolant are reported in Table 14.7. 1. Calculate the minimum flow rate through the guide tube required for control rod levitation. Assume turbulent flow in the annulus between the guide tube and the control rod slug. Use a friction factor equal to 0.017, and form loss coefficients equal to 0.25 and 1.0 for the flow contraction and expansion at the bottom and top of the control rod slug, respectively. Ignore the presence of the stop. Assume steady state. 2 Now consider the operation of this control rod at reduced flow conditions, when it is fully

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inserted in the core. In this situation, the control rod is exposed to a high neutron flux, and so significant heat is generated due to (n, α) reactions on the B4C. Because the absorption cross section of B4C is so high, the neutrons only penetrate a few microns beneath the control rod surface. In fact, for all practical purposes, we can assume that the volumetric heat generation rate within the control rod is zero, and describe the situation simply by means of a surface heat flux. Calculate the maximum temperature in the control rod, assuming that the heat flux at its surface has a cosine axial shape with an average of 80 kW/m2. At the conditions of interest, here the coolant flow rate, inlet temperature, and pressure are 0.3 kg/s, 284 °C, and 15.5 MPa, respectively. Answers: 1. ṁ = 0.54 kg/s 2. To,max = 298.3°C

Figure 14.18 The flow-levitated control rod system (there is a mechanical stop to ensure that the control rod is not ejected by the flow).

TABLE 14.7 Property Data for Problem 14.9 Properties of Liquid Watera Density Viscosity Specific heat Thermal conductivity Properties of Boron Carbide

730 kg/m3 9 × 10−5Pa s 5500 J/kgK 0.56 W/mK

Density

2500 kg/m3

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Specific heat Thermal conductivity a

950 J/kgK 35 W/mK

Assumed constant over the temperature and pressure range of interest.

PROBLEM 14.9 SOLUTION Flow-Levitated Control Rod in a PWR (Chapters 8, 9, 10, and Section 14.5) The following parameters are given by the problem statement. – control rod slug diameter, D = 0.02 m – control rod slug length, L = 3.7 m – – – – – – – –

guide tube inner diameter, D0 = 0.026 m coolant density, ρl = 730 kg/m3 coolant viscosity, µl = 9×10−5 Pa s coolant specific heat, cp,l = 5500 J/kg K coolant conductivity, kl = 0.56 W/m K control rod slug density, ρCR = 2500 kg/m3 control rod slug specific heat, cp,CR = 950 J/kg K control rod slug conductivity, kCR = 35 W/m K

1. Calculate the minimum flow rate through the guide tube required for control rod levitation. The minimum flow rate through the guide tube required for control rod levitation is given by a balance between the four forces acting on the control rod slug, i.e., rod weight, pressure on the top and bottom faces of the rod, shear stress due to flow friction. The force balance is as follows:

ρCR

D 2 Lg +

D 2 Pout =

D 2 Pin + τ πDL

(1)

where Pout is the pressure immediately above the slug, Pin is the pressure immediately before the slug, and τ is the shear stress on the lateral surface of the slug, which can be calculated from knowledge of the friction factor, f, as:

τ=

f G2 8 ρ

which is given as Equation 9.6

(2)

where the problem statement also gives f=0.017. G is the (unknown) mass flux in the annulus between the control rod and guide tube. Substituting Equation (2) into Equation (1) and rearranging the various terms, we get: L G2 (3) ρ CR Lg = Pin − Pout + f D 2ρ

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The pressure difference Pin - Pout can be found form the momentum equation for the coolant in the annulus: (4) where it was assumed that the acceleration pressure change is negligible because the fluid is incompressible and there is no net change in flow area. Now De=Do-D=0.6 cm is the annulus equivalent diameter, and again from the problem statement Kin=0.25 and Kout=1.0 are the form loss coefficient for the annulus entrance and exit, respectively. Substituting Equation (4) into Equation (3) and solving for G, we get: (5)

Then the mass flow rate in the guide tube is m = G

( Do2 − D 2 ) ≈ 0.54 kg/s.

