Oil and gas processing plant

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CHAPTER 1: MAJOR DESIGN MDEA Regenerator (T-302)

HUMAM AL-RIKABAWI 15564041

Table of Contents: 1.1

Introduction ................................................................................................................................ 1

1.1.1.

Literal Review, Objectives and Constraints ......................................................................... 1

1.1.2

Process Mechanism ............................................................................................................ 1

1.1.3

Process Description ............................................................................................................. 1

1.2

Process Design ............................................................................................................................ 3

1.2.1

Design Methodology ........................................................................................................... 3

1.2.2

Basis of Design (Material and Energy Balances) ................................................................. 4

1.2.2.1 Material Balance of Regenerator/Rich Amine Solution Rate Calculations ......................... 5 1.2.2.2 Thermal (Energy) Balance of Regenerator .......................................................................... 6 1.2.3

Process Design Calculations ................................................................................................ 9

1.2.3.1 Minimum Number of Stages, Nmin (Using Fenske Equation): ........................................... 11 1.2.3.2 Minimum Reflux Ratio, Rmin (Underwood Equation): ........................................................ 11 1.2.3.3 Theoretical Number of Stages (Erbar-Maddox correlation-graph): ................................. 12 1.2.3.4 Column Efficiency (Using O’connell correlation): ............................................................. 12 1.2.3.5 Feed Point Location (Using Kirkbride Equation): .............................................................. 13 1.2.3.6 Column Diameter Calculations: ........................................................................................ 14 1.2.3.7 Sieve-Plate Hydraulic Design: ........................................................................................... 18 1.2.3.8 Column Height .................................................................................................................. 25 1.2.4

Sizing/Specifications (Piping, Nozzles) .............................................................................. 25

1.3 Operational Design ......................................................................................................................... 26 1.3.1 Control System Design ............................................................................................................. 26 1.3.2 Operating Procedures, Start-Up and Shut-Down .................................................................... 27 1.3.3 Safety Study (HAZOP) on the Regenerator Column ................................................................. 28 1.4 Mechanical Design of Regenerator ................................................................................................. 29 1.4.1 Materials of Construction ........................................................................................................ 29 1.4.2 Pressure Vessel Design Calculations ........................................................................................ 29 1.4.2.1 Maximum Allowable Stress:.............................................................................................. 29 1.4.2.2 Design of Domed Head ..................................................................................................... 30 1.4.2.3 Minimum Wall Thickness of Vessel................................................................................... 31 1.4.2.4 Minimum Cylinder Shell Thickness ................................................................................... 31 1.4.2.5 Dead Weight of the Column (Shell) .................................................................................. 32 1.4.2.6 Total Weight of Plates ....................................................................................................... 32 1.4.2.7 Insulation Weight Calculations ......................................................................................... 32

1.4.2.8 Total Weight of Column (Shell + Plates + Insulation)........................................................ 33 1.4.2.9 Wind Loading Calculations ................................................................................................ 33 1.4.2.10 Stress Analysis ................................................................................................................. 33 1.4.2.11 Elastic Stability (Buckling) Check ..................................................................................... 34 1.4.2.12 Selection of Pressure Relief Valve ................................................................................... 35 1.4.3 Skirt Supports and Foundations Design ................................................................................... 36 1.4.4 Mechanical Drawing ................................................................................................................ 37 1.5 Summary and Review ..................................................................................................................... 38 1.5.1 Equipment Specification/Data Sheet ....................................................................................... 38 1.5.2 Critical Review .......................................................................................................................... 39 1.6 Nomenclature ................................................................................................................................. 40 1.7 References ...................................................................................................................................... 41

List of Figures: Figure 1 Process Flow Diagram of the Gas Amine Treating with the Regenerator to be designed (Chevron1, 2015) .................................................................................................................................... 2 Figure 2 Enthalpy Balance around the Regenerator (Kohl et al, 1997) .................................................. 4 Figure 3 Density of MDEA at different temperatures (Kidnay et al, 2011) ............................................. 5 Figure 4 the specific heat of MDEA at different temperatures (Kidnay et al, 2011) .............................. 6 Figure 5 Average heats of reaction of CO2 in Amine solutions (Kidnay et al, 2011) .............................. 7 Figure 6 latent heat of vaporization of water at 15, 50 psig (Kidnay et al, 2011) .................................. 7 Figure 7 Specific heat of MDEA at different temperatures (Kidnay et al, 2011) .................................... 8 Figure 8 Erbar-Maddox correlation to find theoretical stages (Sinnott et al, 2009) ............................ 12 Figure 9 Flooding velocity and the K1 value of sieve plates (Sinnott et al, 2009) ................................ 16 Figure 10 Relationship between downcomer area and weir length (Sinnott et al, 2009).................... 18 Figure 11 Weep point correlation (Sinnott et al, 2009) ........................................................................ 20 Figure 12 Discharge coefficient, Co and sieve plate (Sinnott et al, 2009) ............................................ 21 Figure 13 Entrainment correlation for sieve plates (Sinnott et al, 2009) ............................................. 24 Figure 14 Schematic drawing of the plate with the specification (Al-Rikabawi, 2015) ........................ 24 Figure 15 Typical maximum allowable stresses of steels (Sinnott et al, 2009) .................................... 30 Figure 16 Maximum Allowable Joint Efficiency (Sinnott et al, 2009) ................................................... 30 Figure 17 Ellipsoidal domed head of the regenerator (Al-Rikabawi, 2015) .......................................... 31 Figure 18 Wall thickness of vessels including 2 mm corrosion allowance (Sinnott et al, 2009) ........... 31 Figure 19 Upwind and Downwind forces on the regenerator (Al-Rikabawi, 2015) .............................. 34

List of Tables: Table 1 Regenerator Feed Composition ................................................................................................. 2 Table 2 Regenerated Lean Amine Composition ...................................................................................... 2 Table 3 Regenerator Operating Conditions ............................................................................................ 2 Table 4 Design Methodology of Amine Regenerator ............................................................................. 3 Table 5 Regenerator Conditions ............................................................................................................. 4 Table 6 Column feed composition .......................................................................................................... 4 Table 7 Regenerator operational conditions .......................................................................................... 9 Table 8 Feed components and compositions (From HYSYS) .................................................................. 9 Table 9 Process design conditions, k values and volatilities ................................................................. 10 Table 10 Plate specifications ................................................................................................................. 24 Table 11 Sizing of pipes going into the regenerator ............................................................................. 25

1.1Introduction 1.1.1. Literal Review, Objectives and Constraints Natural gas extracted from the well has a wide range of acid gas concentrations range from several parts per million to 50% by volume. The feed gas for the LPG plant comes from the Gippsland basin in Victoria, Australia and contains about 11% by weight of CO2. As the feed gas does not contain sulphur content, the only acid gas present in the natural gas is carbon dioxide. Sales gas is required to be sweetened to have a heating value of no less than 46 MJ/m3. The most widely used process to sweeten natural gas is absorption process using alkanol amines like methyl diethanolamine (MDEA) considering the low levels of partial pressure and concentration of acid gas. (Younger, 2004) The objective of amine treatment is to provide sales gas specification required by successfully removing huge percentage of acid gas from natural gas by an absorption stripping process. Considering some process conditions including low concentration of impurities in sour gas, hydrocarbon composition of sour gas, capital cost and operating costs, MDEA is finalized as the solvent for the acid gas removal in the process. The purpose of this report is to define a detail engineering design of MDEA regenerator column (T302) for the stripping of CO2 from absorber rich solution and to provide lean MDEA back to the absorber. The final design of the regenerator column will provide a design which meets all of the design objectives and constraints for the proposed plant. The final design shows a cost-effective solution to separate acid gas from the rich amine solution by keeping the total cost of equipment and operating costs to a minimum.

1.1.2 Process Mechanism The principle of absorption of CO2 by MDEA is governed by the following equations where R1R2R3N represents MDEA where R1=R2=CH2CH2OH and R3=CH3: CO2 + R1R2R3N + H2O ↔ R1R2R3NH+ + HCO3CO2 + OH- ↔ HCO3HCO3- + OH- ↔ CO3- + H2O R1R2R3NH+ + OH- ↔ R1R2R3N + H2O 2H2O ↔ OH- + H3O+ These are reversible reactions proceed to the right at low temperature and to the left at a higher temperature, thus making CO2 to be absorbed at ambient temperature inside the absorption column. In case of regenerator column, at higher temperature the reaction is reversed that is backward reaction is favoured where the carbonate salt formed is decomposed to release the acid gas absorbed. In order to reduce the release of the carbonate salt stringent control of stripper column temperature should be adopted. The first two reactions are slow reactions because carbon dioxide must form carbonic acid with water before reacting with amine. The other 3 reactions which predominate when MDEA is involved Is relatively fast. (Barreau et al, 2006)

1.1.3 Process Description The general process flow diagram of an amine-CO2 removal process is shown as part of the total project PFD in Figure 1. The sour gas enters the plant through a separator called Slug Catcher to remove any free liquid or entrainment solids. The sour gas then enters the bottom of the absorber (T-301) in counter current contact with aqueous lean amine solution. Sweet gas with low CO2 content leaves the top of the absorber and flows to Dehydration unit. The rich solvent containing CO2 from the bottom of the absorber is flashed in a separator (V-301) to recover some absorbed hydrocarbons in the rich solvent and then pass through amine heat exchanger to the regenerator where the acid gas absorbed is stripped off at a very high temperature and low pressure. 1

Figure 1 Process Flow Diagram of the Gas Amine Treating with the Regenerator to be designed (Chevron1, 2015)

The acid gas then leaves the top of the regenerator column. The lean amine from the bottom of the reboiler attached to the stripper flow through amine-amine heat exchanger (E-301) and through a chiller before being introduced back to the top of the absorber. As part of absorbed acid gas is flashed from the heated rich solution on the top tray. The remainder of the rich solution flows downward through the stripper in counter current contact with vapour generated in reboiler; which strips the acid gas from rich solution. The overhead products pass through condenser where the steam is condensed and cooled and returned to the top of the stripper as a reflux, while the acid gas is separated and sent to be recycled back. Table 1 Regenerator Feed Composition

Component Flow Rate (Kg/h) % Mass CO2 112013.1653 6.16 Water 932175.9301 51.26 MDEA 760705.5025 41.83 *Other components were ignored due to their very small percentage in feed. Table 2 Regenerated Lean Amine Composition

Component CO2 Water MDEA

Flow Rate (Kg/h) 219.8866 781860.8619 760705.5025

% Mass 0.01 50.68 49.31

Table 3 Regenerator Operating Conditions

Name Pressure, kPa Temperature, â °C Mass Flow, kg/hr

Feed Stream 250 100 1,818,582

Top Stream 190 110 275795.75

Bottom Stream 220 126.3 1542786.34

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1.2 Process Design 1.2.1 Design Methodology Table 4 Design Methodology of Amine Regenerator

Steps Material and Energy Balances 1. Rich MDEA flow rate calculation 2. Calculate Heat of Solution, Qsol 3. Calculate the Heat of Reaction, Qrx

4. Calculate the heat of vaporization, Qvap 5. Calculate Regenerator Reboiler Duty, Qtotal or QRE 6. Calculate Lean/Rich Amine exchanger duty, QLR 7. Calculate Overhead vapour and Bottom Steam

Description and Equations to be used To be based on the specifications of rich MDEA from the absorber. đ?‘„đ?‘ đ?‘œđ?‘™ = đ?‘šđ?‘ đ?‘œđ?‘™ Ă— đ??śđ?‘? ,đ?‘ đ?‘œđ?‘™ Ă— ∆đ?‘‡ đ?‘„đ?‘…đ?‘‹ = đ?‘€đ?‘Žđ?‘ đ?‘ đ?‘“đ?‘™đ?‘œđ?‘¤ đ?‘œđ?‘“ đ??śđ?‘‚2 Ă— đ??´đ?‘Łđ?‘’đ?‘&#x;đ?‘Žđ?‘”đ?‘’ â„Žđ?‘’đ?‘Žđ?‘Ą đ?‘œđ?‘“ đ?‘&#x;đ?‘’đ?‘Žđ?‘?đ?‘Ąđ?‘–đ?‘œđ?‘› đ?‘œđ?‘“ đ??śđ?‘‚2 đ?‘–đ?‘› đ??´đ?‘šđ?‘–đ?‘›đ?‘’đ?‘ Based on the latent heat of vaporization table at the specified pressures đ?‘„đ?‘…đ??¸ = đ?‘„đ?‘ đ?‘œđ?‘™ + đ?‘„đ?‘…đ?‘‹ + đ?‘„đ?‘‰đ?‘Žđ?‘? đ?‘„đ??żđ?‘… = đ?‘Šđ??ż Ă— đ??śđ?‘ƒđ??ż Ă— (đ?‘‡đ??ľ − đ?‘‡đ??żđ?‘‹ ) đ?‘†đ?‘Ąđ?‘’đ?‘Žđ?‘š đ?‘Žđ?‘Ą đ?‘?đ?‘œđ?‘Ąđ?‘Ąđ?‘œđ?‘š = đ?‘ đ?‘Ąđ?‘’đ?‘Žđ?‘š đ?‘Žđ?‘Ą â„Žđ?‘’đ?‘Žđ?‘‘ + đ?‘ đ?‘Ąđ?‘’đ?‘Žđ?‘š đ?‘”đ?‘’đ?‘›đ?‘’đ?‘&#x;đ?‘Žđ?‘Ąđ?‘’đ?‘‘ đ?‘“đ?‘&#x;đ?‘œđ?‘š â„Žđ?‘’đ?‘Žđ?‘Ąđ?‘–đ?‘›đ?‘” đ?‘ đ?‘œđ?‘™đ?‘˘đ?‘Ąđ?‘–đ?‘œđ?‘›

Process Design 8. Calculate the K-values for the components using Antoine equation 9. Calculate volatility (Îąi) of components

đ??żđ?‘œđ?‘” đ?‘ƒâˆ— =

10. Calculate minimum number of stages, Nmin

log (

11. Minimum Reflux Ratio, Rmin 12. Calculate Reflux Ratio, R 13. Find Theoretical Number of Stages 14. Find Actual Number of Stages 15. Find Feed Point Location, Using Kirkbride equation

đ??´âˆ’đ??ľ đ??ś+đ?‘‡ đ??žđ?‘– đ?›źđ?‘– = đ??žđ??ťđ??ž

đ?‘ min

=

∑

âˆ?đ?‘– đ?‘Ľđ?‘– ,đ?‘‘ = đ?‘…_ min + 1 = 2.02 âˆ?đ?‘– − đ?œƒ đ?‘… = đ?‘…_ minĂ— 1.5 = 3.02

Using Erbar-Maddox graph. N = 16 đ?‘ đ?‘Žđ?‘?đ?‘Ąđ?‘˘đ?‘Žđ?‘™ =

đ?‘ đ?‘Ąâ„Žđ?‘’đ?‘œđ?‘&#x;đ?‘’đ?‘Ąđ?‘–đ?‘?đ?‘Žđ?‘™ = 23 đ?‘ đ?‘Ąđ?‘Žđ?‘”đ?‘’đ?‘ đ??¸đ?‘“đ?‘“đ?‘’đ?‘?đ?‘–đ?‘’đ?‘›đ?‘?đ?‘Ś

đ?‘ đ?‘&#x; đ??ľ đ?‘Ľđ?‘“ ,đ??ťđ??ž đ?‘Ľđ?‘? ,đ??żđ??ž 2 ) ) log ( ) = 0.206 log (( ) ( )( đ?‘ đ?‘ đ??ˇ đ?‘Ľđ?‘“,đ??żđ??ž đ?‘Ľđ?‘‘ ,đ??ťđ??ž = 12

16. Calculate the Column Diameter đ??ˇ=√ 17. Do Sieve –Plate Hydraulic Design to find plate specifications 18. Tower Height

đ?‘Ľđ??żđ??ž đ?‘Ľ ) Ă— ( đ??ťđ??ž ) đ?‘Ľđ??ťđ??ž đ?‘‘ đ?‘Ľđ??żđ??ž đ?‘? = 10 đ?‘™đ?‘œđ?‘” âˆ?đ??żđ??ž

4 Ă— đ??´đ?‘‡ = 9 đ?‘šđ?‘’đ?‘Ąđ?‘&#x;đ?‘’đ?‘ đ?œ‹

Follow the steps in the calculations>> plates spacing 2.2 metre, plate thickness =3mm etc. đ??ťđ?‘’đ?‘–đ?‘”â„Žđ?‘Ą = (đ?‘ đ?‘œ. đ?‘†đ?‘Ąđ?‘Žđ?‘”đ?‘’đ?‘ − 1)(đ?‘‡đ?‘&#x;đ?‘Žđ?‘Ś đ?‘ đ?‘?đ?‘Žđ?‘?đ?‘–đ?‘›đ?‘”) + (đ?‘‡đ?‘&#x;đ?‘Žđ?‘Ś đ?‘ đ?‘?đ?‘Žđ?‘?đ?‘–đ?‘›đ?‘” Ă— 2) + (đ?‘ đ?‘œ. đ?‘†đ?‘Ąđ?‘Žđ?‘”đ?‘’đ?‘ − 1)(đ?‘‡â„Žđ?‘–đ?‘?đ?‘˜đ?‘›đ?‘’đ?‘ đ?‘ đ?‘œđ?‘“ đ?‘?đ?‘™đ?‘Žđ?‘Ąđ?‘’đ?‘ ) = 53 đ?‘šđ?‘’đ?‘Ąđ?‘&#x;đ?‘’

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1.2.2 Basis of Design (Material and Energy Balances) Table 5 Regenerator Conditions

Inlet Temperature (áľ’C) 100

Inlet Pressure (kPa) 250

Feed Molar Flow (kgmol/hr 60820

Feed Mass Flow (Kg/hr) 1818582.099

Table 6 Column feed composition

Feed Component CO2 MDEA Water

(mole fraction) 0.041844 0.104952 0.850698

Figure 2 Enthalpy Balance around the Regenerator (Kohl et al, 1997)

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1.2.2.1 Material Balance of Regenerator/Rich Amine Solution Rate Calculations Outlet Rich Amine Flow Rate (From absorber calculations) = 1822286.955 kg/hr CO2 mole fraction in rich amine (after absorber) = 4.23E-02 CO2 mass fraction in rich amine (after absorber) = 6.22E-02 Total CO2 in rich amine (after absorber) = 1822286.955 Ă— 6.22E − 02 = 1.13E+05 Kg/hr CO2 amount leaves the top of the separator (V-301) = 1378.277555 Kg/hr Total CO2 amount rate fed to regenerator (T-302) = 1.13E + 05 − 1378.277555 = 1.12E+05 Kg/hr CO2 molecular weight = 44.01 g/mol CO2 molar flow rate fed to regenerator (T-302) = 1.12E + 05 á 44.01 = 2.55E + 03 kgmol/hr Actual acid gas amount absorbed by lean amine solution (from absorber calculations) = 0.403606808 Kgmole acid gas/kgmole pure amine (CO2 loading). MDEA molecular weight = 119.163 g/mol: 2.55đ??¸+03

Pure MDEA Mass Flow Rate = Ă— 119.163 = 7.51E+05 Kg/hr 0.403606808 45% weight MDEA solution was used (going into the absorber) Mass flow rate of lean amine to the absorber (45% MDEA) =

7.51E+05 45

Ă— 100 = 1669891.034 Kg/hr

Design margin given for the process (10%) Lean Amine solution mass flow rate (Considering the 10% margin) = 1836880.137 Kg/hr

1669891.034 100

Ă— 100 =

Actual MDEA in the lean solution = 0.45 Ă— 1836880.137 = 826596.0618 Kg/hr H2O weight % in lean solution (going into the absorber) = 55 Actual H2O amount in the lean solution = 0.55 Ă— 1836880.137 = 1010284.076 Kg/hr Actual CO2 amount in the lean amine = 219.9129904 kg/hr Outlet rich amine solution from the absorber = ∑ đ?‘¤đ?‘Žđ?‘Ąđ?‘’đ?‘&#x; đ?‘šđ?‘Žđ?‘ đ?‘ đ?‘“đ?‘™đ?‘œđ?‘¤ đ?‘&#x;đ?‘Žđ?‘Ąđ?‘’ + đ??ˇđ??¸đ??´ đ?‘šđ?‘Žđ?‘ đ?‘ đ?‘“đ?‘™đ?‘œđ?‘¤ đ?‘&#x;đ?‘Žđ?‘Ąđ?‘’ + đ?‘Žđ?‘?đ?‘–đ?‘‘ đ?‘”đ?‘Žđ?‘ đ?‘šđ?‘Žđ?‘ đ?‘ đ?‘“đ?‘™đ?‘œđ?‘¤ đ?‘&#x;đ?‘Žđ?‘Ąđ?‘’ = 219.9129904 + 1010284.076 + 826596.0618 = 1837100.05 Kg/hr Rich amine solution entering the amine regenerator at temperature of 100áľ’C = 212 F Find the density ( r) of MDEA at 100áľ’C and 45% MDEA through the below figure:

Figure 3 Density of MDEA at different temperatures (Kidnay et al, 2011)

5

Therefore the density = 8.22 lb/gal = 984.9732 Kg/m3 The volumetric flow rate of 45% MDEA =

1836880.137 984.9732

= 1864.903672

đ?‘š3 â„Žđ?‘&#x;

= 0.518

đ?‘š3 đ?‘ đ?‘’đ?‘?

1.2.2.2 Thermal (Energy) Balance of Regenerator Regenerator Reboiler Duty (Qtotal) = QRE = QSOL + QRX + QVAP Heat of solution (QSOL) = msolution x Cp,sol x ∆T Temperature at regernator top = 110 áľ’C Temperature at regnerator bottom = 126.3 áľ’C MDEA lean solution mass flow rate = 1836880 kg/hr CO2 mass flow rate in lean amine = 219.913 kg/hr MDEA rich amine solution mass flow rate = 1836880 + 219.913 = 1837099.913 Kg/hr Average regenerator temperature =

110 126.3

2

= 118.15 áľ’đ??ś = 244.67 áľ’đ??š

Get the heat capacity (Cp) of MDEA at the average temperature of 118.15 áľ’C from the graph below:

Figure 4 the specific heat of MDEA at different temperatures (Kidnay et al, 2011)

Therefore, the heat capacity of MDEA at 118.15 degree Celsius = 1.015 Btu/lb-F = 4.25 Kj/Kg-K Heat of solution (QSOL) = 1837099.913 Ă— 4.25 Ă— (126.3 − 110) = 127265096.5 kJ/hr Now find the heat of reaction (Qrx) or CO2 heat of reaction: The average heat of reaction for CO2 for MDEA solution can be found from the following graph

6

Figure 5 Average heats of reaction of CO2 in Amine solutions (Kidnay et al, 2011)

Therefore, the average heat of reaction of CO2 is = 1840 kJ/Kg Qrx = 1840 Ă— 219.913 = 404639.92 kJ/hr Now find the heat of vaporization of water (QVAP): Assume the steam amount used is 3 times that of acid gas Steam mass flow rate = 3 Ă— 219.913 = 659.739 đ?‘˜đ?‘”/â„Žđ?‘&#x; At column pressure of about 15 psig, water latent heat of vaporization HVAP from the following graph: The water latent of vaporization is found to be = 945 BTU/lb at 15 psig and: QVAPORIZATION needs to be converted to Kj/hr as following: đ?‘˜đ?‘” â„Žđ?‘&#x;

Ă—

đ??ľđ?‘‡đ?‘ˆ đ?‘™đ?‘?

Ă—

1đ?‘™đ?‘? 0.453592đ?‘˜đ?‘”

=

945 Ă—659.739Ă—1000 453.594

>>>>QVAP = 1374474.431 kJ/hr

Figure 6 latent heat of vaporization of water at 15, 50 psig (Kidnay et al, 2011)

7

Regenerator Reboiler Duty = QRE = Qtotal 127265096.5 + 404639.92 + 1374474.431 = 1.29E + 08 kJ/hr Saturated steam at 50psig used and its latent heat was extracted from Figure 4 above to give a value of 912 BTU/lb = 2121.31 kJ/Kg. Steam flow rate (Reboiler) =



1.29E+08 2121.31

= 60832.32098 đ?‘˜đ?‘”/â„Žđ?‘&#x;

Lean/Rich Amine Exchanger Duty: QLR = WL * CPL *(TB – TLX)

Regenerator bottom temperature, TB = 126.3 áľ’C = 259.4 F Lean amine leaving exchanger temperature, TLX = 99.79 áľ’C = 211.6 F Average temperature = 113.045 áľ’C = 235.5 F The specific heat of lean amine (CPL) at a temperature of 113.045 áľ’C can be taken from the graph below: CPL at 113.045 áľ’C = 0.98 Btu/lb-F = 4.103 kJ/Kg-K Lean amine mass flow rate, WL = 1836880 Kg/hr QLR = 1836880 Ă— 4.103 Ă— (126.3 − 99.79) QLR = 199798411.1kJ/hr

Figure 7 Specific heat of MDEA at different temperatures (Kidnay et al, 2011)



Vapour Flow Rate at Bottom:

Vapour presents in both top and column bottom, however the vapour at the bottom is more important than that at the top. 

Overhead Vapour:

Assume that column temperature at the top Is 200 áľ’F and at the bottom is 240 áľ’F and also gas compressibility factor is 1 Idea Gas Law 8

Where n2 = CO2 moles + steam moles at the top = 2545.18 + 659.739 = 3.20đ??¸ + 3 V2 = 3.20E + 3 Ă— ( 

101.3

93

189

15

) Ă— 22.4 Ă— ( ) = 238563.3755 đ?‘š3/đ??ťđ?‘&#x;

Bottom Steam

Steam at bottom = steam at the head + steam generated from heating the solution Steam at the head = 659.739 Ă— (

101.3 189

93

) Ă— (22.4) Ă— ( ) = 49108.8209 đ?‘š3 /â„Žđ?‘&#x; 15

Heat amount required for heating the solution = QSOLUTION + QREACTION = 127265096.5 + 404639.92 = 127669736.4 đ?‘˜đ??˝/â„Žđ?‘&#x; Heat of Vaporization = 945 BTU/lb = 2190.07 Kj/Kg Steam Flow Rate = 127669736.4/2190.07 = 58082.65269 kgmole/hr Volumetric Flow Rate = 58082.65269 Ă— (

101.3

115

189

15

) Ă— (22.4) Ă— (

) = 5346242.864 đ?‘š3/đ??ťđ?‘&#x;

Total steam amount at the bottom = 5346242.864 + 49108.8209 = 5395351.685 đ?‘š3/â„Žđ?‘&#x;

1.2.3 Process Design Calculations Table 7 Regenerator operational conditions

Feed Temperature Feed Pressure Distillate Pressure Bottom Pressure

100 â °C 2.5 1.9 2.2

373.15 K Bar Bar Bar

Table 8 Feed components and compositions (From HYSYS)

Component Feed Mole Fraction Distillate Mole Fraction Bottom Mole Fraction

CO2 4.18E-02 0.2301628 1.00E-04

Water 0.8506976 0.7560262 0.8716831

MDEA 0.104952 2.80E-50 0.1282164

The first thing to start with in a process design is the selection of the light and heavy keys, the light key is the component that it is desired to keep out of the bottom product and the heavy key is the component to be kept out of the top product. (Sinnott, 2009) Therefore, the light key was selected to be the CO2 and the heavy key is the MDEA. Afterwards, the Antoine coefficients (A, B and C) were found for each of the components at the feed temperature from (Yaws et al, 2005), then these values were input in Antoine equation for the calculation of the vapour pressure in bar of the components. Then the K-value of each component was found before finally finding calculating the volatility (Îą).

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Table 9 Process design conditions, k values and volatilities

Component A B C T (â °C) Log P P* (mmHg) P* (bar) K-Value Volatility (Îą)

CO2 (LK) 7.58828 861.82 271.883 100 5.270830695 186565.2243 248.7337338 99.49349351 296412.0823

Water 8.05573 1723.64 233.076 100 2.880814966 760.0024033 1.013255477 0.405302191 1207.480631

MDEA (HK) 8.74263 2501.48 179.692 100 -0.20106521 0.629411672 0.000839148 0.000335659 1

Distillate

K-Value Volatility (Îą)

98192.22334 296412.0823

400.0012649 1207.480631

0.331269301 1

Bottom

K-Value Volatility (Îą)

84802.3747 345.4556378 296412.0823 1207.480631 296412.0823

0.286096215 1

FEED

ÎąLK (Average)

As mentioned earlier the Antoine coefficients were found as seen in the table at the required temperature. Then Antoine equation was used to find to find (Log P) of each component: đ??żđ?‘œđ?‘” đ?‘ƒ = đ??´ − (

đ??ľ ) đ?‘‡(đ??ś) + đ??ś

861.82 ) đ?‘‡đ?‘œ đ?‘“đ?‘–đ?‘›đ?‘‘ đ??żđ?‘œđ?‘” đ?‘ƒ đ?‘œđ?‘“ đ??śđ?‘‚2 = 7.58828 − ( 100 + 271.883 đ??żđ?‘œđ?‘” đ?‘ƒ đ?‘œđ?‘“ đ??śđ?‘‚2 = 5.270830695 To find the vapour pressure (P*) of CO2 use: đ?‘ƒâˆ— (đ?‘šđ?‘šđ??ťđ?‘”) = 105.270830695 đ?‘ƒâˆ— = 186565.2243 mmHg Now convert the vapour pressure from mmHg to bar as follows (1 bar = 750.06 mmHg): đ?‘ƒâˆ— =

186565.2243 = 248.7337338 bar 750.06

Now find the K-value as following: đ?‘ƒâˆ— 248.7337338 đ??ž − đ?‘Łđ?‘Žđ?‘™đ?‘˘đ?‘’ = = = 99.49349351 đ?‘ƒđ?‘Ąđ?‘œđ?‘Ąđ?‘Žđ?‘™ (đ??šđ?‘’đ?‘’đ?‘‘) 2.5 Now find the volatility (Îą) as following: đ?‘‰đ?‘œđ?‘™đ?‘Žđ?‘Ąđ?‘–đ?‘™đ?‘–đ?‘Ąđ?‘Ś (Îą)đ?‘œđ?‘“ đ??śđ?‘‚2 =

đ??žđ?‘– 99.49349351 = = 296412.0823 đ??ž(â„Žđ?‘’đ?‘Žđ?‘Łđ?‘Ś đ?‘“đ?‘’đ?‘’đ?‘‘) 0.000335659

The same procedure was followed to calculate the rest values as shown in the table above.

10

1.2.3.1 Minimum Number of Stages, Nmin (Using Fenske Equation): đ?’?đ?’?đ?’ˆ ( đ?‘ľđ?’Žđ?’Šđ?’? =

đ?’™đ?‘łđ?‘˛ đ?’™ ) Ă— ( đ?‘Żđ?‘˛ ) đ?’™đ?‘Żđ?‘˛ đ?’… đ?’™đ?‘łđ?‘˛ đ?’ƒ đ?’?đ?’?đ?’ˆ ÎąLK

Where, (XLK)D is the light key mole fraction in distillate = 0.230162753 (XHK)D is the heavy key mole fraction in distillate = 2.80E-50 (XHK)B is the heavy key mole fraction in bottom = 0.128216426 (XLK)B is the light key mole fraction in bottom = 1.00E-04 Îą is the average volatility of CO2 (LK) in feed, distillate and bottoms = 296412.0823 Inputting the values into the Fenske equation to get: đ?‘ľđ?’Žđ?’Šđ?’? =

đ?’?đ?’?đ?’ˆ (

0.230162753 0.128216426 ) Ă—( ) 2.80E − 50 đ?’… 1.00E − 04 đ?’ƒ = đ?&#x;—. đ?&#x;“ = đ?&#x;?đ?&#x;Ž đ?’”đ?’•đ?’‚đ?’ˆđ?’†đ?’” đ?’?đ?’?đ?’ˆ 296412.0823

Therefore the minimum number of stages equals to 10 stages as per the calculations above.

1.2.3.2 Minimum Reflux Ratio, Rmin (Underwood Equation): The root of the equation was firstly calculated θ using the following equation with an iterative method as following: ∑

đ?›źđ?‘– ∗ đ?‘Ľđ?‘–, đ?‘“ =1−đ?‘ž đ?›źđ?‘– − đ?œƒ

Where the left hand side of the equation must equal to the right hand side by trial and error by changing the θ value to make this work as following: q (From HYSYS) is the liquid ratio in the feed = 0.9529 1-q= 0.0471 After trial and error calculations, the following table was constructed: Îąi * Xi,f Îąi - θ (Îąi * Xi,f)/(Îąi - θ) 1.24E+04 22580.34273 2.80E+08 1027.201 -272624.259 -280039868.1 0.104952 -273830.74 -28739.08149 Sum = 0.04783 For which the root of the equation was estimated to be = θ = 273831.7396 to satisfy the conditions of the equation above. Now using this found value of root of the equation to be input in the following Underwood equation to find the minimum reflux ratio: ∑ Îąi * Xi,d 68223.02086 912.8870012 2.80E-50

đ?œśđ?’Š ∗ đ?’™đ?’Š, đ?’… = đ?‘šđ?’Žđ?’Šđ?’? + đ?&#x;? đ?œśđ?’Š − đ?œ˝ Îąi - θ (Îąi * Xi,d)/(Îąi - θ) 22580.343 3.021345676 -272624.3 -0.003348517 -273830.7 -1.02E-55 Sum = 3.017997159

11

đ?&#x;‘. đ?&#x;Žđ?&#x;?đ?&#x;•đ?&#x;—đ?&#x;—đ?&#x;•đ?&#x;?đ?&#x;“đ?&#x;— = đ??‘ đ??Śđ??˘đ??§ − đ?&#x;? Therefore, the minimum reflux ratio Rmin = 2.02

1.2.3.3 Theoretical Number of Stages (Erbar-Maddox correlation-graph): Firstly, assume that the actual reflux ratio đ?‘š = đ?&#x;?. đ?&#x;“ đ?‘šđ?’Žđ?’Šđ?’? đ?‘… = 1.5 Ă— 2.02 = 3.026995738

Now,

đ?‘… 3.026995738 = = 0.751675923 đ?‘… + 1 3.026995738 + 1 đ?‘…đ?‘š 2.02 = = 0.668654426 đ?‘…đ?‘š + 1 2.02 + 1 Now use the following graph to get the theoretical number of stages:

Figure 8 Erbar-Maddox correlation to find theoretical stages (Sinnott et al, 2009)

From the graph we can get, đ?‘ đ?‘š đ?‘

= 0.62 , đ?‘ =

10 0.62

= 16.13 = 16 đ?‘ đ?‘Ąđ?‘Žđ?‘”đ?‘’đ?‘

Therefore, the theoretical number of stages (N) are 16 stages.

1.2.3.4 Column Efficiency (Using O’connell correlation): đ??¸đ?‘œ = 51 − 32.5 Ă— log(Âľđ?‘Ž đ?›źđ?‘Ž ) Where, Âľa is the molar average liquid viscosity, mNs/m2 Îąa is the average relative volatility of the light key which equals to = 296412.0823 12

The molar average liquid viscosity, Âľa can be found by doing this calculation:

CO2 Water MDEA

Xi,f 4.18E-02 0.850697572 0.104951992

Viscosity, cP 0.0148 0.894 0.15 Îźa =

Xi,f * Viscosity 6.19E-04 0.760523629 0.015742799 7.77E-01

Therefore, Îźa = 7.77E-01 cP or mNs/m2 Assuming the column efficiency is 70% Now find the actual number of stages by using the efficiency: đ??´đ?‘?đ?‘Ąđ?‘˘đ?‘Žđ?‘™ đ?‘ đ?‘˘đ?‘šđ?‘?đ?‘’đ?‘&#x; đ?‘œđ?‘“ đ?‘†đ?‘Ąđ?‘Žđ?‘”đ?‘’đ?‘ =

đ?‘‡â„Žđ?‘’đ?‘œđ?‘&#x;đ?‘’đ?‘Ąđ?‘–đ?‘?đ?‘Žđ?‘™ đ?‘ đ?‘˘đ?‘šđ?‘?đ?‘’đ?‘&#x; đ?‘œđ?‘“ đ?‘†đ?‘Ąđ?‘Žđ?‘”đ?‘’đ?‘ đ??¸đ?‘“đ?‘“đ?‘’đ?‘?đ?‘–đ?‘’đ?‘›đ?‘?đ?‘Ś

đ??´đ?‘?đ?‘Ąđ?‘˘đ?‘Žđ?‘™ đ?‘ đ?‘˘đ?‘šđ?‘?đ?‘’đ?‘&#x; đ?‘œđ?‘“ đ?‘†đ?‘Ąđ?‘Žđ?‘”đ?‘’đ?‘ =

16 = 22.85 = 23 đ?‘ đ?‘Ąđ?‘Žđ?‘”đ?‘’đ?‘ 0.7

Therefore, the actual number of stages including the reboiler is 23 stages.

1.2.3.5 Feed Point Location (Using Kirkbride Equation): đ??żđ?‘œđ?‘” (

đ?‘ đ?‘&#x; đ??ľ đ?‘Ľđ?‘“ ,đ??ťđ??ž đ?‘Ľđ?‘? ,đ??żđ??ž ) = 0.206 đ?‘™đ?‘œđ?‘” [( ) ( )( )] đ?‘ đ?‘ đ??ˇ đ?‘Ľđ?‘“,đ??żđ??ž đ?‘Ľđ?‘‘,đ??ťđ??ž

Where, Nr is the number of stages above the feed Ns is the number of stages below the feed. Including the reboiler B is the molar flow bottom product = 49788.5680 kgmole/hr D is the molar flow of top product = 11036.37002 kgmole/hr Xf,HK is the mole fraction of the heavy key in feed = 0.104951992 Xf,LK is the mole fraction of the light key in the feed = 4.18E-02 Xd,HK is the mole fraction of the heavy key in distillate = 9.06E-05 Xb,LK is the mole fraction of light key in the bottom = 2.01E-05 đ??ľ 49788.5680 = = 4.511317394 đ??ˇ 11036.37002 đ?‘Ľđ?‘“ ,đ??ťđ??ž 0.104951992 ( )= = 2.51 đ?‘Ľđ?‘“,đ??żđ??ž 4.18E − 02 đ?‘Ľđ?‘? ,đ??żđ??ž 2.01E − 05 ( )= = 0.22 đ?‘Ľđ?‘‘,đ??ťđ??ž 9.06E − 05 đ?‘ đ?‘&#x; 0.206Ă—log(( 4.511317394 )Ă—2.51Ă—0.222 ) = 10 = 0.89 đ?‘ đ?‘ đ?‘ = đ?‘ đ?‘&#x; + đ?‘ đ?‘ đ?‘ đ?‘&#x; + đ?‘ đ?‘ = 22 Therefore,

13

Ns (Stripping section stages) = 11.66 = 12 stages Nr (Rectifying section stages) = 10.34 = 10 stages so the feed stage (NS) is at stage 12

1.2.3.6 Column Diameter Calculations: From HYSYS: Feed flow rate (kgmole/hr) = F = 60824.93803 Feed molecular weight = 29.89862642 Top molecular weight = 24.98971635 Bottom molecular weight = 30.9867587 Overall Material Balance: F = B+D Material balance on CO2: đ?‘‹đ?‘?đ?‘œ2 Ă— đ??š = đ?‘‹đ?‘?đ?‘œ2 Ă— đ??ľ + đ?‘‹đ?‘?đ?‘œ2 Ă— đ??ˇ 4.18E − 02 Ă— 60824.93803 = (1.00E − 4 Ă— B) + (0.2301628 Ă— D) D = 11036.37002 kgmole/hr B = 49788.56801 kgmole/hr đ??żđ?‘–đ?‘žđ?‘˘đ?‘–đ?‘‘ đ?‘Žđ?‘?đ?‘œđ?‘Łđ?‘’ đ?‘“đ?‘’đ?‘’đ?‘‘ đ?‘ đ?‘Ąđ?‘Žđ?‘”đ?‘’, đ??ż = đ?‘… Ă— đ??ˇ = 3.026995738 Ă— 11036.37002 = 33407.045 kgmole/h đ??żđ?‘–đ?‘žđ?‘˘đ?‘–đ?‘‘ đ?‘?đ?‘’đ?‘™đ?‘œđ?‘¤ đ?‘“đ?‘’đ?‘’đ?‘‘ đ?‘ đ?‘Ąđ?‘Žđ?‘”đ?‘’ đ??żâ€˛ = (đ?‘… Ă— đ??ˇ) + đ??š = 33407.045 + 60824 = 94231.98303 kgmol/hr đ?‘‰đ?‘Žđ?‘?đ?‘œđ?‘˘đ?‘&#x; đ?‘…đ?‘Žđ?‘Ąđ?‘’ đ?‘‰ = đ??ˇ Ă— (1 + đ?‘…) = 11036.37002 Ă— (1 + 3.026995738) = 44443.41502 kmol/h Physical properties were obtained from HYSYS: Column Top Side: Density of vapour (Ď v) = 1.508563639 kg/m3 Density of liquid (Ď L) = 950.5033851 kg/m3 Surface Tension (Ďƒ) = 0.046191694 N/m Column Bottom Side: Density of vapour (Ď v) = 1.218466705 kg/m3 Density of liquid (Ď L) = 971.7359111 kg/m3 Surface Tension (Ďƒ) = 0.035990365 N/m Assumptions:   

Column efficiency is 70% Pressure drop per plate is 100mm MDEA = 100 mm MDEA Ignore the reboiler

Number of actual trays (ignoring the reboiler) = 22 Column Pressure Drop: ∆đ?‘ƒđ?‘Ą = 9.81 Ă— 10−3 Ă— â„Žđ?‘Ą Ă— đ?œŒđ??ż Where, Ht is the total plate pressure drop, mm liquid = 2200 ÎĄL is the average density of liquid in top and bottom = 961.1196 kg/m3

14

đ?‘ ∆đ?‘ƒđ?‘Ą = 9.81 Ă— 10−3 Ă— 2200 Ă— 961.1196 = 20742.88425 đ?‘ƒđ?‘Ž ( 2 ) = 20.74288425 đ?‘˜đ?‘ƒđ?‘Ž đ?‘š Assumptions: Neglect differences in molecular weights of both liquid and vapour đ??šđ??żđ?‘‰ đ?‘?đ?‘œđ?‘Ąđ?‘Ąđ?‘œđ?‘š =

đ??żđ?‘¤ √đ?œŒđ?‘Ł /đ?œŒđ??ż đ?‘‰đ?‘¤

Where, Lw is the liquid flow below feed (L’) = 94231.98303 kmole/hr Vw is the vapour rate (V) = 44443.41502 kmol/hr đ??šđ??żđ?‘‰ đ?‘?đ?‘œđ?‘Ąđ?‘Ąđ?‘œđ?‘š =

94231.98303 1.218466705 √ = 44443.41502 971.7359111

đ??żđ?‘–đ?‘žđ?‘˘đ?‘–đ?‘‘ − đ?‘‰đ?‘Žđ?‘?đ?‘œđ?‘˘đ?‘&#x; đ?‘“đ?‘™đ?‘œđ?‘¤ đ?‘“đ?‘Žđ?‘?đ?‘Ąđ?‘œđ?‘&#x; (đ??šđ??żđ?‘‰ ,đ??ľđ?‘œđ?‘Ąđ?‘Ąđ?‘œđ?‘š ) = 0.075079888 đ?‘˜đ?‘šđ?‘œđ?‘™đ?‘’/â„Žđ?‘&#x; đ??šđ??żđ?‘‰ đ?‘Ąđ?‘œđ?‘? =

đ??żđ?‘¤ √đ?œŒđ?‘Ł /đ?œŒđ??ż đ?‘‰đ?‘¤

Where: Lw is the liquid flow above feed (L) đ??šđ??żđ?‘‰ đ?‘Ąđ?‘œđ?‘? =

33407.045 1.508563639 √ = 44443.41502 950.5033851

đ??żđ?‘–đ?‘žđ?‘˘đ?‘–đ?‘‘ − đ?‘‰đ?‘Žđ?‘?đ?‘œđ?‘˘đ?‘&#x; đ?‘“đ?‘™đ?‘œđ?‘¤ đ?‘“đ?‘Žđ?‘?đ?‘Ąđ?‘œđ?‘&#x; (đ??šđ??żđ?‘‰ ,đ?‘Ąđ?‘œđ?‘? ) = 0.084468716 đ?‘˜đ?‘šđ?‘œđ?‘™đ?‘’/â„Žđ?‘&#x;  



According to (Sinnott, 2009), K values should be multiplied by the correction factor for any tension surface values except for tension surface of 0.02. For this calculation the plate spacing was assumed to be 2.2 metres (I firstly assumed a value of 0.9 meters but it didn’t meet the condition of the equation of the downcomer back-up, therefore only a spacing of 2.2 meters can achieve this condition. Using the figure below we can find K1,top and K2,bottom

To find K1,top: FLV,top = 0.084468716 Plate spacing = 2.2 metres So from the graph, K1,top = 0.313 (By extrapolation) To find K1,bottom: FLV,bottom = 0.075079888 So from the graph, K1,bottom = 0.28 (By extrapolation)

15

Figure 9 Flooding velocity and the K1 value of sieve plates (Sinnott et al, 2009)

Correction of K-values using surface tension using: đ??ž1 Ă— [

đ?œŽ 0.2 ] 0.02

đ??ž1 , đ?‘‡đ?‘œđ?‘? (đ??´đ?‘“đ?‘Ąđ?‘’đ?‘&#x; đ?‘?đ?‘œđ?‘&#x;đ?‘&#x;đ?‘’đ?‘?đ?‘Ąđ?‘–đ?‘œđ?‘›) = 0.313 Ă— [

0.046191694 0.2 ] = 0.370042082 0.02

0.035990365 0.2 ] = 0.314912052 đ??ž1 ,đ?‘?đ?‘œđ?‘Ąđ?‘Ąđ?‘œđ?‘š (đ??´đ?‘“đ?‘Ąđ?‘’đ?‘&#x; đ?‘?đ?‘œđ?‘&#x;đ?‘&#x;đ?‘’đ?‘?đ?‘Ąđ?‘–đ?‘œđ?‘›) = 0.28 Ă— [ 0.02 Calculate the flooding vapour velocity of both top and bottom side sing Fair equation: đ?‘˘đ?‘“ = đ??ž1 Ă— √

đ?œŒđ??ż − đ?œŒđ?‘Ł đ?œŒđ?‘Ł

Where, Uf is the flooding vapour velocity in m/s, based on the net column sectional area An K1 is the value of the constant after the correction đ??šđ?‘™đ?‘œđ?‘œđ?‘‘đ?‘–đ?‘›đ?‘” đ?‘Łđ?‘Žđ?‘?đ?‘œđ?‘˘đ?‘&#x; đ?‘Łđ?‘’đ?‘™đ?‘œđ?‘?đ?‘–đ?‘Ąđ?‘Ś đ?‘Žđ?‘Ą đ?‘Ąâ„Žđ?‘’ đ?‘Ąđ?‘œđ?‘? (đ?‘˘đ?‘“ ) 950.5033851 − 1.508563639 = 0.370042082 Ă— √ = 9.281137759 đ?‘š/đ?‘ 1.508563639 đ??šđ?‘™đ?‘œđ?‘œđ?‘‘đ?‘–đ?‘›đ?‘” đ?‘‰đ?‘Žđ?‘?đ?‘œđ?‘˘đ?‘&#x; đ?‘‰đ?‘™đ?‘’đ?‘œđ?‘?đ?‘–đ?‘Ąđ?‘Ś đ?‘Žđ?‘Ą đ?‘Ąâ„Žđ?‘’ đ?‘?đ?‘œđ?‘Ąđ?‘Ąđ?‘œđ?‘š (đ?‘˘đ?‘“ ) 971.7359111 − 1.218466705 = 0.314912052 Ă— √ = 8.887592262 đ?‘š/đ?‘ 1.218466705 

Assume design for 60% flooding at maximum flow rate (I started with a 1st initially flooding percentage of 85% but I didn’t meet the condition in downcomer back-up unless the percent flooding is 60%). 16

𝑢𝑓 ,𝑡𝑜𝑝 (𝑓𝑜𝑟 60% 𝑑𝑒𝑠𝑖𝑔𝑛 𝑓𝑎𝑐𝑡𝑜𝑟) = 0.6 × 9.281137759 = 5.568682655 𝑚/𝑠 𝑢𝑓 ,𝑏𝑜𝑡𝑡𝑜𝑚 (𝑓𝑜𝑟 60% 𝑑𝑒𝑠𝑖𝑔𝑛 𝑓𝑎𝑐𝑡𝑜𝑟) = 0.6 × 8.887592262 = 5.332555357 𝑚/𝑠 𝑚3 𝑉𝑎𝑝𝑜𝑟 𝑟𝑎𝑡𝑒 × 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑤𝑒𝑖𝑔ℎ𝑡 𝑉𝑜𝑙𝑢𝑚𝑒𝑡𝑟𝑖𝑐 𝐹𝑙𝑜𝑤 𝑅𝑎𝑡𝑒 ( ) = 𝑠 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑣𝑎𝑝𝑜𝑟 × 3600 𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝑣𝑜𝑙𝑢𝑚𝑡𝑟𝑖𝑐 𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒 𝑎𝑡 𝑡ℎ𝑒 𝑡𝑜𝑝 = = 204.5043794 𝑚3 /𝑠

44443.41502 × 24.98971635 1.508563639 × 3600

𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝑣𝑜𝑙𝑢𝑚𝑡𝑟𝑖𝑐 𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒 𝑎𝑡 𝑡ℎ𝑒 𝑏𝑜𝑡𝑡𝑜𝑚 = = 313.955001 𝑚3 /𝑠

44443.41502 × 30.9867587 1.218466705 × 3600

Now calculate the net area required as following: 𝑁𝑒𝑡 𝐴𝑟𝑒𝑎 𝑅𝑒𝑞𝑢𝑖𝑟𝑒𝑑 = 𝑁𝑒𝑡 𝐴𝑟𝑒𝑎 𝑅𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑎𝑡 𝑡ℎ𝑒 𝑡𝑜𝑝 =

𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝑉𝑜𝑙𝑢𝑚𝑒 𝑅𝑎𝑡𝑒 𝐹𝑙𝑜𝑜𝑑𝑖𝑛𝑔 𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦 204.5043794 = 36.72401393 𝑚2 5.568682655

𝑁𝑒𝑡 𝐴𝑟𝑒𝑎 𝑅𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑎𝑡 𝑡ℎ𝑒 𝑏𝑜𝑡𝑡𝑜𝑚 =

313.955001 = 58.87515085 𝑚2 5.332555357

Now calculate the Column cross sectional area as following: 

Assume the downcomer area as 12% of total area 𝑁𝑒𝑡 𝐴𝑟𝑒𝑎 0.88 36.72401393 𝑇𝑜𝑝 𝐶𝑜𝑙𝑢𝑚𝑛 𝑐𝑟𝑜𝑠𝑠 𝑆𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝐴𝑟𝑒𝑎 = = 41.73183401 𝑚2 0.88 58.87515085 𝐵𝑜𝑡𝑡𝑜𝑚 𝐶𝑜𝑙𝑢𝑚𝑛 𝑐𝑟𝑜𝑠𝑠 𝑆𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝐴𝑟𝑒𝑎 = = 66.90358051 𝑚2 0.88 𝐶𝑜𝑙𝑢𝑚𝑛 𝑐𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑟𝑒𝑎 =

Now find the tower diameter using the following equation: 𝐷𝑇 = √

4 × 𝐴𝑇 𝜋

Where AT is the cross sectional area of the top and bottom parts. 𝑇𝑜𝑝 𝐶𝑜𝑙𝑢𝑚𝑛 𝐷𝑖𝑎𝑚𝑒𝑡𝑒𝑟 = √

4 × 41.73183401 = 7 𝑚𝑒𝑡𝑟𝑒𝑠 3.14

𝐵𝑜𝑡𝑡𝑜𝑚 𝐶𝑜𝑙𝑢𝑚𝑛 𝐷𝑖𝑎𝑚𝑒𝑡𝑒𝑟 = √

4 × 66.90358051 = 9 𝑚𝑒𝑡𝑟𝑒𝑠 3.14

Therefore the column diameter is 9 metres and that is due to the feed flow being high and also the vapour rate is large and the vapour rate has the highest impact on tower diameter. Now find the Liquid Flow Pattern as following: 17

đ?‘€đ?‘Žđ?‘Ľđ?‘–đ?‘šđ?‘˘đ?‘š đ?‘Łđ?‘œđ?‘™đ?‘˘đ?‘šđ?‘’đ?‘Ąđ?‘&#x;đ?‘–đ?‘? đ?‘™đ?‘–đ?‘žđ?‘˘đ?‘–đ?‘‘ đ?‘&#x;đ?‘Žđ?‘Ąđ?‘’ đ?‘™đ?‘–đ?‘žđ?‘˘đ?‘–đ?‘‘ đ?‘“đ?‘™đ?‘œđ?‘¤ đ?‘?đ?‘’đ?‘™đ?‘œđ?‘¤ đ?‘“đ?‘’đ?‘’đ?‘‘ (đ??żâ€˛ ) Ă— đ?‘?đ?‘œđ?‘Ąđ?‘Ąđ?‘œđ?‘š đ?‘šđ?‘œđ?‘™đ?‘’đ?‘?đ?‘˘đ?‘™đ?‘Žđ?‘&#x; đ?‘¤đ?‘’đ?‘–đ?‘”â„Žđ?‘Ą = = 3600 Ă— đ?‘‘đ?‘’đ?‘›đ?‘ đ?‘–đ?‘Ąđ?‘Ś đ?‘œđ?‘“ đ?‘™đ?‘–đ?‘žđ?‘˘đ?‘‘đ?‘– đ?‘Žđ?‘Ą đ?‘?đ?‘œđ?‘Ąđ?‘Ąđ?‘œđ?‘š đ?‘€đ?‘Žđ?‘Ľđ?‘–đ?‘šđ?‘˘đ?‘š đ?‘Łđ?‘œđ?‘™đ?‘˘đ?‘šđ?‘’đ?‘Ąđ?‘&#x;đ?‘–đ?‘? đ?‘™đ?‘–đ?‘žđ?‘˘đ?‘–đ?‘‘ đ?‘&#x;đ?‘Žđ?‘Ąđ?‘’ =

94231.98303 Ă— 30.9867587 = 0.834687149 đ?‘š3 /đ?‘ 3600 Ă— 971.7359111

1.2.3.7 Sieve-Plate Hydraulic Design: Column diameter (DC) = 9 metres Column Area (AC) = 66.90358051 m2 Downcomer Area (Ad) at 12% of column area = 0.12 x 66.90358051 = 8.028429661 m2 Net Area (An) = Ac – Ad = 58.87515085 m2 Active Area (Aa) = Ac – 2xAd = 50.84672118 m2 Assume hole area (Ah) to be 15% of (Aa) as first trial = Ah = 0.15 x 50.84672118 = 7.627008178 m2 To find weir length (Lw): Weir length can be calculated from the graph below by finding: (

đ??´đ?‘‘ 8.028429661 ) Ă— 100 = ( ) Ă— 100 = 12 đ??´đ?‘? 66.90358051

Now from the graph find Iw/DC:

Figure 10 Relationship between downcomer area and weir length (Sinnott et al, 2009)

Therefore, đ??źđ?‘¤ = 0.76 đ??ˇđ?‘? đ?‘Šđ?‘’đ?‘–đ?‘&#x; đ??żđ?‘’đ?‘›đ?‘”đ?‘Ąâ„Ž (đ??żđ?‘¤ ) = 9 Ă— 0.76 = 7.016223974 đ?‘šđ?‘’đ?‘Ąđ?‘&#x;đ?‘’đ?‘

18

1.2.3.7.1 Check Weeping: Assumptions:   

Weir height (Hw) is recommended between 40-50mm but I have chosen 50 mm Hole diameter is preferred to be 5 mm Plate thickness (recommended for stainless steel) = 3mm đ?‘™đ?‘–đ?‘žđ?‘˘đ?‘–đ?‘‘ đ?‘“đ?‘™đ?‘œđ?‘¤ đ?‘?đ?‘’đ?‘™đ?‘œđ?‘¤ đ?‘“đ?‘’đ?‘’đ?‘‘ (đ??żâ€˛ ) Ă— đ?‘šđ?‘œđ?‘’đ?‘™đ?‘?đ?‘˘đ?‘™đ?‘Žđ?‘&#x; đ?‘¤đ?‘’đ?‘–đ?‘”â„Žđ?‘Ą đ?‘œđ?‘“ đ?‘?đ?‘œđ?‘Ąđ?‘Ąđ?‘œđ?‘š đ?‘€đ?‘Žđ?‘Ľđ?‘–đ?‘šđ?‘˘đ?‘š đ??żđ?‘–đ?‘žđ?‘˘đ?‘–đ?‘‘ đ?‘&#x;đ?‘Žđ?‘Ąđ?‘’ = 3600 đ?‘€đ?‘Žđ?‘Ľđ?‘–đ?‘šđ?‘˘đ?‘š đ??żđ?‘–đ?‘žđ?‘˘đ?‘–đ?‘‘ đ?‘&#x;đ?‘Žđ?‘Ąđ?‘’ =



94231.98303 Ă— 30.9867587 = 811.0954776 đ?‘˜đ?‘”/đ?‘ 3600

Assuming 70% turn-down ratio, therefore: đ?‘€đ?‘–đ?‘›đ?‘–đ?‘šđ?‘˘đ?‘š đ?‘™đ?‘žđ?‘˘đ?‘–đ?‘‘ đ?‘&#x;đ?‘Žđ?‘Ąđ?‘’ = 0.7 Ă— đ?‘šđ?‘Žđ?‘Ľđ?‘–đ?‘šđ?‘˘đ?‘š đ?‘™đ?‘–đ?‘žđ?‘˘đ?‘–đ?‘‘ đ?‘&#x;đ?‘Žđ?‘Ąđ?‘’ đ?‘€đ?‘–đ?‘›đ?‘–đ?‘šđ?‘˘đ?‘š đ?‘™đ?‘žđ?‘˘đ?‘–đ?‘‘ đ?‘&#x;đ?‘Žđ?‘Ąđ?‘’ = 0.7 Ă— 811.0954776 = 567.7668343 đ?‘˜đ?‘”/đ?‘

Now to calculate the weir crest (how), use the following Francis weir formula: 2

â„Žđ?‘œđ?‘¤

( ) 3 đ??żđ?‘¤ ) = 750 Ă— ( đ?œŒđ??ż Ă— đ?‘™đ?‘¤

Where, lw is the weir length in metres How is the weir crest, mm liquid Lw is the liquid flow rate,kg/s ÎĄL is the density of liquid at bottom Now calculate the maximum and minimum weir crest as following: 2

( ) 3 811.0954776 ) đ?‘€đ?‘Žđ?‘Ľđ?‘–đ?‘šđ?‘˘đ?‘š đ?‘¤đ?‘’đ?‘–đ?‘&#x; đ?‘?đ?‘&#x;đ?‘’đ?‘ đ?‘Ą (â„Žđ?‘œđ?‘¤ ) = 750 Ă— ( 971.7359111 Ă— 7.016223974 = 181.4156651 đ?‘šđ?‘š đ?‘™đ?‘–đ?‘žđ?‘˘đ?‘–đ?‘‘ 2

( ) 3 567.7668343 ) đ?‘€đ?‘–đ?‘›đ?‘–đ?‘šđ?‘˘đ?‘š đ?‘¤đ?‘’đ?‘–đ?‘&#x; đ?‘?đ?‘&#x;đ?‘’đ?‘ đ?‘Ą (â„Žđ?‘œđ?‘¤ ) = 750 Ă— ( 971.7359111 Ă— 7.016223974 = 143.0233058 đ?‘šđ?‘š đ?‘™đ?‘–đ?‘žđ?‘˘đ?‘–đ?‘‘

đ??´đ?‘Ą đ?‘šđ?‘–đ?‘›đ?‘–đ?‘šđ?‘˘đ?‘š đ?‘&#x;đ?‘Žđ?‘Ąđ?‘’ (â„Žđ?‘¤ + â„Žđ?‘œđ?‘¤ ) = 7.016223974 + 143.0233058 = 193.0233058 đ?‘šđ?‘š đ?‘™đ?‘–đ?‘žđ?‘˘đ?‘–đ?‘‘ Using the graph below, although my (hw + how) is out of the graph range but it is assumed the curve will stay steady until 193, therefore an estimation of K2 will be: đ??ž2 = 31.5

19

Figure 11 Weep point correlation (Sinnott et al, 2009)

Now calculate the minimum design vapour velocity from the formula on the side by Eduljee: �ℎ =

[đ??ž2 − 0.90 Ă— (25.4 − đ?‘‘â„Ž )] (đ?œŒđ?‘Ł )0.5

Where; Uh is the minimum design vapour velocity through the holes (based on hole area), m/s Dh is the hole diameter, mm K2 is the constant found from the figure above đ?‘˘â„Ž (đ?‘šđ?‘–đ?‘›đ?‘–đ?‘šđ?‘˘đ?‘š đ?‘‘đ?‘’đ?‘ đ?‘–đ?‘”đ?‘› đ?‘Łđ?‘’đ?‘™đ?‘œđ?‘?đ?‘–đ?‘Ąđ?‘Ś) =

[31.5 − 0.9 Ă— (25.4 − 5)] = 11.90387977 đ?‘š/đ?‘ 1.2184667050.5

Now calculate the actual minimum vapour velocity: đ??´đ?‘?đ?‘Ąđ?‘˘đ?‘Žđ?‘™ đ?‘šđ?‘–đ?‘›đ?‘–đ?‘šđ?‘˘đ?‘š đ?‘Łđ?‘Žđ?‘?đ?‘œđ?‘˘đ?‘&#x; đ?‘Łđ?‘’đ?‘™đ?‘œđ?‘?đ?‘–đ?‘Ąđ?‘Ś = = 28.81450965 đ?‘š/đ?‘

đ?‘€đ?‘–đ?‘›đ?‘šđ?‘˘đ?‘š đ?‘‰đ?‘Žđ?‘?đ?‘œđ?‘˘đ?‘&#x; đ?‘…đ?‘Žđ?‘Ąđ?‘’ 0.7 Ă— 313.955001 = đ??´â„Ž 7.627008178

Therefore, weeping will not occur due to the actual minimum vapour velocity of 28.81 is higher than the design velocity of 11.90, above the weep point. 1.2.3.7.2 Plate Pressure Drop (Initially guessed was 100m liquid): Dry plate drop: đ?‘€đ?‘Žđ?‘Ľđ?‘–đ?‘šđ?‘˘đ?‘š đ?‘Łđ?‘Žđ?‘?đ?‘œđ?‘˘đ?‘&#x; đ?‘Łđ?‘’đ?‘™đ?‘œđ?‘?đ?‘–đ?‘Ąđ?‘Ś đ?‘Ąâ„Žđ?‘&#x;đ?‘œđ?‘˘đ?‘”â„Ž â„Žđ?‘œđ?‘™đ?‘’đ?‘ (đ?‘˘â„Ž ,max ) đ?‘šđ?‘Žđ?‘Ľđ?‘–đ?‘šđ?‘˘đ?‘š đ?‘Łđ?‘œđ?‘™đ?‘’đ?‘šđ?‘Ąđ?‘&#x;đ?‘–đ?‘? đ?‘“đ?‘™đ?‘œđ?‘¤ đ?‘&#x;đ?‘Žđ?‘Ąđ?‘’ đ?‘Žđ?‘Ą đ?‘?đ?‘œđ?‘Ąđ?‘Ąđ?‘œđ?‘š = â„Žđ?‘œđ?‘™đ?‘’ đ?‘Žđ?‘&#x;đ?‘’đ?‘Ž

20

313.955001 = 41.16358521 đ?‘š/đ?‘ 7.627008178 Using the figure below, the orifice coefficient Co can be found to calculate the dry plate pressure drop as following: đ?‘€đ?‘Žđ?‘Ľđ?‘–đ?‘šđ?‘˘đ?‘š đ?‘Łđ?‘Žđ?‘?đ?‘œđ?‘˘đ?‘&#x; đ?‘Łđ?‘’đ?‘™đ?‘œđ?‘?đ?‘–đ?‘Ąđ?‘Ś đ?‘Ąâ„Žđ?‘&#x;đ?‘œđ?‘˘đ?‘”â„Ž â„Žđ?‘œđ?‘™đ?‘’đ?‘ (đ?‘˘â„Ž ,max

)=

đ?‘ƒđ?‘™đ?‘Žđ?‘Ąđ?‘’ đ?‘Ąâ„Žđ?‘–đ?‘?đ?‘˜đ?‘›đ?‘’đ?‘ đ?‘ 3 = = 0.6 â„Žđ?‘œđ?‘™đ?‘’ đ?‘‘đ?‘–đ?‘Žđ?‘šđ?‘’đ?‘Ąđ?‘’đ?‘&#x; 5 đ??´â„Ž đ??´â„Ž 7.627008178 = = = 0.15 = 0.15 Ă— 100 = 15 đ??´đ?‘? đ??´đ?‘Ž 50.84672118 Going to the graph now to get the Co, đ?‘‚đ?‘&#x;đ?‘–đ?‘“đ?‘–đ?‘? đ??śđ?‘œđ?‘’đ?‘“đ?‘“đ?‘–đ?‘?đ?‘–đ?‘’đ?‘›đ?‘Ą, đ??śđ?‘œ = 0.775 Now use the following formula to find dry plate drop (hd): â„Žđ?‘‘ = 51 Ă— (

đ?‘˘â„Ž 2 đ?œŒđ?‘Ł ) Ă—( ) đ??śđ?‘œ đ?œŒđ??ż

41.16358521 2 1.218466705 ) ×( ℎ� = 51 × ( ) 0.775 971.7359111 ℎ� = 180.4091547 �� ������

Figure 12 Discharge coefficient, Co and sieve plate (Sinnott et al, 2009)

Now find the residual head, hr using the following equation:

21

â„Žđ?‘&#x; =

12.5 Ă— 103 12.5 Ă— 103 = = 12.86357729 đ?œŒđ??ż 971.7359111

Now calculate the total pressure drop (ht) using the following equation: â„Žđ?‘Ą = â„Žđ?‘‘ + (â„Žđ?‘¤ + â„Žđ?‘œđ?‘¤ ) + â„Žđ?‘&#x; â„Žđ?‘Ą = 180.4091547 + (193.0233058) + 12.86357729 đ?‘‡đ?‘œđ?‘Ąđ?‘Žđ?‘™ đ?‘?đ?‘&#x;đ?‘’đ?‘ đ?‘ đ?‘˘đ?‘&#x;đ?‘’ đ?‘‘đ?‘&#x;đ?‘œđ?‘? (â„Žđ?‘Ą ) = 386.2960379 đ?‘šđ?‘š đ?‘™đ?‘–đ?‘žđ?‘˘đ?‘–đ?‘‘ This calculated total pressure drop of plate is higher than the assumed value of 100m at the beginning, also this plate pressure drop will only effect the column pressure drop calculation for Dpt. Therefore the new actual column pressure drop will be: ∆đ?‘?đ?‘Ą = 9.81 Ă— 10−3 Ă— â„Žđ?‘Ą Ă— đ?œŒđ??ż (đ?‘Žđ?‘Łđ?‘’đ?‘&#x;đ?‘Žđ?‘”đ?‘’ đ?‘–đ?‘› đ?‘Ąđ?‘œđ?‘? đ?‘Žđ?‘›đ?‘‘ đ?‘?đ?‘œđ?‘Ąđ?‘Ąđ?‘œđ?‘š) Where, Ht is the total plate pressure drop, mm liquid for 22 trays = 8498.512833 mm liquid. đ?‘ đ?‘’đ?‘¤ đ?‘?đ?‘œđ?‘™đ?‘˘đ?‘šđ?‘› đ?‘?đ?‘&#x;đ?‘’đ?‘ đ?‘ đ?‘˘đ?‘&#x;đ?‘’ đ?‘‘đ?‘&#x;đ?‘œđ?‘?, ∆đ?‘ƒđ?‘Ą = 9.81 Ă— 10−3 Ă— 8498.512833 Ă— 961.12 đ?‘ đ?‘ đ?‘’đ?‘¤ đ?‘?đ?‘œđ?‘™đ?‘˘đ?‘šđ?‘› đ?‘?đ?‘&#x;đ?‘’đ?‘ đ?‘ đ?‘&#x;đ?‘’ đ?‘‘đ?‘&#x;đ?‘œđ?‘?, ∆đ?‘ƒđ?‘Ą = 80128.93998 đ?‘ƒđ?‘Ž ( 2 ) = 80.12893998 đ?‘˜đ?‘ƒđ?‘Ž đ?‘š 1.2.3.7.3 Downcomer liquid back-up check: Height of the bottom edge of the apron above the plate = hap = hw – (5 to 10 mm) >>Sinnott Take as 10 mm: â„Žđ?‘Žđ?‘? = 50 − 10 = 40 đ?‘šđ?‘š Area under apron, đ??´đ?‘&#x;đ?‘’đ?‘Ž đ?‘˘đ?‘›đ?‘‘đ?‘’đ?‘&#x; đ?‘Žđ?‘?đ?‘&#x;đ?‘œđ?‘› = đ??´đ?‘Žđ?‘? = đ??źđ?‘¤ Ă— â„Žđ?‘Žđ?‘? = 40 Ă— (1000) Ă— 7.016223974 = 0.280648959 đ?‘š2 As (Aap) area is less than downcomer area (Ad) use the following equation: â„Žđ?‘‘đ?‘? = 166 Ă— (

đ??żđ?‘¤đ?‘‘ 2 ) đ?œŒđ??ż Ă— đ??´đ?‘š

Where, Lwd is the maximum liquid rate in kg/s = 811.0954776 kg/s Am = Area under apron â„Žđ?‘‘đ?‘? = 166 Ă— (

2 811.0954776 ) 971.7359111 Ă— 0.280648959

â„Žđ?‘‘đ?‘? = 493.705971 đ?‘šđ?‘š = 0.493705971 đ?‘š Now back-up in downcomer: â„Žđ?‘? = (â„Žđ?‘¤ + â„Žđ?‘œđ?‘¤ ) + â„Žđ?‘Ą + â„Žđ?‘‘đ?‘? Where, ht is the total plate pressure drop = 386.2960379 mm liquid hdc is the downcomer height = 493.705971 mm đ??ˇđ?‘œđ?‘¤đ?‘?đ?‘œđ?‘šđ?‘’đ?‘&#x; đ?‘?đ?‘Žđ?‘?đ?‘˜ − đ?‘˘đ?‘? (â„Žđ?‘? ) = (50 + 181.4156651) + 386.2960379 + 493.705971 22

â„Žđ?‘? = 1111.417674 đ?‘šđ?‘š = 1.111417674 đ?‘š Now check if the equation is satisfied its conditions: â„Žđ?‘? < (0.5 Ă— (đ?‘?đ?‘™đ?‘Žđ?‘Ąđ?‘’ đ?‘ đ?‘?đ?‘Žđ?‘?đ?‘–đ?‘›đ?‘” + đ?‘¤đ?‘’đ?‘–đ?‘&#x; â„Žđ?‘’đ?‘–đ?‘”â„Žđ?‘Ą)) 0.5 Ă— (đ?‘?đ?‘™đ?‘Žđ?‘Ąđ?‘’ đ?‘ đ?‘?đ?‘Žđ?‘?đ?‘–đ?‘›đ?‘” + đ?‘¤đ?‘’đ?‘–đ?‘&#x; â„Žđ?‘’đ?‘–đ?‘”â„Žđ?‘Ą) = 0.5 Ă— (2.2 + 50 Ă— 10−3 ) = 1.125 Therefore, â„Žđ?‘? = 1.111417674 đ?‘–đ?‘ đ?‘™đ?‘œđ?‘¤đ?‘’đ?‘&#x; đ?‘Ąâ„Žđ?‘Žđ?‘› 1.125 (đ??śđ?‘œđ?‘›đ?‘‘đ?‘–đ?‘Ąđ?‘–đ?‘œđ?‘›đ?‘ đ?‘šđ?‘’đ?‘Ą) Now check Residence Time, tr: đ?‘Ąđ?‘&#x; =

đ??´đ?‘‘ â„Žđ?‘?đ?‘? đ?œŒđ??ż đ??żđ?‘¤đ?‘‘

Where, Tr is the residence time in seconds Hbc is the clear liquid back up in metres đ?‘…đ?‘’đ?‘ đ?‘–đ?‘‘đ?‘’đ?‘›đ?‘?đ?‘’ đ?‘‡đ?‘–đ?‘šđ?‘’, đ?‘Ąđ?‘&#x; =

8.028429661 Ă— 1.111417674 Ă— 971.7359111 = 10.69015933 đ?‘ đ?‘’đ?‘?đ?‘œđ?‘›đ?‘‘đ?‘ 811.0954776

The residence time is over 3 seconds therefore this is satisfactory as per Sinnott. 1.2.3.7.4 Number of Holes: Assume area of one hole is 0.00001964 m2 đ?‘ đ?‘˘đ?‘šđ?‘?đ?‘’đ?‘&#x; đ?‘œđ?‘“ đ??ťđ?‘œđ?‘™đ?‘’đ?‘ = đ?‘ đ?‘˘đ?‘šđ?‘?đ?‘’đ?‘&#x; đ?‘œđ?‘“ đ??ťđ?‘œđ?‘™đ?‘’đ?‘ =

đ??ťđ?‘œđ?‘™đ?‘’ đ??´đ?‘&#x;đ?‘’đ?‘Ž, đ??´â„Ž đ??´đ?‘&#x;đ?‘’đ?‘Ž đ?‘œđ?‘“ đ?‘œđ?‘›đ?‘’ â„Žđ?‘œđ?‘™đ?‘’

7.627008178 = 388340.5386 â„Žđ?‘œđ?‘™đ?‘’đ?‘ 0.00001964

1.2.3.7.5 Check Entrainment: �� =

đ?‘€đ?‘Žđ?‘Ľ đ?‘‰đ?‘œđ?‘™đ?‘˘đ?‘šđ?‘’đ?‘Ąđ?‘&#x;đ?‘–đ?‘? đ??šđ?‘™đ?‘œđ?‘¤ đ?‘…đ?‘Žđ?‘Ąđ?‘’ đ?‘Žđ?‘Ą đ?‘?đ?‘œđ?‘Ąđ?‘Ąđ?‘œđ?‘š đ?‘ đ?‘’đ?‘Ą đ??´đ?‘&#x;đ?‘’đ?‘Ž đ?‘…đ?‘’đ?‘žđ?‘˘đ?‘–đ?‘&#x;đ?‘’đ?‘‘

�� =

313.955001 đ?‘š = 5.332555357 58.87515085 đ?‘

Now Flooding Percent, đ??šđ?‘™đ?‘œđ?‘œđ?‘‘đ?‘–đ?‘›đ?‘” đ?‘ƒđ?‘’đ?‘&#x;đ?‘?đ?‘’đ?‘›đ?‘Ą = (

�� 5.332555357 ) × 100 = 60% ) × 100 = ( �� 8.887592262

FLV = 0.075079888 Using the below graph, we can find the fractional entrainment to equal = 0.015

23

Figure 13 Entrainment correlation for sieve plates (Sinnott et al, 2009)

1.2.3.7.6 Plate Specification: Table 10 Plate specifications

PLATE SPECIFICATION Plates Number Plate Inside Diameter (m) Hole Size (mm) Hole Pitch (mm) Total Holes Active Holes Blanking Area Turn-down Plater Material Downcomer Material Plate Spacing Plate Thickness Plate Pressure Drop (mm liquid)

1 9 5 12.55 388340.5386 70% of max rate Stainless Steel Stainless Steel 2.2 3 mm 386.2960379

Figure 14 Schematic drawing of the plate with the specification (Al-Rikabawi, 2015)

24

1.2.3.8 Column Height đ??ťđ?‘’đ?‘–đ?‘”â„Žđ?‘Ą = (đ?‘ đ?‘œ đ?‘œđ?‘“ đ?‘ đ?‘Ąđ?‘Žđ?‘”đ?‘’đ?‘ − 1) Ă— (đ?‘‡đ?‘&#x;đ?‘Žđ?‘Ś đ?‘ đ?‘?đ?‘Žđ?‘?đ?‘–đ?‘›đ?‘”) + (đ?‘‡đ?‘&#x;đ?‘Žđ?‘Ś đ?‘ đ?‘?đ?‘Žđ?‘?đ?‘–đ?‘›đ?‘” Ă— 2) + (đ?‘ đ?‘œ đ?‘œđ?‘“ đ?‘ đ?‘Ąđ?‘Žđ?‘”đ?‘’đ?‘ ) Ă— (đ?‘‡â„Žđ?‘–đ?‘?đ?‘˜đ?‘›đ?‘’đ?‘ đ?‘ đ?‘œđ?‘“ đ?‘?đ?‘™đ?‘Žđ?‘Ąđ?‘’đ?‘ ) đ??ťđ?‘’đ?‘–đ?‘”â„Žđ?‘Ą = (23 − 1) Ă— (2.2) + (2.2 Ă— 2) + (23) Ă— (3 Ă— 10−3 ) = 52.86 = 53 đ?‘šđ?‘’đ?‘Ąđ?‘&#x;đ?‘’đ?‘ Therefore the height of the column is 53 metres.

1.2.4 Sizing/Specifications (Piping, Nozzles) The sizing of the main nozzles or pipes going into the regenerator column were sized base on the equations introduced by Sinnott and Towler in the chapter Pipe Sizing Selection. The following equation was used to calculate the optimum diameter of pipes of feed, top, bottoms, reflux to column and boil up to the regenerator: đ?‘‘đ?‘– , đ?‘œđ?‘?đ?‘Ąđ?‘–đ?‘šđ?‘˘đ?‘š = 0.465 Ă— đ??ş 0.43 Ă— đ?œŒâˆ’0,31 Where, di, optimum is the pipe optimum diameter in metre

G is the fluid flow rate in kg/s

Υ is the fluid density in Kg/m3 









Sizing of Feed pipe Mass flow rate of feed = 505.15 kg/s Density of feed = 49.35 kg/m3 đ?‘‘đ?‘– , đ?‘œđ?‘?đ?‘Ąđ?‘–đ?‘šđ?‘˘đ?‘š = 0.465 Ă— 505.150.43 Ă— 49.35 −0,31 = 2.01 đ?‘š Sizing of Feed to Condenser pipe Mass flow rate of top product = 674.93 kg/s Density = 1.18 kg/m3 đ?‘‘đ?‘– , đ?‘œđ?‘?đ?‘Ąđ?‘–đ?‘šđ?‘˘đ?‘š = 0.465 Ă— 674.93 0.43 Ă— 1.18−0,31 = 7.27 đ?‘š Sizing of Feed to Reboiler pipe Mass flow rate of bottoms = 1102.10 kg/s Density = 953.63 kg/m3 đ?‘‘đ?‘– , đ?‘œđ?‘?đ?‘Ąđ?‘–đ?‘šđ?‘˘đ?‘š = 0.465 Ă— 1102.10 0.43 Ă— 953.63 −0,31 = 1.127 đ?‘š Sizing of Reflux to Column pipe Mass flow rate of reflux = 598.32 kg/s Density = 950.5 kg/m3 đ?‘‘đ?‘– , đ?‘œđ?‘?đ?‘Ąđ?‘–đ?‘šđ?‘˘đ?‘š = 0.465 Ă— 598.32 0.43 Ă— 950.5 −0,31 = 0.86 đ?‘š Sizing of Boil Up to Column pipe Mass flow rate of boil up = 673.55 kg/s Density = 1.218 kg/m3 đ?‘‘đ?‘– , đ?‘œđ?‘?đ?‘Ąđ?‘–đ?‘šđ?‘˘đ?‘š = 0.465 Ă— 673.55 0.43 Ă— 1.218 −0,31 = 7.2 đ?‘š Table 11 Sizing of pipes going into the regenerator

Pipe Optimum Diameter, m Standard Diameter, inch

Feed (3-8) 2.01

Top 7.27

Bottom 1.127

Reflux 0.86

Boil Up 7.2

36

36

36

36

36

25

1.3 Operational Design 1.3.1 Control System Design C/W inlet PSV (Pressure Relief Valve)

FC

TI

CO2 to flare Reflux Drum

Condenser(E-303)

LG

LC TC

PG

FC

#1

TI Regenerator (T-302)

PG

LP Steam

TI

Rich MDEA

#22

TG

TI

LG

Reboiler (E-304)

PDI P

PC Steam Condensate

Steam Trap

CAPSTONE ENERGY

Lean MDEA

Client: Chevron Distillation Column

DWG No. PID 101

Sheet: 2/2

DWG: Humam

Date: 01/11/2015

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The typical control system in the distillation column: Level control Level control used where liquid-liquid or liquid vapour interface is hold. Level controller used in two positions in the regenerator column: 1. at the bottom of distillation column the level controller is used which is called (LG), 2. At the top of the regenerator column (LG) the level controller used to monitor the interface between the liquid and vapour inventories at the column sump (Bottom) and reflux drum (Top). Flow Control (FC) the function of flow control is attached at the top of distillation column is to control the liquid flow coming from the reflux drum to achieve the desired split between distillate. The other flow controller is to control the flow of CO2 vapour product in case of pressure stabilization in the column. Pressure Control (PC) Pressure control used at the top of distillation column to cool down the temperature of the vapour produced at the top of distillation column. Pressure controller usually connected to the condenser to control the vapour inventory in the column. Temperature Control (TC) temperature control used to control the flow of the vapour coming from the reboiler at the bottom of the distillation column by either varying the boil-up or bottoms flow hence varying the temperature to control the flow of the column.

1.3.2 Operating Procedures, Start-Up and Shut-Down 



Start Up 1. Check the equipment before entering the feed 2. Clean all unwanted materials from the bottom 3. Check the pressure and make sure that required pressure has met 4. Check the operation of the condenser and the utility used (Water) 5. Check the operation of the reboiler and its utility (LP Steam) 6. Start entering the feed and gradually increase the flow rate 7. Control heating and cooling while increasing the flow rate Shut Down 1. Reduce the flow rate gradually at low rate 2. Shut down the heating and cooling utilities 3. Drain any liquid or vapour available in the column 4. Reduce the pressure of the vessel gradually till it reaches atmospheric pressure 5. Check the temperature of the vessel and make sure it is around normal conditions

Emergency Shut Down and Maintenance The maintenance of the vessel should be scheduled regularly to maintain the vessel to work at its operating conditions. Maintenance includes replacement of some parts in the vessel like (Trays, Bolts). Some problems appear to be in distillation column like: Weeping, Flooding and foaming. In case of any emergency the following step should be followed: 1. Shut down the feed entering the column 2. Shut down the utilities in the condenser and reboiler

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1.3.3 Safety Study (HAZOP) on the Regenerator Column Key Word

Deviation

No Flow

Consequence  Poor quality of products  High acid gas emission levels  Valve over heat  Loss of production  Loss of profit

Action Required  Install low level alarm on LIC at the base of the column  Install kick-back on pumps  Regular inspection on transfer lines and seals plant emergency shutdown procedures  Install high level alarm on LIC at the base of the column  Install low level alarm on LIC  Install temperature sensors

 Higher production but weaker product  Pump Failure  Possible higher CO2 due to lower absorption and stripping

 Line subjected to full delivery pressure  Possible line fracture or flange leakage  Explosions

 Install kick-back on pumps  Install a pressure gauge upstream of the delivery  Provide thermal expansion relief in the valve section

 Decreased regeneration lower product make

 Install low level alarm on LIC at the base of the column

Less Temperature

 Isolation valve is closed in error while pump running  Thermal expansion on the isolation valve section (fire)  Flange leakage or valve stub blanked but leaking  Less feed temperature from the absorber  Unexpected weather conditions

 Increased CO2 concentrations in product  Less efficient separation occurs  Product not meeting specs

 Install low temperature alarm on LIC at the base of the column  Regular inspection on transfer lines and seals.

Maintenance

 Equipment failure  Flange leak

 Process stoppage  Breach of safety regulations

 Ensure all pipes and fittings are constructed of the right materials and are stress relived

More Temperature More Of

More Pressure

Less Flow

Other

Possible Cause Pump failure Valve fails shut Pipe Fracture Power Failure

 Higher humidity in feed  High Feed rate  Higher feed rates causing larger heats of reaction

More Flow

Less Of

   

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1.4 Mechanical Design of Regenerator 1.4.1 Materials of Construction The materials of construction is an important thing to consider at the design stage of the distillation column so that the column can withstand a specific pressure, stresses and temperature as well as the most critical thing which is the corrosion. The material of construction for the column parts are as following:   



The regenerator will be constructed and designed from a stainless steel of 18CR, 8Ni due to the fluid having CO2 corrosive gas which may cause corrosion issues over the short term. The column support (Skirt) is constructed from plain carbon steel as it is only there to support the column without contacting the corrosive fluids. The regenerator plates are constructed from stainless steel of thickness of 3mm and that gives the plate a long life span before it corrodes due to the ongoing contact of liquid and vapour containing CO2 gas. The downcomer is constructed from a stainless steel due to the same reason as plate.

It might be costly to choose the stainless steel material over the carbon steel but stainless steel will give a better and longer protection against corrosion over the life time of the plant. (Sinnott et al, 2009)

1.4.2 Pressure Vessel Design Calculations Column operating pressure = 220 kPa= 2.2 bar (Absolute) Column Operating temperature = 110 ⠰C = 230 F 

The operating temperature of the column should be the minimum working temperature in the system (Condenser Temperature).

Design pressure should be above the operating pressure by 5-10% of the operating gauge pressure (Sinnott et al, 2009). Using the following formula with 10% safety factor: đ??ˇđ?‘’đ?‘ đ?‘–đ?‘”đ?‘› đ?‘ƒđ?‘&#x;đ?‘’đ?‘ đ?‘ đ?‘˘đ?‘&#x;đ?‘’, đ?‘ƒđ?‘– = đ??śđ?‘œđ?‘™đ?‘˘đ?‘šđ?‘› đ?‘?đ?‘&#x;đ?‘’đ?‘ đ?‘ đ?‘˘đ?‘&#x;đ?‘’ đ?‘–đ?‘› đ?‘?đ?‘Žđ?‘&#x; Ă— 1.1 đ??ˇđ?‘’đ?‘ đ?‘–đ?‘”đ?‘› đ?‘ƒđ?‘&#x;đ?‘’đ?‘ đ?‘ đ?‘˘đ?‘&#x;đ?‘’, đ?‘ƒđ?‘– = (2.2) Ă— 1.1 = 2.42 đ?‘?đ?‘Žđ?‘&#x; = 0.242 đ?‘ /đ?‘šđ?‘š2

1.4.2.1 Maximum Allowable Stress: The material of the column was chosen to be stainless steel 18CR, 8NI, the maximum allowable stress at 230 F for this material was extracted from Table (13.2) in Sinnott book for the typical maximum allowable stresses.

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Figure 15 Typical maximum allowable stresses of steels (Sinnott et al, 2009)

By interpolation between temperatures of 100 and 300 Fahrenheit on the table, was found to be: đ?‘€đ?‘Žđ?‘Ľđ?‘–đ?‘šđ?‘˘đ?‘š đ??´đ?‘™đ?‘™đ?‘œđ?‘¤đ?‘Žđ?‘?đ?‘™đ?‘’ đ?‘†đ?‘Ąđ?‘&#x;đ?‘’đ?‘ đ?‘ (đ?‘†) đ?‘Žđ?‘Ą 230 đ??š = 16750 đ?‘ƒđ?‘ đ?‘– = 115.4871847 đ?‘ /đ?‘šđ?‘š2 The column diameter = 9 m The column height = 53 m

1.4.2.2 Design of Domed Head According to (Sinnott et al, 2009), ellipsoidal head is the most economical and the majority of ellipsoidal heads are constructed with a major and minor axis ratio of 2:1. The minimum thickness can be estimated by the following equation: đ?‘Ą=

đ?‘ƒđ?‘– Ă— đ??ˇđ?‘– +đ??ś 2đ?‘†đ??¸ − 0.2đ?‘ƒđ?‘–

Where, Pi is the design pressure estimated previously = 0.242 N/mm2 Di is the inside diameter of the column which equals to 9 metres = 9000 mm S is the maximum allowable stress estimated above = 115.4871847 N/mm2 E is the welded joint efficiency = 0.85 for double welded butt joint (estimated from the below table) C is the corrosion allowance which assumed to be = 4mm due to the more severe conditions where severe corrosion is expected due to the presence of CO2 in column.

Figure 16 Maximum Allowable Joint Efficiency (Sinnott et al, 2009)

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Now calculate the thickness of the domed head: đ?‘Ą=

0.242 Ă— 9000 +4 2 Ă— 115.4871847 Ă— 0.85 − 0.2 Ă— 0.242 đ?‘Ą = 15 đ?‘šđ?‘š

Therefore, the thickness of domed head of the column is 15 mm thick as can be shown in the figure below:

Figure 17 Ellipsoidal domed head of the regenerator (Al-Rikabawi, 2015)

1.4.2.3 Minimum Wall Thickness of Vessel The minimum practical wall thickness of pressure vessels depends on the vessel diameter as per (Sinnott). As a general guide the wall thickness of any vessel should not be less than the values given in the table below.

Figure 18 Wall thickness of vessels including 2 mm corrosion allowance (Sinnott et al, 2009)

As the regenerator diameter of 9 meters is not listed in the table above, therefore it was assumed by extrapolation that the wall thickness of the regenerator for a 9 metres equals to: đ?‘€đ?‘–đ?‘›đ?‘–đ?‘šđ?‘˘đ?‘š đ?‘Šđ?‘Žđ?‘™đ?‘™ đ?‘‡â„Žđ?‘–đ?‘?đ?‘˜đ?‘›đ?‘’đ?‘ đ?‘ đ?‘œđ?‘“ đ?‘…đ?‘’đ?‘”đ?‘’đ?‘›đ?‘’đ?‘&#x;đ?‘Žđ?‘Ąđ?‘œđ?‘&#x; = 36 đ?‘šđ?‘š (đ??źđ?‘›đ?‘?đ?‘™đ?‘˘đ?‘‘đ?‘–đ?‘›đ?‘” 4 đ?‘šđ?‘š đ?‘?đ?‘œđ?‘&#x;đ?‘&#x;đ?‘œđ?‘ đ?‘–đ?‘œđ?‘› đ?‘Žđ?‘™đ?‘™đ?‘œđ?‘¤đ?‘Žđ?‘›đ?‘?đ?‘’)

1.4.2.4 Minimum Cylinder Shell Thickness It can be found using the following equation: đ?‘Ą= đ?‘Ą=

đ?‘ƒđ?‘– Ă— đ??ˇđ?‘– +đ??ś 2đ?‘†đ??¸ − 1.2đ?‘ƒđ?‘–

0.242 Ă— 9000 +4 2 Ă— 115.4871847 Ă— 0.85 − 1.2 Ă— 0.242 đ?‘Ą = 15.11 đ?‘šđ?‘š

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1.4.2.5 Dead Weight of the Column (Shell) The following equation can be used: đ?‘Šđ?‘Ł = 240đ??śđ?‘¤ đ??ˇđ?‘š (đ??ťđ?‘Ł + 0.8đ??ˇđ?‘š )đ?‘Ą Where, Cw is a factor to account for the weight of nozzles, manways, internal supports etc. and it can be taken as = 1.15 for distillation columns. Dm is the mean diameter of the vessel and can be calculated as following: = (đ??ˇđ?‘– + đ?‘Ą Ă— 10−3 ) = (9 + 36 Ă— 10−3 ) = 9.26787365 đ?‘šđ?‘’đ?‘Ąđ?‘&#x;đ?‘’ Hv is the height or length between tangent lines (length of cylindrical section) = 22 x 2.2 = 48.4 metre đ?‘Šđ?‘Ł = 240 Ă— 1.15 Ă— 9.26787365 Ă— (48.4 + 0.8 Ă— 9.26787365) Ă— 36 = 5139692.791 đ?‘ Therefore, the shell weight of the vessel is đ?&#x;“đ?&#x;?đ?&#x;‘đ?&#x;—đ?&#x;”đ?&#x;—đ?&#x;?. đ?&#x;•đ?&#x;—đ?&#x;? đ?‘ľ or 5139.692791 kN.

1.4.2.6 Total Weight of Plates Plate area can be calculated as following: đ?œ‹ 3.14 ) Ă— 92 = 63.585 đ?‘š2 đ?‘ƒđ?‘™đ?‘Žđ?‘Ąđ?‘’ đ??´đ?‘&#x;đ?‘’đ?‘Ž = ( ) Ă— đ??ˇđ?‘–2 = ( 4 4 Contacting plates (steel) typical loading = 1.2 kN/m2 đ?‘Šđ?‘’đ?‘–đ?‘”â„Žđ?‘Ą đ?‘œđ?‘“ đ?‘Ąâ„Žđ?‘’ đ?‘?đ?‘™đ?‘Žđ?‘Ąđ?‘’ đ?‘–đ?‘›đ?‘?đ?‘™đ?‘˘đ?‘‘đ?‘–đ?‘›đ?‘” đ?‘™đ?‘–đ?‘žđ?‘˘đ?‘–đ?‘‘ đ?‘œđ?‘› đ?‘–đ?‘Ą = đ?‘ƒđ?‘™đ?‘Žđ?‘Ąđ?‘’ đ?‘Žđ?‘&#x;đ?‘’đ?‘Ž Ă— đ?‘ƒđ?‘™đ?‘Žđ?‘Ąđ?‘’ đ??żđ?‘œđ?‘Žđ?‘‘đ?‘–đ?‘›đ?‘” đ?‘Šđ?‘’đ?‘–đ?‘”â„Žđ?‘Ą đ?‘œđ?‘“ đ?‘?đ?‘™đ?‘Žđ?‘Ąđ?‘’ đ?‘–đ?‘›đ?‘?đ?‘˘đ?‘‘đ?‘–đ?‘›đ?‘” đ?‘™đ?‘–đ?‘žđ?‘˘đ?‘–đ?‘‘ đ?‘œđ?‘› đ?‘–đ?‘Ą = 63.585 Ă— 1.2 = 76.302 đ?‘˜đ?‘ Therefore, the total weight for 22 plates is: đ?‘ťđ?’?đ?’•đ?’‚đ?’? đ?’˜đ?’†đ?’Šđ?’ˆđ?’‰đ?’• đ?’?đ?’‡ đ?’•đ?’‰đ?’† đ?’‘đ?’?đ?’‚đ?’•đ?’†đ?’” = đ?&#x;?đ?&#x;? Ă— đ?&#x;•đ?&#x;”. đ?&#x;‘đ?&#x;Žđ?&#x;? = đ?&#x;?đ?&#x;”đ?&#x;•đ?&#x;–. đ?&#x;”đ?&#x;’đ?&#x;’ đ?’Œđ?‘ľ

1.4.2.7 Insulation Weight Calculations A mineral wool is used for the insulation with a density of = 130 kg/m3 Thickness of the wool = 75 mm Approximate volume of insulator can be found as following: đ?‘‰đ?‘œđ?‘™đ?‘˘đ?‘šđ?‘’ đ?‘œđ?‘“ đ??źđ?‘›đ?‘ đ?‘˘đ?‘™đ?‘Žđ?‘Ąđ?‘œđ?‘&#x; = đ?œ‹ Ă— 2 Ă— 48.4 Ă— 75 Ă— 10−3 = 22.7964 đ?‘š3 Insulator weight is calculated as following: đ??źđ?‘›đ?‘ đ?‘˘đ?‘™đ?‘Žđ?‘Ąđ?‘œđ?‘&#x; đ?‘Šđ?‘’đ?‘–đ?‘”â„Žđ?‘Ą = đ?‘‰đ?‘œđ?‘™đ?‘˘đ?‘šđ?‘’ đ?‘œđ?‘“ đ?‘–đ?‘›đ?‘ đ?‘˘đ?‘™đ?‘Žđ?‘Ąđ?‘œđ?‘&#x; Ă— 130 Ă— 9.81 đ??źđ?‘›đ?‘ đ?‘˘đ?‘™đ?‘Žđ?‘Ąđ?‘œđ?‘&#x; đ?‘Šđ?‘’đ?‘–đ?‘”â„Žđ?‘Ą = 22.7964 Ă— 130 Ă— 9.81 = 29072.24892 đ?‘ đ?‘’đ?‘¤đ?‘œđ?‘Ąđ?‘› Double this to allow for fittings, Therefore, the weight of the insulation is: đ?‘°đ?’?đ?’”đ?’–đ?’?đ?’‚đ?’•đ?’Šđ?’?đ?’? đ?‘žđ?’†đ?’Šđ?’ˆđ?’‰đ?’• = đ?&#x;? Ă— đ?&#x;?đ?&#x;—đ?&#x;Žđ?&#x;•đ?&#x;?. đ?&#x;?đ?&#x;’đ?&#x;–đ?&#x;—đ?&#x;? = đ?&#x;“đ?&#x;–đ?&#x;?đ?&#x;’đ?&#x;’. đ?&#x;’đ?&#x;—đ?&#x;•đ?&#x;–đ?&#x;’ đ?‘ľ = đ?&#x;“đ?&#x;–. đ?&#x;?đ?&#x;’đ?&#x;’đ?&#x;’đ?&#x;—đ?&#x;•đ?&#x;–đ?&#x;’ đ?’Œđ?‘ľ

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1.4.2.8 Total Weight of Column (Shell + Plates + Insulation) đ?‘‡đ?‘œđ?‘Ąđ?‘Žđ?‘™ đ?‘¤đ?‘’đ?‘–đ?‘”â„Žđ?‘Ą = đ?‘Šđ?‘’đ?‘–đ?‘”â„Žđ?‘Ą đ?‘œđ?‘“ đ?‘†â„Žđ?‘’đ?‘™đ?‘™ + đ?‘Šđ?‘’đ?‘–đ?‘”â„Žđ?‘Ą đ?‘œđ?‘“ đ?‘ƒđ?‘™đ?‘Žđ?‘Ąđ?‘’đ?‘ + đ?‘Šđ?‘’đ?‘–đ?‘”â„Žđ?‘Ą đ?‘œđ?‘“ đ??źđ?‘›đ?‘ đ?‘˘đ?‘™đ?‘Žđ?‘Ąđ?‘–đ?‘œđ?‘› đ?‘‡đ?‘œđ?‘Ąđ?‘Žđ?‘™ đ?‘¤đ?‘’đ?‘–đ?‘”â„Žđ?‘Ą = 5139.692791 + 1678.644 + 58.14449784 đ?‘ťđ?’?đ?’•đ?’‚đ?’? đ?‘žđ?’†đ?’Šđ?’ˆđ?’‰đ?’• = đ?&#x;”đ?&#x;–đ?&#x;•đ?&#x;”. đ?&#x;’đ?&#x;–đ?&#x;?đ?&#x;?đ?&#x;–đ?&#x;— đ?’Œđ?‘ľ

1.4.2.9 Wind Loading Calculations Assume dynamic wind pressure (Pw) as = 1280 N/m2 (Corresponding to 160 kph (100mph)) Mean diameter (Effective diameter) including insulation can be found as: đ??ˇđ?‘’đ?‘“đ?‘“ = đ??źđ?‘›đ?‘Ąđ?‘’đ?‘&#x;đ?‘›đ?‘Žđ?‘™ đ??ˇđ?‘–đ?‘Žđ?‘šđ?‘’đ?‘Ąđ?‘’đ?‘&#x; Ă— (đ??źđ?‘›đ?‘Ąđ?‘’đ?‘&#x;đ?‘›đ?‘Žđ?‘™ đ??ˇđ?‘–đ?‘Žđ?‘šđ?‘’đ?‘Ąđ?‘’đ?‘&#x; Ă— ((đ??śđ?‘œđ?‘&#x;đ?‘&#x;đ?‘œđ?‘ đ?‘–đ?‘œđ?‘› đ??´đ?‘™đ?‘™đ?‘œđ?‘¤đ?‘Žđ?‘›đ?‘?đ?‘’ + đ?‘‡â„Žđ?‘–đ?‘?đ?‘˜đ?‘›đ?‘’đ?‘ đ?‘ đ?‘œđ?‘“ đ??źđ?‘›đ?‘ đ?‘˘đ?‘™đ?‘Žđ?‘Ąđ?‘–đ?‘œđ?‘›) Ă— 10−3 )) đ??¸đ?‘“đ?‘“đ?‘’đ?‘?đ?‘Ąđ?‘–đ?‘Łđ?‘’ đ??ˇđ?‘–đ?‘Žđ?‘šđ?‘’đ?‘Ąđ?‘’đ?‘&#x; = đ??ˇđ?‘’đ?‘“đ?‘“ = 9 Ă— (9 Ă— ((4 + 75) Ă— 10−3 )) = 9.961191668 đ?‘šđ?‘’đ?‘Ąđ?‘&#x;đ?‘’đ?‘ Now find the loading per length, Fw: đ??šđ?‘¤ = đ?‘ƒđ?‘¤ Ă— đ??ˇđ?‘’đ?‘“đ?‘“ Where, Pw is the wind pressure đ??šđ?‘¤ = 1280 Ă— 9.961191668 = 12750.32534 đ?‘ /đ?‘š Now find the bending moment at bottom tangent line, Mx: đ?‘€đ?‘Ľ =

đ?‘Š Ă— đ?‘Ľ2 2

Where, x is the distance measured from free end = 48.4 metres W is the load per unit length = 12750.32534 N/m đ??ľđ?‘’đ?‘›đ?‘‘đ?‘–đ?‘›đ?‘” đ?‘€đ?‘œđ?‘šđ?‘’đ?‘›đ?‘Ą đ?‘Žđ?‘Ą đ?‘?đ?‘œđ?‘Ąđ?‘Ąđ?‘œđ?‘š, đ?‘€đ?‘Ľ =

12750.32534 Ă— (48.4)2 = 14934201.06 đ?‘ đ?‘š 2

1.4.2.10 Stress Analysis Pressure stresses: đ?‘ƒ Ă— đ??ˇđ?‘– 0.242 Ă— 9 Ă— 1000 đ?‘ = = 15.51467655 4Ă—đ?‘Ą 4 Ă— 36 đ?‘šđ?‘š2 đ?‘ƒ Ă— đ??ˇđ?‘– 0.242 Ă— 9 Ă— 1000 đ??śđ?‘–đ?‘&#x;đ?‘?đ?‘˘đ?‘šđ?‘“đ?‘’đ?‘&#x;đ?‘’đ?‘›đ?‘Ąđ?‘–đ?‘Žđ?‘™ đ?‘†đ?‘Ąđ?‘&#x;đ?‘’đ?‘ đ?‘ = đ?œŽâ„Ž = = = 31.0293531 đ?‘ /đ?‘šđ?‘š2 2Ă—đ?‘Ą 2 Ă— 36 đ??żđ?‘œđ?‘›đ?‘”đ?‘–đ?‘Ąđ?‘˘đ?‘‘đ?‘Žđ?‘™ đ?‘†đ?‘Ąđ?‘&#x;đ?‘’đ?‘ đ?‘ = đ?œŽđ??ż =

Dead Weight Stresses, Ďƒw: đ?œŽđ?‘¤ =

đ?‘Šđ?‘Ł đ?œ‹ Ă— (đ??ˇđ?‘– + đ?‘Ą) Ă— đ?‘Ą

Where, Wv is the total weight in Newton = 6876.481289 x 1000 N 6876.481289 x 1000 đ?‘ đ?œŽđ?‘¤ = = 6.563779941 (đ??śđ?‘œđ?‘šđ?‘?đ?‘&#x;đ?‘’đ?‘ đ?‘ đ?‘–đ?‘Łđ?‘’) đ?œ‹ Ă— (9000 + 36) Ă— 36 đ?‘šđ?‘š2

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Bending Stresses: 𝐷𝑜 = (𝐶𝑜𝑙𝑢𝑚𝑛 𝐷𝑖𝑎𝑚𝑒𝑡𝑒𝑟 𝑖𝑛 𝑚𝑚) + (2 × 𝑀𝑖𝑛𝑚𝑢𝑚 𝑊𝑎𝑙𝑙 𝑇ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠) = 9000 + (2 × 36) = 9303.87365 𝑚𝑚 𝜋 3.14 ) × (9303.873654 − 90004 ) = 1.12484𝐸 + 13 𝑚𝑚4 𝐼𝑣 = ( ) × (𝐷𝑜4 − 𝐷𝑖4 ) = ( 64 64 𝑀 𝐷𝑖 14934201.06 9000 )×( 𝑇ℎ𝑒 𝑏𝑒𝑛𝑑𝑖𝑛𝑔 𝑠𝑡𝑟𝑒𝑠𝑠 = 𝜎𝑏 = ± ( ) × ( + 𝑡) = ± ( + 36) 𝐼𝑣 2 1.12484𝐸 + 13 2 𝑁 = ±6.176277127 𝑚𝑚2 σw is compressive therefore negative 𝜎𝑧 (𝑈𝑝𝑤𝑖𝑛𝑑) = 𝐿𝑜𝑛𝑔𝑖𝑡𝑢𝑑𝑎𝑙 𝑆𝑡𝑟𝑒𝑠𝑠 − 𝐷𝑒𝑎𝑑 𝑊𝑒𝑖𝑔ℎ𝑡 𝑆𝑡𝑟𝑒𝑠𝑠𝑒𝑠 + 𝐵𝑒𝑛𝑑𝑖𝑛𝑔 𝑆𝑡𝑟𝑒𝑠𝑠 = 15.51467655 − 6.563779941 + 6.176277127 = 𝟏𝟓. 𝟏𝟐𝟕𝟏𝟕𝟑𝟕𝟒 𝑵/𝒎𝒎𝟐

𝜎𝑧 (𝐷𝑜𝑤𝑛𝑤𝑖𝑛𝑑) = 𝐿𝑜𝑛𝑔𝑖𝑡𝑢𝑑𝑎𝑙 𝑆𝑡𝑟𝑒𝑠𝑠 − 𝐷𝑒𝑎𝑑 𝑊𝑒𝑖𝑔ℎ𝑡 𝑆𝑡𝑟𝑒𝑠𝑠𝑒𝑠 − 𝐵𝑒𝑛𝑑𝑖𝑛𝑔 𝑆𝑡𝑟𝑒𝑠𝑠 = 15.51467655 − 6.563779941 − 6.176277127 = −𝟐. 𝟕𝟕𝟒𝟔𝟏𝟗𝟒𝟖𝟐 𝑵/𝒎𝒎𝟐

Figure 19 Upwind and Downwind forces on the regenerator (Al-Rikabawi, 2015)

(𝐶𝑖𝑟𝑐𝑢𝑚𝑓𝑒𝑟𝑛𝑡𝑖𝑎𝑙 𝑆𝑡𝑟𝑒𝑠𝑠𝑒𝑠 − (𝐷𝑜𝑤𝑛𝑤𝑖𝑛𝑑 𝑠𝑡𝑟𝑒𝑠𝑠)) = (31.0293531 − (−2.774619482)) = 28.25473362 𝑁/𝑚𝑚2 This value of 28.25473362 is lower than the maximum allowable stress (S) of 115.4871847 which means that the column is withstanding the maximum allowable stress and it is not going to explode.

1.4.2.11 Elastic Stability (Buckling) Check

𝑡 36 ) 𝐶𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝐵𝑢𝑐𝑘𝑙𝑖𝑛𝑔 𝑆𝑡𝑟𝑒𝑠𝑠 = 𝜌𝑐 = (2 × 104 ) × ( ) = (2 × 104 ) × ( 𝐷𝑜 9303.87365 = 𝟕𝟕. 𝟑𝟖𝟕𝟏𝟐𝟏𝟒𝟒 𝑵/𝒎𝒎𝟐

The maximum compressive stress will occur when vessel is not under pressure: 𝐷𝑒𝑎𝑑 𝑊𝑒𝑖𝑔ℎ𝑡 𝑆𝑡𝑟𝑒𝑠𝑠 + 𝐵𝑒𝑛𝑑𝑖𝑛𝑔 𝑆𝑡𝑟𝑒𝑠𝑠 = 6.563779941 + 6.176277127 = 𝟏𝟐. 𝟕𝟒𝟎𝟎𝟓𝟕𝟎𝟕 This value of 12.74005707 well below critical buckling stress. So the design is satisfactory. 34

1.4.2.12 Selection of Pressure Relief Valve The purpose of the pressure relief valve is to protect the pressure vessels from exceeding their design pressures by more than a pre-determined fixed amount of pressure. Different types of safety valves are used in the industry for various purposes. The intention of a safety valve is to protect the pressure vessel from being damaged or exploded by regulating the pressure inside the column, there are two common types of safety valves which are standard design and balanced bellows design. (Whitesides, 2012) The following calculations are done to determine the type of the safety valve to be used at this regenerator top: Firstly the orifice effective area should be calculated using the following equation: đ??´=

đ?‘Š Ă— √đ?‘‡đ?‘? đ??śđ??žđ?‘ƒ1 đ??žđ?‘? √đ?‘€

Where, W is the flow rate in lb/hr which equals to 5416054.93 (from HYSYS) T is the inlet vapour temperature in Fahrenheit which equals to 709 (From HYSYS) Z is the compressibility which is taken as 1 C is the flow constant determined by the ratio of specific heats, can be used as 346 for CO2 K is the coefficient of discharge which equals to 0.975 P1 is the upstream pressure which equals to 29.73+2.973+15 = 47.703 Psia Kb is the correction factor due to the back pressure, can be taken as 1 M is the molecular weight of the fluid which equals to 18.6 (Taken from HYSYS)

đ?‘‡â„Žđ?‘’ đ?‘œđ?‘&#x;đ?‘–đ?‘“đ?‘–đ?‘? đ?‘’đ?‘“đ?‘“đ?‘’đ?‘?đ?‘Ąđ?‘–đ?‘Łđ?‘’ đ?‘Žđ?‘&#x;đ?‘’, đ??´ =

5416054.93 Ă— √709 Ă— 1 346 Ă— 0.975 Ă— 47.703 Ă— 1√18.6

= 2076.34 đ?‘–đ?‘›2 = 1.33 đ?‘š2

Now going back to the table of the standard nozzle orifice data from (Whitesides, 2012), we can see that the maximum orifice area on the table is 26 inch2 and the area I calculated is much bigger and that is due to the big flow rate. The size designation of the pressure safety relief valve can be assumed to be T. Spring-Loaded Relief Valve can be used here as it is the most commonly used device due to its extensive advantages.

35

1.4.3 Skirt Supports and Foundations Design Assume straight cylindrical skirt for plain carbon steel with base angle θs of 90⁰. Young’s modulus E = 200000 N/mm2 𝜋 𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝑑𝑒𝑎𝑑 𝑤𝑒𝑖𝑔ℎ𝑡 𝑙𝑜𝑎𝑑 𝑜𝑛 𝑡ℎ𝑒 𝑠𝑘𝑖𝑟𝑡 = ( × 𝐷𝑖𝑎𝑚𝑒𝑡𝑒𝑟 2 × 𝐻𝑒𝑖𝑔ℎ𝑡) × 1000 × 9.81 4 𝜋 𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝑑𝑒𝑎𝑑 𝑤𝑒𝑖𝑔ℎ𝑡 𝑙𝑜𝑎𝑑 𝑜𝑛 𝑡ℎ𝑒 𝑠𝑘𝑖𝑟𝑡 = ( × 92 × 53) × 1000 × 9.81 = 34785178.61 𝑁 4 = 34785.17861 𝑘𝑁 𝑇𝑜𝑡𝑎𝑙 𝐶𝑜𝑙𝑢𝑚𝑛 𝑊𝑒𝑖𝑔ℎ𝑡 = 𝑉𝑒𝑠𝑠𝑒𝑙 𝐷𝑒𝑎𝑑 𝑊𝑒𝑖𝑔ℎ𝑡 + 𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝐷𝑒𝑎𝑑 𝑊𝑒𝑖𝑔ℎ𝑡 = 34785.17861 + 6876.481289 = 41661.6599 𝑘𝑁 𝑊𝑖𝑛𝑑 𝐿𝑜𝑎𝑑𝑖𝑛𝑔, 𝐹𝑤 =

𝐹𝑤 𝑖𝑛 𝑁 12750.32534 ) = 12.75032534 𝑘𝑁/𝑚 =( 1000 1000

𝐵𝑒𝑛𝑑𝑖𝑛𝑔 𝑚𝑜𝑚𝑒𝑛𝑡 𝑎𝑡 𝑏𝑎𝑠𝑒 𝑜𝑓 𝑠𝑘𝑖𝑟𝑡, 𝑀𝑥 =

𝑤 × 𝑥2 = 2

Assume the height of the skirt is 3 metres. Distance between free end top is 56 𝐵𝑒𝑛𝑑𝑖𝑛𝑔 𝑚𝑜𝑚𝑒𝑛𝑡 𝑎𝑡 𝑏𝑎𝑠𝑒 𝑜𝑓 𝑠𝑘𝑖𝑟𝑡 = 19992.51013 𝑘𝑁𝑚 Skirt thickness (assumed as same to vessel bottom section) = 36 mm 𝑆𝑘𝑖𝑟𝑡 𝑏𝑒𝑛𝑑𝑖𝑛𝑔 𝑠𝑡𝑟𝑒𝑠𝑠 = 𝜎𝑏𝑠 = (4 × 𝑀𝑠 )/(𝜋 × (𝐷𝑠 + 𝑡𝑠 )𝐷𝑠 𝑡𝑠 𝑆𝑘𝑖𝑟𝑡 𝑏𝑒𝑛𝑑𝑖𝑛𝑔 𝑠𝑡𝑟𝑒𝑠𝑠 = 𝜎𝑏𝑠 = (4 × 19992.51013 × 1000 × 1000)/(𝜋 × (9000 + 36) × 9000 × 36 = 8.268471062 𝑁/𝑚𝑚2 𝑂𝑝𝑒𝑟𝑎𝑡𝑖𝑛𝑔 𝑆𝑘𝑖𝑟𝑡 𝐷𝑒𝑎𝑑 𝑊𝑒𝑖𝑔ℎ𝑡 𝑆𝑡𝑟𝑒𝑠𝑠 = 𝜎𝑤𝑠 ,𝑜𝑝𝑒𝑟𝑎𝑡𝑖𝑛𝑔 = =

6876.481289 × 1000 3.14 × ((9000) + 36) × 36

𝑇𝑒𝑠𝑡 𝑆𝑘𝑖𝑟𝑡 𝐷𝑒𝑎𝑑 𝑊𝑒𝑖𝑔ℎ𝑡 𝑆𝑡𝑟𝑒𝑠𝑠 = 𝜎𝑤𝑠 ,𝑡𝑒𝑠𝑡 = = 39.76713614 𝑁/𝑚𝑚2

𝑤 𝜋(𝐷𝑠 + 𝑡𝑠 ) × 𝑡𝑠

= 6.563779941 𝑁/𝑚𝑚2

𝑤 41661.6599 × 1000 = 𝜋(𝐷𝑠 + 𝑡𝑠 ) × 𝑡𝑠 3.14 × ((9000) + 36) × 36

𝑀𝑎𝑥𝑖𝑚𝑢𝑚 (𝐶𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑣𝑒) = 𝜎𝑠 = 𝑆𝑘𝑖𝑟𝑡 𝐵𝑒𝑛𝑑𝑖𝑛𝑔 𝑆𝑡𝑟𝑒𝑠𝑠 + 𝑇𝑒𝑠𝑡 𝑠𝑘𝑖𝑟𝑡 𝑑𝑒𝑎𝑑 𝑠𝑡𝑟𝑒𝑠𝑠 = 8.268471062 + 39.76713614 = 48.0356072 𝑁/𝑚𝑚2 𝑀𝑎𝑥𝑖𝑚𝑢𝑚 (𝑇𝑒𝑛𝑠𝑖𝑙𝑒) = 𝜎𝑠 = 𝑆𝑘𝑖𝑟𝑡 𝐵𝑒𝑛𝑑𝑖𝑛𝑔 𝑆𝑡𝑟𝑒𝑠𝑠 + 𝑜𝑝𝑒𝑟𝑎𝑡𝑖𝑛𝑔 𝑠𝑘𝑖𝑟𝑡 𝑑𝑒𝑎𝑑 𝑠𝑡𝑟𝑒𝑠𝑠 = 8.268471062 + 6.563779941 = 14.832251 𝑁/𝑚𝑚2 Assume joint efficiency, E as 85% = 0.85 𝑆𝑠 × 𝐸 × 𝑆𝑖𝑛∅ = 115.4871847 × 0.85 × 𝑆𝑖𝑛(90) = 87.75838411 So σs (Tensile) is lower than 87.75838411. 𝑡𝑠𝑘 36 ) × sin(90) = 87.15424709 0.125 × 𝐸𝑦 × ( ) × 𝑆𝑖𝑛(∅) = 0.125 × 2000 × ( 𝐷𝑠 9000 And σs (Compressive) is lower than 87.15424709. Both criteria are satisfied Adding 2 mm for corrosion gives a design thickness of a skirt of = 38 mm. 36

1.4.4 Mechanical Drawing 9m

A 4m

B

0.85 m

1 2 3

4 5 6 7 8

9 10 11

12

2.01 m

C 53 m

13

14 15

Service: Regenerator (T-302). Number of Stages: 22 (Excl. Reboiler). Inside Diameter: 9 m. Length (Tangent to Tangent: 53 m. Top Operating Temperature: 110 C. Bottom Operating Temp: 126.3 C. Operating Pressure: 220 kPa. Design Pressure: 132 kPa. Plate Spacing: 2.2 m. Weir Height: 0.05 m. Corrosion Allowance: 0.003 m.

16 17 18

19 20

21 D

4m

22 E

1.1 m 0.038 m

F

Nozzles: A: Over Head Vapour. B: Reflux. C: Feed. D: Boil Up. E: Manhole. F: Bottoms Product.

Skirt Height = 3 m

CAPSTONE ENERGY Client: Chevron Distillation Column DWG No. PID 100

Sheet: 1/2

DWG: Humam Date: 01/11/2015

37

1.5 Summary and Review 1.5.1 Equipment Specification/Data Sheet

38

1.5.2 Critical Review Looking back at the design process for the amine regenerator, I faced few issues which were solved later after the consultation with my supervisor and my group members. These issues started at the first stage of choosing the best design methodology of the amine regenerator. I thought this type of distillation columns need to be designed in a special way as it has three different components with different boiling points. I ended up doing the material and energy balance of the column as I think this is a very important thing for any chemical engineer to apply for any chemical design process. Then I had to design the column using the short cut method suggested by Sinnott with the hydraulic plate design and the mechanical design. Assuming a reasonable values throughout the design process was taking a great care of as assuming a value that doesn’t show consistency will make the design unreliable. Most of the data were extracted from HYSYS software but it doesn’t always have the values I wanted or it shows wrong data that I needed to go to the literature and take them from there, one of these data were the Kvalues for volatility calculations. K-values can be found in two ways the first one through HYSYS and the second one through Antoine Equation, HYSYS showed a really big numbers for K-values which I didn’t rely on instead I went to the literature and looked for the Antoine Coefficients to apply in Antoine Equation and get the right value of K-values as I did at the first stage of the design process. When I reached the stage of sizing the regenerator like finding the column diameter and height, I was kind of shocked when I saw that my diameter is 9 meters and my height is 53 metres (which still below the maximum typical column height of 175 ft). I was uncomfortable with these sizes and I had to go to my supervisor to check on this with him, I learnt through him that because I am having a big flow rate going into my column, the size of the column will be really big to meet the required separation of components. That made me happy and confident at the same time and I learnt a new information as well. This design project of the major component has made me improve many skills within the engineering practice including software’s required to be used by chemical engineers in the industry such as HYSYS and Excel software as well as I improved my professional writing by writing this report to the standards. Now I feel I can apply the skills and techniques I learnt throughout the design of this column when I work in a company in the future. I will always read more and practice designing all the types of the distillation columns to enhance my chemical engineering skills which will add a big advantage to my resume. The design process took pretty long time to finish and I feel I did good job with it and the time I spent really worth it as I learnt new information and methods used in the industry to design such an important equipment like the distillation column which I have been studying only the thermodynamics aspect of it up to the 3rd year of my study at Curtin.

39

1.6 Nomenclature Symbol QRE QSOL QRX QVAP Cp,sol TB WL QLR P* Îą Nmin XLK (XHK)B Ć&#x; Rmin N Eo Âľa B D Nr Ns L L’ V Ďƒ Ď v ∆đ?‘ƒđ?‘Ą

Meaning Reboiler Duty in Watt Heat of Solution, kj/hr Heat of Reaction, kj/kg Heat of Vaporization, kj/hr Specific heat of solution, kj/kg-k Regenerator Bottom Temperature, °C Lean Amine Mass Flow Rate, kg/hr Lean/Rich Amine Exchanger Duty, Watt Vapour Pressure, bar Volatility Minimum Number of Stages Mole Fraction of Light Key Mole Fraction of Heavy Key in Bottoms Root of Equation Minimum Reflux Ratio Actual Number of Stages Column Efficiency, % Molar Average liquid Viscosity, mNs/m2 Molar Flow of Bottom Products, kgmole/hr Molar Flow of Top Product, kgmole/hr Rectifying Section Stages Stripping Section Stages Liquid Above Feed Stage, kgmole/hr Liquid Below Feed Stage, kgmole/hr Vapour Rate, kmol/hr Surface Tension, N/m Density of Vapour, kg/m3 Column Pressure Drop, kPa

Symbol Ht đ??šđ??żđ?‘‰ K1 Uf AT DT Ad Aa Lw Ah Hw How Dh Tr Di,optimum G Pi S

Meaning Total Plate Pressure Drop, mm liquid Liquid Vapour Flow Factor, kmole/hr Correction Factor for Surface Tension Flooding Vapour Velocity, m/s Total Area, m2 Column Diameter, m Downcomer Area, m2 Active Area, m2 Weir Length, m Hole Area, m2 Weir Height, m Weir Crest, m Hole Diameter, mm Residence Time, Sec Optimum Pipe Diameter, m Fluid Flow Rate, kg/s Design Pressure, N/mm2 Maximum Allowable Stress, N/mm2

E

Welded Joint Efficiency, %

C

Corrosion Allowance, mm

Wv Pw Mx đ?œŽđ??ż đ?œŽâ„Ž Ďƒw đ?œŒđ?‘? Fw

Weight of Vessel, kN Dynamic Wind Pressure, N/m2 Bending Moment at Bottom, Nm Longitudal Stress, N/mm2 Circumferential Stress, N/mm2 Dead Weight Stress, N/mm2 Critical Buckling Stress, N/mm2 Wind Loading, kN/m

40

1.7 References Barreau, A, E.Blanchon le Bouelec, K.N. Habchi Tounsi, P.Mougin and F.Lecomte. 2006. “Absorption of H2S and CO2 in Alkanolamine Aqueous Solution.” Oil and Gas Science and Technology. 61(2006): 345-361. Kidnay, Arthur J, William R.Parrish, and Daniel G.McMcartney. 2011. Fundamentals of Natural Gas Processing. New York: Taylor and Francis Group. Kohl, Arthur, and Richard Nielsen. 1997. Gas Purification. Oxford: Elsevier. Sinnott, Ray, and Gavin Towler. 2009. Chemical Engineering Design. Oxford: ButterworthHeinemann. Whitesides, Randall W. 2012. “Selection and Sizing of Pressure Relief Valves.” PhD diss., PDH Online. Yaws, Carl L, Prasad K Narasimhan, and Chaitanya Gabbula. 2005. Yaws’ Handbook of Antoine Coefficients for Vapor Pressure. New York: Knovel. Younger, A.H. 2004. Natural Gas Processing Principles and Technology-Part II. Alberta: Thimm Engineering Inc.

41

CHAPTER 2: MINOR DESIGN Shell & Tube Heat Exchanger (E-301)

HUMAM AL-RIKABAWI 15564041

Table of Contents: 2.1

Introduction ......................................................................................................................................................1

2.2 Process Design .........................................................................................................................................................3 2.2.1 Design Methodology ........................................................................................................................................3 2.2.2 Design Calculations ..........................................................................................................................................4 2.2.2.1 Define Specification and Physical Properties ............................................................................................4 2.2.2.2 Overall Heat Transfer Coefficient Assuming, Uo,ass ...................................................................................4 2.2.2.3 Heat Exchanger Type and Dimensions ......................................................................................................5 2.2.2.4 Heat Transfer Area ....................................................................................................................................6 2.2.2.5 Layout and Tube Size ................................................................................................................................6 2.2.2.6 Number of Tubes ......................................................................................................................................6 2.2.2.7 Bundle and Shell Diameter .......................................................................................................................7 2.2.2.8 Tube Side Heat Transfer Coefficient .........................................................................................................7 2.2.2.9 Shell Side Heat Transfer Coefficient..........................................................................................................9 2.2.2.10 Overall Heat Transfer Coefficient, Uo, calc ..............................................................................................10 2.2.2.11 Pressure Drop (Tube Side) ....................................................................................................................11 2.2.2.12 Pressure Drop (Shell Side) .....................................................................................................................12 2.2.2.13 Summary of the process design ............................................................................................................13 2.2.3 Sizing/Specification of nozzles, pipes .............................................................................................................13 2.3 Operational Design ................................................................................................................................................14 2.3.1 Control System Design ...................................................................................................................................14 2.3.2 Start Up and Shut Down .................................................................................................................................16 2.3.3 Safety Study (HAZID) ......................................................................................................................................17 2.4 Mechanical Design .................................................................................................................................................18 2.4.1 Materials of Construction...............................................................................................................................18 2.4.2 Shell & Tube Mechanical Calculations............................................................................................................18 2.4.2.1 Wall Thickness.........................................................................................................................................18 2.4.2.2 Selection of Pressure Relief System ........................................................................................................18 2.4.3 Mechanical Drawing .......................................................................................................................................19 2.5 Summary and Review ............................................................................................................................................20 2.5.1 Equipment Specification/Data Sheet .............................................................................................................20 2.5.2 Critical Review ................................................................................................................................................21 2.6 Nomenclature ........................................................................................................................................................22 2.7 References .............................................................................................................................................................22

List of Figures: Figure 1 Shell and Tube heat exchanger configuration (Dreamstime, 2015) ................................................................1 Figure 2 the PFD shows the shell and tube exchanger location to be designed (Al-Rikabawi, 2015) ...........................2 Figure 3 Temperature profile of hot and cold streams (Al-Rikabawi, 2015) .................................................................2 Figure 4 Typical overall coefficients for amine solvents (Sinnott, 2009) .......................................................................4 Figure 5 Temperature correction factor of 1 shell pass and even tube passes (Sinnott, 2009) ....................................6 Figure 6 Constants to use in the bundle diameter equation (Sinnott, 2009) ................................................................7 Figure 7 Tube side heat-transfer factor (Sinnott, 2009) ................................................................................................8 Figure 8 Shell-side heat transfer factor (Sinnott, 2009) ..............................................................................................10 Figure 9 Tube-side friction factor (Sinnott, 2009) .......................................................................................................11 Figure 10 Shell side friction factor (Sinnott, 2009) ......................................................................................................12

List of Tables: Table 1 Specifications of hot stream (Lean Amine) .......................................................................................................4 Table 2 Specifications of cold stream (Rich Amine) .......................................................................................................4 Table 3 Summary of thermal design calculations ........................................................................................................13 Table 4 Control System Design Details ........................................................................................................................15

2.1 Introduction Heat exchanger is defined as a device to cool down or heat up a fluid from one stream to another by exchanging heat from one fluid to another. In this design, to increase the temperature of the rich amine feed going into the amine regenerator, a heat exchanger of the type shell and tube is used for this purpose. Assumptions were made in regard to the allocation of fluids in the exchanger as following:  

Rich amine containing CO2 to pass thorugh the tubes due to the ease of cleaning/chanigng of tubes when corrosion takes place. Lean amine without CO2 to pass through the shell.

Figure 1 below shows a typical configuration of a shell and tube heat exchanger with the tube and shell inlets and outlets. The curved arrows represent the number of passes of the shell fluid around the tubes.

Figure 1 Shell and Tube heat exchanger configuration (Dreamstime, 2015)

Shell and tube heat exchanger was selected because of the following reasons: 1- Big Surface Area 2- Small Volume Consumed 3- Good Mechanical Layout 4- Easy Maintenance 5- Wide Range of Material of Construction The basic design objective in designing Shell and Tube Heat Exchanger is calculating the duty of it as the following equation: đ?‘„ = đ?‘ˆ Ă— đ??´ Ă— ∆đ?‘‡đ?‘š In this project, the shell and tube heat exchanger (E-301) shown in Figure 2 was used to rise the temperature of the stream (3-6) containing rich amine before entering the regenerator by exchanging the heat with the hot stream (3-10) coming out of the bottom of the amine regenerator. The resulted streams from this heat transfer process will be heated stream (3-7) going directly into the regenerator column, the second stream is (3-11) where the temperature has reduced due to the heat transfer.

1

Figure 2 the PFD shows the shell and tube exchanger location to be designed (Al-Rikabawi, 2015)

As a summary of this heat transfer process between the streams, Figure 3 below was created to show the temperature profile between the hot and cold streams. Stream (3-6) coming from the CO2 absorber and containing rich amine of the absorbed CO2 has a temperature of 67.80 Celsius and its temperature was increased to 90 Celsius as in stream (3-7). Stream (3-10) of lean amine is hot as it is coming out of the bottom of the regenerator and its temperature is 126.3 Celsius, it has exchanged the heat and resulted in decreasing its temperature down to 99.79 Celsius as in stream (3-11) recycled back to the top of the absorber.

Figure 3 Temperature profile of hot and cold streams (Al-Rikabawi, 2015)

2

2.2 Process Design 2.2.1 Design Methodology Step

Description, Equation or Result

1. Find specifications

Using HYSYS to find T, P and Duty

2. Collect physical properties

HYSYS or literature to find density etc.

3. Assume overall coefficient Uo,ass

Assumed (150 W/m2 C) for amine solutions (Organic Solvents)

4. Decide number of shell and tube passes, ∆Tlm , correction factor F, and ∆Tm.

One shell and two tube passes, ∆Tlm = 34.1 °C, Ft = 0.91, ∆Tm = 31.031.

5. Find heat transfer area required Ao

Ao = q/Uo,ass*∆Tm >>> Ao = 8963 m2

6. Decide type, tube size, material layout and assign fluids to shell or tube side

Tube type to be stainless steel due to the rich amine (containing CO2) passing through the tubes and lean amine through shell side.

7. Calculate number of tubes 8. Calculate shell inside diameter, Ds 9. Estimate tube-side heat transfer coefficient 10. Decide baffle spacing and estimate shellside heat transfer coefficient 11. Calculate overall heat transfer coefficient including fouling factors, Uo,calc

12. Estimate tube-side pressure drop

13. Estimate shell-side pressure drop

No. of tubes = Ao/area of one tube > = 190 tube Ds= Bundle Diameter, Db + Shell Clearance = 3037.44+144 = 3181.7 mm Hi = 1115.8 W/m2 ⁰C Using Kern method: Baffle Spacing = 318.1755 mm Shell side coefficient, Hs = 158.175 W/m2 ⁰C Uo,calc = 105.1269212 W/m2 C. Check (105.1269212 – 150)/ 150 = 29.9%, it is between 0 and 30%. Jump to step 12. Tube Pressure Drop = ∆Pt = 0.28717 bar It is below the specification pressure drop of 0.7 bar. Therefore acceptable. Shell Pressure Drop = ∆Ps = 7.69427E-06 bar. It is below the specification pressure drop of 0.7 bar. Therefore acceptable.

3

2.2.2 Design Calculations 2.2.2.1 Define Specification and Physical Properties The following set of data (specifications and properties) were extracted from HYSYS: Table 1 Specifications of hot stream (Lean Amine)

Hot Stream Flow Rate Temperature Specific Heat, Cp Thermal Conductivity Density Viscosity

3-10

126.3 3.751 0.2274 971.6 0.6181

Mean 1,543,000 429 113.045 3.67 0.2283 982.3 0.8265

3-11

99.79 3.586 0.2291 993 1.035

Unit Kg/hr Kg/s °C kJ/Kg-°C W/m-K Kg/m3 cP or mN sm-2

Table 2 Specifications of cold stream (Rich Amine)

Cold Stream Flow Rate

3-6

Mean 3-7 Unit 1,819,000 Kg/hr 505 Kg/s Temperature 67.80 78.9 90 °C Specific Heat, Cp 3.281 3.39 3.493 kJ/Kg-°C Thermal Conductivity 0.2460 0.2418 0.2375 W/m-K Density 1066 816.25 566.5 Kg/m3 Viscosity 2.273 1.857 1.441 cP or mN sm-2 Volumetric Flow 0.619023311 m3/s Duty taken from HYSYS was found to be 1.502E8 kJ/hr which equals to 41,722 kW. Heat transfer rate (Duty) = Mass flow rate in kg/s * mean heat capacity*(Tin-Tout) So Duty = 429 * 3.67 * (126.3 – 99.79) = 41,738 kW that slight difference in the value of the duty when calculated manually is due to the variation of heat capacity at different temperatures. Therefore the duty of this heat exchanger will be taken as 41,722 kW.

2.2.2.2 Overall Heat Transfer Coefficient Assuming, Uo,ass Assume initial overall heat transfer coefficient Uo, the assumed overall coefficient for this type of heat transfer was based on research of the typical ranges of overall coefficients for organic solvents such as amines as shown in below figure:

Figure 4 Typical overall coefficients for amine solvents (Sinnott, 2009)

4

As can be seen from the above table, the typical range of overall coefficients of amines is 100-300 W/m2â °C, therefore the assumed overall transfer coefficients Uo,assumed chosen is 150 W/m2 C.

2.2.2.3 Heat Exchanger Type and Dimensions The preferred number of tube passes is an even number, therefore, start with one shell and two tube passes. Now calculate the log mean temperature difference using the following equation: ∆đ?‘‡đ?‘™đ?‘š (đ??żđ?‘’đ?‘Žđ?‘› đ?‘Žđ?‘šđ?‘–đ?‘›đ?‘’ đ?‘–đ?‘›đ?‘™ đ?‘Ąđ?‘’đ?‘šđ?‘? − đ?‘…đ?‘–đ?‘?â„Ž đ?‘Žđ?‘šđ?‘–đ?‘›đ?‘’ đ?‘œđ?‘˘đ?‘Ą đ?‘Ąđ?‘’đ?‘šđ?‘?) − (đ??żđ?‘’đ?‘Žđ?‘› đ?‘Žđ?‘šđ?‘–đ?‘›đ?‘’ đ?‘œđ?‘˘đ?‘Ą đ?‘Ąđ?‘’đ?‘šđ?‘? − đ?‘…đ?‘–đ?‘?â„Ž đ?‘Žđ?‘šđ?‘–đ?‘›đ?‘’ đ?‘–đ?‘›đ?‘™ đ?‘Ąđ?‘’đ?‘šđ?‘?) = (đ??żđ?‘’đ?‘Žđ?‘› đ?‘Žđ?‘šđ?‘–đ?‘›đ?‘’ đ?‘–đ?‘›đ?‘™ đ?‘Ąđ?‘’đ?‘šđ?‘? − đ?‘…đ?‘–đ?‘?â„Ž đ?‘Žđ?‘šđ?‘–đ?‘›đ?‘’ đ?‘œđ?‘˘đ?‘Ą đ?‘Ąđ?‘’đ?‘šđ?‘?) ln( (đ??żđ?‘’đ?‘Žđ?‘› đ?‘Žđ?‘šđ?‘–đ?‘›đ?‘’ đ?‘œđ?‘˘đ?‘Ą đ?‘Ąđ?‘’đ?‘šđ?‘? − đ?‘…đ?‘–đ?‘?â„Ž đ?‘Žđ?‘šđ?‘–đ?‘›đ?‘’ đ?‘–đ?‘›đ?‘™ đ?‘Ąđ?‘’đ?‘šđ?‘?) ∆Tlm =

(126.3 − 90) − (99.79 − 67.80) = 34.1 °C 126.3 − 90 ln ( ) 99.79 − 67.80

To find correction factor we need to find R and S according to the following formulas: đ?‘…=

(đ??żđ?‘’đ?‘Žđ?‘› đ?‘Žđ?‘šđ?‘–đ?‘›đ?‘’ đ?‘–đ?‘›đ?‘™ đ?‘Ąđ?‘’đ?‘šđ?‘? − đ??żđ?‘’đ?‘Žđ?‘› đ?‘Žđ?‘šđ?‘–đ?‘›đ?‘’ đ?‘œđ?‘˘đ?‘Ą đ?‘Ąđ?‘’đ?‘šđ?‘?) (đ?‘…đ?‘–đ?‘?â„Ž đ?‘Žđ?‘šđ?‘–đ?‘›đ?‘’ đ?‘œđ?‘˘đ?‘Ą đ?‘Ąđ?‘’đ?‘šđ?‘? − đ?‘…đ?‘–đ?‘?â„Ž đ?‘Žđ?‘šđ?‘–đ?‘›đ?‘’ đ?‘–đ?‘›đ?‘™ đ?‘Ąđ?‘’đ?‘šđ?‘?) đ?‘…=

�=

(126.3 − 99.79) = 1.19414 (90 − 67.80)

(đ?‘…đ?‘–đ?‘?â„Ž đ?‘Žđ?‘šđ?‘–đ?‘›đ?‘’ đ?‘œđ?‘˘đ?‘Ą đ?‘Ąđ?‘’đ?‘šđ?‘? − đ?‘…đ?‘–đ?‘?â„Ž đ?‘Žđ?‘šđ?‘–đ?‘›đ?‘’ đ?‘–đ?‘›đ?‘™ đ?‘Ąđ?‘’đ?‘šđ?‘?) (đ??żđ?‘’đ?‘Žđ?‘› đ?‘Žđ?‘šđ?‘–đ?‘›đ?‘’ đ?‘–đ?‘›đ?‘™ đ?‘Ąđ?‘’đ?‘šđ?‘? − đ?‘…đ?‘–đ?‘?â„Ž đ?‘Žđ?‘šđ?‘–đ?‘›đ?‘’ đ?‘–đ?‘›đ?‘™ đ?‘Ąđ?‘’đ?‘šđ?‘?) đ?‘†=

(90 − 67.80) = 0.3795 (126.3 − 67.80)

Then using the figure below, starting the line from the x-axis at S=0.38 and going vertically upwards until intersecting the curve of R= 1.2 then drawing a horizontal line towards the y-axis and the intersection point is the correction factor which equals to 0.91. Ft = 0.91 To find the mean temperature difference, ∆Tm use the following equation: ∆đ?‘‡đ?‘š = đ??šđ?‘Ą ∗ ∆đ?‘‡đ?‘™đ?‘š

∆đ?‘‡đ?‘š = 0.91 ∗ 34.1

∆đ?‘‡đ?‘š = 31.031

5

Figure 5 Temperature correction factor of 1 shell pass and even tube passes (Sinnott, 2009)

2.2.2.4 Heat Transfer Area Find the Heat transfer area:

A° = A° =

Duty in Watt Assumed overall coefficientĂ—âˆ†Tm

41722000 = 8963.61 m2 150 Ă— 31.031

2.2.2.5 Layout and Tube Size A split-ring floating head exchanger was chosen due to the better efficiency, ease of cleaning and can handle high volume of fluids. Stainless steel pipe is used as it is a corrosion resistant material to CO2 corrosion because of the fluid going through the tubes is rich amine which contains carbon dioxide. Therefore as mentioned earlier rich amine will pass through the tubes and lean amine will pass through the shell. Using a pipe with an outside diameter of 150 mm with inside diameter of 90 mm for a better heat transfer, 100 meter long tube was chosen with a triangular pitch of 160 mm. These pipes diameters and pitch sizing were selected after the final modification of the design to meet all the design equation conditions including the overall heat transfer coefficient and the pressure drops.

2.2.2.6 Number of Tubes The number of tubes was then calculated as following: đ?‘ đ?‘œ. đ?‘‡đ?‘˘đ?‘?đ?‘’đ?‘ =

đ??´đ?‘œ đ??´đ?‘&#x;đ?‘’đ?‘Ž đ?‘œđ?‘“ đ?‘œđ?‘›đ?‘’ đ?‘Ąđ?‘˘đ?‘?đ?‘’

Where Ao is the heat transfer area calculated in step 5. Where area of a tube = Pi x tube outside diameter x Length of tube Area of one tube (neglecting thickness of tube sheets) = Pi x (150*10^-3) x 100 = 47.1 m2 đ?‘ đ?‘œ. đ?‘‡đ?‘˘đ?‘?đ?‘’đ?‘ =

8963.61 = 190.3 47.1 6

This number was then rounded down to 190 tube. As an initial guess, two passes are chosen for the exchanger, giving 95 tube for each pass. Using this number of tubes, the tube side velocity is checked as follows: �� =

đ?‘Ł đ?‘Žđ?‘&#x;đ?‘’đ?‘Ž đ?‘?đ?‘’đ?‘&#x; đ?‘?đ?‘Žđ?‘ đ?‘

Where đ?‘Ł is the volumetric flow rate of rich amine (tube side) which equals = 0.6190 m3/s. Area per pass can be found as follows: Tube cross-sectional area = (Pi/4)*(90*10-3)2 = 0.0063585 m2 so area per pass = 95 x 0.0063585 = 0.605043722 m2 Now 0.6190 đ?‘š đ?‘˘đ?‘Ą = = 1.023 0.605043722 đ?‘ This velocity is satisfactory as it lies between the typical speed of 1 and 2 m/s as per (Sinnott, 2009).

2.2.2.7 Bundle and Shell Diameter From the table in figure below, for triangular pitch and 2 tube passes, K1 = 0.249, n1 = 2.207.

Figure 6 Constants to use in the bundle diameter equation (Sinnott, 2009)

Now the bundle diameter can be calculated as follows: 1

đ?‘›đ?‘˘đ?‘šđ?‘?đ?‘’đ?‘&#x; đ?‘œđ?‘“ đ?‘Ąđ?‘˘đ?‘?đ?‘’đ?‘ đ?‘›1 ) đ??ˇđ?‘? = đ?‘‚đ??ˇ ( đ??ž1 1

190 2.207 ) đ??ˇđ?‘? = 150 ( = 3037.445 đ?‘šđ?‘š (3 đ?‘š) 0.249 A clearance of 144.3mm is required for a split-ring floating head exchanger. Therefore, the total inside diameter of the shell will be: đ??ˇđ?‘ = 3037.445 + 144.3 = 3181.75 đ?‘šđ?‘š = 3.1817 đ?‘š

2.2.2.8 Tube Side Heat Transfer Coefficient To calculate the tube side heat transfer coefficient, the Reynolds number, Prandtl number and the Nusselt number must be calculated as follows: đ?‘…đ?‘’đ?‘Ą =

đ?‘Łđ?‘Ą Ă— đ?‘‘đ?‘– Ă— đ?œŒ đ?œ‡đ?‘Ą

Where, vt is the velocity of the tube side = 1.023 m/s di is the inside diameter of the tube = 90 mm 7

đ?œŒ is the density of the tube side fluid = 816.25 kg/m3 đ?œ‡đ?‘Ą is the viscosity of the tube side fluid = 1.857*10-3 in N sm-2 đ?‘…đ?‘’đ?‘Ą =

1.023 Ă— 90 ∗ 10−3 Ă— 816.25 = 40473.8 0.001857 đ??śđ?‘? ( đ?‘ƒđ?‘&#x;đ?‘Ą =

đ?‘— . đ??ś) Ă— đ?œ‡đ?‘Ą đ?‘˜đ?‘” đ?‘˜

Where, Cp is the specific heat of the tube side fluid = 3.39*1000 j/kg C đ?œ‡đ?‘Ą is the viscosity of the tube side = 1.857*10-3 N sm-2 k is the thermal conductivity of the tube side = 0.2418 W/m-K đ?‘ƒđ?‘&#x;đ?‘Ą =

(3.39 Ă— 1000) Ă— 0.001857 = 26.03 0.2418 đ??ż 100000 = = 1111 đ?‘‘đ?‘– 90

Then using the graph below, the tube-side heat transfer factor jh can be found.

Figure 7 Tube side heat-transfer factor (Sinnott, 2009)

jh is estimated from the graph to be = 0.0035 Nu (Nusselt number) = jh Ă— Ret Ă— (Pr)0.33 đ?‘ đ?‘˘ = 0.0035 Ă— 40473.8 Ă— (26.03)0.33 = 415.31 Now the tube side heat transfer coefficient can be calculated from the below equation: đ?‘˜ â„Žđ?‘– = đ?‘ đ?‘˘ Ă— ( ) đ?‘‘đ?‘– â„Žđ?‘– = 415.31 Ă—

0.2418 đ?‘Š = 1115.8 2 90 ∗ 10−3 đ?‘š đ??ś

This value is clearly good if the overall transfer coefficient is 150 W/m2C. 8

2.2.2.9 Shell Side Heat Transfer Coefficient Kern method is used to estimate the heat transfer coefficient of the shell side. As first trial take the baffle spacing to be đ??ˇđ?‘ 3181.755 = = 318.1755 đ?‘šđ?‘š 10 10 đ??´đ?‘ =

(đ?‘?đ?‘Ą − đ?‘‘đ?‘œ )đ??ˇđ?‘ đ?‘™đ?‘? đ?‘?đ?‘Ą

Where, pt is the triangular pitch size in mm = 160 do is the outside diameter = 150 mm Ds is the shell diameter = 3181.755 mm lb is the baffle spacing = 318.1755 mm đ??´đ?‘ =

(160 − 150)đ?‘Ľ3181.755 Ă— 318.1755 = 63272.31011 đ?‘šđ?‘š2 = 63.2723 đ?‘š2 160

The equivalent diameter, de can be calculated as the following equation: đ?‘‘đ?‘’ = đ?‘‘đ?‘’ =

1.1 (1602 − 0.917 Ă— 1502 ) = 36.428 đ?‘šđ?‘š 150

Volumetric flow rate on shell-side =

So shell-side velocity = us =

1.1 2 (đ?‘? − 0.917đ?‘‘đ?‘œ2 ) đ?‘‘đ?‘œ đ?‘Ą

0.436 63.2723

(1543000/3600) 982.3

= 0.0068961

= 0.436

đ?‘š3 đ?‘

đ?‘š đ?‘

Now the Reynolds number of the shell side can be calculated as following: đ?‘…đ?‘’ =

982.3 đ?‘Ľ 0.0068961 đ?‘Ľ36.42833 Ă— 10−3 = 298.57 0.8265 Ă— 10−3

And now Prandtle number should be calculated as following: Pr =

3.67đ?‘Ľ103 đ?‘Ľ 0.8265 Ă— 10−3 = 13.286 0.2283

Use segmental baffles with a 15% cut. From the graph below, jn = 0.036.

9

Figure 8 Shell-side heat transfer factor (Sinnott, 2009)

If the viscosity correlation is ignored, the shell side heat transfer coefficient is found from: â„Žđ?‘ = (((

â„Žđ?‘ = (((

đ?‘‡â„Žđ?‘’đ?‘&#x;đ?‘šđ?‘Žđ?‘™ đ??śđ?‘œđ?‘›đ?‘‘đ?‘˘đ?‘?đ?‘Ąđ?‘–đ?‘Łđ?‘–đ?‘Ąđ?‘Ś ) Ă— 1000) Ă— đ?‘—đ?‘› Ă— đ?‘…đ?‘’ Ă— (pr 0.33 )) đ??¸đ?‘žđ?‘˘đ?‘–đ?‘Łđ?‘Žđ?‘™đ?‘’đ?‘›đ?‘Ą đ?‘‘đ?‘–đ?‘Žđ?‘šđ?‘’đ?‘Ąđ?‘’đ?‘&#x;

0.2283 ) Ă— 1000) Ă— 0.036 Ă— 298.57 Ă— (13.2860.33 )) 36.428333

đ?‘†â„Žđ?‘’đ?‘™đ?‘™ − đ?‘ đ?‘–đ?‘‘đ?‘’ â„Žđ?‘’đ?‘Žđ?‘Ą đ?‘Ąđ?‘&#x;đ?‘Žđ?‘›đ?‘ đ?‘“đ?‘’đ?‘&#x; đ?‘?đ?‘œđ?‘’đ?‘“đ?‘“đ?‘–đ?‘?đ?‘–đ?‘’đ?‘›đ?‘Ą = â„Žđ?‘ = 158.175 đ?‘Š/đ?‘š2 đ??ś

2.2.2.10 Overall Heat Transfer Coefficient, Uo, calc The overall heat transfer coefficient can now be found using the following formula: đ?‘‘ đ?‘‘đ?‘œ ln ( 0 ) 1 1 1 đ?‘‘đ?‘– = + đ?‘“đ?‘œđ?‘˘đ?‘™đ?‘–đ?‘›đ?‘”đ?‘œ + + + đ?‘“đ?‘œđ?‘˘đ?‘™đ?‘–đ?‘›đ?‘”đ?‘– đ?‘ˆđ?‘œ â„Žđ?‘œ 2đ?‘˜đ?‘¤ â„Žđ?‘– Where, Uo is the overall coefficient based on the outside area of the tube, W/m2 C Ho is the outside fluid film coefficient, W/m2 C Foulingo is the outside dirt coefficient (fouling factor), W/m2 C Do is the tube outside diameter, m Di is the tube inside diameter, m Kw is the thermal conductivity of the tube wall material, W/m2 C Hi is the inside fluid film coefficient, W/m2C Foulingi is the inside dirt coefficient, W/m2C

10

150 ) 150 Ă— 10−3 Ă— ln ( 1 1 150 1 90 + ( =( + 0.00035) ( )+ + 0.00035) đ?‘ˆđ?‘œ 1115.807117 90 2 Ă— 50.2 158.1751599 1 = 0.0095121311 đ?‘ˆđ?‘œ đ?‘źđ?’? = đ?&#x;?đ?&#x;Žđ?&#x;“. đ?&#x;?đ?&#x;?đ?&#x;”đ?&#x;—đ?&#x;?đ?&#x;?đ?&#x;? đ?‘ž/đ?’Žđ?&#x;? â °đ?‘Ş This overall heat transfer coefficient is acceptable as it lies within the range between 0-30 % as per the condition of the equation before step 12: The condition is: 0 <

đ?‘ˆđ?‘œ ,đ?‘?đ?‘Žđ?‘™đ?‘? −đ?‘ˆđ?‘œ ,đ?‘Žđ?‘ đ?‘ đ?‘ˆđ?‘œ ,đ?‘Žđ?‘ đ?‘

< 30%

đ?‘ˆđ?‘œ ,đ?‘?đ?‘Žđ?‘™đ?‘? − đ?‘ˆđ?‘œ ,đ?‘Žđ?‘ đ?‘ 105.1269212 − 150 = = 29.9% đ?‘ˆđ?‘œ ,đ?‘Žđ?‘ đ?‘ 150 0 < 29.9 < 30% Therefore the assumption of the assumed 150 W/m2 C is a safe assumption as it gave a final calculated overall heat transfer coefficient that lies within the range and that is acceptable. Now move to the calculation of pressure drops of both shell and tube.

2.2.2.11 Pressure Drop (Tube Side) The tube side pressure drop can be calculated as following: Number of tubes = 190 tube Number of tube passes = 2 Tube Internal diameter = 90 Tube side velocity, ut = 1.023105 m/s Reynolds number in the tubing, Re = 40473.80593 from the following figure, the tube side friction factor jf can be found:

Figure 9 Tube-side friction factor (Sinnott, 2009)

11

The tube side friction factor, jf is estimated to be = 0.0035. đ??ż đ?œ‡ đ?œŒ Ă— đ?‘˘đ?‘Ą2 ∆đ?‘ƒđ?‘Ą = đ?‘ đ?‘? [8 Ă— đ?‘—đ?‘“ Ă— ( ) Ă— ( ) + 2.5] Ă— đ?‘‘đ?‘– đ?œ‡đ?‘¤ 2 Where, ∆đ?‘ƒđ?‘Ą Is the tube side pressure drop, N/m2 (Pa) Np is the number of tube side passes Ut is the tube side velocity, m/s L is the length of one tube ∆đ?‘ƒđ?‘Ą = 2 Ă— (8 Ă— 0.0035 Ă— (

100000 816.25 Ă— (1.0231052 ) ) + 2.5) Ă— ( ) 90 2

∆đ?‘ƒđ?‘Ą = 28717.49498

đ?‘ = đ?&#x;Ž. đ?&#x;?đ?&#x;–đ?&#x;•đ?&#x;?đ?&#x;•đ?&#x;’đ?&#x;—đ?&#x;“ đ?’ƒđ?’‚đ?’“ đ?‘š2

Therefore the tube side pressure drop equals to 0.28717495 bar which is less than the specification pressure drop of 0.7 bar that means the assumptions made are correct.

2.2.2.12 Pressure Drop (Shell Side) Now find out the pressure drop of the shell side: Find the shell side friction factor from the following graph at the shell side Reynolds number of 298.57 and a baffle cut of 15 percent:

Figure 10 Shell side friction factor (Sinnott, 2009)

Therefore the shell side friction factor is estimated to be, jf = 0.15 12

Neglecting the viscosity correction term, the shell side pressure drop can be found using the following: đ?‘†â„Žđ?‘’đ?‘™đ?‘™ đ?‘‘đ?‘–đ?‘Žđ?‘šđ?‘’đ?‘Ąđ?‘’đ?‘&#x; đ?‘–đ?‘› đ?‘š đ?‘Ąđ?‘˘đ?‘?đ?‘’ đ?‘™đ?‘’đ?‘›đ?‘”đ?‘Ąâ„Ž đ?‘–đ?‘› đ?‘š )Ă—( ) đ??¸đ?‘žđ?‘˘đ?‘–đ?‘Łđ?‘Žđ?‘™đ?‘’đ?‘›đ?‘Ą đ?‘‘đ?‘–đ?‘Žđ?‘šđ?‘’đ?‘Ąđ?‘’đ?‘&#x; đ?‘–đ?‘› đ?‘š đ?‘?đ?‘Žđ?‘“đ?‘“đ?‘™đ?‘’ đ?‘ đ?‘?đ?‘Žđ?‘?đ?‘–đ?‘›đ?‘” (đ?‘ â„Žđ?‘’đ?‘™đ?‘™ đ?‘ đ?‘–đ?‘‘đ?‘’ đ?‘Łđ?‘’đ?‘™đ?‘œđ?‘?đ?‘–đ?‘Ąđ?‘Ś 2 ) Ă— (đ??ˇđ?‘’đ?‘›đ?‘ đ?‘–đ?‘Ąđ?‘Ś đ?‘œđ?‘“ đ?‘ â„Žđ?‘’đ?‘™đ?‘™ đ?‘“đ?‘™đ?‘˘đ?‘–đ?‘‘ Ă— ) 2

∆đ?‘ƒđ?‘ = 8 Ă— đ??śđ?‘œđ?‘&#x;đ?‘&#x;đ?‘’đ?‘?đ?‘Ąđ?‘–đ?‘œđ?‘› đ?‘“đ?‘Žđ?‘?đ?‘Ąđ?‘œđ?‘&#x; Ă— (

∆đ?‘ƒđ?‘ = 8 Ă— 0.15 Ă— (

(0.0068961322 ) 3.181755745 100 ) Ă— (982.3 Ă— )Ă—( ) 36.428333 318.1755745 2

đ?‘†â„Žđ?‘’đ?‘™đ?‘™ đ?‘ đ?‘–đ?‘‘đ?‘’ đ?‘?đ?‘&#x;đ?‘’đ?‘ đ?‘ đ?‘˘đ?‘&#x;đ?‘’ đ?‘‘đ?‘&#x;đ?‘œđ?‘? = ∆đ?‘ƒđ?‘ = 0.769426736

đ?‘ = đ?&#x;•. đ?&#x;”đ?&#x;—đ?&#x;’đ?&#x;?đ?&#x;•đ?‘Ź − đ?&#x;Žđ?&#x;” đ?’ƒđ?’‚đ?’“ đ?‘š2

Therefore the shell side pressure drop is also lower than the specification pressure drop of 0.7 bar from HYSYS. That means the shell pressure drop is within the specification. And the design assumptions are viable and correct.

2.2.2.13 Summary of the process design Table 3 Summary of thermal design calculations

Tube Side Material Inside diameter (ID), mm Outside diameter (OD), mm Length, m Total number of tubes Heat transfer coefficient, W/m2 C Pressure drop, bar

Stainless Steel 304 90 150 100 190 1115.8 0.28717 Shell Side

Material Inside diameter, mm Baffle Spacing, mm Baffle cut, % Heat transfer coefficient, W/m2 C Pressure drop, bar

Carbon Steel 3182 318 15 158.17 7.69427E-06

Overall Assumed overall heat transfer coefficient, W/m C Calculated overall heat transfer coefficient, W/m2C 2

150 105.127

2.2.3 Sizing/Specification of nozzles, pipes The sizes of the nozzles and pipes going into and leaving the heat exchanger will be the same sizes as the ones calculated for the regenerator (T-302) due to the mass rate inside the pipe are the same in both streams of the regenerator bottom and top as the pipe diameter depends basically on mass flow rate.

13

2.3 Operational Design 2.3.1 Control System Design Lean Amine (Outlet)

FT3

TA2 FA1

TT2

PSV (Pressure Relief Valve) +

PC2

FC1

PA2

FT2

PT2

TT5

Rich Amine (Outlet)

E-301

Rich Amine (Inlet)

FT4

TT3

FA2

TA3

PA1

FT1

FB

PC1

PT1

TT4

PSD

+

TA1

TT1

Lean Amine (Inlet)

14

The control system design or piping & instrumentation diagram is an important aspect at the design phase. This diagram includes information on the equipment, instruments, valves, sensors and alarms that control the total system. These details help vastly in the control of the system whenever risk and hazard are expected to take place at any moment during the operating of the heat exchanger, the above diagram shows the control system design applied at the shell and tube heat exchanger (E-301) and the below table explains what each symbol on the diagram means with description and purposes. Table 4 Control System Design Details

Symbol FB

Indication Feedback

PSD FT1

Pressure System Diaphragm Flow Transmitter

FT2 FT3 FT4 FA1

Flow Transmitter Flow Transmitter Flow Transmitter Flow Alarm

FA2

Flow Alarm

PT1 PT2 PA1

Pressure Transmitter Pressure Transmitter Pressure Alarm

PA2

Pressure Alarm

PC1

Pressure Controller

PC2

Pressure Controller

TT1 TT2 TT3 TT4 TT5 TA1

Temperature Transmitter Temperature Transmitter Temperature Transmitter Temperature Transmitter Temperature Transmitter Temperature Alarm

TA2

Temperature Alarm

TA3

Temperature Alarm

PSV

Pressure Relief Valve

Description Through which the disturbance of the system at the Rich Amine Outlet can be measured. Controls the opening and closing of pressure valve. Provide electrical Information of Rich Amine (Inlet) stream flow. Electrical Information of Cold Rich Amine inlet stream flow. Electrical information of Lean Amine Outlet stream flow. Electrical Information of Rich Amine Outlet Provide warning to operators if flow exceeds operating conditions at Lean Amine Outlet for troubleshooting. Provide warning to operators if flow exceeds operating conditions at the Rich Amine Outlet. Electrical information of Lean Amine inlet pressure. Electrical information of Rich Amine inlet pressure. Provide warning to operator if pressure exceeds operating conditions at the Lean Amine inlet. Provide alarm to operator if pressure exceeds operating conditions at the inlet of Rich Amine. Controls the flow rate of the lean amine inlet after receiving signal from PA1. Controls the flow rate of the Rich amine inlet after receiving signal from PA2. Electrical information of the Lean Amine inlet temperature. Electrical information of the Lean Amine outlet temperature Electrical information of the Rich Amine outlet temperature Electrical information of the shell side temperature Electrical information of the tube side temperature Provides signal to operators when the Lean Amine inlet temperature deviates from the operational conditions Provides signal to operators when the Lean Amine outlet temperature deviates from operational conditions Provides signal to operators when the Rich Amine outlet temperature deviates from the operation conditions To ensure the pressure inside the exchanger will not rise to unsafe level to protect from internal failure.

15

2.3.2 Start Up and Shut Down Start-Up To start the shell and tube heat exchanger, some procedures need to be known and followed in order to mitigate the risk of having damage to the equipment. (Thulukkanam, 2014) 1. The temperature of the hot stream (Lean Amine Inlet) has to be increased gradually at the start to avoid possible cracking in the equipment due to the sharp change of temperature (Thermal Stress). 2. Pressure to be increased gradually until it reaches 10% above the operational design pressure then it has to be decreased to the normal operating pressure. 3. The flow rates of the hot and cold stream to be increased gradually.

Shut-Down To shut down the shell and tube heat exchanger for any reason like process conditions deviation, standard procedures need to be applied in order to avoid damage to equipment. (Thulukkanam, 2014) 1. The flow of hot stream has to be switched off first. 2. Then the flow of the cold stream to be gradually reduced to match the hot fluid. 3. The temperature and pressure of both hot and cold stream to be decreased gradually until reaching the atmospheric temperature and pressure. 4. The flow rate of the cold stream to be switched off gradually.

16

2.3.3 Safety Study (HAZID) The safety study is an important tool to consider for chemical engineers, the following study was made after referring to (Seligmann, 2015) Project: Big Gas Project (BGP) – Shell & Tube Heat Exchanger (E-301) HAZID Lead: Humam Al-Rikabawi

Date: 04-11-2015 HAZID Scribe: Humam Al-Rikabawi

Explosion

Corrosion

Noise

High Pressure High Temperature Power Failure

Possible Causes

The heat exchanger operational pressure exceeds the specification. High CO2 concentration in tube and lack of maintenance Operation of heat exchanger

Rupture and Damage to the exchanger

Pressure exceeds design pressure, block of outlets Temperature control loss

Rupture of pipes and exchanger, loss of production and cracking Workers hearing problems, fatigue to vessel Breakage of heat exchanger, loss of profit Damage to heat exchanger, explosion

Power generation malfunction

Loss of production, loss of profit

Possible

Unlikely

Rare

Possible

CONSECUENCE SEVERITY

Hazard

LIELIHOOD

RISK ASSESSMENT HAZARD CONSEQUENCES

Major

Minor

Low

Major

RECOMMENDATIONS

ACTION (By)

Install the pressure relief valve (PSV), Pressure alarms and regular inspection

Operators

RISK RANKING

EXTREME

LOW

LOW

EXTREME

Likely

Low

MEDERATE

Possible

Moderate

HIGH

Ensure reasonable corrosion allowance for walls and regular maintenance Hearing Protection, mitigate the source of noise Pressure Relieve Valve (PSV), high pressure alarms and control High temperature alarms, emergency cooling system Backup power generators and UPS

Operators

H&S Advisor

Operators Operators Operators

17

2.4 Mechanical Design 2.4.1 Materials of Construction The material selection for the heat exchanger is an important to consider at the design stage of the shell and tube heat exchanger especially if there is a corrosive fluid going through the exchanger. The material selection was mainly studied for the shell and the tube parts as these two parts are the main parts that will be in contact with the exchanged fluids. Therefore, the selection of the shell side was mainly of material of a carbon steel of a grade A285, that selection was easy to choose as the fluid going through the shell side is a non-corrosive fluid that does not have CO2 in it which is called lean amine that mainly consists of a hot stream of only water and MDEA. The carbon steel should be a fair selection for such type of normal fluids. The material selection for the tube side had to be a stainless steel due to the fluid going through the tubes being corrosive as it is a rich MDEA cold solution containing of CO2, Water and MDEA. Therefore, a stainless steel is the best selection for such type of fluids as the CO 2 will cause corrosion problems if the wrong type of material chosen. Stainless steel can withstand corrosion like the CO2 corrosions.

2.4.2 Shell & Tube Mechanical Calculations 2.4.2.1 Wall Thickness The thickness of the tube can be calculated through the inside and outside diameters from the process design, therefore: đ?‘‡đ?‘˘đ?‘?đ?‘’ đ?‘¤đ?‘Žđ?‘™đ?‘™ đ?‘Ąâ„Žđ?‘–đ?‘?đ?‘˜đ?‘›đ?‘’đ?‘ đ?‘ = đ?‘‚đ?‘˘đ?‘Ąđ?‘ đ?‘–đ?‘‘đ?‘’ đ?‘‘đ?‘–đ?‘Žđ?‘šđ?‘’đ?‘Ąđ?‘’đ?‘&#x; − đ??źđ?‘›đ?‘ đ?‘–đ?‘‘đ?‘’ đ??ˇđ?‘–đ?‘Žđ?‘šđ?‘’đ?‘Ąđ?‘’đ?‘&#x; = 150đ?‘šđ?‘š − 90đ?‘šđ?‘š = đ?&#x;”đ?&#x;Žđ?’Žđ?’Ž The shell wall thickness can be calculated as following: The heat exchanger operational pressure at the shell side is taken from HYSYS as 2.2 bar The design pressure should be 10% above the operational pressure, therefore: đ??ˇđ?‘’đ?‘ đ?‘–đ?‘”đ?‘› đ?‘?đ?‘&#x;đ?‘’đ?‘ đ?‘ đ?‘˘đ?‘&#x;đ?‘’ = đ?‘ƒđ?‘– = 2.2 đ?‘?đ?‘Žđ?‘&#x; Ă— 1.1 = 2.42 đ?‘?đ?‘Žđ?‘&#x; = 0.242 đ?‘ /đ?‘šđ?‘š2 The temperature T at the shell side is taken as 126.3 â °C which equals to 259.4 â °F. Shell inside diameter, (Di) is taken from the process design as 3181.755745 mm the maximum allowable joint efficiency, E = 1 the maximum allowable stress, (S) for a carbon steel shell was found from the literature at that temperature to be 12900 psi = 88.9 N/mm2 The equation used to calculate the wall thickness of the shell is the cylindrical section thickness equation introduced from ASME BPV Code as following: đ?‘ƒđ?‘– Ă— đ??ˇđ?‘– 0.242 Ă— 3181.755745 đ?‘Šđ?‘Žđ?‘™đ?‘™ đ?‘‡â„Žđ?‘–đ?‘?đ?‘˜đ?‘›đ?‘’đ?‘ đ?‘ đ?‘œđ?‘“ đ?‘Ąâ„Žđ?‘’ đ?‘†â„Žđ?‘’đ?‘™đ?‘™ = đ?‘Ą = = (2 Ă— đ?‘† Ă— đ??¸) − 1.2 Ă— đ?‘ƒđ?‘– (2 Ă— 88.9 Ă— 1) − 1.2 Ă— 0.242 = đ?&#x;”. đ?&#x;‘đ?&#x;‘đ?’Žđ?’Ž đ?‘–đ?‘›đ?‘?đ?‘™đ?‘˘đ?‘‘đ?‘–đ?‘›đ?‘” 2 đ?‘šđ?‘š đ?‘?đ?‘œđ?‘&#x;đ?‘&#x;đ?‘œđ?‘ đ?‘–đ?‘œđ?‘› đ?‘Žđ?‘™đ?‘™đ?‘œđ?‘¤đ?‘Žđ?‘›đ?‘?đ?‘’

2.4.2.2 Selection of Pressure Relief System The pressure relief valve is an important system to install on any pressure vessel including heat exchangers to prevent catastrophic failure of vessels by providing a mean of relieving of the over pressure that occurs inside the vessel and makes a balance of the pressure inside. (Sinnott, 2009). The relief system is considered for the current heat exchanger (E-301) by installing the PSV on the shell’s body of the exchanger as can be seen from the P&ID with the type of Spring-Loaded Relief Valve as it is the mostly used type of relief valves that withstands a good load. Moreover, relief valve is important to Shell and Tube exchangers as the popular practice is to put the higher pressure fluid through the tube and that will save cost for the construction of the shell. The size selection was based on calculations in excel spreadsheet called “E-301� design tab “Mechanical Design� with size designation of T. 18

2.4.3 Mechanical Drawing

Rich MDEA (In)

Lean MDEA (In)

Baffle Spacing = 0.318 m

Baffle Height = 2.72 m

3.2 m

Thickness of Shell = 0.00633 m

Lean MDEA (Out) Rich MDEA (Out)

100 m

19

2.5 Summary and Review 2.5.1 Equipment Specification/Data Sheet

Shell and Tube Heat Exchanger

Project Name: BIG GAS PROJECT Project Number: Chevron 1 Sheet No. 1 REV By Date APVD By Humam 05/11/15 Capstone

Date 05/11/15

Owner’s Name: Humam Al-Rikabawi Plant Location: Longford, VIC, Australia Case Description: Sales Gas & LPG Plant Equipment Number: E-301 Description: Heat Exchanger Data Per Unit

Fluid Total fluid flow (kg/h) Vapour fraction Density (kg/m3) Viscosity (mNsm-2) Specific heat (kJ/kg°C) Thermal conductivity (W/m°C) Temperature (°C) Pressure (bar) Pressure drop allowed (bar) Pressure drop calculated (bar) Flow velocity (m/s) No. of passes Film transfer coefficient (W/m2°C) Ft factor Effective mean temperature difference (°C) Construction & Materials Tube material No. of tubes 190 Pitch (mm) Length (m) 100 O.D (mm) Design pressure (bar) 6.05 (+10% Factor) Shell material Length (m) Baffle material

110 Carbon steel

Design pressure (bar)

O.D (mm) Baffle type

2.42 (+10% Factor)

Shell Side In Out Lean MDEA 1,543,000 0 0 971.6 993.1 0.6181 1.035 3.751 3.586 0.2274 0.2291 126.3 99.79 2.2 1.5 0.7 7.69427E-06 0.006896

Tube Side Out Rich MDEA 1,819,000 0 0.0045 1066 566.5 2.273 1.441 3.281 3.493 0.246 0.2375 67.8 90 6.2 5.5 0.7 0.28717 1.023105

1 158.17

2 1115.8

In

0.91 31.03 Stainless Steel 160 Arrangement 150 I.D (mm) Design temp. (°C) 90 Carbon steel 3188.3 Segmental

I.D (mm) Baffle spacing (mm) Design temp. (°C) 126.3

Triangular 90

3182 318

20

2.5.2 Critical Review The design of this shell and tube heat exchanger was done based on a common technique used by the industry which is explained briefly in (Sinnott, 2009). I started the design project of the heat exchanger in parallel to the major equipment design and that was of a good benefit for me as I thought doing that would save me time later. I started the first step of the exchanger design by making a decision on the fluid type that has to go through the tubes and shell, I had discussion with my group members on what is the best to do for fluids allocation in the exchanger. From my literature review and personal knowledge I thought that the corrosive fluid should go inside the tubes but then one of my group members said that actually the fluid with the higher pressure should go in the tubes, then I had to make a decision on allocating the corrosive fluid (Rich Cold Amine) through the tubes as it contains CO 2 which made it easy selection as tubes can be changed when corrosion takes place. After I was done with doing the design calculations on excel software, my journey was started to optimize the design to meet the design equations conditions like the tube velocity which had to be between 1-2 m/s and overall heat transfer coefficient and the pressure drops. I firstly didn’t meet all of the conditions of the design equations but I kept on going until the final calculations of the pressure drop of the shell part, my main objective was to make the final overall heat transfer coefficient within 30% difference to the assumed one as well as keeping an eye on the pressure drop of both tube and shell side which have to be not maximum than 0.7 bar which is the allowable pressure drop. It was interesting for me how excel is an essential and a very useful tool to be used at the design calculation stage because if something goes wrong along the way of the design then all I need to do is just inputting the new value and everything else will be changed accordingly. Because of my mass flow rates of both streams are high, it was kind of hard for me to optimize the calculations to the point where it would be within the range of the standard values, for example I had to set the tube length to be 100m long to meet the requirements and conditions of the equations especially the overall heat transfer coefficient. This design part could’ve been optimized by choosing a smaller length with changing something else like diameter of the tube and especially the outside diameter as I found out that it will greatly impact the overall heat transfer coefficient. Anyhow, I was happy to meet all the conditions and that my overall heat transfer coefficient lies within the range. The design can be optimized by decreasing the tube length to 25 meters but that will results in a much lower tube heat transfer coefficient but with less pressure drop along the tube with more drop on the shell side, it will also increase the number of tubes and give a bigger shell diameter as can be seen from the calculation In the excel spread sheet of (E-301) tab “Suggested Design” comparing to the used design. The tube overall heat transfer coefficient was chosen over all of the other aspects with 100m tube to be used. I learnt so many things that I wasn’t aware of for the design of the shell and tube heat exchanger, I realized that the main goal behind designing the exchanger is to maximize the heat transferred from the hot to the cold fluid. I also felt that the operational design is important to control the whole process of the heat exchanger in case of hazard which can be mitigated through installing alarms and controllers etc. like how it was done in the HAZID part. I think that now I feel confident about the design of the shell and tube heat exchanger and that will definitely help me in my future career if I get to work as a design engineer for such type of equipment.

21

2.6 Nomenclature Symbol

Meaning

Symbol

Meaning

Q

Heat Load, Watt

Np

Number of Tube Passes

U

Overall Heat Transfer Coefficient, W/m2⁰C

Ht

Tube Side Heat Transfer Coefficient, W/m2 ⁰C

Ao

Heat Transfer Area, m2

Hs

Th,i Th,o

Inlet Temperature of Hot Stream, ⁰C Outlet Temperature of Hot Stream,⁰C

E S

Shell Side Heat Transfer Coefficient, W/m2 ⁰C the maximum allowable joint efficiency Maximum Allowable Stress, N/mm2

Tc,i Tc,o

Inlet Temperature of Cold Stream, ⁰C Outlet Temperature of Cold Stream, ⁰C

Pi Pt

Design Pressure, N/mm2 Tube Pitch, m

∆Tlm

Log Mean Temperature Difference, ⁰C

Db

Bundle Diameter, m

Ft

Temperature Correction Factor

Ds

Shell Diameter, m

Cp ῤ

Specific Capacity, Kj/kg.⁰C Denity, kg/m3

Ut Us

Tube Side Velocity, m/s Shell-Side Velocity, m/s

K

Thermal Conductivity, W/m ⁰C

Nu

Nusselt Number

μ L

Viscosity, cP or mNsm-2 Length of the heat exchanger tube, m

Re Lb

Reynolds Number Baffle Spacing, m

Nt Di

Number of Tubes Inner Diameter, m

As ∆Pt

Cross-flow area , m2 Tube-Side Pressure Drop, bar

Do

Outside Diameter, m

∆Ps

Shell-Side Pressure Drop, bar

2.7 References Sinnott, Ray, and Gavin Towler. 2009. Chemical Engineering Design. Oxford: ButterworthHeinemann. Thulukkanam, Kuppan. 2014. Heat Exchanger Design Handbook. UK: Taylor & Francis. Seligmann, Ben. 2015. “Lecture 9: Quantitative Risk Assessment.” Power Point Lecture notes. www.lms.curtin.edu.au. Dreamstime. 2015. Schematic of Shell and Tube Heat Exchanger. http://www.dreamstime.com/stockphoto-shell-tube-heat-exchanger-schematic-image31635090.

22

CHAPTER 3: MAJOR DESIGN TEG ABSORBER COLUMN (T-401)

TAN QI JIE 16242652

Executive Summary This chapter focuses on the design of the TEG absorber column (T-401). Hydrocarbon gas is contacted with triethylene glycol (here on referred to as TEG) in an absorber column in order to remove the water. After extensive calculations, the column was designed to have a diameter of 4.1 meters, with a shell thickness of 125 mm. It has 28 trays, with a plate spacing of 0.5 meters and holes that are 5 mm in diameter. The detailed design, along with the process controls and operational p can be found in these chapters. The whole column and its supports are made of stainless steel 347. Included in this design are the control systems, HAZOP analysis, as well as the design of the auxiliary components such as the valves and pipes. The pipes that I have opted to use are 10’’ carbon steel API 5L Grade A pipes, supported every 6 meters. The control systems have been designed in accordance with the API RP 14c, which is an industry standard. I am confident in my design and believe that it is more than capable of achieving the required specifications. The data sheet for the designed column can be found at the end of this report, detailing the major specifications associated with the design.

Contents List of Figures ......................................................................................................................................... iii List of Tables .......................................................................................................................................... iv 3.0

Introduction ................................................................................................................................ 1

3.1.1 Objectives .................................................................................................................................... 1 3.1.2 Methodology ............................................................................................................................... 1 3.2

Process Design ............................................................................................................................ 3

3.3

Operational Design - Column Internal Selection......................................................................... 4

3.3.1 Tray Selection .............................................................................................................................. 4 3.3.2 Column Calculations .................................................................................................................... 5 3.3.3 Assumptions ................................................................................................................................ 5 3.3.4 Calculating the number of plates ................................................................................................ 5 3.3.5 Calculating the column diameter ................................................................................................ 6 3.3.6 Weir Dimensions ......................................................................................................................... 7 3.3.7 Hole Dimensions.......................................................................................................................... 8 3.3.8 Checking for weeping .................................................................................................................. 8 3.3.9 Plate Pressure Drops ................................................................................................................. 10 3.3.10 Check Entrainment.................................................................................................................... 12 3.3.11 Calculation of the perforated area............................................................................................ 13 3.3.12 Number of holes required......................................................................................................... 15 3.4

Vessel Structural Calculations ................................................................................................... 15

3.4.1

Thickness calculations .............................................................................................................. 15

3.4.2

Loading on the vessel ............................................................................................................... 16

3.4.3

Analysis of stresses ................................................................................................................... 17

3.4.4

Checking for elastic buckling .................................................................................................... 19

3.4.5

Support structure calculations ................................................................................................. 19

3.4.6

Base ring and anchor bolts ....................................................................................................... 20

3.4.7

Design summary ....................................................................................................................... 22

3.5

Piping Design ............................................................................................................................. 23

3.6

Valve Design .............................................................................................................................. 24

3.7

Mechanical Design .................................................................................................................... 25

3.8

Process Control ......................................................................................................................... 27

3.9

Sensors ...................................................................................................................................... 28

3.9.1

Valves ....................................................................................................................................... 29 i

3.10

Emergency Shut Down .............................................................................................................. 29

3.11

Risk Analysis .............................................................................................................................. 29

3.12

Operating Procedure................................................................................................................. 31

3.12.1 Startup...................................................................................................................................... 31 3.12.2 Normal operation ..................................................................................................................... 32 3.12.3 Commissioning ......................................................................................................................... 32 3.12.4 Shutdown ................................................................................................................................. 33 3.12.5 Hydrocarbon Leak .................................................................................................................... 33 3.12.6 Fire............................................................................................................................................ 33 3.12.7 Maintenance ............................................................................................................................ 34 3.13

Data Sheet ................................................................................................................................. 34

3.14

Critical Review ........................................................................................................................... 35

3.15

References ................................................................................................................................ 36

ii

List of Figures Figure 1: Dehydration Process Flow Diagram ......................................................................................... 3 Figure 2: Relationship between downcomer area and weir length [5] .................................................... 8 Figure 3: Weep- point correlation [5] ....................................................................................................... 9 Figure 4: Discharge Coefficent[5] ........................................................................................................... 10 Figure 5: Flooding velocity [5] ................................................................................................................ 12 Figure 6: Entrainment correlation for sieve plates[5] ............................................................................ 13 Figure 7: Relationship between angle subtended by chord, chord height and chord length [5]........... 14 Figure 8: Relationship between hole area and pitch [5] ........................................................................ 15 Figure 9: Overall Column 3-D Model .................................................................................................... 26 Figure 10: Column Section View ........................................................................................................... 26 Figure 11: Plate Model .......................................................................................................................... 27 Figure 12: TEG Absorber PNID .............................................................................................................. 28

iii

List of Tables Table 1: Difference between packed and plate columns [5] .................................................................... 4 Table 2: Comparision of different plate types [5] .................................................................................... 4 Table 3: Data retrieved from HYSYS........................................................................................................ 5 Table 4: Design Summary...................................................................................................................... 22 Table 5: Piping steel grades and their respective allowable stress[2] ................................................... 23 Table 6: Pipe Standard Diameters ........................................................................................................ 23 Table 7: Standard Nozzle Orifice Data[6] ............................................................................................... 25 Table 8: HAZOP ..................................................................................................................................... 29

iv

3.0 Introduction An important part in the hydrocarbon refinement process is the dehydration step. The presence of water can cause many major issues such as the formation of hydrates and corrosion. This study focuses on the design of the dehydration column. Water is removed by contacting the hydrocarbon stream with tri-ethylene glycol (also known as TEG) in an absorber column. Due to the critical nature of this process, it is of the utmost importance that the column is well designed.

3.1.1 Objectives The aim of this project is to conduct a full in-depth design of the TEG absorber column, starting from the design of the internals, to the supports and foundations and ending with the process control system and the risk assessment associated with the column.

3.1.2 Methodology The flowchart below illustrates the design methodology that I undertook during this process. Caluclate the maximum and minimum vapour and liquid flows for the turn-down ratio required

Collect the system properties and estimate them if required

Select a trial plate spacing

Estimate the column diameter based on flooding considerations

Decide the liquid flow arrangement

1

Make a trial plate layout, including downcomer area, active area, hole area, hole size and weir height

Check the weeping rate

Check the plate pressure drop

Check the down-comer back-up

Decide on plate layout details such as calming zones and unperforated areas. Check hole pitch.

Recalculate the percentage flooding based on the chosen column diameter

Check entrainment. If it’s too high, change the diameter and restart again

2

Optimize design

Finalize design

3.2 Process Design The overall process flow diagram can be found in either Volume 1 or 3. In this section, I will only be focusing on the design of the TEG absorption column. TEG was selected for its stability and high boiling points. There were many things to consider when designing the system but the one with the greatest priority was safety. Many processes and sensors were put in place to either eliminate risk or reduce it as much as reasonably possible. Figure 1 shows the process flow for the dehydration subprocess.

Figure 1: Dehydration Process Flow Diagram

After exiting the acid gas removal system, the hydrocarbon stream enters the TEG dehydration column (T-401) at the bottom and bubbles to the top through the plates. The TEG enters the column from the top and slowly flows to the bottom, contacting the hydrocarbon stream on the way down and removing most of the water. The dehydrated hydrocarbons then flows through a heat exchanger (E-405) and enters the mercury removal system. The rich TEG that exits the bottom of the column gets heated up to 100 ËšC, enters the regeneration column (T-402) and the water is removed from the TEG. The waste water exits from the top of the column and is sent to the water treatment system while the TEG exits through the bottom, flows through the heat exchanger. It gets mixed 3

with make-up water, pressurized and enters a heat exchanger before being recycled back into the absorber.

3.3 Operational Design - Column Internal Selection There are 2 different types of column internals possible: packed and plates. Table 1 compares their differences. Table 1: Difference between packed and plate columns [5]

Packed Not suitable for very low liquid rates Packing should be considered for small diameter columns (less than 0.6m) because plates would be difficult to install Pressure drop per equilibrium stage (HETP) can be lower for packing than plates More suitable for foaming systems Will be cheaper than the plate equivalent if dealing with corrosive liquids Liquid holdup is appreciable lower

Plate Can handle a wider range of liquid and gas flowrates Efficiency of a plate column can be predicted with greater certainty Can be design with greater assurance than packed columns If liquid causes fouling, it is easier to make provision for cleaning Easier to make provision for the withdrawal of side streams from plate columns Easier to make provision for cooling

I chose to go with a plate design in the end due to its greater certainty in design. Also, as the system is assumed to be non-foaming (as mentioned in the basis of design), thus pushing us to select the plate column. Lastly, the difference between maximum and minimum flowrates is quite large, securing my decision in selecting the plate column.

3.3.1 Tray Selection The next major decision to make in the design of the column was the choice of column internals. There are 3 types of plates available: bubble-cap, sieve and valve. Table 2 outlines the differences between them based on several factors. Table 2: Comparision of different plate types [5]

Factor Cost Capacity Operating range

Efficiency Pressure Drop

Reason Bubble-caps> sieve>valve plates at a ratio of 3:1.5:1 Little difference in capacity between the 3 The range of vapour and liquid rates over which the plate will operate satisfactorily will be largest when using a bubble-cap plate and the smallest when using a sieve plate. Valve plates are designed to be more flexible than sieve plates but at a lower cost than bubble-caps No notable difference between all 3 plates Sieve plates give the lowest pressure drop, bubble-caps the most and valve plates somewhere in the middle

4

I opted for the sieve plates due to its cost and suitability for the process. Being substantially cheaper than the bubble-cap tray, it is able to deal with the constraints of our process. In addition, Sinnott [5] recommended the use of bubble-cap trays only when dealing with low vapour rates. Therefore, we could easily eliminate the use of bubble-cap trays.

3.3.2 Column Calculations After completing the HYSYS simulation, I was able to use the data to calculate the number of trays required. I selected methane as my low key and water as my high key. Table 3 outlines the data I used, which was taken from Aspen HYSYS. Table 3: Data retrieved from HYSYS

Variable (xLK)distillate (XHK)distillate (xLK)bottoms (xHK)bottoms ( xHK)feed ( xLK)feed Ď v Ď L ÎąLK Bottoms Flowrate Distillate Flowrate Pressure Vw

Value 0.74138 4.6e-7 0.01063 0.477 0.0038 0.73886 75.75 kg/m3 1107 kg/m3 1.89 25084.57 kmol/h 25172.77 kmol/h 62 bar 173.05 kg/s

3.3.3 Assumptions We had to make several assumptions over the course of our calculations. Firstly, the turn-down ratio was assumed to be 70%. I also assumed there were no energy losses in the system and that the fluid was non-foaming. The last assumption I made was that all the holes would be free of deposits at all time, therefore there would be negligible pressure drop as a result of fouling. However, in the real world, this would be an impossibility.

3.3.4 Calculating the number of plates The minimum number of plates was calculated using the Fenske Equation, đ?‘Ľ đ?‘Ľ log (đ?‘Ľ đ??żđ??ž ) ( đ?‘Ľđ??ťđ??ž ) đ??ťđ??ž đ?‘‘ đ??żđ??ž đ?‘? đ?‘ = log đ?›źđ??żđ??ž =

0.74138 0.477 log (4.6đ?‘’ − 7) (0.01063) log 1.89 = 28.4 đ?‘ đ?‘Ąđ?‘Žđ?‘”đ?‘’đ?‘ ≈ 29 đ?‘ đ?‘Ąđ?‘Žđ?‘”đ?‘’đ?‘

But since there is no reboiler for the absorber column, đ?‘ đ?‘˘đ?‘šđ?‘?đ?‘’đ?‘&#x; đ?‘œđ?‘“ đ?‘ đ?‘Ąđ?‘Žđ?‘”đ?‘’đ?‘ = 29 − 1 5

= 28 đ?‘ đ?‘Ąđ?‘Žđ?‘”đ?‘’đ?‘ To calculate the optimal feed stage, đ?‘ đ?‘… đ??ľ đ?‘Ľđ??š,đ??ťđ??ž đ?‘Ľđ??ľ,đ??żđ??ž log ( ) = 0.206log[( )( )( ) đ?‘ đ?‘† đ??ˇ đ?‘Ľđ??š,đ??żđ??ž đ?‘Ľđ??ˇ,đ??ťđ??ž

2

Where NR=number of stages above the feed, including any partial condenser And NS= number of stages below the feed, including the reboiler 25084.57 0.0038 0.01063 2 = 0.206log[( )( )( ) ] 25172.7 0.73886 4.6đ?‘’ − 7 đ?‘ đ?‘… log ( ) = 1.315 đ?‘ đ?‘† đ?‘ đ?‘… = 20.6 đ?‘ đ?‘† đ?‘ đ?‘… + đ?‘ đ?‘† = 28 đ?‘ đ?‘† = 28 − đ?‘ đ?‘… đ?‘ đ?‘† = 1.29 ≈2 Therefore the column will have 28 stages, with the feed entering in the second stage.