4 Note that the Reynolds number in the annulus is Re=GDe/µ≈167,000, therefore the assumption of turbulent flow is accurate. Finally, it is interesting to note that friction contributes to levitation of the control rod slug in two ways, i.e., it creates the shear stress applied to the lateral area of the control rod, and it decreases the pressure Pout applied to the top face of the control rod. This explains why the friction factor is present twice in Equation (5). 2. Calculate the maximum temperature in the control rod. In this section of the problem the following coolant properties are used: Tin = 284ºC P = 15.5 MPa ṁ = 0.3 kg/s Because the volumetric heat generation rate is zero, the temperature is constant along the radial coordinate within the control rod. That is, the solution of the heat conduction equation within the control rod is T(r)=To, where To is the temperature at the surface of the control rod. Because the heat flux and the coolant bulk temperature vary along the axial direction, To will also vary axially, and must be found from knowledge of the surface heat flux distribution and heat transfer coefficient in the annulus. The mass flux is: (6) The equivalent diameter, as given in Part 1 is: De = Do - D = 0.6 m

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The inlet enthalpy and the saturated liquid enthalpy are: (8) (9) The exit enthalpy is given by the following energy balance: ℎ𝑒𝑒𝑒𝑒 = ℎ𝑖𝑖𝑖𝑖 +

𝑞𝑞�″𝐿𝐿𝐿𝐿𝐿𝐿 𝑚𝑚̇

= 1253 +

80(3.7)𝜋𝜋(0.02) 0.3

= 1315 kJ/kg

(10)

The exit enthalpy is lower than hf. Therefore, the coolant is single phase. The Reynolds number is: (11) The Prandtl number is: (12) The flow is turbulent, and the Dittus-Boelter/McAdams relation, Equation 10.91, may be used to obtain the hat transfer coefficient since entry region effects can be neglected. Nu = 0.023Re0.8Pr0.4 = 0.023(9.22×104)0.8(0.8839)0.4 ≈ 205

= h

kl Nu 0.56(205) 19.1 kW m 2 K = = −3 6 ×10 De

(13) (14)

As stated in the problem question the heat flux at the control rod surface has a cosine axial shape where:

q′′( z ) =

′′ cos(π z / L) qav

′′ =80 kW/m2 and the origin of the axial coordinate z was taken at the control rod with qav midplane. Therefore, the linear power profile is: π 2D ′′ cos(π z / L) = qo′ cos(π z / L) q ' ( z ) = q′′( z )πD = qav (15) 2 π 2D ′′ =7.89 kW/m is the linear power at the midplane. Set in these terms, the qav where qo′ ≡ 2 problem is very similar to the calculation of the maximum outer temperature of the cladding in a fuel rod. The solution to this problem is in Section 14.5 of Chapter 14. Therefore, the location at which the surface temperature has a maximum, zmax, can be found directly from Equation 14.22b:

zmax =

 DLh  tan −1   π  m c p  L

(16)

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And the maximum surface temperature To,max can be found substituting Equation (16) into Equation 14.19 (taking the extrapolation length Le equal to the physical length of the control rod, L): (17)

where Tin is 284°C.

PROBLEM 14.10 QUESTION Thermal Behavior of a Plate Fuel Element Following a Loss of Coolant (Chapters 8, 10, and Section 14.6) A reactor fuel assembly of the MIT research reactor is made up of plate elements as shown in Figure 14.19 (only 5 of 13 elements are shown). Suppose the flow channel between plates 2 and 3 is blocked at the inlet (Figure 14.20). What is the axial location of the maximum fuel temperature in plate 3? Solve this in the following steps: (Steps A and B can be solved independently of each other). A. Find Tw(z) where Tw, is the element 3 surface temperature on the cooled side (RHS). B. Find TFuelLHS(z)−Tw(z) where TFuelLHS(z) is the element 3 surface temperature on the insulated side (LHS). C. Find the axial location of the maximum TFuelLHS(z). In solving this problem, you can make the following assumptions: – All heat transfer through the fuel element is radial, that is, there is no axial heat transfer within the fuel element. – All of the energy generated in plate 3 flows radially to the right to the coolant channel between elements 3 and 4, that is, the left side of element 3 has an insulated boundary (see Figure 14.21). – For simplicity, we neglect the clad and take the elements as only composed of a metallic fuel. – Assume the flow is fully developed. Operating Conditions: –