3.3.5 Calculating the column diameter The column diameter is governed by the vapour flowrate and the vapour velocity must be less than that which will cause excessive liquid entrainment or a high pressure drop. The vapour velocity can be estimated by the Lowenstein equation as calculated below. � ̂� =

(−0.171đ?‘™đ?‘Ą2

(đ?œŒđ??ż − đ?œŒđ?‘‰ ) 1/2 + 0.27đ?‘™đ?‘Ą − 0.047) [ ] đ?œŒđ?‘‰

1107 − 75.75 1/2 = [(−0.171 ∗ 0.5 ) + (0.27 ∗ 0.5) − 0.047)] ∗ [ ] 75.75 2

= 0.1669 đ?‘š/đ?‘ To calculate the column diameter, 4đ?‘‰đ?‘¤ đ??ˇđ?‘? = √ đ?œ‹đ?œŒđ?‘Ł đ?‘˘Ě‚đ?‘Ł 4 ∗ 173.05 =√ đ?œ‹ ∗ 1107 ∗ 0.1169 = 4.17 đ?‘š

6

Where Vw= max V rate (kg/s) Ď v= density of vapour

With a column diameter, the column area can now be calculated đ??´đ?‘? =

đ?œ‹đ?‘‘2 4

đ?œ‹ ∗ 4.172 = 4 = 13.65 đ?‘š2 Downcomer area is assumed to be 12% of the total area đ??´đ?‘‘ = 0.12 ∗ 13.6 = 1.63 đ?‘š2 Therefore the net area, which is the area available for vapour-liquid disengagement, is đ??´đ?‘› = đ??´đ?‘? − đ??´đ?‘‘ = 13.65 − 1.63 = 12.02 đ?‘š2 Active area is calculated to be đ??´đ?‘Ž = đ??´đ?‘? − 2đ??´đ?‘‘ = 13.65 − (2 ∗ 1.63) = 10.39 đ?‘š2 As a first estimate, the hole area is assumed to be 10% of the active area đ??´â„Ž = 0.1 ∗ 10.39 = 1.039 đ?‘š2

3.3.6 Weir Dimensions For columns operating above atmospheric pressure, Sinnott [5] recommends a weir height of 40-50 mm. The higher the weir height, the higher the efficiency. However, this efficiency will come at the cost of an increased plate pressure drop. For an initial estimate, I selected a weir height of 50 mm.

7

Figure 2: Relationship between downcomer area and weir length [5]

The relationship between downcomer area and weir length is graphed in Figure 2. For our case, Ad/Ac was calculated to be 12%, giving us a value of approximately 0.72 for Iw/Dc. As a result, Iw was calculated to be 3 meters.

3.3.7 Hole Dimensions Sinnott [5] states that the hole sizes may vary from 2.5 – 12 mm. However, the preferred size is 5 mm. Larger holes may be used for fouling systems but since our system is assumed to be non-fouling, there is no such need. Because it is cheaper, I opted for the holes to be punched, as opposed to drilling. The only downside to punching is that the hole sizes are limited by the plate material. For carbon steel, the minimum hole size is equal to the plate thickness. Since I will be using a 5 mm carbon steel plate, I chose a 5 mm hole size.

3.3.8 Checking for weeping Weeping occurs when the leakage of liquid through the plate holes is excessive and is a result of low vapour velocity. The hole area has to be such that at the lowest operating rate, the vapour flow velocity is still above the weep point. Eduljee’s equation can be used to calculate the minimum vapour velocity, as shown below. đ?‘˘ĚŒâ„Ž =

[đ??ž2 − 0.9(25.4 − đ?‘‘â„Ž ) (đ?œŒđ?‘Ł )1/2

K2 can be obtained from Figure 3, which illustrates the correlation between K2 and liquid height.

8

Figure 3: Weep- point correlation [5]

But first, we would have to calculate liquid height (hw+how). As mentioned before, we assumed that the column operates with a turn-down ratio of 70%. Therefore, the minimum liquid rate = 0.7 ∗ 3.76 (from HYSYS) = 2.63 đ?‘˜đ?‘”/đ?‘ đ?‘€đ?‘–đ?‘›đ?‘–đ?‘šđ?‘˘đ?‘š â„Žđ?‘œđ?‘¤ = 750 ∗ [

đ??żđ?‘¤ 2/3 = 750 [ ] đ?œŒđ??ż đ?‘™đ?‘¤

2.63 2/3 ] (1106 ∗ 3

= 1.57 ∗ 10−4 đ?‘šđ?‘š đ?‘™đ?‘–đ?‘žđ?‘˘đ?‘–đ?‘‘ Where Lw= liquid flow rate, in kg/s lw= weir length, in meters how= weir crest, in mm liquid đ?‘€đ?‘Žđ?‘Ľđ?‘–đ?‘šđ?‘˘đ?‘š â„Žđ?‘œđ?‘¤ = 750 ∗ [

đ??żđ?‘¤ 2/3 = 750 ∗ [ ] đ?œŒđ??ż đ?‘™đ?‘¤

3.76 2/3 ] (1106 ∗ 3

= 3.21 ∗ 10−4 đ?‘šđ?‘š đ?‘™đ?‘–đ?‘žđ?‘˘đ?‘–đ?‘‘ â„Žđ?‘¤ + â„Žđ?‘œđ?‘¤ đ?‘Žđ?‘Ą đ?‘Ąâ„Žđ?‘’ đ?‘šđ?‘–đ?‘›đ?‘–đ?‘šđ?‘˘đ?‘š đ?‘&#x;đ?‘Žđ?‘Ąđ?‘’ = 50 + 3.21 ∗ 10−4 ≈ 50 đ?‘šđ?‘š đ?‘™đ?‘–đ?‘žđ?‘˘đ?‘–đ?‘‘

Therefore from Figure 3, K2 = 30 and đ?‘˘ĚŒâ„Ž =

[đ??ž2 − 0.9(25.4 − đ?‘‘â„Ž ) (đ?œŒđ?‘Ł )1/2

9

=

[30 − 0.9(25.4 − 5) (75.75)1/2 = 1.288 đ?‘š/đ?‘

đ??´đ?‘?đ?‘Ąđ?‘˘đ?‘Žđ?‘™ đ?‘šđ?‘–đ?‘›đ?‘–đ?‘šđ?‘˘đ?‘š đ?‘Łđ?‘Žđ?‘?đ?‘œđ?‘˘đ?‘&#x; đ?‘&#x;đ?‘Žđ?‘Ąđ?‘’ = =

đ?‘€đ?‘–đ?‘›đ?‘–đ?‘šđ?‘˘đ?‘š đ?‘Łđ?‘Žđ?‘?đ?‘œđ?‘˘đ?‘&#x; đ?‘&#x;đ?‘Žđ?‘Ąđ?‘’ đ??´â„Ž

0.7 ∗ 2.12 1.035

= 1.42 đ?‘š/đ?‘ The actual minimum vapour rate is higher than that which will cause weeping (1.42>1.28), therefore it is acceptable.

3.3.9 Plate Pressure Drops To calculate the dry plate pressure drop, I first had to calculate the minimum vapour velocity through the holes �̂ℎ =

2.12 1.035

= 2.048 đ?‘š/đ?‘

Figure 4: Discharge Coefficent[5]

For a plate thickness and hole diameter ratio of 1, and Ah/Ap of 0.1, the orifice coefficient (Co) can be read from Figure 4 as 0.84. Therefore, the pressure drop through the dry plate can be calculated as đ?‘˘â„Ž 2 đ?œŒđ?‘Ł â„Žđ?‘‘ = 51 ( ) đ??śđ?‘œ đ?œŒđ?‘™ 10

2.048 2 75.75 = 51 ( ) ( ) 0.84 1106 = 22.38 đ?‘šđ?‘š đ?‘™đ?‘–đ?‘žđ?‘˘đ?‘–đ?‘‘ Residual head is calculated as â„Žđ?‘&#x; = =

12.5 ∗ 103 đ?œŒđ?‘™

12.5 ∗ 103 1106

= 11.3 đ?‘šđ?‘š đ?‘™đ?‘–đ?‘žđ?‘˘đ?‘–đ?‘‘ Therefore total pressure drop â„Žđ?‘Ą = 22.38 + 11.3 + 50 = 83.68 đ?‘šđ?‘š đ?‘™đ?‘–đ?‘žđ?‘˘đ?‘–đ?‘‘ đ?‘Ąâ„Žđ?‘’đ?‘&#x;đ?‘’đ?‘“đ?‘œđ?‘&#x;đ?‘’ đ?‘Žđ?‘?đ?‘?đ?‘’đ?‘?đ?‘Ąđ?‘Žđ?‘?đ?‘™đ?‘’ To calculate downcomer pressure loss, first take hap as hw – 10 = 40 mm Therefore the area under the apron, đ??´đ?‘Žđ?‘? = 3 ∗ 40 ∗ 10−3 = 0.12 đ?‘š3 Since Aap < Ad, 2

â„Žđ?‘‘đ?‘?

đ??żđ?‘¤đ?‘‘ = 166 ( ) đ?œŒđ?‘™ đ??´đ?‘Žđ?‘?

2 3.76 = 166 ∗ ( ) 1106 ∗ 0.12

= 4.7 đ?‘šđ?‘š Back-up in downcomer â„Žđ?‘? = (â„Žđ?‘¤ + â„Žđ?‘œđ?‘¤ ) + â„Žđ?‘Ą + â„Žđ?‘‘đ?‘? = 50 + 83.68 + 4.7 ≈ 139 đ?‘šđ?‘š = 0.139 đ?‘š 1 (đ?‘?đ?‘™đ?‘Žđ?‘Ąđ?‘’ đ?‘ đ?‘?đ?‘Žđ?‘?đ?‘–đ?‘›đ?‘” + đ?‘¤đ?‘’đ?‘–đ?‘&#x; â„Žđ?‘’đ?‘–đ?‘”â„Žđ?‘Ą) = 0.275 đ?‘š 2 Since hb < ½ (plate spacing + weir height), plate spacing is acceptable. To check residence time, đ?‘Ąđ?‘&#x; = =

đ??´đ?‘‘ â„Žđ?‘?đ?‘? đ?œŒđ?‘™ đ??żđ?‘¤đ?‘‘

1.63 ∗ 0.139 ∗ 1106 3.76 11

= 66.6 đ?‘ đ?‘’đ?‘?đ?‘œđ?‘›đ?‘‘đ?‘ Since tr> 3 seconds, residence time is acceptable

3.3.10 Check Entrainment Finally, to check the entrainment �� =

2.12 12.02

= 0.176 đ?‘š/đ?‘ đ??šđ??żđ?‘‰ =

=

đ??żđ?‘¤ đ?œŒđ?‘Ł √ đ?‘‰đ?‘¤ đ?œŒđ?‘™

3.76 75.75 √ 173 1106

= 5.9 ∗ 10−3

Figure 5: Flooding velocity [5]

With FLV in hand, K1 can now be estimated from Figure 5 and it came out to be 0.1. Knowing K1, uf can be found đ?œŒđ?‘™ − đ?œŒđ?‘Ł đ?‘˘đ?‘“ = đ??ž1 √ đ?œŒđ?‘Ł 1106 − 75.75 = 0.1 ∗ √ 75.75 = 0.35 12

Using uf, the percentage flooding can be calculated by % đ?‘“đ?‘™đ?‘œđ?‘œđ?‘‘đ?‘–đ?‘›đ?‘” = =

�� ��

0.176 0.35

= 49 % Figure 6 uses FLV and percentage flooding to determine fractional entrainment, represented by the symbol Ďˆ. Therefore, when it is 49% flooded and FLV is 0.1, Ďˆ comes out to be 0.035. This is well below 0.1.

Figure 6: Entrainment correlation for sieve plates[5]

Since flooding is well below the design figure of 85, the column diameter can be reduced. However, this would increase the pressure drop.

3.3.11 Calculation of the perforated area đ?‘™đ?‘¤ 3 = đ??ˇđ?‘? 4.17 ≈ 0.72 Ď´c, read off Figure 7, comes out to be 92Ëš and Lh/DC = 0.15.

13

Figure 7: Relationship between angle subtended by chord, chord height and chord length [5]

𝐴𝑛𝑔𝑙𝑒 𝑠𝑢𝑏𝑡𝑒𝑛𝑑𝑒𝑑 𝑏𝑦 𝑒𝑑𝑔𝑒 𝑜𝑓 𝑝𝑙𝑎𝑡𝑒 = 180 − 92 = 88° 𝑀𝑒𝑎𝑛 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑢𝑛𝑝𝑒𝑟𝑓𝑜𝑟𝑎𝑡𝑒𝑑 𝑒𝑑𝑔𝑒 𝑠𝑡𝑟𝑖𝑝𝑠 = (𝐷𝑐 − 50 ∗ 10−3 )𝜋 = (4.17 − 50 ∗ 10−3 )𝜋 ∗

∡ 180

88 180

= 6.327 𝑚 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑢𝑛𝑝𝑒𝑟𝑓𝑜𝑟𝑎𝑡𝑒𝑑 𝑒𝑑𝑔𝑒 𝑠𝑡𝑟𝑖𝑝 = 6.327 ∗ 50 ∗ 10−3 = 0.316 𝑚2 𝑀𝑒𝑎𝑛 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑐𝑎𝑙𝑚𝑖𝑛𝑔 𝑧𝑜𝑛𝑒 = 𝑤𝑒𝑖𝑟 𝑙𝑒𝑛𝑔𝑡ℎ + 𝑤𝑖𝑑𝑡ℎ 𝑜𝑓 𝑢𝑛𝑝𝑒𝑟𝑓𝑜𝑟𝑎𝑡𝑒𝑑 𝑠𝑡𝑟𝑖𝑝 = 3 + 50 ∗ 10−3 = 3.05 𝑚 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑐𝑎𝑙𝑚𝑖𝑛𝑔 𝑧𝑜𝑛𝑒 = 2 ∗ (3.05 ∗ 50 ∗ 10−3 ) = 0.305 𝑚2

𝑇𝑜𝑡𝑎𝑙 𝑎𝑟𝑒𝑎 𝑓𝑜𝑟 𝑝𝑒𝑟𝑓𝑜𝑟𝑎𝑡𝑖𝑜𝑛 = 𝐴𝑎 − 0.305 − 0.316 = 10.39 − 0.305 − 0.316 = 9.769 𝑚2 𝐴ℎ 1.039 = 𝐴𝑝 9.769 14

= 0.106

Figure 8: Relationship between hole area and pitch [5]

From Figure 8, lp/dh is read as 2.9 which is satisfactory as it is between 2.5 and 4.0.

3.3.12 Number of holes required đ??´đ?‘&#x;đ?‘’đ?‘Ž đ?‘œđ?‘“ 1 â„Žđ?‘œđ?‘™đ?‘’ =

đ?œ‹đ?‘‘2 4

= 1.964 ∗ 10−5 đ?‘š2 đ?‘‡â„Žđ?‘’đ?‘&#x;đ?‘’đ?‘“đ?‘œđ?‘&#x;đ?‘’ đ?‘›đ?‘˘đ?‘šđ?‘?đ?‘’đ?‘&#x; đ?‘œđ?‘“ â„Žđ?‘œđ?‘™đ?‘’đ?‘ =

1.039 1.96 ∗ 10−5

= 52902 â„Žđ?‘œđ?‘™đ?‘’đ?‘

3.4 Vessel Structural Calculations In my first trial, I selected the first went with carbon steel but found that it was not able to deal with the large amount of internal stresses and therefore I selected stainless steel mixed with 18 Cr, 10 Ni and Cb (also known as Stainless Steel 347). Its density was found to be 8800 kg/m3. [3] I also chose to use double butt joints with a full radioscopic exam for the entire construction. As a result, S = 117.8 and E = 1. As a safety net, the design pressure of the system is calculated to be 10% higher than gauge pressure. Therefore, the design pressure comes out at 68.2 bar, which converts to 6.82 N/mm2.

3.4.1 Thickness calculations In order to deal with the large amount of internal forces, I opted to use a hemispherical head. The thickness of the head can be calculated by

15

đ?‘Ą= =

đ?‘ƒđ?‘– đ??ˇđ?‘– 4đ?‘†đ??¸ − 0.4đ?‘ƒđ?‘–

6.82 ∗ 4.1 ∗ 103 (4 ∗ 117.8 ∗ 1) − (0.4 ∗ 6.82) = 59.68 đ?‘šđ?‘š

Where S is the maximum allowable stress and is a result of selecting SS 347 as the construction material, obtained from the Sinnott book. E represents the welded-joint efficiency and is equal to 1 as a result of selecting the double butt joint. The thickness of the shell can be calculated as đ?‘Ą= =

đ?‘ƒđ?‘– đ??ˇđ?‘– 2đ?‘†đ??¸ − 1.2đ?‘ƒđ?‘–

6.82 ∗ 4.1 ∗ 103 (2 ∗ 117.8 ∗ 1) − (1.2 ∗ 6.82) = 122.9 đ?‘šđ?‘š ≈ 123 đ?‘šđ?‘š

Adding in corrosion allowance, đ?‘Ąâ„Žđ?‘–đ?‘?đ?‘˜đ?‘›đ?‘’đ?‘ đ?‘ đ?‘&#x;đ?‘’đ?‘žđ?‘˘đ?‘–đ?‘&#x;đ?‘’đ?‘‘ = 123 + 2 = 125 đ?‘šđ?‘š

3.4.2 Loading on the vessel The dead weight of the vessel and its contents is calculated by đ?‘Šđ?‘Ł = đ??śđ?‘¤ đ?œ‹đ?œŒđ?‘š đ??ˇđ?‘š đ?‘”(đ??ťđ?‘Ł + 0.8đ??ˇđ?‘š )đ?‘Ą ∗ 10−3 = 1.15 ∗ đ?œ‹ ∗ 8800 ∗ 4.1 ∗ 9.81 ∗ [30 + (0.8 ∗ 4.1)] ∗ 125 ∗ 10−3 = 5230201.8 đ?‘ = 5230 đ?‘˜đ?‘ Where Wv is the total weight of the shell Cw is the coefficient to account for the accessories Hv is the height of the cylindrical section g is the gravitational constant t is the wall thickness Dm is the mean diameter of the vessel Ď m is the density of the material To calculate the weight of the plates

16

đ??´đ?‘&#x;đ?‘’đ?‘Ž đ?‘œđ?‘“ đ?‘œđ?‘›đ?‘’ đ?‘?đ?‘™đ?‘Žđ?‘Ąđ?‘’ = =

đ?œ‹đ?‘‘2 4

đ?œ‹ ∗ 4.12 4

= 13.2 đ?‘š2 đ?‘Šđ?‘’đ?‘–đ?‘”â„Žđ?‘Ą đ?‘œđ?‘“ đ?‘œđ?‘›đ?‘’ đ?‘?đ?‘™đ?‘Žđ?‘Ąđ?‘’ & đ?‘–đ?‘Ąđ?‘ đ?‘?đ?‘œđ?‘›đ?‘Ąđ?‘’đ?‘›đ?‘Ąđ?‘ = 1.2 ∗ 13.2 = 15.84 đ?‘˜đ?‘ đ?‘Šđ?‘’đ?‘–đ?‘”â„Žđ?‘Ą đ?‘œđ?‘“ 30 đ?‘?đ?‘™đ?‘Žđ?‘Ąđ?‘’đ?‘ = 15.84 ∗ 30 = 475 đ?‘˜đ?‘ Therefore đ?‘‡đ?‘œđ?‘Ąđ?‘Žđ?‘™ đ?‘¤đ?‘’đ?‘–đ?‘”â„Žđ?‘Ą đ?‘œđ?‘“ đ?‘Ąâ„Žđ?‘’ đ?‘?đ?‘œđ?‘™đ?‘˘đ?‘šđ?‘› = đ??ˇđ?‘’đ?‘Žđ?‘‘ đ?‘¤đ?‘’đ?‘–đ?‘”â„Žđ?‘Ą + đ?‘?đ?‘™đ?‘Žđ?‘Ąđ?‘’đ?‘ đ?‘Žđ?‘›đ?‘‘ đ?‘?đ?‘œđ?‘›đ?‘Ąđ?‘’đ?‘›đ?‘Ąđ?‘ = 5230 + 475 = 5705 đ?‘˜đ?‘ To calculate wind loading 2 đ?‘ƒđ?‘¤ = 0.07 ∗ đ?‘˘đ?‘¤

= 0.07 ∗ 63.462 = 281.9 đ?‘ /đ?‘š2 Where uw is the maximum velocity of the wind (obtained from the wind rose in the basis of design) The load per unit length is calculated by đ?‘Š = đ?‘ƒđ?‘¤ ∗ đ??ˇđ?‘’đ?‘“đ?‘“ = 281.9 ∗ (4.1 + 0.4) = 1268.55 đ?‘ /đ?‘š Where Deff is the effective diameter of the column after taking into account ladders and manways Bending moment exerted by the wind đ?‘€đ?‘Ľ = =

đ?‘Šđ?‘Ľ 2 2

1268.55 ∗ 302 2

= 570847.5 đ?‘ đ?‘š

3.4.3 Analysis of stresses At the bottom tangent line, Pressure stresses:

17

đ?œŽđ??ż =

đ?‘ƒđ?‘– đ??ˇđ?‘– 4đ?‘Ą

6.82 ∗ 4.1 ∗ 103 = 4 ∗ 125 = 56.87 đ?‘ /đ?‘šđ?‘š2

đ?œŽâ„Ž = =

đ?‘ƒđ?‘– đ??ˇđ?‘– 2đ?‘Ą

(6.82 ∗ 4.1 ∗ 103 ) 2 ∗ 125

= 113.7 đ?‘ /đ?‘šđ?‘š2

Dead weight stress: đ?œŽđ?‘¤ = =

đ?‘Šđ?‘§ đ?œ‹(đ??ˇđ?‘– + đ?‘Ą)đ?‘Ą

5705 ∗ 103 đ?œ‹(4.1 ∗ 103 + 125) ∗ 125 = 3.43 đ?‘ /đ?‘šđ?‘š2

Bending stress: đ??ˇđ?‘œ = 4.1 ∗ 103 + (2 ∗ 125) = 4350 đ?‘šđ?‘š đ??źđ?‘Ł = =

đ?œ‹ (đ??ˇ 4 − đ??ˇđ?‘–4 ) 64 đ?‘œ

đ?œ‹ (43504 − 41004 ) 64 = 3.7 ∗ 1012 đ?‘šđ?‘š4 đ?œŽđ?‘? = Âą

đ?‘€ đ??ˇđ?‘– ( + đ?‘Ą) đ??źđ?‘‰ 2

570847.5 ∗ 103 4100 =Âą ( + 125) 3.7 ∗ 1012 2 = Âą0.335 đ?‘ /đ?‘šđ?‘š2 Therefore resultant longitudinal stress: đ?œŽđ?‘§ = đ?œŽđ??ż + đ?œŽđ?‘¤ Âą đ?œŽđ?‘? Since Ďƒw is compressive, it is negative đ?œŽđ?‘§ (đ?‘˘đ?‘?đ?‘¤đ?‘–đ?‘›đ?‘‘) = 56.87 − 3.43 + 0.335 = 53.775 18

đ?œŽđ?‘§ (đ?‘‘đ?‘œđ?‘¤đ?‘›đ?‘¤đ?‘–đ?‘›đ?‘‘) = 56.87 − 3.43 − 0.335 = 53.105 The greatest difference between the principal stresses is on the downwind side therefore = 113.7 − 53.105 = 60.6 đ?‘ /đ?‘šđ?‘š2

3.4.4 Checking for elastic buckling Critical buckling stress: đ?œŽđ?‘? = 2 ∗ 104 (

đ?‘Ą ) đ??ˇđ?‘œ

125 = 2 ∗ 104 ( ) 4350 = 574.7 đ?‘ /đ?‘šđ?‘š2 The maximum compressive stress will occur when the vessel is not under pressure. Therefore đ?‘€đ?‘Žđ?‘Ľ đ?‘ đ?‘Ąđ?‘&#x;đ?‘’đ?‘ đ?‘ = 3.43 + 0.335 = 3.76 Which is much lower than the critical buckling stress therefore the design is acceptable.

3.4.5 Support structure calculations There are 3 main types of supports we could use: saddle, legs or skirts. However, due to the vertical orientation of our vessel, saddle supports are easily eliminated. Support legs can also be eliminated due to the weight and height of our column. Therefore the only support available is the skirt type. The maximum dead weight occurs during hydrotesting when the vessel is full of water. đ??ˇđ?‘’đ?‘Žđ?‘‘ đ?‘¤đ?‘’đ?‘–đ?‘”â„Žđ?‘Ą đ?‘œđ?‘“ đ?‘Łđ?‘’đ?‘ đ?‘ đ?‘’đ?‘™ đ?‘¤â„Žđ?‘’đ?‘› đ?‘–đ?‘Ą đ?‘–đ?‘ đ?‘“đ?‘˘đ?‘™đ?‘™ đ?‘œđ?‘“ đ?‘¤đ?‘Žđ?‘Ąđ?‘’đ?‘&#x; = =

đ?œ‹ ∗ đ?‘‘đ?‘–2 ∗ â„Ž ∗ đ?‘‘đ?‘’đ?‘›đ?‘ đ?‘–đ?‘Ąđ?‘Ś đ?‘œđ?‘“ đ?‘¤đ?‘Žđ?‘Ąđ?‘’đ?‘&#x; ∗ đ?‘”đ?‘&#x;đ?‘Žđ?‘Łđ?‘–đ?‘Ąđ?‘Ś 4

đ?œ‹ ∗ 4.12 ∗ 30 ∗ 1000 ∗ 9.81 4 = 3885.5 đ?‘˜đ?‘

đ?‘‡đ?‘œđ?‘Ąđ?‘Žđ?‘™ đ?‘¤đ?‘’đ?‘–đ?‘”â„Žđ?‘Ą = đ??ˇđ?‘’đ?‘Žđ?‘‘ đ?‘¤đ?‘’đ?‘–đ?‘”â„Žđ?‘Ą đ?‘œđ?‘“ đ?‘Łđ?‘’đ?‘ đ?‘ đ?‘’đ?‘™ đ?‘“đ?‘˘đ?‘™đ?‘™ đ?‘œđ?‘“ đ?‘¤đ?‘Žđ?‘Ąđ?‘’đ?‘&#x; + đ?‘¤đ?‘’đ?‘–đ?‘”â„Žđ?‘Ą đ?‘œđ?‘“ đ?‘Łđ?‘’đ?‘ đ?‘ đ?‘’đ?‘™ = 3885.5 + 5705 = 9590.5 đ?‘˜đ?‘ Wind loading was calculated previously as 1.268 kN/m Therefore, đ?‘€đ?‘?đ?‘Žđ?‘ đ?‘’ =

1.268 ∗ 332 2 19

= 690.4 đ?‘˜đ?‘ đ?‘š For the first trial, I selected the skirt thickness to be the same as the thickness of the shell. đ?œŽđ?‘?đ?‘ = =

4đ?‘€đ?‘ đ?œ‹(đ??ˇđ?‘ + đ?‘Ąđ?‘ đ?‘˜ )đ?‘Ąđ?‘ đ?‘˜ đ??ˇđ?‘

4 ∗ 690.4 ∗ 103 ∗ 103 đ?œ‹ ∗ (4100 + 125) ∗ 125 ∗ 4100 = 0.4 đ?‘ /đ?‘šđ?‘š2

đ?œŽđ?‘¤đ?‘ (đ?‘Ąđ?‘’đ?‘ đ?‘Ą) =

9590 ∗ 103 đ?œ‹ ∗ (4100 + 125) ∗ 125

= 5.78 đ?‘ /đ?‘šđ?‘š2 5705 ∗ 103 đ?œ‹ ∗ (4100 + 125) ∗ 125 ∗ 4100

đ?œŽđ?‘¤đ?‘ (đ?‘œđ?‘?đ?‘’đ?‘&#x;đ?‘Žđ?‘Ąđ?‘–đ?‘›đ?‘”) =

= 3.43 đ?‘ /đ?‘šđ?‘š2 đ?‘€đ?‘Žđ?‘Ľđ?‘–đ?‘šđ?‘˘đ?‘š đ?œŽĚ‚đ?‘ (đ?‘?đ?‘œđ?‘šđ?‘?đ?‘&#x;đ?‘’đ?‘ đ?‘ đ?‘–đ?‘Łđ?‘’) = 0.4 + 5.78 = 6.18 đ?‘ /đ?‘šđ?‘š2 In order for the design to be considered satisfactory, đ?œŽĚ‚đ?‘ ≯ đ?‘“đ?‘ đ??˝đ?‘ đ?‘–đ?‘›đ?œƒ 6.18 ≯ 1 ∗ 165đ?‘ đ?‘–đ?‘›90 6.18 ≯ 165 Therefore, the design is satisfactory. Adding an additional 2 mm for corrosion brings the skirt thickness to 127 mm.

3.4.6 Base ring and anchor bolts Take pitch circle diameter as 4.2 m Therefore đ??śđ?‘–đ?‘&#x;đ?‘?đ?‘˘đ?‘šđ?‘“đ?‘’đ?‘&#x;đ?‘’đ?‘›đ?‘?đ?‘’ đ?‘œđ?‘“ đ?‘Ąâ„Žđ?‘’ đ?‘?đ?‘œđ?‘™đ?‘Ą đ?‘?đ?‘–đ?‘&#x;đ?‘?đ?‘™đ?‘’ = 4200đ?œ‹ đ?‘ đ?‘˘đ?‘šđ?‘?đ?‘’đ?‘&#x; đ?‘œđ?‘“ đ?‘?đ?‘œđ?‘™đ?‘Ąđ?‘ đ?‘&#x;đ?‘’đ?‘žđ?‘˘đ?‘–đ?‘&#x;đ?‘’đ?‘‘ đ?‘Žđ?‘Ą đ?‘&#x;đ?‘’đ?‘?đ?‘œđ?‘šđ?‘šđ?‘’đ?‘›đ?‘‘đ?‘’đ?‘‘ đ?‘?đ?‘œđ?‘™đ?‘Ą đ?‘ đ?‘?đ?‘Žđ?‘?đ?‘–đ?‘›đ?‘” =

4200đ?œ‹ 600

= 21.9 Rounded up to the closest multiple of 4 = 24 bolts Take bolt stress as 125 N/mm2 Ms was calculated before at 690.4 kNm and W is taken at the operational value at 5705 kN. đ??´đ?‘? =

1 4đ?‘€đ?‘ − đ?‘Š] [ đ?‘ đ?‘? đ?‘“đ?‘? đ??ˇđ?‘?

20

=

1 4 ∗ 690.4 ∗ 103 [ − 5705] 24 ∗ 125 4.2 = 217.2 đ?‘šđ?‘š2

Where Ab is the area of one bolt at the root of the thread in mm2 Nb is the number of bolts fb is the maximum allowable bolt stress in N/mm2 Ms is the bending moment at the base in Nm W is the weight of the vessel in N Db is the bolt circle diameter 217.2 ∗ 4 đ??ľđ?‘œđ?‘™đ?‘Ą đ?‘&#x;đ?‘œđ?‘œđ?‘Ą đ?‘‘đ?‘–đ?‘Žđ?‘šđ?‘’đ?‘Ąđ?‘’đ?‘&#x; = √ đ?œ‹ = 16.6 đ?‘šđ?‘š đ?‘Ąâ„Žđ?‘’đ?‘&#x;đ?‘’đ?‘“đ?‘œđ?‘&#x;đ?‘’ đ?‘Žđ?‘?đ?‘?đ?‘’đ?‘?đ?‘Ąđ?‘Žđ?‘?đ?‘™đ?‘’ Total compressive load on the base ring per unit length 4đ?‘€đ?‘ đ?‘Š đ??šđ?‘? = [ 2 + ] đ?œ‹đ??ˇđ?‘ đ?œ‹đ??ˇđ?‘ 4 ∗ 690.4 ∗ 103 5705 =[ + ] 2 đ?œ‹ ∗ 4.2 đ?œ‹ ∗ 4.2 = 50.26 ∗ 103 đ?‘ /đ?‘š Where Fb is the compressive load on the base ring Ds is the skirt diameter

Assume the bearing pressure is 5 N/mm2 đ??żđ?‘? = =

đ??šđ?‘? ∗ 103 đ?‘“đ?‘?

50.26 ∗ 103 5 ∗ 103

= 10.05 đ?‘šđ?‘š Since its small, the skirt doesn’t have to be flared. Therefore the skirt diameter can be taken as 4.2 m đ??ľđ?‘œđ?‘™đ?‘Ą đ?‘ đ?‘?đ?‘Žđ?‘?đ?‘–đ?‘›đ?‘” =

đ?œ‹ ∗ 4100 24 đ?‘?đ?‘œđ?‘™đ?‘Ąđ?‘

= 536 đ?‘šđ?‘š đ?‘Ąâ„Žđ?‘’đ?‘&#x;đ?‘’đ?‘“đ?‘œđ?‘&#x;đ?‘’ đ?‘ đ?‘Žđ?‘Ąđ?‘–đ?‘ đ?‘“đ?‘Žđ?‘?đ?‘Ąđ?‘œđ?‘&#x;đ?‘Ś đ?‘Šđ?‘–đ?‘‘đ?‘Ąâ„Ž đ?‘&#x;đ?‘’đ?‘žđ?‘˘đ?‘–đ?‘&#x;đ?‘’đ?‘‘ = đ??żđ?‘&#x; + đ?‘Ąđ?‘ + 50 đ?‘šđ?‘š = 150 + 125 + 50 21

= 325 đ?‘šđ?‘š Actual bearing pressure on concrete foundation: đ?‘“đ?‘?′ =

đ??šđ?‘? đ?‘¤đ?‘–đ?‘‘đ?‘Ąâ„Ž đ?‘&#x;đ?‘’đ?‘žđ?‘˘đ?‘–đ?‘&#x;đ?‘’đ?‘‘

=

50.26 ∗ 103 325 ∗ 103

= 0.15 đ?‘ /đ?‘šđ?‘š2 3đ?‘“đ?‘?′ đ?‘Ąđ?‘? = đ??żđ?‘&#x; √ đ?‘“đ?‘&#x; (3 ∗ 0.15) = 150√ 140 = 8.5 đ?‘šđ?‘š Where fr is the allowable design stress in the ring material and is typically 140 N/mm2

3.4.7 Design summary Table 4 outlines the key aspects of my absorber design Table 4: Design Summary

Aspect Plate type Column Diameter Shell thickness Tray Spacing Plate Thickness Column material Number of plates Plate pressure drop Number of holes Hole size Hole pitch Hole construction method Height of the column Weir Height Weir Length Turn down ratio Column Head Type Head thickness Support Type Skirt Thickness Skirt material Type of weld

Value Column Design Valve tray 4.1 m 125 mm 0.5 m 5 mm Stainless steel 347 28 stages 83.68 mm liquid 52902 holes 5 mm 12.5 mm Punched 30 meters 50 mm 3 meters 70% Hemispherical 59.68 mm Vessel Support Design Skirt 125 mm Stainless steel 347 Double butt joint with full radioscopic exam 22

Type of bolt

M56

3.5 Piping Design There are a wide array of pipes we could choose, with varying inner and outer diameters as well steel grades. In essence, there are 3 types of steel grades we could select. The main difference between the 3 is their allowable stress. The steel grades and respective allowable stresses are presented in Table 5. Table 5: Piping steel grades and their respective allowable stress[2]

Steel Grade Spec. API 5L Grade A Spec. API 5L Grade B Spec. API 5LX Grade X-42

Allowable Stress (S) 25,500 psi 29,750 psi 35,700 psi

It is vital to ensure that the pipes will be able to deal with the pressures encountered during the operation in order to prevent pipe rupture. The equation used to calculate the allowable working pressure is as follows. đ?‘ƒ=

1.75đ?‘†(đ?‘Ą − 0.057) đ??ˇ + 0.04 − 0.7đ?‘Ą

Where P is the allowable working pressure in psi, t is the wall thickness in inches, D is the outside diameter in inches S is the allowable stress in psi Because pipes come in standard sizes, the allowable pressures were calculated and presented in Table 6. I chose to use Grade A pipes because of its cost effectiveness. Also, the pressures dealt with in this system do not justify the use of Grade X-42 pipes, as can be seen in the following table. Table 6: Pipe Standard Diameters

Pipe Outer Diameter (in)

10.75

12.75

Pipe Inner Diameter (in) 10.312 10.25 10.192 10.136 10.02 12.312 12.25 12.188 12.126 12 11.874

Thickness (in) 0.438 0.5 0.562 0.624 0.688 0.438 0.5 0.562 0.624 0.75 0.876

Allowable Stress (bar) 111.7429853 130.4670007 149.3473407 168.3859712 188.2069029 93.84033296 109.4915987 125.2524533 141.1240519 173.7253251 206.7988788 23

14

11.75 13.562 13.5 13.438 13.376 13.312 13.25 13.124 13

1 0.438 0.5 0.562 0.624 0.688 0.75 0.876 1

239.8185887 85.299 99.494 113.780 128.156 143.093 157.658 187.546 217.347

As the process only has a maximum design pressure of 68 bar, I would be able to use any pipe size greater than 14 inches. Weight was not much of a concern as the plant is land based. Therefore, the next criteria of importance was the flowrates in the pipes. The American Petroleum Institute’s criteria for sizing pipelines is guided by the empirical formula as shown below.[1] �� =

đ?‘? √đ?œŒđ?‘š

Where Ve is the erosional velocity in ft/s, c is an empirical constant, Ď m is the density of the fluid in lbs/ft3 For solid free fluids and pipes that are in continuous service, the constant “câ€? will have a value of 100. Therefore, for our process, the erosional velocity comes out at 11.48 ft/s (3.505 m/s). By selecting a 10.75 outer and 10.312 inch inner diameter pipe, the flow velocity for our system is 6.748*10-6 m/s, more than satisfying our design requirements. A pipe with a 10.312 inch diameter results in a schedule number of 30. In summary, the main pipelines connected to the absorber column will be Schedule 30 pipes with an internal diameter of 10.312 inch, an outer diameter of 10.75 inch and made of API 5L Grade A steel. Even though the plant is built onshore, the pipes would still have to be supported. The maximum allowable length of unsupported pipe can be calculated by đ?‘† = 6.6√đ?‘ƒ = 6.6 ∗ √10.75 = 21.6 đ?‘“đ?‘Ą Where S is the span length in feet and P is the outside diameter of the pipe in inches Therefore, the pipe will have to be supported every 6 meters.

3.6 Valve Design The control valve on line 4-5 can be sized using the following equation:

24

đ??´=

=

đ?‘Šâˆšđ?‘‡đ?‘? đ??śđ??žđ?‘ƒ1 đ??žđ?‘? √đ?‘€

1.37 ∗ 106 √589 ∗ 0.8002 348 ∗ 0.975 ∗ 1002.2 ∗ 1 ∗ √23.91 = 3.743 đ?‘–đ?‘›2

Where W is the flow rate in lb/hr T is the inlet temperature in R Z is the compressibility factor C is the flow constant P1 is the upstream pressure in psi Kb is the correction factor due to backpressure M is the molecular weight and A is the valve effective orifice area in in2

Therefore, the control valve on line 4-5 will require a valve with an effective orifice area of 3.7 inch2. From Table 7, such a valve would require a M valve. Table 7: Standard Nozzle Orifice Data[6]

Size Designation D E F G H J K L M N P Q R T

Orifice Area, in2 0.11 0.196 0.307 0.503 0.785 1.28 1.84 2.85 3.6 4.34 6.38 11.05 16.00 26.00

3.7 Mechanical Design The design was modelled in Autodesk in order to get a better visualization of what the column looked like with its 28 trays and skirt support, which can be seen in Figure 9.

25

Figure 9: Overall Column 3-D Model

Figure 10 gives a better view of the column internals.

Figure 10: Column Section View

26

1900 holes

3m

0.5 m 4.1 m

Figure 11: Plate Model

Figure 11 is an example of one plate in the column without the support column. Each plate has approximately 1900 holes and a weir length of 3 meters. In order to maintain a good seal, the sides of the plate will be welded to the walls of the column, thereby strengthening the plate. The plate supports are there as an added precaution as they do not interfere much with the column operation. The plate will be the most difficult component in the column to manufacture due to the large number of holes required. Good accuracy has to be achieved as well when cutting the disks for the trays but should not be an issue when using computer aided manufacturing methods. If the disks are not cut accurately, we run the risk of not being having a good seal.

3.8 Process Control In order to maximise process efficiency and safety, the system is controlled by various instruments and sensors. Figure 12 is the process and instrument diagram (PNID) for the absorber. It was designed with the American Petroleum Institute RP 14C in mind. The biggest concern for this process is the risk of fire and explosion due to the large amount of hydrocarbons flowing through it at all times. If left unchecked, the risk of fatality would increase exponentially.

27

Figure 12: TEG Absorber PNID

3.9 Sensors An important sensor attached to the absorber is PIT 001. It constantly monitors the vessel to ensure that it does not exceed the design specifications. There are 2 phases to this alarm. As the operating pressure of the vessel is 62 bar, the first alarm would trip when the vessel hits 65 bar, activating PSHH 001. This causes an alarm to go off in the control room, alerting the shift engineers. In the meantime, valve-403 will open, directing some hydrocarbons back to M-101 and mixes the sweet dry hydrocarbons back with the feed exiting the slug catcher. However, if the pressure still continues to rise, PSHH-002 will then trip and alert the shift engineers. Valve-404 will open and send all the hydrocarbons to the flare, rapidly depressurizing the column. Triggering PSHH-002 will also shut down Pump 4.1 so that the flow of TEG to the column would be stopped. Level sensors also constantly monitor the column, ensuring that it does not overfill. The level transmitter (LIT) will feedback to Pump 4.1, controlling the flow of TEG into the column. Maintaining an optimal level in the column is vital for an efficient operation. Lastly, temperature sensors also aid in the efficient operation of the column. TIT-401 feeds information back and controls E-401. If the temperature of the feed exceeds 50ËšC, the absorption of water starts to decrease, thus increasing the water content in stream 4-5. 28

3.9.1 Valves Check valves can be found on the major lines, streams 4-3, 4-5 and 4-7 in order to prevent backflow and ensure that equipment do not get damaged. In addition, valves-401, 403, 405 and 406 are designed to be fail closed so that in case of a malfunction, all the hydrocarbons would be flared. As such, valve-404 is designed to be fail open.

3.10 Emergency Shut Down In the event of an emergency shut down, valve-401, 403, and 405 will close and valve-404 opens. This sends all the hydrocarbons to the flare. These emergency shut down protocols will also be tied to the other sections of the plant, shutting down the major and more sensitive pieces of equipment such as the fractionation columns.

3.11 Risk Analysis A HAZOP was conducted for the absorber column based on the guidelines set forth by the State of New South Wales Department of Planning [4] and the results are as presented in table 8. Table 8: HAZOP

Item TEG Absorber

Deviation

Causes

Consequence

Low Flow Rate

Pipelines feeding into the absorber might be partially blocked due to poor maintenance

Weeping occurs resulting in a reduction in efficiency

High Flow Rate

Residence time for gas is too long

Potential to exceed vessel design pressures.

Safeguards

Recommendation

Pressure sensors constantly monitor absorber, sending the gas to flare in case pressure exceeds safety margins

2) Ensure absorber is sized to deal with maximum flowrate

3) Ensure that the flare line is appropriately sized

Reverse Flow

Presure in the absorber is greater than that in the pipes

A pressure buildup would occur in the pipes, causing a rupture

Install a check valve to prevent reverse flow

29

Sensors would monitor both the pressure in the column, as well as the pipeline. As soon as the pressure in the vessel exceeds the pressure in the pipe, the gas in the column is flared and the pressure lowered

High Temperat ure

High Pressure

Gas could exit from the amine at an elevated temperature.

Pressure could buildup in the column for a multitude of reasons such as lack of maintenance

Tensile strength of the vessel might be compromised if temperature exceeds design limits

The column could explode, spilling hydrocarbons everywhere.

Cooler (E502?!?) cools the gas before it enters the TEG absorber.

Temperature sensors in the TEG absorber also constantly monitor the vessel Pressure sensors constantly monitor absorber, sending the gas to flare in case pressure exceeds safety margins

4) The material used to construct the vessel should be able to deal with temperatures much higher than the standard operating conditions

4) The material used to construct the vessel should be able to deal with pressures much higher than the standard operating conditions

30

Power Failure

Power failures could occur in the event of a large storm where the electrical lines could be downed.

The flow of TEG would stop but the flow of gas would continue. The gas could cause a buildup of pressure in the column, or it could continue into the rest of the plant where the high water content could damage sensitive equipment down the process.

The safety valve on the flare line is designed to be fail open, so that the column would be purged of hydrocarbons in case of an emergency

Ensure that the safety valve on the flare line is "fail open" while the rest of the safety valves around the column are fail close, so that the hydrocarbons only have one line to escape

Loss of containm ent

The seals could fail or the pipes crack after an extended period of time if no maintenance is carried out

Hydrocarbons would spill out, posing a huge fire risk

Periodically conduct maintenance checks

Designate an area around the column where no ignition sources and open flames are allowed Surround the column with a firefighting system

3.12 Operating Procedure It is important to set out a procedure for the operators to follow in order to mitigate the risks of oversight. Laying out an operating procedure would also give the operators a plan to follow in case of an emergency.

3.12.1 Startup When the column is started up, the lead engineers are to make sure that this operating procedure is followed. 1) Ensure all safety systems are fully operational 2) Do a final sweep in the column to check if there are any foreign objects left behind. The lead engineer is to take charge and restrict the number of personnel entering the column. 3) Other crew members are to visually inspect the external surface of the column to check for any possible issues 31

4) Once the internal inspection is done, the lead engineer is to ensure the column is free of personnel 5) Purge the column with inert gas 6) Inform all personnel that the column is soon to be operational and to report any issues when spotted 7) Ensure that all the relevant valves have been opened 8) Start filling the column with TEG 9) Allow the circulation of TEG to stabilize 10) Introduce the flow of hydrocarbons into the column and slowly increase it 11) Monitor the column’s sensor and gauge 1 hour to make sure its fully functional and there are no issues 12) Inform the next shift of the column start up and occasionally check on the column

3.12.2 Normal operation Under normal operation, the system will be self-monitoring and constantly adjusting its own parameters to deal with the changes. The operators need only to stay in the control room and monitor any alarms that occurs and deal with them accordingly.

3.12.3 Commissioning The following steps are to be followed when commissioning the column. Hydrostatic tests will be performed to ensure it is mechanically sound. 1) Start up all safety systems 2) The commissioning engineers are to do a sweep, ensuring all the valves and machinery are properly tagged and tested. They are to ensure all the lines are connected correctly as well. 3) Do a final sweep in the column to check if there are any foreign objects left behind. The lead engineer is to take charge and restrict the number of personnel entering the column. 4) Other crew members are to visually inspect the external surface of the column to check for any possible issues 5) Once the internal inspection is done, the lead engineer is to ensure the column is free of personnel 6) Ensure that all the required valves are closed and labelled as such. Leave the valves at the top of the column open so that air may escape 7) Start filling the column with water 8) Once the column is full, shut off the water pumps 9) Check the column for any leaks 10) After it has been confirmed that the column is water tight, it can be drained 11) Next, close all valves and fill the column with compressed air until goes beyond the design limits. 12) Monitor the column and confirm that the first relief valves trip when the pressure reaches 65 bar 13) Maintain 65 bar for a minute to check if the valve will shut 14) Continue pressurizing the vessel and check that the second relief valve operates once the column reaches 68 bar, sending it all to the flare. 15) Once it has all been done, the commissioning engineers are to log it. Labels are to be placed on the equipment, noting down the name of the engineer which conducted the commissioning.

32

3.12.4 Shutdown When the column is started up, the lead engineers are to make sure that this operating procedure is followed. 1) 2) 3) 4) 5) 6) 7) 8) 9)

Slowly reduce the flow of hydrocarbons into the column Once the valves are shut, place tags on them to warn people not to open it back up Monitor the surrounding pipelines to ensure that the pressure is not building up elsewhere Once all hydrocarbon flow has ceased, next reduce the flow of TEG into the column Once the valves are shut, place tags on them to warn people not to open it back up After the column is empty, purge the column with inert gas Wait 1 hour for the column to inert gas to disperse and the column to reach ambient temperature Place signs around the column warning people of that the column is non-operational Maintenance works can now be carried out on the column

3.12.5 Hydrocarbon Leak In case of a hydrocarbon leak, the lead engineers are to make sure that this operating procedure is followed. 1) In case of emergency, all personnel are to move to the evacuation area 2) Alert the local authorities 3) Engineers in the control room are to ensure that the emergency shut down devices have kicked in and the flow of hydrocarbons have stopped 4) Send all hydrocarbons to the flare 5) Visually inspect that the flare is working 6) Ensure that the firefighting systems have started. If not, engineers in the control room have to manually start it 7) Allow the system to slowly stabilize and after which, clean up the hydrocarbon spill. Ensure that no ignition sources are allowed near the spill. Only relevant personnel will be allowed into the incident area 8) Once the area has been cleaned up, inspect the plant to find out where and why the spill occurred

3.12.6 Fire In case of a fire, the lead engineers are to make sure that this operating procedure is followed. 1) In case of emergency, all personnel are to move to the evacuation area 2) Alert the local authorities 3) Engineers in the control room are to ensure that the emergency shut down devices have kicked in and the flow of hydrocarbons have stopped 4) Send all hydrocarbons in the system to the flare 5) Visually inspect that the flare is working 6) Set up a safety perimeter and ensure only the relevant personnel into the area 7) Ensure that the firefighting systems have started. If not, engineers in the control room have to manually start it 8) Once the fire has been put out, allow the people back into the area to investigate the cause of the fire

33

3.12.7 Maintenance The column should be routinely cleaned during the planned system shut-down which occurs once a year. During this time, personnel would have to enter the column to clean any fouled surfaces and ensure the integrity of the vessel’s internal components. When doing so, the start-up and shut-down procedure should be followed. During maintenance, the following things should be checked for 1) Column integrity. The engineers should visually inspect the plates to ensure that they are not damaged in any way 2) Fouling. The engineers would have to take note of any fouling that occurs and clean them up if possible. 3) Corrosion. The engineers should systematically check the column for any rusty surfaces, removing and treating the surfaces as required. 4) Leaks. The column should be checked for leaks to ensure the structural integrity. 5) Any anomalies spotted should be logged by the safety engineers and investigated.

3.13 Data Sheet

Column Tray Data Sheet Operating Data Tower Inside Diameter Total Trays in Section Tray Space

Equipment No.

T-401

Description

Absorber Column

Top

Bottom 4.1 m

Rate (kg/hr) Density (kg/m3) Pressure Temperature

28 Trays 0.5 m Internal Conditions at Tray Number Vapour to Tray 623000 75.75 62 40

623000 75.75 62 40

Rate (kg/hr) Density (kg/m3) Temperature Viscosity

Liquid From Tray 3.76 1106 40 11.44

3.76 1106 40 11.44

Technical/ Mechanical Data Tower Manhole Inside Diameter Tray Material Tray Thickness Cap Material

1m Carbon Steel 5 mm SS 347 34

Nuts and Bolts Material Support Ring Material Corrosion Allowance Tower Attachments Trays Numbered from Date of Enquiry Order No. Manufacturer

Remarks Prepared Approved Checked Equipment Number Project Number

Carbon Steel Carbon Steel 2 mm 10 inche pipes and sensors

1/11/2015 1

Top Date of Order Drg No. Capstone Energy

5/11/2015 1

1/11/2015 3/11/2015 Joel 1 Joel Tan Tan 6/11/2015 By Appr Date Capstone Company Energy 1 Address 103 Kent Street, Cannington

6/11/2015 Date Rev T-405

3.14 Critical Review Due to the lack of data, there were many assumptions we had to make throughout the course of this design, therefore limiting the accuracies of this design. However, it has to be noted that the assumptions were realistic and is in line with the recommendations of common literature. Despite this, I have a few reservations about the work that I have done. The first point of discussion would be the column diameter. Based on the data that I have collected, the column diameter was calculated as 4.1 meters and after taking into account the required thickness of the walls and including allowances for corrosion, the outer diameter of the column is 4.2 meters. This seems too large of a number and could easily be reduced by splitting the gas and using 2 absorber columns instead. However, this would have a knock on effect on the capital expenditure of the plant, as well as increase utilities and maintenance. Therefore, an economic evaluation of this alternative should be conducted before commencing a full design. Unfortunately, I was unable to perform such an analysis due to time constraints. That being said, I have full confidence in the work I have performed thus far as I believe the assumptions I have made were in line with literature. The absorption kinetics and efficiency of the column can only be verified once the column is operational. Further simulations could be done regarding the type of column used. There is potential for the use of packed columns instead and in-depth simulations could be done comparing the use of the 2

35

different types of columns. However, this would require specialist software, of which I have no access to. There are many good features in my column design. Safety is the first and foremost consideration of my design and the process controls reflect this. I had set the pressure tolerances quite low so that the column would never encounter a significant over-pressurization. That being said, I understand that the internal pressure will occasionally exceed the operational pressures due to several factors and therefore have made provisions to prevent all the contents from flaring at every instance. I believe that the limits I have set are realistic and yet still within the safety boundaries as the vessel would be able to withstand a lot more stress. . Admittedly, the design can be further optimized. Because the calculated flooding is much lower than the design value, I could have reduced the column diameter further and recalculated since the pressure drops at the moment are minimal. However, due to time constraints, I had to move on. All in all, I feel that this column was designed well and am proud to stand behind the work that I have done.

3.15 References 1)

Brenton S. McLaury, S. A. S. (2000). "An Alternate Method to API RP 14E for Predicting Solids Erosion in Multiphase Flow." Journal of Energy Resources Technology 122: 115-122.

2)

McAllister, E. W. (2005). 3 - Pipe Design. Pipeline Rules of Thumb Handbook (Sixth Edition). E. W. McAllister. Burlington, Gulf Professional Publishing: 85-115.

3)

Online, M. S. "Stainless Steel 347." 20/10/2015, from http://www.suppliersonline.com/propertypages/347.asp.

4)

Planning, S. o. N. S. W. D. o. (2011). HAZOP Guidelines. D. o. Planning. 8.

5)

Sinnott, R. K. (2005). Chemical Engineering Design: Chemical Engineering, Elsevier Science.

6)

Whitesides, R. W. (2012) Selection and Sizing of Pressure Relief Valves.

36

CHAPTER 4: MINOR DESIGN SHELL AND TUBE CONDENSER (E-503)

TAN QI JIE 16242652

Executive Summary This chapter focuses on the design of the condenser (E-503) attached to the first fractionation column. Designed as a shell and tube heat exchanger, it uses 1,1,1,2-Tetrafluoroethane (also known as R-134a) as the refrigerant. With the hydrocarbons flowing in the tubes and the refrigerant in the shell, it exchanges 8.3667 kJ of heat per hour. This condenser is particularly important as it determines the sales gas composition. If it fails, our product might not be able to meet the pipeline specifications, as well as cause many other operational hazards. With a shell diameter of 1.6 meters, the condenser has 4162 tubes, undergoing 4 passes. Each tube is 5 meters long, with a triangular pitch of 23.85 mm and has an inner and outer diameter of 14.83 and 19.05 mm respectively. This report also details the control systems attached to the condenser. Pressure sensors constantly monitor the heat exchanger to ensure that it does not over-pressurize and check valves are used to prevent backflow. Also contained in this report is a HAZID evaluation which can be found in Appendix A.

Contents List of Figures .......................................................................................................................................... ii List of Tables .......................................................................................................................................... iii 4.0

Introduction ................................................................................................................................ 1

4.0.1 Objective ..................................................................................................................................... 1 4.0.2 Methodology ............................................................................................................................... 1 4.1

Process Design ............................................................................................................................ 2

4.2

Condenser Design ....................................................................................................................... 3

4.2.1 Refrigerant Selection ................................................................................................................... 3 4.2.2 Allocation of fluids....................................................................................................................... 3 4.2.3 Orientation .................................................................................................................................. 4 4.2.4 Thermal Design............................................................................................................................ 4 4.2.5 Mechanical Drawings ................................................................................................................ 11 4.3

Process Control ......................................................................................................................... 12

4.3

HAZID ........................................................................................................................................ 13

4.4

Operational Procedure.............................................................................................................. 13

4.5

Critical Summary ....................................................................................................................... 14

4.6

References ................................................................................................................................ 14

4.7

Data Sheet ................................................................................................................................. 15

Appendix A ............................................................................................................................................ 17

i

List of Figures Figure 1: Condenser design methodology .............................................................................................. 1 Figure 2: PFD of fractionation section .................................................................................................... 2 Figure 3: Temperature correction factor for 1 shell pass and 2 or more even tube passes [3] ............... 6 Figure 4: Constants K1 and n1 for bundle and shell calculation[3] ........................................................... 7 Figure 5:Tube side heat transfer factor[3] ............................................................................................... 8 Figure 6: Shell side heat transfer factor[3] ............................................................................................. 10 Figure 7: Sketch of Condenser .............................................................................................................. 11 Figure 8: Process control system for the condenser ............................................................................ 12

ii

List of Tables Table 1: Common Refrigerants ............................................................................................................... 3 Table 2: R-134a Properties [1][2][4] ............................................................................................................ 3 Table 3: Allocation of fluids to shell or tube[3] ........................................................................................ 4 Table 4: Data Retrieved From HYSYS ...................................................................................................... 4 Table 5: Table of assumptions ................................................................................................................ 5

iii

4.0 Introduction One of the most important heat exchangers in the process is the condenser on the first fractionation column. If this column fails to perform, the efficiency of the system would plummet and consequently, the sales gas production would drop.

4.0.1 Objective The aim of this project is to conduct a minor design of the condenser attached to the first fractionation column. Only a brief HAZID will be conducted, as well as implementing a control system design and writing up key aspects of the start-up and shut-down procedures. A full thermal design on the condenser will be performed, as well as deciding on the materials of construction

4.0.2 Methodology Figure 1 illustrates the design methodology that I followed during this process.

Figure 1: Condenser design methodology[3]

1

4.1 Process Design The first fractionation column separates the methane and ethane from the rest of the hydrocarbons in the feed stream. Because of the product specifications on the LPG product, we did not think it fesible to separate the methane from the ethane, only to have to mix it back together. The system was designed with safety as the utmost priority and is reflected in the process controls. As we will be dealing with a large amount of hydrocarbons, the risk of fire and explosions is high and this translates into an increased risk of fatality.

Figure 2: PFD of fractionation section

The fractionation column operates under the principle of distillation, separating the methane and ethane from the rest of the hydrocarbons using the differences in boiling points. The role of the condenser is to chill the hydrocarbons in order to allow this to happen. As a result, good design and operation of this condenser is of the utmost importance. Figure 2 is the process flow diagram of the fractionation subsection. As the feed enters this portion of the process, it goes through a turboexpander which expands and cools the gas down. After this, it passes through a gas-gas heat exchanger where it is cooled by the overheads from the first fractionation column (T-501) and V-501. It then goes through a cooler, reducing the temperature to 50ËšC and exits as a 2 phase fluid where it is separated in V-501. The liquids finally enter T-501 where the methane and ethane get separated from the rest of the hydrocarbon stream. The propane and heavier components exit through the bottom of the column, passes through a cooler and enter the second fractionation column (T-502) where the propane and butane are separated from the condensates. The condensates get recycled back to the condensate stabilization section. The gas from the top of V-501 gets heated up in the gas-gas heat exchanger and is mixed with the methane and ethane. The sales gas gets compressed and cooled to pipeline specification before entering the Victorian pipe network. During the course of our simulation, we compared the power difference between straight refrigeration and the use of a turboexpander. It was found that the turboexpander reduces the power requirements by 500 kW, making it more cost effective.

2

4.2 Condenser Design There are a number of design aspects that will have to be tackled. This includes but is not limited to choice of coolant, orientation of condenser, construction materials and the thermal design.

4.2.1 Refrigerant Selection There are a multitude of refrigerant available in the market these days. Table 1 lists some of the common refrigerants and their properties. Table 1: Common Refrigerants

Refrigerant Number R-728 R-290 R-717 R-170 R-600 R-134a

Name

Molecular Mass Nitrogen 28.0134 Propane 44.097 Ammonia 17.02 Ethane 30.07 n-Butane 58.12 Tetrafluoroethane 102.03

Boiling Point at 14.7 psia (˚F) -320.4 -44 -28 -127 31.2 -15

Freezing Point at 14.7 psia (˚F) -346 -309.8 -107.9 -218 -217 -142

Though it is possible to separate propane or n-butane for use as a refrigerant, we deemed it economically unfeasible as we would have had to design and build an extra fractionation column. It would have been much cheaper to buy refrigerants from the market instead. From the table above, I chose to use R-134a. It is a colourless gas that is non-flammable at atmospheric pressure. I also excluded ammonia as it is more hazardous than R-134a. Table 2 lists the physical and chemical properties of R-134a. Table 2: R-134a Properties [1][2][4]

Property Appearance Odour Flammability Boiling point Melting point Liquid Density Specific gravity Vapour pressure Critical temperature Critical pressure Thermal Conductivity Heat of vaporization Specific heat capacity Viscosity

Information/ Value Clear gas (liquefied under pressure) Slight ethereal odour Non-flammable -26.4˚C -101˚C 1206 kg/m³ 1.1 to 1.21 665 kPa @ 25˚C 100.6 ˚C 4.060 kPa 0.092 W/mK 198.6 kJ/kg 1.418 kJ/kg K 0.202 cP

4.2.2 Allocation of fluids With the coolant selected, they can now be allocated to the shell or tubes. Table 3 outlines some guidelines to aid in the decision making.

3

Table 3: Allocation of fluids to shell or tube[3]

Factor Corrosion

Fouling Fluid Temperature Operating Pressure

Pressure Drop Viscosity

Stream Flow-rate

Allocation The more corrosive fluid should be in the tubes, reducing the cost of expensive alloy or clad components The fluid with the greater tendency to foul is to be in the tubes as it will be easier to clean The fluid with the higher temperature should be in the tubes, reducing the need for lagging The fluid with the higher operating pressure should be in the tubes as high pressure tubes are cheaper high pressure shells The fluid with the lower allowable pressure drop should be in the tubes. A higher heat transfer coefficient can be achieved by allocating the more viscous fluid to the shell if the flow is turbulent. If turbulent flow cannot be achieved, then it is better to allocate the fluid in the tubes. Placing the fluid with the lower flow-rate in the shell will give the most economical design

Because the hydrocarbon stream has a higher flowrate, has a greater tendency to foul and has a greater fluid temperature and operating pressure, I decided to put the R-134a in the shell and the hydrocarbon in the tubes.

4.2.3 Orientation There are 4 possible condenser configurations. It could either be vertical or horizontal, with condensation occurring in either the shell or the tubes. I opted to use a vertical condenser, with the condensation outside of the tubes, thereby making use of the gas’ natural tendency to flow upwards and thus reducing the pump power required.

4.2.4 Thermal Design Table 4 outlines the important data that was used during the design of the condenser obtained from HYSYS. Table 4: Data Retrieved From HYSYS

Variable Hot fluid temperature in (T1) Hot fluid temperature out (T2) Heat flow Hydrocarbon volumetric flow Viscosity of hydrocarbon stream Specific heat of hydrocarbon stream Specific heat capacity of R-134a Mass flow of hydrocarbon Density of hydrocarbons

Value -30.94˚C -49.15˚C 8.366 *107 kJ/h 1003 m3/h 9.478 *10-3 cP 2.132 kJ/kg ˚C 0.88 kJ/kg K 3.549*105 kg/h 75.75 kg/m3

4

In addition, table 5 lists the assumptions I had to make during this design. The cold fluid temperatures were selected by allowing a temperature approach of 20ËšC Table 5: Table of assumptions

Variable Cold fluid temperature in (t1) Cold fluid temperature out (t2) Fouling factor of hydrocarbon Fouling factor of refrigerant

Value -89ËšC -49ËšC 0.0002 0.0002

As for the materials of construction, I decided to use aluminium for the tubes due to its good thermal conductivity and carbon steel for the shell because of its hardy nature and ease of construction. To begin the thermal design of the condenser, I first had to calculate the amount of refrigerant required. đ?‘š1 đ??śđ?‘?1 ∆đ?‘‡ = đ?‘š2 đ??śđ?‘?2 ∆đ?‘‡ 3.549 ∗ 105 ∗ 2.132 ∗ (−30.94 − −49.15) = đ?‘š2 ∗ 1.418 ∗ (224 − 184) đ?‘š2 = 242922

đ?‘˜đ?‘” â„Ž

Next, we had find the log mean temperature difference ∆đ?‘‡đ?‘™đ?‘š =

=

(đ?‘‡1 − đ?‘Ą2 ) − (đ?‘‡2 − đ?‘Ą1 ) đ?‘‡ −đ?‘Ą ln [đ?‘‡1 − đ?‘Ą2 ] 2 1

[(−30.94 − −69) − (−49.15 − −89)] (−30.94 − −69) đ?‘™đ?‘› (−49.15 − −89) =

−1.79 −0.04

= 38.94°đ??ś Therefore đ?‘…= =

đ?‘‡1 − đ?‘‡2 đ?‘Ą2 − đ?‘Ą1

−30.94 − −49.15 −69 − −89 = 0.9105

And �= =

đ?‘Ą2 − đ?‘Ą1 đ?‘‡1 − đ?‘Ą1

−69 − −89 −30.94 − −89 5

= 0.3444 For the first iteration, I decided to use 1 shell and 4 tube passes. Having an even number of passes would simplify the pipework in the condenser. From Figure 3, Ft is found to be approximately 0.92

Figure 3: Temperature correction factor for 1 shell pass and 2 or more even tube passes [3]

As a result ∆đ?‘‡đ?‘š = đ??šđ?‘Ą ∆đ?‘‡đ?‘™đ?‘š = 0.95 ∗ 38.94 = 37.33Ëšđ??ś In my first iteration, I chose U to be 500 W/ m2C. Therefore the area required is calculated to be đ??´= =

đ?‘„ đ?‘ˆâˆ†đ?‘‡đ?‘š

23250đ?‘’3 500 ∗ 37.33

= 1245.4 đ?‘š2 Where Q is the heat transferred per unit time in watts U is the overall heat transfer coefficient in W/m2C A is the heat transfer area in m2

I chose to use a split-ring floating head heat exchanger for ease of maintenance and as a first trial, I selected a tube with a 19.05 mm outer and 14.83 mm inner diameter. It was to be 5 m long on a triangular 23.81 mm pitch. Therefore,

6

đ??´đ?‘&#x;đ?‘’đ?‘Ž đ?‘œđ?‘“ đ?‘œđ?‘›đ?‘’ đ?‘Ąđ?‘˘đ?‘?đ?‘’ =

đ?œ‹đ?‘‘2 4

= 0.2992 đ?‘š2 1245.4 0.2992

đ?‘ đ?‘˘đ?‘šđ?‘?đ?‘’đ?‘&#x; đ?‘œđ?‘“ đ?‘Ąđ?‘˘đ?‘?đ?‘’đ?‘ = ≈ 4163 đ?‘Ąđ?‘˘đ?‘?đ?‘’đ?‘ Since there are 4 passes, tubes per pass = 1040.75 To check for tube-side velocity,

đ?‘‡đ?‘˘đ?‘?đ?‘’ đ?‘?đ?‘&#x;đ?‘œđ?‘ đ?‘ − đ?‘ đ?‘’đ?‘?đ?‘Ąđ?‘–đ?‘œđ?‘›đ?‘Žđ?‘™ đ?‘Žđ?‘&#x;đ?‘’đ?‘Ž =

đ?œ‹ ∗ đ??ź. đ??ˇ 2 4

= 0.0001727 đ?‘š2 đ??´đ?‘&#x;đ?‘’đ?‘Ž đ?‘?đ?‘’đ?‘&#x; đ?‘?đ?‘Žđ?‘ đ?‘ = 1040.75 ∗ 0.0001727 = 0.18 đ?‘‡đ?‘˘đ?‘?đ?‘’ đ?‘ đ?‘–đ?‘‘đ?‘’ đ?‘Łđ?‘’đ?‘™đ?‘œđ?‘?đ?‘–đ?‘Ąđ?‘Ś =

đ?‘‰đ?‘œđ?‘™đ?‘˘đ?‘šđ?‘’đ?‘Ąđ?‘&#x;đ?‘–đ?‘? đ??šđ?‘™đ?‘œđ?‘¤ đ?‘œđ?‘“ đ??ťđ?‘Śđ?‘‘đ?‘&#x;đ?‘œđ?‘?đ?‘Žđ?‘&#x;đ?‘?đ?‘œđ?‘›đ?‘ đ??´đ?‘&#x;đ?‘’đ?‘Ž =

0.27 0.18

= 1.5 đ?‘š/đ?‘ Since it is between 1 and 2 m/s, it is acceptable and the next step is to calculate the bundle and shell diameter. From Figure 4, K1 and n1 for my design is 0.175 and 2.285 respectively.

Figure 4: Constants K1 and n1 for bundle and shell calculation[3]

Therefore, the diameter of the bundle is 1

đ?‘ đ?‘Ą đ?‘›1 đ??ˇđ?‘? = đ??ˇđ?‘œ ( ) đ??ž1 1

4163 2.285 = 19.05 ∗ ( ) 0.175 7

1567.1 đ?‘šđ?‘š The typical clearance for a split-ring floating head heat exchanger is 56 mm. Therefore, the diameter of the shell is đ??ˇđ?‘ = đ??ˇđ?‘? + đ?‘?đ?‘™đ?‘’đ?‘Žđ?‘&#x;đ?‘Žđ?‘›đ?‘?đ?‘’ = 1567.1 + 56 = 1623.1 đ?‘šđ?‘š To calculate the tube side heat coefficient, đ?‘…đ?‘’ = =

đ?œŒđ?‘Łđ?‘‘ đ?œ‡

75.75 ∗ 1.5 ∗ 14.83 ∗ 10−3 9.478 ∗ 10−3 = 178.0 đ?‘ƒđ?‘&#x; =

=

đ??śđ?‘? đ?œ‡ đ?‘˜đ?‘“

2.132 ∗ 103 ∗ 9.478 ∗ 10−3 2.24 ∗ 10−2 = 902.1 đ??ż 5000 = đ?‘‘đ?‘– 14.83 = 337

From Figure 5, jh = 3*10-3

Figure 5:Tube side heat transfer factor[3]

đ?‘ đ?‘˘ = đ?‘—â„Ž đ?‘…đ?‘’đ?‘ƒđ?‘&#x; 0.33 = 3 ∗ 10−3 ∗ 178.0 ∗ 902.10.33 8

= 5.04

Therefore, â„Žđ?‘– = đ?‘ đ?‘˘ ∗ = 5.04 ∗

đ?‘˜đ?‘“ đ?‘‘đ?‘–

2.241 ∗ 10−2 14.83 ∗ 10−3

= 7.62

đ?‘Š đ?‘š2 đ??ś

To calculate the shell side coefficient, I used Kern’s method. For the first iteration, I used a baffle spacing of Ds/5. đ??´đ?‘ =

=

(đ?‘?đ?‘Ą − đ?‘‘đ?‘œ )đ??ˇđ?‘ đ?‘™đ?‘? đ?‘?đ?‘Ą

[(23.81 − 19.05) ∗ 1913.5 ∗

1623.1 ] 5

23.81 = 105339.6 đ?‘šđ?‘š2 = 0.146 đ?‘š2

Where pt is the tube pitch Do is the tube outside diameter in meters Ds is the shell inside diameter in meters Lb is the baffle spacing in meters đ?‘‘đ?‘’ = =

1.1 2 (đ?‘? − 0.917đ?‘‘đ?‘œ2 ) đ?‘‘đ?‘œ đ?‘Ą

1.1 (23.812 − 0.917 ∗ 19.052 ) 19.05 = 13.52 đ?‘šđ?‘š

đ?‘‰đ?‘œđ?‘™đ?‘˘đ?‘šđ?‘’đ?‘Ąđ?‘&#x;đ?‘–đ?‘? đ?‘“đ?‘™đ?‘œđ?‘¤ đ?‘&#x;đ?‘Žđ?‘Ąđ?‘’ đ?‘œđ?‘“ đ?‘… − 134đ?‘Ž = = 0.055

đ?‘š3 đ?‘

đ?‘†â„Žđ?‘’đ?‘™đ?‘™ đ?‘ đ?‘–đ?‘‘đ?‘’ đ?‘Łđ?‘’đ?‘™đ?‘œđ?‘?đ?‘–đ?‘Ąđ?‘Ś = = 0.522 đ?‘…đ?‘’ =

242922 1 ∗ 3600 1206

0.055 0.105

đ?‘š đ?‘

1208 ∗ 0.522 ∗ 13.52 ∗ 10−3 0.202

9

= 42.21 đ?‘ƒđ?‘&#x; =

1.418 ∗ 0.202 0.0824 = 3.47

In order to minimize the pressure drop, I opted to use a segmental baffle with a 25% cut

Figure 6: Shell side heat transfer factor[3]

And from Figure 6, the heat transfer factor, jh, was read to be 2.5*10-2 As a result, the shell side heat transfer coefficient was calculated to be 1

đ?‘˜đ?‘“ đ?‘—â„Ž đ?‘…đ?‘’đ?‘ƒđ?‘&#x; 3 â„Žđ?‘ = đ?‘‘đ?‘’ =(

0.082 ∗ 103 ) ∗ 2.5 ∗ 10−2 ∗ 42.21 ∗ 3.470.33 13.52 = 9.523

Therefore the overall coefficient is calculated to be đ?‘‘ đ?‘‘đ?‘œ ln ( đ?‘œ ) đ?‘‘ 1 1 1 1 đ?‘‘đ?‘œ 1 đ?‘‘đ?‘– đ?‘œ = + + + ∗ + ∗ đ?‘ˆđ?‘œ â„Žđ?‘œ â„Žđ?‘œđ?‘‘ 2đ?‘˜đ?‘¤ đ?‘‘đ?‘– â„Žđ?‘–đ?‘‘ đ?‘‘đ?‘– â„Žđ?‘– 19.05 −3 1 19.05 19.05 ∗ 10 ln (14.83) 1 =( + 0.0002) ∗ + + + 0.0002 7.623 14.83 2 ∗ 45 9.523 1 = 0.27 đ?‘ˆđ?‘œ 10

đ?‘ˆđ?‘œ = 3.65 đ?‘Š/đ?‘š2 đ??ś Where Uo is the overall coefficient ho is the outside fluid film coefficient hi is the inside fluid film coefficient hod is the outside dirt coefficient (fouling factor) hid is the inside dirt coefficient kw is the thermal conductivity of the tube wall which is 245 W/mC for aluminium tubes di is the tube inside diameter do is the outside tube diameter The last thing before revising the design is to check the pressure drop in the tube side đ??ż đ?œŒđ?‘˘đ?‘Ą2 ∆đ?‘ƒđ?‘Ą = đ?‘ đ?‘? [8đ?‘—đ?‘“ ( ) + 2.5] đ?‘‘đ?‘– 2 = 4[8 ∗ 2.5 ∗ 10−2 ∗ (

5000 (75.75 ∗ 1.52 ) ) + 2.5] 14.83 2

= 23832.4

đ?‘ đ?‘š2

= 0.2 đ?‘?đ?‘Žđ?‘&#x; Because the assumed U and the calculated U is not between a 30% margin, I recalculated using the new U. However, despite running through several iterations on Excel, I was unable to get the values to converge.

4.2.5 Mechanical Drawings

Figure 7: Sketch of Condenser

Figure 7 is a sketch of the condenser with a summary of the important dimensions annotated. Because it is a split ring floating head heat exchanger, it will be possible to separate the shell from the tubes at the tube-sheet for maintenance whenever required.

11

4.3 Process Control

Figure 8: Process control system for the condenser

The primary focus of the control system is to prevent the over-pressurization of the condenser. As such, a pressured sensor, PIT-001 constantly monitors both the shell and the tubes. If it starts to exceed the design specifications, alarm PSHH-501 will trip, alerting the personnel in the control room. The control system will also automatically close valve-513 and open valve-512, sending all the hydrocarbons to the flare. Unlike the absorber section, I chose have an option to send the hydrocarbons back to M-101 because the methane and ethane are more refined. Therefore, flaring it would produce less emissions. The column also has a pressure sensor monitoring it, which will also trip valves-512 and 513 in case of an over-pressurization incident. Therefore if either the column or the condenser exceeds their design pressures, the entire column will shut down and its contents flared.

12

The condenser also has a temperature sensor, TI-504. It is used to control the flow of refrigerant, which in turn manipulates the exit temperature of the hydrocarbons. The operators will also be able to monitor this information from the control room. A check valve is installed at the exit of the condenser in order to prevent backflow and damage to the equipment. Other valves of importance are valves 512 and 513, which are designed to be fail open and fail close respectively. This is so that in case of a malfunction, the only line for the hydrocarbons to enter would be the flare line. It might seem as if the condenser is not very well-protected. However, because the column itself is well-protected by its own set of sensors, any major incident would by stopped there before even reaching the heat exchanger.

4.3 HAZID A HAZID risk assessment was performed on the condenser to ensure that all aspects of risk pertaining to the equipment have been reduced as much as possible. This can found in Appendix A.

4.4 Operational Procedure Start-up is usually considered one of the most hazardous times of an equipment as many things could go wrong due to oversight such as not opening valves that are required. In order to ensure that the condenser is safely operated, the lead engineer is to ensure that these procedures are adhered to. 1) 2) 3) 4) 5) 6) 7) 8)

Ensure all safety systems are fully operational Visually inspect the heat exchanger to ensure that there are no cracks Inform the personnel in the area that the column is starting up soon Open all the relevant valves and remove their respective warning labels Purge the system with inert gas to ensure the pipework is not blocked Start circulating nitrogen in the shell Once the system has stabilized, allow hydrocarbons to flow in the tubes Monitor the system at all times using the sensors and gauges, ready to shut the system down in case of any incidences

Though shut-down is not as hazardous, it is important to ensure that procedures are followed to ensure the equipment is not mistakenly switched on. 1) Slowly reduce the flow of hydrocarbons in the tubes and place warning labels on the valves that have been shut 2) Once the flow of hydrocarbons has stopped, start reducing the flow of nitrogen in the shell, once again placing warning labels on the valves that have been shut 3) Wait 30 min for the remaining hydrocarbons to disperse 4) Place signs around the column indicating that the column and condenser are nonoperational 5) After waiting 30 minutes, qualified personnel may start working on the condensers 6) The operators are to enter it shut-down of the column in a log-book and the next shift informed so as to ensure the column is not mistakenly switched on

13

4.5 Critical Summary There are many different limitations associated with this design that arise due to the assumptions that I have had to make which then reduce the accuracy of my design. One example would be the fouling factor. Even though they were retrieved from trusted literature, fouling is the condenser would change depending on the situation and how long it has been since the last maintenance. It is also unclear how much fouling the hydrocarbons and refrigerant would cause. Because of the multitude of choices available in the industry with regards to refrigerants, there could be a better choice than R-134a. But due to the time constraints, I was unable to spend more time researching the different types of refrigerants. This would have a significant impact on the efficiency of the condenser, as well as pipe work. In addition, the Further simulation work can be done on the fluid flow within the condenser. Turbulence in the shell can be a good thing as it would increase the heat transfer coefficient. However, the calculations performed during this design of the condenser might not be a good reflection of that. Therefore, it might be more beneficial, as well as accurate, to model it in specialist software. During the design of the condenser, I found that the variable with one of the greatest impact on heat transfer coefficient is the fluid’s flow rate and thermal conductivity. Fouling, though important, might not be a big factor in reducing the efficiency since the fouling factors are quite. Although they are assumed values, the values I selected were in line with trusted literature. Even though my assumed values were respectable, I do have some reservations about my work as they do not accurately reflect the situation as the number of tubes and the shell diameter seems a bit too large. In addition, due to time constraints, I have been unable to get the overall heat transfer coefficient to converge. As such, more work would have to been done in order to ascertain the root of the problem. Perhaps the initial assumption that the HYSYS simulation is reliable was wrong. Despite this, there are a multitude of good points in my design. I believe that the split ring, floating head type heat exchanger is a good choice as it is easy to construct, clean and maintain. And as can be seen from the process control systems, my design prioritises safety above all else. The next step in design if there were more time would be to calculate and design the supports for the condenser. However, because it would be bolted onto the column it might cause considerable torque to the column which its supports would have to deal with.

4.6 References 1) 2)

3) 4)

BOC (2015). R-134a Safety Data Sheet. Refrigeration, I. I. o. "Thermophyscal properties of refrigerants: R134a ". Retrieved 19/10/2015, 2015, from http://www.iifiir.org/userfiles/file/webfiles/summaries/tabl_r134a_en.pdf. Sinnott, R. K. (2005). Chemical Engineering Design: Chemical Engineering, Elsevier Science. Toolbox, T. E. "Refrigerants - Physical Properties." Retrieved 20/10/2015, 2015, from http://www.engineeringtoolbox.com/refrigerants-d_902.html.

14

4.7 Data Sheet

Heat exchanger data sheet

Equipment Tag No. E-503 Description Condenser

Operating data

Size

Shells per unit Surface per unit

Fluid Circulating Total Fluid Entering Vapour Liquid Steam Water Non-condensables Mol Wt Vapour Mol Wt Non-condensables Viscosity Latent Heat

1245.4 m^2

Split Ring, Floating Head

Type Horizontal No. of connected units 1 in Series Surface 0.105 m2 per shell Performance of one unit Shell Side R-134a (Refrigerant) 242922 kg/h In Out R-134a (Refrigerant) 102.03 0.202 198.6

1

In

Tube Side Hydrocarbons 3.59e5 kg/h Out Hydrocarbons 23.89 0.00948 N/A 15

Specific Heat Thermal Conductivity Temperature Operating Pressure (bar) Velocity No. of passes Pressure Drop Fouling Resistance Heat Exchanged

allow

1 bar

1.34 0.0824 -80 20

2.132 2.24E-02 -60 24

1

4

calc 0.0002

Service Equipment No. Project No.

allow

1 bar

calc 0.0002

0.2 bar

8.366e7 kj/h Construction of one shell 68 bar 70 bar -80 -80

Design Pressure Test Pressure Design Temperature Š Metal Temperature Tubes Shell Baffles Cross Type of Joint Design Code Prepared Checked Approved

0.5 bar

OD

19.05mm

1/11/2015 4/11/2015 7/11/2015 Date Rev

ID ID

1 Joel Tan Appr

14.83 mm Length 5m 1623 mm Clearance 56 mm 324 mm Double butt weld AS 4041

Pitch

23.85 mm

7/11/2015 Date

E-504 BGP

16

Appendix A

1

Natural disasters

HAZARD DESCRIPTION / HAZARDOUS EVENT

Extreme temperatures

HAZARD CONSEQUENCES

-

-

2

Natural disaster

Earthquake resulting in a hydrocarbon leak

-

3

Loss of Containment - Leakage from pipe/flange

Explosive hazards

-

Temperature exceeding design limits Explosion Product specifications not met Loss of property Hydrocarbon leak Potential for fatality Explosions Fire Loss of production Property damage Hydrocarbon leak Potential for fatality

PREVENTION / DETECTION / BARRIES (Existing)

-

RISK ASSESSMENT CONSECUENCE SEVERITY

HAZARD CATEGORY (GUIDEWORD)

LIELIHOOD

HAZARD ID

Temperature controller of feed stream, reboiler and condenser

Rare

Modera te

Include a safety factor during the design of the column

Rare

Modera te

Flow sensors to measure any deviation. If a leak is detected, the pump will shut down Staff to patrol the area to

Unlikely

Critical

RECOMMENDATIONS

ACTION (By)

None

N/A

None

N/A

-

Company

RISK RANKING

Low

-

-

-

Extreme

-

Regular maintenance Isolation of ignition sources

-

17

4

External Effects

Structural failure

-

Explosions Loss of containment Property damage

-

-

Process Upsets 5

6

7

Environmental Impact

Human Factors

Corrosion/ero sion leading to structural failure

Hydrocarbon / Chemical leak

Improper/ inadequate training and apathetic staff

-

-

Loss of structural integrity Hydrocarbon leak Fire Explosion Explosions Fire Loss of life Loss of property Operation failure Loss of life Property damage Explosion

-

-

-

spot any leaks Install a mercury removal unit to remove any chance of a corrosion attack Selecting the proper materials during the design phase Selecting the proper materials during the design phase

Unlikely

Critical

High

-

unlikely

Major

high

-

Sensors and gauges to monitor any leaks

Unlikely

Training Put procedures in place to ensure the process is

Likely

Critical

Extreme

-

Critical

Extreme

-

Conduct regular maintenance Isolation of ignition sources

Design Engineers and operators

Corrosive resistive materials Maintenance / inspection

Design engineers and operators

Conduct regular maintenance and inspection Ignition source isolation Proper training programs Increase safety awareness

Design engineers and operators Company Operators

18

-

8

Start-up/shut down

Blocked pipes resulting in overpressure and ultimately leading to explosions

-

Explosions Overpressure Loss of property

-

-

9

Power failure

Loss of safety systems

-

Process operating under inappropriate conditions resulting in damage to property

-

done correctly Ensure staff do not work long hours continuously Put procedures in place to ensure the process is done correctly Ensure that there are no foreign objects blocking the pipes before starting machinery Ensure vital safety equipment are fail safe.

Unlikely

Critical

High

-

-

Unlikely

Critical

High

-

Increase safety Operators awareness of the and process personnel engineers Slowly increase the flow and pressures and stop if an anomaly occurs

Periodically test safety systems to ensure their operational capabilities

Design and process engineers

19

CHAPTER 5: MAJOR DESIGN Triethylene Glycol (TEG) Regenerator Column (T-402)

SHIRJAN THAPA 16154359

Contents List of Figures ......................................................................................................................................... iv List of Tables .......................................................................................................................................... iv 5.1 Introduction ...................................................................................................................................... 1 5.1.1 Equipment processing objective and constraints ...................................................................... 1 5.1.2 Design scope .............................................................................................................................. 1 5.2 Process design ................................................................................................................................... 2 5.2.1 Design methodology .................................................................................................................. 2 5.2.2 Minimum Number of Stages ...................................................................................................... 3 5.2.3 Minimum Reflux ratio (Rmin) ....................................................................................................... 3 5.2.4 Theoretical number of stages (N) .............................................................................................. 4 5.2.5 Theoretical feed point location .................................................................................................. 5 5.2.6 Efficiency .................................................................................................................................... 5 5.2.7 Actual number of stages (Nactual) ................................................................................................ 6 5.2.8 Actual feed point location .......................................................................................................... 6 5.2.9 Column diameter ....................................................................................................................... 6 5.2.9.1 Liquid-vapour flow factor (FLV) ............................................................................................ 7 5.2.9.2 Correction for surface tensions .......................................................................................... 7 5.2.9.3 Maximum volumetric flow rate .......................................................................................... 8 5.2.9.4 Net area required ................................................................................................................ 8 5.2.9.5 Column cross sectional area ............................................................................................... 8 5.2.9.6 Calculate column diameter ................................................................................................. 8 5.2.9.7 Liquid Flow Pattern ............................................................................................................. 9 5.2.10 Provisional Plate Design ........................................................................................................... 9 5.2.11 Check Weeping....................................................................................................................... 10 5.2.12 Plate pressure Drop................................................................................................................ 11 5.2.12.1 Dry plate drop ................................................................................................................. 11 5.2.12.2 Residual head .................................................................................................................. 12 5.2.12.3 Total plate pressure drop................................................................................................ 12 5.2.13 Downcomer liquid backup ..................................................................................................... 12 5.2.13.1 Downcomer pressure loss ............................................................................................... 12 5.2.13.2 Backup in downcomer .................................................................................................... 12 5.2.13.3 Check residence time ...................................................................................................... 13 5.2.14 Check entrainment ................................................................................................................. 13 5.2.15 Column height ........................................................................................................................ 13 5.2.16 Trial layout.............................................................................................................................. 13 i

5.2.17 Perforated area ...................................................................................................................... 14 5.2.18 Number of holes..................................................................................................................... 16 5.2.19 Plate specification .................................................................................................................. 16 5.2.20 Pipe size selection .................................................................................................................. 17 5.2.21 Summary of column design .................................................................................................... 17 5.3 Operational design .......................................................................................................................... 17 5.3.1 Control system design .............................................................................................................. 17 5.3.2 Operating procedures .............................................................................................................. 18 5.3.2.1 Commissioning procedure ................................................................................................ 18 5.3.2.2 Pre-start-up procedure ..................................................................................................... 18 5.3.2.3 Start-up procedure............................................................................................................ 18 5.3.2.4 Shutdown procedure ........................................................................................................ 19 5.3.2.5 Normal operating procedure ............................................................................................ 19 5.3.2.6 Maintenance ..................................................................................................................... 19 5.3.2 Safety study .............................................................................................................................. 20 5.3.3.1 Hazard and operability (HAZOP) ....................................................................................... 20 5.4 Mechanical design .......................................................................................................................... 22 5.4.1 Materials of construction ......................................................................................................... 22 5.4.2 Design pressure ........................................................................................................................ 22 5.4.3 Design temperature ................................................................................................................. 22 5.4.4 Corrosion allowance ................................................................................................................. 22 5.4.6 Actual wall thickness ................................................................................................................ 23 5.4.6.1 Cylindrical section ............................................................................................................. 23 5.4.6.2 Torispherical head............................................................................................................. 23 5.4.7 Dead weight of vessel .............................................................................................................. 23 5.4.7.1 Shell ................................................................................................................................... 23 5.4.7.2 Plates ................................................................................................................................. 24 5.4.7.3 Insulation .......................................................................................................................... 24 5.4.7.4 Platform ............................................................................................................................ 24 5.4.7.5 Access ladder .................................................................................................................... 24 5.4.8 Wind loading ............................................................................................................................ 24 5.4.9 Bending moment (Mx) .............................................................................................................. 24 5.4.10 Stress analysis ........................................................................................................................ 25 5.4.10.1 Pressure stresses............................................................................................................. 25 5.4.10.2 Bending stresses ............................................................................................................. 25 5.4.10.3 The resultant longitudinal stress (ﾏホ) .............................................................................. 25 ii

5.4.10.4 Buckling ........................................................................................................................... 26 5.4.10 Skirt support ........................................................................................................................... 26 5.4.11 Pressure relief ........................................................................................................................ 28 5.4.11.1 Condenser ....................................................................................................................... 28 5.4.11.2 Reboiler ........................................................................................................................... 29 5.4.11 Reinforcement........................................................................................................................ 29 5.4.12 Mechanical drawing ............................................................................................................... 30 5.5 Summary and review ...................................................................................................................... 30 5.5.1 Equipment specifications ......................................................................................................... 30 5.5.2 Critical review ........................................................................................................................... 30 5.6 Nomenclature ................................................................................................................................. 30 References ............................................................................................................................................ 31 Appendix A ............................................................................................................................................ 32 Appendix B ............................................................................................................................................ 33 Appendix C ............................................................................................................................................ 34 Appendix D ............................................................................................................................................ 35

iii

List of Figures Figure 1: TEG regenerator column (T-402) ............................................................................................. 1 Figure 2: Erbar-Maddox correlation[8]..................................................................................................... 4 Figure 3: Flooding velocity, sieve plates ................................................................................................. 7 Figure 4: Selection of liquid-flow arrangement ...................................................................................... 9 Figure 5: Relation between downcomer area and weir length .............................................................. 9 Figure 6: Weep-point correlation ......................................................................................................... 10 Figure 7: Discharge coefficient, sieve plates ......................................................................................... 11 Figure 8: Entrainment correlation for seive plates[8] ............................................................................ 14 Figure 9: Relation between angle subtended by chord, chord height and chord length [8] ................. 15 Figure 10: Relation between hole area and pitch[8] .............................................................................. 15 Figure 11: Plate specification ................................................................................................................ 16 Figure 12: Stress analysis ...................................................................................................................... 26 Figure 13: Piping & Instrumentation Diagram (P&ID) for TEG regeneration column........................... 32 Figure 14: Mechanical Drawing of TEG regenerator column ................................................................ 33

List of Tables Table 1: Materials stream for T-402 ....................................................................................................... 1 Table 2: Compositions of components in T-402 ..................................................................................... 2 Table 3: K-value and relative volatility of components .......................................................................... 3 Table 4: LHS value for underwood equation .......................................................................................... 4 Table 5: Plate data ............................................................................................................................... 16 Table 6: Parameters for pipe size selection .......................................................................................... 17 Table 7: Column data ............................................................................................................................ 17 Table 8: Types of controllers[1] .............................................................................................................. 18 Table 9: Column deviations................................................................................................................... 20 Table 10: HAZOP study.......................................................................................................................... 21 Table 11: Column data required for mechanical design ....................................................................... 22 Table 12: Wall thickness dependent on vessel diameter[8] .................................................................. 22 Table 13: Standard nozzle orifice data [9] .............................................................................................. 28 Table 14: Data sheet ............................................................................................................................. 34

iv

5.1 Introduction 5.1.1 Equipment processing objective and constraints Triethylene glycol (TEG) regenerator column (T-402) is chosen to be designed as a major equipment. The column mainly separates water (H2O) and TEG. Water is sent to be further treated before releasing it into the environment and TEG is recycled back to the absorber to carry out the dehydration of hydrocarbons. Table 1, gives the over view of the materials stream. While there are various components going into the column, only three components are considered for the design propose as those three components are present in feed, distillate and bottom unlike the other components which aren’t present in bottom. The three components are H2O, Methyl diethanolamine (MDEA) and TEG. The light component is water and the heavy component is TEG. The composition of components are shown in table 2.

Figure 1: TEG regenerator column (T-402)

5.1.2 Design scope The report covers the general design of the specified column illustrated in figure 1. Firstly the process design of the column is covered based on the design procedure for multistage distillation column. Then the operational design is discussed which deals with control system design, operating procedures and safety study. Lastly mechanical design of the vessel is covered where vessel thickness is calculated and stress analysis is performed and few other factors are calculated as well. Table 1: Materials stream for T-402

Temperature (â °C) Pressure (bar) Molar Flow (kgmole/h) Mass Flow (kg/h)

Feed 120 1.1 179.5807 15200.5465

Distillate 102.0013 1.013 89.4469 1671.9117

Bottom 288.0812 1.033 90.1338 13528.6348 1

Liquid Volume Flow (m3/h) Heat Flow (kcal/h)

13.8074 -92683516.5588

1.8172 -21028548.5004

11.9903 -62076473.5110

Table 2: Compositions of components in T-402

Component Water (H2O) Methyl diethanolamine (MDEA) TEG

Molar Flow (kgmole/h) 85.1267 0.1741 89.9810

Mole Fraction (%), x 0.47403 0.00097 0.501062

Distillate (d)

H2O MDEA TEG

85.1167 0.0018 0.0295

0.951589 0.000021 0.00033

Bottom (b)

H2O MDEA TEG

0.01 0.1723 89.9185

0.000111 0.001911 0.997978

FEED (f)

5.2 Process design 5.2.1 Design methodology The methodology covered in process design of the regenerator column is as follows: 1. Specify product specifications and operating condition 2. Determine the K value followed by relative volatility in order to calculate minimum number of stages 3. Calculate minimum reflux ratio with the aid of Underwood equation 4. Calculate reflux ratio 5. Using Erbar-Maddox correlation determine theoretical number of stages 6. Calculate efficiency 7. Find actual number of stages 8. Calculate column diameter 8.1 Perform mass balance 8.2 Obtain physical properties 8.3 Determine liquid vapour flow factor 8.4 Assume plate spacing and perform correction for surface tension 8.5 Assume flooding, determine maximum volumetric flow rate and net area 8.6 Assume downcomer area, calculate cross sectional area 8.7 Determine column diameter 9. Provisional plate design 9.1 Assume hole area 10. Check weeping 11. Determine plate pressure drop 12. Downcomer liquid backup 12.1 Downcomer pressure loss 12.2 Backup in downcomer 12.3 Check if backup in downcomer is acceptable, if not go back to step 8.4 12.4 Check residence time 2

13. Check entrainment 13.1 Determine percentage flooding, if higher than assumed then go back to step 8.5 13.2 Determine entrainment, if more than 0.1 design is optimum 14. Calculate column height 15. Plate trial layout 16. Perforated area 16.1 Check if assumed hole area is acceptable, if not go back to step 9.1 17. Number of holes 18. Pipe size selection

5.2.2 Minimum Number of Stages In order to calculate the minimum number of stages, Fenske equation[8] is used and it is stated as follows. đ?‘ đ?‘šđ?‘–đ?‘›

đ?‘Ľ đ?‘Ľ đ?‘™đ?‘œđ?‘”[ đ??żđ??ž ]đ?‘‘ [ đ??ťđ??ž ]đ?‘? đ?‘Ľđ??ťđ??ž đ?‘Ľđ??żđ??ž = đ?‘™đ?‘œđ?‘” đ?›źđ??żđ??ž

Where, Nmin = Minimum number of stages including reboiler xi = concentration or mole fraction of a component LK = Light Key HK = Heavy Key Îą = average relative volatility d = distillate b = bottom To calculate ÎąLK, thermal conductivity (K-value) of light key component is required. K-value of the components are extracted from Aspen HYSYS. đ??žđ?‘Žđ?‘Łđ?‘” =

đ??žđ?‘ đ?‘Ąđ?‘Žđ?‘”đ?‘’ 1 + đ??žđ?‘&#x;đ?‘’đ?‘?đ?‘œđ?‘–đ?‘™đ?‘’đ?‘&#x; 2 đ?›źđ?‘– =

đ??žđ?‘– đ??žđ??ťđ??ž

Table 3: K-value and relative volatility of components

Component H2O MDEA TEG

Kstage 1 1.447 0.03624 0.001156

Kreboiler 33.15 5.17 0.9884

đ?‘ đ?‘šđ?‘–đ?‘›

Kavg 17.2985 2.60312 0.494778

Îąi 34.96214 5.261188 1

0.951589 0.997978 đ?‘™đ?‘œđ?‘”[ ∗ ] 0.00033 0.000111 = đ?‘™đ?‘œđ?‘” 34.96214 đ?‘ đ?‘šđ?‘–đ?‘› = 4.8029 ≈ 5

5.2.3 Minimum Reflux ratio (Rmin) Underwood equation[8] as shown below, is used to calculate the minimum reflux ratio.

3

đ?›´

đ?›źđ?‘– đ?‘Ľđ?‘– đ?‘‘ = đ?‘…đ?‘šđ?‘–đ?‘› + 1 đ?›źđ?‘– − đ?œƒ

Where the following equation is used to calculate θ and the value of θ has to be between the values of Îą of light and heavy key, đ?›´

đ?›źđ?‘– đ?‘Ľđ?‘– đ?‘“ đ?›źđ?‘– − đ?œƒ

=1−đ?‘ž

q = 0.8789 Now, solving the above equation, θ = 2.2916 (the value meets the requirement). Table 4: LHS value for underwood equation

Component H2O MDEA TEG Total

(ιi*xid)/(ιi-θ) 1.018336016 3.72055E-05 -0.000255497 1.018117724

Then, 1.0181 = đ?‘…đ?‘šđ?‘–đ?‘› + 1 ∴ đ?‘…đ?‘šđ?‘–đ?‘› = 0.0181

5.2.4 Theoretical number of stages (N) đ?‘…đ?‘’đ?‘“đ?‘™đ?‘˘đ?‘Ľ đ?‘“đ?‘™đ?‘œđ?‘¤đ?‘&#x;đ?‘Žđ?‘Ąđ?‘’ đ?‘?đ?‘œđ?‘šđ?‘–đ?‘›đ?‘” đ?‘“đ?‘&#x;đ?‘œđ?‘š đ?‘?đ?‘œđ?‘›đ?‘‘đ?‘’đ?‘›đ?‘ đ?‘’đ?‘&#x; đ??šđ?‘’đ?‘’đ?‘‘ đ?‘“đ?‘™đ?‘œđ?‘¤ đ?‘&#x;đ?‘Žđ?‘Ąđ?‘’ đ?‘”đ?‘œđ?‘–đ?‘›đ?‘” đ?‘–đ?‘›đ?‘Ąđ?‘œ đ?‘?đ?‘œđ?‘›đ?‘‘đ?‘’đ?‘›đ?‘ đ?‘’đ?‘&#x; 289 đ?‘˜đ?‘”đ?‘šđ?‘œđ?‘™đ?‘’/â„Ž đ?‘…= = 1.4530 198.9 đ?‘˜đ?‘”đ?‘šđ?‘œđ?‘™đ?‘’/â„Ž

đ?‘…đ?‘’đ?‘“đ?‘™đ?‘˘đ?‘Ľ đ?‘&#x;đ?‘Žđ?‘Ąđ?‘–đ?‘œ (đ?‘…) =

Figure 2: Erbar-Maddox correlation[8]

4

đ?‘… = 0.5923 đ?‘…+1 đ?‘…đ?‘šđ?‘–đ?‘› = 0.0178 đ?‘…đ?‘šđ?‘–đ?‘› + 1 Therefore, from the Erbar-Maddox correlation shown above in figure 2, Nm/N is approximately 0.82 and the value of Nm is 5 (calculated above as Nmin) Hence, theoretical number of stages including reboiler are 6.

5.2.5 Theoretical feed point location Empirical equation by Kirkbride[8] is used to determine the feed point location. The equation is stated as follows: 2 đ?‘Ľđ?‘? đ?‘ đ?‘&#x; đ??ľ đ?‘Ľđ?‘“ đ?‘™đ?‘œđ?‘” [ ] = 0.206 đ?‘™đ?‘œđ?‘” [( ) ( đ??ťđ??ž ) ( đ??żđ??ž ) ] đ?‘ đ?‘ đ??ˇ đ?‘Ľđ?‘“ đ?‘Ľđ?‘‘ đ??ťđ??ž đ??żđ??ž

Where, Nr = number of stages above the feed Ns = number of stages below the feed B = molar flow of bottoms D = molar flow of distillate đ?‘ đ?‘&#x; 90.1338 0.5011 0.0001 2 ) ] đ?‘™đ?‘œđ?‘” [ ] = 0.206 đ?‘™đ?‘œđ?‘” [ Ă— Ă—( đ?‘ đ?‘ 89.4469 0.47403 0.0003 đ?‘ đ?‘&#x; = 0.8275 đ?‘ đ?‘ Theoretical number of stages excluding the reboiler = 5 đ?‘ đ?‘&#x; + đ?‘ đ?‘ = 5 đ?‘ đ?‘ = 5 − đ?‘ đ?‘&#x; = 5 − 0.8275đ?‘ đ?‘ đ?‘ đ?‘ = 2.74 Therefore, the theoretical feed point location is stage number 3.

5.2.6 Efficiency Overall efficiency of the column is determined using the O’Connell equation[8] where the column efficiency is related to the liquid viscosity and the relative volatility of the light key at average column temperature. đ??¸đ?‘œ = 51 − 32.5 đ?‘™đ?‘œđ?‘”(Âľđ?‘Ž đ?›źđ?‘Ž ) where, Eo = efficiency Âľa = molar average liquid viscosity, mNs/m2 Îąa = average relative volatility of the light key Now, Âľa = 0.00195 mNs/m2 (extracted from HYSYS) Îąa = 34.9621

5

đ??¸đ?‘œ = 51 − 32.5 đ?‘™đ?‘œđ?‘”(0.00195 ∗ 34.9622) đ??¸đ?‘œ = 88.91 %

5.2.7 Actual number of stages (Nactual) đ?‘ đ?‘Žđ?‘?đ?‘Ąđ?‘˘đ?‘Žđ?‘™ =

đ?‘‡â„Žđ?‘’đ?‘œđ?‘&#x;đ?‘–đ?‘Ąđ?‘–đ?‘?đ?‘Žđ?‘™ đ?‘›đ?‘˘đ?‘šđ?‘?đ?‘’đ?‘&#x; đ?‘œđ?‘“ đ?‘ đ?‘Ąđ?‘Žđ?‘”đ?‘’đ?‘ 6 = = 6.75 đ??¸đ?‘“đ?‘“đ?‘–đ?‘?đ?‘–đ?‘’đ?‘›đ?‘?đ?‘Ś 0.8891

∴ Nactual including the reboiler is 7 stages

5.2.8 Actual feed point location The same equation used to calculate theoretical feed point location is used. 2 đ?‘ đ?‘&#x; đ??ľ đ?‘Ľđ?‘“ đ??ťđ??ž đ?‘Ľđ?‘? đ??żđ??ž đ?‘™đ?‘œđ?‘” [ ] = 0.206 đ?‘™đ?‘œđ?‘” [( ) ( )( ) ] đ?‘ đ?‘ đ??ˇ đ?‘Ľđ?‘“ đ?‘Ľđ?‘‘ đ??ťđ??ž đ??żđ??ž

So, đ?‘ đ?‘&#x; = 0.8275 đ?‘ đ?‘ Actual number of stages excluding the reboiler is 6 đ?‘ đ?‘&#x; + đ?‘ đ?‘ = 6 đ?‘ đ?‘ = 6 − đ?‘ đ?‘&#x; = 6 − 0.8275đ?‘ đ?‘ đ?‘ đ?‘ = 3.28 Hence, the actual feed point location is also stage 3.

5.2.9 Column diameter Molar feed flow rate (kgmole/hr) = 179.5807 Feed molecular weight = 84.64 Overall mass balance: D + B = 179.5807 Mass balance on light key (Water): đ?‘Ľđ?‘¤đ?‘Žđ?‘Ąđ?‘’đ?‘&#x; đ??ˇ + đ?‘Ľđ?‘¤đ?‘Žđ?‘Ąđ?‘’đ?‘&#x; đ??ľ = đ?‘Ľđ?‘¤đ?‘Žđ?‘Ąđ?‘’đ?‘&#x; đ??š 0.9516đ??ˇ + 0.0001đ??ľ = 0.4740đ??š Therefore, D = 89.4468 kgmole/hr and B = 90.1338 kgmole/hr Vapour rate (V): V= D(1 + R) = 89.4468(1 + 1.4530) = 219.4131 kgmole/hr Since the feed is saturated liquid: Liquid flow above feed, L = RD = 1.4530 Ă— 89.4468 = 129.9663 kgmole/hr Liquid flow below feed, L’ = RD + F = 129.9663 + 179.5807 = 309.5495 kgmole/hr Physical properties obtained from HYSYS: Above feed stage, vapour density (Ď V) = 0.6119 kg/m3 liquid density (Ď L) = 125 kg/m3 surface tension (Ďƒ) = 0.0015 N/m Molecular weight (MW) = 18.69 Below feed stage, Ď v = 0.35 kg/m3 Ď L = 876.1 kg/m3 Ďƒ = 0.25 N/m MW = 150.1

6

5.2.9.1 Liquid-vapour flow factor (FLV) đ??šđ??żđ?‘‰ =

đ??żđ?‘Š đ?œŒđ?‘‰ √ đ?‘‰đ?‘Š đ?œŒđ??ż

where, Lw = L for FLV top and L’ for FLV bottom Vw = V (Vapour rate) đ??šđ??żđ?‘‰ đ?‘Ąđ?‘œđ?‘? =

129.9663 0.6119 √ = 0.0414 219.4131 125

đ??šđ??żđ?‘‰ đ?‘?đ?‘œđ?‘Ąđ?‘Ąđ?‘œđ?‘š =

309.5495 0.35 √ = 0.0282 219.4131 876.1

Figure 3: Flooding velocity, sieve plates

Assume the plate spacing to be 0.9m. Then from figure 3 obtain the top and base K1 values. top K1 = 0.14 base K1 = 0.14

5.2.9.2 Correction for surface tensions đ??ž1 Ă— [

đ?œŽ 0.2 ] 0.2

where, K1 = constant obtained from figure 2 0.0015 0.2 ] = 0.0526 đ?‘Ąđ?‘œđ?‘? đ??ž1 = 0.14 Ă— [ 0.2 0.25 0.2 ] = 0.1464 đ?‘?đ?‘Žđ?‘ đ?‘’ đ??ž1 = 0.14 Ă— [ 0.2 7

đ?‘˘đ?‘“ = đ??ž1 √

đ?œŒđ??ż − đ?œŒđ?‘‰ đ?œŒđ?‘‰

where, uf = vapour velocity (m/s) at flooding point based on net area đ?‘Ąđ?‘œđ?‘? đ?‘˘đ?‘“ = 0.0526 √

125 − 0.6119 = 0.7502 đ?‘š/đ?‘ 0.6119

đ?‘?đ?‘Žđ?‘ đ?‘’ đ?‘˘đ?‘“ = 0.1464 √

876.1 − 0.35 = 7.3226 đ?‘š/đ?‘ 0.35

Design for 90% flooding at maximum flow rate: đ?‘Ąđ?‘œđ?‘? đ?‘˘đ?‘› = 0.7502 Ă— 0.9 = 0.6752 đ?‘š/đ?‘ đ?‘?đ?‘Žđ?‘ đ?‘’ đ?‘˘đ?‘› = 7.3226 Ă— 0.9 = 6.5904 đ?‘š/đ?‘

5.2.9.3 Maximum volumetric flow rate

đ?‘‰ Ă— đ?‘€đ?‘Š đ?œŒđ?‘Ł

đ?‘Ąđ?‘œđ?‘? =

219.4131 Ă— 18.69 = 1.8616 đ?‘š/đ?‘ 0.6119 Ă— 3600

đ?‘?đ?‘Žđ?‘ đ?‘’ =

219.4131 Ă— 150.1 = 17.4312đ?‘š/đ?‘ 0.35 Ă— 3600

5.2.9.4 Net area required đ?‘‰đ?‘œđ?‘™đ?‘˘đ?‘šđ?‘’đ?‘Ąđ?‘&#x;đ?‘–đ?‘? đ?‘“đ?‘™đ?‘œđ?‘¤ đ?‘&#x;đ?‘Žđ?‘Ąđ?‘’ đ?‘˘đ?‘› đ?‘Ąđ?‘œđ?‘? =

1.8616 = 2.7571 đ?‘š2 0.6752

đ?‘?đ?‘Žđ?‘ đ?‘’ =

17.4312 = 2.6450 đ?‘š2 6.5904

As a first trial take down comer area as 8% of total

5.2.9.5 Column cross sectional area 2.7571 = 2.9969 đ?‘š2 0.92

đ?‘Ąđ?‘œđ?‘? =

đ?‘?đ?‘Žđ?‘ đ?‘’ =

2.6450 = 2.8749đ?‘š2 0.92

5.2.9.6 Calculate column diameter đ?‘Ąđ?‘œđ?‘? = √

2.9969 Ă— 4 = 1.9539 đ?‘š đ?œ‹

đ?‘?đ?‘Žđ?‘ đ?‘’ = √

2.8749 Ă— 4 = 1.9137 đ?‘š đ?œ‹

8

Since, the diameter of base is smaller than the diameter of top so use the same diameter above and below the feed which is 1.9539 m. This also reduces the perforated area for plates above feed.

Figure 4: Selection of liquid-flow arrangement

Figure 5: Relation between downcomer area and weir length

5.2.9.7 Liquid Flow Pattern đ?‘€đ?‘Žđ?‘Ľđ?‘–đ?‘šđ?‘˘đ?‘š đ?‘Łđ?‘œđ?‘™đ?‘˘đ?‘šđ?‘’đ?‘Ąđ?‘&#x;đ?‘–đ?‘? đ?‘™đ?‘–đ?‘žđ?‘˘đ?‘–đ?‘‘ đ?‘&#x;đ?‘Žđ?‘Ąđ?‘’ = đ?‘€đ?‘Žđ?‘Ľđ?‘–đ?‘šđ?‘˘đ?‘š đ?‘Łđ?‘œđ?‘™đ?‘˘đ?‘šđ?‘’đ?‘Ąđ?‘&#x;đ?‘–đ?‘? đ?‘™đ?‘–đ?‘žđ?‘˘đ?‘–đ?‘‘ đ?‘&#x;đ?‘Žđ?‘Ąđ?‘’ =

đ??żâ€˛ Ă— đ?‘€đ?‘Šđ?‘?đ?‘Žđ?‘ đ?‘’ đ?œŒđ??żđ?‘?đ?‘Žđ?‘ đ?‘’

309.5469 Ă— 150.1 = 0.0098 đ?‘š3 â „đ?‘ 3600 Ă— 876.1

Using the column diameter and liquid flow pattern, figure 4 is used to determine liquid-flow arrangement which is cross flow (single pass).

5.2.10 Provisional Plate Design Column diameter, Dc = 1.9539 m Column area, Ac = 2.9969 m2 Downcomer area, Ad = 0.08 × 2.9969 = 0.2398 m2 Net area, An = Ac – Ad = 2.9969 – 0.2398 = 2.7571 m2 Active area, Aa = Ac – 2Ad = 2.9969 – 0.4796 = 2.5174 m2 Hole area, Ah take 10% Aa as first trial = 0.1 × 2.5174 = 0.2517 m2 Weir length, lw: From figure 5, lw/Dc = 0.72 So, lw = 0.72 × 1.9539 = 1.4068 m Take, weir height (hw) = 50 mm Hole diameter (dh) = 5 mm Plate thickness = 5 mm (carbon steel)[8]

9

5.2.11 Check Weeping đ?‘€đ?‘Žđ?‘Ľđ?‘–đ?‘šđ?‘˘đ?‘š đ?‘™đ?‘–đ?‘žđ?‘˘đ?‘–đ?‘‘ đ?‘&#x;đ?‘Žđ?‘Ąđ?‘’ = đ??żâ€˛ Ă— đ?‘€đ?‘Šđ?‘?đ?‘Žđ?‘ đ?‘’ =

309.5569 Ă— 150.1 = 8.6071 đ?‘˜đ?‘”â „đ?‘ 3600

đ?‘€đ?‘–đ?‘›đ?‘–đ?‘šđ?‘˘đ?‘š đ?‘™đ?‘–đ?‘žđ?‘˘đ?‘–đ?‘‘ đ?‘&#x;đ?‘Žđ?‘Ąđ?‘’, đ?‘Žđ?‘Ą 70% đ?‘Ąđ?‘˘đ?‘&#x;đ?‘›đ?‘‘đ?‘œđ?‘¤đ?‘› = 0.7 Ă— 8.6071 = 6.0250 đ?‘˜đ?‘”â „đ?‘ Now, to estimate the height of the liquid crest over the weir, Francis weir equation [8] shown below can be used. đ??żđ?‘¤ 2/3 ] â„Žđ?‘œđ?‘¤ = 750 [ đ?œŒđ??ż đ?‘™đ?‘¤ where, lw = weir length (m) how = height of liquid crest over downcomer weir (mm liquid) Lw = liquid flow rate (kg/s) 2â „3 8.6071 ] = 27.4016 đ?‘šđ?‘š đ?‘™đ?‘–đ?‘žđ?‘˘đ?‘–đ?‘‘ 876.1 Ă— 1.4068 2â „3 6.0250 ] = 750 [ = 22.0122 đ?‘šđ?‘š đ?‘™đ?‘–đ?‘žđ?‘˘đ?‘–đ?‘‘ 876.1 Ă— 1.4068

đ?‘€đ?‘Žđ?‘Ľđ?‘–đ?‘šđ?‘˘đ?‘š â„Žđ?‘œđ?‘¤ = 750 [ đ?‘€đ?‘–đ?‘›đ?‘–đ?‘šđ?‘˘đ?‘š â„Žđ?‘œđ?‘¤

đ??´đ?‘Ą đ?‘šđ?‘–đ?‘›đ?‘–đ?‘šđ?‘˘đ?‘š đ?‘&#x;đ?‘Žđ?‘Ąđ?‘’ â„Žđ?‘¤ + â„Žđ?‘œđ?‘¤ = 50 + 22.0122 = 72.0122 đ?‘šđ?‘š Figure 6 below is used to estimate the value of K2, which is a constant, dependent on the depth of clear liquid on the plate.

Figure 6: Weep-point correlation

đ??ž2 = 30.65 The correlation by Eduljee[8], shown below, is used to calculate the minimum design vapour velocity. đ?‘˘â„Ž =

[đ??ž2 − 0.9(25.4 − đ?‘‘â„Ž )] (đ?œŒđ?‘Ł )1/2

where, uh = minimum vapour velocity through the holes (m/s) dh = hole diameter (mm) K2 = constant from figure 5 10

[30.65 − 0.9(25.4 − 5)] = 20.7739 đ?‘š/đ?‘ 0.350.5 đ?‘šđ?‘–đ?‘›đ?‘–đ?‘šđ?‘˘đ?‘š đ?‘Łđ?‘Žđ?‘?đ?‘œđ?‘˘đ?‘&#x; đ?‘&#x;đ?‘Žđ?‘Ąđ?‘’ 0.7 Ă— 17.4312 đ??´đ?‘?đ?‘Ąđ?‘˘đ?‘Žđ?‘™ đ?‘šđ?‘–đ?‘›đ?‘–đ?‘šđ?‘˘đ?‘š đ?‘Łđ?‘Žđ?‘?đ?‘œđ?‘˘đ?‘&#x; đ?‘Łđ?‘’đ?‘™đ?‘œđ?‘?đ?‘–đ?‘Ąđ?‘Ś = = = 48.4703 đ?‘š/đ?‘ đ??´â„Ž 0.2517 The calculated value is acceptable as it is above the weep point. đ?‘˘â„Ž (đ?‘šđ?‘–đ?‘›) =

5.2.12 Plate pressure Drop 5.2.12.1 Dry plate drop Maximum vapour velocity through holes: �ℎ (���) =

đ?‘€đ?‘Žđ?‘Ľđ?‘–đ?‘šđ?‘˘đ?‘š đ?‘Łđ?‘œđ?‘™đ?‘˘đ?‘šđ?‘’đ?‘Ąđ?‘&#x;đ?‘–đ?‘? đ?‘“đ?‘™đ?‘œđ?‘¤ đ?‘&#x;đ?‘Žđ?‘Ąđ?‘’đ?‘?đ?‘Žđ?‘ đ?‘’ 17.4312 = = 69.2433 đ?‘šâ „đ?‘ đ??´â„Ž 0.2517

Figure 7: Discharge coefficient, sieve plates

With the aid of figure 7, orifice coefficient (CO) is estimated. đ?‘ƒđ?‘™đ?‘Žđ?‘Ąđ?‘’ đ?‘Ąâ„Žđ?‘–đ?‘?đ?‘˜đ?‘›đ?‘’đ?‘ đ?‘ 5 = =1 â„Žđ?‘œđ?‘™đ?‘’ đ?‘‘đ?‘–đ?‘Žđ?‘šđ?‘’đ?‘Ąđ?‘’đ?‘&#x; 5 đ??´â„Ž = 0.1 đ??´đ?‘Ž Therefore, CO = 0.84 đ?‘˘â„Ž 2 đ?œŒđ?‘Ł â„Žđ?‘‘ = 51 [ ] đ??śđ?‘‚ đ?œŒđ??ż â„Žđ?‘‘ = 51 [

69.2433 2 0.35 ] = 138.4463 �� ������ 0.84 876.1

11

5.2.12.2 Residual head â„Žđ?‘&#x; = â„Žđ?‘&#x; =

12.5 Ă— 103 đ?œŒđ??ż

12.5 × 103 = 14.2678 �� ������ 876.1

5.2.12.3 Total plate pressure drop â„Žđ?‘Ą = â„Žđ?‘‘ + (â„Žđ?‘¤ + â„Žđ?‘œđ?‘¤ ) + â„Žđ?‘&#x; where, ht = total plate pressure drop, mm liquid â„Žđ?‘Ą = 138.4463 + (50 + 27.4016) + 14.2678 = 230.1157 đ?‘šđ?‘š đ?‘™đ?‘–đ?‘žđ?‘˘đ?‘–đ?‘‘ Column pressure drop = 9.81 Ă— 10−3 ht Ď L Ă— đ?‘ đ?‘‚. đ?‘œđ?‘“ đ?‘ đ?‘Ąđ?‘Žđ?‘”đ?‘’đ?‘ where, Ď L = density of liquid, kg/m3 Ď L = (125 + 876.1) / 2 = 500.55 kg/m3 Column pressure drop = 9.81 Ă— 10−3 Ă— 230.6352 Ă— 6 Ă— 500.55 = 6779.7546 Pa

5.2.13 Downcomer liquid backup 5.2.13.1 Downcomer pressure loss â„Žđ?‘Žđ?‘? = â„Žđ?‘¤ − 10 đ??´đ?‘Žđ?‘? = â„Žđ?‘Žđ?‘? đ?‘™đ?‘¤ đ??żđ?‘¤đ?‘‘ 2 ] â„Žđ?‘‘đ?‘? = 166 [ đ?œŒđ??ż đ??´đ?‘š â„Žđ?‘? = (â„Žđ?‘¤ + â„Žđ?‘œđ?‘¤ ) + â„Žđ?‘Ą + â„Žđ?‘‘đ?‘? where, hap = the height of the bottom edge of the apron above the plate (mm) Aap = the clearance area under the downcomer (m2) hdc = head loss in the downcomer (mm) Lwd = liquid flow rate in downcomer (kg/s) Am = either Ad or Aap, whichever is smaller (m2) hd = downcomer backup, measure form plate surface (mm) â„Žđ?‘Žđ?‘? = 50 − 10 = 40 đ?‘šđ?‘š đ??´đ?‘Žđ?‘? = 0.04 Ă— 1.4068 = 0.0563 đ?‘š2 â„Žđ?‘‘đ?‘?

2 8.6071 ] = 5.0598 đ?‘šđ?‘š = 166 [ 876.1 Ă— 0.0563

đ?‘ đ?‘Žđ?‘Ś 6 đ?‘šđ?‘š

5.2.13.2 Backup in downcomer â„Žđ?‘? = (50 + 27.4016) + 230.1157 + 6 = 313.5173 đ?‘šđ?‘š Now, to check if the downcomer backup is acceptable use the following correlation: 12

1 (đ?‘?đ?‘™đ?‘Žđ?‘Ąđ?‘’ đ?‘ đ?‘?đ?‘Žđ?‘?đ?‘–đ?‘›đ?‘” + đ?‘¤đ?‘’đ?‘–đ?‘&#x; â„Žđ?‘’đ?‘–đ?‘”â„Žđ?‘Ą) 2 1 313.5173 < (0.9 Ă— 103 + 50) 2 â„Žđ?‘? <

313.5173 < 475 Therefore, backup in downcomer is acceptable.

5.2.13.3 Check residence time đ?‘Ąđ?‘&#x; =

đ??´đ?‘‘ â„Žđ?‘?đ?‘? đ?œŒđ??ż đ??żđ?‘¤đ?‘‘

where, tr = residence time, s hbc = clear liquid back-up, m đ?‘Ąđ?‘&#x; =

0.2398 Ă— 0.3135 Ă— 876.1 = 7.6510 đ?‘ 8.6071

5.2.14 Check entrainment �� =

đ?‘€đ?‘Žđ?‘Ľđ?‘–đ?‘šđ?‘˘đ?‘š đ?‘Łđ?‘œđ?‘™đ?‘˘đ?‘šđ?‘’đ?‘Ąđ?‘&#x;đ?‘–đ?‘? đ?‘“đ?‘™đ?‘œđ?‘¤ đ?‘&#x;đ?‘Žđ?‘Ąđ?‘’ đ?‘ đ?‘’đ?‘Ą đ?‘Žđ?‘&#x;đ?‘’đ?‘Ž đ?‘˘đ?‘› đ?‘?đ?‘’đ?‘&#x;đ?‘?đ?‘’đ?‘›đ?‘Ąđ?‘Žđ?‘”đ?‘’ đ?‘“đ?‘™đ?‘œđ?‘œđ?‘‘đ?‘–đ?‘›đ?‘” = đ?‘˘đ?‘“

where, un = vapour velocity based on net area �� =

17.4312 = 6.3222 đ?‘š/đ?‘ 2.7571

đ?‘?đ?‘’đ?‘&#x;đ?‘?đ?‘’đ?‘›đ?‘Ąđ?‘Žđ?‘”đ?‘’ đ?‘“đ?‘™đ?‘œđ?‘œđ?‘‘đ?‘–đ?‘›đ?‘” =

6.3222 = 86.34 % 7.3226

To check entrainment, figure 8 is used. FLV = 0.0282 and percentage flooding is 86.34% hence, fractional entrainment (Ń°) is 0.125 kg/kg gross liquid flow. The Ń° values is above 0.1, which suggests that the design is optimum.

5.2.15 Column height đ??ťđ?‘’đ?‘–đ?‘”â„Žđ?‘Ą = (đ?‘ đ?‘Žđ?‘?đ?‘Ąđ?‘˘đ?‘Žđ?‘™ − 1)(đ?‘?đ?‘™đ?‘Žđ?‘Ąđ?‘’ đ?‘ đ?‘?đ?‘Žđ?‘?đ?‘–đ?‘›đ?‘”) + (đ?‘?đ?‘™đ?‘Žđ?‘Ąđ?‘’ đ?‘ đ?‘?đ?‘Žđ?‘?đ?‘–đ?‘›đ?‘” Ă— 2) + (đ?‘ đ?‘Žđ?‘?đ?‘Ąđ?‘˘đ?‘Žđ?‘™ − 1)(đ?‘?đ?‘™đ?‘Žđ?‘Ąđ?‘’ đ?‘Ąâ„Žđ?‘–đ?‘?đ?‘˜đ?‘›đ?‘’đ?‘ đ?‘ )

đ??ťđ?‘’đ?‘–đ?‘”â„Žđ?‘Ą = 6 Ă— 0.9 + 0.9 Ă— 2 + 6 Ă— 0.005 = 7.23 đ?‘š

5.2.16 Trial layout Using cartridge-type construction: Unperforated strip around plate edge = 50 mm Calming zones = 50 mm

13

Figure 8: Entrainment correlation for seive plates[8]

5.2.17 Perforated area From figure 9, at lw/Dc = 0.72, θc = 93° Angle subtended by the edge of the plate = 180 – 93 = 87° Mean length, unperforated edge strip = (Dc – 50 × 10-3)π × 87/180 = 2.8895 m Area of unperforated edge strips = 50 × 10-3 × 2.8895 = 0.1445 m2 Mean length, calming zone = weir length + width of unperforated = 1.4068 + 0.05 = 1.4568 m Area of calming zones = 2(1.4568 × 0.05) = 0.1457 m2 Total area for perforations, Ap = Aa – 0.1445 – 0.1457 = 2.2272 m2 Ah/Ap = 0.2517/2.2272 = 0.1130 m2 From figure 10, lp/dh is 2.8 when Ah/Ap is 0.1130. The value of lp/dh is satisfactory as it is between 2.5 to 4. 14

Figure 9: Relation between angle subtended by chord, chord height and chord length [8]

Figure 10: Relation between hole area and pitch[8]

15

5.2.18 Number of holes đ??´đ?‘&#x;đ?‘’đ?‘Ž đ?‘œđ?‘“ đ?‘œđ?‘›đ?‘’ â„Žđ?‘œđ?‘™đ?‘’ = đ??´đ?‘&#x;đ?‘’đ?‘Ž đ?‘œđ?‘“ đ?‘œđ?‘›đ?‘’ â„Žđ?‘œđ?‘™đ?‘’ =

đ?œ‹ Ă— 0.0052 = 1.9625 Ă— 10−5 đ?‘š2 4

đ?‘ đ?‘˘đ?‘šđ?‘?đ?‘’đ?‘&#x; đ?‘œđ?‘“ â„Žđ?‘œđ?‘™đ?‘’đ?‘ = đ?‘ đ?‘˘đ?‘šđ?‘?đ?‘’đ?‘&#x; đ?‘œđ?‘“ â„Žđ?‘œđ?‘™đ?‘’đ?‘ =

đ?œ‹ Ă— đ?‘‘â„Ž 2 4

đ??´â„Ž đ??´đ?‘&#x;đ?‘’đ?‘Ž đ?‘œđ?‘“ đ?‘œđ?‘›đ?‘’ â„Žđ?‘œđ?‘™đ?‘’

0.2517 = 12827.4020 ≈ 12828 1.9625 Ă— 10−5

Therefore, number of holes are 12828.

5.2.19 Plate specification

Figure 11: Plate specification

Figure 11 above shows the final plate specification. Table 5 below contains some information on plate specification. Table 5: Plate data

Type Plate internal diameter Weir length Weir height Hole diameter Plate thickness Plate material Weir crest Turn-down Plate spacing Plate pressure drop Backup in downcomer Residence time Hole pitch No. of holes

Seive plate 1.95 m 1.41 m 50 mm 5 mm 5 mm Carbon steel 27.40 mm liquid 70% of max rate 0.9 m 230 mm liquid = 2.26 kPa 313.52 mm 7.65 s 12.5 mm 12828

16

5.2.20 Pipe size selection The internal diameter of the pipes are calculated using the equation below which is for carbon steel pipes. The values used in the equation are extracted from HYSYS and are shown in table 6 below. Table 6: Parameters for pipe size selection

3

Density (kg/m ) Mass flow rate (kg/hr)

Feed 23.16 15200.5465

To condenser 0.6007 3594

To reboiler 880.5 56230

đ?‘‘đ?‘– , đ?‘œđ?‘?đ?‘&#x;đ?‘–đ?‘šđ?‘˘đ?‘š đ?‘“đ?‘œđ?‘&#x; đ?‘?đ?‘Žđ?‘&#x;đ?‘?đ?‘œđ?‘› đ?‘ đ?‘Ąđ?‘’đ?‘’đ?‘™ đ?‘?đ?‘–đ?‘?đ?‘’ = 0.534đ??ş 0.43 đ?œŒâˆ’0.3 where, di = internal diameter (mm) G = mass flow rate (kg/s) đ?‘‘đ?‘– đ?‘œđ?‘“ đ?‘“đ?‘’đ?‘’đ?‘‘ = 0.534 Ă—

15200.54360.43 Ă— 23.16−0.3 = 0.3865 đ?‘š = 386.5 đ?‘šđ?‘š 3600

đ?‘‘đ?‘– đ?‘œđ?‘“ đ?‘?đ?‘œđ?‘›đ?‘‘đ?‘’đ?‘›đ?‘ đ?‘’đ?‘&#x; đ?‘–đ?‘›đ?‘™đ?‘’đ?‘Ą = 0.543 Ă—

35640.43 Ă— 0.6007−0.3 = 0.6218 đ?‘š = 621.8 đ?‘šđ?‘š 3600

562300.43 đ?‘‘đ?‘– đ?‘œđ?‘“ đ?‘&#x;đ?‘’đ?‘?đ?‘œđ?‘–đ?‘™đ?‘’đ?‘&#x; đ?‘–đ?‘›đ?‘™đ?‘’đ?‘Ą = 0.543 Ă— Ă— 880.5−0.3 = 0.2277 đ?‘š = 227.7 đ?‘šđ?‘š 3600

5.2.21 Summary of column design Table 7: Column data

Minimum number of trays (including reboiler) Minimum reflux ratio Actual reflux ratio Theoretical number of stages (including reboiler) Efficiency Actual number of stages (including reboiler) Actual feed point location Column diameter Column height

5 0.0181 1.4530 6 88.91 % 7 3 1.95 m 7.23 m

5.3 Operational design 5.3.1 Control system design Controlling the process variables can achieve a safe operation process and help control the quality of the products. Correction to the operating conditions of the vessel can be made referring to the detection of any disturbances [5]. Several regulator arrangement are available to be used in a distillation column, such as cascade control, ratio control, inferential control and selective control as shown in table 8[1). Control system in TEG regenerator column ensures the maximum separation of water and TEG as the bottom product (Lean TEG) is sent back to the absorber to dehydrate the hydrocarbons, thus ratio control would be the best option for T-402. Piping and Instrumentation Diagram (P&ID) for T-402 is shown is Appendix A, figure 13. 17

Table 8: Types of controllers [1]

Types of control Inferential Control Cascade control Selective control Ratio control

Description Measures process variables such as temperature, pressure and flow to conclude the difficult values to measure Sleeve controller’s set point is measured by using the output from main controller Selects from 2 input signals controlled by operator Even if the variables are varying the ratio between selected two or more variables do not change

5.3.2 Operating procedures 5.3.2.1 Commissioning procedure To ensure a safe commissioning of the column the following steps should be actioned as mistakes could have been made during the construction phase. 1. 2. 3. 4. 5. 6. 7.

Match up the TEG P&ID against the mechanical construction of the column The interior of the column should be clean Hazard and operability study must be conducted Ensure that the measuring devices shown in P&ID are tested prior to installation If any deviations are made during construction phase, ensure it meets the specifications All personal must be trained and aware of the risks involved for their and plant safety All personal must be in constant communication with each other to ensure that everyone is kept up to date

5.3.2.2 Pre-start-up procedure Safe operation of the vessel can be ensured by following the pre-start-up procedure listed below prior to actioning the start-up procedure. 1. 2. 3. 4. 5.

Any piping connected to the vessel and the vessel itself must be dry Flanges need to be tightened and the drains need to be shut Make sure that utilities such as heating and cooling mediums and electricity supply is ready Ensure that the vessel and its interiors are installed correctly Reboilers and condensers should be in good working condition, having passed the inspection

5.3.2.3 Start-up procedure After the previous two procedures have been completed, the following steps need to be followed to start-up TEG regeneration column: 1. 2. 3. 4.

Power supply to the vessel and its peripherals should be turned on The measuring devices show in P&ID should be relayed back to the control room Ensure that feed is connected to T-402 Level and temperature controllers should be controlled manually until the vessel reaches its normal operating condition 5. Control the inlet of the column, progressively increasing it till the maximum mass flow rate is reached 6. Operate the column at 75% of its designed pressure 7. At this point, start the condenser and reboiler. When the level controller attached to the vessel starts showing specified reading move onto the next step 18

8. Check the level inside the reflux drum 9. As the column and its peripherals reach operating condition, change all controller mode to auto 10. Analyse the flowrate and concentration of the distillate and bottoms 11. Monitor the vessel to look for signs of leakage 12. The operation of the vessel should be under surveillance to ensure it does not deviate from its specification

5.3.2.4 Shutdown procedure Typically there are planned and unplanned shutdowns. While planned shutdowns are occur for maintenance and inspection propose, unplanned shutdown are carried out during emergencies. 5.3.2.4.1 Planned shutdown To carry out a planned shutdown of a TEG regenerator column the following steps are actioned. 1. 2. 3. 4. 5. 6. 7. 8. 9.

Slowly decrease the mass flow rate to the inlet of the vessel Condenser temperature is raised Reflux ratio is increased to ensure the contents of the column remain in the column Shut the inlet valve to stop inlet flow when the vessel temperature is above the boiling point of heavy key Monitor differential pressure transmitter to ensure that the column pressure does not exceed design pressure Turn off the condenser Make sure that the vessel is steadily depressurised to flare Stop the power supply Drain the vessel of its contents

5.3.2.4.2 Unplanned shutdown The following procedures should be followed in case of unplanned shutdown for the TEG regenerator column: 1. 2. 3. 4. 5. 6. 7. 8.

Clove the inlet valve Detach the outlets to ensure the product does not get contaminated Turn the safety faucet on, if needed Stop the operation of reboiler and condenser Make sure that the vessel is steadily depressurised to flare Detect the cause of irregularity Ensure that the problem is fixed as well as any damages caused by it Once it is safe to operate the vessel again, follow the pre-start-up and start-up procedures

5.3.2.5 Normal operating procedure TEG regenerator column’s normal operating procedure are as follows 1. Routinely check the components and concentrations of the distillate and bottom 2. Ensure the operating parameters such as temperature, pressure and flow are met 3. Check equipments for faults such as leakage

5.3.2.6 Maintenance Inspect the column regularly and conduct a test every 2 years. The performance and efficiency of the vessel must be logged before and after inspection/maintenance. Following steps act as basic maintenance guide for the vessel. 19

1. All controllers and transmitters attached to the vessel and it’s peripherals have to be in good working condition 2. All the faucets and safety valves also need to be in good repair to handle emergencies

5.3.2 Safety study Safety of the plant and personals should always be a priority when designing and operating any vessel. To be more specific for TEG regenerator column, pressure and temperature deviations can hugely affect the safety of operators, plane and the vessel itself. During the process design of the vessel, it was ensured that the plate spacing met the design specification and flooding of the plate was considered to avoid uncontrolled loss of containment.

5.3.3.1 Hazard and operability (HAZOP) Hazard and Operability (HAZOP) analysis is a methodical way of detecting possible threats in an industrial plant and its process units. For TEG regeneration column, some probable deviations could occur in the column, reboiler, reflux drum and reflux pump. They are highlighted in table 9 with the corrective action for each problem to ensure wellbeing of the TEG regenerator column throughout its operation. Moreover, HAZOP is completed to recognize the likely exposures related to the column operation in order to comprehend the origins and significances of abnormalities through the operation processes as shown in table 10. Table 9: Column deviations

Problem Flooding in the vessel

Possible cause Increased inlet mass flow rate

Increased reflux ratio

Deviation in inlet temperature Reboiler level above normal

Contains of the reflux drum below normal level Reboiler level above normal

Incorrect differential pressure transmitter (DPT) interpretation Level controller or transmitter malfunction Increased inlet mass flow rate Reboiler heating medium not in balance Malfunction of medium inlet valve

Feed flow loss

Reboiler pipes spoiled on the outside or inside Inlet flow controller malfunction

Corrective Action • Decrease inlet flow • Ensure inlet and outlet flows are in equilibrium • Decrease inlet flow • Ensure analyser controller and transmitter connected to the distillate are functioning • Maintain temperature regulation in the vessel • Regulate inlet temperature • Decrease inlet flow • Ensure the reboiler flow is accurate • Regulate reboiler level • Confirm DPT analysis • Regulate vessel temperature • Physically regulate the level • Replace or repair the level controller/transmitter • Decrease inlet flow • Decrease inlet flow • Regulate the flow of medium into the reboiler • Open the bypass to maintain the medium level in reboiler • Repair or replace the faucet • Replace or clean the pipes • Replace or repair flow controller 20

Table 10: HAZOP study

Parameter Pressure

Flow

Level

Temperature

Deviation Low

Cause • Vapour content of the vessel is low • Pressure controller and/or transmitter malfunction

Result • Vessel flooding • Loss of production

High

• Distillate tubes clogged • Condenser malfunction • Reboiler energy increased

• Vessel is flooded • Vessel pressure increased substantially

Low

• Piping leak • Pipe clogged • Contains of the reflux drum below normal level

• Loss of production

High

• Inlet control valve failure

No Flow

• No inlet flow • Pipe clogged

• Loss of production • Loss of production

Low

• Reboiler energy increased

High

• Bottom product pipe clogged • Condenser energy intake increased • Plate perforation clogged • Reboiler malfunction • Condenser energy intake increased

Low

High

• Condenser malfunction • Reboiler energy intake increased

• Damage to reboiler due to high energy intake • Vessel level low • Loss of production • Flooded vessel

• Weeping inside the column • Loss of production • Loss of production

Required action • Replace pressure controller and/or transmitter • Ensure DPT is functional • Inspect the operation on regular basis • Install pressure alarms • Install pressure relief valve • Inspect the operation on regular basis • Install low flow alarm at the inlet • Inspect the operation on regular basis • Install high flow alarm on the inlet line • Inspect the operation on regular basis • Install low level indicator and alarm • Inspect the operation on regular basis • Install high level indicator and alarm • Inspect the operation on regular basis • Install temperature indicator and alarm • Inspect the operation on regular basis • Install temperature indicator and alarm. • Inspect the operation on regular basis

21

5.4 Mechanical design The mechanical design of the pressure vessel is created based on the parameters of the vessel attained from process design. The main parameters used for the design are listed in table 11. The vessel is designed following the Coiler and Pressure Vessel (BPV) Code set by American Society of Mechanical Engineers (ASME). Table 11: Column data required for mechanical design

Length of cylindrical section Internal diameter No. of stages (excluding reboiler) Thickness of insulation

7.23 m 1.95 m 6 40 mm

5.4.1 Materials of construction TEG regenerator column will be constructed using carbon steel based on the operating conditions and specifications. The suitability of carbon steel to the process environment is also considered when selecting it. As the acid is removed before the dehydration stage, there is no need to select high corrosion resistant material such as stainless steel as it is more expensive. The method of fabrication is wielding and the purposed material is suitable for it as well. The standards used for material of construction in the calculations are from the guidelines provided by ASME BPV code.

5.4.2 Design pressure In order to design the vessel to withstand maximum pressure, the column needs to designed at 10% above the normal working pressure as suggested by American Petroleum Institute (API) Recommended Practice (RP) 520[8]. Hence, column design pressure = 1.1 × Pressurefeed = 1.1 × 1.1 = 1.21 bar = 17.55 psia = 0.121 N/mm2

5.4.3 Design temperature The minimum temperature in the column is the distillate temperature at 102°C which is equal to 215.6°F.

5.4.4 Corrosion allowance As the material being used is carbon-steel, 2 mm thickness needs to be added to the thickness of the material to account for corrosion as suggested by ASME BPV Code Sec. VIII D.1[8] 5.4.5 Minimum practical wall thickness Table 12: Wall thickness dependent on vessel diameter [8]

Vessel diameter (m) 1 1 to 2 2 to 2.5 2.5 to 3 3.0 to 3.5

Minimum thickness (mm) 5 7 9 10 12

Vessel diameter determines the minimum wall thickness of the vessel. The minimum thickness shown in table 8 takes into account the 2mm corrosion allowance. As the vessel diameter is 1.95 m, the minimum wall thickness has to be 7 mm as shown in table 12 under ASME BPV Code Sec. VIII D.1.

22

5.4.6 Actual wall thickness 5.4.6.1 Cylindrical section đ?‘Ą=

đ?‘ƒđ?‘– đ??ˇđ?‘– + đ?‘?đ?‘œđ?‘&#x;đ?‘œđ?‘ đ?‘ đ?‘–đ?‘œđ?‘› đ?‘Žđ?‘™đ?‘™đ?‘œđ?‘¤đ?‘Žđ?‘›đ?‘?đ?‘’ 2đ?‘†đ??¸ − 1.2đ?‘ƒđ?‘–

Where, t = thickness of plate or shell (mm) S = maximum allowable stress of the given material at design temperature (N/mm2) Pi = internal pressure (N/mm2) Di = internal diameter (mm) E = joint efficiency, wielded joint For carbon steel at 102°C, S = 12900 psi = 88.9 N/mm2[8] E = 1 for full double-welded butt joint or equivalent[8]

đ?‘Ą=

0.121 Ă— 1.95 Ă— 103 + 2 = 3.3304 đ?‘šđ?‘š 2 Ă— 88.9 Ă— 1 − 1.2 Ă— 0.121

The calculated wall thickness is lower than the suggested minimum wall thickness for a vessel with diameter of 1.95m. Hence, the wall thickness of 7 mm will be used in the design of the vessel.

5.4.6.2 Torispherical head For vessels with pressure rating less than 10 bar, the most economical selection of domed head is torispherical head[8]. 0.885đ?‘ƒđ?‘– đ?‘…đ?‘? + đ?‘?đ?‘œđ?‘&#x;đ?‘œđ?‘ đ?‘ đ?‘–đ?‘œđ?‘› đ?‘Žđ?‘™đ?‘™đ?‘œđ?‘¤đ?‘Žđ?‘›đ?‘?đ?‘’ đ?‘†đ??¸ − 0.1đ?‘ƒđ?‘–

đ?‘Ą= Where, Rc = crown radius (mm) Rc = Di = 1.95 m = 1.95 Ă— 103 mm

0.885 Ă— 0.121 Ă— 1.95 Ă— 103 đ?‘Ą= + 2 = 2.1412 đ?‘šđ?‘š 88.9 Ă— 1 − 0.1 Ă— 0.121 The head thickness is less than 7 mm. so, take the head thickness to be 7 mm as well. To avoid buckling, in torispherical head the ratio of the knuckle and crown radius has to be 0.6 or above[8]. So, take knuckle radius to be 0.06 of crown radius. Thus, knuckle radius is 0.117 m.

5.4.7 Dead weight of vessel The weight of vessel shell, plates, insulation, platform and access ladder are considered in determining the dead weight of the TEG regenerator column.

5.4.7.1 Shell The following equation is used to determine an approximate weight of shell of a steel vessel that has domed end and the thickness of the wall is constant for the whole length of the vessel7]. đ?‘Šđ?‘Ł = 240đ??śđ?‘¤ đ??ˇđ?‘š (đ??ťđ?‘Ł + 0.8đ??ˇđ?‘š )đ?‘Ą where, Wv = total weight of the shell minus the internal contents (N) Cw= factor to account for the weight of nozzles, manways, internal supports etc for distillation column Cw = 1.15 23

Dm = mean diameter of the vessel (m) = Di + t Ă— 10-3 Hv = height of the cylindrical section (m) đ??ˇđ?‘š = 1.95 + 7 Ă— 10−3 = 1.9609 đ?‘šđ?‘š đ?‘Šđ?‘Ł = 240 Ă— 1.15 Ă— 1.9609(7.23 + 0.8 Ă— 1.9609) Ă— 7.23 = 34428.5908 đ?‘ = 34.4286 đ?‘˜đ?‘

5.4.7.2 Plates Weight of steel plates including liquid on it per area is 1.2 kN/m2 [8] đ?‘ƒđ?‘™đ?‘Žđ?‘Ąđ?‘’ đ?‘Žđ?‘&#x;đ?‘’đ?‘Ž = đ?‘ƒđ?‘™đ?‘Žđ?‘Ąđ?‘’ đ?‘Žđ?‘&#x;đ?‘’đ?‘Ž =

đ?œ‹ Ă— đ??ˇđ?‘– 2 4

đ?œ‹ Ă— 1.952 = 2.9865 đ?‘š2 4

đ?‘¤đ?‘’đ?‘–đ?‘”â„Žđ?‘Ą đ?‘œđ?‘“ đ?‘Ž đ?‘?đ?‘™đ?‘Žđ?‘Ąđ?‘’ đ?‘–đ?‘›đ?‘?đ?‘™đ?‘˘đ?‘‘đ?‘–đ?‘›đ?‘” đ?‘™đ?‘–đ?‘žđ?‘˘đ?‘–đ?‘‘ đ?‘œđ?‘› đ?‘–đ?‘Ą ≈ 1.2 Ă— 2.9865 = 3.5838 đ?‘˜đ?‘

5.4.7.3 Insulation Mineral wool density = 130 kg/m3[8] Mineral wool thickness = 40 mm đ??´đ?‘?đ?‘?đ?‘&#x;đ?‘œđ?‘Ľđ?‘–đ?‘šđ?‘Žđ?‘Ąđ?‘’ đ?‘Łđ?‘œđ?‘™đ?‘˘đ?‘šđ?‘’ đ?‘œđ?‘“ đ?‘–đ?‘›đ?‘ đ?‘˘đ?‘™đ?‘Žđ?‘Ąđ?‘œđ?‘&#x; = đ?œ‹ Ă— 1.95 Ă— 7.23 Ă— 10−3 Ă— 40 = 1.7752 đ?‘š3 đ?‘Šđ?‘’đ?‘–đ?‘”â„Žđ?‘Ą = 3.3285 Ă— 130 Ă— 9.81 = 2263.9156 đ?‘ = 2.2639 đ?‘˜đ?‘ Approximately double of the calculated weight is need to allow for fittings and other components. Hence, the weight of insulation is 8 kN.

5.4.7.4 Platform Weight of steel platforms in vertical columns per area is 1.7 kN/m2[8]. đ?‘Šđ?‘’đ?‘–đ?‘”â„Žđ?‘Ą đ?‘œđ?‘“ đ?‘?đ?‘™đ?‘Žđ?‘Ąđ?‘“đ?‘œđ?‘&#x;đ?‘š = 2.9565 Ă— 1.7 = 5.0770 đ?‘˜đ?‘

5.4.7.5 Access ladder Caged ladder will be installed on the side of the vessel to access it. The weight of steel caged ladder is 360 N/m length. ����ℎ� =

360 Ă— 7.23 = 2.6028 đ?‘˜đ?‘ 1000

∴ Total dead weight = 34.4285 + 21.5026 + 8 + 5.0770 + 2.6028 = 71.6110 kN

5.4.8 Wind loading The vessel is designed to withstand the wind speed of 160 km/hr. At this wind speed, dynamic wind pressure is 1280 N/m2[8]. Mean diameter, including insulation = 1.95 + 2(7 Ă— 40 ) Ă— 10−3 = 2.0479 m đ?‘Šđ?‘–đ?‘›đ?‘‘ đ??żđ?‘œđ?‘Žđ?‘‘đ?‘–đ?‘›đ?‘” (đ??šđ?‘¤ ) = 1280 Ă— 2.0479 = 2621.2969 đ?‘ â „đ?‘š = 2.6213 đ?‘˜đ?‘ /đ?‘š

5.4.9 Bending moment (Mx) �� =

đ?‘Šđ?‘–đ?‘›đ?‘‘ đ??żđ?‘œđ?‘Žđ?‘‘đ?‘–đ?‘›đ?‘” Ă— đ??ťđ?‘’đ?‘–đ?‘”â„Žđ?‘Ą đ?‘œđ?‘“ đ?‘Ąâ„Žđ?‘’ đ?‘Łđ?‘’đ?‘ đ?‘ đ?‘’đ?‘™ 2 2

�� =

2621.2969 Ă— 7.232 = 68511.3957 đ?‘ đ?‘š 2

24

5.4.10 Stress analysis 5.4.10.1 Pressure stresses đ?œŽâ„Ž =

đ?‘ƒđ??ˇđ?‘– 2đ?‘Ą

đ?œŽđ??ż =

đ?‘ƒđ??ˇđ?‘– 4đ?‘Ą

where, Ďƒh = Circumferential stress (N/mm2) ĎƒL = Longitudinal stress (N/mm2) đ?œŽâ„Ž = đ?œŽđ??ż =

0.121 Ă— 1.95 Ă— 103 = 16.8872 đ?‘ /đ?‘šđ?‘š2 2 Ă—7 0.121 Ă— 1.95 Ă— 103 = 8.4436 đ?‘ /đ?‘šđ?‘š2 4Ă—7

Dead weight stress: đ?œŽđ?‘¤ =

đ?‘Šđ?‘Ł đ?œ‹(đ??ˇđ?‘– + đ?‘Ą)đ?‘Ą

where, Ďƒw = stress due to weight loss of vessel (N/mm2)

đ?œŽđ?‘¤ =

71.6110 Ă— 103 = 1.6607 đ?‘ â „đ?‘šđ?‘š2 (đ?‘?đ?‘œđ?‘šđ?‘?đ?‘&#x;đ?‘’đ?‘ đ?‘ đ?‘–đ?‘Łđ?‘’) đ?œ‹(1.95 Ă— 103 + 7)7

5.4.10.2 Bending stresses đ??źđ?‘Ł =

đ?œ‹ (đ??ˇ 4 − đ??ˇđ?‘– 4 ) 64 đ?‘œ

đ??ˇđ?‘œ = đ??ˇđ?‘– + 2đ?‘Ą đ?œŽđ?‘? = Âą

đ?‘€ đ??ˇđ?‘– ( + đ?‘Ą) đ??źđ?‘Ł 2

where, Iv = Second moment of area of vessel (mm4) Do = outside diameter (mm) Ďƒb = bending stress (N/mm2) đ??ˇđ?‘œ = 1.95 Ă— 103 + 2 Ă— 7 = 1967.8882 đ?‘šđ?‘š đ??źđ?‘Ł = đ?œŽđ?‘? = Âą

đ?œ‹ (1967.8882 − 1.95 Ă— 103 ) = 2.07 Ă— 1010 đ?‘šđ?‘š4 64

68511.3957 Ă— 103 1.95 Ă— 103 ( + 7) = Âą3.2756 đ?‘ /đ?‘šđ?‘š2 2.07 Ă— 1010 2

5.4.10.3 The resultant longitudinal stress (Ďƒz) đ?œŽđ?‘§ = đ?œŽđ??ż + đ?œŽđ?‘¤ Âą đ?œŽđ?‘? Ďƒw is compressive, hence it is negative đ?œŽđ?‘§ (đ?‘˘đ?‘?đ?‘¤đ?‘–đ?‘›đ?‘‘) = 8.4436 − 1.6607 + 3.2756 = 10.0585 đ?‘ /đ?‘šđ?‘š2 25

đ?œŽđ?‘§ (đ?‘‘đ?‘œđ?‘¤đ?‘›đ?‘¤đ?‘–đ?‘›đ?‘‘) = 8.4436 − 1.6607 − 3.2756 = 3.5073 đ?‘ /đ?‘šđ?‘š2 Since there is no stress due to torque, the main stress to account for will be Ďƒz and Ďƒh. The resultant longitudinal stress for upwind and downwind are both positive which suggests that they are both in tension and there is no compressive stress. Hence, the highest value of longitudinal stress will be accounted for when designing the vessel in order for it to withstand maximum longitudinal stress at tension (upwind) as, shown in figure 12.

Figure 12: Stress analysis

The difference between the principal stresses = 16.8872 – 10.0585 = 6.8286 N/mm2. This is below the value of maximum allowable design stress of 88.9 N/mm2.

5.4.10.4 Buckling đ?œŽđ?‘? = 2 Ă— 104 (

đ?‘Ą ) đ??ˇđ?‘œ

where, Ďƒc = critical buckling stress (N/mm2) 7 ) = 7.1142 đ?‘ /đ?‘šđ?‘š^2 đ?œŽđ?‘? = 2 Ă— 104 ( 1967.8882 When the vessel is not under pressure, it will be subjected to maximum compressive stress (Ďƒw + Ďƒb) which is 4.9362 N/mm2. The design of the vessel is acceptable as the maximum compressive stress is below the critical buckling stress.

5.4.10 Skirt support The skirt experiences maximum dead weight when the column is full of water. For carbon steel, a straight cylindrical skirt (θs) = 90° Young’s modulus at ambient air = 200,000 N/mm2 S = 88.9 N/mm2 Ď water = 1000 kg/m3 Assume skirt thickness to be 1 m đ??´đ?‘?đ?‘?đ?‘&#x;đ?‘œđ?‘Ľđ?‘–đ?‘šđ?‘Žđ?‘Ąđ?‘’ đ?‘¤đ?‘’đ?‘–đ?‘”â„Žđ?‘Ą = (

đ?œ‹ Ă— 1.95 Ă— 7.23) 1000 Ă— 9.81 = 108.8421 đ?‘˜đ?‘ 4 26

đ?‘‡đ?‘œđ?‘Ąđ?‘Žđ?‘™ đ?‘¤đ?‘’đ?‘–đ?‘”â„Žđ?‘Ą = 108.8421 + 71.6110 = 180.4531 đ?‘˜đ?‘ đ??ľđ?‘’đ?‘›đ?‘‘đ?‘–đ?‘›đ?‘” đ?‘šđ?‘œđ?‘šđ?‘’đ?‘›đ?‘Ą đ?‘Žđ?‘Ą đ?‘Ąâ„Žđ?‘’ đ?‘?đ?‘Žđ?‘ đ?‘’ đ?‘œđ?‘“ đ?‘Ąâ„Žđ?‘’ đ?‘ đ?‘˜đ?‘–đ?‘&#x;đ?‘Ą = Take the thickness of the skirt to be 3mm đ?œŽđ?‘?đ?‘ =

2.6213 Ă— (7.23 + 1)2 = 88.7740 đ?‘˜đ?‘ đ?‘š 2

4đ?‘€đ?‘ đ?œ‹(đ??ˇđ?‘ + đ?‘Ąđ?‘ đ?‘˜ ) đ?‘Ąđ?‘ đ?‘˜ đ??ˇđ?‘

where, Ďƒbs = bending stress in the skirt (N/mm2) Ms = maximum bending moment (Nmm) Ds = Inside diameter of the skirt (mm) tsk = skirt thickness (mm) 4 Ă— 88.7740 Ă— 103 Ă— 103 đ?œŽđ?‘?đ?‘ = = 9.8539 đ?‘ /đ?‘šđ?‘š2 đ?œ‹(1.95 Ă— 103 + 3)3 Ă— 1.95 Ă— 103 đ?œŽđ?‘¤đ?‘ =

đ?‘Šđ?‘Ł đ?œ‹(đ??ˇđ?‘ + đ?‘Ąđ?‘ đ?‘˜ )đ?‘Ąđ?‘ đ?‘˜

where, Ďƒws = the dead weight stress in the skirt (N/mm2) Wv = total weight of the vessel and contents (N) đ?œŽđ?‘¤đ?‘ (đ?‘Ąđ?‘’đ?‘ đ?‘Ą) =

180.4531 Ă— 103 = 9.7842 đ?‘ /đ?‘šđ?‘š2 đ?œ‹(1.95 Ă— 103 + 3)3

đ?œŽđ?‘¤đ?‘ (đ?‘œđ?‘?đ?‘’đ?‘&#x;đ?‘Žđ?‘Ąđ?‘–đ?‘›đ?‘”) =

71.6110 Ă— 103 = 3.8827 đ?‘ /đ?‘šđ?‘š2 đ?œ‹(1.95 Ă— 103 + 3)3

Ďƒws (test) refers to dead weight stress in the skirt during hydraulic testing i.e. when the column is full of water. đ?‘€đ?‘Žđ?‘Ľđ?‘–đ?‘šđ?‘˘đ?‘š đ?œŽđ?‘ (đ?‘Ąđ?‘’đ?‘›đ?‘ đ?‘–đ?‘™đ?‘’) = đ?œŽđ?‘?đ?‘ − đ?œŽđ?‘¤đ?‘ (đ?‘œđ?‘?đ?‘’đ?‘&#x;đ?‘Žđ?‘Ąđ?‘–đ?‘›đ?‘”) đ?‘€đ?‘Žđ?‘Ľđ?‘–đ?‘šđ?‘˘đ?‘š đ?œŽđ?‘ (đ?‘?đ?‘œđ?‘šđ?‘?đ?‘&#x;đ?‘’đ?‘ đ?‘ đ?‘–đ?‘Łđ?‘’) = đ?œŽđ?‘?đ?‘ + đ?œŽđ?‘¤đ?‘ (đ?‘Ąđ?‘’đ?‘ đ?‘Ą) where, Ďƒs = stress in skirt support (N/mm2) đ?‘€đ?‘Žđ?‘Ľđ?‘–đ?‘šđ?‘˘đ?‘š đ?œŽđ?‘ (đ?‘Ąđ?‘’đ?‘›đ?‘ đ?‘–đ?‘™đ?‘’) = 9.8539 − 3.8827 = 5.9711 đ?‘ /đ?‘šđ?‘š2 đ?‘€đ?‘Žđ?‘Ľđ?‘–đ?‘šđ?‘˘đ?‘š đ?œŽđ?‘ (đ?‘?đ?‘œđ?‘šđ?‘?đ?‘&#x;đ?‘’đ?‘ đ?‘ đ?‘–đ?‘Łđ?‘’) = 9.8539 + 9.7842 = 19.6382 đ?‘ /đ?‘šđ?‘š2 If the stress in skirt support meets the following criteria, then the skirt thickness is acceptable. đ?œŽđ?‘ (đ?‘Ąđ?‘’đ?‘›đ?‘ đ?‘–đ?‘™đ?‘’) ≤ đ?‘†đ?‘ đ??¸đ?‘ đ?‘–đ?‘›đ?œƒđ?‘ đ?‘Ąđ?‘ đ?‘˜ đ?œŽđ?‘ (đ?‘?đ?‘œđ?‘šđ?‘?đ?‘&#x;đ?‘’đ?‘ đ?‘ đ?‘–đ?‘Łđ?‘’) < 0.125 đ??¸đ?‘Œ ( ) đ?‘ đ?‘–đ?‘›đ?œƒđ?‘ đ??ˇđ?‘ Ss = S = 88.9 N/mm2 E = 0.85 for spot double-welded butt joint EY = Young’s modulus = 200,000 N/mm2 đ?œŽđ?‘ (đ?‘Ąđ?‘’đ?‘›đ?‘ đ?‘–đ?‘™đ?‘’) ≤ 88.9 Ă— 0.85 Ă— sin90 5.9711 ≤ 67.5549

27

đ?œŽđ?‘ (đ?‘?đ?‘œđ?‘šđ?‘?đ?‘&#x;đ?‘’đ?‘ đ?‘ đ?‘–đ?‘Łđ?‘’) < 0.125 Ă— 200000 Ă—

3 Ă— đ?‘ đ?‘–đ?‘›90 1.95 Ă— 103

19.6382 < 34.3161

Since the criteria is met by both tensile and compressive stress in skirt support the assumptions of skirt height being 1m and thickness being 3 mm are correct. This is the lowest height and thickness of skirt possible for given condition. If it is any higher, the criteria would not be met. Now, 2 mm corrosion allowance is added to skirt thickness giving the new thickness of 5mm.

5.4.11 Pressure relief 5.4.11.1 Condenser The following equation is used to calculate the effective area of the nozzle orifice required for pressure relief in a condenser as the product is mainly water vapour [9]. đ??´=

đ?‘Š 51.5đ?‘ƒ1 đ??žđ??žđ?‘› đ??žđ?‘ â„Ž

Where, A = valve orifice area (in2) W = Flow (lb/hr) P1 = design pressure (psia) K = Coefficient of discharge obtainable from valve manufacturer Kn = Correction factor for saturated stream at set pressures > 1500 psia Ksh = Correction factor due to the degree of superheat in steam So, W = 1671.9117 kg/hr = 3685.9299 lb/hr P1 = 17.55 psia K = 0.975 for many nozzle-type valves [9] Kn + Ksh = 1 as P1 < 1500 psia đ??´=

3685.9229 = 4.1828 đ?‘–đ?‘›2 51.5 Ă— 17.55 Ă— 0.975 Ă— 1 Ă— 1

Table 13: Standard nozzle orifice data [9]

Nozzle Orifice Areas Size Designation Orifice Area, in2 D 0.110 E 0.196 F 0.307 G 0.503 H 0.785 J 1.280 K 1.840 L 2.850 M 3.600 N 4.340 P 6.380 Q 11.050 R 16.000 28

T

26.000

From Table 13, smallest standard size designation that has an area equal to or greater than 4.1828 in2 is “N�. Hence, “N� orifice will be installed in condenser.

5.4.11.2 Reboiler The following equation is used to calculate the effective area of the nozzle orifice required for pressure relief in a reboiler as the product is in liquid form [9]. đ??´=

đ?‘„√đ??ş

27.2đ??žđ?‘? đ??žđ?‘¤ đ??žđ?‘Ł √đ?›Ľđ?‘ƒ For design pressure margin of less than 25%: đ??žđ?‘? = −0.0014(%đ?‘šđ?‘Žđ?‘&#x;đ?‘”đ?‘–đ?‘›)2 + 0.073(%đ?‘šđ?‘Žđ?‘&#x;đ?‘”đ?‘–đ?‘›) + 0.016 where, Q = Flow (gpm) G = Specific gravity of water at reboiler temperature Kp = Correction factor for relieving capacity vs. lift for relief valves in liquid service Kw = Correction factor due to back pressure for use with balanced bellows valves Kv = Correction factor for viscosity ΔP = Differential pressure (psig), (design pressure – back pressure) So, Q = 13528.6348 kg/h = 59.56 gpm G = 0.72 at 288áľ’C[3] Kv = 1 Kw = 1 as there is no back pressure ΔP = 17.55 psig đ??žđ?‘? = −0.0014 Ă— 102 + 0.073 Ă— 10 + 0.016 = 0.606 đ??´=

59.56√0.72 27.2 Ă— 0.606 Ă— 1 Ă— 1 Ă— √17.55

= 0.7319 đ?‘–đ?‘›2

From Table 13, smallest standard size designation that has an area equal to or greater than 0.7319 in2 is “H�. Hence, “H� orifice will be installed in reboiler.

5.4.11 Reinforcement Stress is higher than normal around the openings of the vessel structure. Hence around such area the wall thickness must be increased in order to reimburse for that. The equal area method is one of the method can be used to calculated the amount of reinforcement required [8]. The diameter of reinforcement is calculated by: đ?‘‘đ?‘&#x; = 1.5 Ă— đ??ˇđ?‘– where, dr = reinforcement diameter đ?‘‘đ?‘&#x; = 1.5 Ă— 1.95 = 2.9308 đ?‘š

29

5.4.12 Mechanical drawing Figure 14 in Appendix B illustrates the mechanical drawing of the TEG regenerator column based on process and mechanical design. The scale size of the drawing is 1:50 metric.

5.5 Summary and review The report contains a design of TEG regenerator column involving process, operational and mechanical design. Process design began with methodology and moved on to calculations to find the sizing of major components of the column. Second part deals with control system design, operation procedures and safety study. In the mechanical design the material of construction is selected and also contains vessel design, dimensions of supports and mechanical drawing.

5.5.1 Equipment specifications Based on the chemical and mechanical design calculation, a data sheet for TEG regeneration column is in Appendix C, table 14.

5.5.2 Critical review The TEG regenerator column was designed as a multicomponent distillation column as there are three components that are available in all three streams of the column, the feed, distillate and bottoms. The values such as feed flowrate, composition and molar flow rate were extracted from Aspen HYSYS as the simulation was run using this software. It was not possible to calculate the value of density, which is required for calculations, due to the unknown number of variables. Hence I was limited to extract the value from HYSYS, which has a degree of uncertainty. The design methodology is based on text book named Chemical Engineering Design by Ray Sinnott and Gavin Towler. The process in the book is well structured and easy enough to follow which gives values with low uncertainty of error. It was found that, while there are lots of variables in play during process design of the column. It was also discovered that mechanical design mainly depends on the column height, diameter, number of stages and operating pressure and temperature. Optimised value of plate spacing and flooding are shown in the calculations in order to get an optimised column height and diameter. Number of holes and skirt height and thickness are also optimised. Furthermore economic study is required to understand if the optimised column design is economically viable.

5.6 Nomenclature Appendix D has a nomenclature for the symbols used for calculations throughout this report with dimensions in mass, length and time (MLTϴ).

30

References 1. Chopey (1996). Instrumentation and process control / edited by Nicholas P. Chopey and the staff of Chemical Engineering. New York, NY : McGraw-Hill 2. Dow Ethylene Glycols. 2015. Dow Chemical Company. Accessed November 1, http://www.dow.com/ethyleneglycol/about/properties.htm 3. Engineeringtoolbox.com,. 2015. 'Water - Temperature And Specific Gravity'. http://www.engineeringtoolbox.com/water-temperature-specific-gravityd_1179.html. 4. Kidnay, A. J. 2011. Fundamentals of Natural Gas Processing. Edited by William Rutledge Parrish and Daniel G. McCartney. 2nd ed. / Arthur J. Kidnay, William R. Parrish, Daniel G. McCartney.. ed. Boca Raton, FL: Boca Raton, FL : CRC Press. 5. Kister, H. Z. (1992). Distillation design / Henry Z. Kister. New York, NY : McGraw-Hill. 6. Kletz, T. A. (1999). Hazop and Hazan : identifying and assessing process industry hazards/ Trevor Kletz. Rugby, Warwickshire, Rugby, Warwickshire : Institution of Chemical Engineers. 7. Mycheme, “Types of Vessel Head,� (blog), March 4, 2013, http://www.mycheme.com/typesof-vessel-head/ 8. Towler, Gavin, and Ray Sinnott. 2013. Chemical engineering design : principles, practice, and economics of plant and process design. Oxford ; Waltham, MA : ButterworthHeinemann. 9. Whitesides, Randall W. 2012. "Selection and Sizing of Pressure Relief Valves." PDH Online Course M112. http://www.pdhonline.org/courses/m112/Selection%20and%20Sizing%20of%20Pre ssure%20Relief%20Valves.pdf

31

Appendix A PC

Coolant E-403 PT LT

Reflux Drum LC FC

FT

FC

AT

X

FC

Feed

FT

Water Vapour

AC

FT

T-402 DPT

TT

TC

FC

LT

FT

Steam

E-404 LC

Lean TEG FT FC

Figure 13: Piping & Instrumentation Diagram (P&ID) for TEG regeneration column

Index: T-402 -> TEG regeneration column E-403 -> Condenser E-404 -> Reboiler AC: Analyser controller FC: Flow controller LC: Level controller PC: Pressure controller TC: Temperature controller DPT: Differential pressure transmitter

AT: Analyser transmitter FT: Flow transmitter LT: Level transmitter PT: Pressure transmitter TT: Temperature transmitter Pressure relief valve 32

Appendix B I.D: 1.95 m O.D: 1.97 m

(To Condenser) 0.62 m

(From Reflux) 0.23 m

1

2 (Feed) 0.39 m

3

7.23 m

4 0.9 m 5

40 mm

50 mm (Boilup) 0.23 m

6

Skirt height 1m

(To Reboiler) 0.23 m

Figure 14: Mechanical Drawing of TEG regenerator column

Sizing scale 1:50 metric 33

Appendix C Table 14: Data sheet

Project Name: BIG GAS PROJECT Project Number: Chevron 1 Sheet No. 1 REV By Date APVD By Triethylene glycol (TEG) Regenerator Column Owner’s Name: Plant Location: Longford, VIC, Australia Case Description: Sales Gas & LPG Processing Plant Equipment Number: T-402 Description: Vertical Column Process Data No. of stages Feed stage Reflux ratio Efficiency Total vapour flow (kg/h) Total liquid flow (kg/h) Construction & Materials No. of stages (excluding reboiler) Shell material Shell internal diameter (m) Shell outside diameter with corrosion allowance (m) Shell height (m) Shell thickness (mm) Design temperature (⁰C) Design pressure (bar) Skirt height (m) Skirt thickness with corrosion allowance (mm) Tray type No. of trays No. of holes per tray Tray diameter (m) Hole diameter (mm) Hole pitch (mm) Area of one hole (m2) Total hole area (m2) Tray thickness (mm) Tray spacing (m) Weir height (mm) Weir length (mm) Downcomer clearance (mm) Pressure drop per plate (bar)

Date

6 3 1.453 88.91 450.7 1.475 × 104 6 Carbon steel 1.95 1.97 7.23 7 101 1.21 1 3 Sieve 6 12828 1.95 5 12.5 1.9625 × 10-5 0.2517 5 0.9 50 1.41 313.52 0.0226

34

Appendix D µa A Aa Aap Ac Ad Ah Am An Ap b B CO Cw d D Dc dh Di Dm Do dr Ds E Eo EY FLV Fw G G hap hb hbc hd hd hdc HK how hr ht Hv hw Iv K K K1 K2

Molar average liquid viscosity Valve orifice area Active area Clearance area under apron Column area Downcomer cross-sectional area Total hole area Either downcomer cross-sectional area or clearance area under apron, whichever is smaller Net area Perforated area Bottom Moles of bottom product per unit time Orifice coefficient Weight factor Distillate Moles of distillate per unit time Column diameter Hole diameter Internal diameter Mean diameter of the vessel Outside diameter Reinforce diameter Skirt internal diameter Joint efficiency, wielded joint Overall column efficiency Young’s Modulus Liquid-vapour flow factor Wind pressure times mean diameter Mass flow rate Specific gravity of water at reboiler temperature Apron clearance Height of liquid backed up in downcomer Clear liquid back-up Dry-plate pressure drop, head of liquid Downcomer backup Head loss in downcomer Heavy key Height of liquid crest over downcomer weir Plate residual pressure drop, head of liquid Total plate pressure drop, head of liquid Height of cylindrical section Weir height Second moment of area of vessel Constant Coefficient of discharge obtainable from valve manufacturer Constant Constant

Dimension in MLTϴ LT-1 L2 L2 L2 L2 L2 L2 L2 L2 L2 MT-1 MT-1 L L L L L L L ML-1T-2 MT-2 MT-1 L L L L L L L L L L L L4 -

35

Kn Kp Ksh Kv Kw L L’ LK lp Lw lw Ms MW Mx N Nactual Nmin Nr Ns P1 Pi q Q R Rc Rmin S Ss t tr tsk uf uh un un V Vw W Wv xi α αa αHK αLK ΔP θ θs ρ

Correction factor for saturated stream at set pressures > 1500 psia Correction factor for relieving capacity vs. lift for relief valves in liquid service Correction factor due to the degree of superheat in steam Correction factor for viscosity Correction factor due to back pressure for use with balanced bellows valves Liquid flow above feed Liquid flow below feed Light key Pitch of holes (distance between centres) Liquid mass flow rate Weir length Maximum bending moment Molecular weight Bending moment at point x from free end of the column Theoretical number of stages Actual number of stages Minimum number of stages including reboiler Number of stages above feed Number of stages below feed Design pressure Internal pressure Heat to vaporize one mol feed divided by molar latent heat Flow Reflux ratio Crown radius Minimum reflux ratio maximum allowable stress Maximum allowable stress for skirt material minimum wall thickness required to resist internal pressure Residence time Skirt thickness Vapour velocity at flooding point based on net area Vapour velocity through holes Vapour velocity based on net cross-sectional area Vapour velocity based on net area Vapour flow rate mols per unit time Vapour mass flow rate Flow Total weight of the shell minus the internal contents Concentration or mole fraction of a component Relative volatility Average relative volatility of light key Relative volatility of heavy key Relative volatility of light key Differential pressure Root of equation Angle Density

MT-1 MT-1 L MT-1 L ML2T-2 MT-1 ML2T-2 ML-1T-2 ML-1T-2 MT-1 L ML-1T-2 ML-1T-2 L T L LT-1 LT-1 LT-1 LT-1 MT-1 MT-1 MT-1 MLT-2 ML-1T-2 ML-3 36

ρL ρV σ σb σbs σc σh σh σs σw σws σz Ѱ

Liquid density Vapour density Surface tension Bending stress Bending stress in the skirt Critical bulking stress Circumferential stress Longitudinal stress Stress in skirt support Stress due to the weight of the vessel Dead weight stress in the skirt Axial stress in vessel Fractional entrainment

ML-3 ML-3 MT-2 ML-1T-2 ML-1T-2 ML-1T-2 ML-1T-2 ML-1T-2 ML-1T-2 ML-1T-2 ML-1T-2 ML-1T-2 -

37

CHAPTER 6: MINOR DESIGN Shell and Tube Heat Exchanger (E-402)

SHIRJAN THAPA 16154359

Contents List of Figures ......................................................................................................................................... iii List of Tables .......................................................................................................................................... iii 6.1 Introduction ...................................................................................................................................... 1 6.1.1 Equipment processing objective and constraints ...................................................................... 1 6.1.2 Design scope .............................................................................................................................. 1 6.2 Process design ................................................................................................................................... 2 6.2.1 Design methodology .................................................................................................................. 2 6.2.2 Basis of design ............................................................................................................................ 2 6.2.3 Overall heat transfer coefficient ................................................................................................ 3 6.2.4 Exchanger type and dimensions ................................................................................................ 3 6.2.5 Heat transfer area ...................................................................................................................... 4 6.2.6 Layout and tube size .................................................................................................................. 4 6.2.7 Number of tubes ........................................................................................................................ 5 6.2.8 Bundle and shell diameter ......................................................................................................... 6 6.2.9 Tube side heat transfer coefficient ............................................................................................ 6 6.2.10 Shell side heat transfer coefficient .......................................................................................... 8 6.2.11 Overall coefficient .................................................................................................................... 9 6.2.12 Pressure drop ........................................................................................................................... 9 Summary ........................................................................................................................................... 10 6.3 Operational design .......................................................................................................................... 11 6.3.1 Control system design .............................................................................................................. 11 6.3.2 Operating procedures .............................................................................................................. 11 6.3.2.1 Pre-start-up procedure ..................................................................................................... 11 6.3.2.2 Start-up procedure............................................................................................................ 11 6.3.2.3 Shutdown procedure ........................................................................................................ 11 6.3.3 Safety study .............................................................................................................................. 12 6.3.3.1 Hazard identification (HAZID) ........................................................................................... 12 6.4 Mechanical design .......................................................................................................................... 13 6.4.1 Material of construction .......................................................................................................... 13 6.4.2 Design pressure ........................................................................................................................ 13 6.4.3 Design temperature ................................................................................................................. 13 6.4.4 Corrosion allowance ................................................................................................................ 13 6.4.5 Wall thickness .......................................................................................................................... 13 6.4.5.1 Shell ................................................................................................................................... 13 6.4.5.2 Tube .................................................................................................................................. 14 i

6.4.6 Pressure relief .......................................................................................................................... 14 6.4.6 Mechanical drawing ................................................................................................................. 15 6.5 Summary and review ...................................................................................................................... 15 6.5.1 Equipment specification .......................................................................................................... 15 6.5.2 Critical review .......................................................................................................................... 15 6.6 Nomenclature ................................................................................................................................. 16 References ............................................................................................................................................ 17 Appendix A ............................................................................................................................................ 18 Appendix B ............................................................................................................................................ 19 Appendix C ............................................................................................................................................ 20 Appendix D ............................................................................................................................................ 21

ii

List of Figures Figure 1: Process flow diagram of E-402 ................................................................................................. 1 Figure 2: Temperature correction factor: one shell, two or more even tube passes[4] .......................... 4 Figure 3: Shell-bundle clearance[4] .......................................................................................................... 6 Figure 4: Tube-side heat transfer factor[4] .............................................................................................. 7 Figure 5: Shell side heat transfer factor, segmental baffles[4] ................................................................ 8 Figure 6: Tube side friction factor[4] ...................................................................................................... 10 Figure 7: Piping & Instrumentation Diagram (P&ID) for Shell and Tube heat exchanger .................... 18 Figure 8: Mechanical drawing of Shell and Tube heat exchanger ........................................................ 19

List of Tables Table 1: Materials stream for E-402 ....................................................................................................... 1 Table 2: Physical properties of Lean and Rich TEG ................................................................................. 2 Table 3: Recommended dimensions for steel tubes[4] ............................................................................ 5 Table 4: Summary of proposed process design .................................................................................... 10 Table 5: HAZID study ............................................................................................................................. 12 Table 6: Standard nozzle orifice data[6] ................................................................................................. 15 Table 7: Data sheet ............................................................................................................................... 20

iii

6.1 Introduction 6.1.1 Equipment processing objective and constraints Shell and tube heat exchanger (E-402) is chosen to be designed as a major component. The heat exchanger uses heat transfer principle to heat up rich triethylene glycol (TEG) which is the bottom product of the dehydration absorber and cool down lean TEG that’s coming out as a bottom product of TEG regenerator column. As rich TEG is dirtier than lean TEG, rich TEG goes into the tube for ease of cleaning and maintenance. Hence lean TEG goes into shell. The required pressure drop in both shell and tube are 0.8 bar. Table 1 shows the materials stream for the heat exchanger.

Shell-side outlet (Lean TEG)

Tube-side Inlet (Rich TEG)

Tube-side outlet (Rich TEG)

Shell-side Inlet (Lean TEG) Figure 1: Process flow diagram of E-402

6.1.2 Design scope The report covers the general design of the specified shell and tube heat exchanger illustrated in figure 1. Firstly the process design of the column is covered based on the design procedure for shell and tube heat exchanger. Then the operational design is discussed which deals with control system design, operating procedures and safety study. Lastly mechanical design of the heat exchanger is covered where vessel thickness and pressure is determined. Table 1: Materials stream for E-402

Lean TEG inlet Vapour Fraction 0 Temperature (áľ’C) 288.0812 Pressure (bar) 1.033 Molar Flow (kgmole/h) 90.1338 Mass Flow (kg/h) 13528.6348 Liquid Volume Flow 11.9903 (m3/h) Heat Flow (kcal/h) -14836649.6280

Lean TEG outlet 0 205.3148 0.833 90.1338 13528.6348 11.9903

Rich TEG inlet 0.0228 48.6361 1.3 179.5807 15200.5465 13.8074

Rich TEG outlet 0.1211 120 1.1 179.5807 15200.5465 13.8074

-15720437.0867

-23035704.0174

-22151916.5587 1

6.2 Process design 6.2.1 Design methodology The methodology covered in process design of the shell and tube heat exchanger is as follows: 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11.

Collect physical properties Assume the value of overall heat transfer coefficient Decide number of shell and tube passes Calculate log mean temperature difference with correction factor Calculate heat transferred per unit time followed by the calculation of heat transfer area Decide the tube dimensions, material, layout Assign fluid to shell and tube Calculate number of tubes and tube side velocity Calculate shell diameter Estimate tube side heat transfer coefficient Decide baffle spacing, calculate shell side velocity then estimate shell side heat transfer coefficient 12. Calculate overall heat transfer coefficient 13. Check if the calculated overall coefficient is acceptable using the following equation đ?‘ˆđ?‘œ,đ?‘Žđ?‘ đ?‘ − đ?‘ˆđ?‘œ,đ?‘?đ?‘Žđ?‘™đ?‘? < 30% đ?‘ˆđ?‘œ,đ?‘Žđ?‘ đ?‘ If calculated overall coefficient is not acceptable, go back to step 2 and use the calculated coefficient as new assumed coefficient If it is acceptable, estimate shell and tube pressure drop If pressure drop is not acceptable, go back to step 6, otherwise check if the design can be further optimised If further optimisation is not possible, accept the design 0 <

14. 15. 16. 17.

6.2.2 Basis of design Table 2 contains the physical properties of the inlet and outlet streams of shell and tube used for designing shell and tube heat exchanger. Table 2: Physical properties of Lean and Rich TEG

Lean TEG Temperature (áľ’C) Specific heat (kJ/kgáľ’C) Thermal conductivity (W/máľ’C) Density (kg/m3) Viscosity (mNsm-2)

Inlet 288.0812 3.5697 0.16 876.0639 0.2153

Outlet 205.3148 3.0499 0.18 964.7584 0.5536

Average 246.6980 3.3098 0.17 920.4112 0.3845

Rich TEG Temperature (áľ’C) Specific heat (kJ/kgáľ’C) Thermal conductivity (W/máľ’C) Density (kg/m3) Viscosity (mNsm-2)

Inlet 48.6361 2.5970 0.24 155.8674 9.4381

Outlet 120.0000 2.8645 0.24 23.1577 1.9605

Average 84.3180 2.7307 0.24 89.5126 5.6993 2

6.2.3 Overall heat transfer coefficient For a shell and tube heat exchanger where both the hot and cold fluid is an organic solvent the overall heat transfer coefficient lies in the range 100 to 300 W/m2áľ’C[4]. So assumed heat transfer coefficient (Uo,ass) is 140 W/m2áľ’C.

6.2.4 Exchanger type and dimensions The preferred arrangement for shell and tube heat exchange is one where there are even number of tube passes. This positions the inlet and outlet nozzles at the same end of the exchange, simplifying the pipe arrangement. Assume 1 shell pass and 2 tube passes. Hot fluid is entering into the tube and cold fluid is entering the shell. ∆đ?‘‡đ?‘™đ?‘š =

(đ?‘‡1 − đ?‘Ą2 ) − (đ?‘‡2 − đ?‘Ą1 ) (đ?‘‡ − đ?‘Ą2 ) đ?‘™đ?‘› 1 (đ?‘‡2 − đ?‘Ą1 ) đ?‘…=

đ?‘‡1 − đ?‘‡2 đ?‘Ą2 − đ?‘Ą1

�=

đ?‘Ą2 − đ?‘Ą1 đ?‘‡1 − đ?‘Ą1

where, ΔTlm = log mean temperature difference (áľ’C) T1 = hot fluid temperature, inlet (áľ’C) T2 = hot fluid temperature, outlet (áľ’C) t1 = cold fluid temperature, inlet (áľ’C) t2 = cold fluid temperature, outlet (áľ’C) R = shell side fluid flow times the fluid mean specific heat, divided by the tube side fluid flow rate times the tube side fluid specific heat S = measure of the temperature efficiency of the exchanger ∆đ?‘‡đ?‘™đ?‘š =

(288.0812 − 120) − (205.3148 − 48.6361) = 162.3132áľ’đ??ś 288.0812 − 120 đ?‘™đ?‘› 205.3148 − 48.6361 đ?‘…=

288.0812 − 205.3148 = 1.1598 120 − 48.6361

�=

120 − 48.6361 = 0.2980 28.0812 − 48.6361

From figure 2, log mean temperature difference correction factor (Ft) is 0.97. ∆đ?‘‡đ?‘š = 0.97 Ă— 162.3132 = 157.4438

3

Figure 2: Temperature correction factor: one shell, two or more even tube passes[4]

6.2.5 Heat transfer area đ?‘„ = đ?‘šđ??śđ?‘? đ?›Ľđ?‘‡ đ??´=

đ?‘„ đ?‘ˆđ?›Ľđ?‘‡đ?‘š

where, Q = heat transferred per unit time (W) m = mass flow (kg/s) Cp = Specific heat (kJ/kgᴟC) ΔT = temperature difference (ᴟC) A = heat transfer area (m2) Take the values for Lean TEG to find Q. �=

13528.6348 Ă— 3.3098 Ă— (288.0812 − 205.3148) = 1029.4543 đ?‘˜đ?‘Š = 1029454.3 đ?‘Š 3600 đ??´=

1029454.3 = 46.7039 đ?‘š2 140 Ă— 157.4439

6.2.6 Layout and tube size Split-ring floating head exchanger will be used as it is efficient and low maintenance i.e. easy to clean. Both rich TEG and lean TEG aren’t corrosive and the operating pressure of the heat exchanger isn’t very high either so carbon steel will be used as a material of constriction for both tubes and shell. Rich TEG will be passed through the tubes as it is dirtier than lean TEG. This results in a low cost exchanger.

4

Table 3: Recommended dimensions for steel tubes[4]

Outside diameter (mm) 16 19 25 32 38 50

Wall thickness (mm) 1.2 1.7 1.7 1.7 1.7 -

2.1 2.1 2.1 2.1 2.1 2.1

2.8 2.8 2.8 2.8 2.8

3.4 3.4 3.4 3.4

As recommended in table 3, the outside diameter of the tubes 20.5 mm, wall thickness is 1.7 mm and internal diameter is 17.1 mm. The length of the tube is chosen to be 5 m as it is most commonly used[4]. The tubes are arranged in a triangular pitch as it gives higher transfer rate. The tube pitch (pt) is 23.625 mm which is 1.25 times the outside diameter.