pressure, P = 55 psi (0.379 MPa)

–

inlet temperature, Tin = 123.8 F (51 °C)

–

mass flow rate between plates 3 and 4, ṁ = 0.32 kg/s

–

peak volumetric heat generation rate, 𝑞𝑞0‴ = (8.54 × 105 )cos � 𝐿𝐿 � kW/m3

𝜋𝜋𝜋𝜋

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Geometry: –

axial heated length, L = 23 in (58.42 cm)

–

channel width, s = 0.098 in (0.249 cm)

–

plate width, t = 0.030 in (0.0762 cm)

–

plate depth, w = 2.082 in (5.288 cm)

Properties: –

coolant specific heat, cpι = 4.181 kJ/kg K

–

coolant conductivity, kι = 0.644 W / m K

–

coolant viscosity, μι = 544×10−6 Pa s

–

coolant density, ρι = 987.2 kg/m3

–

Prandtl number, Pr = 3.597

–

fuel conductivity, kfuel = 41.2 W/m K

Answers: 3q‴ wtL

πz

0 �sin � L � + 1� + A. Tw (z) = Tin + 2πm ̇ c ,ι p

q‴(z)t2

B. TFuelLHS (z) − Tw (z) = 2k

C. z = 2.7 cm

πz L

tq‴ 0 cos h

fuel

FIGURE 14.19 MIT research reactor fuel plate elements (only 5 of 13 elements are shown).

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FIGURE 14.20 Fuel plate elements with flow blockage between 2 and 3.

FIGURE 14.21 Geometry of fuel plate elements.

PROBLEM 14.10 SOLUTION Thermal Behavior of a Plate Fuel Element Following a Loss of Coolant (Chapters 8, 10, and Section 14.6) The following parameters are given by the problem statement: – – – – – – – – – –

inlet temperature, Tin = 51°C mass flow rate between plates 3 and 4, ṁ = 0.32 kg/s peak volumetric heat generation rate, 𝑞𝑞0‴ = (8.54 × 108 ) W/m3 axial heated length, L = 0.5842 m channel width, s = 0.00249 m plate width, t = 0.000762 m plate depth, w = 0.05288 m coolant specific heat, cp,ι = 4181 J/kg K coolant conductivity, kι = 0.644 W/m K coolant viscosity, μι = 544×10−6 Pa s

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– – –

coolant density, ρι = 987.2 kg/m3 Prandtl number, Pr = 3.597 fuel conductivity, kfuel = 41.2 W/m K

A. Find Tw(z), where Tw is the element 3 surface temperature on the cooled side (RHS). The volumetric heat generation rate is radially constant and axially variable with a cosine distribution. 𝜋𝜋𝜋𝜋 𝑞𝑞 ″ (𝑧𝑧) = 𝑞𝑞0‴ cos � � (1) 𝐿𝐿 The heat produced by a slice of fuel of thickness dz around the axial position z satisfies the following balance. 𝑞𝑞 ″ (𝑧𝑧)𝑤𝑤 𝑑𝑑𝑑𝑑 = 𝑞𝑞 ‴ (𝑧𝑧) 𝑡𝑡 𝑤𝑤 𝑑𝑑𝑑𝑑

(2)

𝑞𝑞 ″ (𝑧𝑧) = 𝑞𝑞 ‴ (𝑧𝑧) 𝑡𝑡 = 𝑡𝑡𝑡𝑡0‴ cos �

(3)

Therefore, the axial heat flux distribution is:

𝜋𝜋𝜋𝜋 � 𝐿𝐿

Let us assume, for simplicity, that the coolant channel between elements 3 and 4 removes all the heat generated by element 3 and half of the heat generated by element 4. Therefore, for that coolant channel, the peak linear heat generation rate is: 3 3 ′ (4) 𝑞𝑞0,𝑐𝑐ℎ34 = 𝑤𝑤𝑞𝑞0″ (𝑧𝑧) 𝑤𝑤𝑞𝑞0‴ 2 2

The coolant bulk temperature, for the channel between elements 3 and 4, is given by Equation 14.14: ′ 𝑞𝑞0,𝑐𝑐ℎ34 𝜋𝜋𝜋𝜋 3𝑤𝑤𝑞𝑞0‴ 𝐿𝐿 𝜋𝜋𝜋𝜋 𝑇𝑇𝑚𝑚 (𝑧𝑧) = 𝑇𝑇𝑖𝑖𝑖𝑖 �1 + sin � �� = 𝑇𝑇𝑖𝑖𝑖𝑖 + �1 + sin � �� (5) 𝑚𝑚̇𝑐𝑐𝑝𝑝 , 𝜄𝜄𝜄𝜄 𝐿𝐿 2𝑚𝑚̇𝑐𝑐𝑝𝑝 , 𝜄𝜄𝜄𝜄 𝐿𝐿 The equivalent diameter of the coolant channel, taking into account the heated perimeter, is:

The mass flux is:

𝐷𝐷𝑒𝑒 = 𝐺𝐺 =

4𝑤𝑤𝑤𝑤 = 2𝑠𝑠 = 2(0.00249) = 0.00498 m 2𝑤𝑤

𝑚𝑚̇ 0.32 = = 2430 kg/m2 s 𝑤𝑤𝑤𝑤 0.05288(0.00249)

(6)

(7)

The Reynolds number for that coolant channel is: Re =

𝐺𝐺𝐷𝐷𝑒𝑒 2430(0.00498) = = 2.225 × 104 𝜇𝜇𝜇𝜇 544 × 10−6

(8)

The flow is turbulent, and the Dittus-Boelter/McAdams relation, Equation 10.91, may be used to calculate the heat transfer coefficient. Nu = 0.023Re0.8 Pr 0.4 = 0.023(2.225 × 104 )0.8 (3.597)0.4 = 115.3 ℏ=

𝑘𝑘𝑘𝑘Nu 0.644(115.3) W = = 14.91 × 103 2 𝐷𝐷𝑒𝑒 0.00498 m K 398

(9) (10)

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Chapter 14 - Single Heated Channel: Steady-State Analysis

The wall temperature is calculated rearranging Equation 14.17: 𝑇𝑇𝑤𝑤 (𝑧𝑧) = 𝑇𝑇𝑚𝑚 (𝑧𝑧)

𝑞𝑞 ″(𝑧𝑧) ℎ

𝜋𝜋𝜋𝜋 𝑡𝑡𝑡𝑡0‴ cos � � 3𝑞𝑞0‴ 𝑤𝑤𝑤𝑤𝑤𝑤 𝜋𝜋𝜋𝜋 𝐿𝐿 �1 + sin � �� + = 𝑇𝑇𝑖𝑖𝑖𝑖 + 𝑚𝑚̇𝑐𝑐𝑝𝑝 , 𝜄𝜄𝜄𝜄 𝐿𝐿 ℎ

(11)

B. Find TFuelLHS(z) – Tw(z), where TFuelLHS(z) is the element 3 surface temperature on the insulated side (LHS). The volumetric heat generation rate q‴(z) is radially constant, so the heat transfer through plate 3 may be calculated integrating Equation 8.30 as follows. 𝑘𝑘