6.2.7 Number of tubes đ??´đ?‘&#x;đ?‘’đ?‘Ž đ?‘œđ?‘“ 1 đ?‘Ąđ?‘˘đ?‘?đ?‘’ = đ?œ‹ Ă— đ?‘‘đ?‘œ Ă— đ?‘™đ?‘Ą đ?‘ đ?‘˘đ?‘šđ?‘?đ?‘’đ?‘&#x; đ?‘œđ?‘“ đ?‘Ąđ?‘˘đ?‘?đ?‘’đ?‘ =

đ??´ đ??´đ?‘&#x;đ?‘’đ?‘Ž đ?‘œđ?‘“ 1 đ?‘Ąđ?‘˘đ?‘?đ?‘’

Where, do = tube outside diameter (m) lt = tube length (m) đ??´đ?‘&#x;đ?‘’đ?‘Ž đ?‘œđ?‘“ 1 đ?‘Ąđ?‘˘đ?‘?đ?‘’ = đ?œ‹ Ă— đ?‘ đ?‘˘đ?‘šđ?‘?đ?‘’đ?‘&#x; đ?‘œđ?‘“ đ?‘Ąđ?‘˘đ?‘?đ?‘’đ?‘ (đ?‘ đ?‘Ą ) =

20.5 Ă— 5 = 0.3220 đ?‘š2 1000

46.7039 = 145.0373 ≈ 150 0.3220

For 2 tube pass, đ?‘‡đ?‘˘đ?‘?đ?‘’đ?‘ đ?‘?đ?‘’đ?‘&#x; đ?‘?đ?‘Žđ?‘ đ?‘ =

150 = 75 2

Velocity of the tube: đ?‘‡đ?‘˘đ?‘?đ?‘’ đ?‘?đ?‘&#x;đ?‘œđ?‘ đ?‘ đ?‘ đ?‘’đ?‘?đ?‘Ąđ?‘–đ?‘œđ?‘›đ?‘Žđ?‘™ đ?‘Žđ?‘&#x;đ?‘’đ?‘Ž =

đ?œ‹ Ă— đ?‘‘đ?‘– 2 4

Where, di = internal diameter of tube (m) đ?‘‡đ?‘˘đ?‘?đ?‘’ đ?‘?đ?‘&#x;đ?‘œđ?‘ đ?‘ đ?‘ đ?‘’đ?‘?đ?‘Ąđ?‘–đ?‘œđ?‘›đ?‘Žđ?‘™ đ?‘Žđ?‘&#x;đ?‘’đ?‘Ž =

đ?œ‹ 17.1 2 ) = 0.0002297 đ?‘š2 Ă—( 4 1000

đ??´đ?‘&#x;đ?‘’đ?‘Ž đ?‘?đ?‘’đ?‘&#x; đ?‘?đ?‘Žđ?‘ đ?‘ = 0.0002297 Ă— 75 = 0.01722 đ?‘š2 đ?‘‰đ?‘œđ?‘™đ?‘˘đ?‘šđ?‘’đ?‘Ąđ?‘&#x;đ?‘–đ?‘? đ?‘“đ?‘™đ?‘œđ?‘¤ đ?‘&#x;đ?‘Žđ?‘Ąđ?‘’ = đ?‘‰đ?‘œđ?‘™đ?‘˘đ?‘šđ?‘’đ?‘Ąđ?‘&#x;đ?‘–đ?‘? đ?‘“đ?‘™đ?‘œđ?‘¤ đ?‘&#x;đ?‘Žđ?‘Ąđ?‘’ đ?‘Ąđ?‘˘đ?‘?đ?‘’ đ?‘ đ?‘–đ?‘‘đ?‘’ =

đ?‘šđ?‘Žđ?‘ đ?‘ đ?‘“đ?‘™đ?‘œđ?‘¤ đ?‘&#x;đ?‘Žđ?‘Ąđ?‘’ đ?‘‘đ?‘’đ?‘›đ?‘ đ?‘–đ?‘Ąđ?‘Ś

15200.5465 1 Ă— = 0.04717 đ?‘š3 /đ?‘ 3600 89.5126

đ?‘‡đ?‘˘đ?‘?đ?‘’ đ?‘ đ?‘–đ?‘‘đ?‘’ đ?‘Łđ?‘’đ?‘™đ?‘œđ?‘?đ?‘–đ?‘Ąđ?‘Ś, đ?‘˘đ?‘Ą =

0.04717 = 2.7386 đ?‘š/đ?‘ 0.01722 5

6.2.8 Bundle and shell diameter

Figure 3: Shell-bundle clearance[4] 1

đ?‘ đ?‘Ą đ?‘›1 đ??ˇđ?‘? = đ?‘‘đ?‘œ ( ) đ??ž1 where, Db = bundle diameter (mm) K1 = constant for the equation n1 = index for the equation For tubes arranged in triangular pitch with 2 passes, K1 is 0.249 and n1 is 2.207[4]. 1

150 2.207 ) đ??ˇđ?‘? = 20.5 ( = 372.6776 đ?‘šđ?‘š 0.249 Using the bundle diameter, for split-ring floating head exchanger, typical clearance is 54 mm (figure 3). đ?‘†â„Žđ?‘’đ?‘™đ?‘™ đ?‘–đ?‘›đ?‘ đ?‘–đ?‘‘đ?‘’ đ?‘‘đ?‘–đ?‘Žđ?‘šđ?‘’đ?‘Ąđ?‘’đ?‘&#x; (đ??ˇđ?‘ ) = 372.6776 + 54 = 426.6776 đ?‘šđ?‘š

6.2.9 Tube side heat transfer coefficient đ?‘…đ?‘’ =

đ?œŒđ?‘˘đ?‘‘đ?‘– Âľ

Pr =

đ??śđ?‘? Âľ đ?‘˜đ?‘“

6

Where, Re = Reynolds number Ď = density (kg/m3) Âľ = viscosity (Ns/m2) Pr = Prandtl number Cp = Specific heat (J/kgá´źC) kf = thermal conductivity (W/má´źC) đ?‘…đ?‘’ =

89.5126 Ă— 2.7386 Ă— 17.1 Ă— 10−3 = 735.5073 5.6993 Ă— 10−3

Pr =

2.7307 Ă— 103 Ă— 5.6993 Ă— 10−3 = 64.8475 0.24 đ?‘™đ?‘Ą 5000 = = 292.3977 đ?‘‘đ?‘– 17.1

Now, negelecting the viscosity correction: đ?‘ đ?‘˘ = đ?‘—â„Ž đ?‘…đ?‘’đ?‘ƒđ?‘&#x; 0.33 đ?‘ đ?‘˘ =

â„Žđ?‘– đ?‘‘đ?‘– đ?‘˜đ?‘“

where, Nu = Nusselt number jh = heat transfer factor hi = film transfer coefficient inside a tube (W/m2á´źC) kf = thermal conductivity of fluid (W/má´źC) From figure 4, hear transfer factor for tubes is 0.0035.

Figure 4: Tube-side heat transfer factor[4]

đ?‘ đ?‘˘ = 0.0035 Ă— 735.5073 Ă— 64.84750.33 = 10.1995 â„Žđ?‘– =

10.1995 Ă— 0.24 = 143.1511 đ?‘Š/đ?‘š2 á´źđ??ś 17.1 Ă— 10−3 7

6.2.10 Shell side heat transfer coefficient Assume baffle spacing (lb) to be Ds/5 which is 85.3355. So take baffle spacing to be 100 mm which should provide good heat transfer without high pressure drop. đ??´đ?‘&#x;đ?‘’đ?‘Ž đ?‘œđ?‘“ đ?‘ â„Žđ?‘’đ?‘™đ?‘™, đ??´đ?‘ = đ??´đ?‘ =

đ?‘?đ?‘Ą − đ?‘‘đ?‘œ Ă— đ??ˇđ?‘ Ă— đ?‘™đ?‘? đ?‘?đ?‘Ą

23.75 − 20.5 Ă— 426.6776 Ă— 100 = 8533.5529 đ?‘šđ?‘š2 = 0.008533 đ?‘š2 25.625 đ?‘‘đ?‘’ =

1.1 2 (đ?‘? − 0.917đ?‘‘đ?‘œ 2 ) đ?‘‘đ?‘œ đ?‘Ą

where, de = equivalent diameter (mm) đ?‘‘đ?‘’ =

1.1 (25.6252 − 0.917 Ă— 20.52 ) = 14.9816 đ?‘šđ?‘š 20.5

đ?‘‰đ?‘œđ?‘™đ?‘˘đ?‘šđ?‘’đ?‘Ąđ?‘&#x;đ?‘–đ?‘? đ?‘“đ?‘™đ?‘œđ?‘¤ đ?‘&#x;đ?‘Žđ?‘Ąđ?‘’ đ?‘ â„Žđ?‘’đ?‘™đ?‘™ đ?‘ đ?‘–đ?‘‘đ?‘’ =

13528.6348 1 Ă— = 0.004083 đ?‘š3 /đ?‘ 3600 920.4112

đ?‘†â„Žđ?‘’đ?‘™đ?‘™ đ?‘ đ?‘–đ?‘‘đ?‘’ đ?‘Łđ?‘’đ?‘™đ?‘œđ?‘?đ?‘–đ?‘Ąđ?‘Ś, đ?‘˘đ?‘ = đ?‘…đ?‘’ =

0.004083 = 0.4785 đ?‘š/đ?‘ 0.008533

920.4112 Ă— 0.4785 Ă— 14.9816 Ă— 10−3 = 17160.2586 0.3845 Pr =

3.3098 Ă— 103 Ă— 0.3845 Ă— 10−3 = 7.4853 0.17

Using segmental baffles with 25% cut gives good heat transfer coefficient without too large pressure drop. So, from figure 5, the heat transfer factor for shell is 0.0047.

Figure 5: Shell side heat transfer factor, segmental baffles[4]

8

đ?‘ đ?‘˘ = 0.47 Ă— 17160.2586 Ă— 7.4853 = 156.7150 â„Žđ?‘ =

đ?‘ đ?‘˘ Ă— đ?‘˜đ?‘“ đ?‘‘đ?‘’

Where, hs = film transfer coefficient inside a shell 156.7150 Ă— 0.17 = 1778.2839 đ?‘Š/đ?‘š2 á´źđ??ś 14.9816 Ă— 10−3

â„Žđ?‘ =

6.2.11 Overall coefficient đ?‘‘ đ?‘‘đ?‘œ ln ( đ?‘œ ) đ?‘‘ 1 1 1 1 đ?‘‘đ?‘œ 1 đ?‘‘đ?‘– đ?‘œ = + + + Ă— + Ă— đ?‘ˆđ?‘œ â„Žđ?‘ â„Žđ?‘œđ?‘‘ 2đ?‘˜đ?‘¤ đ?‘‘đ?‘– â„Žđ?‘–đ?‘‘ đ?‘‘đ?‘– â„Žđ?‘– where, Uo = overall coefficient based on the outside area of the tube (W/m2á´źC) hod = inside fouling factor (W/m2á´źC) hid = outside fouling factor (W/m2á´źC) kw = thermal conductivity of tube wall material (W/má´źC) Fouling factor of TEG is 0.00035 (W/m2á´źC)-1[5] and since there is TEG in both shell and tubes hod is equal to hid. The thermal conductivity of carbon steel is 0.00035 W/má´źC[4]. 20.5 ) 20.5 Ă— 10−3 ln ( 1 1 1 17.1 + 1920.5 Ă— 0.00035 + 20.5 Ă— = + 0.00035 + đ?‘ˆđ?‘œ 1778.2839 2 Ă— 54 17.1 17.1 143.1511 đ?‘ˆđ?‘œ = 102.6596 đ?‘Š/đ?‘š2 á´źđ??ś Check if Uo is acceptable: 0 < 0 <

đ?‘ˆđ?‘œ,đ?‘Žđ?‘ đ?‘ − đ?‘ˆđ?‘œ,đ?‘?đ?‘Žđ?‘™đ?‘? Ă— 100% < 30% đ?‘ˆđ?‘œ,đ?‘Žđ?‘ đ?‘

140 − 102.6596 Ă— 100% < 30% 140 0 < 26.67% < 30%

Hence, the calculated overall heat transfer coefficient is acceptable.

6.2.12 Pressure drop đ?‘™đ?‘Ą đ?œŒđ?‘˘đ?‘Ą 2 ∆đ?‘ƒđ?‘Ą = đ?‘ đ?‘? (8đ?‘—đ?‘“ ( ) + 2.5) đ?‘‘đ?‘– 2 Where, ΔPt = tube side pressure drop (N/m2) Np = number of tube side passes jf = friction factor As the tube side Reynolds number is 735.50, friction factor from figure 6 is 0.01.

9

5000 89.5126 Ă— 2.73862 đ?‘ ) + 2.5) đ?›Ľđ?‘ƒđ?‘Ą = 2 (8 Ă— 0.01 ( = 17382.2134 2 = 0.17 đ?‘?đ?‘Žđ?‘&#x; 17.1 2 đ?‘š Similarly shell side pressure drop is 0.15 bar.

Figure 6: Tube side friction factor[4]

Summary Table 4: Summary of proposed process design

Head type No. of shell pass No. of tube pass Material of tubes No. of tube Length of tube Outside diameter of tube Inside diameter of tube Heat transfer area Internal diameter of shell Baffle spacing Tube side coefficient Shell side coefficient Assumed overall heat transfer coefficient Calculated overall heat transfer coefficient Fouling factor of TEG Pressure drop, tube side Pressure drop, shell side

Split ring floating head 1 2 Carbon steel 150 5m 20.5 mm 17.1 mm 46.70 m2 0.43 m 100 mm, 25% cut 143.15 W/m2á´źC 1778.28 W/m2á´źC 140 W/m2á´źC 102.67 W/m2á´źC 0.00035 (W/m2á´źC)-1 0.17 bar, +0.01 for nozzles; specified 0.2 bar overall. 0.15 bar, +0.01 for nozzles; specified 0.2 bar overall.

10

6.3 Operational design 6.3.1 Control system design Controlling the process variables can achieve a safe operation process and help control the parameters of outlet stream. It can also help detect any variations in pressure and temperature in the heat exchanger. This helps keep the variables in check and within range for the heat exchanger to function at normal to optimum level. There is a temperature transmitter connected to the heat exchanger which in turn controls the flow in to the tube and shell. Flow control, flow transmitter, temperature control, pressure control and pressure transmitter are also included in the control system of the heat exchanger. They can be seen in the Piping and Instrumentation Diagram (P&ID) for E-402 in figure 7 in Appendix A.

6.3.2 Operating procedures 6.3.2.1 Pre-start-up procedure Safe operation of the heat exchanger can be ensured by following the pre-start-up procedure listed below prior to actioning the start-up procedure. 1. 2. 3. 4.

The heat exchanger and attached pipes should be dry Make sure all drains and vents are closed and ensure all flanges are tightened All utilities should be lined-up All instrumentations should be fitted and lined-up

6.3.2.2 Start-up procedure After the pre-start-up procedure is completed, the following steps should be followed to start-up the shell and tube heat exchanger. 1. 2. 3. 4.

Turn on the power of the equipment Open the rich TEG inlet valve and also open the vent to remove air from the system Once the tube side is full of rich TEG, close the vent When the heat exchanger reaches operating temperature, open the inlet valve of shell side to let lean TEG in.

6.3.2.3 Shutdown procedure Typically there are planned and unplanned shutdowns. While planned shutdowns are occur for maintenance and inspection propose, unplanned shutdown are carried out during emergencies. 6.3.2.3.1 Planned shutdown The following steps are taken to ensure the safe shutdown for the proposed heat exchanger. 1. 2. 3. 4. 5.

Close the tube inlet Close the shell inlet Depressurize the heat exchanger Turn off all utilities Drain any liquid from shell and tube

6.3.2.3.2 Unplanned shutdown The following procedures should be followed in case of unplanned shutdown for the pure argon distillation column: 1. Close the feed line 2. Isolate the outlets to avoid contaminating the product 11

3. 4. 5. 6. 7.

Open the safety valve, if necessary Depressurize the heat exchanger Detect the cause of irregularity Ensure that the problem is fixed as well as any damages caused by it Once it is safe to operate the vessel again, follow the pre-start-up and start-up procedures

6.3.3 Safety study The proposed shell and tube heat exchanger is designed for safety against uncontrolled loss of containment caused by leaks or mechanical failures. The heat exchanger is also protected against electrical power and pressure deviations. All the safety factures and failures need to be considered when designing and operating the heat exchanger no matter how small they are as failure to recognise them can lead to injury to personal or even loss of life.

6.3.3.1 Hazard identification (HAZID) Table 5 below shows the hazards identified along with its consequences and what precautions can be actioned to keep it from occurring. Table 5: HAZID study

Risk assessment

Recommendation

Extreme heat

• Outlet temperatures affected

• Temperature controller of heat exchanger

Rare

Minor

Low

• None

Sensor failure

• Explosion • Level errors • Overheating

• Multiple sensor

Unlikely

Major

High

• Regular maintenance of sensors

Pressure deviations

• Loss of production • Low efficiency

• Pressure and flow controllers

Unlikely

Moderate

Medium

Structural failure

• Loss of containment • Property damage • Loss of production

• regulation of pressure and temperature

Rare

Critical

High

• Account for baffle spacing • Design the heat exchanger to handle pressure 10% above the operating pressure • Materials of construction should meet the specification • Regular maintenance

Risk ranking

Prevention/ Detection

Consequence severity

Hazard consequences

Likelihood

Hazard identified

12

Corrosion

• Ruptures and crack • structural integrity compromised • Build-up of contaminant

• Acid gas removal unit

Unlikely

Major

High

Inadequate training

• Operation failure • Injury to personal • loss of production

• Controller • Training • operational procedure

Likely

Moderate

High

• Corrosion allowance considered for materials • Maintenance and inspection • Corrosive contaminant removed in acid gas removal unit • Proper training programs • Increase safety awareness

6.4 Mechanical design The mechanical design of the pressure vessel is created based on the parameters of the vessel obtained from process design.

6.4.1 Material of construction As both rich and light TEG are not corrosive, carbon steel will be used for the construction of both shell and tube.

6.4.2 Design pressure In order to design the heat exchanger to withstand maximum pressure, heat exchanger needs to design 10% above the normal working pressure as suggested by American Petroleum Institute (API) Recommended Practice (RP) 520[4]. Hence, heat exchanger shell-side design pressure = 1.1 Ă— Pressureoperating = 1.1 Ă— 1.3 = 1.43 bar. So, design pressure is 0.143 N/mm2. Similarly, heat exchanger tube-side design pressure is 1.1363 bar or 0.1136 N/mm2.

6.4.3 Design temperature The minimum temperature in shell-side and tube-side are 48.63á´źC and 205.31á´źC respectively.

6.4.4 Corrosion allowance As the material being used is carbon-steel, 2 mm thickness needs to be added to the thickness of the material to account for corrosion as suggested by ASME BPV Code Sec. VIII D.1[4].

6.4.5 Wall thickness 6.4.5.1 Shell đ?‘Ą=

đ?‘ƒđ?‘– đ??ˇđ?‘– + đ?‘?đ?‘œđ?‘&#x;đ?‘œđ?‘ đ?‘ đ?‘–đ?‘œđ?‘› đ?‘Žđ?‘™đ?‘™đ?‘œđ?‘¤đ?‘Žđ?‘›đ?‘?đ?‘’ 2đ?‘†đ?‘ đ??¸ − 1.2đ?‘ƒđ?‘–

where, t = thickness of plate or shell (mm) 13

Ss = maximum allowable stress of the given material at design temperature (N/mm2) Pi = internal pressure (N/mm2) Di = internal diameter (mm) E = joint efficiency, wielded joint For carbon steel at 48.63°C, S = 12900 psi = 88.9 N/mm2[4] E = 1 for full double-welded butt joint or equivalent[4] Di = 399.4085 mm Pi = 0.1163 N/mm2 đ?‘Ą=

0.143 Ă— 426.6776 + 2 = 2.3435 đ?‘šđ?‘š 2 Ă— 88.9 Ă— 1 − 1.2 Ă— 0.143

Therefore, the wall thickness of the shell is 2.34 mm, this is lower than required thickness for a shell with an internal diameter of almost 400 mm. For a carbon steel shell with an internal diameter ranging from 330 to 580 mm, the minimum thickness should be 9.5 mm with corrosion allowance [4]. Hence, 9.5 mm will be the shell diameter for mechanical design purpose.

6.4.5.2 Tube As estimated from table 3 in process design, the wall thickness of the tube is 1.7 mm which already accounts for corrosion allowance.

6.4.6 Pressure relief The following formula is used to calculate the effective area of the nozzle orifice required for pressure relief in the shell as its fluid is in liquid form [6]. đ??´=

đ?‘„√đ??ş

27.2đ??žđ?‘? đ??žđ?‘¤ đ??žđ?‘Ł √đ?›Ľđ?‘ƒ For design pressure margin of less than 25%: đ??žđ?‘? = −0.0014(%đ?‘šđ?‘Žđ?‘&#x;đ?‘”đ?‘–đ?‘›)2 + 0.073(%đ?‘šđ?‘Žđ?‘&#x;đ?‘”đ?‘–đ?‘›) + 0.016 where, A = valve orifice area (in2) Q = Flow (gpm) G = Specific gravity of water at maximum shell temperature Kp = Correction factor for relieving capacity vs. lift for relief valves in liquid service Kw = Correction factor due to back pressure for use with balanced bellows valves Kv = Correction factor for viscosity ΔP = Differential pressure (psig), (design pressure – back pressure) So, Q = 13528.6348 kg/h = 59.56 gpm G = 0.72 at 288áľ’C Kv = 1 Kw = 1 as there is no back pressure ΔP = 20.74 psig đ??žđ?‘? = −0.0014 Ă— 102 + 0.073 Ă— 10 + 0.016 = 0.606 đ??´=

59.56√0.72 27.2 Ă— 0.606 Ă— 1 Ă— 1 Ă— √20.74

= 0.6732 đ?‘–đ?‘›2

14

Table 6: Standard nozzle orifice data[6]

Nozzle Orifice Areas Size Designation Orifice Area, in2 D 0.110 E 0.196 F 0.307 G 0.503 H 0.785 J 1.280 K 1.840 L 2.850 M 3.600 N 4.340 P 6.380 Q 11.050 R 16.000 T 26.000 From Table 6, smallest standard size designation that has an area equal to or greater than 0.6732 in2 is “H”. Hence, “H” orifice will be installed in shell.

6.4.6 Mechanical drawing Figure 8 in Appendix B illustrates the mechanical drawing of the shell and tube heat exchanger based on process and mechanical design.

6.5 Summary and review The report contains a design of shell and tube heat exchanger involving process, operational and mechanical design. Process design began with methodology and moved on to calculations to find the tube and shell sizing, overall heat transfer coefficient and pressure drop. Second part deals with control system design, operation procedures and safety study. In the mechanical design the material of construction is selected and also contains vessel design parameters such as design temperature and pressure and thickness of the vessel is calculated.

6.5.1 Equipment specification Based on the chemical and mechanical design calculation, a data sheet for shell and tube heat exchanger is in Appendix C, table 7.

6.5.2 Critical review The shell and tube heat exchanger is designed in order to reduce the power load of the plant. Rich Teg is cold and needs to be heated up while the lean TEG is hot and needs to be cooled down. The two fluids are used to transfer heat between one another. The physical properties required for the calculation are extracted from Aspen HYSYS. The first assumption made in the design process was the overall heat transfer coefficient. TEG being organic solvent, its overall coefficient lies within the range of 100 to 300 W/m2ᵒC. The assumed value lies within this range. As the final calculated value of overall coefficient also lies within this range with 15

less than 30% deviation from the assumed value, it is acceptable. The calculated coefficient also met the pressure drop for both shell and tube. In terms of future work required for the design of the purposed shell and tube heat exchanger, the design can further be modified to produce a design with better heat transfer coefficient and pressure drop which at the same time is more economical. The design methodology is based on text book named Chemical Engineering Design by Ray Sinnott and Gavin Towler. The process in the book is well structured and easy enough to follow which gives values with low uncertainty of error.

6.6 Nomenclature Appendix D has a nomenclature for the symbols used for calculations throughout this report with dimensions in mass, length and time (MLTϴ).

16

References 1. Engineeringtoolbox.com,. 2015. 'Water - Temperature And Specific Gravity'. http://www.engineeringtoolbox.com/water-temperature-specific-gravityd_1179.html. 2. Kidnay, A. J. 2011. Fundamentals of Natural Gas Processing. Edited by William Rutledge Parrish and Daniel G. McCartney. 2nd ed. / Arthur J. Kidnay, William R. Parrish, Daniel G. McCartney.. ed. Boca Raton, FL: Boca Raton, FL : CRC Press. 3. Thermal conductivity of Materials and Gases. 2015. The Engineering Toolbox. http://www.engineeringtoolbox.com/thermal-conductivity-d_429.html 4. Towler, Gavin, and Ray Sinnott. 2013. Chemical engineering design : principles, practice, and economics of plant and process design. Oxford ; Waltham, MA : ButterworthHeinemann. 5. Typical Fouling Factors. 2015. Engineering page. http://www.engineeringpage.com/technology/thermal/fouling_factors.html 6. Whitesides, Randall W. 2012. "Selection and Sizing of Pressure Relief Valves." PDH Online Course M112. http://www.pdhonline.org/courses/m112/Selection%20and%20Sizing%20of%20Pre ssure%20Relief%20Valves.pdf

17

Appendix A

Shell Inlet

Tube Inlet FT

FT FC

FC PC

PT

FT

FT

FC

Shell Outlet TC

TT

FC

Tube Outlet TT

TC

Figure 7: Piping & Instrumentation Diagram (P&ID) for Shell and Tube heat exchanger

Index: FC: Flow controller PC: Pressure controller TC: Temperature controller

FT: Flow transmitter PT: Pressure transmitter TT: Temperature transmitter

Pressure relief valve

18

Appendix B 7m 5m 100 mm

O.D: 445.68 mm I.D: 426.68 mm

Figure 8: Mechanical drawing of Shell and Tube heat exchanger

19

Appendix C Table 7: Data sheet

Project Name: BIG GAS PROJECT Project Number: Chevron 1 Sheet No. 1 REV By Date APVD By

Date

Shell and Tube Heat Exchanger Owner’s Name: Plant Location: Longford, VIC, Australia Case Description: Sales Gas & LPG Processing Plant Equipment Number: E-402 Description: Heat Exchanger Data Per Unit

Fluid Total fluid flow (kg/h) Vapour fraction Density (kg/m3) Viscosity (mNsm-2) Specific heat (kJ/kg°C) Thermal conductivity (W/m°C) Temperature (°C) Pressure (bar) Pressure drop allowed (bar) Pressure drop calculated (bar) Flow velocity (m/s) No. of passes Film transfer coefficient (W/m2°C) Ft factor Effective mean temperature difference (°C) Construction & Materials Tube material No. of tubes 150 Pitch (mm) Length (m) 5 O.D (mm) Design pressure (bar) 1.14 Shell material Length (m) 7 Baffle Carbon steel material Design pressure (bar) 1.43

O.D (mm) Baffle type

Shell Side In Out Lean TEG 13528.63 0 0 876.06 964.76 0.23 0.55 3.57 3.05 0.16 0.18 288.08 205.31 1.033 0.833 0.2 0.17 2.74

Tube Side In Out Rich TEG 15200.55 0.0228 0.1211 155.87 23.15 9.43 1.96 2.60 2.86 0.24 0.24 48.64 120 1.3 1.1 0.2 0.15 0.48

1 1778.28

2 143.15 0.7 157.44

Carbon steel 23.63 Arrangement Triangular 21.5 I.D (mm) 17.1 Design temp. (°C) 205.31 Carbon steel 445.68 I.D (mm) 25 % cut Baffle segmental spacing (mm) Design temp. (°C) 48.63

426.68 100

20

Appendix D µ A A As At Cp Db de di Di do Ds E Ft G hi hi hid hod hs jf jh K1 kf Kp Kv kw Kw lb lt m n1 Np Nu Pi Pr Q Q R Re S Ss t T1 t1 T2

Viscosity Heat transfer area (m2) Valve orifice area Area of shell Area of tube Specific heat (kJ/kgᴼC) Bundle diameter Equivalent diameter Internal diameter of tube Internal diameter Tube outside diameter Shell inside diameter Joint efficiency Log mean temperature difference correction factor Specific gravity of water at maximum shell temperature Film transfer coefficient inside a tube Film transfer coefficient inside a tube Outside fouling factor Inside fouling factor Film transfer coefficient inside a shell Friction factor Heat transfer factor Constant in equation to find bundle diameter Thermal conductivity Correction factor for relieving capacity vs. lift for relief valves in liquid service Correction factor for viscosity Thermal conductivity of tube wall material Correction factor due to back pressure for use with balanced bellows valves Baffle spacing Tube length Mass flow (kg/s) Index used in equation to find bundle diameter Number of tube side passes Nusselt number Internal pressure Prandtl number Heat transferred per unit time (W) Flow Shell side fluid flow times the fluid mean specific heat, divided by the tube side fluid flow rate times the tube side fluid specific heat Reynolds number Measure of the temperature efficiency of the exchanger Maximum allowable stress Thickness of shell or plate Hot fluid temperature, inlet Cold fluid temperature, inlet Hot fluid temperature, outlet

Dimension in MLTϴ ML-1T-1 L2 L2 L2 L2 L2T-2ϴ-1 L L L L L L MT-3ϴ-1 MT-3ϴ-1 MT-3ϴ-1 MT-3ϴ-1 MT-3ϴ-1 MLT-3ϴ-1 MLT-3ϴ-1 L L MT-1 ML-1T-2 ML2T-3 MT-1 ML-1T-2 L ϴ ϴ ϴ 21

t2 Uo Uo,ass Uo,calc us ut ΔP ΔPt ΔT ΔTlm ρ

Cold fluid temperature, outlet Overall heat transfer coefficient based on tube outside area Assumed overall heat transfer coefficient Calculated overall heat transfer coefficient Shell side velocity Tube side velocity Differential pressure Tube side pressure drop Temperature difference (ᴼC) Log mean temperature difference Density

ϴ MT-3ϴ-1 MT-3ϴ-1 MT-3ϴ-1 LT-1 LT-1 ML-1T-2 ML-1T-2 ϴ ϴ ML-3

22

CHAPAPTER 7: MAJOR DESIGN Distillation Column (T-501)

Abdullah AL Kindi 15404435

Contents List of Figures ................................................................................................................................... iii List of Tables .................................................................................................................................... iii Introduction ...................................................................................................................................... 1 Equipment Processing Objectives and Constraints ......................................................................... 1 Design Scope ................................................................................................................................. 1 7.1. Process Design ............................................................................................................................ 2 7.1.1

Methodology.................................................................................................................. 2

7.1.2

Minimum number of stages (Nmin): ................................................................................. 3

7.1.3

Minimum Reflux Ratio .................................................................................................... 3

7.1.4

Actual Reflux Ratio ......................................................................................................... 4

7.1.5

Theoretical Number of Stages: ....................................................................................... 4

7.1.6

Feed Point Location ........................................................................................................ 5

7.1.7

Efficiency........................................................................................................................ 6

7.1.8

Real Number of Trays ..................................................................................................... 6

7.1.9

Column Diameter ........................................................................................................... 6

7.1.10

Liquid Flow Pattern ........................................................................................................ 9

7.1.11

Provisional Plate design .................................................................................................. 9

7.1.12

Weeping....................................................................................................................... 10

7.1.13

Plate Pressure Drop ...................................................................................................... 11

7.1.14

Check Entrainment ....................................................................................................... 13

7.1.15

Perforated Area............................................................................................................ 14

7.1.16

Number of Holes .......................................................................................................... 15

7.1.17

Column Height ............................................................................................................. 16

7.2.0

Operational Design........................................................................................................... 17

7.2.1

Control System Objectives ............................................................................................ 17

7.2.2

P&ID............................................................................................................................. 18

7.2.3

Operating Procedures .................................................................................................. 19

7.2.3.1

Start-up and Shut-down ........................................................................................... 19

7.2.3.2

Maintenance ............................................................................................................ 19

7.2.4 7.3.0

Safety Study ................................................................................................................. 20 Mechanical Design ........................................................................................................... 22

7.3.1

Design Pressure............................................................................................................ 22

7.3.2

Cylindrical Section ........................................................................................................ 22

7.3.3

Type of Head (Domed) ................................................................................................. 22

7.3.3.1

Standard Dished Head Torisphere............................................................................. 23 i

7.3.3.2

Standard Ellipsoidal Head ......................................................................................... 23

7.3.3.3

Hemispherical Head.................................................................................................. 23

7.3.4

Dead Weight ................................................................................................................ 23

7.3.4.1

Weight of the Vessel................................................................................................. 23

7.3.4.1

Weight of the Plates ................................................................................................. 23

7.3.4.2

Weight of Insulation ................................................................................................. 24

7.3.5

Wind Loading ............................................................................................................... 24

7.3.6

Stress Analysis .............................................................................................................. 24

7.3.6.1

Dead weight stress ................................................................................................... 24

7.3.6.2

Bending stresses....................................................................................................... 24

7.3.6.3

The resultant longitudinal stress ............................................................................... 25

7.3.7

Supporting Skirt Design ................................................................................................ 25

7.3.8

Pressure Relief ............................................................................................................. 26

7.3.9

Mechanical Drawing ..................................................................................................... 28

7.4.0

Summary and Review ....................................................................................................... 29

7.4.1

Equipment Specifications / data sheet .......................................................................... 29

7.4.2

Critical Review .............................................................................................................. 30

Nomenclature ................................................................................................................................. 31 References ...................................................................................................................................... 33

ii

List of Figures Figure 1: PFD for demethanizer ......................................................................................................... 1 Figure 2: Methodology ...................................................................................................................... 2 Figure 3: Erbar-Maddox correlation (Erbar and Maddox, 1961).......................................................... 5 Figure 4: Flooding velocity, seive plates (Sinnot and Towler, 2009) .................................................... 7 Figure 5: Flooding velocity, seive plates (Sinnot and Towler, 2009) .................................................... 8 Figure 6 Relation between downcomer area and weir length (Sinnot and Towler, 2009) ................. 10 Figure 7 Weep point correlation (Sinnot and Towler, 2009) ............................................................. 11 Figure 8: Discharge coefficients, sieve plates (Liebson et al., 1957) .................................................. 12 Figure 9: Entrainment correlation for sieve plates (Fair, 1961) ......................................................... 13 Figure 10: Relation between angle subtended by chord, chord height and chord length (Sinnot and Towler, 2009) .................................................................................................................................. 14 Figure 11: Relation between hole area and pitch (Sinnot and Towler, 2009) .................................... 15 Figure 12: Plate layout ..................................................................................................................... 15 Figure 13: PI&D of the demethanizer V-501 ..................................................................................... 18 Figure 14: Mechanical Drawing ........................................................................................................ 28

List of Tables Table 1: Relative Volatilities Calculations ........................................................................................... 4 Table 2: HAZOP analysis for V-501 ................................................................................................... 20

iii

Introduction Equipment Processing Objectives and Constraints A distillation column is a piece of equipment that is used in the industry to separate mixtures of fluids. The main driving mechanism of the separation is the volatility of the components entering the feed. In this project a detailed design of a distillation column (Demethanizer) which separates light hydrocarbons (Methane and Ethane) as distillate. The bottoms of the column will be propane and heavier hydrocarbons which will be sent to the second part of the fractionation which is the debutanizer. A mixture of sweet hydrocarbons will enter the column at a rate of 1.955×105 kg/h at 30 o C. It is important to mention that most of the equations and the methodology has been adopted from Sinnot and Towler chemical engineering design (2009).

Design Scope This project will include:  

Process Design: In this section all of the design calculations will be carried out. Sizing of the equipment will also be determined in this section Operational Design: This section will cover the start-up and shut-down procedures. It will also cover the safety study i.e. a detailed HAZOP analysis for operating this particular distillation column. Mechanical Design: This section will cover a detailed calculations and analysis of the mechanical aspect of the column. A detailed design of the column’s shell and supports. Stress analysis will also carried out to find out the stress and forces acting on the column. This is done to achieve and inherently safer design.

Figure 1: PFD for demethanizer

1

7.1. Process Design 7.1.1 Methodology 1) Find duty and specs required 8) Calculation of area height and diameter

14) Stress Analysis

2) Find physical properties e.g. relative volatilities 9) Selection the material of construction e.g. carbon steel, aluminium

3) Calculate minimum number of plates 10) Selection of trays appropriate to the process 4) Calculate minimum reflux ratio 11) Optimization

5) Calculation of minimum reflux ratio 12) Design cannot be optimized? anymore? 6) Find out the exact number of stages needed using computer aided programmes

13) Mechanical Design

7) Find Feed Tray

Figure 2: Methodology

2

7.1.2 Minimum number of stages (Nmin): To calculate the minimum number of stages Fenske equation has been used and it is as follows

đ?‘ đ?‘šđ?‘–đ?‘›

đ?‘Ľ đ?‘Ľ đ?‘™đ?‘œđ?‘” [đ?‘Ľ đ??żđ??ž ] [ đ?‘Ľđ??ťđ??ž ] đ??ťđ??ž đ?‘‘ đ??żđ??ž đ?‘? = đ?‘™đ?‘œđ?‘”đ?›źđ??żđ??ž

Nmin= Minimum number of stages. xLK= Light key concentration. xHK= Heavy key concentration. ÎąLK= KLK/KHK = Relative volatility.

đ?›źđ??żđ??ž =

đ?‘ đ?‘šđ?‘–đ?‘›

4.977

= 20.5681 0.2420 0.6672 0.476 đ?‘™đ?‘œđ?‘” [0.0001 Ă— 0.0001 ] = = 6 đ?‘ đ?‘Ąđ?‘Žđ?‘”đ?‘’đ?‘ đ?‘¤đ?‘–đ?‘Ąâ„Ž đ?‘Ąđ?‘œđ?‘Ąđ?‘Žđ?‘™ đ?‘&#x;đ?‘’đ?‘“đ?‘™đ?‘˘đ?‘Ľ đ?‘™đ?‘œđ?‘”(20.5681)

7.1.3 Minimum Reflux Ratio ∑

đ?›źđ?‘– đ?‘Ľđ?‘–,đ?‘‘ = đ?‘…đ?‘šđ?‘–đ?‘› + 1 đ?›źđ?‘– − đ?œƒ

Rmin= Minimum reflux ratio xi,d= concentration of the component i in the distillate. θ= Root of the equation. θ is found using trial and error using the following equation ∑

đ?›źđ?‘– đ?‘Ľđ?‘–,đ?‘“ = 1−đ?‘ž đ?›źđ?‘– − đ?œƒ

q= depends on the condition of the feed and in this case it is considered to be 1. Hence the equation will equal to 0. Trial and error was used using Excel to find the value of θ. Table 1 shows the complete method on how the values were obtained. The equation was equated to 0.0062 (close enough). The value of θ is 2.071 and the minimum reflux ratio turned out to be 0.2139.

3

Table 1: Relative Volatilities Calculations Component CO2

KStage 1 1.5211

KReboiler 4.2829

Kave 2.9020

Îąi 11.9911

x,if 0.0315

(ιi.xi,f/ιi -θ) 0.0381

xi,d 0.0471

(ιi.xi,d/ιi -θ) 0.056927426

Nitrogen

12.4953

10.7061

11.6007

47.9347

0.0010

0.0010

0.0015

0.001537163

Methane

3.9737

5.9817

4.9777

20.5681

0.4466

0.4966

0.6672

0.741900043

Ethane

0.6042

2.6035

1.6038

6.6272

0.1934

0.2813

0.2842

0.413356841

Propane

0.1508

1.3910

0.7709

3.1854

0.1358

0.3882

0.0000

0.000142256

i-Butane

0.0566

0.8771

0.4668

1.9290

0.0654

-0.8885

0.0000

-8.31289E-07

n-Butane

0.0386

0.7479

0.3933

1.6251

0.0751

-0.2738

0.0000

-2.55303E-08

i-Pentane

0.0147

0.4693

0.2420

1.0000

0.0233

-0.0218

0.0000

-6.13688E-12

n-Pentane

0.0105

0.4162

0.2134

0.8817

0.0157

-0.0117

0.0000

-4.44994E-13

n-Hexane

0.0030

0.2361

0.1196

0.4941

0.0083

-0.0026

0.0000

-5.93212E-17

n-Heptane

0.0009

0.1371

0.0690

0.2852

0.0020

-0.0003

0.0000

-1.59532E-31

Mcyclohexane

0.0012

0.1409

0.0710

0.2936

0.0012

-0.0002

0.0000

-5.8062E-21

Toulene

0.0013

0.1349

0.0681

0.2814

0.0003

0.0000

0.0000

-1.54412E-20

n-Octane

0.0003

0.0804

0.0403

0.1666

0.0003

0.0000

0.0000

-2.85377E-21

Σ(ιi.xif/ιi -θ)

0.00618

Σ (ιi.xi,d/ιi -θ)

1.213862873

7.1.4 Actual Reflux Ratio The actual reflux ratio is simply calculated using the following equation: đ?‘…đ?‘’đ?‘“đ?‘™đ?‘˘đ?‘Ľ đ?‘&#x;đ?‘’đ?‘Ąđ?‘˘đ?‘&#x;đ?‘›đ?‘’đ?‘‘ đ?‘Ąđ?‘œ đ?‘Ąâ„Žđ?‘’ đ?‘?đ?‘œđ?‘™đ?‘˘đ?‘šđ?‘› đ??šđ?‘™đ?‘œđ?‘¤ đ?‘ đ?‘’đ?‘›đ?‘Ą đ?‘Ąđ?‘œ đ?‘Ąâ„Žđ?‘’ đ?‘?đ?‘œđ?‘›đ?‘‘đ?‘’đ?‘›đ?‘ đ?‘œđ?‘&#x; 7203 đ?‘˜đ?‘”đ?‘šđ?‘œđ?‘™/â„Ž đ?‘… = 14410 đ?‘˜đ?‘”đ?‘šđ?‘œđ?‘™/â„Ž = đ?&#x;Ž. đ?&#x;’đ?&#x;—đ?&#x;—đ?&#x;–đ?&#x;•đ?&#x;?

đ?‘…=

7.1.5 Theoretical Number of Stages: Calculating the number of stages require the use of the Erbar-Maddox chart (Figure 3) đ?‘… 0.499871 = = 0.333 đ?‘… + 1 0.499871 + 1 đ?‘…đ?‘šđ?‘–đ?‘› 0.2139 = = 0.176 đ?‘…đ?‘šđ?‘–đ?‘› + 1 0.2139 + 1 đ?‘ đ?‘šđ?‘–đ?‘› = 0.56 đ?‘ đ?‘ = 10.7 = đ?&#x;?đ?&#x;? đ?’”đ?’•đ?’‚đ?’ˆđ?’†đ?’”

4

Figure 3: Erbar-Maddox correlation (Erbar and Maddox, 1961)

7.1.6 Feed Point Location The feed point location is determined by using Kirkbride equation: 2

đ?‘ đ?‘&#x; đ??ľ đ?‘Ľđ?‘“,đ??ťđ??ž đ?‘Ľđ?‘?,đ??żđ??ž đ?‘™đ?‘œđ?‘” [ ] = 0.206 log [( ) ( )( ) ] đ?‘ đ?‘ đ??ˇ đ?‘Ľđ?‘“,đ??żđ??ž đ?‘Ľđ?‘‘,đ??ťđ??ž 2

đ?‘ đ?‘&#x; 3559 0.0233 1.87 Ă— 10−12 đ?‘™đ?‘œđ?‘” [ ] = 0.206 đ?‘™đ?‘œđ?‘” [( )( )( ) ] đ?‘ đ?‘ 7203 0.4466 6 Ă— 10−13 đ?‘ đ?‘&#x; [ ] = 0.7518 đ?‘ đ?‘ đ?‘ đ?‘&#x; + đ?‘ đ?‘ = 10 đ?‘ đ?‘&#x; = 0.7518đ?‘ đ?‘ đ?‘ đ?‘ = 10 − 0.7518đ?‘ đ?‘ 10 đ?‘ đ?‘ = = 5.69 = 6 1.7581

Feed stage will be the sixth stage.

5

7.1.7 Efficiency The number of trays calculated in the previous sections is a theoretical number. To calculate the real number of trays for multicomponent distillation column the efficiency is first determined using O’Connell’s correlation: đ??¸đ?‘œ = 51 − 3.25 log(đ?œ‡đ?‘Ž đ?›źđ?‘Ž ) Âľa = the molar average liquid viscosity, mNs/m2 Îąa = average relative volatility of the light key. đ??¸đ?‘œ = 51 − 3.25 log(0.1142 Ă— 20.5681) = 38.95%

7.1.8 Real Number of Trays Real number of trays is then calculated using this equation: đ?‘‡â„Žđ?‘’đ?‘œđ?‘&#x;đ?‘’đ?‘Ąđ?‘–đ?‘?đ?‘Žđ?‘™ đ?‘ đ?‘˘đ?‘šđ?‘?đ?‘’đ?‘&#x; đ?‘œđ?‘“ đ?‘‡đ?‘&#x;đ?‘Žđ?‘Śđ?‘ đ??¸đ?‘“đ?‘“đ?‘–đ?‘?đ?‘–đ?‘’đ?‘›đ?‘?đ?‘Ś 11 đ?‘…đ?‘’đ?‘Žđ?‘™ đ?‘ đ?‘˘đ?‘šđ?‘?đ?‘’đ?‘&#x; đ?‘œđ?‘“ đ?‘‡đ?‘&#x;đ?‘Žđ?‘Śđ?‘ = = 28 đ?‘‡đ?‘&#x;đ?‘Žđ?‘Śđ?‘ 0.3895 đ?‘…đ?‘’đ?‘Žđ?‘™ đ?‘ đ?‘˘đ?‘šđ?‘?đ?‘’đ?‘&#x; đ?‘œđ?‘“ đ?‘‡đ?‘&#x;đ?‘Žđ?‘Śđ?‘ =

7.1.9 Column Diameter Before determining the column diameter a series of calculations need to be done. Overall mass balance (Equation 1) đ??ˇ + đ??ľ = 10762

đ?‘˜đ?‘”đ?‘šđ?‘œđ?‘™ â„Ž

Mass balance on methane (Equation 2) 0.6672đ??ˇ + 0.0001đ??ľ = 4805.9

đ?‘˜đ?‘”đ?‘šđ?‘œđ?‘™ â„Ž

Evaluating D using equation 1 and 2 đ?‘˜đ?‘”đ?‘šđ?‘œđ?‘™ â„Ž đ?‘˜đ?‘”đ?‘šđ?‘œđ?‘™ 0.0001đ??ˇ + 0.0001đ??ľ = 1.0762 â„Ž 4804.83238 đ??ˇ= 0.6671 đ?‘˜đ?‘”đ?‘šđ?‘œđ?‘™ đ??ˇ = 7202.55 â„Ž đ?‘˜đ?‘”đ?‘šđ?‘œđ?‘™ đ??ľ = 3559.45 â„Ž 0.6672đ??ˇ + 0.0001đ??ľ = 4805.9

Vapour rate calculations đ?‘‰ = đ??ˇ(1 + đ?‘…) đ?‘‰ = 7202.55(1 + 0.5) = 10803.8

đ?‘˜đ?‘”đ?‘šđ?‘œđ?‘™ â„Ž

Liquid flowrate above the feed đ??żđ?‘Žđ?‘?đ?‘œđ?‘Łđ?‘’ đ?‘Ąâ„Žđ?‘’ đ?‘“đ?‘’đ?‘’đ?‘‘ = đ?‘…đ??ˇ = 3601.3

đ?‘˜đ?‘”đ?‘šđ?‘œđ?‘™ â„Ž

Liquid flowrate below the feed 6

đ??żđ?‘?đ?‘’đ?‘™đ?‘œđ?‘¤ đ?‘Ąâ„Žđ?‘’ đ?‘“đ?‘’đ?‘’đ?‘‘ = đ?‘…đ??ˇ + đ??š = 14363.3

đ?‘˜đ?‘”đ?‘šđ?‘œđ?‘™ â„Ž

Column Diameter calculations đ??šđ??żđ?‘‰ =

đ??żđ?‘¤ đ?œŒđ?‘Ł √ đ?‘‰đ?‘¤ đ?œŒđ??ż

Lw = liquid mass flowrate, kg/s Vw = vapour mass flowrate, kg/s Ď v = density of vapour, kg/m3 Ď L = density of liquid, kg/m3 đ??šđ??żđ?‘‰đ?‘Ąđ?‘œđ?‘? =

3601.3 2.994 √ = 0.0264 10803.8 468.7

đ??šđ??żđ?‘‰đ?‘?đ?‘œđ?‘Ąđ?‘Ąđ?‘œđ?‘š =

14363.3 5.613 √ = 0.1498 10803.8 442.2

The FLV values is then used to determine the K values using the graphs below

Figure 4: Flooding velocity, seive plates (Sinnot and Towler, 2009)

7

Figure 5: Flooding velocity, seive plates (Sinnot and Towler, 2009)

đ?‘‡đ?‘œđ?‘? đ??ž1 = 0.14 đ??ľđ?‘œđ?‘Ąđ?‘Ąđ?‘œđ?‘š đ??ž1 = 0.12 Correction for surface tension đ??ž1 (đ?‘?đ?‘œđ?‘&#x;đ?‘&#x;đ?‘’đ?‘?đ?‘Ąđ?‘’đ?‘‘) = (

đ?œŽ 0.2 ) Ă— đ??ž1 0.02

0.00833 0.2 đ?‘‡đ?‘œđ?‘? đ??ž1 = ( ) Ă— 0.14 = 0.118 0.02 0.0.002925 0.2 đ??ľđ?‘œđ?‘Ąđ?‘Ąđ?‘œđ?‘š đ??ž1 = ( ) Ă— 0.12 = 0.0817 0.02 Flooding velocity is calculated using the following equation đ?œŒđ??ż − đ?œŒđ?‘Ł đ?‘˘đ?‘“ = đ??ž1 √ đ?œŒđ?‘Ł uf = flooding vapour verlocity, m/s, based on the net column cross sectional Area A n 468.7 − 2.994 đ?‘˘đ?‘“ (đ?‘‡đ?‘œđ?‘?) = 0.118√ = 1.479 2.994 442.2 − 5.631 đ?‘˘đ?‘“ (đ??ľđ?‘œđ?‘Ąđ?‘Ąđ?‘œđ?‘š) = 0.0817√ = 0.721 5.631 Design for 85% flooding at the maximum flowrate đ?‘‡đ?‘œđ?‘? đ?‘˘đ?‘› = 1.479 Ă— 0.85 = 1.257 đ??ľđ?‘œđ?‘Ąđ?‘Ąđ?‘œđ?‘š đ?‘˘đ?‘› = 0.721 Ă— 0.85 = 0.6124 Maximum volumetric flowrate đ?‘‡đ?‘œđ?‘? =

10803.8 Ă— 21.37 = 21.814 đ?‘š 3 /đ?‘ 2.994 Ă— 3600

8

đ??ľđ?‘œđ?‘Ąđ?‘Ąđ?‘œđ?‘š =

10803.8 Ă— 54.94 = 29.37 đ?‘š 3 /đ?‘ 5.631 Ă— 3600

Net Area required 21.814 = 17.35 đ?‘š 2 1.257 29.93 đ??ľđ?‘œđ?‘Ąđ?‘Ąđ?‘œđ?‘š = = 47.96 đ?‘š 2 0.6124 đ?‘‡đ?‘œđ?‘? =

Downcomer is then taken as 12% from the total area 17.35 = 19.718 đ?‘š 2 0.88 47.96 đ??ľđ?‘œđ?‘Ąđ?‘Ąđ?‘œđ?‘š = = 54.504 đ?‘š 2 0.88 đ?‘‡đ?‘œđ?‘? =

Column Diameter 19.718 Ă— 4 đ?‘‡đ?‘œđ?‘? đ??ˇđ?‘–đ?‘Žđ?‘šđ?‘’đ?‘Ąđ?‘’đ?‘&#x; = √ = 5.01 đ?‘š đ?œ‹ 54.504 Ă— 4 đ??ľđ?‘œđ?‘Ąđ?‘Ąđ?‘œđ?‘š đ??ˇđ?‘–đ?‘Žđ?‘šđ?‘’đ?‘Ąđ?‘’đ?‘&#x; = √ = 8.33 đ?‘š đ?œ‹

7.1.10 Liquid Flow Pattern đ?‘€đ?‘Žđ?‘Ľđ?‘–đ?‘šđ?‘˘đ?‘š đ?‘Łđ?‘œđ?‘™đ?‘˘đ?‘šđ?‘’đ?‘Ąđ?‘&#x;đ?‘–đ?‘? đ?‘™đ?‘–đ?‘žđ?‘˘đ?‘–đ?‘‘ đ?‘&#x;đ?‘Žđ?‘Ąđ?‘’ =

đ??żâ€˛ Ă— đ?‘€đ?‘Šđ?‘?đ?‘Žđ?‘ đ?‘’ 14363.3 Ă— 54.94 đ?‘š3 = = 0.4957 3600 Ă— đ?œŒđ??ż,đ?‘?đ?‘Žđ?‘ đ?‘’ 3600 Ă— 468.7 đ?‘

7.1.11 Provisional Plate design Column diameter = 8.33 metres Column Area = Ac = 54.5 m2 Downcomer Area đ??´đ?‘‘ = 0.12 Ă— 54.5 = 6.54 đ?‘š 2 Net Area đ??´đ?‘› = đ??´đ?‘? − đ??´đ?‘‘ = 54.5 − 6.54 = 47.96 đ?‘š 2 Active Area đ??´đ?‘Ž = đ??´đ?‘? − 2đ??´đ?‘‘ = 54.5 − (2 Ă— 6.54) = 41.423 đ?‘š 2 First trial for the hole area will be 10% đ??´â„Ž = 0.1 Ă— đ??´đ?‘Ž = 4.14 đ?‘š 2 Weir length

9

Figure 6 Relation between downcomer area and weir length (Sinnot and Towler, 2009)

đ?‘™đ?‘¤ = đ??ˇđ?‘? Ă— 0.74 = 8.33 Ă— 0.76 = 6.33 đ?‘šđ?‘’đ?‘Ąđ?‘&#x;đ?‘’đ?‘

Standard Sizes Taken from Sinnot and Towler Weir height Hole Diameter Plate thickness

50 mm 5 mm 5 mm

7.1.12 Weeping Maximum Liquid flowrate 14363.3 Ă— 54.94 đ?‘˜đ?‘” = 219.2 3600 đ?‘ đ?‘€đ?‘–đ?‘›đ?‘–đ?‘šđ?‘˘đ?‘š đ?‘™đ?‘–đ?‘žđ?‘˘đ?‘–đ?‘‘ đ?‘&#x;đ?‘Žđ?‘Ąđ?‘’, 70% đ?‘Ąđ?‘˘đ?‘&#x;đ?‘›đ?‘‘đ?‘œđ?‘¤đ?‘› = 0.7 Ă— 219.2 = 153.4 đ?‘€đ?‘Žđ?‘Ľđ?‘–đ?‘šđ?‘˘đ?‘š â„Žđ?‘œđ?‘¤

2

2

2

2

đ?‘˜đ?‘” đ?‘

3 3 đ??żđ?‘¤ 219.2 = 750 ( ) = 750 Ă— ( ) = 137.27 đ?‘šđ?‘š đ?‘™đ?‘–đ?‘žđ?‘˘đ?‘–đ?‘‘ đ?œŒđ?‘Ł Ă— đ?‘™đ?‘¤ 442.2 Ă— 6.33

3 3 đ??żđ?‘¤ 153.44 đ?‘€đ?‘–đ?‘›đ?‘–đ?‘šđ?‘˘đ?‘š â„Žđ?‘œđ?‘¤ = 750 ( ) = 750 Ă— ( ) = 108.22 đ?‘šđ?‘š đ?‘™đ?‘–đ?‘žđ?‘˘đ?‘–đ?‘‘ đ?œŒđ?‘Ł Ă— đ?‘™đ?‘¤ 442.2 Ă— 6.33 đ??´đ?‘Ą đ?‘šđ?‘–đ?‘›đ?‘–đ?‘šđ?‘˘đ?‘š đ?‘&#x;đ?‘Žđ?‘Ąđ?‘’ â„Žđ?‘¤ + â„Žđ?‘œđ?‘¤ = 50 + 108.22 = 158.22 đ?‘šđ?‘š đ?‘™đ?‘–đ?‘žđ?‘˘đ?‘–đ?‘‘

10

Using the graph on figure 5

Figure 7 Weep point correlation (Sinnot and Towler, 2009)

The value of the x axis is out of range so by approximating K2 is found to be 31.8

Next the minimum vapour velocity is calculated using the following equation �ℎ =

[đ??ž2 − 0.9(25.4 − đ?‘‘â„Ž )] 1 (đ?œŒđ?‘Ł )2

=

31.8 − 0.9(25.4 − 5) 1 5.631 ⠄2

đ??´đ?‘?đ?‘Ąđ?‘˘đ?‘Žđ?‘™ đ?‘šđ?‘–đ?‘›đ?‘–đ?‘šđ?‘˘đ?‘š đ?‘Łđ?‘Žđ?‘?đ?‘œđ?‘˘đ?‘&#x; đ?‘Łđ?‘’đ?‘™đ?‘œđ?‘?đ?‘–đ?‘Ąđ?‘Ś =

= 5.63

đ?‘š đ?‘

đ?‘šđ?‘–đ?‘›đ?‘–đ?‘šđ?‘˘đ?‘š đ?‘Łđ?‘Žđ?‘?đ?‘œđ?‘˘đ?‘&#x; đ?‘&#x;đ?‘Žđ?‘Ąđ?‘’ 0.7 Ă— 29.37 đ?‘š = = 4.963 đ??´â„Ž 4.14 đ?‘

The hole diameter is then reduced to be 3% of the Active Area, Aa, đ??´â„Ž = 0.03 Ă— 41.423 = 1.24 đ?‘š 2 0.7 Ă— 29.37 đ?‘š đ?‘ đ?‘’đ?‘¤ đ?‘Łđ?‘Žđ?‘?đ?‘œđ?‘˘đ?‘&#x; đ?‘Łđ?‘’đ?‘™đ?‘?đ?‘œđ?‘–đ?‘Ąđ?‘Ś = = 16.55 1.24 đ?‘

7.1.13 Plate Pressure Drop Dry plate pressure drop Maximum vapour velocity through the holes is calculated using the following equation: �ℎ (maximum) =

29.37 = 23.64 đ?‘š/đ?‘ 1.24

Pressure drop through the dry plate can be estimated by using the following equation đ?‘˘â„Ž 2 đ?œŒđ?‘Ł â„Žđ?‘‘ = 51 [ ] đ??ś0 đ?œŒđ??ż

11

C0 = orifice coefficient. This is obtained from the graph in figure 6

Figure 8: Discharge coefficients, sieve plates (Liebson et al., 1957)

23.64 2 5.631 ] = 594.5 đ?‘šđ?‘š 0.78 442.2 The residual head drop is calculated using the following equation â„Žđ?‘‘ = 51 [

â„Žđ?‘&#x; =

12.5 Ă— 103 12.5 Ă— 103 = = 28.27 đ?œŒđ??ż 442.2

Total pressure drop â„Žđ?‘Ą = â„Žđ?‘‘ + (â„Žđ?‘¤ + â„Žđ?‘œđ?‘¤ ) + â„Žđ?‘&#x; = 594.5 + 28.27 + 50 + 108.22 = 781.01 đ?‘šđ?‘š Downcomer pressure loss To calculate the downcomer pressure loss the hap needs to be calculated as follows â„Žđ?‘Žđ?‘? = â„Žđ?‘¤ − 10 = 40 đ?‘šđ?‘š đ??´đ?‘?đ?‘&#x;đ?‘œđ?‘› đ?‘Žđ?‘&#x;đ?‘’đ?‘Ž = đ??´đ?‘Žđ?‘? = 6.33 Ă— 40 Ă— 10−3 = 0.253 đ?‘š 2 If Aap < Ad then downcomer pressure can be calculated using the following formula 2 đ??żđ?‘¤đ?‘‘ 2 219.2 ] = 166 [ ] = 63.61 đ?‘šđ?‘š â„Žđ?‘‘đ?‘? = 166 [ đ?œŒđ??ż đ??´đ?‘š 442.2 Ă— 0.253 Am = Either Ad or Aap (Whichever is smaller)

Back-up in downcomer â„Žđ?‘? = 50 + 63.61 + 33.54 + 12.58 = 151.02 đ?‘šđ?‘š Since 151.02mm < 0.5 (plate spacing + weir height)

12

đ?‘Ąđ?‘&#x; =

đ??´đ?‘‘ â„Žđ?‘?đ?‘? đ?œŒđ??ż 6.54 Ă— 151.02 Ă— 10−3 Ă— 442.2 = = 20 đ?‘ đ?‘’đ?‘?đ?‘œđ?‘›đ?‘‘đ?‘ đ??żđ?‘¤đ?‘“ 219.2

7.1.14 Check Entrainment �� =

29.37 đ?‘š = 0.6124 47.96 đ?‘

đ?‘?đ?‘’đ?‘&#x;đ?‘?đ?‘’đ?‘›đ?‘Ąđ?‘Žđ?‘”đ?‘’ đ?‘œđ?‘“ đ?‘“đ?‘™đ?‘œđ?‘œđ?‘‘đ?‘–đ?‘›đ?‘” =

0.6124 = 85% 0.72

Using the bottom FLV to find Ďˆ from the graph below Ψ = 0.02

Figure 9: Entrainment correlation for sieve plates (Fair, 1961)

13

7.1.15 Perforated Area From the graph shown below Lh/Dc is 0.8 and θ=0.99o

Figure 10: Relation between angle subtended by chord, chord height and chord length (Sinnot and Towler, 2009)

The angle subtended by the edge of the plate = 180 - 99 = 81o Mean length unperforated edge strips = (8.33 – 50 × 10-3)π × 81/180 = 11.7 m Area of the unperforated edge strips = 50 × 10-3 × 11.7 = 0.585 m2 Mean length of the calming zone, approximately = weir length + width of the unperforated strip = 6.33 + 0.05 = 6.38 m Area of the calming zone = 2(6.38 × 0.05) = 0.638 m2 Total area of perforations Ap = 41.42 – 0.585 – 0.638 = 39.78 m2 Ah/Ap = 4.14/39.78 = 0.104. Using the graph in figure 9 lp/dh = 2.85.

14

Figure 11: Relation between hole area and pitch (Sinnot and Towler, 2009)

7.1.16 Number of Holes Area of one hole = 1.964 × 10-5

8.33 m

6.33 m

Number of holes = Ah/1.964 × 10-5 = 4.142/1.964 × 10-5=210,896

Figure 12: Plate layout

15

7.1.17 Column Height The height of the column is determined by simple equation as follows 𝐻𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝐶𝑜𝑙𝑢𝑚𝑛 = (𝑁𝑜. 𝑆𝑡𝑎𝑔𝑒𝑠 − 1)(𝑡𝑟𝑎𝑦 𝑠𝑝𝑎𝑐𝑖𝑛𝑔) + (𝑡𝑟𝑎𝑦 𝑠𝑝𝑎𝑐𝑖𝑛𝑔 × 2) + (𝑁𝑜. 𝑆𝑡𝑎𝑔𝑒𝑠 − 1)(𝑡ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠𝑜𝑓 𝑝𝑙𝑎𝑡𝑒𝑠) 𝐻𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑐𝑜𝑙𝑢𝑚𝑛 = (28 − 1)(0.9) + (0.9 × 2) + (28 − 1)(0.05)

𝐻𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝐶𝑜𝑙𝑢𝑚𝑛 = 27.45 𝑚

16

7.2.0 Operational Design 7.2.1 Control System Objectives Following the process design, the operational design should be carried out. In this chapter a detailed piping instrumentation diagram is constructed to identify the all the controls needed to operate a demethanizer column in the most effective and safe way. Various controllers and transmitters were used in this column which can be seen from the PI&D in figure 11. The inlet feed will have a power valve that will be controlled by a FC (Flow Controller) based on the signal transmitted by the FT (Flow Transmitter). This is to ensure that the feed rate is maintained and any abnormalities occurring upstream can be dealt with. Along the column a DPT (Difference Pressure Transmitter) is installed, this is to ensure that the pressure in the column is maintained and doesn’t get out of range specified (21 bar -23 bar). At the bottom of the column an LT (Level Transmitter) is installed to detect the liquid accumulated at the bottom of the column. This transmitter communicates with the level controller which in turn connected with the flow controller in the condensate stream. The more liquid is accumulated the more the flow controller at the condensate stream will open up. Given that, flow transmitter installed just before monitors the flow and keeps the operator informed of the amount of condensate coming out from the reboiler. On the right hand of the column a TT (Temperature Transmitter) is installed, which is connected to a TC (Temperature Controller) and the temperature controller is connected to the flow controller of the steam stream. This is the most important part of the column. The temperature is measured inside the column and it sends a signal to the steam stream to set up the temperature at the reboiler. The flow transmitter in the steam stream monitors the amount of steam entering the reboiler. At the top of the column the vapour will enter the condenser and the reflux drum and whatever condensates will be sent back to the column as reflux. There is a power control valve in the reflux stream with a flow controller to control the flow and a flow transmitter to monitor the flow. There is also an AT (Analyser Transmitter) that analyses the liquid composition at the top of the column and transmits an electric signal to the AC (Composition Analyser) to control the condenser temperature range. The AT also transmits to the flow controller in the reflux stream. On the top of the condenser there is a coolant stream entering the condenser and it has the typical flow controller and the flow transmitter to monitor the feed of the coolant entering the condenser. Finally in the distillate stream there are also a flow controller and transmitter to monitor the flow which in this case is the flow of the sales gas.

FC FT LC LT AT AC TT TC

Flow Controller Flow Transmitter Level Controller Level Transmitter Analyser Transmitter Component Analyser Temperature Transmitter Temperature Controller

17

7.2.2 P&ID

Figure 13: PI&D of the demethanizer V-501

18

7.2.3 Operating Procedures 7.2.3.1 Start-up and Shut-down This may be considered to be one of the most important parts of designing any equipment. There are few issues that can be encountered and therefore before starting to pump in the process fluids. There are many ways to plan a start-up and shut-down procedures. The most common procedures are as follows (Kister, 1990): 1) Line Blowing: This considered to be a common prestart-up practice that allows to get rid of any constructions debris. Air or nitrogen is blown into the column at relatively high pressures. However creating a large pressure that is above the design value can cause damage to the trays by deforming them or even breaking them. 2) Pressurising and Depressurising: This practice is performed in both start-ups and shut-downs to ensure that there are no leaks in the column. It is also used to remove the inert gas or air before the start-up and is used to remove any residual process gas in shut-downs. 3) Purging: V-501 is a distillation column that separates the light and heavy hydrocarbons. Both are considered to be flammable liquids. Purging is used in both start-up and shut-down to ensure that there is no air in the column that can cause fire. 4) Leak Testing: In this procedure all drains and vents are closed, the column is pressurised with inert gas (normally nitrogen). Then soap solution is applied at each flange, vent, drain, joint or valve to check for leaks. Other devices like ultrasonic leak detector are also used. 5) Washing: Distillation columns are normally washed prior to start up. One of the reasons is to remove and scale, mud or any construction leftovers. It is also used as method to cool the column. Moreover to coat internal with inhibitive paints that can prevent corrosion. 6) Solvent Testing: Usually just before commissioning the plant, the column is injected with a solvent that has similar properties to the process gas intended to be separated. The reason being is check the performance of the distillation column. The operations mentioned above may initially seem costly to perform but it is how most industrial companies do before they start the commercial production. It essentially troubleshoot the column before facing the actual problem and ensure that the column will deliver its optimum performance. Based on the performance reports the engineers will be able to foresee the difficulties that the column may encounter in the lifetime of the plant, and in turn they can provide the solutions. It is important to note that these operations are used for troubleshooting and may seem harmless, but poor planning and executing of these operations may cause serious accidents and damage to the column. This is specially the case for columns with trays such as this one.

7.2.3.2 Maintenance Every equipment in a process must undergo regular inspection and maintenance. Distillation columns are one of the most expensive piece of equipment in any plant. It is very costly to operate since it uses a massive amount of energy. Therefore it is crucial that it is maintained and kept in its best shape throughout the lifespan of the plant. Distillation column V-501 has a small percentage of carbon dioxide in the feed stream, which is considered to be acidic gas that may cause corrosion. Careful inspection of corrosion must be made when maintenance is due. All trays, caps joints must be checked for any corrosion traces. Another aspect that needs to be examined is the physical conditions of the trays. Since V-501 is operating at high pressures, fluctuations of pressure may damage the trays, flanges or joints. Moreover, leak test should be performed ensure that there are no leakage of process gases. Leakage of these flammable gases can cause fatal accidents and disruption in the production

19

7.2.4 Safety Study Table 2: HAZOP analysis for V-501

Parameters

Guideword

Possible Cause (s) Deviation  Stop of production  Upstream pumps failure

Consequence  Loss of production and profit  Damage of downstream equipment

Safeguard (Existing)  Flow Transmitter at the feed

1

Flow

No

2

Flow

Less

 Upstream pump failure  Blockage due to scaling  Leakage

 Loss of production and profit  Release of toxic hydrocarbons  Fire/explosion

 Flow transmitter at the feed

3

Flow

More

 Production increase

 Increase in pressure.  Damage in the column e.g. deformation of trays  Column shell rupture  Release of toxic hydrocarbons

 Flow transmitter at the feed

Recommendations

Action (By)

 Immediate action should be taken to ensure the shutdown of the downstream equipment to avoid damage  Immediate action should be taken to ensure the shutdown of the downstream equipment to avoid damage.  Regular maintenance and check for leaks  Use the flow controller to control the amount of feed entering the column.

 Control engineer/operator

 Control engineer/operator  Maintenance engineer

 Control engineer/operator

20

 Fire/explosion  Loss of production  Off spec product  Damage of joints upstream

4

Pressure

Low

 Production decrease

 Difference pressure transmitter installed along the column  Flow transmitter at the feed  Flow transmitter at the feed stream  Pressure difference transmitter along the column

 Regular maintenance of the joints and pipelines

 Maintenance engineer

5

Pressure

High

 Increase in production  Clogging and scaling in pipelines in up

 Rupture of column shell  Damage of trays  Loss of production  Release of toxic hydrocarbons  Fire/explosion

 Shut-down procedure must be done as soon as there is a dramatic pressure increase.

 Control engineer/ operator

6

Temperature

high

 Change in atmospheric conditions (summer)

 Off spec product

 Temperature transmitter and temperature controller installed at the column

 Control engineer/ operator

 High recovery of heavy hydrocarbons in the

 Level transmitter at the bottom of the column

 Temperature controller is connected with the flow controller at the steam stream that goes to the reboiler.  Transmitter at the bottom of the column sends signal to the flow controller in the distillate stream to allow higher flows.

7

Level

High

 Change in the upstream compositions

 Control engineer /operator

21

7.3.0 Mechanical Design This section will cover the mechanical design of the distillation column. A detailed design that will include the thickness of the shell, type of head, thickness of the head, Stress analysis‌etc. Moreover wind loading has also been taken into consideration. Mechanical design is an important step to consider before the manufacturing of the equipment. After process design, the operating conditions of the column are known and as per the requirement the mechanical design will have to operate at the specified conditions. The material of construction of the distillation column shell will be carbon steel, which is the most commonly used material in the process industry due to its high strength and resistance to corrosion.

7.3.1 Design Pressure Column Length Column Diameter Number of stages

27.45 8.33 27

The pressure of the column was obtained from the simulation HYSYS and it was 23 bar. The pressure needs to be converted to N/mm2. The pressure also needs to be 10% above the gauge pressure. = (23 -1) Ă— 1.1 = 24.2 bar = 2.42 N/mm2

7.3.2 Cylindrical Section In this section the thickness of the shell will be calculated using the following equation đ?‘Ą=

đ?‘ƒđ?‘– đ??ˇđ?‘– 2đ?‘†đ??¸ − 1.2đ?‘ƒđ?‘–

t = thickness of the shell in mm Pi = pressure of the vessel in N/mm2 Di = diameter of the vessel in mm S = maximum allowable stress of carbon Steel at -64 degrees Celsius in N/mm2 E = weld joint efficiency đ?‘Ą=

2.42 Ă— 8.33 Ă— 103 = 106 đ?‘šđ?‘š (2 Ă— 96.5 Ă— 1) − (1.2 Ă— 2.42)

Then 2 mm is added for corrosion allowance = 106 + 2 = 108 mm

7.3.3 Type of Head (Domed) There are three types of domed heads; torisphere, standard ellipsoidal and hemispherical. This section will include the calculations of the thickness of the head that will be chosen. The smallest thickness will be most favourable because it will be the most economical since it will need the least amount of material to construct it.

22

7.3.3.1 Standard Dished Head Torisphere đ?‘Ą=

0.885đ?‘ƒđ?‘– đ?‘…đ?‘? đ?‘†đ??¸ − 0.1đ?‘ƒđ?‘–

Rc = Crown radius = Di đ?‘Ą=

0.885 Ă— 2.42 Ă— 8.33 Ă— 103 = 185 đ?‘šđ?‘š (96.5 Ă— 1) − (0.1 Ă— 2.42)

7.3.3.2 Standard Ellipsoidal Head

đ?‘ƒđ?‘– đ??ˇđ?‘– 2đ?‘†đ??¸ − 0.2đ?‘ƒđ?‘– 2.42 Ă— 8.33 Ă— 103 đ?‘Ą= = 105 đ?‘šđ?‘š 2(96.5 Ă— 1) − 0.2(2.42) đ?‘Ą=

7.3.3.3 Hemispherical Head

đ?‘ƒđ?‘– đ??ˇđ?‘– 4đ?‘†đ??¸ − 0.4đ?‘ƒđ?‘– 2.42 Ă— 8.33 Ă— 103 đ?‘Ą= = 52 đ?‘š 4(96.5 Ă— 1) − 0.4(2.42) đ?‘Ą=

According the calculations above the hemispherical head has the smallest thickness it is the most economical way of constructing the head of the column.

7.3.4 Dead Weight In this section the weight of the shell, weight of the plates inside the column and the weight of the insulation will be calculated.

7.3.4.1 Weight of the Vessel First the shell weight is calculated using this equation đ?‘Šđ?‘Ł = 240đ??śđ?‘¤ đ??ˇđ?‘š (đ??ťđ?‘Ł + 0.8đ??ˇđ?‘š )đ?‘Ą Wv = Total weight of the shell excluding internal fittings, such as plates in N Cw = factor to account for the weight of the nozzles, manways, internal supports, etc taken as 1.15 for distillation columns Dm = mean diameter of the vessel = (Di + t Ă— 10-3) in m Hv = height (length of the shell) in m đ?‘Šđ?‘Ł = 240 Ă— 1.15 Ă— (8.33 + 108 Ă— 10−3 )(8.33 + 0.8(8.33 + 108 Ă— 10−3 )) Ă— 108 Ă— 10−3 = 8605483 đ?‘

7.3.4.1 Weight of the Plates To calculate the weight of the plates the area of one plate must be calculated đ?œ‹ Ă— 8.332 = 54.5 đ?‘š 2 4 đ?‘Šđ?‘’đ?‘–đ?‘”â„Žđ?‘Ą đ?‘œđ?‘“ đ?‘œđ?‘›đ?‘’ đ?‘?đ?‘™đ?‘Žđ?‘Ąđ?‘’ = 1.2 Ă— 54.5 Ă— 103 = 65426 đ?‘ đ?‘Šđ?‘’đ?‘–đ?‘”â„Žđ?‘Ą đ?‘œđ?‘“ đ?‘Žđ?‘™đ?‘™ đ?‘?đ?‘™đ?‘Žđ?‘Ąđ?‘’đ?‘ = 65426 Ă— 27 = 1766489 đ?‘ đ??´đ?‘&#x;đ?‘’đ?‘Ž đ?‘œđ?‘“ đ?‘œđ?‘›đ?‘’ đ?‘?đ?‘™đ?‘Žđ?‘Ąđ?‘’ =

23

7.3.4.2 Weight of Insulation The insulation used for this column is mineral wool. For distillation columns with operating temperatures between -150 to 150 oF mineral wool is the most suitable type of insulation (PetroChemical, 2011). đ?‘˜đ?‘” đ?‘€đ?‘–đ?‘›đ?‘’đ?‘&#x;đ?‘Žđ?‘™ đ?‘¤đ?‘œđ?‘œđ?‘™ đ?‘‘đ?‘’đ?‘›đ?‘ đ?‘–đ?‘Ąđ?‘Ś = 130 3 đ?‘š đ?œ‹ đ??´đ?‘?đ?‘?đ?‘&#x;đ?‘œđ?‘Ľđ?‘–đ?‘šđ?‘Žđ?‘Ąđ?‘’ đ?‘Łđ?‘œđ?‘™đ?‘˘đ?‘šđ?‘’ đ?‘œđ?‘“ đ?‘–đ?‘›đ?‘ đ?‘˘đ?‘™đ?‘Žđ?‘Ąđ?‘–đ?‘œđ?‘› = Ă— 8.332 Ă— 27.45 Ă— 75 Ă— 10−3 = 112 đ?‘š 3 4 đ?‘Šđ?‘’đ?‘–đ?‘”â„Žđ?‘Ą đ?‘œđ?‘“ đ?‘–đ?‘›đ?‘ đ?‘˘đ?‘™đ?‘Žđ?‘Ąđ?‘–đ?‘œđ?‘› = 112 Ă— 130 Ă— 9.81 = 165117 đ?‘ The weight of the insulation is then doubled to count for the fittings đ?‘Šđ?‘’đ?‘–đ?‘”â„Žđ?‘Ą đ?‘œđ?‘“ đ?‘–đ?‘›đ?‘ đ?‘˘đ?‘™đ?‘Žđ?‘Ąđ?‘–đ?‘œđ?‘› đ?‘¤đ?‘–đ?‘Ąâ„Ž đ?‘“đ?‘–đ?‘Ąđ?‘Ąđ?‘–đ?‘›đ?‘”đ?‘ = 165117 Ă— 2 = 330234 Total weight = 10702206 N = 10.7 MN

7.3.5 Wind Loading Wind loading is only considered for long vertical columns same as the one present in this project. This kind of vessel acts like a cantilever beam undergoing a wind load. Taking the dynamic pressure as 1280 N/m2 which corresponds to 160 kph of wind speed đ?‘€đ?‘’đ?‘Žđ?‘› đ?‘‘đ?‘–đ?‘Žđ?‘šđ?‘’đ?‘Ąđ?‘’đ?‘&#x; đ?‘–đ?‘›đ?‘?đ?‘™đ?‘˘đ?‘‘đ?‘–đ?‘›đ?‘” đ?‘–đ?‘›đ?‘ đ?‘˘đ?‘™đ?‘Žđ?‘Ąđ?‘–đ?‘œđ?‘› = 8.33 + 2(0.108 + 0.075) = 8.7 đ?‘š đ?‘ đ?‘Šđ?‘–đ?‘›đ?‘‘ đ?‘™đ?‘œđ?‘Žđ?‘‘đ?‘–đ?‘›đ?‘” (đ?‘?đ?‘’đ?‘&#x; đ?‘™đ?‘–đ?‘›đ?‘’đ?‘Žđ?‘&#x; đ?‘šđ?‘’đ?‘Ąđ?‘&#x;đ?‘’ ) = đ?‘Š = đ?‘ƒđ?‘¤ đ??ˇđ?‘’đ?‘“đ?‘“ = 1280 Ă— 8.7 = 11131 đ?‘š đ?‘Šđ?‘Ľ 2 11131 Ă— 272 đ??ľđ?‘’đ?‘›đ?‘‘đ?‘–đ?‘›đ?‘” đ?‘šđ?‘œđ?‘šđ?‘’đ?‘›đ?‘Ą = đ?‘€đ?‘Ľ = = = 4193576 đ?‘ đ?‘š 2 2

7.3.6 Stress Analysis At the bottom tangent line The longitudinal circumferential stresses are to be calculated using the following equations đ?œŽđ??ż =

đ?‘ƒđ??ˇđ?‘– 2.42 Ă— 8330 = = 46.64 đ?‘ /đ?‘šđ?‘š 2 4đ?‘Ą 4 Ă— 108

đ?œŽâ„Ž =

đ?‘ƒđ??ˇđ?‘– 2.42 Ă— 8330 = = 93.29 đ?‘ /đ?‘šđ?‘š 2 2đ?‘Ą 2 Ă— 108

7.3.6.1 Dead weight stress Dead weight stress is the compressive stress that occurs due to the weight of the column and it can be calculated using the following equation đ?œŽđ?‘¤ =

đ?‘Šđ?‘§ 1766489 đ?‘ = = 3.73 (đ?‘?đ?‘œđ?‘šđ?‘?đ?‘&#x;đ?‘’đ?‘ đ?‘ đ?‘–đ?‘Łđ?‘’) đ?œ‹(đ??ˇđ?‘– + đ?‘Ą)đ?‘Ą đ?œ‹(8330 + 108) Ă— 108 đ?‘šđ?‘š 2

7.3.6.2 Bending stresses đ??ˇđ?‘œ = 8330 + 2 Ă— 108 = 8546 đ?‘šđ?‘š đ?œ‹ đ?œ‹ 4 (85464 − 83304 ) = 2.55 Ă— 1013 đ?‘šđ?‘š 4 (đ??ˇ0 − đ??ˇđ?‘–4 ) = đ??źđ?‘Ł = 64 64

24

đ?œŽđ?‘? = Âą

đ?‘€ đ??ˇđ?‘– 4193576 Ă— 1000 8330 đ?‘ ( + đ?‘Ą) = Âą ( + 108) = 0.703 13 đ??źđ?‘Ł 2 2.55 Ă— 10 2 đ?‘šđ?‘š 2

7.3.6.3 The resultant longitudinal stress

đ?‘ đ?‘šđ?‘š 2 đ?‘ đ?œŽđ?‘§ (đ?‘‘đ?‘œđ?‘¤đ?‘›đ?‘¤đ?‘–đ?‘›đ?‘‘) = đ?œŽđ??ż + đ?œŽđ?‘¤ Âą đ?œŽđ?‘? = 46.64 + 3.73 − 0.703 = 42.2 đ?‘šđ?‘š 2 Biggest difference between the stresses will be on the upwind side đ?œŽđ?‘§ (đ?‘˘đ?‘?đ?‘¤đ?‘–đ?‘›đ?‘‘ ) = đ?œŽđ??ż + đ?œŽđ?‘¤ Âą đ?œŽđ?‘? = 46.64 + 3.73 + 0.703 = 43.6

93.29 − 43.6 = 49.7

đ?‘ đ?‘šđ?‘š 2

Check Buckling To calculate the critical buckling stress the following equation is used đ?œŽđ?‘? = 2 Ă— 104 (

đ?‘Ą 108 đ?‘ ) = 2 Ă— 104 ( ) = 252.9 đ??ˇ0 8546 đ?‘šđ?‘š 2

đ?‘€đ?‘Žđ?‘Ľđ?‘–đ?‘šđ?‘˘đ?‘š đ?‘?đ?‘œđ?‘šđ?‘?đ?‘&#x;đ?‘’đ?‘ đ?‘ đ?‘–đ?‘Łđ?‘’ đ?‘ đ?‘Ąđ?‘&#x;đ?‘’đ?‘ đ?‘ = 0.703 + 3.73 = 4.43

đ?‘ đ?‘šđ?‘š 2

Compressive stress is way lower than the critical buckling stress therefore the design is considered satisfactory and safe.

7.3.7 Supporting Skirt Design The supporting skirt is a cylindrical shell that is welded to the base of the column. This support can come in many shapes and configurations. In this project the skirt support is chosen and calculations will be made to understand what is the minimum thickness required to handle the weight of the column. The skirt height is taken as 3 metres which will be added to the height of the vessel in the calculations The maximum dead weight is approximately calculated by filling the whole vessel with water. đ?œ‹ đ?‘Šđ?‘’đ?‘–đ?‘”â„Žđ?‘Ą đ?‘œđ?‘“ đ?‘Łđ?‘’đ?‘ đ?‘ đ?‘’đ?‘™ đ?‘“đ?‘–đ?‘™đ?‘™đ?‘’đ?‘‘ đ?‘¤đ?‘–đ?‘Ąâ„Ž đ?‘¤đ?‘Žđ?‘Ąđ?‘’đ?‘&#x; = ( Ă— 8.332 Ă— 27.45) Ă— 1000 Ă— 9.81 = 14677061 đ?‘ 4 đ?‘Šđ?‘’đ?‘–đ?‘”â„Žđ?‘Ą đ?‘œđ?‘“ đ?‘Ąâ„Žđ?‘’ đ?‘Łđ?‘’đ?‘ đ?‘ đ?‘’đ?‘™ = 10702206 đ?‘ đ?‘‡đ?‘œđ?‘Ąđ?‘Žđ?‘™ đ?‘¤đ?‘’đ?‘–đ?‘”â„Žđ?‘Ą = 14677061 + 10702206 = 2.54 Ă— 107 đ?‘ 30.452 đ??ľđ?‘’đ?‘›đ?‘‘đ?‘–đ?‘›đ?‘” đ?‘šđ?‘œđ?‘šđ?‘’đ?‘›đ?‘Ą đ?‘Žđ?‘Ą đ?‘?đ?‘Žđ?‘ đ?‘’ đ?‘œđ?‘“ đ?‘Ąâ„Žđ?‘’ đ?‘ đ?‘˜đ?‘–đ?‘&#x;đ?‘Ą = 11131 Ă— = 5160293 đ?‘ đ?‘š 2 The thickness of the skirt is taken as the same thickness of the vessel which is 108 mm. First the bending stress (Ďƒbs) is calculated using the following equation đ?œŽđ?‘?đ?‘ =

4đ?‘€đ?‘ 4 Ă— 1000 Ă— 5160293 đ?‘ = = 0.865 đ?œ‹(đ??ˇđ?‘ + đ?‘Ąđ?‘ đ?‘˜ )đ?‘Ąđ?‘ đ?‘˜ đ??ˇđ?‘ đ?œ‹(8330 + 108) Ă— 108 Ă— 8330 đ?‘šđ?‘š 2

Ďƒbs = bending stress in N/mm2 Ms= bending moment at base of the skirt a in N/mm2 Ds= skirt diameter in mm tsk = skirt thickness in mm

25

The dead weight in the skirt Ďƒbs is then calculated using the following equation đ?‘Šđ?‘Ł 2.54 Ă— 107 đ?‘ = = 0.0011 đ?œ‹(đ??ˇđ?‘ + đ?‘Ąđ?‘ đ?‘˜ )đ?‘Ąđ?‘ đ?‘˜ đ??ˇđ?‘ đ?œ‹ (8330 + 108) Ă— 108 Ă— 8330 đ?‘šđ?‘š 2 7 đ?‘Šđ?‘Ł 10.7 Ă— 10 đ?‘ đ?œŽđ?‘¤đ?‘ (đ?‘œđ?‘?đ?‘’đ?‘&#x;đ?‘Žđ?‘Ąđ?‘–đ?‘›đ?‘”) = = = 0.00045 đ?œ‹(đ??ˇđ?‘ + đ?‘Ąđ?‘ đ?‘˜ )đ?‘Ąđ?‘ đ?‘˜ đ??ˇđ?‘ đ?œ‹(8330 + 108) Ă— 108 Ă— 8330 đ?‘šđ?‘š 2 đ?œŽđ?‘¤đ?‘ (đ?‘Ąđ?‘’đ?‘ đ?‘Ą) =

The test condition is calculated for the vessel when it is filled with water this is normally called the hydraulic test. The skirt has to meet two conditions for it to be safe and ready to operate. These two conditions are as follows đ?‘Ąđ?‘ đ?‘˜ ) đ?‘ đ?‘–đ?‘›đ?œƒđ?‘ đ??ˇđ?‘ đ?œŽđ?‘ (đ?‘Ąđ?‘’đ?‘›đ?‘ đ?‘–đ?‘™đ?‘’) < đ?‘†đ?‘ đ??¸đ?‘ đ?‘–đ?‘› đ?œƒđ?‘

đ?œŽđ?‘ (đ?‘?đ?‘œđ?‘šđ?‘?đ?‘&#x;đ?‘’đ?‘ đ?‘ đ?‘–đ?‘Łđ?‘’) < 0.125đ??¸đ?‘Œ (

EY = Young’s Modulus which is taken as 200,000 N/mm2 Ss = Maximum allowable design stress in N/mm2 θs = base angle of the conical skirt taken as 90 o

đ?‘ đ?‘šđ?‘š 2 đ?‘ đ?‘€đ?‘Žđ?‘Ľđ?‘–đ?‘šđ?‘˘đ?‘š đ?œŽđ?‘ (đ?‘Ąđ?‘’đ?‘›đ?‘ đ?‘–đ?‘™đ?‘’) = 0.865 − 0.00045 = 0.864623 đ?‘šđ?‘š 2 đ?‘Ąđ?‘ đ?‘˜ 108 0.125đ??¸đ?‘Œ ( ) đ?‘ đ?‘–đ?‘›đ?œƒđ?‘ = 0.125 Ă— 200000 Ă— Ă— sin(90) = 324.3 đ??ˇđ?‘ 8330 đ?‘†đ?‘ đ??¸đ?‘ đ?‘–đ?‘› đ?œƒđ?‘ = 96.5 Ă— 1 Ă— sin(90) = 82.025 0.866135 < 324.3 0.86423 < 82.025 đ?‘€đ?‘Žđ?‘Ľđ?‘–đ?‘šđ?‘˘đ?‘š đ?œŽđ?‘ (đ?‘?đ?‘œđ?‘šđ?‘?đ?‘&#x;đ?‘’đ?‘ đ?‘ đ?‘–đ?‘Łđ?‘’ ) = 0.865 + 0.0011 = 0.866135

The maximum tensile and the compressive stresses are way less than the criteria specified which makes the specification

7.3.8 Pressure Relief For the pressure relief valve calculation is adopted from Whitesides booklet (Whitesides, 2012)using the following equation Condenser pressure relief system đ??´=

đ?‘Šâˆšđ?‘‡đ?‘? đ??śđ??žđ?‘ƒ1 đ??žđ?‘? √đ?‘€

= 11.14 đ?‘–đ?‘›2

The closest size designation will be R at 16 in2 A = orifice area in in2 W = Flow, lb/hr T = Temperature in oR 26

Z = compressibility factor C = Flow constant determined by the ratio of specific heats K = Coefficient of discharge P1 = upstream pressure in psia Kb = Correction factor due to back pressure M = molecular weight Reboiler pressure relief system For this part a different equation is used since the feed is in the liquid phase. The following equation is used đ??´=

đ?‘„ √đ??ş 27.2đ??žđ?‘? đ??žđ?‘¤ đ??žđ?‘Ł √∆đ?‘ƒ

= 1.78 đ?‘–đ?‘›2

The closest size designation will be K at 1.8 in2. It can be seen that the size of the condenser orifice is quite large. This is due to the high mass flow flowing as distillate from the demethanizer column. This makes it a subject to high pressures. Q = Flow in gpm Kp = Correction factor for relieving capacity vs lift for relief valves in liquid phase Kw = Correction factor due to back pressure Kv = Correction factor for viscosity ΔP = Differential Pressure in psia

27

7.3.9 Mechanical Drawing

Figure 14: Mechanical Drawing

28

7.4.0 Summary and Review 7.4.1 Equipment Specifications / data sheet Company Name

Plant location: Longford, Victoria, Australia Equipment to be designed: Demethanizer Distillation Column Equipment label: V-501 Process Data Operating temperature Pressure Density Top Section Mass flowrate Liquid Surface Tension Liquid Dynamic Viscosity Operating temperature Pressure Density Bottom Section Mass flowrate Liquid Surface Tension Liquid Dynamic Viscosity Construction and Material Shell Material Shell length Shell Thickness Type of head Thickness of head Number of Trays Pipe material Feed tray Tray spacing Weir height Weir length Area of hole Number of holes Tray thickness

Project Name: Big Gas Project

-49.15 oC 21 bar 2.994 kg/m3 1.539 × 103 kg/h 8.33 dyne/cm 0.009151 cP 95.32 oC 23 bar 442.2 kg/m3 1.955 × 103 kg/h 2.925 dyne/cm 0.07596 cP Carbon Steel 27.45 m 108 × 10-3 m Hemispherical 52 × 10-3 m 27 (without reboiler) Stainless steel 6th stage 0.9 m 50 × 10-3 m 6.33 m 1.964 × 10-5 m 210,896 5 × 10-3 m

29

7.4.2 Critical Review Looking back at when we started the design project, with absolutely no clue of the magnitude of this task I can say that I have learnt a lot in this project. Applying all what is learnt in the past 4 years is most interesting. The simulations on HYSYS made the job so much easier, I can only imagine how hard and time consuming it could have been in the past. The good points in the project is that it is focused and straight to the point with no side tracking. It could have been in more details if the time allocated to do it was longer. In terms of limitations, lack of actual field experience can be clearly noticed when reading this report. Field experience can be a huge advantage while doing this project. It can save time and it can help with obtaining better assumptions. One example can be taken is the start-up and shutdown procedures. Prior to researching the topic my previous impression was after constructing the tower they just start the operations. Optimisation is almost always possible. Optimisation can be done in terms of the recovery factors or even the economic aspect. There is always room for improvements, but the time allocated can be tight and providing the optimum design do take a lot of time. Further simulations can be used to achieve a better separation for the light and heavy hydrocarbons. However this will need more time and maybe advanced skills in HYSYS. Different simulation programs can also be used, this will give the possibility of comparing and finding which one is more suitable and reliable as a design. The things I would do differently are rely more on books rather than rushing to journals and articles in the internet. Books are the best sources of information. It is more focused on the topic and does not side track.

30

Nomenclature ¾a = the molar average liquid viscosity, mNs/m2 Aa = Active Area m2 Ad = Downomer Area m2 Ah = hole area in m2 Cw = factor to account for the weight of the nozzles, manways, internal supports, etc taken as 1.15 for distillation columns Di = diameter of the vessel in mm Dm = mean diameter of the vessel in m Ds= skirt diameter in mm E = weld joint efficiency EY = Young’s Modulus which is taken as 200,000 N/mm2 FC = flow controller hap = apron area pressure loss in mm hb = pressure drop in Back-up in downcomer in mm hd = Pressure drop through the dry plate in mm hdc = downcomer pressure drop in mm hr = residual head drop in mm ht = Total pressure drop in mm Hv = height (length of the shell) in m Lw = liquid mass flowrate, kg/s lw = weir length in m Ms= bending moment at base of the skirt a in N/mm2 Nmin= Minimum number of stages. Pi = pressure of the vessel in N/mm2 Rc = Crown radius in mm Rmin= Minimum reflux ratio S = maximum allowable stress of carbon Steel at -64 degrees Celsius in N/mm2 Ss = Maximum allowable design stress in N/mm2 t = thickness of the shell in mm tr = residence time in s tsk = skirt thickness in mm 31

uf = flooding vapour verlocity, m/s, based on the net column cross sectional Area An uh= Maximum vapour velocity through the holes in m/s Vw = vapour mass flowrate, kg/s Wv = Total weight of the shell excluding internal fittings, such as plates in N xHK= Heavy key concentration. xi,d= concentration of the component i in the distillate. xLK= Light key concentration. αa = average relative volatility of the light key. αLK= KLK/KHK = Relative volatility. θ= Root of the equation. θs = base angle of the conical skirt taken as 90 o ρL = density of liquid, kg/m3 ρv = density of vapour, kg/m3 σb = bending stress N/mm2 σc = buckling stress in N/mm2 σh = circumferential stress N/mm2 σL = longitudinal stress N/mm2 σz = resultant longitudinal stress in N/mm2

32

References Engineering ToolBox. 2015. Temperature and Allowable Stresses for Pipes. http://www.engineeringtoolbox.com/temperature-allowable-stresses-pipes-d_1338.html Kister, Henry Z. 1990. Distillation Operation. https://books.google.com.au/books/about/Distillation_Operation.html?id=XG4MCrEBEB8C PetroChemical. 2011. Manufacturing of Petrochemical Product. http://petrochemical2011.blogspot.com.au/2011/03/insulation.html Sinnott, Ray K., Towler, Gavin. 2009. Chemical Engineering Design.