𝜕𝜕𝜕𝜕(𝑧𝑧, 𝑥𝑥) + 𝑞𝑞 ‴ (𝑧𝑧)𝑥𝑥 = 𝐶𝐶1 𝑑𝑑𝑑𝑑

(12)

where C1 is a constant that can be determined applying a boundary condition. In this case, the boundary condition is zero heat flux on the left side of the plate, at x = 0. 𝑘𝑘

𝜕𝜕𝜕𝜕(𝑧𝑧, 𝑥𝑥) |𝑥𝑥 = 0 = 0 ⇒ 𝐶𝐶1 = 0 𝜕𝜕𝜕𝜕

(13)

The heat transfer equation is therefore:

𝜕𝜕𝜕𝜕(𝑧𝑧, 𝑥𝑥) + 𝑞𝑞 ‴ (𝑧𝑧)𝑥𝑥 = 0 𝜕𝜕𝜕𝜕

𝑘𝑘

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Integrating between the left and the right of plate 3, and assuming constant fuel conductivity: 𝑘𝑘𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 �𝑇𝑇𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹 (𝑧𝑧) − 𝑇𝑇𝑤𝑤 (𝑧𝑧)� = 𝑞𝑞 ‴ (𝑧𝑧) 𝑇𝑇𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹 (𝑧𝑧) − 𝑇𝑇𝑤𝑤 (𝑧𝑧) = 𝑞𝑞 ‴ (𝑧𝑧)

𝑡𝑡 2 2

𝑡𝑡 2 2𝑘𝑘𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓

(15)

(16)

C. Find the axial location of the maximum TFuelLHS(z). Combining the results of part A and part B 𝑇𝑇𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹 (𝑧𝑧) = 𝑇𝑇𝑤𝑤 (𝑧𝑧) + 𝑞𝑞 ‴ (𝑧𝑧)

𝑡𝑡 2 2𝑘𝑘𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓

𝜋𝜋𝜋𝜋 𝑡𝑡𝑡𝑡0‴ cos � 𝐿𝐿 � 3𝑞𝑞0‴ 𝑤𝑤𝑤𝑤𝑤𝑤 𝜋𝜋𝜋𝜋 𝜋𝜋𝜋𝜋 𝑡𝑡 2 ‴ = 𝑇𝑇𝑖𝑖𝑖𝑖 + �1 + sin � �� + + 𝑞𝑞0 cos � � ̇ 𝑐𝑐𝑝𝑝 , 𝜄𝜄𝜄𝜄 𝐿𝐿 ℎ 𝐿𝐿 2𝑘𝑘𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 2𝑚𝑚

= 51 +

(17)

3(8.54 × 10 )(0.05288)(0.000762)(0.5842) 𝜋𝜋𝜋𝜋 �1 + sin � �� 2(0.32)4181𝜋𝜋 0.5842

𝜋𝜋𝜋𝜋 2 � 0.5842 + 8.54 × 108 cos � 𝜋𝜋𝜋𝜋 � 0.000762 0.5842 2(41.2)

(0.000762)(8.54 × 108 ) cos � 14.91 × 103

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Chapter 14 - Single Heated Channel: Steady-State Analysis

= 58.17 + 49.65 cos(5.378𝑧𝑧) + 7.174 sin(5.378𝑧𝑧)

The plot of this function is shown in Figure SM-14.4:

FIGURE SM-14.4 LHS fuel temperature as a function of z

The location of the maximum wall temperature can be determined taking the first derivative of the axial distribution function and setting it to zero. 𝑑𝑑𝑑𝑑𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹 (𝑧𝑧) = 38.58 cos(5.378𝑧𝑧) − 267.0 sin(5.378𝑧𝑧) = 0 𝑑𝑑𝑑𝑑 38.58 cos(5.378𝑧𝑧) = 267.0 sin(5.378𝑧𝑧)

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tan(5.378𝑧𝑧) = 0.1445

Solving for z: 𝑧𝑧 =

1 arctan(0.1445) = 0.027 m 5.378

400

(19)

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