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CHAPTER 8: MINOR DESIGN Shell & Tube Heat Exchanger (E-508)

Abdullah AL Kindi 15404435

Contents List of Figures .................................................................................................................................... ii List of Tables ..................................................................................................................................... ii Introduction ...................................................................................................................................... 1 Design Scope ..................................................................................................................................... 1 8.1

Process Design ....................................................................................................................... 2

8.1.1

Methodology.................................................................................................................. 2

8.1.2

Physical Properties ......................................................................................................... 3

8.1.4

Finding Heat Transfer Area ............................................................................................. 4

8.1.5

Layout of the Heat Exchanger ......................................................................................... 5

8.1.6

Number of Tubes............................................................................................................ 5

8.1.8

Tube Side Heat Transfer Coefficient................................................................................ 6

8.1.7

Shell-Side Heat Transfer Coefficient................................................................................ 8

8.1.8

Overall Heat Coefficient ................................................................................................. 8

8.1.9

Pressure Drop ................................................................................................................ 9

8.2

Operational Design ................................................................................................................ 9

8.2.1

Control System Design.................................................................................................... 9

8.2.2

Operating Procedures .................................................................................................. 10

8.2.2.1

Start-up and Shut-Down ........................................................................................... 10

8.2.2.2

Maintenance ............................................................................................................ 10

8.2.3 8.3

Safety Study ................................................................................................................. 11

Mechanical Design ............................................................................................................... 13

8.3.1

Vessel Design ............................................................................................................... 13

8.3.2

Pressure Relief System ................................................................................................. 13

8.3.3 Mechanical Drawing .................................................................................................... 14 8.4

Summary and Review........................................................................................................... 15

8.4.1

Equipment Specification / Data Sheet ........................................................................... 15

8.4.2

Critical Review .............................................................................................................. 15

Nomenclature ................................................................................................................................. 16 References ...................................................................................................................................... 17

i

List of Figures Figure 1: PFD for the heat exchanger ................................................................................................. 1 Figure 2: Methodology ...................................................................................................................... 2 Figure 3: Temperature correcting factor (Sinnot and Towler, 2009) ................................................... 4 Figure 4: Bundle diameter (Sinnot and Towler, 2009) ........................................................................ 6 Figure 5: Tube side heat transfer factor (Sinnot and Towler, 2009) .................................................... 7 Figure 6: Shell side heat transfer factor (Sinnot and Towler, 2009) .................................................... 9 Figure 7: P&ID for heat exchanger ................................................................................................... 10 Figure 8: Mechanical drawing of heat exchanger ............................................................................. 14

List of Tables Table 1 Physical Properties ................................................................................................................ 3 Table 2 Flowrates .............................................................................................................................. 3 Table 3: Layout and tube size............................................................................................................. 5 Table 4: HAZID study ....................................................................................................................... 11

ii

Introduction Heat exchanger is a piece of equipment that is used in the process industry to cool or heat fluid flow rates. This is done by the exchange of heat from a cooler or a hotter fluid. The most common heat exchanger used is a shell and tube heat exchanger. In this project, a minor design will be carried out to calculate the process and mechanical requirement of cooler to cool a sales gas flow from 11.88 degrees Celsius to -10 degrees Celsius. The sales gas will contain mostly methane and ethane gas flowing at 4.258 × 105 kg/h, since it needs to be cooled down to sub zero temperature the most common coolant used in the process industry is ethylene glycol. Ethylene glycol is normally mixed with water at different percentages. In this project 40% glycol water solution is used. The reason for using this particular percentage it can operate at low as -30 degrees Celsius and it is the least viscous. The higher the concentration of glycol the more viscous it becomes and hence will create a large pressure drop due to the low Reynold number it can achieve.

Figure 1: PFD for the heat exchanger

Design Scope In this project the calculations will include       

The calculations of the number of tubes, shell passes Calculations of the pressure drops across the heat exchanger Mechanical Design of the shell and tube heat exchanger. Process and Instrumentation diagram. HAZID safety study Start-up and shut-down procedures along with maintenance procedures. Summary and the data sheet of the equipment.

1

8.1 Process Design 8.1.1 Methodology 1) Find the requirements and duty of the HX

8) Find shell side HT coefficient 2) Choose coolant and find physical properties

9) Find the overall HT coefficient 3) Assume an initial value of Uo

No

10) Pressure drop < 0.56

4) Calculate Tlm, Ft and Tm

Yes 11) Mechanical Design

5) Find HT Area, tube size and area, volumetric flow

6) Find Bundle Shell diameter

7) Find Tube side HT coefficient

Figure 2: Methodology

2

8.1.2 Physical Properties Table 1 Physical Properties

Feed

Inlet

Mean

Outlet

11.88

0.94

-10

Specific heat (kJ/kgmole. C)

42.12

42.17

42.22

Thermal conductivity (W/m.K)

0.03163

0.03028

0.02893

Density (kg/m )

23.32

24.36

25.4

Viscosity (mN sm-2)

0.00001139

1.1E-05

0.00001068

Ethylene Glycol (40%)

Outlet

Mean

Inlet

-10

-20

-30

Specific heat (kJ/kgmole. C)

3.48

3.48

3.48

Thermal conductivity (W/m.K)

0.39253

0.39253

0.39253

Density (kg/m3)

1074.6

1074.6

1074.6

0.015

0.015

0.015

Temperature (oC) o

3

Temperature (oC) o

-2

Viscosity (mN sm ) Table 2 Flowrates

Feed Flowrate Ethylene Glycol Flowrate Duty

428500 kg/h 5644784 kg/h 109132 kw

In this heat exchanger the flowrate of the ethylene glycol was unknown so as the inlet and the outlet temperatures. The temperatures were assumed as seen in Table 1. The flowrate was calculated using energy balance since the duty of bother inlets and outlet should equal to each other. This was done using the following equation đ?‘„ = đ?‘šđ?‘?đ?‘?đ?›Ľđ?‘‡ m = the mass flowrate in kg/s cp = the specific heat in kJ/kgmole.oC ΔT = difference between inlet and outlet temperatures in oC After obtaining the physical properties seen above the log mean temperature using the following equation.

���� =

(đ?‘‡1 − đ?‘Ą2 ) − (đ?‘‡2 − đ?‘Ą1 ) = 20.93 °đ??ś (đ?‘‡ − đ?‘Ą ) ln (đ?‘‡1 − đ?‘Ą2 ) 2 1

3

Then the values of R and S are found using the following equations đ?‘…=

(đ?‘‡1 − đ?‘‡2 ) = 1.094 (đ?‘Ą2 − đ?‘Ą1 )

�=

(đ?‘Ą2 − đ?‘Ą1 ) = 0.478 (đ?‘‡1 − đ?‘Ą1 )

These values are then used to find the temperature correcting factor Ft in Figure 2 shown below

Figure 3: Temperature correcting factor (Sinnot and Towler, 2009)

The fouling factor from the graph is 0.83 which is then used to find the true temperature difference using the following equation Δđ?‘‡đ?‘š = đ??šđ?‘Ą đ?‘‡đ?‘™đ?‘š = 17.37 ℃

8.1.4 Finding Heat Transfer Area Next step is assuming a value for the overall heat transfer coefficient. A typical value for this kind of shell and tube exchanger will be 100 W/m2 oC. After that the heat transfer area is found by rearranging the following equation đ?‘„ = đ?‘ˆđ?›Ľđ?‘‡đ?‘š đ?‘„ đ??´= = 62834 đ?‘š 2 đ?‘ˆđ?›Ľđ?‘‡đ?‘š

4

8.1.5 Layout of the Heat Exchanger In this section an initial value of the length, inside diameter and outside of the tube. The value which was chose first was 6 meters for the length, 18 mm for the outside diameter and 16mm for the inside diameter. The calculations of the process was then continued until reaching the pressure drop of the tube side. The pressure drop calculated was very high and by following the methodology in Figure 1 the values for the length, inside diameter and outside diameter of the chosen as follows Table 3: Layout and tube size

Length of tube chosen Outside diameter Inside diameter Number of passes

25 m 0.0262 mm 0.02618 mm 4

8.1.6 Number of Tubes To calculate the number of tubes, first the area of one tube needs to be calculated as follows đ??´đ?‘&#x;đ?‘’đ?‘Ž đ?‘œđ?‘“ đ?‘œđ?‘›đ?‘’ đ?‘Ąđ?‘˘đ?‘?đ?‘’ = đ?œ‹ Ă— 0.02618 Ă— 25 = 2.06 đ?‘š 2 To calculate the number of tubes, the area of the heat transfer is divided by the area of one tube đ?‘›đ?‘˘đ?‘šđ?‘?đ?‘’đ?‘&#x; đ?‘œđ?‘“ đ?‘Ąđ?‘˘đ?‘?đ?‘’đ?‘ =

62834 = 30531 đ?‘Ąđ?‘˘đ?‘?đ?‘’đ?‘ 2.06

Since the design selected has 4 passes the number of tubes in each pass is 30531 = 7633 đ?‘Ąđ?‘˘đ?‘?đ?‘’đ?‘ 4 After that the tube velocity is calculated as followed đ?œ‹ (0.02618)2 = 0.0005384 đ?‘š 2 4 đ??´đ?‘&#x;đ?‘’đ?‘Ž đ?‘?đ?‘’đ?‘&#x; đ?‘?đ?‘Žđ?‘ đ?‘ = 7633 Ă— 0.0005384 = 4.11 đ?‘š 2 kg 5644784 h 1 đ?‘š3 đ?‘‰đ?‘œđ?‘™đ?‘˘đ?‘šđ?‘’đ?‘Ąđ?‘&#x;đ?‘–đ?‘? đ?‘“đ?‘™đ?‘œđ?‘¤đ?‘&#x;đ?‘Žđ?‘Ąđ?‘’ = Ă— = 1.46 đ?‘ kg đ?‘ 3600 1074.6 3 â„Ž m 1.46 đ?‘š đ?‘‡đ?‘˘đ?‘?đ?‘’ đ?‘ đ?‘–đ?‘‘đ?‘’ đ?‘Łđ?‘’đ?‘™đ?‘œđ?‘?đ?‘–đ?‘Ąđ?‘Ś = = 0.356 4.11 đ?‘ đ?‘‡đ?‘˘đ?‘?đ?‘’ đ?‘?đ?‘&#x;đ?‘œđ?‘ đ?‘ đ?‘ đ?‘’đ?‘?đ?‘Ąđ?‘–đ?‘œđ?‘›đ?‘Žđ?‘™ đ?‘Žđ?‘&#x;đ?‘’đ?‘Ž =

8.1.7 Bundle and Shell Diameter Bundle diameter is calculated using the following equation đ??ˇđ?‘? = đ?‘‘đ?‘œ (

đ?‘ đ?‘Ą 1/đ?‘›1 ) = 5.15 đ?‘š đ??ž1

do = outside diameter in mm Nt = number of tubes K1 = values obtained from Sinnot and Towler , 0.175 n1 = values obtained from Sinnot and Towler, 2.285

5

Figure 4: Bundle diameter (Sinnot and Towler, 2009)

Since the value of Db is out of range the value of clearance is extrapolated to be 218 mm (0.218 m). Shell diameter is then found by adding up the two value đ??ˇđ?‘ = 5.15 + 0.218 = 5.37 đ?‘š

8.1.8 Tube Side Heat Transfer Coefficient There are number of values that need to be calculated so the value of the heat transfer coefficient is obtained. First the Reynold number is calculated đ?‘…đ?‘’ =

đ?œŒđ?‘˘đ?‘‘đ?‘– = 665 đ?œ‡

Re = Reynolds number Ď = density of the ethylene glycol in kg/m3 u = velocity of the fluid in the tube in m/s di = inside diameter in m Âľ = dynamic viscosity of ethylene glycol in mNs sm-2 After that the prandtl number needs to be evaluated using the following equation đ?‘ƒđ?‘&#x; =

đ??śđ?‘? đ?œ‡ = 133 đ?‘˜đ?‘“

Pr = Prandtl number 6

Cp= Heat Capacity in kJ/kgmole.oC Âľ = dynamic viscosity of ethylene glycol in mNs sm-2 kf = thermal conductivity of ethylene glycol in W/m.K đ??ż = 954.93 đ?‘‘đ?‘– Using the values obtained above jh is found from Figure 4

Figure 5: Tube side heat transfer factor (Sinnot and Towler, 2009)

The value of L/di is out of the range so it was jh was approximated to be 0.0015 Nusselt number and the heat transfer coefficient of the tube is calculated using the following đ?‘ đ?‘˘ =

â„Ž đ?‘– đ?‘‘đ?‘– = đ?‘—đ??ť đ?‘…đ?‘’đ?‘ƒđ?‘&#x; 0.33 = 5.01 đ?‘˜đ?‘“ â„Žđ?‘– = 75.21 đ?‘Š/đ?‘š 2 ℃

7

8.1.7 Shell-Side Heat Transfer Coefficient The same method is used to find out the coefficient in the shell side 1

đ?‘ đ?‘Ą đ?‘›1 đ??ˇđ?‘? = đ?‘‘đ?‘œ ( ) = 5.16 đ?‘š đ??ž1 đ??ˇđ?‘ = 5.38 đ?‘š The baffle spacing is taken as Ds/5 which will equal to 1.075 m and the tube pitch is taken as 1.25 Ă— outside diameter which will give 0.0328 m. Then the shell area is calculated (đ?‘?đ?‘Ą − đ?‘‘đ?‘œ )đ??ˇđ?‘ đ?‘™đ??ľ = 1.16 đ?‘š 2 đ?‘?đ?‘Ą 1.10 2 (đ?‘?đ?‘Ą − 0.917đ?‘‘đ?‘œ2 ) = 0.018 đ?‘š đ?‘‘đ?‘’ = đ?‘‘đ?‘œ đ??´đ?‘ =

As = Shell area pt = tube pitch in m lB = Baffle spacing in m de = equivalent diameter in m The volumetric flowrate is then found using the same equation used before kg 425800 h 1 đ?‘š3 đ?‘‰đ?‘œđ?‘™đ?‘˘đ?‘šđ?‘’đ?‘Ąđ?‘&#x;đ?‘–đ?‘? đ?‘“đ?‘™đ?‘œđ?‘¤đ?‘&#x;đ?‘Žđ?‘Ąđ?‘’ = Ă— = 4.86 đ?‘ kg đ?‘ 3600 â„Ž 24.36 3 m 4.86 đ?‘š đ?‘†â„Žđ?‘’đ?‘™đ?‘™ đ?‘ đ?‘–đ?‘‘đ?‘’ đ?‘Łđ?‘’đ?‘™đ?‘?đ?‘œđ?‘–đ?‘Ąđ?‘Ś = = 4.2 1.16 đ?‘ đ?œŒđ?‘˘đ?‘‘đ?‘– đ?‘…đ?‘’ = = 172619 đ?œ‡ đ??śđ?‘? đ?œ‡ đ?‘ƒđ?‘&#x; = = 15.37 đ?‘˜đ?‘“ hs = 1955 W/m2.oC

8.1.8 Overall Heat Coefficient

đ?‘‘ đ?‘‘đ?‘œ đ?‘™đ?‘› ( đ?‘‘đ?‘œ ) đ?‘‘ 1 1 1 1 1 đ?‘œ đ?‘– = + + + ( + ) đ?‘ˆđ?‘œ â„Žđ?‘œ â„Žđ?‘œđ?‘‘ 2đ?‘˜đ?‘¤ đ?‘‘đ?‘– â„Žđ?‘–đ?‘‘ â„Žđ?‘– đ?‘ˆđ?‘œ = 70.2 đ?‘Š/đ?‘š 2 ℃

Uo = the overall coefficient based on the outside area of the tube, W/m2 oC ho = outside fluid film coefficient, W/m2 oC hi = inside fluid film coefficient, W/m2 oC hod = outside dirt coefficient (fouling factor), W/m2 oC hid = inside dirt coefficient, W/m2 oC kw = thermal conductivity of the tube wall material, W/m oC di = tube inside diameter, m 8

do = tube outside diameter, m This value is considered acceptable since it is within 30% range of the initial

Figure 6: Shell side heat transfer factor (Sinnot and Towler, 2009)

8.1.9 Pressure Drop The pressure drop is has to be under the required value of 0.56 bar many trials and errors are made to meet the is criterion đ??ż đ?œ‡ −đ?‘š đ?œŒđ?‘˘đ?‘Ą2 đ?›Ľđ?‘ƒđ?‘Ą = đ?‘ đ?‘? [8đ?‘—đ?‘“ ( ) ( ) + 2.5] = 0.472 đ?‘?đ?‘Žđ?‘&#x; đ?‘‘đ?‘– đ?œ‡đ?‘š 2 The pressure drop obtained is less than the value simulated on HYSYS which makes it an acceptable design for this particular shell and heat exchanger

8.2 Operational Design 8.2.1 Control System Design The main objective of the control system in the heat exchanger is to monitor the temperature, flowrate and the pressure differences around the shell. Figure 7 below shows where each transmitter and controller is installed.

9

Figure 7: P&ID for heat exchanger

Entering the tub, two transmitters and one controllers are installed. First FT (Flow Transmitter) is installed to monitor the flow of the ethylene glycol (cold fluid). The other transmitter TT (Temperature Transmitter) is connected after the valve. FC (Flow Controller) is installed to adjust the amount of ethylene glycol entering the tube side of the heat exchanger. The hot fluid which is the feed mixture of light hydrocarbons enters in the shell side. The same configuration of transmitters is installed. To monitor and control the feed and temperature going in the shell. One more transmitter is installed along the heat exchanger which is the DPT (Differential Pressure Transmitter) this is to measure the pressure drop in the shell.

8.2.2 Operating Procedures 8.2.2.1 Start-up and Shut-Down In a shell and tube heat exchanger the start-up procedure should be carried in a way that it no differential expansion occurring. The colder fluid is first introduced in the equipment then the hotter fluid is gradually introduced to avoid thermal shock. Moreover, while introducing the fluids to the equipment, all the vents must be open to ensure that all the air is purged out and the shell and tubes are completely filled with the fluids. On the other hand, the shut-down procedure is also done to avoid the thermal shock that will lead to the expansion. The hot and cold fluid should be gradually drained from the shell and tube respectively. Normally the hot fluid is drained first followed by the hot fluid.

8.2.2.2 Maintenance Heat exchangers are very dangerous equipment if left unmaintained. Daily checking for leakage should be carried out. Regular maintenance should be done to prevent fatigue and corrosion. This is not only dangerous because it might leakage and spillage it will also affect the heat transfer between fluids. Any scales or slug visible on the tubes should be cleaned immediately to ensure the equipment is operated at its optimum capacity. Operators should be aware of any abnormalities in the values of pressure and temperature in both inlets and outlets. A log should be kept for the star-up, shut-down and emergency procedures

10

8.2.3 Safety Study Table 4: HAZID study

1

2

3

HAZARD DESCRIPTION / HAZARDOUS EVENT

HAZARD CONSEQUENCES

PREVENTION / DETECTION / BARRIES (Existing)

Natural disasters

 Hurricane or unusual weather conditions

Equipment / Instrumentation Malfunction

 Transmitter failure  Upstream increase in production

Process Upsets

 Pressure deviation

N/A  Damage to the equipment  Spillage  Release of toxic hydrocarbons  Affect flora and fauna  Fire /explosion Flow transmitter  Rupture of shell or pipe  Spillage  Release of toxic hydrocarbons  Fire/explosion Pressure  Loss of transmitter production  Products off spec

RISK ASSESSMENT CONSECUENCE SEVERITY

HAZARD CATEGORY (GUIDEWORD)

LIELIHOOD

HAZA RD ID

RECOMMENDATION S

ACTION (By)

RISK RANKING

Possible

Severe

Extreme

 Proper training for the operators and employees to deal with such emergency situations

 Operating company

Unlikely

Severe

High

 Regular maintenance and check-up on control systems

 Control Engineer

Unlikely

Severe

High

 Regular maintenance and check ups on the control system

 Control Engineer

11

4

Composition Problems

 Dewpointing

5

Utility Failures

 Power failure

6

External effects

 Fatigue/ cracking

 Rupture of shell or tubes  Spillage  Temperature  Corrosion transmitter  Formation of hydrates  Blockage and scaling  Shell and tube rupture N/A  Loss of production  Off spec product  Need to do shut down and start-up procedures N/A  Shell and tube damage  Loss of production  Spillage  Fire/ explosion  Release of toxic hydrocarbon

Possible

Major

High

 Constant checkups of the duty of the HX  Maintenance of sensors

 Operating company  Control engineer

Rare

Minor

Low

 Provide an emergence power back up

 Operating company

Rare

medium

High

 Regular maintenance and check-up of the equipment

 Maintena nce Engineer

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8.3 Mechanical Design 8.3.1 Vessel Design The mechanical design is a very important step into designing a heat exchanger. For the minor part of the project the mechanical design will only cover the wall thickness of the shell and the pressure relief system. The calculations below shows the how is the thickness of the shell is calculated. The same calculations used for the major design is used here. The operating pressure of the heat exchanger is 26.56 bar. To find the shell length, the tube length is taken and one metre is added to each side. So the shell length will be 27 metres. Shell diameter was calculated in the process calculations part to be 5.4 metres. đ?‘‡đ?‘Žđ?‘˜đ?‘’ 10% đ?‘Žđ?‘?đ?‘œđ?‘Łđ?‘’ đ?‘Ąâ„Žđ?‘’ đ?‘œđ?‘?đ?‘’đ?‘&#x;đ?‘Žđ?‘Ąđ?‘–đ?‘›đ?‘” đ?‘?đ?‘&#x;đ?‘’đ?‘ đ?‘ đ?‘˘đ?‘&#x;đ?‘’ = 26.56 đ?‘?đ?‘Žđ?‘&#x; Ă— 1.1 = 29.216 đ?‘?đ?‘Žđ?‘&#x; 26.216 đ?‘?đ?‘Žđ?‘&#x; = 2.8116 đ?‘ƒđ?‘– đ??ˇđ?‘– 2.8116 Ă— 5.4 đ?‘‡â„Žđ?‘–đ?‘?đ?‘˜đ?‘›đ?‘’đ?‘ đ?‘ đ?‘œđ?‘“ đ?‘Ąâ„Žđ?‘’ đ?‘ â„Žđ?‘’đ?‘™đ?‘™ = đ?‘Ą = = = 79.67 đ?‘šđ?‘š 2đ?‘†đ??¸ − 1.2đ?‘ƒđ?‘– (2 Ă— 96.5 Ă— 1) − (1.2 Ă— 2.8116) ≈ 80 đ?‘šđ?‘š đ??śđ?‘œđ?‘&#x;đ?‘&#x;đ?‘œđ?‘ đ?‘–đ?‘œđ?‘› đ?‘Žđ?‘™đ?‘™đ?‘œđ?‘¤đ?‘Žđ?‘›đ?‘?đ?‘’ = 80 + 2 = 82 đ?‘šđ?‘š S = maximum allowable stress of carbon Steel in N/mm2 E = weld joint efficiency

8.3.2 Pressure Relief System To calculate the pressure relief system the equation acquired below (Whitesides, 2012) determine the area of the orifice in the valve needed đ??´=

đ?‘Šâˆšđ?‘‡đ?‘? đ??śđ??žđ?‘ƒ1 đ??žđ?‘? √đ?‘€

=

938800 Ă— √471.88 Ă— 0.9125 348 Ă— 0.975 Ă— 438.42 Ă— 1 Ă— √18.98

= 30.1 đ?‘–đ?‘›2

The closest value is the set at the size design of T which has an area of 26 in2. However it is smaller therefore it will not be acceptable. The reason behind this is the massive flowrate of the feed going to the shell, which turned out to be outside the range of the values obtained in the data sheet for the pressure relief valve calculations. A = area of the orifice in in2 W = flowrate in lb/hr T = temperature in oR Z = Compressibility factor C = flow constant determined by the ratio of specific heats K = coefficient of discharge taken as 0.975 P1= Flow Pressure in psig Kb = Correction factor due to back pressure

13

8.3.3 Mechanical Drawing

Figure 8: Mechanical drawing of heat exchanger

14

8.4 Summary and Review 8.4.1 Equipment Specification / Data Sheet Company Name

Plant location: Longford, Victoria, Australia Equipment to be designed: Heat Exchanger (Cooler) Equipment label: E-508

Project Name: Big Gas Project

Process Data Tube (Ethylene Glycol)

Shell (Light Hydrocarbon)

Operating temperature Density Mass flowrate Liquid Dynamic Viscosity Operating temperature Density Mass flowrate Liquid Dynamic Viscosity

Construction and Material Shell material Shell length Shell thickness Tube outside diameter Tube inside diameter Length of tube Number of passes

-30 - -10 oC 1074.6 kg/m3 5.64× 106 kg/h 0.015 cP 11.88 - -10 oC 24.36 kg/m3 4.258 × 105 kg/h 0.01139 cP Carbon Steel 27.45 m 82 × 10-3 m 26.2 × 10-3 m 26.18 × 10-3 m 25 m 4

8.4.2 Critical Review The initial plan was to create a shell and tube heat exchanger with water as a coolant fluid. However, the temperature needed at the outlet of the feed was sub zero temperatures. At that moment I started to worry if the simulation was done wrong. I decided to seek my supervisor about this issue and he advised me to look into different glycols. They are commonly used as coolants to cool fluids to sub zero temperatures. Ethylene glycol is used in this project which is used to cool fluids down to –20 degrees Celsius. Other glycols like TEG are used to cool down fluids up to -40 degrees Celsius. The second issue encountered was the high viscosity of the coolant, since the concentration of the coolant I chose was 60% glycol in water. This gave a very low Reynolds number, and hence didn’t give the amount of overall heat transfer coefficient. Later I decided to use the 40% glycol solution since it is less viscous and would give a higher Reynolds number. Again for the minor, lack of field experience played a negative role on the progress of this report. Literature review was essential before commencing the calculations on the process design.

15

Nomenclature µ = dynamic viscosity of ethylene glycol in mNs sm-2 As = Shell area cp = the specific heat in kJ/kgmole.oC Cp= Heat Capacity in kJ/kgmole.oC de = equivalent diameter in m di = inside diameter in m do = outside diameter in mm E = weld efficiency hi = inside fluid film coefficient, W/m2 oC hid = inside dirt coefficient, W/m2 oC ho = outside fluid film coefficient, W/m2 oC hod = outside dirt coefficient (fouling factor), W/m2 oC hs = 1955 W/m2.oC K1 = values obtained from Sinnot and Towler , 0.175 kf = thermal conductivity of ethylene glycol in W/m.K kw = thermal conductivity of the tube wall material, W/m oC lB = Baffle spacing in m m = the mass flowrate in kg/s n1 = values obtained from Sinnot and Towler, 2.285 Nt = number of tubes Pr = Prandtl number pt = tube pitch in m Re = Reynolds number S = maximum allowable stress of carbon Steel in N/mm2 u = velocity of the fluid in the tube in m/s Uo = the overall coefficient based on the outside area of the tube, W/m2 oC ΔT = difference between inlet and outlet temperatures in oC ρ = density of the ethylene glycol in kg/m3

16

References Engineering ToolBox. 2015. Ethylene Glycol Heat-Transfer Fluid. http://www.engineeringtoolbox.com/ethylene-glycol-d_146.html MEGlobal. 2008. Ethylene Glycol: Product Guide. http://www.meglobal.biz/media/product_guides/MEGlobal_MEG.pdf Sinnott, Ray K., Towler, Gavin. 2009. Chemical Engineering Design. Thulukkanam, Kuppan. 2013. Heat Exchanger Design Handbook. https://books.google.com.au/books/about/Heat_Exchanger_Design_Handbook_Second_Ed. html?id=ZsU5A1mANWUC&redir_esc=y Whitesides, Randall W. 2012. Selection and Sizing of Pressure Relief Valves. PDHonline course M112 (3PDH)

17

CHAPTER 9: MAJOR DESIGN De-Butanizer: Column (T-502)

SAMAH ZAKI NAJI ALRASHID 17113924

Contents Introduction .............................................................................................................................................. 9.1-Minimum Number of Stages.......................................................................................................... 1-2 9.2-Minimum Number of stages .......................................................................................................... 2-3 9.3-Optimum Number of Stages .......................................................................................................... 3-4 9.4- Tray Efficiency ............................................................................................................................... 4-5 9.5-Optimum Feed Plate Location .......................................................................................................... 5 9.6-Flowrate Balance .......................................................................................................................5-6-7 9.7-Column Height ................................................................................... Error! Bookmark not defined. 9.8-Internal Design.................................................................................................................................. 7 9.8.1-Plate Selection Types .................................................................................................................. 7-8 9.8.2-Column Diameter...................................................................................................................8-9-10 9.8.3-Sieve Tray Design ......................................................................................................................... 10

9.8.3.1-The Flow Pattern ...................................................................................................................... 10 9.8.3.2-Weir Length Estimation ............................................................................................................ 10

9.8.3.3-Weir Height .............................................................................................................................. 11 9.8.3.4-Hole Pitch..................................................................................... Error! Bookmark not defined. 9.8.3.5-Check Weeping .................................................................................................................... 11-12 9.8.3.6-Pressure Drop for Plate Column .....................................................................................12-13-14 9.8.3.7-Down Comer Liquid Backup................................................................................................. 14-15

9.8.3.8-Down Comer Residence Time .................................................................................................. 15 9.8.3.9-Check Entrainment .............................................................................................................. 15-16 9.8.3.10-Trail Layout ........................................................................................................................ 16-17 9.9-Tray to Tray Calculation.................................................................................................. 17-18-19-20 9.10-Mechanical Design ........................................................................................................................ 20

9.10.1-Material Selection ..................................................................................................................... 20 9.10.2-Shell Thickness...................................................................................................................... 20-21 9.10.3- Head Design and Closures ........................................................................................................ 21 9.10.4-Pipe Sizing ............................................................................................................................. 21-22 9.10.5-Combined Loading ..................................................................................................................... 22 9.10.5.1-Dead Weight Calculation ..............................................................................................22-23-24

9.10.5.2-Wind Loading ......................................................................................................................... 24 9.10.5.3-Earthquake Loading ................................................................................................................ 24 1

9.10.6-Stress Analysis....................................................................................................................... 25-26

9.10.7-Vessel Support Design .............................................................................................................. 26 9.10.7.1-Skirt Design ........................................................................................................................ 26-27 9.10.7.2-Base Ring and Anchor BoltDesign ..................................................................................... 27-28 9.10.8-Flage Selection........................................................................................................................... 28 9.10.9- Gaskets Selection ................................................................................................................. 28-29 9.10.10-Flange Face Selection .............................................................................................................. 29 9.10.11-Flange Standards ..................................................................................................................... 29 9.10.12-Pressure Releif System ....................................................................................................... 29-30 9.10.13-Manhole Design ....................................................................................................................... 30 9.10.14-Auxiliary Equipment ........................................................................................................... 30-31 9.11-Operational Procedure ................................................................................................................. 31 9.11.1-Startup Procedure ................................................................................................................ 31-32 9.11.2-Shutdown Procedure ................................................................................................................. 32 9.11.3-Emergyncy Shutdown ................................................................................................................ 32 9.11.4-Maintenance .............................................................................................................................. 33 9.11.5-Plot Plan ..................................................................................................................................... 33

9.12-Safety Hazard................................................................................................................................ 34 9.12.1-Hazard Study ............................................................................................................................. 34 9.12.2-Hazop Study............................................................................................................................... 35 9.13-Process Control System ........................................................................................................... 36-37 9.14-Specification Sheet ....................................................................................................................... 38

9.15- Mechanical Drawing .................................................................................................................... 39 9.16-Critical Design Review................................................................................................................... 40 References ................................................................................................................................................

2

Introduction The objective of the Debutanizer column is to separate the remaining light hydrocarbons (ethane, propane, and butane) from the condensate hydrocarbons (pentane and heavier) to lower the RVP of the condensate and to produce LPG for utilisation in the plant (as a fuel for electrical energy production). A portion of the produced LPG can also be reinjected into the final sales gas stream (provided that the dew point specification and other specification associated with the final product are not breached).

Design Methodology The scope for the design of the Debutanizer column is illustrated below: -

Mass balance on the column (đ?‘ đ?‘’đ?‘’ đ?‘ đ?‘’đ?‘?đ?‘Ąđ?‘–đ?‘œđ?‘› 1). Calculation of the minimum number of stages (đ?‘ đ?‘’đ?‘’ đ?‘ đ?‘’đ?‘?đ?‘Ąđ?‘–đ?‘œđ?‘› 2). Calculation of the minimum reflux ratio (đ?‘ đ?‘’đ?‘’ đ?‘ đ?‘’đ?‘?đ?‘Ąđ?‘–đ?‘œđ?‘›3). Calculation of the optimum number of stages(đ?‘ đ?‘’đ?‘’ đ?‘ đ?‘’đ?‘?đ?‘Ąđ?‘–đ?‘œđ?‘› 4). The tray efficiency and the actual number of stages(đ?‘ đ?‘’đ?‘’ đ?‘ đ?‘’đ?‘?đ?‘Ąđ?‘–đ?‘œđ?‘› 5). The feed stage location (đ?‘ đ?‘’đ?‘’ đ?‘ đ?‘’đ?‘?đ?‘Ąđ?‘–đ?‘œđ?‘› 6). Calculation of the column height(đ?‘ đ?‘’đ?‘’ đ?‘ đ?‘’đ?‘?đ?‘Ąđ?‘–đ?‘œđ?‘› 7). Selection of the tray type(đ?‘ đ?‘’đ?‘’ đ?‘ đ?‘’đ?‘?đ?‘Ąđ?‘–đ?‘œđ?‘› 8.1). Sieve tray design including(đ?‘ đ?‘’đ?‘’ đ?‘ đ?‘’đ?‘?đ?‘Ąđ?‘–đ?‘œđ?‘› 8.2) : flow pattern, weir length estimation, weir height, hole pitch, column diameter, and check of weeping. Pressure drop for plate and column(đ?‘ đ?‘’đ?‘’ đ?‘ đ?‘’đ?‘?đ?‘Ąđ?‘–đ?‘œđ?‘› 8.3). Down comer liquid backup(đ?‘ đ?‘’đ?‘’ đ?‘ đ?‘’đ?‘?đ?‘Ąđ?‘–đ?‘œđ?‘› 8.4). Down comer residual time(đ?‘ đ?‘’đ?‘’ đ?‘ đ?‘’đ?‘?đ?‘Ąđ?‘–đ?‘œđ?‘› 8.5). The checking of entrainment(đ?‘ đ?‘’đ?‘’ đ?‘ đ?‘’đ?‘?đ?‘Ąđ?‘–đ?‘œđ?‘› 8.6). Trial layout(đ?‘ đ?‘’đ?‘’ đ?‘ đ?‘’đ?‘?đ?‘Ąđ?‘–đ?‘œđ?‘› 8.7). Tray to tray calculation(đ?‘ đ?‘’đ?‘’ 9). Mechanical design including(đ?‘ đ?‘’đ?‘’ đ?‘ đ?‘’đ?‘?đ?‘Ąđ?‘–đ?‘œđ?‘› 10): material selection, wall Column Thickness, head design, platforms and ladders, pressure relief system, and pipes sizing. Full stress analysis (đ?‘ đ?‘’đ?‘’ đ?‘ đ?‘’đ?‘?đ?‘Ąđ?‘–đ?‘œđ?‘› 10.6). Vessel sport design including(đ?‘ đ?‘’đ?‘’ đ?‘ đ?‘’đ?‘?đ?‘Ąđ?‘–đ?‘œđ?‘› 10.7): Skirt design, and Base and a chart bolt design. Flange selection(đ?‘ đ?‘’đ?‘’ đ?‘ đ?‘’đ?‘?đ?‘Ąđ?‘–đ?‘œđ?‘› 10.8). Gaskets selection(đ?‘ đ?‘’đ?‘’ đ?‘ đ?‘’đ?‘?đ?‘Ąđ?‘–đ?‘œđ?‘› 10.9). Flange face selection(đ?‘ đ?‘’đ?‘’ đ?‘ đ?‘’đ?‘?đ?‘Ąđ?‘–đ?‘œđ?‘› 10.10). Pressure relief system(đ?‘ đ?‘’đ?‘’ đ?‘ đ?‘’đ?‘?đ?‘Ąđ?‘–đ?‘œđ?‘› 10.11). Mechanical Drawing (see section 15) Design of the process control and P&ID.

3

Process Flow Diagram:

cond

To condenser

Reflux

Feed: Liquid phase Temperature: 31c Pressure: 2300kpa Molar flow 3196 kgmole/h Heat flow:-4.531e+008 kJ/h

Reflux Drum

5-14 Feed (from Demethanizer bottom)

Distillate (LPG product)

Top: Liquid phase Temperature: 72.08c Pressure: 2100kpa Molar flow 2119 kgmole/h Heat flow:-2.6366e+008 kJ/h

5-13

Boilup

To Reboiler

5-15

Bottom: Liquid phase Temperature: 140.2c Bottom product Pressure: 2300kpa To condensate Molar flow 1078 kgmole/h stabilization Heat flow:-1.546e+008 kJ/h

Title: Distillation Column to Produce LPG

Drawing Title: PFD Vessel Tag no: T-502 Designed by: Samah Zaki Alrashid

Approved by: Dr.Gia Pham

4

9.1-Minimum Number of Stages It can be used Fenske’s equation to calculate the minimum number of stages. Fenske’s equation is: 𝑥 𝑥 ln[( 𝐷,𝐿𝐾 )∗( 𝐵,𝐻𝐾 )]

𝑁𝑚𝑖𝑛 =

𝑥𝐷,𝐻𝐾

𝑥𝐵,𝐿𝐾

ln 𝛼𝑎𝑣

(Kister, 1992)

𝛼𝐿𝐾/𝐻𝐾 : 𝑡ℎ𝑒 𝑣𝑜𝑙𝑎𝑡𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 𝑙𝑖𝑔ℎ𝑡 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 𝑡𝑜 𝑡ℎ𝑒 ℎ𝑒𝑎𝑣𝑖𝑒𝑠𝑡 𝑜𝑛𝑒, 𝑖𝑡 𝑐𝑎𝑛 𝑏𝑒 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒𝑑 𝑏𝑦: 𝟏 − 𝑈𝑠𝑒 𝐴𝑛𝑡𝑜𝑛𝑖𝑜 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑡𝑜 𝑒𝑠𝑡𝑖𝑚𝑎𝑡𝑒 𝑡ℎ𝑒 𝑠𝑎𝑡𝑢𝑟𝑎𝑡𝑖𝑜𝑛 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑎𝑡 304.14𝐾: 𝐄𝐭𝐡𝐚𝐧𝐞: ln 𝑃 𝑠𝑎𝑡 = 𝐴 −

𝐵 𝑇+𝐶

= 9.27428 −

1582.178 (Stichlmair, 1998) 304.15−13.762

𝑃 𝑠𝑎𝑡 = 45.9 𝑏𝑎𝑟 𝐏𝐫𝐨𝐩𝐚𝐧𝐞: 1872.824

ln 𝑃 𝑠𝑎𝑡 = 9.10434 − 304.15−25.101 (Stichlmair, 1998) 𝑃 𝑠𝑎𝑡 = 10.9 𝑏𝑎𝑟 i-Butane: 2133.243

ln 𝑃 𝑠𝑎𝑡 = 9.15169 − 304.15−28.162 (Stichlmair, 1998) 𝑃 𝑠𝑎𝑡 = 4.15𝑏𝑎𝑟 n-Butane: 2154.897

ln 𝑃 𝑠𝑎𝑡 = 9.05814 − 304.15−34.420 (Stichlmair, 1998) 𝑃 𝑠𝑎𝑡 = 2.9126𝑏𝑎𝑟 i-Pentane: 1040.73

log 𝑃 𝑠𝑎𝑡 = 6.83315 − 31+235.445 (Grote, 2009) 𝑃 𝑠𝑎𝑡 = 1.13𝑏𝑎𝑟 n-Pentane: ln 𝑃 𝑠𝑎𝑡 = 9.21312 −

2477.075 (Stichlmair, 1998) 304.15−39.945

𝑃 𝑠𝑎𝑡 = 0.85𝑏𝑎𝑟 n-Hexane: 2738.418

ln 𝑃 𝑠𝑎𝑡 = 9.29213 − 304.15−46.870 (Stichlmair, 1998) 𝑃 𝑠𝑎𝑡 = 0.26𝑏𝑎𝑟

5

n-Heptane: 2911.320

ln đ?‘ƒ đ?‘ đ?‘Žđ?‘Ą = 9.25363 − 304.15−56.510 (Stichlmair, 1998) đ?‘ƒ đ?‘ đ?‘Žđ?‘Ą = 0.082 đ?‘?đ?‘Žđ?‘&#x; Mcyclohexane: 1270.763

log đ?‘ƒ đ?‘ đ?‘Žđ?‘Ą = 6.823 − 31+221.416 (Grote, 2009) đ?‘ƒ đ?‘ đ?‘Žđ?‘Ą = 0.08đ?‘?đ?‘Žđ?‘&#x; Toluene: 3090.783

ln đ?‘ƒ đ?‘ đ?‘Žđ?‘Ą = 9.38490 − 304.15−53.960 (Stichlmair, 1998) đ?‘ƒ đ?‘ đ?‘Žđ?‘Ą = 0.05đ?‘?đ?‘Žđ?‘&#x; n-octane: 3128.752

ln đ?‘ƒ đ?‘ đ?‘Žđ?‘Ą = 9.34011 − 304.15−63.295 (Stichlmair, 1998) đ?‘ƒ đ?‘ đ?‘Žđ?‘Ą = 0.026đ?‘?đ?‘Žđ?‘&#x; E-benzene: log đ?‘ƒ đ?‘ đ?‘Žđ?‘Ą = 6.95719 −

1424.255 (Grote, 2009) 31+213.206

đ?‘ƒ đ?‘ đ?‘Žđ?‘Ą = 0.017đ?‘?đ?‘Žđ?‘&#x; đ?&#x;? − đ??śđ?‘Žđ?‘™đ?‘?đ?‘˘đ?‘™đ?‘Žđ?‘Ąđ?‘’ đ?‘Ąâ„Žđ?‘’ đ??ž − đ?‘Łđ?‘Žđ?‘™đ?‘˘đ?‘’ đ?‘?đ?‘Ś: B đ?‘ƒđ?‘ đ?‘Žđ?‘Ą đ?‘–

=23barđ??žđ?‘– = đ?‘ƒ

đ?‘Ąđ?‘œđ?‘Ąđ?‘Žđ?‘™

đ?‘˛ − đ?’—đ?’‚đ?’?đ?’–đ?’† đ??¸đ?‘Ąâ„Žđ?‘Žđ?‘›đ?‘’ đ?‘ƒđ?‘&#x;đ?‘œđ?‘?đ?‘Žđ?‘›đ?‘’ i-butane n-butane i-pentane n-pentane n-hexane n-heptane Mcyclohexane Toluene n-octane E-benzene

Feed at P=23bar 1.99 0.47 0.126 0.18 0.049 0.037 0.011 0.0035 0.0034 0.0021 0.0011 0.00072

Top(distillate) at P=21bar 2.19 0.52 0.14 0.19 0.05 0.04 0.012 0.0039 0.0038 0.0023 0.0012 0.0008

Bottom at P=23bar 1.99 0.47 0.126 0.18 0.049 0.037 0.011 0.0035 0.0034 0.0021 0.0011 0.00072

Table 1: K value of each component at the feed, top, and bottom. đ?&#x;‘ − đ??śđ?‘Žđ?‘™đ?‘?đ?‘˘đ?‘™đ?‘Žđ?‘Ąđ?‘’ đ?‘Ąâ„Žđ?‘’ đ?‘?đ?‘œđ?‘šđ?‘?đ?‘œđ?‘›đ?‘’đ?‘›đ?‘Ą đ?‘Łđ?‘œđ?‘™đ?‘Žđ?‘Ąđ?‘–đ?‘™đ?‘–đ?‘Ąđ?‘Ś đ?‘?đ?‘Ś: Heavy key component is (i-pentane) and light key component is (n-butane) 6

�� =

đ??žđ?‘– đ??žâ„Žđ?‘’đ?‘Žđ?‘Łđ?‘Ś đ?‘˜đ?‘’đ?‘Ś

Alpha values from Alpha component/Alpha HK (i-Pentane)

Feed

đ??¸đ?‘Ąâ„Žđ?‘Žđ?‘›đ?‘’ đ?‘ƒđ?‘&#x;đ?‘œđ?‘?đ?‘Žđ?‘›đ?‘’ i-butane n-butane i-pentane n-pentane n-hexane n-heptane Mcyclohexane Toluene n-octane E-benzene

40.6 9.59 2.57 3.67 1 0.755 0.22 0.0714 0.069 0.043 0.022 0.015

Top(distillate)

43.8 10.4 2.8 3.8 1 0.8 0.24 0.078 0.076 0.046 0.024 0.016

Bottom

40.6 9.59 2.57 3.67 1 0.755 0.22 0.0714 0.069 0.043 0.022 0.015

Average (F,B,and D)

41.66 9.86 2.64 3.713 1 0.77 0.23 0.073 0.071 0.044 0.022 0.0153

Table 2: Alpha value of each component at the feed, top, and bottom. đ?‘ đ?‘œđ?‘¤, đ?‘–đ?‘Ą đ?‘?đ?‘Žđ?‘› đ?‘?đ?‘’ đ?‘“đ?‘œđ?‘˘đ?‘›đ?‘‘ đ?‘Ąâ„Žđ?‘’ đ?‘šđ?‘–đ?‘›đ?‘–đ?‘šđ?‘˘đ?‘š đ?‘›đ?‘˘đ?‘šđ?‘?đ?‘’đ?‘&#x; đ?‘œđ?‘“ đ?‘ đ?‘Ąđ?‘Žđ?‘”đ?‘’đ?‘ đ?‘?đ?‘Ś: đ?‘Ľ đ?‘Ľ ln[( đ??ˇ,đ??żđ??ž )∗( đ??ľ,đ??ťđ??ž )]

đ?‘ đ?‘šđ?‘–đ?‘› =

đ?‘Ľđ??ˇ,đ??ťđ??ž

đ?‘Ľđ??ľ,đ??żđ??ž

ln ���

(Kister, 1992)

3

đ?›źđ?‘Žđ?‘Ł = √3.67 ∗ 3.8 ∗ 3.67

đ?›źđ?‘Žđ?‘Ł = 3.71 đ?‘ đ?‘šđ?‘–đ?‘›

0.1606 0.2058 ln [( )∗( )] 0.0065 0.2575 = ln 3.71

Nmin= 2.27 (say 3) đ?‘ťđ?’‰đ?’†đ?’“đ?’†đ?’‡đ?’?đ?’“đ?’†, đ?’•đ?’‰đ?’† đ?’Žđ?’Šđ?’?đ?’Šđ?’Žđ?’–đ?’Ž đ?’?đ?’–đ?’Žđ?’ƒđ?’†đ?’“ đ?’?đ?’‡ đ?’”đ?’•đ?’‚đ?’ˆđ?’†đ?’” đ?’–đ?’?đ?’…đ?’†đ?’“ đ?’•đ?’?đ?’•đ?’‚đ?’? đ?’“đ?’†đ?’‡đ?’?đ?’–đ?’™ đ?’Šđ?’” đ?&#x;‘ đ?’”đ?’•đ?’‚đ?’ˆđ?’†đ?’”. Note: Kister (page. 106) suggest to calculate the minimum number of stages by Fenske’s equation, it can be used the average volatility of light key at the top and the bottom or, it can be used the average volatility of the light key at the feed, top, and the bottom, therefore, I have used the average volatility of the light key at the feed, top, and the bottom condition.

9.2-Minimum Reflux Ratio The estimation of minimum reflux ratio will calculated by using Underwood’s equations: ∑

đ?›źđ?‘– ∗đ?‘Ľđ??ˇđ?‘– đ?›źđ?‘– −đ?œƒ

− 1 = đ?‘…đ?‘šđ?‘–đ?‘› (Kister, 1992)

1-Calculate the value of đ?œ˝, this value should be between the light component volatility and the heavy component volatility, by: ∑

đ?›źđ?‘– ∗đ?‘Ľđ?‘“đ?‘– đ?›źđ?‘– −đ?œƒ

= 1 − đ?‘ž (Kister, 1992)

đ?›źđ?‘– : đ?‘?đ?‘œđ?‘šđ?‘?đ?‘œđ?‘›đ?‘’đ?‘›đ?‘Ą đ?‘Łđ?‘œđ?‘™đ?‘Žđ?‘Ąđ?‘–đ?‘™đ?‘–đ?‘Ąđ?‘Ś. 7

đ?‘Ľđ?‘“đ?‘– : đ?‘šđ?‘œđ?‘™đ?‘’ đ?‘“đ?‘&#x;đ?‘Žđ?‘?đ?‘Ąđ?‘–đ?‘œđ?‘› đ?‘œđ?‘“ đ?‘Ąâ„Žđ?‘’ đ?‘?đ?‘œđ?‘šđ?‘?đ?‘œđ?‘›đ?‘’đ?‘›đ?‘Ą đ?‘–đ?‘› đ?‘Ąâ„Žđ?‘’ đ?‘“đ?‘’đ?‘’đ?‘‘. đ?‘ž: đ?‘Ąâ„Žđ?‘’ đ?‘Žđ?‘šđ?‘œđ?‘˘đ?‘›đ?‘Ą đ?‘œđ?‘“ đ?‘™đ?‘–đ?‘žđ?‘˘đ?‘–đ?‘‘ đ?‘–đ?‘› đ?‘“đ?‘’đ?‘’đ?‘‘ = 1 (đ?‘Žđ?‘™đ?‘™ đ?‘™đ?‘–đ?‘žđ?‘˘đ?‘–đ?‘‘). đ?‘Şđ?’?đ?’Žđ?’‘đ?’?đ?’?đ?’†đ?’?đ?’• đ??¸đ?‘Ąâ„Žđ?‘Žđ?‘›đ?‘’ đ?‘ƒđ?‘&#x;đ?‘œđ?‘?đ?‘Žđ?‘›đ?‘’ i-butane n-butane i-pentane n-pentane n-hexane n-heptane Mcyclohexane Toluene n-octane E-benzene

Feed (stream 5-13) 0.0149 04416 0.1654 0.1879 0.0808 0.0585 0.0352 0.0085 0.0053 0.0008 0.0011 0.0001

Top(distillate) stream(5-14) 0.0205 0.6 0.211 0.1606 0.0064 0.0013 0.001 0.0001 0.0001 0 0 0

Bottom Stream(5-15) 0 0.0445 0.1601 0.2575 0.2058 0.1547 0.1187 0.0331 0.0179 0.0024 0.0048 0.0004

Table 3: Mole fraction of each component at the feed, top, and bottom. đ?‘€đ?‘–đ?‘?đ?‘&#x;đ?‘œđ?‘ đ?‘œđ?‘“đ?‘Ą đ??¸đ?‘Ľđ?‘?đ?‘’đ?‘™ đ?‘¤đ?‘Žđ?‘ đ?‘˘đ?‘ đ?‘’đ?‘‘ đ?‘Ąđ?‘œ đ?‘?đ?‘Žđ?‘™đ?‘?đ?‘˘đ?‘™đ?‘Žđ?‘Ąđ?‘’ đ?œ˝ , ,and it was determined that đ?‘Ąâ„Žđ?‘’ đ?‘Łđ?‘Žđ?‘™đ?‘˘đ?‘’ đ?‘œđ?‘“ đ?œ˝ đ?’Šđ?’” đ?&#x;?. đ?&#x;Žđ?&#x;–đ?&#x;•đ?&#x;’đ?&#x;?đ?&#x;– đ?‘¤â„Žđ?‘–đ?‘?â„Ž đ?‘–đ?‘ đ?‘?đ?‘’đ?‘Ąđ?‘¤đ?‘’đ?‘’đ?‘› đ?‘›_đ?‘?đ?‘˘đ?‘Ąđ?‘Žđ?‘›đ?‘’ (đ??żđ??ž) đ?‘Łđ?‘œđ?‘™đ?‘Žđ?‘Ąđ?‘–đ?‘™đ?‘–đ?‘Ąđ?‘Ś đ?‘Žđ?‘›đ?‘‘ đ?‘–_đ?‘?đ?‘’đ?‘›đ?‘Ąđ?‘Žđ?‘›đ?‘’ (đ??ťđ??ž) đ?‘Łđ?‘œđ?‘™đ?‘Žđ?‘Ąđ?‘–đ?‘™đ?‘–đ?‘Ąđ?‘Ś. đ?‘°đ?’• đ?’Šđ?’” đ?’Šđ?’? đ?’‚đ?’„đ?’„đ?’†đ?’‘đ?’•đ?’‚đ?’ƒđ?’?đ?’† đ?’“đ?’‚đ?’?đ?’ˆđ?’† (Kister, 1992). 2-Calculate the minimum reflux ratio by: ∑

đ?›źđ?‘– ∗đ?‘Ľđ??ˇđ?‘– đ?›źđ?‘– −đ?œƒ

− 1 = đ?‘…đ?‘šđ?‘–đ?‘› (Kister, 1992)

đ?‘Ľđ??ˇđ?‘– : đ?‘Ąâ„Žđ?‘’ đ?‘šđ?‘œđ?‘™đ?‘’ đ?‘“đ?‘&#x;đ?‘Žđ?‘?đ?‘Ąđ?‘–đ?‘œđ?‘› đ?‘?đ?‘œđ?‘šđ?‘?đ?‘œđ?‘›đ?‘’đ?‘›đ?‘Ą đ?‘Žđ?‘Ą đ?‘Ąâ„Žđ?‘’ đ?‘Ąđ?‘œđ?‘?. Microsoft Excel was used to calculate and the value of đ?‘šđ?’Žđ?’Šđ?’? = đ?&#x;Ž. đ?&#x;?đ?&#x;?đ?&#x;“đ?&#x;’ So, the minimum reflux ratio for this column is đ?&#x;Ž. đ?&#x;?đ?&#x;?đ?&#x;“đ?&#x;’.

9.3-Optimum Number of stages After the minimum reflux ratio and the minimum number of stages have been estimated, now it is possible to calculate the actual number of stages by using Gilliland’s correlations. To start with, we have to assume the reflux ratio and find đ?‘Œ by Molokanov’s correlation: (1+54.4đ?‘‹)(đ?‘‹âˆ’1)

đ?‘Œ = 1 − exp((11+117.2đ?‘‹)(đ?‘‹ 0.5 ) (Dutta, 2007) Where: đ?‘‹=

đ?‘… − đ?‘…đ?‘šđ?‘–đ?‘› đ?‘…+1

đ?‘Œ=

đ?‘ − đ?‘ đ?‘šđ?‘–đ?‘› đ?‘ +1

After finding đ?‘Œ, it is easily to find the actual number of stage. 8

As the reflux ratio increases, the number of trays and thus the capital cost of the column decreases. However, as a trade-off, an increase in reflux ratio will also increase the vapor rate within the tower, thus increasing expenses such as condensers and reboilers. Most columns are designed to operate between 1.2 and 1.5 times the minimum reflux ratio because this is approximately the region of minimum operating cost (Stichlmair, 1998). Assessed Reflux 1.1đ?‘…đ?‘šđ?‘–đ?‘› 1.2đ?‘…đ?‘šđ?‘–đ?‘› 1.3đ?‘…đ?‘šđ?‘–đ?‘› 1.4đ?‘…đ?‘šđ?‘–đ?‘› 1.5đ?‘…đ?‘šđ?‘–đ?‘› 1.6đ?‘…đ?‘šđ?‘–đ?‘› 1.7đ?‘…đ?‘šđ?‘–đ?‘› 1.8đ?‘…đ?‘šđ?‘–đ?‘› 2đ?‘…đ?‘šđ?‘–đ?‘› 2.5đ?‘…đ?‘šđ?‘–đ?‘›

Reflux Ratio X Y 0.23694 0.018626611 0.25848 0.035423686 0.28002 0.051655443 0.30156 0.067349949 0.3231 0.082533444 0.34464 0.097230485 0.36618 0.111464082 0.38772 0.125255815 0.4308 0.151593514 0.5385 0.210984725

0.666509 0.628444 0.606316 0.58816 0.571759 0.5565 0.542155 0.528609 0.503623 0.451141

Number of stages 6.997166 6.382768 6.080222 5.85626 5.67027 5.509585 5.368292 5.242758 5.0292 4.643921

Table 4: Reflux Ration and Number of stages. After plotting the actual number of stages against reflux ratio as shown below, draw several lines, the line that has the shortest distance through the intersection with the curved will determine the number of stages and it wasđ?&#x;?. đ?&#x;?. Therefore, the actual reflux ratio (đ?‘…) is almostđ?&#x;Ž. đ?&#x;?đ?&#x;“đ?&#x;–đ?&#x;“and the number of stages is almost 6.382768 stages( can say 7 stages) Number of Stage vs Reflux Ratio 8

Number of Trays

7 6 5 4 3 2 1 0 0

0.1

0.2

0.3

0.4

0.5

0.6

Reflux Ratio Figure (1): Optimum Reflux ratio determination Approximate Method.

9.4-Tray Efficiency There are many correlations to calculate the tray efficiency such as O′Connell’s correlation, Drickamer Bradford’s correlation, and Lockett’s Correlation. 1-Lockett’s Correlation: đ??¸đ?‘œ = 0.492 ∗ (đ?œ‡đ?‘™ ∗ đ?›ź)−0.245 (Kister, 1992) 9

Where: 𝜇𝑙,𝑎𝑣 : 𝑡ℎ𝑒 𝑚𝑜𝑙𝑎𝑟 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑙𝑖𝑞𝑢𝑖𝑑 𝑣𝑖𝑠𝑐𝑜𝑠𝑖𝑡𝑦, 𝑐𝑃 𝛼𝑎 : 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑟𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝑣𝑜𝑙𝑎𝑡𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 𝑡ℎ𝑒 𝑙𝑖𝑔ℎ𝑡 𝑘𝑒𝑦 𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑟𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝑣𝑜𝑙𝑎𝑡𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 𝑡ℎ𝑒 𝑙𝑖𝑔ℎ𝑡 𝑘𝑒𝑦, 𝛼𝑎 = 3.71 The molar average liquid viscosity =0.1416 cP 𝐸𝑜 = 0.492 ∗ (𝜇𝑙 ∗ 𝛼)−0.245 = 57.60% 2-Drickamer Bradford’s correlation: 𝐸𝑜 = 0.133 − 0.668 ∗ log 𝜇 = 70% (Seader, 1998)

3-O′ Connell’s correlation: 𝐸𝑜 = 0.503 ∗ (𝜇𝑙 ∗ 𝛼)−0.226 = 58.17% (Wankat, 1988) Based on above calculation, it can be assumed that the efficiency, 𝑬𝒐 = 𝟓𝟕. 𝟔𝟎%. Now, it is possible to calculate the actual number of plates: 𝑵𝒂𝒄𝒕𝒖𝒂𝒍 =

𝑻𝒉𝒆𝒐𝒓𝒊𝒕𝒊𝒄𝒂𝒍 𝒏𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒑𝒍𝒂𝒕𝒆𝒔 𝟕 = = 𝟏𝟑 𝒔𝒕𝒂𝒈𝒆𝒔 𝒆𝒇𝒇𝒊𝒄𝒊𝒆𝒏𝒄𝒚 𝟎. 𝟓𝟕𝟔

9.5-Optimum Feed Plate Location: The total number of stages are 13, therefore: 𝑁𝑟 + 𝑁𝑠 = 13

𝒆𝒒𝒖𝒂𝒕𝒊𝒐𝒏 . 𝑨 2

𝑁𝑟 𝐵 𝑥𝑓,𝐻𝐾 𝑥𝐵,𝐿𝐾 =[ ∗ ∗( ) ]0.206 𝑁𝑠 𝐷 𝑥𝑓,𝐿𝐾 𝑥𝐷,𝐻𝐾 Where: 𝑁𝑟 : 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑠𝑡𝑎𝑔𝑒𝑠 𝑎𝑏𝑜𝑣𝑒 𝑡ℎ𝑒 𝑓𝑒𝑒𝑑 𝑝𝑙𝑎𝑡𝑒. 𝑁𝑠 : 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑠𝑡𝑎𝑔𝑒𝑠 𝑏𝑒𝑙𝑜𝑤 𝑡ℎ𝑒 𝑓𝑒𝑒𝑑 𝑝𝑙𝑎𝑡𝑒. 0.206

𝑁𝑟 1078 0.0808 0.25752 =[ ∗ ∗ ] 𝑁𝑠 2119 0.1879 0.0065 𝑁𝑟 = 3.33 𝑁𝑠

= 1.816 𝒆𝒒𝒖𝒂𝒕𝒊𝒐𝒏 . 𝑩

𝑁𝑠 = ≡ 3𝑠𝑡𝑎𝑔𝑒𝑠 𝑁𝑟 = 9.99 ≡ 10 𝑠𝑡𝑎𝑔𝑒𝑠 This gives the feed plat is 𝟑 from the bottom and 𝟏𝟎 from the top. 9.6-Flowrate Balance on the Column The feed to the distillation is 100% liquid, the molar flow rate in the feed provided by Hysys which is 𝑘𝑔𝑚𝑜𝑙𝑒 3196 ℎ𝑟 . 𝑇ℎ𝑒 𝑜𝑣𝑒𝑟𝑎𝑙𝑙 𝑏𝑎𝑙𝑎𝑛𝑐𝑒 𝑔𝑖𝑣𝑒𝑠:

10

đ?‘­ =đ?‘Ť+đ?‘Š

đ?’†đ?’’đ?’–đ?’‚đ?’•đ?’Šđ?’?đ?’?. đ?‘¨

đ?‘Šâ„Žđ?‘’đ?‘&#x;đ?‘’: đ??š: đ?‘šđ?‘œđ?‘™đ?‘Žđ?‘&#x; đ?‘“đ?‘™đ?‘œđ?‘¤ đ?‘&#x;đ?‘Žđ?‘Ąđ?‘’ đ?‘–đ?‘› đ?‘Ąâ„Žđ?‘’ đ?‘“đ?‘’đ?‘’đ?‘‘.

đ??ˇ: đ?‘šđ?‘œđ?‘™đ?‘Žđ?‘&#x; đ?‘“đ?‘™đ?‘œđ?‘¤ đ?‘&#x;đ?‘Žđ?‘Ąđ?‘’ đ?‘–đ?‘› đ?‘Ąâ„Žđ?‘’ đ?‘Ąđ?‘œđ?‘?.

đ??ľ: đ?‘šđ?‘œđ?‘™đ?‘Žđ?‘&#x; đ?‘“đ?‘™đ?‘œđ?‘¤ đ?‘&#x;đ?‘Žđ?‘Ąđ?‘’ đ?‘–đ?‘› đ?‘Ąâ„Žđ?‘’ đ?‘?đ?‘œđ?‘Ąđ?‘Ąđ?‘œđ?‘š. đ??ľđ?‘Žđ?‘™đ?‘Žđ?‘›đ?‘?đ?‘’ đ?‘“đ?‘œđ?‘&#x; đ?‘›_đ?‘?đ?‘˘đ?‘Ąđ?‘Žđ?‘›đ?‘’ đ?‘Žđ?‘&#x;đ?‘œđ?‘˘đ?‘›đ?‘‘ đ?‘Ąâ„Žđ?‘’ đ?‘?đ?‘œđ?‘™đ?‘˘đ?‘šđ?‘›: đ?‘­ ∗ đ?’™đ?’‡ = đ?‘Ť ∗ đ?’™đ?‘Ť + đ?‘Š ∗ đ?’™đ?‘Š

đ?’†đ?’’đ?’–đ?’‚đ?’•đ?’Šđ?’?đ?’?. đ?‘Š

đ?‘Ľđ??ˇ : đ?‘šđ?‘œđ?‘™. đ?‘“đ?‘&#x;đ?‘Žđ?‘?đ?‘Ąđ?‘–đ?‘œđ?‘› đ?‘œđ?‘“ đ?‘›_đ?‘?đ?‘˘đ?‘Ąđ?‘Žđ?‘›đ?‘’ đ?‘–đ?‘› đ?‘Ąđ?‘œđ?‘? = 0.1606 đ?‘Ľđ??ľ : đ?‘šđ?‘œđ?‘™. đ?‘“đ?‘&#x;đ?‘Žđ?‘?đ?‘Ąđ?‘–đ?‘œđ?‘› đ?‘œđ?‘“ đ?‘›_đ?‘?đ?‘˘đ?‘Ąđ?‘Žđ?‘›đ?‘’ đ?‘–đ?‘› đ?‘Ąâ„Žđ?‘’ đ?‘?đ?‘œđ?‘Ąđ?‘Ąđ?‘œđ?‘š = 0.2575 đ?‘Ľđ?‘“ : đ?‘šđ?‘œđ?‘™. đ?‘“đ?‘&#x;đ?‘Žđ?‘?đ?‘Ąđ?‘–đ?‘œđ?‘› đ?‘œđ?‘“ đ?‘›_đ?‘?đ?‘˘đ?‘Ąđ?‘Žđ?‘›đ?‘’ đ?‘–đ?‘› đ?‘Ąâ„Žđ?‘’ đ?‘“đ?‘’đ?‘’đ?‘‘ = 0.1879 đ??ľđ?‘Ś đ?‘ đ?‘œđ?‘™đ?‘Łđ?‘–đ?‘›đ?‘” đ?‘Ąâ„Žđ?‘’đ?‘ đ?‘’ đ?‘Ąđ?‘¤đ?‘œ đ?‘’đ?‘žđ?‘˘đ?‘Žđ?‘Ąđ?‘–đ?‘œđ?‘›: đ?‘Ť = đ?&#x;?đ?&#x;?đ?&#x;—đ?&#x;“. đ?&#x;“đ?&#x;–

đ?’Œđ?’ˆđ?’Žđ?’?đ?’?đ?’† đ?’‰đ?’“

đ?‘Š = đ?&#x;—đ?&#x;Žđ?&#x;Ž. đ?&#x;’đ?&#x;?

đ?’Œđ?’ˆđ?’Žđ?’?đ?’?đ?’† đ?’‰đ?’“

9.7-Column height: The tower height can be related to the number of trays in the column. The following formula assumes that a spacing of two feet between trays will be sufficient including additional five to ten feet at both ends of the tower. This includes a fifteen percent excess allowance of space (Douglas, 1988). It can be estimated the height by: đ??ťđ?‘Ąđ?‘œđ?‘¤đ?‘’đ?‘&#x; = (đ?‘ đ?‘Žđ?‘?đ?‘Ąđ?‘˘đ?‘Žđ?‘™ − 1) ∗ đ?‘™đ??ľ + (đ?‘ đ?‘‡ + đ?‘ đ??ľ ) + (đ?‘ đ?‘Žđ?‘?đ?‘Ąđ?‘˘đ?‘Žđ?‘™ − 1) ∗ đ?‘Ąđ?‘ đ?‘™đ??ľ : đ?‘Ąđ?‘&#x;đ?‘Žđ?‘Ś đ?‘ đ?‘?đ?‘Žđ?‘?đ?‘–đ?‘›đ?‘”, đ?‘–đ?‘Ą đ?‘–đ?‘ đ?‘ đ?‘?đ?‘’đ?‘?đ?‘“đ?‘–đ?‘’đ?‘‘ đ?‘Ąđ?‘œ đ?‘?đ?‘’ 0.6096 (Kister, 1992) đ?‘ đ?‘‡ : đ?‘Ąâ„Žđ?‘’ đ?‘ đ?‘?đ?‘Žđ?‘?đ?‘’ đ?‘Žđ?‘Ą đ?‘Ąâ„Žđ?‘’ đ?‘Ąđ?‘œđ?‘? đ?‘œđ?‘“ đ?‘Ąâ„Žđ?‘’ đ?‘?đ?‘œđ?‘™đ?‘˘đ?‘šđ?‘›, đ?‘–đ?‘Ą đ?‘ đ?‘?đ?‘’đ?‘?đ?‘–đ?‘“đ?‘–đ?‘’đ?‘‘ đ?‘Ąđ?‘œ đ?‘?đ?‘’ 1.5 đ?‘š (Kister, 1992). đ?‘ đ??ľ : đ?‘Ąâ„Žđ?‘’ đ?‘ đ?‘?đ?‘Žđ?‘?đ?‘’ đ?‘Žđ?‘Ą đ?‘Ąâ„Žđ?‘’ đ?‘?đ?‘œđ?‘Ąđ?‘Ąđ?‘œđ?‘š đ?‘œđ?‘“ đ?‘Ąâ„Žđ?‘’ đ?‘?đ?‘œđ?‘™đ?‘˘đ?‘šđ?‘›, đ?‘–đ?‘Ą đ?‘–đ?‘ đ?‘ đ?‘?đ?‘’đ?‘?đ?‘–đ?‘“đ?‘–đ?‘’đ?‘‘ đ?‘Ąđ?‘œ đ?‘?đ?‘’ 2 đ?‘š (Kister, 1992) đ?‘Ąđ?‘ : đ?‘Ąâ„Žđ?‘’ đ?‘Ąâ„Žđ?‘–đ?‘?đ?‘˜đ?‘›đ?‘’đ?‘ đ?‘ đ?‘œđ?‘“ đ?‘Ąâ„Žđ?‘’ đ?‘Ąđ?‘&#x;đ?‘Žđ?‘Ś , đ?‘–đ?‘Ą đ?‘–đ?‘ đ?‘ đ?‘?đ?‘’đ?‘?đ?‘–đ?‘“đ?‘–đ?‘’đ?‘‘ đ?‘Ąđ?‘œ đ?‘?đ?‘’ 2.5 đ?‘šđ?‘š (Kister, 1992) đ??ťđ?‘Ąđ?‘œđ?‘¤đ?‘’đ?‘&#x; = (13 − 1) ∗ 0.6096 + (1.5 + 2) + (13 − 1) ∗ 0.0025 đ?‘Żđ?’•đ?’?đ?’˜đ?’†đ?’“ = đ?&#x;?đ?&#x;Ž. đ?&#x;–đ?&#x;’ đ?’Ž

9.8-Internal Design 9.8.1-Plate Type Selection Before 1960s, the most commonly trays used in the distillation column was bubble-cup tray. After that, it replaced by other types which are the sieve and valve trays (Sinnott, 2005).Currently, the most popular types are the sieve and valve trays, at the same time; the bubble-cup tray is used only for special cases. The bubble-cup tray is a flat perforated plate with risers around the holes and caps. The caps have rectangular or triangular slots cut around its side. The gas flows up through the perforations, rises, turns around and passes out through the slots of the cup directly into the liquid, producing a "froth regime". The caps are well sealed to the plate surface, preventing the liquid from draining at low gas flow rate. Bubble cap trays are best suited for applications with low liquid flows and/or high 11

turndown ratios. In terms of capacity, however, they are slightly lower than valve or sieve trays. They are also the most expensive tray option. Sieve trays are made from a flat perforated plate which allows the passage of vapor through the liquid. They are the most economical tray option when low turndown is required. They have better anti-fouling characteristics and lower pressure drop than valve or bubble cap trays. Sieve trays are simple and easy to fabricate, therefore, they are relatively inexpensive. In valve trays, perforations are covered by lift able caps. Vapor flows lifts the caps, thus self creating a flow area for the passage of vapor. The lifting cap directs the vapor to flow horizontally into the liquid, thus providing better mixing than is possible in sieve trays. Comparison of the common tray types (Kister, 1992): Option Tray Name 1 The bubble-cup 2 Sieve trays 3 valve trays

Cost High đ??żđ?‘œđ?‘¤ Higher than Sieve

Pressure Drop High Moderate Moderate to High

Capacity Moderately High High High to very High

Rank 3đ?‘&#x;đ?‘‘ 1đ?‘ đ?‘Ą 2đ?‘›đ?‘‘

Therefore, based on the above table, the most suitable tray for our distillation is Sieve trays. 8.2-Column diameter The first part for any design of the internal column is the estimation of the column diameter. It will be calculated both the top and the bottom diameter and the choice will be the biggest of them to fit with the rest of the column. â—? Estimate the liquid-vapor flooding factor for both top and bottom: Note: all densities and flowrates from Hysys. To condenser Vapor phase Liquid phase Mass density[kg/m3] 49.07 443.3 Molecular weight 66.92 Molar flow[kgmole/h] VD=2542 LD=423.7

đ??šđ??żđ?‘‰ =

đ??ż đ?œŒđ?‘Ł ∗√ đ?‘‰ đ?œŒđ?‘™

đ??šđ?‘œđ?‘&#x; đ?‘Ąâ„Žđ?‘’ đ?‘Ąđ?‘œđ?‘? đ?‘œđ?‘“ đ?‘Ąâ„Žđ?‘’ đ?‘?đ?‘œđ?‘™đ?‘˘đ?‘šđ?‘›:

đ??šđ??żđ?‘‰ =

đ??šđ??żđ?‘‰ =

đ??żđ??ˇ đ?œŒđ?‘Ł ∗√ đ?‘‰đ??ˇ đ?œŒđ?‘™

423.7 49.07 ∗√ 2542 443.3

đ??šđ??żđ?‘‰ = 0.055 đ??šđ?‘œđ?‘&#x; đ?‘Ąâ„Žđ?‘’ đ?‘?đ?‘œđ?‘Ąđ?‘Ąđ?‘œđ?‘š đ?‘œđ?‘“ đ?‘Ąâ„Žđ?‘’ đ?‘?đ?‘œđ?‘™đ?‘˘đ?‘šđ?‘›: To Reboiler Vapor phase Mass density[kg/m3] 64.82 Molecular weight 62.34 Molar flow[kgmole/h] VB=4650

Liquid phase 437.1 LB=1078 12

𝐹𝐿𝑉 =

1078 64.82 ∗ √ 4650 437.1

𝐹𝐿𝑉 = 0.09 All the densities and molar flow provided by Hysys. ● Estimate the flooding velocity by: 𝑢𝑓𝑙𝑜𝑜𝑑 = 𝐾 ∗ √

𝜌𝐿 −𝜌𝑉 𝜌𝑉

(Wankat, 1988)

𝑇ℎ𝑒 𝑓𝑎𝑐𝑡𝑜𝑟 𝐾 𝑏𝑦 𝑓𝑖𝑔𝑢𝑟𝑒 11.27 (Sinnott, 2005): 𝐾𝑡𝑜𝑝 𝑤𝑖𝑡ℎ 𝐹𝐿𝑉,𝑡𝑜𝑝 = 0.11

And

𝐾𝑏𝑡𝑡𝑜𝑚 𝑤𝑖𝑡ℎ 𝐹𝐿𝑉,𝑏𝑜𝑡𝑡𝑜𝑚 = 0.1

443.3 − 49.07 𝑚 𝑇ℎ𝑒 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑓 𝑓𝑙𝑜𝑜𝑑𝑖𝑛𝑔 𝑖𝑛 𝑡𝑜𝑝 = 0.11 ∗ √ = 0.31 49.07 𝑠

𝑇ℎ𝑒 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑓 𝑓𝑙𝑜𝑜𝑑𝑖𝑛𝑔 𝑖𝑛 𝑏𝑜𝑡𝑡𝑜𝑚 = 0.1 ∗ √

437.1 − 64.82 𝑚 = 0.238 64.82 𝑠

𝐴𝑠𝑠𝑢𝑚𝑒 𝑡ℎ𝑎𝑡 𝑡ℎ𝑒 𝑑𝑒𝑠𝑖𝑔𝑛 𝑓𝑜𝑟 85% 𝑓𝑙𝑜𝑜𝑑𝑖𝑛𝑔:

𝑚 𝑇ℎ𝑒 𝑎𝑐𝑡𝑢𝑎𝑙 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑓 𝑓𝑙𝑜𝑜𝑑𝑖𝑛𝑔 𝑖𝑛 𝑡𝑜𝑝 = 0.312 ∗ 0.85 = 0.2652 𝑠 𝑚 𝑇ℎ𝑒 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑓 𝑓𝑙𝑜𝑜𝑑𝑖𝑛𝑔 𝑖𝑛 𝑏𝑜𝑡𝑡𝑜𝑚 = 0.144 ∗ 0.85 = 0.203 𝑠 ● Estimate the cross section area of the column for both the top and the bottom: (

𝑉

)∗𝑀𝑤

3600 𝐼𝑡 𝑐𝑎𝑛 𝑏𝑒 𝑒𝑠𝑡𝑖𝑚𝑎𝑡𝑒𝑑 𝑡ℎ𝑒 𝑐𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝑎𝑟𝑒𝑎 = 𝜌 ∗𝑎𝑐𝑡𝑢𝑎𝑙 (Wankat, 1988) 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑣 2542 (3600) ∗ 46.34 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑡𝑜𝑝 = = 2.514 𝑚2 49.07 ∗ 0.2652 4650 (3600) ∗ 62.34 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑜𝑡𝑡𝑜𝑚 = = 6.09 𝑚2 64.82 ∗ 0.203 𝐶𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒 𝑡ℎ𝑒 𝑛𝑒𝑡 𝑎𝑟𝑒𝑎 𝑤𝑖𝑡ℎ 12% 𝑓𝑜𝑟 𝑑𝑜𝑤𝑛𝑐𝑜𝑚𝑒𝑟 𝑎𝑟𝑒𝑎: 2.514 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑡𝑜𝑝 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑜𝑙𝑢𝑚𝑛 = = 2.857 𝑚2 (1 − 0.12) 10.15 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑜𝑡𝑡𝑜𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑜𝑙𝑢𝑚𝑛 = = 6.93 𝑚2 (1 − 0.12) 𝑁𝑜𝑤 𝑡ℎ𝑒 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑜𝑙𝑢𝑚𝑛 𝑐𝑎𝑛 𝑏𝑒 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒𝑑:

4 ∗ 𝐴𝑡𝑜𝑝 𝐷𝑖𝑎𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 𝑡ℎ𝑒 𝑡𝑜𝑝 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑜𝑙𝑢𝑚𝑛 = √ = 1.91 𝑚 𝜋 4 ∗ 𝐴𝑏𝑜𝑡𝑡𝑜𝑚 𝐷𝑖𝑎𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑜𝑡𝑡𝑜𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑜𝑙𝑢𝑚𝑛 = √ = 2.97 𝑚 𝜋 The choice is the highest diameter between the top and the bottom; therefore, the top diameter is the column diameter: 𝑻𝒉𝒆 𝒅𝒊𝒂𝒎𝒆𝒕𝒆𝒓 𝒐𝒇 𝒕𝒉𝒆 𝒄𝒐𝒍𝒖𝒎𝒏 = 𝟐. 𝟗𝟕 𝒎

9.8.3-Sieve Tray Design 9.8.3.1-The Flow Pattern 13

It can be used figure 11.28 (Sinnott, 2005) and/or table 8.2 (Ludwig E. E., 1999) to determine basic suggestion about the flow type. However, first should be calculated the maximum volumetric flow rate by: đ?‘€đ?‘Žđ?‘Ľđ?‘–đ?‘šđ?‘˘đ?‘š đ?‘‰đ?‘œđ?‘™đ?‘˘đ?‘šđ?‘’đ?‘Ąđ?‘&#x;đ?‘–đ?‘? đ??šđ?‘™đ?‘œđ?‘¤ =

đ?‘€đ?‘Žđ?‘Ľđ?‘–đ?‘šđ?‘˘đ?‘š đ?‘‰đ?‘œđ?‘™đ?‘˘đ?‘šđ?‘’đ?‘Ąđ?‘&#x;đ?‘–đ?‘? đ??šđ?‘™đ?‘œđ?‘¤ =

đ??ż ∗ đ?‘€đ?‘¤ đ?œŒđ??ż,đ?‘?đ?‘œđ?‘Ąđ?‘Ąđ?‘œđ?‘š ∗ 3600

1078 ∗ 66.92 đ?‘š3 = 0.045 437.1 ∗ 3600 đ?‘

đ??šđ?‘œđ?‘&#x; đ?‘Ąâ„Žđ?‘–đ?‘ đ?‘Łđ?‘œđ?‘™đ?‘˘đ?‘šđ?‘’đ?‘Ąđ?‘&#x;đ?‘–đ?‘? đ?‘“đ?‘™đ?‘œđ?‘¤ đ?‘Žđ?‘›đ?‘‘ đ?‘Ąâ„Žđ?‘’ đ?‘?đ?‘œđ?‘™đ?‘˘đ?‘šđ?‘› đ?‘‘đ?‘–đ?‘Žđ?‘šđ?‘’đ?‘Ąđ?‘’đ?‘&#x;, đ?‘Ąâ„Žđ?‘’ đ?‘“đ?‘™đ?‘œđ?‘¤ đ?‘?đ?‘Žđ?‘Ąđ?‘Ąđ?‘’đ?‘&#x;đ?‘› đ?‘–đ?‘ đ?’ˆđ?’‚đ?’”đ?’Œđ?’†đ?’• đ?’…đ?’?đ?’–đ?’ƒđ?’?đ?’† đ?’‘đ?’‚đ?’”đ?’”đ?’†đ?’”. 9.8.3.2-Weir Length Estimation đ??ľđ?‘Žđ?‘ đ?‘’đ?‘‘ đ?‘œđ?‘› đ?‘“đ?‘–đ?‘”đ?‘˘đ?‘&#x;đ?‘’ 11.31 (Sinnott, 2005) , đ?‘¤đ?‘’ đ?‘›đ?‘’đ?‘’đ?‘‘ đ?‘Ąđ?‘œ đ?‘?đ?‘Žđ?‘™đ?‘?đ?‘˘đ?‘™đ?‘Žđ?‘Ąđ?‘’ đ?‘Ąâ„Žđ?‘’ đ?‘&#x;đ?‘Žđ?‘Ąđ?‘–đ?‘œ đ?‘?đ?‘’đ?‘Ąđ?‘¤đ?‘’đ?‘’đ?‘› đ?‘Ąâ„Žđ?‘’ đ?‘‘đ?‘œđ?‘¤đ?‘›đ?‘?đ?‘œđ?‘šđ?‘’đ?‘&#x; đ?‘Žđ?‘&#x;đ?‘’đ?‘Ž đ?‘Žđ?‘›đ?‘‘ đ?‘Ąâ„Žđ?‘’ đ?‘?đ?‘œđ?‘™đ?‘˘đ?‘šđ?‘› đ?‘Žđ?‘&#x;đ?‘’đ?‘Ž. đ??ťđ?‘œđ?‘¤đ?‘’đ?‘Łđ?‘’đ?‘&#x;, đ?‘¤đ?‘’ â„Žđ?‘Žđ?‘Łđ?‘’ đ?‘Žđ?‘™đ?‘&#x;đ?‘’đ?‘Žđ?‘‘đ?‘Ś đ?‘šđ?‘’đ?‘›đ?‘Ąđ?‘–đ?‘œđ?‘›đ?‘’đ?‘‘ đ?‘Ąâ„Žđ?‘Žđ?‘Ą đ?‘Ąâ„Žđ?‘’ đ?‘‘đ?‘œđ?‘¤đ?‘›đ?‘?đ?‘œđ?‘šđ?‘’đ?‘&#x; đ?‘Žđ?‘&#x;đ?‘’đ?‘Ž đ?‘–đ?‘ 12% đ?‘“đ?‘&#x;đ?‘œđ?‘š đ?‘Ąâ„Žđ?‘’ đ?‘?đ?‘œđ?‘™đ?‘˘đ?‘šđ?‘› đ?‘Žđ?‘&#x;đ?‘’đ?‘Ž. đ??śđ?‘œđ?‘™đ?‘˘đ?‘šđ?‘› đ??´đ?‘&#x;đ?‘’đ?‘Ž =

đ?œ‹ đ?œ‹ ∗ đ??ˇđ?‘?đ?‘œđ?‘™đ?‘˘đ?‘šđ?‘› 2 = ∗ 2.972 = 6.92 đ?‘š2 4 4

đ??ˇđ?‘œđ?‘¤đ?‘›đ?‘?đ?‘œđ?‘šđ?‘’đ?‘&#x; đ??´đ?‘&#x;đ?‘’đ?‘Ž = 0.12 ∗ 6.92 = 0.83 đ?‘š2 Now from this ratio and figure đ?&#x;?đ?&#x;?. đ?&#x;‘đ?&#x;? (Sinnott, 2005), the

đ?‘™đ?‘¤ đ??ˇđ?‘?

= 0.75:

đ?‘™đ?‘¤ = 0.75 ∗ 2.97 = 2.22 đ?‘š The width of each zone is usually made the same; recommended value with diameter above 1.5 m is 100mm, therefore, the width of the zone will be 100 mm (Sinnott, 2005). 9.8.3.3-Weir Height The height of the weir increase the plate efficiency as well as pressure drop of the plate, therefore, it is important to specify suitable one. For columns operating above atmospheric pressure the weir heights will normally be between 40 mm to 90 mm (1.5 to 3.5 in.) and the optimum range is between 40 to 50 mm (Sinnott, 2005). Therefore, I will useđ?&#x;“đ?&#x;Ž đ?’Žđ?’Ž. Other parameters: đ?‘ đ?‘’đ?‘Ą đ??´đ?‘&#x;đ?‘’đ?‘Ž = đ??śđ?‘œđ?‘™đ?‘˘đ?‘šđ?‘› đ??´đ?‘&#x;đ?‘’đ?‘Ž − đ??ˇđ?‘œđ?‘¤đ?‘›đ?‘?đ?‘œđ?‘šđ?‘’đ?‘&#x; đ??´đ?‘&#x;đ?‘’đ?‘Ž đ??´đ?‘›đ?‘’đ?‘Ą = 6.92 − 0.83 = 6.09 đ?‘š2 đ??´đ?‘?đ?‘Ąđ?‘–đ?‘Łđ?‘’ đ??´đ?‘&#x;đ?‘’đ?‘Ž = đ??śđ?‘œđ?‘™đ?‘˘đ?‘šđ?‘› đ??´đ?‘&#x;đ?‘’đ?‘Ž − 2 ∗ đ??ˇđ?‘œđ?‘¤đ?‘›đ?‘?đ?‘œđ?‘šđ?‘’đ?‘&#x; đ??´đ?‘&#x;đ?‘’đ?‘Ž đ??´đ?‘Žđ?‘?đ?‘Ąđ?‘–đ?‘Łđ?‘’ = 6.92 − 2 ∗ 0.83 = 5.26 đ?‘š2 đ??ťđ?‘œđ?‘™đ?‘’ đ??´đ?‘&#x;đ?‘’đ?‘Ž = 5% ∗ đ?‘Žđ?‘?đ?‘Ąđ?‘–đ?‘Łđ?‘’ đ?‘Žđ?‘&#x;đ?‘’đ?‘Ž

(Sinnott, 2005)

đ??´â„Žđ?‘œđ?‘™đ?‘’ = 0.05 ∗ 5.26 = 0.263 đ?‘š2 The hole sizes used vary from 2.5 to 12 mm, the selection will be 5 mm (Sinnott, 2005). The plate thickness will specify to be 2.5 mm. 14

9.8.3.4-Hole pitch The hole pitch (distance between the hole centres) 𝑙𝑃 should not be less than 2.0 hole diameters, and the normal range will be 2.5 to 4.0 diameters. Within this range the pitch can be selected to give the number of active holes required for the total hole area specified. Square and equilateral triangular patterns are used; triangular is preferred (Sinnott, 2005). 9.8.3.5- Check Weeping The lower limit of the operating range occurs when liquid leakage through the plate holes becomes excessive. This is known as the weep point (Sinnott, 2005). I will follow Coulson and Richardson’s vol. 6 to check the weeping: 𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝐿𝑖𝑞𝑢𝑖𝑑 𝑟𝑎𝑡𝑒 =

𝐿𝑖𝑞𝑢𝑖𝑑 𝑏𝑒𝑙𝑜𝑤 𝑓𝑒𝑒𝑑 1078 ∗ 66.92 𝑘𝑔 = = 20.03 3600 3600 𝑠

𝑀𝑎𝑥 𝑣𝑜𝑙𝑢𝑚𝑒𝑡𝑟𝑖𝑐 𝑓𝑙𝑜𝑤𝑟𝑎𝑡𝑒 =

20.03 𝑚3 = 0.046 437.1 𝑠

𝑀𝑖𝑛𝑖𝑚𝑢𝑚 𝑙𝑖𝑞𝑢𝑖𝑑 𝑟𝑎𝑡𝑒, 𝑎𝑡 70 𝑝𝑒𝑟𝑐𝑒𝑛𝑡 𝑡𝑢𝑟𝑛 − 𝑑𝑜𝑤𝑛 = 0.7 ∗ 20.03 = 14.03 𝑀𝑖𝑛 𝑣𝑜𝑙𝑢𝑚𝑒𝑡𝑟𝑖𝑐 𝑓𝑙𝑜𝑤𝑟𝑎𝑡𝑒 = ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑙𝑖𝑞𝑢𝑖𝑑 𝑐𝑟𝑒𝑠𝑡 = 750 ∗ (

𝑘𝑔 𝑠

14.03 𝑚3 = 0.032 437.1 𝑠

𝑣𝑜𝑙𝑢𝑚𝑒𝑡𝑟𝑖𝑐 𝑓𝑙𝑜𝑤𝑟𝑎𝑡𝑒 2 )3 𝑤𝑒𝑖𝑟 𝑙𝑒𝑛𝑔𝑡ℎ

(Sinnott, 2005) 2

0.032 3 𝑀𝑖𝑛. ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑙𝑖𝑞𝑢𝑖𝑑 𝑐𝑟𝑒𝑠𝑡(𝑎𝑡 70% 𝑑𝑜𝑤𝑛) = 750 ∗ ( ) = 44.42 𝑚𝑚 2.22 2

0.046 3 𝑀𝑎𝑥. ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑙𝑖𝑞𝑢𝑖𝑑 𝑐𝑟𝑒𝑠𝑡(𝑎𝑡 70% 𝑑𝑜𝑤𝑛) = 750 ∗ ( ) = 56.58 𝑚𝑚 2.22 𝐴𝑡 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑟𝑎𝑡𝑒 = 𝑚𝑎𝑥. ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑙𝑖𝑞𝑢𝑖𝑑 𝑐𝑟𝑒𝑠𝑡 + 𝑤𝑒𝑖𝑟 ℎ𝑒𝑖𝑔ℎ𝑡 𝐴𝑡 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑟𝑎𝑡𝑒, ℎ𝑤 + ℎ𝑤𝑜 = 56.58 + 50 = 106.58 𝑚𝑚 𝐵𝑎𝑠𝑒𝑑 𝑜𝑛 𝑡ℎ𝑖𝑠 𝑣𝑎𝑙𝑢𝑒 𝑎𝑛𝑑 𝒇𝒊𝒈𝒖𝒓𝒆 𝟏𝟏. 𝟑𝟎 (Sinnott, 2005), 𝐾1 = 31.1 𝐴𝑡 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 𝑟𝑎𝑡𝑒 = 𝑚𝑖𝑛. ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑙𝑖𝑞𝑢𝑖𝑑 𝑐𝑟𝑒𝑠𝑡 + 𝑤𝑒𝑖𝑟 ℎ𝑒𝑖𝑔ℎ𝑡 𝐴𝑡 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 𝑟𝑎𝑡𝑒, ℎ𝑤 + ℎ𝑤𝑜 = 44.42 + 50 = 94.42 𝑚𝑚 𝐵𝑎𝑠𝑒𝑑 𝑜𝑛 𝑡ℎ𝑖𝑠 𝑣𝑎𝑙𝑢𝑒 𝑎𝑛𝑑 𝒇𝒊𝒈𝒖𝒓𝒆 𝟏𝟏. 𝟑𝟎 (Sinnott, 2005), 𝐾2 = 30.8 𝑁𝑜𝑤, 𝑖𝑡 𝑖𝑠 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑡𝑜 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒 𝑡ℎ𝑒 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 𝑣𝑎𝑝𝑜𝑟 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑏𝑦: 𝑢ℎ =

𝐾2 − 0.9 ∗ (25.4 − 𝑑ℎ ) 𝜌𝑣,𝑏𝑒𝑙𝑜𝑤 𝑡ℎ𝑒 𝑓𝑒𝑒𝑑 1/2

𝑊ℎ𝑒𝑟𝑒: 𝑑ℎ = ℎ𝑜𝑙𝑒 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟, 5𝑚𝑚. 𝑢ℎ =

30.8 − 0.9 ∗ (25.4 − 5) 1 (64.82)2

= 1.55

𝑚 𝑠

15

𝑉𝑎𝑝𝑜𝑟 𝐹𝑙𝑜𝑤𝑟𝑎𝑡𝑒 𝑎𝑡 𝑡𝑜𝑝 =

𝑉𝑎𝑝𝑜𝑟 𝑟𝑎𝑡𝑒 𝑎𝑡 𝑡𝑜𝑝 2542 ∗ 46.34 𝑚3 = = 0.667 3600 ∗ 𝜌𝑣𝑎𝑝𝑜𝑟 𝑎𝑡 𝑡𝑜𝑝 3600 ∗ 49.07 𝑠

𝑉𝑎𝑝𝑜𝑟 𝐹𝑙𝑜𝑤𝑟𝑎𝑡𝑒 𝑎𝑡 𝑏𝑜𝑡𝑡𝑜𝑚 =

𝑉𝑎𝑝𝑜𝑟 𝑟𝑎𝑡𝑒 𝑎𝑡 𝐵𝑜𝑡𝑡𝑜𝑚 4650 ∗ 62.34 𝑚3 = = 1.24 3600 ∗ 𝜌𝑣𝑎𝑝𝑜𝑟 𝑎𝑡 𝑏𝑜𝑡𝑡𝑜𝑚 3600 ∗ 64.82 𝑠

𝐴𝑐𝑡𝑢𝑎𝑙 𝑣𝑎𝑝𝑜𝑟 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑎𝑡 𝑡ℎ𝑒 𝑏𝑜𝑡𝑡𝑜𝑚 = 𝐴𝑐𝑡𝑢𝑎𝑙 𝑣𝑎𝑝𝑜𝑟 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑎𝑡 𝑡ℎ𝑒 𝑡𝑜𝑝 =

0.7 ∗ 𝑉𝑜𝑙𝑢𝑚𝑒𝑡𝑟𝑖𝑐𝑠𝑡𝑟𝑖. 0.7 ∗ 1.24 𝑚 = = 3.3 𝐴ℎ𝑜𝑙𝑒 0.263 𝑠

0.7 ∗ 𝑉𝑜𝑙𝑢𝑚𝑒𝑡𝑟𝑖𝑐𝑟𝑒𝑐𝑡. 0.7 ∗ 0.667 𝑚 = = 1.78 𝐴ℎ𝑜𝑙𝑒 0.263 𝑠

It can be seen that both actual vapor velocities at the top and the bottom will be well above weep point which is means weeping will not occur.

9.8.3.6- Plate and Column Pressure drop Calculation The pressure drop over the plates is an important design consideration. There are two main sources of pressure loss: that due to vapor flow through the holes, and that due to the static head of liquid on the plate (Sinnott, 2005). The estimation of the dry plate pressure by: 𝑢ℎ 𝜌𝑣 𝐷𝑟𝑦 𝑝𝑙𝑎𝑡𝑒 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑑𝑟𝑜𝑝 = 51 ∗ ( )2 ∗ 𝐶𝑜 𝜌𝑙 I have calculated the velocities of vapor at the top and the bottom in weep checking section, 3.3 𝑚 therefore, 𝑢ℎ , 𝑎𝑐𝑡𝑢𝑎𝑙 𝑣𝑎𝑝𝑜𝑟 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑏𝑜𝑡𝑡𝑜𝑚 = 0.7 = 4.71 𝑠 𝑢ℎ , 𝑎𝑐𝑡𝑢𝑎𝑙 𝑣𝑎𝑝𝑜𝑟 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑡𝑜𝑝 =

1.78 𝑚 = 2.54 0.7 𝑠

𝐵𝑦 𝑢𝑠𝑖𝑛𝑔 𝒇𝒊𝒈𝒖𝒓𝒆 𝟏𝟏. 𝟑𝟒 (Sinnott, 2005), 𝑤𝑖𝑡ℎ 𝑡ℎ𝑒 𝑎𝑠𝑠𝑢𝑚𝑝𝑡𝑖𝑜𝑛 𝑡ℎ𝑎𝑡

𝐴ℎ 𝐴𝑝

𝑎𝑙𝑚𝑜𝑠𝑡 𝑒𝑞𝑢𝑎𝑙 𝑡𝑜

𝐴ℎ , 𝐴𝑎𝑐𝑡𝑖𝑣𝑒

𝑖𝑡 𝑐𝑎𝑛 𝑏𝑒 𝑓𝑜𝑢𝑛𝑑 𝐶𝑜 = 0.62. 2.54 2 49.07 𝐷𝑟𝑦 𝑝𝑙𝑎𝑡𝑒 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑑𝑟𝑜𝑝 𝑡𝑜𝑝, ℎ𝑑 = 51 ∗ ( ) ∗ = 94.75 𝑚𝑚 0.62 443.3 4.71 2 64.82 𝐷𝑟𝑦 𝑝𝑙𝑎𝑡𝑒 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑑𝑟𝑜𝑝 𝑏𝑜𝑡𝑡𝑜𝑚, ℎ𝑑 = 51 ∗ ( ) ∗ = 436.47 𝑚𝑚 0.62 437.1 𝑊ℎ𝑖𝑙𝑒, 𝑡ℎ𝑒 𝑟𝑒𝑠𝑖𝑑𝑢𝑎𝑙 ℎ𝑒𝑎𝑑 𝑐𝑎𝑛 𝑏𝑒 𝑒𝑠𝑡𝑖𝑚𝑎𝑡𝑒𝑑 𝑏𝑦: ℎ𝑟 =

12.5∗ 103 𝜌𝐿

(Sinnott, 2005) ℎ𝑟 𝑓𝑜𝑟 𝑡𝑜𝑝 =

12.5 ∗ 103 = 28.19𝑚𝑚 443.3

ℎ𝑟 𝑓𝑜𝑟 𝑏𝑜𝑡𝑡𝑜𝑚 =

12.5 ∗ 103 = 28.59𝑚𝑚 437.1

𝑇ℎ𝑒 𝑡𝑜𝑡𝑎𝑙 𝑝𝑙𝑎𝑡𝑒 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑑𝑟𝑜𝑝 = ℎ𝑟 + ℎ𝑑 + ℎ𝑤𝑒𝑖𝑟 + ℎ𝑤𝑒𝑖𝑟,𝑚𝑎𝑥 Where: ℎ𝑤𝑒𝑖𝑟 ∶ 𝑤𝑒𝑖𝑟 ℎ𝑖𝑒𝑔ℎ𝑡, 𝑖𝑡 𝑤𝑎𝑠 𝑐ℎ𝑜𝑠𝑒𝑛 𝑡𝑜 𝑏𝑒 50 𝑚𝑚. 16

ℎ𝑤𝑒𝑖𝑟,𝑚𝑎𝑥 : 𝑀𝑎𝑥. ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑙𝑖𝑞𝑢𝑖𝑑 𝑐𝑟𝑒𝑠𝑡, 𝑖𝑡 𝑤𝑎𝑠 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒𝑑 𝑖𝑛 𝐶ℎ𝑒𝑐𝑘 𝑊𝑒𝑒𝑝𝑖𝑛𝑔 𝑇ℎ𝑒 𝑡𝑜𝑡𝑎𝑙 𝑝𝑙𝑎𝑡𝑒 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑑𝑟𝑜𝑝 𝑡𝑜𝑝 = 28.19 + 94.75 + 50 + 56.58 = 229.52 𝑚𝑚 𝑇ℎ𝑒 𝑡𝑜𝑡𝑎𝑙 𝑝𝑙𝑎𝑡𝑒 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑑𝑟𝑜𝑝 𝑏𝑜𝑡𝑡𝑜𝑚 = 28.59 + 436.47 + 50 + 44.42 = 559.48 𝑚𝑚 𝑇ℎ𝑒 𝑡𝑜𝑡𝑎𝑙 𝑝𝑙𝑎𝑡𝑒 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑑𝑟𝑜𝑝 𝑡𝑜𝑝 = 0.229 ∗ 9.81 ∗ 443.3 = 0.998 𝑘𝑝𝑎/𝑠𝑡𝑎𝑔𝑒 𝑇ℎ𝑒 𝑡𝑜𝑡𝑎𝑙 𝑝𝑙𝑎𝑡𝑒 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑑𝑟𝑜𝑝 𝑏𝑜𝑡𝑡𝑜𝑚 = 0.55948 ∗ 9.81 ∗ 437.1 = 2.399 𝑘𝑝𝑎/𝑠𝑡𝑎𝑔𝑒 𝐴𝑠 𝐼 𝑠ℎ𝑜𝑤𝑒𝑑 𝑎𝑏𝑜𝑣𝑒 𝑡ℎ𝑎𝑡 𝑡ℎ𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑠𝑡𝑎𝑔𝑒𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑟𝑒𝑐𝑡𝑖𝑓𝑦𝑖𝑛𝑔 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝑖𝑠 10 𝑎𝑛𝑑 3 𝑖𝑛 𝑡ℎ𝑒 𝑠𝑡𝑟𝑖𝑝𝑝𝑖𝑛𝑔 𝑠𝑒𝑐𝑡𝑖𝑜𝑛. 𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒, 𝑡ℎ𝑒 𝑡𝑜𝑡𝑎𝑙 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑐𝑎𝑛 𝑏𝑒 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒𝑑 𝑏𝑦: 𝑇ℎ𝑒 𝑡𝑜𝑡𝑎𝑙 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑑𝑟𝑜𝑝 = 10 𝑠𝑡𝑎𝑔𝑒 ∗ 0.998

𝑘𝑝𝑎 𝑘𝑝𝑎 + 3 𝑠𝑡𝑎𝑔𝑒 ∗ 2.399 𝑠𝑡𝑎𝑔𝑒 𝑠𝑡𝑎𝑔𝑒

𝑇ℎ𝑒 𝑡𝑜𝑡𝑎𝑙 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑜𝑙𝑢𝑚𝑛 = 17.17 𝑘𝑝𝑎 = 0.17 𝑏𝑎𝑟.

9.8.3.7-Down comer liquid backup The down comer area and plate spacing must be such that the level of the liquid and froth in the down comer is well below the top of the outlet weir on the plate above. The column is likely to flood if the level rises above the outlet weir. The main reason behind the liquid backup in the down comer is the pressure drop over the plate.

Downcomer Back-up The down comer liquid backup can be estimated by: ℎ𝑏𝑎𝑐𝑘𝑢𝑝 = ℎ𝑤𝑒𝑖𝑟 + ℎ𝑐𝑟𝑒𝑠𝑡 𝑜𝑣𝑒𝑟 𝑤𝑒𝑖𝑟 + ℎ𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 + ℎ𝑙𝑜𝑠𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑑𝑜𝑤𝑛𝑐𝑜𝑚𝑒𝑟 𝑁𝑜𝑤, 𝐼 𝑛𝑒𝑒𝑑 𝑡𝑜 𝑒𝑠𝑡𝑖𝑚𝑎𝑡𝑒 𝑡ℎ𝑒 ℎ𝑙𝑜𝑠𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑑𝑜𝑤𝑛𝑐𝑜𝑚𝑒𝑟 𝑎𝑛𝑑 𝐼 𝑎𝑙𝑟𝑒𝑎𝑑𝑦 ℎ𝑎𝑣𝑒 𝑒𝑠𝑡𝑖𝑚𝑎𝑡𝑒𝑑 ℎ𝑤𝑒𝑖𝑟 , ℎ𝑐𝑟𝑒𝑠𝑡 𝑜𝑣𝑒𝑟 𝑤𝑒𝑖𝑟 , 𝑎𝑛𝑑 ℎ𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 . ℎ𝑙𝑜𝑠𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑑𝑜𝑤𝑛𝑐𝑜𝑚𝑒𝑟 𝑐𝑎𝑛 𝑏𝑒 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒𝑑 𝑏𝑦: 𝐿𝑤 2 ℎ𝑙𝑜𝑠𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑑𝑜𝑤𝑛𝑐𝑜𝑚𝑒𝑟 = 166 ∗ ( ) 𝜌𝑙 ∗ 𝐴𝑚 𝐴𝑑 ∶ 𝑇ℎ𝑒 𝑑𝑜𝑤𝑛𝑐𝑜𝑚𝑒𝑟 𝑎𝑟𝑒𝑎.

,

𝐴𝑎𝑝 : 𝑇ℎ𝑒 𝑐𝑙𝑒𝑎𝑟𝑎𝑛𝑐𝑒 𝑎𝑟𝑒𝑎 𝑢𝑛𝑑𝑒𝑟 𝑡ℎ𝑒 𝑑𝑜𝑤𝑛𝑐𝑜𝑚𝑒𝑟.

17

𝐴𝑎𝑝 = ℎ𝑎𝑝 ∗ 𝑙𝑤 Where ℎ𝑎𝑝 is height of the bottom edge of the apron above the plate. This height is normally set at 5 𝑡𝑜 10 𝑚𝑚, I chose 𝟏𝟎 𝒎𝒎: ℎ𝑎𝑝 = ℎ𝑤 − 10 = 50 − 10 = 40 𝑚𝑚 𝐴𝑎𝑝 = 40 ∗ 10−3 ∗ 2.22 = 0.089 𝑚2 𝑎𝑛𝑑 𝐴𝑑 = 0.83 𝑚2 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒𝑑 𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒, 𝐴𝑎𝑝 𝑤𝑎𝑠 𝑐ℎ𝑜𝑠𝑒𝑛 𝑎𝑠 𝑠𝑚𝑎𝑙𝑙𝑒𝑟 𝑎𝑟𝑒𝑎: 𝐿

ℎ𝑙𝑜𝑠𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑑𝑜𝑤𝑛𝑐𝑜𝑚𝑒𝑟 = 166 ∗ (𝜌 ∗𝐴𝑤 )2 (Sinnott, 2005) 𝑙

𝑎𝑝

2 1078 ℎ𝑙𝑜𝑠𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑑𝑜𝑤𝑛𝑐𝑜𝑚𝑒𝑟 = 166 ∗ ( ) = 1465.64 𝑚𝑚 437.1 ∗ 0.83

ℎ𝑏𝑎𝑐𝑘𝑢𝑝 = 50 + 44.42 + 559.48 + 1465.64 = 2.12 𝑚 𝑇𝑜 𝑐ℎ𝑒𝑐𝑘 𝑡ℎ𝑎𝑡 𝑜𝑢𝑟 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒𝑑 𝑏𝑎𝑐𝑘𝑢𝑝𝑠 𝑙𝑒𝑠𝑠 𝑡ℎ𝑎𝑛 𝑡ℎ𝑒 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑎𝑙𝑙𝑜𝑤𝑎𝑛𝑐𝑒 𝑤ℎ𝑖𝑐ℎ 𝑖𝑠 0.5 ∗ (𝑡𝑟𝑎𝑦 𝑠𝑝𝑎𝑐𝑖𝑛𝑔 + 𝑤𝑒𝑖𝑟 ℎ𝑖𝑒𝑔ℎ𝑡) > 0.367 This shows that the plate is acceptable as the downcomer backup is less.

9.8.3.8-Down comer Residence Time The recommended residence time is above three seconds and it can be calculated by: 𝑡𝑟𝑒𝑠𝑖𝑑𝑒𝑛𝑐𝑒 =

𝐴𝑑𝑜𝑤𝑛𝑐𝑜𝑚𝑒𝑟 ∗ ℎ𝑙𝑖𝑞𝑢𝑖𝑑𝑏𝑎𝑐𝑘𝑢𝑝 ∗ 𝜌𝑙 𝐿

𝑡𝑟𝑒𝑠𝑖𝑑𝑒𝑛𝑐𝑒 𝑎𝑡 𝑡ℎ𝑒 𝑡𝑜𝑝 =

0.83 ∗ 2.12 ∗ 443.3 = 55.64 𝑠𝑒𝑐𝑜𝑛𝑑𝑠 14.02

𝑡𝑟𝑒𝑠𝑖𝑑𝑒𝑛𝑐𝑒 𝑎𝑡 𝑡ℎ𝑒 𝑏𝑜𝑡𝑡𝑜𝑚 =

0.83 ∗ 2.12 ∗ 437.1 = 38.39 𝑠𝑒𝑐𝑜𝑛𝑑𝑠 20.03

So, the residence times for the top and the bottom above three seconds, therefore this will prevent the liquid carry under.

9.8.3.9-Check the Entrainment Entrainment is a result of high vapor flow rates and refers to the liquid carried up by vapor to the tray above. It is unfavorable as tray efficiency is reduced. To check it starts with: 𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑓𝑙𝑜𝑜𝑑𝑖𝑛𝑔 = 𝑢𝑓 = 𝐾1 ∗ √

𝜌𝑙 −𝜌𝑣 𝜌𝑣

𝑎𝑐𝑡𝑢𝑎𝑙 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑏𝑎𝑠𝑒𝑑 𝑜𝑛 𝑡ℎ𝑒 𝑛𝑒𝑡 𝑎𝑟𝑒𝑎 𝑢𝑓

𝑒𝑞. 11.81 (Sinnott, 2005)

𝑎𝑐𝑡𝑢𝑎𝑙 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑏𝑎𝑠𝑒𝑑 𝑜𝑛 𝑡ℎ𝑒 𝑛𝑒𝑡 𝑎𝑟𝑒𝑎, 𝐴𝑎𝑐 = 𝐴𝑎𝑐 𝑎𝑡 𝑡ℎ𝑒 𝑡𝑜𝑝 =

𝑣𝑎𝑝𝑜𝑟 𝑣𝑜𝑙𝑢𝑚𝑒𝑡𝑟𝑖𝑐 𝑓𝑙𝑜𝑤𝑟𝑎𝑡𝑒 𝑛𝑒𝑡 𝑎𝑟𝑒𝑎

0.667 𝑚 = 0.11 6.09 𝑠 18

𝐴𝑎𝑐 𝑎𝑡 𝑡ℎ𝑒 𝑏𝑜𝑡𝑡𝑜𝑚 =

1.24 𝑚 = 0.203 6.09 𝑠

𝐼 𝑎𝑙𝑟𝑒𝑎𝑑𝑦 ℎ𝑎𝑣𝑒 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒𝑑 𝑢𝑓 𝑓𝑜𝑟 𝑏𝑜𝑡ℎ 𝑡ℎ𝑒 𝑡𝑜𝑝 𝑎𝑛𝑑 𝑡ℎ𝑒 𝑏𝑜𝑡𝑡𝑜𝑚 . 𝑎𝑛𝑑 𝑡ℎ𝑒𝑦𝑤𝑒𝑟𝑒 0.31 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑡𝑜𝑝 𝑎𝑛𝑑 0.238 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑏𝑜𝑡𝑡𝑜𝑚. 𝑁𝑜𝑤, 𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑓𝑙𝑜𝑜𝑑𝑖𝑛𝑔 𝑐𝑎𝑛 𝑏𝑒 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒𝑑: 𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑓𝑙𝑜𝑜𝑑𝑖𝑛𝑔 𝑎𝑡 𝑡ℎ𝑒 𝑡𝑜𝑝 = 𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑓𝑙𝑜𝑜𝑑𝑖𝑛𝑔 𝑎𝑡 𝑡ℎ𝑒 𝑏𝑜𝑡𝑡𝑜𝑚 =

0.203 ∗ 0.238

0.11 ∗ 100% = 36% 0.31

100% = 85%

Now from figure 11.29 (Sinnott, 2005) ,and these flooding percentages, it can be estimated the entrainment fraction: 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑡𝑜𝑝, 𝑡ℎ𝑒 𝑒𝑛𝑡𝑟𝑎𝑖𝑚𝑒𝑛𝑡 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 = 0.0026 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑏𝑜𝑡𝑡𝑜𝑚, 𝑡ℎ𝑒 𝑒𝑛𝑡𝑟𝑎𝑖𝑚𝑒𝑛𝑡 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 = 0.055 Therefore, if check figure 11.29 , it can be seen clearly that the higher value of the entrainment fraction is 0.1, and based on our calculation both the top and the bottom below this value. Therefore, no modification need on the plate.

9.8.3.10-Trail Layout Cartridge will use as the construction type, this one allow 50 mm unperforated strip round plate edge; 50 mm wide calming zones (Sinnott, 2005). 𝑏𝑦 𝑢𝑠𝑖𝑛𝑔 𝒇𝒊𝒈𝒖𝒓𝒆 𝟏𝟏. 𝟑𝟐 (Sinnott, 2005) 𝑤𝑖𝑡ℎ

𝑙𝑤 𝐷𝑐

= 0.75, 𝑖𝑡 𝑐𝑎𝑛 𝑏𝑒 𝑒𝑣𝑎𝑙𝑢𝑎𝑡𝑒𝑑 𝜃𝑐 = 95°.

𝑇ℎ𝑒 𝑎𝑛𝑔𝑙𝑒 𝑠𝑢𝑏𝑡𝑒𝑛𝑑𝑒𝑑 = 180 − 𝜃𝑐 = 85° 𝑀𝑒𝑎𝑛 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑢𝑛𝑝𝑒𝑟𝑓𝑜𝑟𝑎𝑡𝑒𝑑 𝑒𝑑𝑔𝑒 𝑠𝑡𝑟𝑖𝑝𝑠 = (𝐷𝑐 − 𝑠𝑡𝑟𝑖𝑝 𝑒𝑑𝑔𝑒) ∗ 𝜋 ∗

𝑎𝑛𝑔𝑙𝑒 𝑠𝑢𝑏𝑡𝑒𝑛𝑑𝑒𝑑 180

𝑠𝑡𝑟𝑖𝑝 𝑒𝑑𝑔𝑒, 𝑟𝑒𝑐𝑜𝑚𝑚𝑒𝑛𝑑𝑒𝑑 = 50𝑚𝑚. 𝑀𝑒𝑎𝑛 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑢𝑛𝑝𝑒𝑟𝑓𝑜𝑟𝑎𝑡𝑒𝑑 𝑒𝑑𝑔𝑒 𝑠𝑡𝑟𝑖𝑝𝑠 = (2.97 − 0.05) ∗ 𝜋 ∗

85 = 4.33 𝑚 180

𝐴𝑟𝑒𝑎 𝑜𝑓 𝑢𝑛𝑝𝑒𝑟𝑓𝑜𝑟𝑎𝑡𝑒𝑑 𝑒𝑑𝑔𝑒 𝑠𝑡𝑟𝑖𝑝𝑠 = 𝑚𝑒𝑎𝑛 𝑙𝑒𝑛𝑔𝑡ℎ ∗ 𝑠𝑡𝑟𝑖𝑝 𝑒𝑑𝑔𝑒 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑢𝑛𝑝𝑒𝑟𝑓𝑜𝑟𝑎𝑡𝑒𝑑 𝑒𝑑𝑔𝑒 𝑠𝑡𝑟𝑖𝑝𝑠 = 4.33 ∗ 0.05 = 0.216 𝑚2 𝑀𝑒𝑎𝑛 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑐𝑎𝑙𝑚𝑖𝑛𝑔 𝑧𝑜𝑛𝑒𝑠 = 𝑤𝑒𝑖𝑟 𝑙𝑒𝑛𝑔𝑡ℎ + 𝑤𝑖𝑑𝑡ℎ 𝑜𝑓 𝑢𝑛𝑝𝑒𝑟𝑓𝑜𝑟𝑎𝑡𝑒𝑑 𝑠𝑡𝑟𝑖𝑝 𝑊𝑖𝑑𝑡ℎ 𝑜𝑓 𝑢𝑛𝑝𝑒𝑟𝑓𝑜𝑟𝑎𝑡𝑒𝑑 𝑠𝑡𝑟𝑖𝑝, 𝑟𝑒𝑐𝑜𝑚𝑚𝑒𝑛𝑑𝑒𝑑 = 50 𝑚𝑚. 𝑀𝑒𝑎𝑛 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑐𝑎𝑙𝑚𝑖𝑛𝑔 𝑧𝑜𝑛𝑒𝑠 = 2.22 + 0.05 = 2.27 𝑚 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑐𝑎𝑙𝑚𝑖𝑛𝑔 𝑧𝑜𝑛𝑒𝑠 = 2 ∗ (𝑐𝑎𝑙𝑚𝑖𝑛𝑔 𝑚𝑒𝑎𝑛 𝑙𝑒𝑛𝑔𝑡ℎ ∗ 0.05) = 0.23 𝑚2 𝑇𝑜𝑡𝑎𝑙 𝑎𝑟𝑒𝑎 𝑝𝑒𝑟𝑓𝑜𝑟𝑎𝑡𝑒𝑑 = 𝑎𝑐𝑡𝑖𝑣𝑒 𝑎𝑟𝑒𝑎 − 𝑢𝑛𝑝𝑒𝑟𝑓𝑜𝑟𝑎𝑡𝑒𝑑 𝑎𝑟𝑒𝑎 𝑒𝑑𝑔𝑒 − 𝑐𝑙𝑎𝑚𝑖𝑛𝑔 𝑧𝑜𝑛𝑒 𝑎𝑟𝑒𝑎 𝐴𝑝 = 5.26 − 0.216 − 0.23 = 4.814 𝑚2 Now the fraction of perforated area: 19

đ??´â„Žđ?‘œđ?‘™đ?‘’ đ??´đ?‘?đ?‘’đ?‘&#x;đ?‘“đ?‘œđ?‘&#x;đ?‘Žđ?‘Ąđ?‘’đ?‘‘

=

0.263 = 0.0546 4.814

đ?‘™

đ??ľđ?‘Ś đ?‘˘đ?‘ đ?‘–đ?‘›đ?‘” đ?’‡đ?’Šđ?’ˆđ?’–đ?’“đ?’† đ?&#x;?đ?&#x;?. đ?&#x;‘đ?&#x;‘ (Sinnott, 2005), đ??ˇđ?‘?đ?‘–đ?‘Ąđ?‘?â„Ž = 3.65, đ?‘–đ?‘Ą đ?‘–đ?‘ đ?‘ đ?‘Žđ?‘Ąđ?‘–đ?‘ đ?‘“đ?‘Žđ?‘?đ?‘Ąđ?‘œđ?‘&#x;đ?‘Ś, đ?‘?đ?‘’đ?‘?đ?‘Žđ?‘˘đ?‘ đ?‘’, đ?‘–đ?‘Ą đ?‘–đ?‘ đ?‘–đ?‘› â„Žđ?‘œđ?‘™đ?‘’

đ?‘Ąâ„Žđ?‘’ đ?‘&#x;đ?‘’đ?‘?đ?‘œđ?‘šđ?‘šđ?‘’đ?‘›đ?‘‘đ?‘’đ?‘‘ đ?‘&#x;đ?‘Žđ?‘”đ?‘’ 2.5 đ?‘Ąđ?‘œ 4 (Sinnott, 2005). đ??ťđ?‘œđ?‘™đ?‘’đ?‘ đ?‘?đ?‘–đ?‘Ąđ?‘?â„Ž = 3.65 ∗ 2.5 = 9.125 đ?‘šđ?‘š, đ?‘?đ?‘–đ?‘Ąđ?‘?â„Ž đ?‘Ąđ?‘Śđ?‘?đ?‘’ đ?‘Ąđ?‘&#x;đ?‘–đ?‘Žđ?‘›đ?‘”đ?‘˘đ?‘™đ?‘Žđ?‘&#x; đ??´đ?‘&#x;đ?‘’đ?‘Ž đ?‘œđ?‘“ đ?‘œđ?‘›đ?‘’ â„Žđ?‘œđ?‘™đ?‘’ = đ?‘ đ?‘˘đ?‘šđ?‘?đ?‘’đ?‘&#x; đ?‘œđ?‘“ â„Žđ?‘œđ?‘™đ?‘’đ?‘ =

đ?œ‹ đ?œ‹ ∗ đ??ˇâ„Žđ?‘œđ?‘™đ?‘’ 2 = ∗ 0.0052 = 0.00002 đ?‘š2 4 4

đ?‘Žđ?‘&#x;đ?‘’đ?‘Ž â„Žđ?‘œđ?‘™đ?‘’ 0.263 â„Žđ?‘œđ?‘™đ?‘’đ?‘ = = 13150 đ?‘Žđ?‘&#x;đ?‘’đ?‘Ž đ?‘œđ?‘“đ?‘ đ?‘–đ?‘›đ?‘”đ?‘™đ?‘’ â„Žđ?‘œđ?‘™đ?‘’đ?‘ 0.00002 đ?‘Ąđ?‘&#x;đ?‘Žđ?‘Ś

9.9-Tray to Tray Calculation Tray to tray calculation were completed utilizing a bottom up approach position to determine the compositions of the components on each of the 3 bottoms-most plates. These calculations were completed utilizing the equation of stripping section: đ?‘‰ (đ??ľ) ∗ (đ?‘Ś)đ?‘– đ?‘šâˆ’1 + đ?‘Ľđ??ľđ?‘– đ?‘Śđ?‘– ( )đ?‘š = đ?›źđ?‘– [ ] đ?‘‰ đ?‘Śâ„Ž ( ) ∗ (đ?‘Śâ„Ž )đ?‘šâˆ’1 + đ?‘Ľđ??ľâ„Ž đ??ľ đ??´đ?‘Ą đ?‘‡đ?‘?đ?‘œđ?‘Ąđ?‘Ąđ?‘œđ?‘š = 127.8℃ , đ?‘Ąâ„Žđ?‘’ đ?‘Łđ?‘œđ?‘™đ?‘Žđ?‘Ąđ?‘–đ?‘™đ?‘–đ?‘Ąđ?‘–đ?‘’đ?‘ đ?‘œđ?‘“ đ?‘Ąâ„Žđ?‘’ đ?‘?đ?‘œđ?‘šđ?‘?đ?‘œđ?‘›đ?‘’đ?‘›đ?‘Ąđ?‘ âˆś â—?The volatilities of the component at đ?‘‡đ?‘?đ?‘œđ?‘Ąđ?‘Ąđ?‘œđ?‘š = 127.8℃ can be calculated as I have shown in the calculation of the minimum number of stages(đ?‘ đ?‘’đ?‘’ 1), therefore: Tray one đ?’‚đ?’• đ?&#x;?đ?&#x;?đ?&#x;•. đ?&#x;–℃: đ??śđ?‘œđ?‘šđ?‘?đ?‘œđ?‘›đ?‘’đ?‘›đ?‘Ą

����

đ?‘˛ đ?’—đ?’‚đ?’?đ?’–đ?’†

đ??¸đ?‘Ąâ„Žđ?‘Žđ?‘›đ?‘’

179.1

7.786

14.41

đ?‘ƒđ?‘&#x;đ?‘œđ?‘?đ?‘Žđ?‘›đ?‘’

61.64

2.68

4.96

i-butane

30.8

1.34

2.48

n-butane

24.02

1.04

1.93

i-pentane

12.38

0.54

1

n-pentane

10.5

0.45

0.83

n-hexane

4.74

0.21

0.38

n-heptane

2.23

0.097

0.18

Mcyclohexane

2.03

0.088

0.16

Toluene

1.61

0.07

0.13

n-octane

1.07

0.045

0.08

Alpha value

20

E-benzene

0.8

0.035

0.06

Table 5: K & Alpha values of each component at 𝑇𝑏𝑜𝑡𝑡𝑜𝑚 = 127.8℃. 𝑁𝑜𝑤, 𝑡ℎ𝑒 𝑎𝑠𝑠𝑢𝑚𝑝𝑡𝑖𝑜𝑛 𝑡𝑜 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒 𝑡ℎ𝑒 𝑚𝑜𝑙. 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑎𝑡 𝑒𝑎𝑐ℎ 𝑠𝑡𝑎𝑔𝑒 𝑖𝑠 𝑡ℎ𝑎𝑡 𝑡ℎ𝑒 𝑣𝑜𝑙𝑎𝑡𝑖𝑙𝑖𝑡𝑖𝑒𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡𝑠 𝑎𝑟𝑒 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡. By using this correlation 𝑉 (𝐵) ∗ (𝑦)𝑖 𝑚−1 + 𝑥𝐵𝑖 𝑦𝑖 ( )𝑚 = 𝛼𝑖 [ ] 𝑉 𝑦ℎ (𝐵) ∗ (𝑦ℎ )𝑚−1 + 𝑥𝐵ℎ V/B = 4774/1078 provided from Hysys

𝐶𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡

i

XBi

Ym-1

Yi / Yh

𝐸𝑡ℎ𝑎𝑛𝑒 𝑃𝑟𝑜𝑝𝑎𝑛𝑒

0 0.0445

14.41 4.96

0.0001288 0.1073

i-butane

0.1601

2.48

n-butane

0.2575

i-pentane

Yi

4.35013E-05 0.1151453

8.848E-06 0.02342008

0.2261

0.514656265

0.10467897

1.93

0.3344

0.989881248

0.20133779

0.2058

1

0.159

1.00000289

0.20339649

n-pentane

0.1547

0.83

0.1108

0.854532122

0.17380833

n-hexane

0.1187

0.38

0.00048

0.349432275

0.07107309

n-heptane

0.0331

0.18

0.00785

0.414339129

0.08427488

Mcyclohexane

0.0179

0.16

0.00477

0.268041567

0.05451856

Toluene

0.0024

0.13

0.007233

0.291074532

0.05920337

n-octane

0.0048

0.08

0.00067

0.106698545

0.02170205

E-benzene

0.0004

0.06

0.0000659

0.012671948

0.00257742

Table 6: Mole fraction of vapour at Tray one 𝑇𝑏𝑜𝑡𝑡𝑜𝑚 = 127.8℃. For heavy key Xh=0.2058

𝑦𝑖 ⁄𝑦 𝑦𝑖 = 𝑦 ℎ ∑ 𝑖⁄𝑦ℎ And 𝑦ℎ =

1 𝑦 𝑖 ∑ ⁄𝑦ℎ

Then

𝑦ℎ = 0.20339

Tray two 𝒂𝒕 𝟏𝟏𝟖. 𝟐℃: 21

 Alpha value

𝐶𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡

𝑷𝒔𝒂𝒕

𝑲 𝒗𝒂𝒍𝒖𝒆

𝐸𝑡ℎ𝑎𝑛𝑒

160.7

7.056

15.68

𝑃𝑟𝑜𝑝𝑎𝑛𝑒

53.71

2.36

5.24

i-butane

26.48

1.16

2.57

n-butane

20.5

0.9

2

i-pentane

10.35

0.45

1

n-pentane

8.7

0.38

0.84

n-hexane

3.82

0.167

0.37

n-heptane

1.75

0.077

0.17

Mcyclohexane

1.61

0.071

0.157

Toluene

1.25

0.055

0.122

n-octane

0.82

0.036

0.08

E-benzene

0.61

0.026

0.057

Table 7: K & Alpha values of each component at 𝑇𝑏𝑜𝑡𝑡𝑜𝑚 = 118.2℃. 𝑁𝑜𝑤, 𝑡ℎ𝑒 𝑎𝑠𝑠𝑢𝑚𝑝𝑡𝑖𝑜𝑛 𝑡𝑜 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒 𝑡ℎ𝑒 𝑚𝑜𝑙. 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑎𝑡 𝑠𝑒𝑐𝑜𝑛𝑑 𝑠𝑡𝑎𝑔𝑒 𝑖𝑠 𝑡ℎ𝑎𝑡 𝑡ℎ𝑒 𝑣𝑜𝑙𝑎𝑡𝑖𝑙𝑖𝑡𝑖𝑒𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡𝑠 𝑎𝑟𝑒 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡. By using this correlation 𝑉 ( ) ∗ (𝑦)𝑖 𝑚−1 + 𝑥𝐵𝑖 𝑦𝑖 𝐵 ( )𝑚 = 𝛼𝑖 [ ] 𝑉 𝑦ℎ (𝐵) ∗ (𝑦ℎ )𝑚−1 + 𝑥𝐵ℎ V/B = 4789/1078 provided from Hysys.

𝐶𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡

i

XBi

Ym-1

Yi / Yh

Yi

𝐸𝑡ℎ𝑎𝑛𝑒

0

15.68

8.848E-06

𝑃𝑟𝑜𝑝𝑎𝑛𝑒

0.0445

5.24

0.02342008

2.25966E-06 0.025552884

2.10025E-07 0.002375024

i-butane

0.1601

2.57

0.10467897

0.219259565

0.020379177

n-butane

0.2575

2

0.20133779

0.519180012

0.048255415

i-pentane

0.2058

1

0.20339649

1.000001702

0.092945599

n-pentane

0.1547

0.84

0.17380833

0.99458991

0.092442598

n-hexane

0.1187

0.37

0.07107309

1.058393626

0.098372862

22

n-heptane

0.0331

0.17

0.08427488

2.160658787

0.200823384

Mcyclohexane

0.0179

0.157

0.05451856

1.493328746

0.138798099

Toluene

0.0024

0.122

0.05920337

1.960992525

0.182265315

n-octane

0.0048

0.08

0.02170205

1.140397137

0.105994715

E-benzene

0.0004

0.057

0.00257742

0.187399187

0.0174179

Table 8: Mole fraction of vapour at Tray Two 𝑇𝑏𝑜𝑡𝑡𝑜𝑚 = 118.2℃. 𝑦𝑖 =

𝑦𝑖 ⁄𝑦ℎ 𝑦 ∑ 𝑖⁄𝑦ℎ

And 𝑦ℎ =

1 ∑𝑦𝑖⁄𝑦ℎ

Then

𝑦ℎ = 0.093

Tray three 110.2℃:  Alpha value

𝐶𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡

𝑷𝒔𝒂𝒕

𝑲 𝒗𝒂𝒍𝒖𝒆

𝐸𝑡ℎ𝑎𝑛𝑒

146.8

6.5

16.66

𝑃𝑟𝑜𝑝𝑎𝑛𝑒

48.05

2.13

5.46

i-butane

23.2

1.028

2.64

n-butane

17.7

0.785

2.013

i-pentane

8.85

0.39

1

n-pentane

7.39

0.33

0.85

n-hexane

3.16

0.14

0.35

n-heptane

1.41

0.06

0.15

Mcyclohexane

1.31

0.058

0.15

Toluene

0.99

0.044

0.113

n-octane

0.64

0.028

0.071

E-benzene

0.47

0.021

0.054

Table 9: K & Alpha values of each component at 𝑇𝑏𝑜𝑡𝑡𝑜𝑚 = 110.2℃.

𝑁𝑜𝑤, 𝑡ℎ𝑒 𝑎𝑠𝑠𝑢𝑚𝑝𝑡𝑖𝑜𝑛 𝑡𝑜 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒 𝑡ℎ𝑒 𝑚𝑜𝑙. 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑎𝑡 𝑡ℎ𝑖𝑟𝑑 𝑠𝑡𝑎𝑔𝑒 𝑖𝑠 𝑡ℎ𝑎𝑡 𝑡ℎ𝑒 𝑣𝑜𝑙𝑎𝑡𝑖𝑙𝑖𝑡𝑖𝑒𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡𝑠 𝑎𝑟𝑒 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡. By using this correlation 𝑉 ( ) ∗ (𝑦)𝑖 𝑚−1 + 𝑥𝐵𝑖 𝑦𝑖 𝐵 ( )𝑚 = 𝛼𝑖 [ ] 𝑉 𝑦ℎ (𝐵) ∗ (𝑦ℎ )𝑚−1 + 𝑥𝐵ℎ V/B = 4753/1078 provided from Hysys 23

đ??śđ?‘œđ?‘šđ?‘?đ?‘œđ?‘›đ?‘’đ?‘›đ?‘Ą

ď Ąi

XBi

Ym-1

Yi / Yh

Yi

đ??¸đ?‘Ąâ„Žđ?‘Žđ?‘›đ?‘’

0

16.66

đ?‘ƒđ?‘&#x;đ?‘œđ?‘?đ?‘Žđ?‘›đ?‘’

0.0445

5.46

2.10025E-07 0.002375024

9.06182E-08 0.016414092

1.93463E-09 0.000350429

i-butane

0.1601

2.64

0.020379177

0.154356848

0.003295407

n-butane

0.2575

2.013

0.048255415

0.380861377

0.008131114

i-pentane

0.2058

1

0.092945599

1.003628276

0.021426735

n-pentane

0.1547

0.85

0.092442598

1.078474976

0.023024658

n-hexane

0.1187

0.35

0.098372862

2.573258264

0.054937196

n-heptane

0.0331

0.15

0.200823384

9.983461843

0.213139664

Mcyclohexane

0.0179

0.15

0.138798099

6.845928134

0.146155596

Toluene

0.0024

0.113

0.182265315

11.62893692

0.248269362

n-octane

0.0048

0.071

0.105994715

10.84134223

0.231454787

E-benzene

0.0004

0.054

0.0174179

2.330652655

0.049757742

Table 10: Mole fraction of vapour at Tray Three đ?‘‡đ?‘?đ?‘œđ?‘Ąđ?‘Ąđ?‘œđ?‘š = 110.2℃. đ?‘Śđ?‘– =

đ?‘Śđ?‘– â „đ?‘Śâ„Ž ∑đ?‘Śđ?‘–â „đ?‘Śâ„Ž

And đ?‘Śâ„Ž =

1 ∑đ?‘Śđ?‘–â „đ?‘Śâ„Ž

đ?‘Śâ„Ž = 0.0214

9.10-Mechanical Design 9.10.1-Materials Selection Distillation columns are constructed from plain carbon steels; low and high alloy steels, other alloys, clad plate, and reinforced plastics. Selection of a suitable material must take into account the suitability of the material for fabrication (particularly welding) as well as the compatibility of the material with the process environment. The distillation column design codes and standards include lists of acceptable materials; in accordance with the appropriate material standards (Sinnott, 2005). As we showed in memo 2, our selection for the construction material of the distillation columns is Carbon Steel . 9.10.2-Shell Thickness The column material used Carbon Steel and as I showed before that the maximum tension stress can đ?‘ be estimated by extrapolation at the design temperature. The tension stress is165 đ?‘šđ?‘š2: Materials Stainless Steel (18Cr/9Ni, Ti stabilised)

0 to 50 165

Design stress at temperature °C (N/mm2) 100 150 200 250 300 350 400 450

500

150

115

140

135

130

130

125

120

120

24

đ?‘ƒđ?‘‘đ?‘’đ?‘ đ?‘–đ?‘”đ?‘› = 1.15 ∗ 23 = 26.45 đ?œŽâ„Ž =

đ?‘ƒđ?‘‘đ?‘’đ?‘ đ?‘–đ?‘”đ?‘› ∗đ??ˇđ?‘–đ?‘›đ?‘Ąđ?‘’đ?‘&#x;đ?‘›đ?‘Žđ?‘™ 2∗đ?‘Ą

This gives that: đ?‘Ą=

đ?‘ƒ ∗ đ??ˇđ?‘–đ?‘›đ?‘Ąđ?‘’đ?‘&#x;đ?‘›đ?‘Žđ?‘™ 2 ∗ đ?œŽâ„Ž − đ?‘ƒ

đ?‘‡â„Žđ?‘’ đ?‘‘đ?‘’đ?‘ đ?‘–đ?‘”đ?‘› đ?‘?đ?‘&#x;đ?‘’đ?‘ đ?‘ đ?‘˘đ?‘&#x;đ?‘’, đ?‘ƒ = 1.15 ∗ 5.066 = 5.8259 đ?‘?đ?‘Žđ?‘&#x; đ?‘Ą=

2.645 ∗ 2970 = 23 đ?‘šđ?‘š 2 ∗ 165 − 2.3

Corrosion allowance is generally in the range of 2mm – 4mm (Stichlmair, 1998). 9.10.3-Head Design and Closures There are four typical types of heads of various shapes which are used for the end of a cylindrical vessel including:  Flat plates and formed flat heads  Hemispherical heads  Ellipsoidal heads  Torispherical heads Flat plates are usually used as covers for manways and are used as channel covers for heat exchangers. Formed flat ends are sometimes known as ‘flange-only’ heads. This type of head is the cheapest type to manufacture as fabrication costs are low, however, they are limited to low pressures and small diameter vessels. Ellipsoidal head are the most commonly used type for vessels with operating pressures up to 15 bars. They can be used with vessels with higher pressures. The hemispherical head type has the strongest shape and is able to withstand double the pressure of a tori spherical head which is of the same thickness. For this column, ellipsoidal was chosen. And the thickness of the head can be calculated by: đ?‘Ą=

đ?‘Ą=

đ?‘ƒđ?‘‘đ?‘’đ?‘ đ?‘–đ?‘”đ?‘› ∗ đ??ˇđ?‘–đ?‘›đ?‘Ąđ?‘’đ?‘&#x;đ?‘›đ?‘Žđ?‘™ 2 ∗ đ?œŽâ„Ž − 0.2đ?‘ƒđ?‘‘đ?‘’đ?‘ đ?‘–đ?‘”đ?‘›

2.645 ∗ 2970 = 23 đ?‘šđ?‘š 2 ∗ 165 − 0.2 ∗ 2.3

9.10.4-Pipe Sizing To calculate the diameter of the pipe, we need to assume the type of flow, therefore, based on the feed flow rate, the type of flow is turbulent. After this assumption we can use the flowing equation to calculate the optimum diameter: đ??ˇđ?‘œđ?‘?đ?‘Ąđ?‘–đ?‘šđ?‘˘đ?‘š =

0.363 ∗ đ?‘š0.45 ∗ đ?œ‡0.025 đ?œŒ0.32

To the feed section:

25

đ??ˇđ?‘œđ?‘?đ?‘Ąđ?‘–đ?‘šđ?‘˘đ?‘š =

0.363 ∗ 48.940.45 ∗ (1.4 ∗ 10−4 )0.025 551.20.32 đ??ˇđ?‘œđ?‘?đ?‘Ąđ?‘–đ?‘šđ?‘˘đ?‘š = 0.22 đ?‘š

To the top section: đ??ˇđ?‘œđ?‘?đ?‘Ąđ?‘–đ?‘šđ?‘˘đ?‘š =

0.363 ∗ 28.890.45 ∗ (7.8 ∗ 10−5 )0.025 = 0.18 đ?‘š 443.30.32

To the bottom section: đ??ˇđ?‘œđ?‘?đ?‘Ąđ?‘–đ?‘šđ?‘˘đ?‘š =

0.363 ∗ 20.330.45 ∗ (7.4 ∗ 10−5 )0.025 = 0.15 đ?‘š 437.10.32

Therefore, the sizing of the pipes will be as specified above. 9.10.5-Combined Loading Design 9.10.5.1-Dead Weight Calculation I will follow (Sinnott, 2005) method to calculate the weight loads in the column: 1- Column Weight: đ?‘Šđ?‘‰ = 240 ∗ đ??śđ?‘‰ ∗ đ?œ‹ ∗ đ??ˇđ?‘š ∗ (đ??ťđ?‘Ąđ?‘Žđ?‘›.đ?‘Ąđ?‘œ đ?‘Ąđ?‘Žđ?‘›đ?‘” + 0.8 ∗ đ??ˇđ?‘š ) ∗ đ?‘Ą ∗ 10−3 Where: đ??ťđ?‘Ąđ?‘Žđ?‘›.đ?‘Ąđ?‘œ đ?‘Ąđ?‘Žđ?‘›đ?‘” : â„Žđ?‘’đ?‘–đ?‘”â„Žđ?‘Ą đ?‘œđ?‘“ đ?‘Ąâ„Žđ?‘’ đ?‘?đ?‘œđ?‘™đ?‘˘đ?‘šđ?‘›. đ??śđ?‘‰ : đ?‘“đ?‘Žđ?‘?đ?‘Ąđ?‘œđ?‘&#x; đ?‘œđ?‘“ đ?‘Ąâ„Žđ?‘’ đ?‘’đ?‘ đ?‘Ąđ?‘–đ?‘šđ?‘Žđ?‘Ąđ?‘–đ?‘œđ?‘› đ?‘œđ?‘“ đ?‘›đ?‘œđ?‘§đ?‘§đ?‘™đ?‘’đ?‘ , đ?‘šđ?‘Žđ?‘›đ?‘¤đ?‘Žđ?‘Śđ?‘ đ?‘Žđ?‘›đ?‘‘ đ?‘–đ?‘›đ?‘Ąđ?‘’đ?‘&#x;đ?‘›đ?‘Žđ?‘™ đ?‘ đ?‘˘đ?‘?đ?‘?đ?‘œđ?‘&#x;đ?‘Ąđ?‘ đ?‘¤đ?‘’đ?‘–đ?‘”â„Žđ?‘Ą, đ?‘&#x;đ?‘’đ?‘?đ?‘œđ?‘šđ?‘šđ?‘’đ?‘›đ?‘‘đ?‘’đ?‘‘ 1.15. đ?‘Ą: đ?‘ â„Žđ?‘’đ?‘™đ?‘™ đ?‘Ąâ„Žđ?‘–đ?‘?đ?‘˜đ?‘›đ?‘’đ?‘ đ?‘ , đ?‘?đ?‘Žđ?‘™đ?‘?đ?‘˘đ?‘™đ?‘Žđ?‘Ąđ?‘’đ?‘‘ đ?‘–đ?‘› đ?‘ đ?‘’đ?‘?đ?‘Ąđ?‘–đ?‘œđ?‘›(10.2) đ?‘Ąđ?‘œ đ?‘?đ?‘’ 23 đ?‘šđ?‘š. đ??ˇđ?‘š : đ?‘šđ?‘’đ?‘Žđ?‘› đ?‘Łđ?‘’đ?‘ đ?‘ đ?‘’đ?‘™ đ?‘‘đ?‘–đ?‘Žđ?‘šđ?‘’đ?‘Ąđ?‘’đ?‘&#x; = đ??ˇđ?‘–đ?‘›đ?‘Ąđ?‘’đ?‘&#x;đ?‘›đ?‘Žđ?‘™ + đ?‘Ą ∗ 10−3 = 2.97 + 23 ∗ 10−3 = 2.99 đ?‘š. đ?‘Šđ?‘‰ = 240 ∗ 1.15 ∗ 2.99 ∗ (10.84 + 0.8 ∗ 2.97) ∗ 23 ∗ 10−3 đ?‘Šđ?‘‰ = 250.8 đ?‘˜đ?‘ 2- The Weight of Plates: đ??´đ?‘&#x;đ?‘’đ?‘Ž đ?‘œđ?‘“ đ?‘œđ?‘›đ?‘’ đ?‘?đ?‘™đ?‘Žđ?‘Ąđ?‘’ =

đ?œ‹ đ?œ‹ ∗ đ??ˇđ?‘?đ?‘œđ?‘™đ?‘˘đ?‘šđ?‘› 2 = ∗ 2.972 = 6.925 đ?‘š2 4 4

đ??´đ?‘&#x;đ?‘’đ?‘Ž đ?‘œđ?‘“ đ?‘Žđ?‘™đ?‘™ đ?‘Ąâ„Žđ?‘’ đ?‘?đ?‘™đ?‘Žđ?‘Ąđ?‘’đ?‘ = 6.925 ∗ 13 = 90.025 đ?‘š2 As suggested by Nelson the multiplier factor to calculate the weight of column internals and liquid đ?‘˜đ?‘ loading on the trays is 1.2 2 (Sinnott, 2005). đ?‘š

Therefore, đ?‘‡đ?‘œđ?‘Ąđ?‘Žđ?‘™ đ?‘¤đ?‘’đ?‘–đ?‘”â„Žđ?‘Ą đ?‘œđ?‘“ đ?‘?đ?‘™đ?‘Žđ?‘Ąđ?‘’đ?‘ = 1.2 ∗ 90.025 = 108.03 đ?‘˜đ?‘ 3- The Insulation Weight: I will choose the mineral wool as insulation with density, đ?œŒđ?‘š = 130 volume of the insulation by:

đ?‘˜đ?‘” đ?‘š3

(Sinnott, 2005) and the

26

𝐼𝑛𝑠𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑉𝑜𝑙𝑢𝑚𝑒 = 𝜋 ∗ 2 ∗ 𝑡𝑖𝑛𝑠𝑢𝑙𝑎𝑡𝑖𝑜𝑛 ∗ 𝐻𝑣 (Sinnott, 2005) 𝑡𝑖𝑛𝑠𝑢𝑙𝑎𝑡𝑖𝑜𝑛 : 𝑡ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠 𝑜𝑓 𝑖𝑛𝑠𝑢𝑙𝑎𝑡𝑖𝑜𝑛, 𝑠𝑝𝑒𝑐𝑖𝑓𝑖𝑒𝑑 75 𝑚𝑚. 𝐼𝑛𝑠𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑉𝑜𝑙𝑢𝑚𝑒 = 𝜋 ∗ 2 ∗ 0.075 ∗ 10.84 = 5.11 𝑚3 𝑊𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑖𝑛𝑠𝑢𝑙𝑎𝑡𝑖𝑜𝑛 = 𝑣𝑜𝑙𝑢𝑚𝑒 ∗ 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 ∗ 𝑎𝑐𝑐𝑒𝑙. 𝑊𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑖𝑛𝑠𝑢𝑙𝑎𝑡𝑖𝑜𝑛 = 6.52 ∗ 130 ∗ 9.81 = 6.52 𝑘𝑁 The value for the weight of insulation has to be doubled in order to allow for fittings: 𝑇ℎ𝑒 𝑡𝑜𝑡𝑎𝑙 𝑤𝑒𝑖𝑔ℎ𝑡 𝑤𝑖𝑡ℎ 𝑓𝑖𝑡𝑡𝑖𝑛𝑔𝑠 = 2 ∗ 𝑊𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑖𝑛𝑠𝑢𝑙𝑎𝑡𝑖𝑜𝑛 = 13.04 𝑘𝑁 4- Caged Ladder Weight and Platforms Weight: Assume the caged ladder is steel with 360 𝑘𝑁/𝑚 is attached to the column with platforms1.7 𝑘𝑁/ 𝑚2 . 𝑇𝑜𝑡𝑎𝑙 𝐶𝑎𝑔𝑒𝑑 𝑙𝑎𝑑𝑑𝑒𝑟 𝑤𝑒𝑖𝑔ℎ𝑡 = 360 ∗ 10.84 = 3.9 𝑘𝑁 𝑇𝑜𝑡𝑎𝑙 𝑝𝑙𝑎𝑡𝑓𝑜𝑟𝑚𝑠 𝑤𝑒𝑖𝑔ℎ𝑡 = 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑝𝑙𝑎𝑡𝑓𝑜𝑟𝑚𝑠 ∗ 𝑐𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝑎𝑟𝑒𝑎 ∗ 1.7 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑝𝑙𝑎𝑡𝑓𝑜𝑟𝑚𝑠 =

𝐻𝑣 10.84 = = 2.7 𝑠𝑎𝑦 3 4 4

This means that 3 𝑚 as height between each two platforms. 𝐶𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝑎𝑟𝑒𝑎 =

𝜋 𝜋 ∗ 𝐷 2 = ∗ 2.972 = 6.924 𝑚2 4 4

𝑇𝑜𝑡𝑎𝑙 𝑝𝑙𝑎𝑡𝑓𝑜𝑟𝑚𝑠 𝑤𝑒𝑖𝑔ℎ𝑡 = 3 ∗ 1.7 ∗ 6.924 = 35.31𝑘𝑁 5- Total Weight: 𝑇𝑜𝑡𝑎𝑙 𝑊𝑒𝑖𝑔ℎ𝑡 = 250.8 + 180.03 + 13.04 + 3.9 + 35.31 = 483.1 𝑘𝑁 𝑻𝒐𝒕𝒂𝒍 𝑾𝒆𝒊𝒈𝒉𝒕 = 𝟓𝟎 𝒕𝒐𝒏𝒏𝒆𝒔 9.10.5.2-Wind Loading Our column height is 10.84 𝑚, therefore, the wind loading is designed to wind speed 160 𝑘𝑚/ℎ. A column must be designed to withstand the highest wind speed that is likely to be encountered at the site during the life of the plant. 𝑃𝑤 = 0.07 𝑢𝑤 2 (Sinnott, 2005) 𝑃𝑤 : 𝑤𝑖𝑛𝑑 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒

𝑁 . 𝑚2

𝑢𝑤 : 𝑤𝑖𝑛𝑑 𝑠𝑝𝑒𝑒𝑑 𝑃𝑤 = 0.07 ∗ 1602 = 1792

𝑘𝑚 . ℎ

𝑁 𝑚2

𝑁𝑜𝑤, 𝑖𝑡 𝑐𝑎𝑛 𝑏𝑒 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒𝑑 𝑡ℎ𝑒 𝑙𝑜𝑎𝑑𝑖𝑛𝑔 𝑏𝑦: 𝐹𝑤 = 𝑃𝑤 ∗ 𝑒𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 (Sinnott, 2005) 𝑒𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 = 𝐷𝑐 + 2 ∗ (𝑡𝑖𝑛𝑠𝑢𝑙𝑙𝑎𝑡𝑖𝑜𝑛 + 𝑡𝑤𝑎𝑙𝑙 ) 𝑒𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 = 2.97 + 2 ∗ (75 + 23) ∗ 10−3 = 3.166 𝑚

27

đ??šđ?‘¤ = 1792 ∗ 3.166 = 5673.47

đ?‘ đ?‘š

đ??´đ?‘ đ?‘ đ?‘˘đ?‘šđ?‘’ đ?‘Ąâ„Žđ?‘’ đ?‘ đ?‘˜đ?‘–đ?‘&#x;đ?‘Ą â„Žđ?‘’đ?‘–đ?‘”â„Žđ?‘Ą 2 đ?‘š, đ?‘&#x;đ?‘’đ?‘?đ?‘œđ?‘šđ?‘šđ?‘’đ?‘›đ?‘‘đ?‘’đ?‘‘ (Wankat, 1988). đ?‘‡â„Žđ?‘’đ?‘&#x;đ?‘’đ?‘“đ?‘œđ?‘&#x;đ?‘’, đ?‘Ąâ„Žđ?‘’ đ?‘Ąđ?‘œđ?‘Ąđ?‘Žđ?‘™ â„Žđ?‘’đ?‘–đ?‘”â„Žđ?‘Ą đ?‘–đ?‘ 12.84 đ?‘š. đ?‘ đ?‘œđ?‘¤, đ?‘Ąâ„Žđ?‘’ đ?‘Ąđ?‘œđ?‘Ąđ?‘Žđ?‘™ đ?‘?đ?‘’đ?‘›đ?‘‘đ?‘–đ?‘›đ?‘” đ?‘šđ?‘œđ?‘šđ?‘’đ?‘›đ?‘Ą đ?‘Žđ?‘Ą đ?‘Ąâ„Žđ?‘’ đ?‘?đ?‘œđ?‘Ąđ?‘Ąđ?‘œđ?‘š đ?‘Ąđ?‘Žđ?‘›đ?‘”đ?‘’đ?‘›đ?‘Ą đ?‘™đ?‘–đ?‘›đ?‘’: đ?‘€đ?‘Ľ =

5673.47 ∗ 2

12.842 = 467680.012 đ?‘ đ?‘š (Sinnott, 2005)

9.10.5.3-Earthquake loading The movement of the earth’s surface during an earthquake produces horizontal shear forces on tall self-supported vessels, the magnitude of which increases from the base upward (Sinnott, 2005). The possibility of earthquake damage may be considered if the site is a Major Hazards installation, in addition to, as we mentioned be for that our project site is not in a seismic zone. 9.10.6-Stress Analysis Primary stresses 1- The longitudinal and circumferential stresses due to pressure (internal or external), given by: đ?‘ƒ ∗ đ??ˇđ?‘–đ?‘›đ?‘Ąđ?‘’đ?‘&#x;đ?‘›đ?‘Žđ?‘™ 1.15 ∗ 2.3 ∗ 2970 đ?œŽđ??ż = = = 85.38đ?‘ /đ?‘šđ?‘š2 4∗đ?‘Ą 4 ∗ 23 đ?œŽâ„Ž =

đ?‘ƒ ∗ đ??ˇđ?‘–đ?‘›đ?‘Ąđ?‘’đ?‘&#x;đ?‘›đ?‘Žđ?‘™ 1.15 ∗ 2.3 ∗ 2970 = = 170.775đ?‘ /đ?‘šđ?‘š2 2∗đ?‘Ą 2 ∗ 23

2- The direct stress đ?œŽđ?‘¤ due to the weight of the vessel by: đ?‘Š 483.1 ∗ 1000 đ?œŽđ?‘¤ = = = 2.24đ?‘ /đ?‘šđ?‘š2 đ?œ‹ ∗ (đ??ˇđ?‘– + đ?‘Ą) ∗ đ?‘Ą đ?œ‹ ∗ (2970 + 23) ∗ 23 3- The bending stresses will be compressive or tensile, depending on location, and 467680.012are given by: đ?‘€ đ??ˇđ?‘– đ?œŽđ?‘? = Âą ∗ ( + đ?‘Ą) đ??źđ?‘Ł 2 đ??źđ?‘Ł =

đ?œ‹ (đ??ˇ 4 − đ??ˇđ?‘– 4 ) 4 đ?‘œ

, đ??ˇđ?‘œ = 2970 + 2 ∗ 3 = 2976đ?‘šđ?‘š

đ??źđ?‘Ł = 4.95 ∗ 1011 đ?‘š4 , and the bending stress, đ?œŽđ?‘? = Âą2 đ?‘ /đ?‘šđ?‘š2 The resultant longitudinal stress is: đ?œŽđ?‘§ = đ?œŽđ??ż + đ?œŽđ?‘¤ Âą đ?œŽđ?‘? đ?œŽđ?‘§,đ?‘˘đ?‘?đ?‘¤đ?‘–đ?‘›đ?‘‘ = 85.38 − 2.24 + 2 = 85.14 đ?‘ /đ?‘šđ?‘š2 đ?œŽđ?‘§,đ?‘‘đ?‘œđ?‘¤đ?‘›đ?‘¤đ?‘–đ?‘›đ?‘‘ = 85.38 − 2.24 − 2 = 81.14 đ?‘ /đ?‘šđ?‘š2 đ?‘‡â„Žđ?‘’ đ?‘‘đ?‘’đ?‘ đ?‘–đ?‘”đ?‘› đ?‘ đ?‘Ąđ?‘&#x;đ?‘’đ?‘ đ?‘ đ?‘’đ?‘Ľđ?‘?đ?‘’đ?‘&#x;đ?‘–đ?‘’đ?‘›đ?‘?đ?‘’đ?‘‘ đ?‘?đ?‘Žđ?‘› đ?‘?đ?‘’ đ?‘?đ?‘Žđ?‘™đ?‘?đ?‘˘đ?‘™đ?‘Žđ?‘Ąđ?‘’đ?‘‘ đ?‘Žđ?‘ đ?‘Ąâ„Žđ?‘’ đ?‘‘đ?‘–đ?‘“đ?‘“đ?‘’đ?‘&#x;đ?‘’đ?‘›đ?‘?đ?‘’ đ?‘?đ?‘’đ?‘Ąđ?‘¤đ?‘’đ?‘’đ?‘› đ?œŽâ„Ž đ?‘Žđ?‘›đ?‘‘ đ?œŽđ?‘§ đ?‘–đ?‘› đ?‘?đ?‘œđ?‘Ąâ„Ž đ?‘?đ?‘Žđ?‘ đ?‘’đ?‘ đ?‘˘đ?‘?đ?‘¤đ?‘–đ?‘›đ?‘‘ đ?‘Žđ?‘›đ?‘‘ đ?‘‘đ?‘œđ?‘¤đ?‘›đ?‘¤đ?‘–đ?‘›đ?‘‘. đ?‘‡â„Žđ?‘’ đ?‘”đ?‘&#x;đ?‘’đ?‘Žđ?‘Ąđ?‘’đ?‘ đ?‘Ą đ?‘‘đ?‘–đ?‘“đ?‘“đ?‘’đ?‘&#x;đ?‘’đ?‘›đ?‘?đ?‘’ đ?‘?đ?‘’đ?‘Ąđ?‘¤đ?‘’đ?‘’đ?‘› đ?‘Ąâ„Žđ?‘’đ?‘š đ?‘¤đ?‘–đ?‘™đ?‘™ đ?‘?đ?‘’ đ?‘Ąâ„Žđ?‘’ đ?‘‘đ?‘’đ?‘ đ?‘–đ?‘”đ?‘› đ?‘ đ?‘Ąđ?‘&#x;đ?‘’đ?‘ đ?‘ đ?‘’đ?‘Ľđ?‘?đ?‘’đ?‘&#x;đ?‘–đ?‘’đ?‘›đ?‘?đ?‘’đ?‘‘. đ?‘‡â„Žđ?‘’ đ?‘‘đ?‘’đ?‘ đ?‘–đ?‘”đ?‘› đ?‘ đ?‘Ąđ?‘&#x;đ?‘’đ?‘ đ?‘ đ?‘’đ?‘Ľđ?‘?đ?‘’đ?‘&#x;đ?‘–đ?‘’đ?‘›đ?‘?đ?‘’đ?‘‘ đ?‘˘đ?‘?đ?‘¤đ?‘–đ?‘›đ?‘‘ = 170.775 − 81.14 = 89.64đ?‘ /đ?‘šđ?‘š2 28

đ?‘‡â„Žđ?‘’đ?‘&#x;đ?‘’đ?‘“đ?‘œđ?‘&#x;đ?‘’, đ?‘Ąâ„Žđ?‘’ đ?‘‘đ?‘’đ?‘ đ?‘–đ?‘”đ?‘› đ?‘ đ?‘Ąđ?‘&#x;đ?‘’đ?‘ đ?‘ đ?‘’đ?‘Ľđ?‘?đ?‘’đ?‘&#x;đ?‘–đ?‘’đ?‘›đ?‘?đ?‘’đ?‘‘ = 89.64 đ?‘ đ?‘Ąđ?‘&#x;đ?‘’đ?‘ đ?‘ đ?‘–đ?‘ 165

đ?‘ , đ?‘Žđ?‘›đ?‘‘ đ?‘Ąâ„Žđ?‘’ đ?‘Žđ?‘?đ?‘Ąđ?‘˘đ?‘Žđ?‘™ đ?‘‘đ?‘’đ?‘ đ?‘–đ?‘”đ?‘› đ?‘šđ?‘š2

đ?‘ . đ?‘°đ?’• đ?’Šđ?’” đ?’‚đ?’„đ?’„đ?’†đ?’ƒđ?’•đ?’‚đ?’ƒđ?’?đ?’†. đ?‘šđ?‘š2

9.10.7-Vessel Support Design 9.10.7.1-Skirt Design Skirt supports are recommended for vertical vessels as they do not impose concentrated loads on the vessel shell; they are particularly suitable for use with tall columns subject to wind loading (Sinnott, 2005). The skirts use to hold the column on the ground and they have to be opening for any pipe connection. Almost, the material of the skirts same as the column material and it is welded to the outside surface of the column. Skirt Thickness Calculation The skirt thickness must be sufficient to withstand the dead-weight loads and bending moments imposed on it by the vessel; it will not be under the vessel pressure. The resultant stresses in the skirt will be: đ?œŽđ?‘Ąđ?‘’đ?‘›đ?‘ đ?‘–đ?‘™đ?‘’ = đ?œŽđ?‘?đ?‘ − đ?œŽđ?‘¤đ?‘

đ?œŽđ?‘?đ?‘œđ?‘šđ?‘?đ?‘&#x;đ?‘’đ?‘ đ?‘ đ?‘–đ?‘Łđ?‘’ = đ?œŽđ?‘?đ?‘ + đ?œŽđ?‘¤đ?‘

And

Where: đ?œŽđ?‘?đ?‘ = đ?‘?đ?‘’đ?‘›đ?‘‘đ?‘–đ?‘›đ?‘” đ?‘ đ?‘Ąđ?‘&#x;đ?‘’đ?‘ đ?‘ đ?‘–đ?‘› đ?‘Ąâ„Žđ?‘’ đ?‘ đ?‘˜đ?‘–đ?‘&#x;đ?‘Ą, đ?œŽđ?‘?đ?‘ =

4 ∗ đ?‘€đ?‘ đ?œ‹ ∗ (đ??ˇđ?‘ + đ?‘Ąđ?‘ ) ∗ đ?‘Ąđ?‘ ∗ đ??ˇđ?‘

đ?œŽđ?‘¤đ?‘ = đ?‘‘đ?‘’đ?‘Žđ?‘‘ đ?‘¤đ?‘’đ?‘–đ?‘”â„Žđ?‘Ą đ?‘ đ?‘Ąđ?‘&#x;đ?‘’đ?‘ đ?‘ đ?‘–đ?‘› đ?‘Ąâ„Žđ?‘’ đ?‘ đ?‘˜đ?‘–đ?‘&#x;đ?‘Ą. đ?œŽđ?‘¤đ?‘ =

đ?‘Š đ?œ‹ ∗ (đ??ˇđ?‘ + đ?‘Ąđ?‘ ) ∗ đ?‘Ąđ?‘

Where: đ?‘€đ?‘ = Maximum bending moment, evaluated at the base of the skirt (due to wind, seismic and eccentric loads). đ?‘Š = Total weight of the vessel and contents đ??ˇđ?‘ = Inside diameter of the skirt, at the base đ?‘Ąđ?‘ = Skirt thickness, first assume it is 6 đ?‘šđ?‘š and let check: đ?œŽđ?‘?đ?‘ =

4 ∗ 467680.012 ∗ 103 đ?‘ = 11.24 đ?œ‹ ∗ (2970 + 6) ∗ 6 ∗ 2970 đ?‘šđ?‘š2

đ?œŽđ?‘¤đ?‘ = đ?œŽđ?‘¤đ?‘ ,đ?‘Ąđ?‘’đ?‘ đ?‘Ą =

483.1 ∗ 103 đ?‘ = 8.7 đ?œ‹ ∗ (2970 + 6) ∗ 6 đ?‘šđ?‘š2 1108.34 ∗ 103 đ?‘ = 19.76 đ?œ‹ ∗ (2970 + 6) ∗ 6 đ?‘šđ?‘š2

đ?‘€đ?‘Žđ?‘Ľđ?‘–đ?‘šđ?‘˘đ?‘š đ?‘‡đ?‘’đ?‘›đ?‘ đ?‘–đ?‘™đ?‘’ đ?‘ đ?‘Ąđ?‘&#x;đ?‘’đ?‘ đ?‘ = 11.24 − 8.7 = 23.92

đ?‘ đ?‘šđ?‘š2 29

đ?‘€đ?‘Žđ?‘Ľđ?‘–đ?‘šđ?‘˘đ?‘š đ??śđ?‘œđ?‘šđ?‘?đ?‘&#x;đ?‘’đ?‘ đ?‘ đ?‘–đ?‘Łđ?‘’ đ?‘ đ?‘Ąđ?‘&#x;đ?‘’đ?‘ đ?‘ = 11.24 + 19.76 = 32

đ?‘ đ?‘šđ?‘š2

Checking: 32 < 160 đ?‘–đ?‘Ą đ?‘–đ?‘ đ?‘‚đ?‘˜ đ?‘ťđ?’‰đ?’†đ?’“đ?’†đ?’‡đ?’?đ?’“đ?’†, đ?’•đ?’‰đ?’† đ?’•đ?’‰đ?’Šđ?’„đ?’Œđ?’?đ?’†đ?’”đ?’” đ?’˜đ?’Šđ?’?đ?’? đ?’ƒđ?’† đ?&#x;” đ?’Žđ?’Ž (đ?’•đ?’‰đ?’† đ?’Žđ?’Šđ?’?đ?’Šđ?’Žđ?’–đ?’Ž đ?’“đ?’†đ?’„đ?’?đ?’Žđ?’Žđ?’†đ?’?đ?’…đ?’†đ?’… đ?’•đ?’‰đ?’Šđ?’Œđ?’?đ?’†đ?’”đ?’” đ?&#x;” đ?’Žđ?’Ž (Sinnott, 2005)). 9.10.7.2-Base ring and Anchor Bolt Design The loads carried by the skirt are transmitted to the foundation slab by the skirt base ring (bearing plate). The moment produced by wind and other lateral loads will tend to overturn the vessel; this will be opposed by the couple set up by the weight of the vessel and the tensile load in the anchor bolts. The anchor bolts are assumed to share the overturning load equally, and the bolt area required is given by: đ??´đ?‘? =

1 đ?‘ đ?‘? ∗đ?‘“đ?‘?

∗[

4∗đ?‘€đ?‘ đ??ˇđ?‘?đ?‘?

− đ?‘¤] (Sinnott, 2005)

Where: đ??´đ?‘? = Area of one bolt at the root of the thread, đ?‘šđ?‘š2 đ?‘ đ?‘? = đ?‘ đ?‘˘đ?‘šđ?‘?đ?‘’đ?‘&#x; đ?‘œđ?‘“ đ??ľđ?‘œđ?‘™đ?‘Ąđ?‘ . đ?‘“đ?‘? = Maximum allowable bolt stress,đ?‘ /đ?‘šđ?‘š2 , typical design value 125 đ?‘ /đ?‘šđ?‘š2 . đ?‘€đ?‘ = Bending (overturning) moment at the base, đ?‘ đ?‘š đ?‘Š = Weight of the vessel đ??ˇđ?‘?đ?‘? = Bolt circle diameter, m đ?‘ đ?‘? =

đ?‘?đ?‘œđ?‘™đ?‘Ą đ?‘&#x;đ?‘–đ?‘›đ?‘” đ?‘?đ?‘œđ?‘™đ?‘Ą đ?‘ đ?‘?đ?‘Žđ?‘?đ?‘–đ?‘›đ?‘”

(Sinnott, 2005)

đ??´đ?‘ đ?‘ đ?‘˘đ?‘šđ?‘’ đ?‘?đ?‘œđ?‘™đ?‘Ą đ?‘ đ?‘?đ?‘Žđ?‘?đ?‘–đ?‘›đ?‘” = 600đ?‘šđ?‘š (Sinnott, 2005) đ?‘Žđ?‘›đ?‘‘ đ?‘Ąâ„Žđ?‘’ đ?‘?đ?‘–đ?‘Ąđ?‘?â„Ž đ?‘?đ?‘œđ?‘™đ?‘Ą đ?‘‘đ?‘–đ?‘Žđ?‘šđ?‘’đ?‘Ąđ?‘’đ?‘&#x; 1.32đ?‘š. đ?‘?đ?‘œđ?‘™đ?‘Ą đ?‘&#x;đ?‘–đ?‘›đ?‘” = đ?œ‹ ∗ 1320 = 4146.9 đ?‘šđ?‘š đ?‘ đ?‘? =

4146.9 600

= 6.912 đ?‘ đ?‘Žđ?‘Ś 8 đ?‘&#x;đ?‘’đ?‘?đ?‘œđ?‘šđ?‘šđ?‘’đ?‘›đ?‘‘đ?‘’đ?‘‘ (Sinnott, 2005) đ??´đ?‘? =

1 4 ∗ 467680.012 ∗[ − 483.1 ∗ 1000] = 934.12 đ?‘šđ?‘š2 8 ∗ 125 1.32 4 ∗ đ??´đ?‘? đ??ľđ?‘œđ?‘™đ?‘Ą đ?‘&#x;đ?‘œđ?‘œđ?‘Ą đ?‘‘đ?‘–đ?‘Žđ?‘šđ?‘’đ?‘Ąđ?‘’đ?‘&#x; = √ = 0.05 đ?‘š đ?œ‹

One of Scheiman guide rules which can be used for the selection of the anchor bolts that bolts smaller than 25 mm (1 in.) diameter should not be used (Sinnott, 2005). Therefore, the bolts diameter is large enough to consider.

30

9.10.8-Flange Selection Several different types of ange are used for various applications. The principal types used in the process industries are: 1. Welding-neck anges. 3. Lap-joint anges.

2. Slip-on anges, hub and plate types. 4. Screwed anges.

5. Blank, or blind, anges.

Welding-neck anges: have a long tapered hub between the ange ring and the welded joint. This gradual transition of the section reduces the disconti- nuity stresses between the ange and branch, and increases the strength of the ange assembly. Welding-neck anges are suitable for extreme service conditions; where the ange is likely to be subjected to temperature, shear and vibration loads. They will normally be speciďŹ ed for the connections and nozzles on process vessels and process equipment (Sinnott, 2005). Therefore, the selection welding-neck flanges, because showed, it is suitable for different rage of temperature. 9.10.9-Gaskets Selection A gasket is a sealing component placed between anges to create a static seal between the two stationary anges of a mechanical assembly and maintain that seal under all design and operating conditions, which may vary depending on changes in pressure and temperature during the lifetime of the flange. Gasket selection depends on process condition maintenance requirements and flange types. Rubber and Elastomers. This group includes natural rubber and the many synthetic grades of elastomer, like neoprene, nitrile, butyl, ethylene propylenediene, styrene butadiene, and Viton. Each elastomer has its own mechanical characteristics and resistance to process media. When used independently, these elastomeric materials are selected for media at lower design temperatures, 3928F (2008C) maximum and low pressures, ASME 150 and 300 class. Elastomers are best suited to transport noncorrosive hydrocarbons and for utility services. 9.10.10-Flange Face Selection Flange faces Flanges are also classiďŹ ed according to the type of ange face used. There are two basic types: 1. Full-faced anges, where the face contacts area extends outside the circle of bolts; over the full face of the ange. 2. Narrow-faced anges, where the face contacts area is located within the circle of bolts. Full face, wide-faced, anges are simple and inexpensive, but are only suitable for low pressures. The gasket area is large, and an excessively high bolt tension would be needed to achieve sufďŹ cient gasket pressure to maintain a good seal at high operating pressures. The raised face, narrow-faced, ange is probably the most commonly used type of ange for process equipment (Sinnott, 2005).

9.10.11-Flange Standards All the flanges in the design will be based on the đ?‘Šđ?‘şđ?&#x;?đ?&#x;“đ?&#x;”đ?&#x;Ž standard. The design of the flange will be design based on the design pressure and temperature. The design pressure for the flanges utilization will be . đ?&#x;“ đ?‘ˇđ?’”đ?’Š , the nearest standard is đ?&#x;–đ?&#x;• đ?‘ˇđ?’”đ?’Š (Sinnott, 2005). 9.10.12-Pressure Relief System Emergency relief in the process industries aims to protect equipment, the environment and operating personnel from abnormal conditions. In addition, pressure relief system is also required 31

on every vessel to prevent overpressure and rupture of the vessel. The type of pressure relief valve chosen was conventional pressure relief valve, because they have high sensitive to any change in the pressure and they have low failure rate (Ludwig E. E., 1999).. The minimum required effective discharge area can be calculated by: (Parry, 1992) đ??´ =

đ?‘„ ∗ √đ??ş 272 ∗ đ??žđ?‘ƒ ∗ đ??žđ?‘Š ∗ đ??žđ?‘‰ ∗ √∆đ?‘ƒ

‌ đ?‘’đ?‘žđ??´

đ?‘„ = đ?‘…đ?‘’đ?‘žđ?‘˘đ?‘–đ?‘&#x;đ?‘’đ?‘‘ đ?‘&#x;đ?‘’đ?‘™đ?‘–đ?‘’đ?‘Łđ?‘–đ?‘›đ?‘” đ?‘?đ?‘Žđ?‘?đ?‘Žđ?‘?đ?‘–đ?‘Ąđ?‘Ś, đ?‘”đ?‘Žđ?‘™đ?‘™đ?‘œđ?‘› đ?‘?đ?‘’đ?‘&#x; đ?‘šđ?‘–đ?‘›, 892 đ?‘?đ?‘&#x;đ?‘œđ?‘Łđ?‘–đ?‘‘đ?‘’đ?‘‘ đ?‘“đ?‘&#x;đ?‘œđ?‘š đ??ťđ?‘Śđ?‘ đ?‘Śđ?‘ . đ??ş = đ?‘†đ?‘?đ?‘’đ?‘?đ?‘–đ?‘“đ?‘–đ?‘? đ??şđ?‘&#x;đ?‘Žđ?‘Łđ?‘–đ?‘Ąđ?‘Ś =

đ?œŒ đ?œŒđ?‘¤đ?‘Žđ?‘Ąđ?‘’đ?‘&#x;

=

551.2 = 0.5512 1000

đ??žđ?‘ƒ = đ??śđ?‘œđ?‘&#x;đ?‘&#x;đ?‘’đ?‘?đ?‘Ąđ?‘–đ?‘œđ?‘› đ??šđ?‘Žđ?‘?đ?‘Ąđ?‘?đ?‘&#x; đ?‘œđ?‘“ đ?‘&#x;đ?‘’đ?‘™đ?‘–đ?‘’đ?‘Łđ?‘–đ?‘›đ?‘” đ?‘?đ?‘Žđ?‘?đ?‘Žđ?‘?đ?‘–đ?‘Ąđ?‘Ś, đ??žđ?‘Ł = Correction factor for viscosity . đ??žđ?‘¤ = Correction factor for back pressure . ∆đ?‘ƒ = Differential pressure (set pressure, psig − back pressure, psig)

Typical values of Kp range from 0.3 for an overpressure of 0%, 1.0 for 25%, and up to 1.1 for an overpressure of 50%. Since the overpressure of Conventional pressure relief valve was is 10% , so use the following equation : For % overpressure < 25, đ??žđ?‘? = −0.0014(%đ?‘œđ?‘Łđ?‘’đ?‘&#x;đ?‘?đ?‘&#x;đ?‘’đ?‘ đ?‘ đ?‘˘đ?‘&#x;đ?‘’)2 + 0.073((%đ?‘œđ?‘Łđ?‘’đ?‘&#x;đ?‘?đ?‘&#x;đ?‘’đ?‘ đ?‘ đ?‘˘đ?‘&#x;đ?‘’) + 0.016 = 0.61 (Parry, 1992). Set pressure, Kpa from Hysys for feed stream is 2300 and 10% for voverpressure so the set pressure will be 33.36 psig Firstly, Select an orifice trial size by setting Kv = 1.0 and using Equation A .Since the back pressure = 0, then Kw = 1.0: đ??´ =

892 ∗ √0.5521 27.2 ∗ 0.61 ∗ 1 ∗ 1 ∗ √33.36

= 6.5 đ?‘–đ?‘›2 (đ?‘¤đ?‘’ đ?‘?đ?‘Žđ?‘› đ?‘ đ?‘Žđ?‘Ś đ?‘–đ?‘Ą đ?‘–đ?‘ đ?‘›đ?‘’đ?‘Žđ?‘&#x; 6.38 đ?‘“đ?‘œđ?‘&#x; (đ?‘ƒ)đ?‘œđ?‘&#x;đ?‘–đ?‘“đ?‘–đ?‘?đ?‘’ đ?‘Ąđ?‘&#x;đ?‘–đ?‘Žđ?‘™ đ?‘ đ?‘–đ?‘§đ?‘’ ) So now From the following Table, it can be seen that an orifice size designation “Pâ€?with an actual area of 6.38 in2 must be used. So by Using the “Pâ€? orifice area, calculate the Reynolds number using the following equation to find đ??žđ?‘Ł : đ?‘…đ?‘’ =

2800đ??şâˆ—đ?‘„ 2800∗0.5512∗892 = =1404956 đ?œ‡đ??´ 01416∗6.92

Then đ??žđ?‘Ł = −0.0077(ln đ?‘…đ?‘’)2 + 0.165((ln đ?‘…đ?‘’) + 0.128 = 0.9 Now compute a corrected orifice effective area based on the now known value of Kv:

32

đ??´ =

892 ∗ √0.5521 27.2 ∗ 0.61 ∗ 1 ∗ 0.9 ∗ √33.36

= 7.9 đ?‘–đ?‘›2

Since the corrected orifice effective area (7.9 in2 ) is greater than the selected trial orifice area (6.38 in2 ), the “Pâ€? orifice is unacceptable. Select the next larger size orifice (Q) with an area of 11.05 in2 for this viscous application (Parry, 1992) 9.10.13-Manhole Design kister (1992) suggested that one manhole for each 30 trays should be used. Therefore, for this distillation with 13 tray, it was used one manhole with size đ?&#x;Ž. đ?&#x;” đ?’Ž. The location of this manhole will be at đ?&#x;“ đ?’Ž height from the top of the column.

9.11-Operational Procedure 9.11.1-Startup Procedures The startup procedure includes a number of points to start: 1- Pre-startup activities cover the completion of all the details of the plant design before the startup part. At the same time, the startup team organizes the safety procedures to be considered by the operations and maintenance departments when performing the field inspections and commissioning equipment. 2- Commissioning the equipment is the functional check of the process up to the point when the actual materials are introduced. Commissioning activities includes: Pressure testing: test the pressure of columns and pipes for mechanical strength and tightness. Piping inspection, pipes should provide adequately for expansion and contraction due to the temperature changes. Cleaning and Flushing: it needs to make sure that no construction debris is left in the pipes of the vessels-welding rods, bolts, gloves, rags ect. Before the flushing is started, check the process thoroughly to ensure; screens have been installed in front of pumps suctions, blinds in front of the equipment such as compressors and turbines and jumper spool pieces to allow for continuity of flow. Utilities usually presents in the first phase of commissioning, as these need to operational first, before the rest of the plant can be commissioned. For example, utilities include supply pressure for all services- steam, cooling water, nitrogen. In addition to, purge/blow out lines to each piece of equipment, and electricity requirements. To sum up, during the Pre-startup and commissioning, safety review should be performed. The operability and reliability of the safety interlock systems, the alarms, the safety relief valves, all these parameters should be considered. 3- Operate the column at total reflux ratio. 4- Provide the cooling and heating mediums for both the condenser and reboiler. 5- Put the column at temperature close to the operating temperature to avoid the thermal shock in the column. 6- Adjust the column control and instrumentation for any unexpected behavior. 9.11.2-Shutdown Procedure The shutdown involves:

33

1- Close the feed valves slowly, this will gives the column time to drain all the liquid in the column. 2- Stop the cooling and heating medium flow to the reboiler and the condenser. 3- Switch of the reboiler and close the valves, after this, the temperature n the column will decrease gradually. 4- Switch off the condenser and close the reflux valves. 5- By venting reduce the pressure of the column to the atmospheric pressure. 6- Open the column vessel to the atmosphere. 9.11.3-Emergency Shutdown The emergency shutdown can be happen as accident or, we have to do it in some situation for a safe operation of the column. There are many situations for the emergency shutdown such as: 1- Power failure: it can be solves this problem by using backup generators with auto start, these generators will help to light the control room and other properties with small consumed of energy. 2- Cooling medium failure: this failure leads to reduce the condenser efficiency as a result; the product will not meet the specifications. To reduce the effect of this problem, we need to switch off the reboiler to prevent the temperature build up in the column and follow the above procedure to shutdown. The emergency shutdown needs some steps: 1- Alarms to warm the staff that there is an emergency shutdown. 2- Prepare any emergency tools such as fire-water to control any fire case, ect. 9.11.4-Maintenance It is essential for reducing the hazards associated with the process. The column maintenance can be supplied only after these steps take place: 1-

Shutdown and closing all valves to ensure that the column and all the pipes has been venting. 2- Pressurize and prepressurize to create safe area for maintenance. 3- Ensure that the maintenance staff has appropriate equipments to do the maintenance. 4- After the maintenance bring the system to work again by using the above startup procedure. 9.11.5-Plot Plan The plot plan was presented to give more understanding about the column and it’s auxiliary equipment. The mechanical drawing was plotted, should be used in conjunction with the plot plan to get more detailed understanding for the process. 1.193m

3m

2.97m

2.5m

Reboiler

1.789m

4.5m

Column

Backup Pump

0.6m

0.8m

0.8m

Main Pump

1m

2.5m 1.5m

34

9.12- Safety Design 9.12.1- Hazard Study: Project: 𝑫𝒊𝒔𝒕𝒊𝒍𝒍𝒂𝒕𝒊𝒐𝒏 𝑪𝒐𝒍𝒖𝒎𝒏 (𝑫𝒆𝒃𝒖𝒕𝒂𝒏𝒊𝒛𝒆𝒓) 𝑮𝒖𝒊𝒅𝒆𝒘𝒐𝒓𝒅

𝑪𝒂𝒖𝒔𝒆𝒔

𝑪𝒐𝒏𝒔𝒆𝒒𝒖𝒆𝒏𝒄𝒆𝒔

𝑺𝒂𝒇𝒆𝒈𝒖𝒂𝒓𝒅𝒔

𝑺𝒆𝒗𝒆𝒓𝒊𝒕𝒚

𝑳𝒊𝒌𝒆𝒍𝒊𝒉𝒐𝒐𝒅

𝑨𝒄𝒕𝒊𝒐𝒏

𝐹𝑖𝑟𝑒

● 𝐸𝑙𝑒𝑐𝑡𝑟𝑖𝑐𝑎𝑙 𝐹𝑎𝑢𝑙𝑡 . ● 𝐻𝑢𝑚𝑎𝑛 𝑒𝑟𝑟𝑜𝑟. ●𝑙𝑖𝑔ℎ𝑡𝑛𝑖𝑛𝑔 𝑠𝑡𝑟𝑖𝑘𝑒𝑠.

● 𝐷𝑎𝑚𝑎𝑔𝑒 𝑖𝑛 𝑡ℎ𝑒 𝑐𝑜𝑙𝑢𝑚𝑛. ● 𝑃𝑒𝑟𝑠𝑜𝑛𝑎𝑙 𝑖𝑛𝑗𝑢𝑟𝑦. ● 𝐹𝑖𝑛𝑎𝑛𝑐𝑖𝑎𝑙 𝑙𝑜𝑠𝑠𝑒𝑠. ● 𝑃𝑟𝑜𝑑𝑢𝑐𝑡𝑖𝑜𝑛 𝑙𝑜𝑠𝑠𝑒𝑠.

● 𝐻𝑖𝑔ℎ 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑎𝑙𝑎𝑟𝑚. ● 𝑅𝑒𝑑𝑢𝑐𝑒 𝑡ℎ𝑒 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 ● 𝐴𝑑𝑒𝑞𝑢𝑎𝑡𝑒 𝑚𝑎𝑖𝑛𝑡𝑒𝑛𝑎𝑛𝑐𝑒. ● 𝐸𝑚𝑒𝑟𝑔𝑒𝑛𝑐𝑦 𝑝𝑟𝑜𝑐𝑒𝑑𝑢𝑟𝑒𝑠 𝑓𝑜𝑟 𝑓𝑖𝑟𝑒. ● 𝐸𝑚𝑒𝑟𝑔𝑒𝑛𝑐𝑦 𝑠ℎ𝑢𝑡𝑑𝑜𝑤𝑛.

𝐻𝑖𝑔ℎ

𝐿𝑖𝑘𝑒𝑙𝑦

● 𝑇𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑎𝑙𝑎𝑟𝑚

𝐿𝑒𝑎𝑘 − 𝑔𝑎𝑠 𝑖𝑠𝑠𝑢𝑒𝑠

● 𝑃𝑖𝑝𝑒𝑙𝑖𝑛𝑒 𝑟𝑢𝑝𝑡𝑢𝑟𝑒. ● 𝐻𝑢𝑚𝑎𝑛 𝑒𝑟𝑟𝑜𝑟. ●𝑂𝑣𝑒𝑟𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒.

● 𝑂𝑓𝑓 𝑠𝑝𝑐𝑖𝑓𝑖𝑐𝑎𝑡𝑖𝑜𝑛 𝑝𝑟𝑜𝑑𝑢𝑐𝑡. ● 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑝𝑟𝑜𝑏𝑙𝑒𝑚𝑠. ● 𝑃𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑖𝑛𝑗𝑢𝑟𝑒𝑠 𝑓𝑜𝑟 𝑝𝑒𝑟𝑠𝑜𝑛𝑛𝑒𝑙.

● 𝑉𝑎𝑝𝑜𝑢𝑟 𝐷𝑒𝑡𝑒𝑐𝑡𝑜𝑟 𝑖𝑛 𝑑𝑖𝑠𝑡𝑖𝑙𝑙𝑎𝑡𝑖𝑜𝑛 𝑠𝑦𝑠𝑡𝑒𝑚. ● 𝑉𝑒𝑛𝑡𝑖𝑙𝑎𝑡𝑖𝑜𝑛 𝑠𝑦𝑠𝑡𝑒𝑚. ● 𝑃𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑖𝑛𝑗𝑢𝑟𝑒𝑠

𝐻𝑖𝑔ℎ

𝐿𝑖𝑘𝑒𝑙𝑦

● 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑎𝑙𝑎𝑟𝑚

𝑈𝑡𝑖𝑙𝑖𝑡𝑦 𝑓𝑎𝑖𝑙𝑢𝑟𝑒

● 𝐹𝑎𝑖𝑙𝑢𝑟𝑒 𝑖𝑛 𝑝𝑢𝑚𝑝𝑠 𝑜𝑓 𝑐𝑙𝑒𝑎𝑛𝑖𝑛𝑔 𝑤𝑎𝑡𝑒𝑟. ● 𝐹𝑎𝑖𝑙𝑢𝑟𝑒 𝑖𝑛 𝑣𝑎𝑙𝑣𝑒𝑠. ● 𝐹𝑎𝑖𝑙𝑢𝑟𝑒 𝑖𝑛 𝑐𝑜𝑛𝑡𝑟𝑜𝑙 𝑠𝑦𝑠𝑡𝑒𝑚.

● 𝐿𝑎𝑐𝑘 𝑜𝑓 𝑜𝑝𝑒𝑟𝑎𝑡𝑖𝑛𝑔 𝑠𝑦𝑠𝑡𝑒𝑚 ● 𝐴𝑏𝑛𝑜𝑟𝑚𝑎𝑙 𝑜𝑝𝑒𝑟𝑎𝑡𝑖𝑛𝑔 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛.

● 𝑅𝑔𝑢𝑙𝑎𝑟 𝑖𝑛𝑠𝑝𝑒𝑐𝑡𝑖𝑜𝑛. ● 𝑅𝑒𝑔𝑢𝑙𝑎𝑟 𝑚𝑎𝑖𝑛𝑡𝑒𝑛𝑎𝑛𝑐𝑒.

𝐿𝑜𝑤

𝑈𝑛𝑙𝑖𝑘𝑒𝑙𝑦

𝑃𝑜𝑤𝑒𝑟 𝑓𝑎𝑖𝑙𝑢𝑟𝑒

● 𝐸𝑙𝑒𝑐𝑡𝑟𝑖𝑐𝑖𝑡𝑦 𝑝𝑟𝑜𝑏𝑙𝑒𝑚𝑠. ● 𝐻𝑢𝑚𝑎𝑛 𝑒𝑟𝑟𝑜𝑟. ● 𝐴𝑏𝑛𝑜𝑟𝑚𝑎𝑙 𝑜𝑝𝑒𝑟𝑎𝑡𝑖𝑛𝑔 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛. ● 𝑅𝑒𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑜𝑟 𝑓𝑎𝑖𝑙𝑢𝑟𝑒.

● 𝑆ℎ𝑢𝑡𝑑𝑜𝑤𝑛. ● 𝑁𝑜 𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠. ● 𝐷𝑖𝑠𝑝𝑒𝑟𝑠𝑎𝑙 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑜𝑛𝑡𝑟𝑜𝑙 𝑠𝑦𝑠𝑡𝑒𝑚.

● 𝑅𝑒𝑔𝑢𝑙𝑎𝑟 𝑖𝑛𝑠𝑝𝑒𝑐𝑡𝑖𝑜𝑛 𝑤𝑖𝑡ℎ 𝑚𝑎𝑖𝑛𝑡𝑒𝑛𝑎𝑛𝑐𝑒. ● 𝑃𝑟𝑜𝑣𝑖𝑑𝑒 𝑝𝑟𝑒𝑐𝑎𝑢𝑡𝑖𝑜𝑛 𝑟𝑒𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑖𝑜𝑛.

𝐻𝑖𝑔ℎ

𝑃𝑜𝑠𝑠𝑖𝑏𝑙𝑒

𝑂𝑝𝑒𝑟𝑎𝑡𝑜𝑟 𝑟𝑖𝑠𝑘

● 𝐸𝑥𝑡𝑟𝑒𝑚𝑒 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛𝑠. ● 𝐻𝑢𝑚𝑚𝑎𝑛 𝑒𝑟𝑟𝑜𝑟𝑠.

● 𝑃𝑒𝑟𝑠𝑜𝑛𝑎𝑙 𝑖𝑛𝑗𝑢𝑟𝑦.

𝑀𝑒𝑑𝑖𝑢𝑚

𝑃𝑜𝑠𝑠𝑖𝑝𝑙𝑒

𝑅𝑒𝑙𝑖𝑒𝑓 𝑣𝑎𝑙𝑣𝑒𝑠: ●𝐹𝑎𝑖𝑙 𝑜𝑝𝑒𝑛 ●𝐹𝑎𝑖𝑙 𝑐𝑙𝑜𝑠𝑒 𝐶𝑜𝑙𝑢𝑚𝑛 𝐿𝑜𝑤 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦

●𝑀𝑒𝑐ℎ𝑎𝑛𝑖𝑐𝑎𝑙 𝐹𝑎𝑖𝑙𝑢𝑟𝑒. ●𝐻𝑢𝑚𝑎𝑛 𝐹𝑎𝑖𝑙𝑢𝑟𝑒.

● ● ● ●

● 𝐸𝑛𝑠𝑢𝑟𝑒 𝑜𝑝𝑒𝑟𝑎𝑡𝑜𝑟𝑠 𝑎𝑟𝑒 𝑡𝑟𝑎𝑖𝑛𝑒𝑑 𝑎𝑛𝑑 𝑞𝑢𝑎𝑙𝑖𝑓𝑖𝑒𝑑 𝑓𝑜𝑟 𝑜𝑝𝑒𝑟𝑎𝑡𝑖𝑜𝑛 ● 𝑅𝑒𝑔𝑢𝑙𝑎𝑟 𝑖𝑛𝑠𝑝𝑒𝑐𝑡𝑖𝑜𝑛 . ● 𝐼𝑛𝑐𝑟𝑒𝑎𝑠𝑒 𝑛𝑜𝑧𝑧𝑙𝑒 𝑠𝑖𝑧𝑒.

𝑀𝑒𝑑𝑖𝑢𝑚

𝑃𝑜𝑠𝑠𝑖𝑏𝑙𝑒

● 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑎𝑙𝑎𝑟𝑚.

𝑀𝑒𝑑𝑖𝑢𝑚

𝑃𝑜𝑠𝑠𝑖𝑏𝑙𝑒

● 𝐶ℎ𝑒𝑐𝑘 𝐶𝑜𝑛𝑡𝑟𝑜𝑙 𝑠𝑦𝑠𝑡𝑒𝑚.

𝑇𝑢𝑏𝑒 𝐹𝑎𝑖𝑙𝑢𝑟𝑒

● 𝐶𝑜𝑟𝑟𝑜𝑠𝑖𝑜𝑛 𝑃𝑟𝑜𝑏𝑙𝑒𝑚𝑠. ● 𝐻𝑖𝑔ℎ 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒. ● 𝐶𝑜𝑟𝑟𝑜𝑠𝑖𝑜𝑛 𝑝𝑟𝑜𝑏𝑙𝑒𝑚𝑠. ● 𝐻𝑖𝑔ℎ 𝑙𝑖𝑞𝑢𝑖𝑑 𝑙𝑒𝑣𝑒𝑙. ● 𝐼𝑛𝑡𝑒𝑟𝑛𝑎𝑙 𝐷𝑎𝑚𝑎𝑔𝑒.

● 𝑃𝑟𝑜𝑑𝑢𝑐𝑡 𝑙𝑜𝑠𝑠.

● 𝑊𝑜𝑟𝑘 𝑡ℎ𝑒 𝑒𝑞𝑢𝑖𝑝𝑚𝑒𝑛𝑡 𝑖𝑛 𝑎 𝑠𝑎𝑓𝑒 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛. ● 𝑅𝑒𝑔𝑢𝑙𝑎𝑟 𝑖𝑛𝑠𝑝𝑒𝑐𝑡𝑖𝑜𝑛. ● 𝑅𝑒𝑔𝑢𝑙𝑎𝑟 𝑡𝑒𝑠𝑡 𝑓𝑜𝑟 𝑒𝑚𝑒𝑟𝑔𝑒𝑛𝑐𝑦 𝑠ℎ𝑢𝑡𝑑𝑜𝑤𝑛. ● 𝑅𝑒𝑔𝑢𝑙𝑎𝑟 𝑖𝑛𝑠𝑝𝑒𝑐𝑡𝑖𝑜𝑛

𝑀𝑒𝑑𝑖𝑢𝑚

𝑃𝑜𝑠𝑠𝑖𝑏𝑙𝑒

● 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑎𝑙𝑎𝑟𝑚

● 𝑅𝑒𝑑𝑢𝑐𝑖𝑛𝑔 𝑡ℎ𝑒 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒. ● 𝑂𝑓𝑓 𝑠𝑝𝑒𝑐. 𝑝𝑟𝑜𝑑𝑢𝑐𝑡.

● 𝑅𝑒𝑔𝑢𝑙𝑎𝑟 𝑖𝑛𝑠𝑝𝑒𝑐𝑡𝑖𝑜𝑛 𝑎𝑛𝑑 𝑚𝑎𝑖𝑛𝑡𝑒𝑛𝑎𝑛𝑐𝑒.

𝐻𝑖𝑔ℎ

𝐿𝑖𝑘𝑒𝑙𝑦

● 𝐶𝑜𝑟𝑟𝑜𝑠𝑖𝑜𝑛 𝑝𝑟𝑜𝑏𝑙𝑒𝑚𝑠. ● 𝐻𝑖𝑔ℎ 𝑙𝑖𝑞𝑢𝑖𝑑 𝑙𝑒𝑣𝑒𝑙. ● 𝐼𝑛𝑡𝑒𝑟𝑛𝑎𝑙 𝐷𝑎𝑚𝑎𝑔𝑒. ●𝑀𝑒𝑐ℎ𝑎𝑛𝑖𝑐𝑎𝑙 𝐹𝑎𝑖𝑙𝑢𝑟𝑒. ●𝐻𝑢𝑚𝑎𝑛 𝐹𝑎𝑖𝑙𝑢𝑟𝑒. ●𝐶𝑜𝑛𝑡𝑟𝑜𝑙 𝑃𝑟𝑜𝑏𝑙𝑒𝑚𝑠.

● 𝑅𝑒𝑑𝑢𝑐𝑖𝑛𝑔 𝑡ℎ𝑒 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒. ● 𝑂𝑓𝑓 𝑠𝑝𝑒𝑐. 𝑝𝑟𝑜𝑑𝑢𝑐𝑡

● 𝑅𝑒𝑔𝑢𝑙𝑎𝑟 𝑖𝑛𝑠𝑝𝑒𝑐𝑡𝑖𝑜𝑛 𝑎𝑛𝑑 𝑚𝑎𝑖𝑛𝑡𝑒𝑛𝑎𝑛𝑐𝑒.

𝐻𝑖𝑔ℎ

𝐿𝑖𝑘𝑒𝑙𝑦

● 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑃𝑟𝑜𝑏𝑙𝑒𝑚𝑠. ●𝐼𝑛𝑡𝑒𝑟𝑛𝑎𝑙 𝑑𝑎𝑚𝑎𝑔𝑒. ● 𝑁𝑜 𝑝𝑟𝑜𝑑𝑢𝑐𝑡.

● 𝑅𝑒𝑔𝑢𝑙𝑎𝑟 𝑖𝑛𝑠𝑝𝑒𝑐𝑡𝑖𝑜𝑛 𝑎𝑛𝑑 𝑚𝑎𝑖𝑛𝑡𝑒𝑛𝑎𝑛𝑐𝑒.

𝑀𝑒𝑑𝑖𝑢𝑚

𝑃𝑜𝑠𝑠𝑖𝑏𝑙𝑒

● 𝑇𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑎𝑙𝑎𝑟𝑚. ● 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑎𝑙𝑎𝑟𝑚. ● 𝐹𝑙𝑜𝑤 𝑎𝑙𝑎𝑟𝑚.

●𝑀𝑒𝑐ℎ𝑎𝑛𝑖𝑐𝑎𝑙 𝐹𝑎𝑖𝑙𝑢𝑟𝑒. ●𝐻𝑢𝑚𝑎𝑛 𝐹𝑎𝑖𝑙𝑢𝑟𝑒. ●𝐶𝑜𝑛𝑡𝑟𝑜𝑙 𝑃𝑟𝑜𝑏𝑙𝑒𝑚𝑠.

● 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑃𝑟𝑜𝑏𝑙𝑒𝑚𝑠. ●𝐼𝑛𝑡𝑒𝑟𝑛𝑎𝑙 𝑑𝑎𝑚𝑎𝑔𝑒. ● 𝑁𝑜 𝑝𝑟𝑜𝑑𝑢𝑐𝑡.

● 𝑅𝑒𝑔𝑢𝑙𝑎𝑟 𝑖𝑛𝑠𝑝𝑒𝑐𝑡𝑖𝑜𝑛 𝑎𝑛𝑑 𝑚𝑎𝑖𝑛𝑡𝑒𝑛𝑎𝑛𝑐𝑒.

𝑀𝑒𝑑𝑖𝑢𝑚

𝑃𝑜𝑠𝑠𝑖𝑏𝑙𝑒

● 𝑇𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑎𝑙𝑎𝑟𝑚. ● 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑎𝑙𝑎𝑟𝑚. ● 𝐹𝑙𝑜𝑤 𝑎𝑙𝑎𝑟𝑚.

𝑅𝑒𝑏𝑜𝑖𝑙𝑒𝑟 𝑖𝑛𝑐𝑙𝑢𝑑𝑒: ● 𝐿𝑒𝑎𝑘 𝑓𝑎𝑖𝑙𝑢𝑟𝑒 ● 𝐹𝑜𝑢𝑙𝑖𝑛𝑔 𝑓𝑎𝑖𝑙𝑢𝑟𝑒 ● 𝐿𝑜𝑤 𝑑𝑢𝑡𝑦 𝑓𝑎𝑖𝑙𝑢𝑟𝑒 𝐶𝑜𝑛𝑑𝑒𝑛𝑠𝑒𝑟 𝑖𝑛𝑐𝑙𝑢𝑑𝑒: ● 𝐿𝑒𝑎𝑘 𝑓𝑎𝑖𝑙𝑢𝑟𝑒 ● 𝐿𝑜𝑤 𝑑𝑢𝑡𝑦 𝑓𝑎𝑖𝑙𝑢𝑟𝑒 𝐹𝑒𝑒𝑑 𝑣𝑎𝑙𝑣𝑒𝑠: ●𝐹𝑎𝑖𝑙 𝑜𝑝𝑒𝑛 ●𝐹𝑎𝑖𝑙 𝑐𝑙𝑜𝑠𝑒 𝑂𝑢𝑡𝑙𝑒𝑡 𝑣𝑎𝑙𝑣𝑒𝑠: ●𝐹𝑎𝑖𝑙 𝑜𝑝𝑒𝑛 ●𝐹𝑎𝑖𝑙 𝑐𝑙𝑜𝑠𝑒

● 𝐴𝑏𝑛𝑜𝑟𝑚𝑎𝑙 𝑜𝑝𝑒𝑟𝑎𝑡𝑖𝑛𝑔 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛. ● 𝑃𝑟𝑜𝑏𝑙𝑒𝑚𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑑𝑒𝑠𝑖𝑔𝑛. ● 𝐼𝑛𝑡𝑒𝑟𝑛𝑎𝑙 𝑑𝑎𝑚𝑎𝑔𝑒.

𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑙𝑜𝑠𝑠. 𝑃𝑟𝑜𝑑𝑢𝑐𝑡 𝐿𝑜𝑠𝑠. 𝐻𝑖𝑔ℎ 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒. 𝑂𝑓𝑓 𝑠𝑝𝑒𝑐. 𝑝𝑟𝑜𝑑𝑢𝑐𝑡

● 𝑇𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑎𝑙𝑎𝑟𝑚

● 𝑇𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑎𝑙𝑎𝑟𝑚

35

9.12.2-HAZOP Study: Project: 𝑫𝒊𝒔𝒕𝒊𝒍𝒍𝒂𝒕𝒊𝒐𝒏 𝑪𝒐𝒍𝒖𝒎𝒏 (𝑫𝒆𝒃𝒖𝒕𝒂𝒏𝒊𝒛𝒆𝒓) 𝑪𝒂𝒖𝒔𝒆𝒔

𝑮𝒖𝒊𝒅𝒆𝒘𝒐𝒓𝒅

𝑫𝒆𝒗𝒊𝒂𝒕𝒊𝒐𝒏

𝑪𝒐𝒏𝒔𝒆𝒒𝒖𝒆𝒏𝒄𝒆𝒔

𝑨𝒄𝒕𝒊𝒐𝒏 𝑹𝒆𝒒𝒖𝒊𝒓𝒆𝒅

𝑇𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒

𝐿𝑜𝑤

● 𝐹𝑎𝑖𝑙𝑢𝑟𝑒 𝑖𝑛 𝑡ℎ𝑒 𝑟𝑒𝑏𝑜𝑖𝑙𝑒𝑟 ● 𝐹𝑎𝑖𝑙𝑢𝑟𝑒 𝑖𝑛 𝑐𝑜𝑛𝑑𝑒𝑛𝑠𝑒𝑟 ● 𝐶𝑜𝑛𝑡𝑟𝑜𝑙𝑙𝑒𝑟 𝑓𝑎𝑖𝑙𝑢𝑟𝑒 ● 𝐹𝑜𝑢𝑙𝑖𝑛𝑔 ● 𝑃𝑖𝑝𝑒 𝑙𝑒𝑎𝑘𝑎𝑔𝑒

● 𝐶𝑜𝑛𝑑𝑒𝑛𝑠𝑒𝑟 𝑜𝑣𝑒𝑟𝑤𝑜𝑟𝑘 ● 𝑅𝑒𝑑𝑢𝑐𝑒 𝑡ℎ𝑒 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 ● 𝑂𝑓𝑓 𝑠𝑝𝑐𝑖𝑓𝑖𝑐𝑎𝑡𝑖𝑜𝑛 𝑝𝑟𝑜𝑑𝑢𝑐𝑡 ● 𝑊𝑒𝑒𝑝𝑖𝑛𝑔 𝑖𝑛 𝑐𝑜𝑙𝑢𝑚𝑛

● 𝑅𝑒𝑔𝑢𝑙𝑎𝑟 𝑖𝑛𝑠𝑝𝑒𝑐𝑡𝑖𝑜𝑛 ● 𝐿𝑜𝑤 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑎𝑙𝑎𝑟𝑚 ● 𝑐𝑜𝑛𝑡𝑟𝑜𝑙 𝑠𝑦𝑠𝑡𝑒𝑚 𝑑𝑒𝑡𝑒𝑐𝑡

𝑇𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒

𝐻𝑖𝑔ℎ

● 𝐿𝑜𝑤 𝑓𝑒𝑒𝑑 𝑓𝑙𝑜𝑤𝑟𝑎𝑡𝑒 ● 𝑆𝑢𝑝𝑝𝑙𝑦 ℎ𝑖𝑔ℎ 𝑑𝑢𝑡𝑦 𝑓𝑜𝑟 𝑟𝑒𝑏𝑜𝑖𝑒𝑟 ● 𝐹𝑎𝑖𝑙𝑢𝑟𝑒 𝑖𝑛 𝑐𝑜𝑛𝑑𝑒𝑛𝑠𝑒𝑟, 𝑚𝑜𝑟𝑒 𝑣𝑎𝑝𝑜𝑟

● 𝐼𝑛𝑡𝑒𝑟𝑛𝑎𝑙 𝑑𝑎𝑚𝑎𝑔𝑒 ● 𝐼𝑛𝑐𝑟𝑒𝑎𝑠𝑒 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 ● 𝑂𝑓𝑓 𝑠𝑝𝑐𝑖𝑓𝑖𝑐𝑎𝑡𝑖𝑜𝑛 𝑝𝑟𝑜𝑑𝑢𝑐𝑡

● 𝑐𝑜𝑛𝑡𝑟𝑜𝑙 𝑠𝑦𝑠𝑡𝑒𝑚 𝑑𝑒𝑡𝑒𝑐𝑡 ● 𝑅𝑒𝑔𝑢𝑙𝑎𝑟 𝑖𝑛𝑠𝑝𝑒𝑐𝑡𝑖𝑜𝑛 ● 𝐻𝑖𝑔ℎ 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑎𝑙𝑎𝑟𝑚

𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒

𝐿𝑜𝑤

● 𝐿𝑒𝑎𝑘𝑎𝑔𝑒 𝑖𝑛 𝑐𝑜𝑙𝑢𝑚𝑛 𝑜𝑟 𝑝𝑖𝑝𝑒𝑠 ● 𝐶𝑜𝑟𝑟𝑜𝑠𝑖𝑜𝑛 𝑜𝑓 𝑝𝑙𝑎𝑡𝑒𝑠 ● 𝐿𝑜𝑤 𝑓𝑒𝑒𝑑 𝑟𝑎𝑡𝑒

● 𝑂𝑓𝑓 𝑠𝑝𝑐𝑖𝑓𝑖𝑐𝑎𝑡𝑖𝑜𝑛 𝑝𝑟𝑜𝑑𝑢𝑐𝑡 ● 𝑊𝑒𝑒𝑝𝑖𝑛𝑔 𝑖𝑛 𝑡ℎ𝑒 𝑐𝑜𝑙𝑢𝑚𝑛

● 𝑅𝑒𝑔𝑢𝑙𝑎𝑟 𝑖𝑛𝑠𝑝𝑒𝑐𝑡𝑖𝑜𝑛 ● 𝐿𝑜𝑤 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑎𝑙𝑎𝑟𝑚

𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒

𝐻𝑖𝑔ℎ

● 𝐵𝑙𝑜𝑐𝑘𝑎𝑔𝑒 𝑖𝑛 𝑡ℎ𝑒 𝑡𝑜𝑝 𝑠𝑡𝑟𝑒𝑎𝑚 ● 𝐵𝑙𝑜𝑐𝑘𝑎𝑔𝑒 𝑖𝑛 𝑐𝑜𝑛𝑑𝑒𝑛𝑠𝑒𝑟 𝑜𝑟 𝑟𝑒𝑏𝑜𝑖𝑙𝑒𝑟

● 𝑅𝑒𝑣𝑒𝑟𝑠𝑒 𝑓𝑙𝑜𝑤 ● 𝑉𝑒𝑠𝑠𝑒𝑙 𝑒𝑥𝑝𝑙𝑜𝑠𝑖𝑜𝑛 ● 𝑂𝑓𝑓 𝑠𝑝𝑒𝑐𝑖𝑓𝑖𝑐𝑎𝑡𝑖𝑜𝑛 𝑝𝑟𝑜𝑑𝑢𝑐𝑡

● 𝐻𝑖𝑔ℎ 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑟𝑒𝑙𝑖𝑒𝑣𝑒 𝑣𝑎𝑙𝑣𝑒 ● 𝐻𝑖𝑔ℎ 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑎𝑙𝑎𝑟𝑚 ● 𝑅𝑒𝑔𝑢𝑙𝑎𝑟 𝑖𝑛𝑠𝑝𝑒𝑐𝑡𝑖𝑜𝑛

𝐹𝑙𝑜𝑤

𝐿𝑜𝑤

● 𝐷𝑒𝑐𝑟𝑒𝑎𝑠𝑒 𝑖𝑛 𝑓𝑒𝑒𝑑 𝑓𝑙𝑜𝑤𝑟𝑎𝑡𝑒 ● 𝐵𝑙𝑜𝑐𝑘𝑎𝑔𝑒 𝑖𝑛 𝑓𝑒𝑒𝑑 𝑝𝑖𝑝𝑒 ● 𝐹𝑜𝑢𝑙𝑖𝑛𝑔

● 𝑇𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑒 ● 𝑂𝑓𝑓 𝑠𝑝𝑒𝑐𝑖𝑓𝑖𝑐𝑎𝑡𝑖𝑜𝑛 𝑝𝑟𝑜𝑑𝑢𝑐𝑡 ● 𝐷𝑎𝑚𝑎𝑔𝑒 𝑜𝑓 𝑖𝑛𝑡𝑒𝑟𝑛𝑎𝑙 𝑐𝑜𝑙𝑢𝑚𝑛

● 𝑅𝑒𝑔𝑢𝑙𝑎𝑟 𝑖𝑛𝑠𝑝𝑒𝑐𝑡𝑖𝑜𝑛 𝑓𝑜𝑟 𝑓𝑒𝑒𝑑 𝑣𝑎𝑙𝑣𝑒𝑠 ● 𝐼𝑛𝑠𝑡𝑎𝑙𝑙 𝑙𝑒𝑣𝑒𝑙 𝑖𝑛𝑑𝑖𝑐𝑎𝑡𝑜𝑟𝑠

𝐹𝑙𝑜𝑤

𝐻𝑖𝑔ℎ

● 𝐹𝑎𝑖𝑙𝑢𝑟𝑒 𝑖𝑛 𝑡ℎ𝑒 𝑓𝑒𝑒𝑑 𝑣𝑎𝑙𝑣𝑒𝑠 ● 𝐼𝑛𝑐𝑟𝑒𝑎𝑠𝑒 𝑖𝑛 𝑝𝑙𝑎𝑛𝑡 𝑡ℎ𝑟𝑜𝑢𝑔ℎ𝑝𝑢𝑡

● 𝐶𝑜𝑙𝑢𝑚𝑛 𝑓𝑙𝑜𝑜𝑑𝑖𝑛𝑔 ● 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑒 ● 𝐶𝑜𝑙𝑢𝑚𝑛 𝑜𝑣𝑒𝑟𝑙𝑜𝑎𝑑

● 𝐼𝑛𝑠𝑡𝑎𝑙𝑙 𝑓𝑙𝑜𝑤 𝑖𝑛𝑑𝑖𝑐𝑎𝑡𝑜𝑟 ● 𝐼𝑛𝑠𝑡𝑎𝑙𝑙 𝑓𝑙𝑜𝑤 𝑎𝑙𝑎𝑟𝑚

𝐹𝑙𝑜𝑤

𝑁𝑜𝑛𝑒

● 𝑃𝑖𝑝𝑒 𝑏𝑙𝑜𝑐𝑘𝑎𝑔𝑒 ● 𝑉𝑎𝑙𝑣𝑒𝑠 𝑓𝑎𝑖𝑙𝑢𝑟𝑒 ● 𝑃𝑖𝑝𝑒 𝑟𝑢𝑝𝑡𝑢𝑟𝑒

● 𝑁𝑜 𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑖𝑜𝑛

● 𝐼𝑛𝑠𝑡𝑎𝑙𝑙 𝑓𝑙𝑜𝑤 𝑖𝑛𝑑𝑖𝑐𝑎𝑡𝑜𝑟 ● 𝐼𝑛𝑠𝑡𝑎𝑙𝑙 𝑓𝑙𝑜𝑤 𝑎𝑙𝑎𝑟𝑚

𝐿𝑒𝑣𝑒𝑙

𝐿𝑜𝑤

● 𝐸𝑥𝑐𝑒𝑠𝑠𝑖𝑣𝑒 𝑏𝑜𝑖𝑙 − 𝑢𝑝 ● 𝐶𝑜𝑛𝑑𝑒𝑛𝑠𝑒𝑟 𝑓𝑎𝑖𝑙𝑢𝑟𝑒 ● 𝑃𝑢𝑚𝑝 𝑓𝑎𝑖𝑙𝑢𝑟𝑒 ● 𝐶𝑜𝑙𝑢𝑚𝑛 𝑙𝑒𝑎𝑘𝑎𝑔𝑒 ● 𝑃𝑖𝑝𝑒 𝑙𝑒𝑎𝑘𝑎𝑔𝑒

● 𝑂𝑓𝑓 𝑠𝑝𝑒𝑐𝑖𝑓𝑖𝑐𝑎𝑡𝑖𝑜𝑛 𝑝𝑟𝑜𝑑𝑢𝑐𝑡 ● 𝐶𝑜𝑙𝑢𝑚𝑛 𝑒𝑥𝑝𝑙𝑜𝑑𝑒𝑠

● 𝐼𝑛𝑠𝑡𝑎𝑙𝑙 𝑓𝑙𝑜𝑤 𝑖𝑛𝑑𝑖𝑐𝑎𝑡𝑜𝑟 ● 𝐼𝑛𝑠𝑡𝑎𝑙𝑙 𝑓𝑙𝑜𝑤 𝑎𝑙𝑎𝑟𝑚 ● 𝑅𝑒𝑔𝑢𝑙𝑎𝑟 𝑖𝑛𝑠𝑝𝑒𝑐𝑡𝑖𝑜𝑛

𝐿𝑒𝑣𝑒𝑙

𝐻𝑖𝑔ℎ

● 𝐵𝑙𝑜𝑐𝑘𝑎𝑔𝑒 𝑖𝑛 𝑏𝑜𝑡𝑡𝑜𝑚 𝑠𝑡𝑟𝑒𝑎𝑚 ● 𝑉𝑎𝑙𝑣𝑒 𝑓𝑎𝑖𝑙𝑢𝑟𝑒 ● 𝐸𝑥𝑐𝑒𝑠𝑠𝑖𝑣𝑒 𝑐𝑜𝑛𝑑𝑒𝑛𝑠𝑒𝑟 𝑑𝑢𝑡𝑦 ● 𝐹𝑎𝑖𝑙𝑢𝑟𝑒 𝑖𝑛 𝑡ℎ𝑒 𝑟𝑒𝑏𝑜𝑖𝑙𝑒𝑟

● 𝑂𝑓𝑓 𝑠𝑝𝑒𝑐𝑖𝑓𝑖𝑐𝑎𝑡𝑖𝑜𝑛 𝑝𝑟𝑜𝑑𝑢𝑐𝑡 ● 𝐹𝑙𝑜𝑜𝑑𝑖𝑛𝑔 𝑖𝑛 𝑡ℎ𝑒 𝑐𝑜𝑙𝑢𝑚𝑛 ● 𝐷𝑒𝑐𝑟𝑒𝑎𝑠𝑒 𝑖𝑛 𝑡ℎ𝑒 𝑑𝑖𝑠𝑡𝑖𝑙𝑙𝑒 𝑟𝑎𝑡𝑒

● 𝐼𝑛𝑠𝑡𝑎𝑙𝑙 𝑓𝑙𝑜𝑤 𝑖𝑛𝑑𝑖𝑐𝑎𝑡𝑜𝑟 ● 𝐼𝑛𝑠𝑡𝑎𝑙𝑙 𝑓𝑙𝑜𝑤 𝑎𝑙𝑎𝑟𝑚 ● 𝑅𝑒𝑔𝑢𝑙𝑎𝑟 𝑖𝑛𝑠𝑝𝑒𝑐𝑡𝑖𝑜𝑛

𝐶𝑜𝑚𝑝𝑜𝑠𝑖𝑡𝑖𝑜𝑛

𝐿𝑜𝑤

● 𝐼𝑛𝑐𝑟𝑒𝑎𝑠𝑒 𝑜𝑟 𝑑𝑐𝑟𝑒𝑎𝑠𝑒 𝑖𝑛 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 ● 𝐻𝑖𝑔ℎ 𝑜𝑟 𝑙𝑜𝑤 𝑓𝑙𝑜𝑤 ● 𝐼𝑛𝑐𝑟𝑒𝑎𝑠𝑒 𝑜𝑟 𝑑𝑐𝑟𝑒𝑎𝑠𝑒 𝑖𝑛 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒

● 𝑂𝑓𝑓 𝑠𝑝𝑒𝑐𝑖𝑓𝑖𝑐𝑎𝑡𝑖𝑜𝑛 ● 𝐼𝑛𝑐𝑜𝑚𝑒 𝑙𝑜𝑠𝑠𝑒𝑠

● 𝑐𝑜𝑚𝑝𝑜𝑠𝑖𝑡𝑖𝑜𝑛 𝑚𝑜𝑛𝑖𝑡𝑜𝑟 ● 𝑅𝑒𝑔𝑢𝑙𝑎𝑟 𝑐𝑜𝑚𝑝𝑜𝑠𝑖𝑡𝑖𝑜𝑛 𝑖𝑛𝑠𝑝𝑒𝑐𝑡𝑖𝑜𝑛

36

9.13-Process Control System and P&ID The basic objective of the control process is to ensure that the column operates as defined by the design. A more advanced objective is that, in addition to this basic requirement, compensation should be made for deviation from the design condition; energy can be saved by carful trimming and sufficiently robust control schemes can be derived to achieve stable control over wide ranges. Process control includes: Mass Balance Feed-back Control: The feedback on the distillation take-off rate must be achieved by measuring the composition. The measurement of the product at the top of the column, and operating the column at the designed reflux specification, the feedback control will adjust to the distillate rate to achieve the mass balance to a high degree of accuracy. Pressure Control: A pressure drop controller can be used to monitor the pressure across the distillation column. The pressure drop controller (PDC) will detect any pressure differences which may occur in the column. This is linked to the steam control valve of the reboiler and this causes the control valve to prevent the flowrate from entering the column from the reboiler. Once the pressure in the column returns to its set point, the PDC will regulate the control valve to open again allowing the flowrate to pass from the reboiler to the column. In some cases, control of the pressure in the distillation can be achieved by using vent line from the condenser that is left open to the atmosphere. Level control: It can be done by using a level controller; it can control the flow of the largest stream and the smallest stream related by ratio control to the larger stream. 1- Satisfaction of constraints For safe, satisfactory operation of the column, certain constraints must be observed. For example, 1- The column shall not flood. 2-Column pressure drop should be high enough to maintain effective column operation, that is, to prevent serious weeping or dumping. 3-The temperature difference in the reboiler should not exceed the critical temperature difference. 4- Column feed rate should not be high as to overload reboiler or condenser heat-transfer capacity. 2- Boilup should not be so high that an increase will cause a decrease in product purity at the top of the column. 3- Column pressure should not exceed a maximum permissible value.

37

Piping and Instrument Diagram

FT

FIC

FT

Flow Transmitter

LC

Level Controller

LG

Level Gauge

To Flare LG

TIC

FIC

V-42

Flow Indicating Controller

PT TT

TG

TT

V-100

Temperature Transmitter

Pressure Transmitter

PT

LC

Condenser

FC

Cold Medium Out PG

Flow Controller

Temperature Indicator

TI

Cold Medium In LT

Level Transmitter

TIC

Temperature Indicating Controller

FIC P-170

Electric Signal

FIC

FT

Data Signal

Gate Valve

Globe Valve

Relief Valve

Angle Valve

Product

Diaphragm Valve

TG

Temperature Gauge

FT FT

FC

Feed FIC TI

Utility Connection

FT

TI

Hot Medium In TI

FC

LT TG

Reboiler

Hot Medium Out

Drain LIC

Bottom Product

Title: Distillation Column to Produce LPG

Owner: Capstone Energy

Drawing Title: P&ID

Vessel Tag no: T-502 Designed by: Samah Zaki Alrashid

Approved by: Dr.Gia Pham

38

9.14-Specification Sheet Design

đ??’đ??Šđ??žđ??œđ??˘đ??&#x;đ??˘đ??œđ??šđ??­đ??˘đ??¨đ??§ đ??’đ??Ąđ??žđ??žđ??­ đ??&#x;đ??¨đ??Ť đ??ƒđ??žđ??›đ??Žđ??­đ??šđ??§đ??˘đ??łđ??žđ??Ť đ??ƒđ??˘đ??Źđ??­đ??˘đ??Ľđ??Ľđ??šđ??­đ??˘đ??¨đ??§ đ??‚đ??¨đ??Ľđ??Žđ??Śđ??§ Note: Initial data provided by simulation program (Hysys) T − 502 Separate light hydrocarbons Date: 06/11/2015 from heavier to produce LPG Properties Specification for each Stream đ?‘†đ?‘Ąđ?‘&#x;đ?‘’đ?‘Žđ?‘š Feed Top Bottom Temperature(℃) 31 72.08 140.2 Pressure(kpa) 2300 2100 2300 đ?‘‡đ?‘œđ?‘Ąđ?‘Žđ?‘™ đ?‘€đ?‘œđ?‘™đ?‘Žđ?‘&#x; đ??šđ?‘™đ?‘œđ?‘¤đ?‘&#x;đ?‘Žđ?‘Ąđ?‘’(đ?‘˜đ?‘”đ?‘šđ?‘œđ?‘™đ?‘’/â„Ž) 3196 2119 1078 Vapor Fraction 0.00 0.00 0.00 Liquid Fraction 1.00 1.00 1.00 đ?‘€đ?‘œđ?‘™đ?‘’đ?‘?đ?‘˘đ?‘™đ?‘Žđ?‘&#x; đ?‘Šđ?‘’đ?‘–đ?‘”â„Žđ?‘Ą 55.11 46.34 66.92 đ??‚đ??¨đ??Ľđ??Žđ??Śđ??§ đ???đ??šđ??Ťđ??šđ??Śđ??žđ??­đ??žđ??Ťđ??Ź Column Height(m) Tray Spacing(m) 10.48 Actual Reflux Ratio 0.2585 0.6096 ID Ref.: Function:

Column Diameter(m)

2.97

Tray Efficiency

57.60%

Plate Type

Sieve

13

Shell Thickness(mm)

23

Head Thickness(mm)

23

0.2154

Head Type

Hole diameter(mm)

5

Weir Height(mm)

50

Plate Thickness(mm)

2.5

No. of Bolts

8

Weir Length (m)

2.22

Hole Patter

Triangular

Bolt Root Area(mm2)

934.12

50

Corrosion Allowance

0

95°

Skirt Thickness(mm)

6

hap Downcomer

40

Turndown Ratio

70%

No. of Tray Minimum Reflux Ratio

Ellipsoidal

No. of Holes

13150 Calm. Zone Width(mm)

Hole Pitch (mm)

9.125

Pressure Releif Area(mm2)

8879.5

Angle Subtended

Insulation Material kg Density of Insulation( 3 ) m đ??šđ?‘&#x;đ?‘Žđ?‘?đ?‘Ąđ?‘–đ?‘œđ?‘›đ?‘Žđ?‘™ đ??¸đ?‘›đ?‘Ąđ?‘&#x;đ?‘Žđ?‘–đ?‘›đ?‘šđ?‘’đ?‘›đ?‘Ą đ?‘‡đ?‘œđ?‘?

Down comer Residence Time Top (Sec) Material of Construction

85°

Value of đ?œƒđ?‘?

đ??ľđ?‘œđ?‘™. đ?‘ƒđ?‘–đ?‘Ąđ?‘?â„Ž đ??ˇđ?‘–đ?‘Žđ?‘šđ?‘’đ?‘Ąđ?‘’đ?‘&#x;(đ?‘š) Mineral Wool 130

0.05

Total Dead Weight(KN)

483.1

Skirt Height(m)

2

0.0026

đ??šđ?‘&#x;đ?‘Žđ?‘?đ?‘Ąđ?‘–đ?‘œđ?‘›đ?‘Žđ?‘™ đ??¸đ?‘›đ?‘Ąđ?‘&#x;đ?‘Žđ?‘–đ?‘›đ?‘šđ?‘’đ?‘›đ?‘Ą đ??ľđ?‘œđ?‘Ąđ?‘Ąđ?‘œđ?‘š

0.0036

55.64

Down comer Residence Time Bottom (Sec)

38.39

đ??śđ?‘Žđ?‘&#x;đ?‘?đ?‘œđ?‘› đ?‘ đ?‘Ąđ?‘’đ?‘’đ?‘™

39

9.15-Mechanical Drawing: Cross View

Front View

0.018 0.018

0.023 0.297

0.343

0.12

0.023

0.06

0.018

0.23

0.23

1.0840

0.23

0.50

0.50

0.12 0.06 0.06

0.05

0.022

0.022

8 9 10 11 12 13

Base Ring Bolts

0.27

0.0132

0.27

1.0840

0.06

1 2 3 4 5 6 7

Top View

0.05

0.023

0.023

0.023

0.023 0.015

0.015

0.297

0.297 0.343

0.343

0.343

Title: Distillation Column to Produce LPG Owner: Capstone Energy

Drawing Title: Mechanical Drawing

Vessel Tag no: T-502 Designed by: Samah Zaki Alrashid

Approved by: Dr.Gia Pham 40

9.16-Summary, Recommendation and Critical Design Review: Section 14 presents all of the results of the distillation column design in a summary form. Essentially the design of the debutanizer involved calculation of the minimum number of trays, minimum reflux ratio, optimum reflux ratio, tray efficiency, actual number of trays, column height, column diameter, optimum feed tray location, detailed plate design, 3tray to tray calculations, full mechanical design of the column, plot plan and mechanical drawings of the column, hazard identification analyses(hazard and hazop) , operating procedures for start-up, shutdown, normal operations, emergency shutdown and maintenance, completion of Piping and Instrumentation, completion of a detailed specification sheet. The summary of the most important results of the design are as follow:     

Column operating reflux ratio=1.2*0.2154=0.2585 Column inside diameter =2.97m Column height=10.84m Column shell thickness= 23mm Column pressure drop=0.17

Recommendations for future aspects are numerous and include things such as detailed costing of the column to determine the economic optimum diameter (to justify the use of the nearest closest standard).

References

Dutta, B. K. (2007). Principles of Mass Transfer and Separation Processes. New Delhi : PrenticeHall of India. Kister, H. Z. (1992). Distillation design. New York: New York : McGraw-Hill. Ludwig, E. E. (1999). Applied process design for chemical and petrochemical plants. Houston : Gulf Pub. Co. Parry, C. F. (1992). Relief systems handbook. Rugby, Warwickshire, UK : Institution of Chemical Engineers ; New York : Distributed exclusively in the USA and Canada by VCH . Seader, J. D. (1998). Separation process principles. New York : Wiley. Sinnott, R. K. (2005). Chemical engineering design. Amsterdam : Elsevier/Butterworth Heinemann. Stichlmair, J. (1998). Distillation : principles and practices. New York : Wiley. Wankat, P. C. (1988). Equilibrium staged separations : separations in chemical engineering. New York : Elsevier.

41

CHAPTER 10: MINOR DESIGN Shell & Tube Heat Exchanger (E-405)

SAMAH ZAKI NAJI ALRASHID 17113924

Contents Introduction .................................................................................................................................................... 10.1-The Design Process ................................................................................................................................. 10.1.1- Brief Description of Process………………………………………………………………………………………………………..…. 10.1.2-Design Specifications and Physical Data………………………………………………………………………………….….…... 10.1.2.1- Hot Stream Properties…………………………………………………………………………………………………………..…... 10.1.2.2-Cold Stream Properties ………………………………………………………………………………………………………….…… 10.1.2.3-Calculation of Heat Transfer………………………………………………………………………………………………….….... 10.2-The Assumption of Overall Heat Transfer Coefficient……………………………………………………………………….. 10.3-Flow Pattern and Mean Temperature Difference………………………………………………………………………………… 10.3.1- Shell Side and Tube Side Fluid Allocation…………………………………………………………………………………….… 10.3.2-Flow pattern……………………………………………………………………………………………………………………………….…. 10.3.3-Mean Temperature Difference…………………………………………………………………………………………………….… 10.4-Choosing Number of Passes……………………………………………………………………………………………………………… 10.5-Heat Exchanger Area………………………………………………………………………………………………………………………… 10.6-Layout and Tube Size………………………………………………………………………………………………………………………… 10.6.1- Number of Tubes………………………………………………………………………………………………………………………….. 10.6.2-Tube Arrangement…………………………………………………………………………………………………………………………. 10.6.3-Bundle and Shell Diameter…………………………………………………………………………………………………………….. 10.6.4-Baffle Size ……………………………………………………………………………………………………………………………………… 10.7-Overall Heat Transfer Calculation…………………………………………………………………………………………………… 10.7.1-Tube Side Coefficient……………………………………………………………………………………………………………………… 10.7.2-Shell Side Coefficient………………………………………………………………………………………………………………………. 10.8-Pressure Drop…………………………………………………………………………………………………………………………………… 10.8.1-Tube Side…………………………………………………………………………………………………………….…………………………. 10.8.2-Shell Side…………………………………………………………………………………………………………….…………………………. 10.9-Viscosity Correction Factor………………………………………………………………………………………………………………… 10.10-Pressure Relief System……………………………………………………………………………………………………………………. 10.11-Mechanical Design ………………………………………………………………………………………………………………………….

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10.12-Operating Design ……………………………………………………………………………………………………………….……… 10.11.1-Startup Procedure ………………………………….………………………………………………………………….…………… 10.11.2-Shutdown Procedure……………………………………………………………………………………………….….…………… 10.11.3- Maintenance ………………………………….…………………………………………………………………………………….… 10.13- Process Control System and P&ID………………….………………………………………………………………………….… Piping and Instrument ………………………………………………………………………………………………………………………….. 10.14-Mechanical Drawing …………………………………………………………………………………………………………………… 10.15-Specification Sheet……………………………………………………………………………………………………………………… 10.16-Hazard and Safety……………………………………………………………………………………………………………………..… Critical Review Design ……………………………………..……………………………………………………………………………...…… References…………………………………………………………………………………………………………………

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Introduction The project is design of shell and tube heat exchanger to condense small amount of TEG to liquid again. I will design shell and tube heat exchanger to work as a recondenser. The properties required to design a heat exchanger were provided by hysys. Using this information and various external sources, a suitable shell and tube heat exchanger design has thus been proposed.

Design scope The scope of shell and tube heat exchanger design as illustrate: -

Design specification and properties data (see section1). The assumption of overall heat transfer coefficient (see section 2). Flow pattern and log mean temperature difference (see3). Choosing the number of passes (see4). Total area of the heat exchanger (see5). Layout and tube size (see6). Overall heat transfer coefficient calculation include: the shell side and tube side (see7). Pressure drop for both shell and tube (see8). Viscosity correlation factor calculation (see9). Pressure relief system on the tube side calculation (see10). Mechanical design including: material selection and shell thickness (see11). Operating design including: startup, shutdown, and maintenance (see12). Process control and P&ID (see13). Safety design (see14).

Process flow diagram

Inlet cold stream at 54.08C

Hot stream at 215.5C

Outlet hot stream at 55C

E-2

Outlet cold stream at57.6C

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10.1-The Design Process 10.1.1- Brief Description of Process The step-by-step process used in this design procedure is shown in Figure 3-1. This systematic approach ensured that minimal errors were made in the calculations, since any modifications made to the variables could be replaced accordingly with ease. A tentative geometry was fixed using information supplied by textbooks which helped determine component specifications and dimensions. The assumed value of an overall heat transfer coefficients (OHTC) and calculated overall heat transfer coefficients (OHTC) were compared to verify initial assumptions. If discrepancies existed, the above had to be recalculated using new assumptions till a reasonable OHTC value was obtained. Pressure drop was then evaluated to ensure the proposed design was well within operating parameters; changes were then made accordingly to the component specifications.

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10.1.2-Design Specifications and Physical Data It is important to gather all known specifications and data before commencing calculations. Furthermore, units for each set of data or properties should be clearly identified and converted to consistent terms to avoid critical calculation errors. These specifications will help determine viable heat exchanger designs and its associated features and components. Note: All the properties provided by hysys. 10.1.2.1- Hot Stream Properties: Values of density, viscosity and heat capacity shall be taken at the average temperature: Th,avg =

Th1 + Th2 215.5 + 55 = = 135.25℃ 2 2

đ?‘˜đ?‘”

Molecular Weight, đ?‘€đ?‘¤đ?‘Ą = 150.1 đ?‘˜đ?‘šđ?‘œđ?‘™

Calculate mass flow rate: ṁh = 1.354 × 104 kg/hr = 3.76 kg/s

đ?‘ťđ?’‰đ?&#x;? 215.5 3.068

đ?‘ťđ?’‰đ?&#x;? 55 2.231

đ?‘ťđ?’‰,đ?’‚đ?’—đ?’ˆ 135.25 2.6495

Thermal conductivity (

0.1799

0.1945

0.18745

đ?‘‰đ?‘–đ?‘ đ?‘?đ?‘œđ?‘ đ?‘ đ?‘–đ?‘Ąđ?‘Ś, đ?‘?đ?‘ƒ Mass density [kg/m3]

0.4933 978.2

11.44 1106

5.97 1042.1

Hot Stream Temperature℃ đ??žđ??˝ Heat capacity (đ?‘˜đ?‘”.đ?‘?)

đ?‘¤ ) đ?‘š2 .đ?‘˜

1.2.2-Cold Stream Properties : Calculate mass flow rate: ṁc = 6.038 Ă— 105 kg/h = 167.72 kg/s đ?‘‡đ??śđ?‘Žđ?‘Ł = Molecular Weight, đ?‘€đ?‘¤đ?‘Ą = 23.45 Cold Stream Temperature℃

54.08 + 57.6 = 55.84℃ 2

đ?‘˜đ?‘” đ?‘˜đ?‘šđ?‘œđ?‘™

đ?‘ťđ?’„đ?&#x;? 54.08

đ?‘ťđ?’„đ?&#x;? 57.6

đ?‘ťđ?’„,đ?’‚đ?’—đ?’ˆ 55.84

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đ??žđ??˝

Heat capacity (đ?‘˜đ?‘”.đ?‘?)

�

Thermal conductivity (đ?‘š2 .đ?‘˜) đ?‘‰đ?‘–đ?‘ đ?‘?đ?‘œđ?‘ đ?‘ đ?‘–đ?‘Ąđ?‘Ś, đ?‘?đ?‘ƒ Mass density [kg/m3]

2.578

2.560

2.578

0.03862

0.03892

0.03877

0.01431 65.98

0.01436 64.24

0.014335 65.11

1.2.3-Calculation of Heat Transfer: The objective of this heat exchanger is to condense small amount of TEG to liquid again, this means that phase change will be on hot side. However, the flow rate of the tube side very small (3.76kg/s) kg

compare to the flowrate in the shell side(167.72 s ), and the heat from the tube side smaller than the heat on the shell side. Therefore, in our calculation the duty will be calculated based on the cold side (shell side). Now, to calculate the overall heat transfer: đ?‘ž = đ?‘šĚ‡đ?‘? đ?‘?đ?‘?đ?‘? (đ?‘‡đ?‘?2 − đ?‘‡đ?‘?1 ) = 167.72 ∗ 2.578 ∗ (57.6 − 54.08) = 1522 đ?‘˜đ?‘¤ đ?‘ťđ?’‰đ?’† đ?’…đ?’–đ?’•đ?’š đ?’?đ?’‡ đ?’•đ?’‰đ?’Šđ?’” đ?’‰đ?’†đ?’‚đ?’• đ?’†đ?’™đ?’„đ?’‰đ?’‚đ?’?đ?’ˆđ?’†đ?’“ đ?’Šđ?’” đ?&#x;?đ?&#x;“đ?&#x;?đ?&#x;? đ?’Œđ?’˜, đ?’‚đ?’?đ?’Žđ?’?đ?’”đ?’• đ?’”đ?’‚đ?’Žđ?’† đ?’‚đ?’” đ?’•đ?’‰đ?’† đ?’—đ?’‚đ?’?đ?’–đ?’† đ?’‘đ?’“đ?’?đ?’—đ?’Šđ?’…đ?’†đ?’… đ?’ƒđ?’š đ?‘Żđ?’šđ?’”đ?’šđ?’”.

10.2-The Assumption of Overall Heat Transfer Coefficient For this heat exchanger, involving liquid and vapor hydrocarbons. Therefore, the calculations will start đ?‘ž đ?‘ž with an OHTC of 850 đ?&#x;? ( Hysys estimate that the overall heat transfer coefficient is almost 841 đ?&#x;? . đ?’Ž .℃

đ?’Ž .℃

10.3-Flow Pattern and Mean Temperature Difference 3.1- Shell Side and Tube Side Fluid Allocation Factor to be considered in the allocation of the two fluids is the fluids temperature. Fluid Temperature: The hot fluid should be placed in the tubes to reduce energy losses and overall cost. It would also be safer as the cold fluid would be surrounding it (Incropera 2007). 3.2-Flow pattern Two different flow patterns exist for this exchanger; these are counter flow and parallel flow. The mean temperature difference for both flows is calculated using the equation: ∆đ?‘‡đ??żđ?‘€ =

∆đ?‘‡1 − ∆đ?‘‡2 ∆đ?‘‡ ln ∆đ?‘‡1 2

A counter flow exchanger can achieve a higher ∆Tlm and hence, a lower heat transfer area. This would optimize the cost of production. The temperature difference is also more uniform compared to parallel flow which would minimize stress in the exchanger from reduced expansions and contractions in the components. This leads to lower maintenance or replacement costs. In addition this uniform difference produces a more uniform rate of heat transfer throughout the exchanger (Coulson 1999). Due to these advantages, a counter flow exchanger is selected. 3.3-Mean Temperature Difference Before starting the log means temperature calculation, as I motioned before that the purpose of this heat exchanger to recondense the vapor to the liquid again. This means that there is phase change in 6|P a g e

the tube side as I allocated, therefore, we have to consider that change in phase. However, the flowrate of the tube side very small (0.244 kg/s) compare to the flowrate in the shell side(105.22 kg/s). Therefore, the effect of phase change in the tube side can be ignored (Kuppan, 2000). ∆Tlm can now be calculated using the above equation. ∆T2 = 215.5 − 57.6 = 157.9℃ ∆T1 = 55 − 54.08 = 0.92℃ ∆Tlm =

∆T1 − ∆T2 157.9 − 0.92 = = 30.5℃ ∆T 157.9 ln ∆T1 ln 0.92 2

Realistically, the flow would not be purely counter flow. Hence, a mean temperature difference has to be calculated using: ∆Tm = F∆Tlm where ∆Tm : mean temperature difference for any exchanger F: correction factor The correction factor is determined by the calculation of parameters R and P and from đ??šđ?‘–đ?‘”. 1. Assumptions  constant OHTC in each pass  constant cross-sectional temperature of the shell-side fluid in any pass  No leakage of fluid between shell passes For four tube passes, one shell pass: P=

Th2 − Th1 55 − 215.5 = = 0.99 Tc1 − Th1 54.08 − 215.5

R=

Tc1 − Tc2 54.08 − 57.6 = = 0.022 Th2 − Th1 55 − 215.5

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(đ?‘­đ?’Šđ?’ˆ. đ?&#x;?)đ?‘Şđ?’?đ?’“đ?’“đ?’†đ?’„đ?’•đ?’Šđ?’?đ?’? đ?‘­đ?’‚đ?’„đ?’•đ?’?đ?’“ đ?‘Şđ?’‰đ?’‚đ?’“đ?’• đ?’‡đ?’?đ?’“ đ?‘­đ?’?đ?’–đ?’“ đ?‘ťđ?’–đ?’ƒđ?’† đ?‘ˇđ?’‚đ?’”đ?’”đ?’†đ?’” (đ?‘°đ?’?đ?’„đ?’“đ?’?đ?’‘đ?’†đ?’“đ?’‚ đ?&#x;?đ?&#x;Žđ?&#x;?đ?&#x;?) From the chart it can be seen thatF ≈ 1. As this value is greater than 0.8, the process is considered efficient. The mean temperature difference can then be evaluated: ∆đ?‘ťđ?’?đ?’Ž,đ?’„đ?’?đ?’“đ?’“đ?’†đ?’„đ?’•đ?’†đ?’… = đ?&#x;‘đ?&#x;Ž. đ?&#x;“ Ă— đ?&#x;? = đ?&#x;‘đ?&#x;Ž. đ?&#x;“℃ , đ?’”đ?’‚đ?’Žđ?’† đ?’—đ?’‚đ?’?đ?’–đ?’† đ?’˜đ?’Šđ?’•đ?’‰ đ?‘Żđ?’šđ?’”đ?’šđ?’”

10.4-Choosing Number of Passes It was decided that calculations should be made following an initial choice of one shell pass and four tube passes. This decision was based on the fact that a higher number of tube passes give a greater flow velocity. The decrease in residence time leads to a higher heat transfer coefficient and increasing heat transfer efficiency. However, a large number of tube passes would also lead to increase the pressure drop – which would affect the required pressure drop limit. A compromise was made which resulted in four tube passes. The configuration of one shell pass is most commonly used in heat exchangers. Multiple shell passes are only used in cases of a temperature cross as the exchanger would not be considered economically viable using one shell pass. A temperature cross occurs when the outlet temperature of the cold fluid exceeds that of the hot fluid (Sinnott & Towler 2008) and is indicated byF < 0.75. Since the calculated value of F is far greater than 0.75, one shell pass is considered sufficient.

10.5-Heat Exchanger Area To optimize the overall exchanger size, the heat transfer area must be calculated. This is done using: q = UAđ?‘œ ∆Tm Ađ?‘œ : Heat transfer area, Uđ?‘Žđ?‘ đ?‘ đ?‘˘đ?‘šđ?‘’đ?‘‘ = 850 W/m2 . ℃, and q = 1522 kW đ??€đ?’? =

đ??Ş đ?&#x;?đ?&#x;“đ?&#x;?đ?&#x;? Ă— đ?&#x;?đ?&#x;Žđ?&#x;‘ = = đ?&#x;“đ?&#x;–. đ?&#x;• đ??Śđ?&#x;? ( đ?‘Żđ?’šđ?’”đ?’šđ?’” đ?’†đ?’”đ?’•đ?’Šđ?’Žđ?’‚đ?’•đ?’†đ?’… đ?&#x;”đ?&#x;Ž. đ?&#x;‘đ?&#x;?) đ??”∆đ??“đ??Ś đ?&#x;–đ?&#x;“đ?&#x;Ž Ă— đ?&#x;‘đ?&#x;Ž. đ?&#x;“

From this value, exchanger specifications can then be determined and optimized.

10.6-Layout and Tube Size To start with a 20 mm Outer Diameter Tube with 2 mm thickness will be used, both of which are standard sizes. Stainless Steel 304 will also be used as the tube material as it should provide sufficient 8|P a g e

thermal conductivity to not impede the Overall Heat Transfer Coefficient. Finally the tube length was set to 6.1đ?‘š, which should be sufficient large to lower the cost and reduce the shell diameter đ?‘Ťđ?’? = đ?&#x;?đ?&#x;Ž đ?’Žđ?’Ž,

đ?‘Ťđ?’Š = đ?&#x;?đ?&#x;” đ?’Žđ?’Ž,

đ?‘ł = đ?&#x;”. đ?&#x;? đ?’Ž

Stainless steel 304 = 21.7 W/m K (Coulson & Richardson’s Chemical Engineering Design) 6.1- Number of Tubes After choosing the dimensions, the area of a single tube and number of tubes can be calculated: Area of single tube = Ď€Dđ?‘œ L = Ď€ Ă— 0.02 Ă— 6.1 = 0.3833m2 Where: đ??ˇđ?‘œ : đ?‘‚đ?‘˘đ?‘Ąđ?‘ đ?‘–đ?‘‘đ?‘’ đ?‘‘đ?‘–đ?‘Žđ?‘šđ?‘’đ?‘Ąđ?‘’đ?‘&#x; (đ?‘š),

đ??ż: đ?‘ƒđ?‘–đ?‘?đ?‘’ đ?‘™đ?‘’đ?‘›đ?‘”đ?‘Ąâ„Ž (đ?‘š)

Number of tubes =

Ađ?‘œ 58.7 = = 152 tubes Ađ?‘Ąđ?‘˘đ?‘?đ?‘’ 0.3833

6.2-Tube Arrangement Three main tube arrangements exist and are detailed below: Triangular arrangement: higher permissible heat transfer and pressure drop Square arrangement: lower heat transfer and pressure drop but easier to clean up. Rotated square arrangement: higher heat transfer and pressure drop compared to square, easy to clean therefore good for heavy fouling fluids. The square arrangement was chosen as it has the highest range. 6.3-Bundle and Shell Diameter The bundle diameter can be calculated using: Dđ?‘? Nđ?‘Ą = đ??ž1 ( )đ?‘›1 Dđ?‘œ Where: đ?‘ đ?‘Ą : Number of tubes Do : Tube outside diameter (mm)

đ??ˇđ?‘? : bundle diameter (mm) K1 and n1 are constants.

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(Table.1) Constants of Equation, Bundle Diameter (Coulson 1983) Since square pitch with four passes was chosen, from đ?‘‡đ?‘Žđ?‘?đ?‘™đ?‘’ 1: đ??ž1 = 0.158,

đ?‘›1 = 2.263

With these values, bundle diameter can be calculated: Dđ?‘? 2.263 152 = 0.158 ( ) 0.02 Dđ?‘? = 0.42 m

(đ?‘­đ?’Šđ?’ˆ. đ?&#x;?) đ?‘şđ?’‰đ?’†đ?’?đ?’? đ?‘Šđ?’–đ?’?đ?’…đ?’?đ?’† đ?‘Şđ?’?đ?’†đ?’‚đ?’“đ?’‚đ?’?đ?’„đ?’† (đ?‘Şđ?’?đ?’–đ?’?đ?’”đ?’?đ?’? đ?&#x;?đ?&#x;—đ?&#x;–đ?&#x;‘) From Figure 2-8, the shell diameter (Ds ) for pull-through floating head is given by: Ds = Dđ?‘? + clearance = 0.42 + 0.09 = 0.51 m 6.4-Baffle Size Baffles are used in the shell across tubes to direct the fluid stream, increase the fluid velocity and hence improve the rate of heat transfer (Coulson 1983). The most commonly used type of baffle is the single segmental baffle as shown in Figure 2-9.

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(𝑭𝒊𝒈. 𝟑) 𝑺𝒆𝒈𝒎𝒆𝒏𝒕𝒂𝒍 𝑩𝒂𝒇𝒇𝒍𝒆 (𝑪𝒐𝒖𝒍𝒔𝒐𝒏 𝟏𝟗𝟖𝟑) The baffle spacing ranges from 0.2 to 1.0 shell diameters (Kuppan, 2000). The flowrate in the shell side is high; therefore, the utilization of the baffle will increase the pressure drop. Therefore, the baffle space will be: 𝑏𝑎𝑓𝑓𝑙𝑒 𝑠𝑝𝑎𝑐𝑖𝑛𝑔 = 0.51 ∗ 0.5 = 0.255 𝑚

10.7-Overall Heat Transfer Calculation 7.1-Tube Side Coefficient 𝑇𝑢𝑏𝑒 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝑎𝑟𝑒𝑎 = 𝑇𝑢𝑏𝑒𝑠 𝑝𝑒𝑟 𝑝𝑎𝑠𝑠 =

𝜋 (0.016)2 = 0.0002𝑚2 4

𝑁𝑡 152 = = 38 𝑡𝑢𝑏𝑒𝑠/𝑝𝑎𝑠𝑠 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑝𝑎𝑠𝑠𝑒𝑠 4

𝑇𝑜𝑡𝑎𝑙 𝑡𝑢𝑏𝑒 𝑠𝑖𝑑𝑒 𝑓𝑙𝑜𝑤 𝑎𝑟𝑒𝑎 = 38 × 0.0002 = 0.0076 𝑚2 𝑀𝑎𝑠𝑠 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑓 ℎ𝑜𝑡 𝑠𝑡𝑟𝑒𝑎𝑚 = 𝑅𝑒 = 𝑢: 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑓 𝑓𝑙𝑢𝑖𝑑,

𝑚̇ 3.76 = = 494.74 𝑘𝑔/𝑚2 𝑠 𝑡𝑜𝑡𝑎𝑙 𝑡𝑢𝑏𝑒 𝑠𝑖𝑑𝑒 𝑎𝑟𝑒𝑎 0.0076

𝑚𝑡𝑢𝑏𝑒 . ∗ 𝐷𝑖 494.74 ∗ 0.016 = = 1.33 ∗ 104 𝜇 5.97 ∗ 10−4

𝑚 𝑠

𝐷𝑖 , 𝑖𝑛𝑛𝑒𝑟 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟, 𝑚𝑚

The tube side coefficient (ht) is given by: ℎ𝑡𝑢𝑏𝑒 =

𝑁𝑢 ∗𝑘 𝐷𝑖

𝜇 0.14 𝑁𝑢 = 𝑗ℎ × 𝑅𝑒 × 𝑃𝑟 0.33 ( ) 𝜇𝑤 𝑵𝒐𝒕𝒆: 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑣𝑖𝑠𝑐𝑜𝑠𝑖𝑡𝑦 𝑡𝑒𝑟𝑚 (

𝜇 0.14 ) , 𝑖𝑛𝑖𝑡𝑖𝑎𝑙𝑙𝑦 𝑎𝑠𝑢𝑚𝑒 𝑖𝑡 𝑒𝑞𝑢𝑎𝑙 1, 𝑎𝑛𝑑 𝐼 𝑤𝑖𝑙𝑙 𝑐ℎ𝑒𝑐𝑘 𝑖𝑡 𝑙𝑎𝑡𝑒𝑟. 𝜇𝑤

𝑃𝑟 =

𝜇𝑐𝑝 5.97 ∗ 10−4 ∗ 2.65 ∗ 103 = = 8.5 𝑘 0.1872

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(𝑭𝒊𝒈. 𝟒) 𝑻𝒖𝒃𝒆 − 𝒔𝒊𝒅𝒆 𝑯𝒆𝒂𝒕 𝑻𝒓𝒂𝒏𝒔𝒇𝒆𝒓 𝑭𝒂𝒄𝒕𝒐𝒓 (𝑪𝒐𝒖𝒍𝒔𝒐𝒏 𝟏𝟗𝟖𝟑) 𝑆𝑖𝑛𝑐𝑒, 𝐿 = 6.1 𝑚, 𝑎𝑛𝑑 𝐷𝑖 = 0.016 𝑚 𝐿 6.1 = = 381.25 𝐷𝑖 0.016 𝑗ℎ may then be obtained using Figure 2-11: 𝑗ℎ = 4 × 10−3 𝑁𝑢 = 𝑗ℎ × 𝑅𝑒 × 𝑃𝑟 0.33 = 4 × 10−3 × 1.33 × 104 × 8.50.33 = 107.8 ℎ𝑡𝑢𝑏𝑒 =

107.8 × 0.1872 = 1261.26 𝑊/𝑚2 . ℃ 0.016

7.2-Shell Side Coefficient The shell diameter was calculated to be 0.51 𝑚 and the recommended pitch is: 𝑃𝑡 = 1.25𝐷𝑜 = 1.25 × 0.02 = 0.025 𝑚 𝐴𝑠 = 𝐴𝑠 =

𝐷𝑠 × (𝑃𝑡 − 𝐷𝑜 ) × 𝐵 𝑃𝑡

0.51 × (0.025 − 0.02) × 0.255 = 0.02601 𝑚2 0.025 𝑚𝑠 =

167.72 = 6448.3 𝑘𝑔/𝑚2 . 𝑠 0.02601

For an equilateral square pitch arrangement: 𝐷𝑒 = 𝑅𝑒 =

1.1 (𝑝 2 − 0.917𝐷𝑜 2 ) = 0.014201 𝑚 𝐷𝑜 𝑡

𝑚𝑠 𝐷𝑒 6448.3 × 0.014201 = = 6.39 × 106 𝜇 1.433 ∗ 10−5 12 | P a g e

B: Baffle spacing(m)

De : hydraulic diameter(m)

(𝑭𝒊𝒈. 𝟓) 𝑺𝒉𝒆𝒍𝒍 − 𝒔𝒊𝒅𝒆 𝒉𝒆𝒂𝒕 𝒕𝒓𝒂𝒏𝒔𝒇𝒆𝒓 𝒇𝒂𝒄𝒕𝒐𝒓𝒔, 𝒔𝒆𝒈𝒎𝒆𝒏𝒕𝒂𝒍 𝒃𝒂𝒇𝒇𝒍𝒆𝒔 (𝑪𝒐𝒖𝒍𝒔𝒐𝒏 𝟏𝟗𝟖𝟑) 𝜇𝑐𝑝 𝑃𝑟 = = 2.55162 𝑘 From (𝑭𝒊𝒈. 𝟔) the shell side heat transfer factor can be estimated: 𝑗ℎ = 6 ∗ 10−4 𝜇 0.14 𝑁𝑢 = 𝑗ℎ × 𝑅𝑒 × 𝑃𝑟 0.33 × ( ) = 6 ∗ 10−4 × 6.39 × 106 × 0.95320.33 = 3773.83 𝜇𝑤 ℎ𝑠 =

𝑁𝑢 ∗ 𝐾 3773.83 ∗ 0.03877 𝑤 = = 10303.63705 2 𝐷𝑒 0.014201 𝑚 . °𝐶

The overall heat transfer coefficient (𝑈𝑜 ) is given by: 1 1 1 𝐷𝑜 = +( )∗ 𝑈𝑜 ℎ𝑠 ℎ𝑖 𝐷𝑖 ℎ𝑠 : 𝑠ℎ𝑒𝑙𝑙 𝑠𝑖𝑑𝑒 𝑐𝑜𝑛𝑣𝑒𝑐𝑡𝑖𝑜𝑛 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 (W/m2 . °C) ℎ𝑖 : 𝑡𝑢𝑏𝑒 𝑠𝑖𝑑𝑒 𝑐𝑜𝑛𝑣𝑒𝑐𝑡𝑖𝑜𝑛 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 (W/m2 . °C) 1 1 1 0.02 = +( ) 𝑈𝑜 10303.63705 1261.26 0.016 𝑈𝑜 = 919 % 𝐷𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 =

𝑤 𝑚2 . ℃

𝑈𝑜 − 𝑈𝑎𝑠𝑠 919 − 850 = × 100% = 8.117% 𝑈𝑎𝑠𝑠 850 13 | P a g e

As the percentage difference is less thanđ?&#x;‘đ?&#x;Ž%, the assumption may be assumed to be fairly accurate and no need to modify the assumption.

10.8-Pressure Drop 8.1-Tube Side The two major sources of pressure drop on the tube side are caused by friction loss in the tubes and tube-nozzle pressure drop. In order to determine the tube side pressure drop the following equation can be used: ∴ ∆đ?‘ƒđ?‘Ą = đ?‘ đ?‘? (4 Ă— đ?‘—đ?‘“ Ă—

đ??ż đ?œ‡ Ă— ( )−0.14 + 1.25) Ă— đ?œŒđ?‘‰ 2 đ?‘‘đ?‘– đ?œ‡đ?‘¤

∆đ?‘ƒ: đ?‘ƒđ?‘&#x;đ?‘’đ?‘ đ?‘ đ?‘˘đ?‘&#x;đ?‘’ đ?‘‘đ?‘&#x;đ?‘œđ?‘? (đ?‘?đ?‘Žđ?‘&#x;đ?‘ )

đ??ż: đ??żđ?‘’đ?‘›đ?‘”đ?‘Ąâ„Ž đ?‘œđ?‘“ đ?‘Ąđ?‘˘đ?‘?đ?‘’ (m)

đ?‘‰: đ??šđ?‘™đ?‘œđ?‘¤ đ?‘Łđ?‘’đ?‘™đ?‘œđ?‘?đ?‘–đ?‘Ąđ?‘Ś (đ?‘š/đ?‘ )

đ?‘—đ?‘“ : đ??šđ?‘&#x;đ?‘–đ?‘?đ?‘Ąđ?‘–đ?‘œđ?‘› đ?‘“đ?‘Žđ?‘?đ?‘Ąđ?‘œđ?‘&#x;

(đ?‘­đ?’Šđ?’ˆ. đ?&#x;”) đ?‘ťđ?’–đ?’ƒđ?’† − đ?’”đ?’Šđ?’…đ?’† đ?’‡đ?’“đ?’Šđ?’„đ?’•đ?’Šđ?’?đ?’? đ?’‡đ?’‚đ?’„đ?’•đ?’?đ?’“đ?’” (đ?‘Şđ?’?đ?’–đ?’?đ?’”đ?’?đ?’? đ?&#x;?đ?&#x;—đ?&#x;–đ?&#x;‘) Therefore, based on Renolad number for the tube side which is 1.33 ∗ 104 and the above figure: đ?‘—đ?‘“ = 4.6 ∗ 10−3 ∆đ?‘ƒđ?‘Ą = đ?‘ đ?‘? (4 Ă— đ?‘—đ?‘“ Ă—

đ??ż + 1.25) Ă— đ?œŒđ?‘‰ 2 đ?‘‘đ?‘–

∆đ?‘ƒđ?‘Ą = 7.773đ?‘˜đ?‘?đ?‘Ž Allow an additional 20 % of total pressure drop to account for pressure drop across inlet and outlet nozzles (Sinnott&Towler 2008): 0.2 Ă— 7.773 + 7.773 = 9.3276Kpa đ?‘şđ?’?, đ?’•đ?’‰đ?’† đ?’•đ?’?đ?’•đ?’‚đ?’? đ?’‘đ?’“đ?’†đ?’”đ?’”đ?’–đ?’“đ?’† đ?’…đ?’“đ?’?đ?’‘ đ?’Šđ?’? đ?’•đ?’‰đ?’† đ?’•đ?’–đ?’ƒđ?’† đ?’”đ?’Šđ?’…đ?’† đ?’Šđ?’” đ?&#x;—. đ?&#x;‘ đ?‘˛đ?’‘đ?’‚. 14 | P a g e

8.2-Shell Side The pressure drop on the shell side is calculated using: ∆đ?‘ƒđ?‘ = 4 Ă— đ?‘—đ?‘“ Ă—

đ??ˇđ?‘ đ??ż Ă— Ă— đ?œŒ Ă— đ?‘‰2 đ??ˇđ?‘’ đ?‘™đ?‘?

Where: đ??ˇđ?‘’ : đ??¸đ?‘žđ?‘˘đ?‘–đ?‘Łđ?‘Žđ?‘™đ?‘’đ?‘›đ?‘Ą đ?‘‘đ?‘–đ?‘Žđ?‘šđ?‘’đ?‘Ąđ?‘’đ?‘&#x; (đ?‘š)

đ??żđ?‘? : đ??ľđ?‘Žđ?‘“đ?‘“đ?‘™đ?‘’ đ?‘ đ?‘?đ?‘Žđ?‘?đ?‘–đ?‘›đ?‘” (đ?‘š)

(đ?‘­đ?’Šđ?’ˆ. đ?&#x;•) đ?‘şđ?’‰đ?’†đ?’?đ?’? đ?’”đ?’Šđ?’…đ?’† đ?’‡đ?’“đ?’Šđ?’„đ?’•đ?’Šđ?’?đ?’? đ?’‡đ?’‚đ?’„đ?’•đ?’?đ?’“đ?’”, đ?’”đ?’†đ?’ˆđ?’Žđ?’†đ?’?đ?’•đ?’‚đ?’? đ?’ƒđ?’‚đ?’‡đ?’‡đ?’?đ?’†đ?’” (đ?‘Şđ?’?đ?’–đ?’?đ?’”đ?’?đ?’? đ?&#x;?đ?&#x;—đ?&#x;–đ?&#x;‘) Therefore, based on Renolad number for the shell side which is 6.39 Ă— 106 and the above figure with 45%: đ?‘—đ?‘“ = 1.7 ∗ 10−2 ∆đ?‘ƒđ?‘ = 4 Ă— đ?‘—đ?‘“ Ă—

đ??ˇđ?‘ đ??ż Ă— Ă— đ?œŒ Ă— đ?‘‰2 đ?‘‘đ?‘’ đ?‘™đ?‘?

∆đ?‘ƒđ?‘ = 37.309 đ?‘˜đ?‘?đ?‘Ž đ?‘şđ?’?, đ?’•đ?’‰đ?’† đ?’‘đ?’“đ?’†đ?’”đ?’”đ?’–đ?’“đ?’† đ?’…đ?’“đ?’?đ?’‘ đ?’?đ?’? đ?’•đ?’‰đ?’† đ?’”đ?’‰đ?’†đ?’?đ?’? đ?’”đ?’Šđ?’…đ?’† đ?’Šđ?’” đ?&#x;‘đ?&#x;•đ?&#x;‘đ?&#x;Žđ?&#x;—. đ?&#x;’đ?&#x;’ đ?’Œđ?’‘đ?’‚.

10.9-Viscosity Correction Factor The viscosity correction factor was neglected when calculating the heat transfer coefficients and pressure drops. To check if the viscosity correction factor will affect the calculations, it will now be evaluated: Estimate wall temperature �� by:

15 | P a g e

(đ?‘‡đ?‘¤ − đ?‘‡â„Ž,đ?‘Žđ?‘Łđ?‘” ) ∗ â„Žđ?‘– = đ?‘ˆđ?‘œ (đ?‘‡đ?‘?,đ?‘Žđ?‘Łđ?‘” − đ?‘‡â„Ž,đ?‘Žđ?‘Łđ?‘” ) (đ?‘‡đ?‘¤ − 135.25) ∗ 1261.26 = 919 ∗ (55.84 − 135.25) đ?‘‡đ?‘¤ = 77.38 ℃ Now, by linear interpolation, it can be found the viscosity at this temperature: The hot stream viscosity at this temperature = 9.9136 đ?‘?đ?‘? đ?œ‡ 0.14 ) đ?œ‡đ?‘¤

∴ Viscosity correction factor = (

5.97 0.14 ) 9.9136

=(

= 0.94

Since this value is close to one, it can be concluded that the exclusion of the viscosity correction factor would not significantly affect the calculations.

10.10-Pressure Relief System Emergency relief in the process industries aims to protect equipment, the environment and operating personnel from abnormal conditions. In addition, pressure relief system is also required on every vessel to prevent overpressure and rupture of the vessel. The type of pressure relief valve chosen was conventional pressure relief valve, because they have high sensitive to any change in the pressure and they have low failure rate (Ludwig E. E., 1999). The material of construction was chosen to be stainless steel. The pressure relief valve will be allocated on the tube side and the minimum required effective discharge area can be calculated by: đ??´ =

13160 ∗ đ?‘Š ∗ √đ?‘‡đ?‘? đ??ś ∗ đ??ž ∗ đ?‘ƒ ∗ đ??žđ?‘? ∗ √đ?‘€đ?‘¤đ?‘Ą

Where: đ?‘Š = đ?‘…đ?‘’đ?‘žđ?‘˘đ?‘–đ?‘&#x;đ?‘’đ?‘‘ đ?‘&#x;đ?‘’đ?‘™đ?‘–đ?‘’đ?‘Łđ?‘–đ?‘›đ?‘” đ?‘?đ?‘Žđ?‘?đ?‘Žđ?‘?đ?‘–đ?‘Ąđ?‘Ś, đ?‘˜đ?‘–đ?‘™đ?‘œđ?‘”đ?‘&#x;đ?‘Žđ?‘šđ?‘ đ?‘?đ?‘’đ?‘&#x; â„Žđ?‘œđ?‘˘đ?‘&#x;, 1.35 ∗ 104

đ?‘˜đ?‘” â„Ž

đ??ś = đ??śđ?‘œđ?‘’đ?‘“đ?‘“đ?‘–đ?‘?đ?‘–đ?‘’đ?‘›đ?‘Ą đ?‘‘đ?‘’đ?‘Ąđ?‘’đ?‘&#x;đ?‘šđ?‘–đ?‘›đ?‘’đ?‘‘ đ?‘“đ?‘&#x;đ?‘œđ?‘š đ?‘Žđ?‘› đ?‘’đ?‘Ľđ?‘?đ?‘&#x;đ?‘’đ?‘ đ?‘ đ?‘–đ?‘œđ?‘› đ?‘œđ?‘“ đ?‘Ąâ„Žđ?‘’ đ?‘&#x;đ?‘Žđ?‘Ąđ?‘–đ?‘œ đ?‘œđ?‘“ đ?‘ đ?‘?đ?‘’đ?‘?đ?‘–đ?‘“đ?‘–đ?‘? â„Žđ?‘’đ?‘Žđ?‘Ąđ?‘ đ?‘œđ?‘“ đ?‘Ąâ„Žđ?‘’ đ?‘”đ?‘Žđ?‘ đ?‘œđ?‘&#x; đ?‘Łđ?‘Žđ?‘?đ?‘œđ?‘&#x; đ?‘Žđ?‘Ą đ?‘ đ?‘Ąđ?‘Žđ?‘›đ?‘‘đ?‘Žđ?‘&#x;đ?‘‘ đ?‘?đ?‘œđ?‘›đ?‘‘đ?‘–đ?‘Ąđ?‘–đ?‘œđ?‘›đ?‘ , 344. đ?‘‡ = đ??´đ?‘?đ?‘ đ?‘œđ?‘™đ?‘˘đ?‘Ąđ?‘’ đ?‘Ąđ?‘’đ?‘šđ?‘?đ?‘’đ?‘&#x;đ?‘Žđ?‘Ąđ?‘˘đ?‘&#x;đ?‘’ đ?‘œđ?‘“ đ?‘Ąâ„Žđ?‘’ đ?‘“đ?‘™đ?‘˘đ?‘–đ?‘‘ đ?‘Žđ?‘Ą đ?‘Ąâ„Žđ?‘’ đ?‘Łđ?‘Žđ?‘™đ?‘Łđ?‘’ đ?‘–đ?‘›đ?‘™đ?‘’đ?‘Ą, đ?‘‘đ?‘’đ?‘”đ?‘&#x;đ?‘’đ?‘’đ?‘ đ??žđ?‘’đ?‘™đ?‘Łđ?‘–đ?‘›, 488.65đ??ž. đ??ž = Effective coefficient of discharge. K = 0.975. đ??žđ?‘? = Capacity correction factor due to back pressure, based on C = 344, it will be 1.27. đ?‘ƒ = Relieving pressure, kpa absolute. đ?‘ƒ = đ?‘Ąâ„Žđ?‘’ đ?‘ đ?‘’đ?‘Ą đ?‘?đ?‘&#x;đ?‘’đ?‘ đ?‘ đ?‘˘đ?‘&#x;đ?‘’ + đ?‘œđ?‘Łđ?‘’đ?‘&#x;đ?‘?đ?‘&#x;đ?‘’đ?‘ đ?‘ đ?‘˘đ?‘&#x;đ?‘’ + đ?‘Žđ?‘Ąđ?‘šđ?‘œđ?‘ đ?‘?â„Žđ?‘’đ?‘&#x;đ?‘–đ?‘? đ?‘?đ?‘&#x;đ?‘’đ?‘ đ?‘ đ?‘˘đ?‘&#x;đ?‘’ 16 | P a g e

đ?‘ƒ = 62.57 + 0.1 ∗ 62.57 + 1 = 69 đ?‘?đ?‘Žđ?‘&#x; = 6900đ?‘˜đ?‘?đ?‘Ž đ?‘€đ?‘¤đ?‘Ą = đ?‘€đ?‘œđ?‘™đ?‘’đ?‘?đ?‘˘đ?‘™đ?‘Žđ?‘&#x; đ?‘¤đ?‘’đ?‘–đ?‘”â„Žđ?‘Ą, 19. 13160 ∗ 1.34 ∗ 104 ∗ √488.65 ∗ 1 đ??´ = = 30.5 đ?‘šđ?‘š2 = 0.05 đ?‘–đ?‘›2 344 ∗ 0.975 ∗ 6900 ∗ 1.27 ∗ √19 The nearest standard area is 0.11 đ?‘–đ?‘›2 and the appropriate valve for this area from (Parry, 1992) is D type.

10.11-Mechanical Design 10.1-Material Selection Due to the low temperature operating the material of construction was chosen to be stainless steel 304. The stainless steel material has high corrosion resistance and excellent low temperature properties. 10.2-Shell Thickness The shell thickness can be calculated by: đ?‘†â„Žđ?‘’đ?‘™đ?‘™ đ?‘Ąâ„Žđ?‘–đ?‘?đ?‘˜đ?‘›đ?‘’đ?‘ đ?‘ =

đ?‘ƒđ??ˇđ?‘’đ?‘ đ?‘–đ?‘”đ?‘› ∗ đ??ˇ 2đ?‘“ − đ?‘ƒ

đ?‘ƒđ??ˇđ?‘’đ?‘ đ?‘–đ?‘”đ?‘› = 1.15 ∗ 9.3 = 10.695 đ?‘?đ?‘Žđ?‘&#x; = 1.0695

đ?‘ đ?‘šđ?‘š2

đ?‘‡đ?‘’đ?‘šđ?‘?đ?‘’đ?‘&#x;đ?‘Žđ?‘Ąđ?‘˘đ?‘&#x;đ?‘’đ??ˇđ?‘’đ?‘ đ?‘–đ?‘”đ?‘› = 1.0695 ∗ 135.25 = 144.6 ℃ The design stress at the design temperature from the table below will be 141.08 đ?‘ /đ?‘šđ?‘š2 . Materials Tensile Design stress at temperature °C (N/mm2) Stainless strength 0 to 100 150 200 250 300 350 400 450 (N/mm2) 50 Steel (18Cr/9Ni, 540 165 150 140 135 130 130 125 120 120 Ti stabilised) (đ?‘ťđ?’‚đ?’ƒđ?’?đ?’†. đ?&#x;?)đ?‘Ťđ?’”đ?’Šđ?’ˆđ?’? đ?’”đ?’•đ?’“đ?’†đ?’”đ?’” đ?’?đ?’‡ đ?‘şđ?’•đ?’‚đ?’Šđ?’?đ?’?đ?’†đ?’”đ?’” đ?‘şđ?’•đ?’†đ?’†đ?’? (Ludwig, 1999) đ?‘†â„Žđ?‘’đ?‘™đ?‘™ đ?‘Ąâ„Žđ?‘–đ?‘?đ?‘˜đ?‘›đ?‘’đ?‘ đ?‘ =

500 115

1.0695 ∗ 510 = 1.94 đ?‘šđ?‘š 2 ∗ 141.08 − 1

The thickness will be 1.94đ?‘šđ?‘š, and no need to corrosion allowance because the material of construction is stainless steel.

17 | P a g e

10.12-Operating Design To operate the heat exchanger safely and efficiently, it is important to make sure the entire of the exchanger is clean to prevent plugging. Additionally, it can be used fresh hot water to clean any soft deposit in the shell side. 11.1-Startup Procedure 1- Open the vent connections. 2- Open the vent and start the cold fluid in the shell side. Ensure that the shell side flooding and close the vent. 3- Gradually introduce the hot fluid to the tube side. Ensure that the tube filled and close the vent. 4- When the fluids exist in the both side, bring the heat exchanger to the operating temperature. 11.2-Shutdown Procedure 1- Shutdown the hot fluid first. 2- Stop the circulating of the cold fluid. 3- Bring the heat exchanger to the atmospheric condition by opening the vents. 11.3- Maintenance Regular inspection and maintenance is necessary to operate the exchanger safely and effectively. The most common problem in all the exchangers are the tube fouling, therefore, it is important to clean the tubes because the fouling is reducing the heat transfer coefficient. The valves must be checked to make sure that they are still work.

10.13- Process Control System and P&ID The basic objective of the control system to make sure that the heat exchanger is working as defined by design. The most important factor in the control of heat exchanger is the outlets temperatures for both the shell and tube sides. Therefore, it is important to choose which stream needs control. In this recondenser, the shell side have high flow rate compare to the tube side, therefore the control will be on the shell side. Check P&ID for brief description.

18 | P a g e

Piping and Instrument FC

TC

TG

FT

Shell Inlet

TG

PG

TG

Tube Inlet TT

Tube outlet

Drain

TG

Shell Outlet

TG

FC

TC

Temperature Gauge

Flow Controller

Temperature Controller

Electric Signal

PG

TI

FT

Pressure Gauge

Temperature Indicator

Flow Transmitter Title:Shell& tube Heat Exchanger.

Globe Valve

Drawing Title: P&ID Gate Valve

Relief Valve

Vessel Tag no: E-405 Designed by: Samah Zaki Alrashid

Approved by: Dr.Gia Pham

19 | P a g e

10.14-Mechanical Drawing

Shell Inlet

Tube Inlet

Tubes

0.194

Cross Section View

Shell Vent

6.1

0.02 m

0.03

Square Pattern 0.51

0.51

Floating Tubes Sheet

0.0050

Shell Cover

0.0050

Floating Head Cover 0.02 m 0.03

0.194

Drain

Foundation

Tube Outlet

Sport Saddle

Tie Rods

Shell Outlet

Title:Shell& tube Heat Exchanger. Shell Diameter: 0.51m

O.D Tube: 20mm

Flow Pattern : Countercurrent

Shell Thickness : 1.94mm

Shell Side Area : 0.02601m2

I.D Tube: 16mm

Tube Length: 4m

Tube Arrangement : Square

Hydraulic Diameter : 0.014201m

Length: 6.1 m

Tubes No: 152

Tube Passes no: Four passes

Tubes Side Area : 0.0076m2

Total Heat Exchanger Area: 58.7m2

Bundle Diameter : 0.42m

Drawing Title: Mechanical Drawing Vessel Tag no: E-405 Designed by: Samah Zaki Alrashid

Approved by: Dr.Gia Pham

20 | P a g e

10.15-Specification Sheet

𝑬𝒒𝒖𝒊𝒑𝒎𝒆𝒏𝒕: 𝑺𝒉𝒆𝒍𝒍 𝒂𝒏𝒅 𝑻𝒖𝒃𝒆 𝑯𝒆𝒂𝒕 𝑬𝒙𝒄𝒉𝒂𝒏𝒈𝒆𝒓 (𝑹𝒆𝒄𝒐𝒏𝒅𝒆𝒏𝒔𝒆𝒓) 𝐹𝑙𝑢𝑖𝑑 𝐴𝑙𝑙𝑜𝑐𝑎𝑡𝑖𝑜𝑛

Shell side

Tube side

𝐼𝑛𝑙𝑒𝑡 𝑇𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒

54.08℃

215.5℃

𝑂𝑢𝑡𝑙𝑒𝑡 𝑇𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒

57.6℃

55℃

𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑇𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒

55.84℃

135.25℃

𝑅𝑒𝑛𝑜𝑙𝑑 𝑁𝑢𝑚𝑏𝑒𝑟

6.39 × 106

1.33 × 104

𝑃𝑟𝑎𝑛𝑑𝑡𝑙 𝑁𝑢𝑚𝑏𝑒𝑟

2.55162

8.5

𝑁𝑢𝑠𝑠𝑢𝑙𝑡 𝑁𝑢𝑚𝑏𝑒𝑟

3773

107.8

𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝐷𝑟𝑜𝑝

37.309 𝑘𝑝𝑎 w 10303 2 m .s

9.3 𝑘𝑝𝑎 𝑤 1261.26 2 𝑚 .𝑠

𝐻𝑒𝑎𝑡 𝑇𝑟𝑎𝑛𝑠𝑓𝑒𝑟 𝐶𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡

𝑫𝒆𝒔𝒊𝒈𝒏 𝑷𝒂𝒓𝒂𝒎𝒆𝒕𝒆𝒓 𝑆ℎ𝑒𝑙𝑙 𝐷𝑖𝑎𝑚𝑒𝑡𝑒𝑟(𝑚)

0.51

ℎ𝑦𝑑𝑟𝑎𝑢𝑙𝑖𝑐 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟, 𝐷𝑒 (𝑚)

0.014201

𝐵𝑢𝑛𝑑𝑙𝑒 𝐷𝑖𝑎𝑚𝑒𝑡𝑒𝑟(𝑚)

0.4

𝑆ℎ𝑒𝑙𝑙 𝑇ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠

1.94𝑚𝑚

𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑇𝑢𝑏𝑒𝑠 𝑃𝑎𝑠𝑠𝑒𝑠

𝑓𝑜𝑢𝑟

𝑃𝑖𝑡𝑐ℎ 𝑇𝑦𝑝𝑒

𝑆𝑞𝑢𝑎𝑟𝑒 𝑃𝑖𝑡𝑐ℎ

𝐵𝑢𝑛𝑑𝑙𝑒 𝐷𝑖𝑎𝑚𝑒𝑡𝑒𝑟(𝑚)

0.42

𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑆ℎ𝑒𝑙𝑙 𝑃𝑎𝑠𝑠𝑒𝑠

𝑜𝑛𝑒

𝑃𝑖𝑡𝑐ℎ(𝑚𝑚)

25

𝐵𝑎𝑓𝑓𝑒𝑙 𝐹𝑎𝑐𝑡𝑜𝑟

1

𝑆ℎ𝑒𝑙𝑙 𝑆𝑖𝑑𝑒 𝐴𝑟𝑒𝑎 m2

0.02601

𝐵𝑎𝑓𝑓𝑒𝑙 𝑙 𝑆𝑝𝑎𝑐𝑖𝑛𝑔(𝑚)

0.0002

𝑇𝑢𝑏𝑒 𝑆𝑖𝑑𝑒 𝐴𝑟𝑒𝑎 m2

0.0076

𝑇𝑢𝑏𝑒 𝐿𝑒𝑛𝑔𝑡ℎ(𝑚)

6.1

𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑇𝑢𝑏𝑒𝑠

152

𝑂𝑢𝑡𝑒𝑟 𝑇𝑢𝑏𝑒 𝐷𝑖𝑎𝑚𝑒𝑡𝑒𝑟(𝑚)

0.02

𝐼𝑛𝑛𝑒𝑟 𝑇𝑢𝑏𝑒 𝐷𝑖𝑎𝑚𝑒𝑟𝑡𝑒𝑟(𝑚)

0.016

𝑇𝑦𝑝𝑒 𝑜𝑓 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑟𝑒𝑙𝑒𝑖𝑓 𝑣𝑎𝑙𝑣𝑒

𝐷 𝑡𝑦𝑝𝑒

𝑇𝑜𝑡𝑎𝑙 ℎ𝑒𝑎𝑡 𝑒𝑥𝑐ℎ𝑎𝑛𝑔𝑒𝑟 𝑎𝑟𝑒𝑎

58.7 m2

𝑇𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝐶𝑜𝑟𝑟𝑒𝑙𝑎𝑡𝑖𝑜𝑛 𝐹𝑎𝑐𝑡𝑜𝑟

1

𝐿𝑜𝑔 𝑚𝑒𝑎𝑛 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 30.5℃

𝑀𝑎𝑡𝑒𝑟𝑖𝑎𝑙 Thermal Conductivity(

𝑇ℎ𝑒 𝐻𝑒𝑎𝑡 𝐸𝑥𝑐ℎ𝑎𝑛𝑔𝑒𝑟 𝑇𝑦𝑝𝑒 𝑇ℎ𝑒 𝑂𝑣𝑒𝑟𝑎𝑙𝑙 𝐻𝑒𝑎𝑡 𝑇𝑟𝑎𝑛𝑠𝑓𝑒𝑟 𝐶𝑜𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑀𝑎𝑡𝑒𝑟𝑖𝑎𝑙 𝑜𝑓 𝑡ℎ𝑒 𝐶𝑜𝑛𝑠𝑡𝑟𝑢𝑐𝑡𝑖𝑜𝑛

w ) m2 . s

0.0387

𝑃𝑢𝑙𝑙 𝑇ℎ𝑟𝑜𝑢𝑔ℎ 𝐹𝑙𝑜𝑎𝑡𝑖𝑛𝑔 𝐻𝑒𝑎𝑑 𝑤 919 2 𝑚 .℃ 𝑆𝑡𝑎𝑖𝑛𝑙𝑒𝑠𝑠 𝑠𝑡𝑒𝑒𝑙 (304) 21 | P a g e

10.16-Hazard and Safety A preliminary hazard analysis was carried out to identify the main hazards and the possible consequence associated with them. The below table shows which are the main hazards of the system.

𝑭𝒂𝒊𝒍𝒖𝒓𝒆 𝑴𝒐𝒅𝒆 𝐹𝑖𝑟𝑒

𝑇𝑢𝑏𝑒 𝐹𝑎𝑖𝑙𝑢𝑟𝑒

𝑪𝒂𝒖𝒔𝒆𝒔 𝑭𝒓𝒆𝒒𝒖𝒆𝒏𝒄𝒚 𝑪𝒐𝒏𝒔𝒆𝒒𝒖𝒆𝒏𝒄𝒚 𝑨𝒄𝒕𝒊𝒐𝒏 ●𝑆𝑝𝑎𝑟𝑘 𝑖𝑛 𝐻𝑒𝑎𝑡 𝑒𝑥𝑐ℎ𝑎𝑛𝑔𝑒𝑟 . 𝑅𝑎𝑟𝑒 ●𝐸𝑞𝑢𝑖𝑏𝑚𝑒𝑛𝑡 𝑑𝑎𝑚𝑎𝑔𝑒. ●𝐻𝑖𝑔ℎ 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑎𝑙𝑎𝑟𝑚. ●𝑃𝑒𝑟𝑠𝑜𝑛𝑎𝑙 𝑒𝑛𝑗𝑢𝑟𝑦. ●𝑆𝑢𝑑𝑑𝑒𝑛 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑟𝑖𝑠𝑒. ●𝑅𝑒𝑑𝑢𝑐𝑒 𝑡ℎ𝑒 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 ●𝐸𝑙𝑒𝑐𝑡𝑟𝑖𝑐𝑎𝑙 𝑓𝑎𝑢𝑙𝑡. ●𝑅𝑒𝑔𝑢𝑙𝑎𝑟 𝑖𝑛𝑠𝑝𝑒𝑐𝑡𝑖𝑜𝑛 ●𝐻𝑢𝑚𝑎𝑛 𝑒𝑟𝑟𝑜𝑟. ●𝐶𝑜𝑟𝑟𝑜𝑠𝑖𝑜𝑛. 𝐹𝑟𝑒𝑐𝑞𝑢𝑒𝑛𝑡 ●𝑂𝑓𝑓 𝑠𝑝𝑐𝑖𝑓𝑖𝑐𝑎𝑡𝑖𝑜𝑛 𝑝𝑟𝑜𝑑𝑢𝑐𝑡●𝑀𝑒𝑡ℎ𝑜𝑑𝑠 𝑎𝑔𝑖𝑎𝑛𝑠𝑡 𝑐𝑜𝑟𝑟𝑜𝑠𝑖𝑜𝑛: 𝑐𝑜𝑎𝑡𝑖𝑛𝑔, 𝑎𝑛𝑑 𝑐𝑎𝑡ℎ𝑜𝑑𝑖𝑐 𝑝𝑟𝑜𝑡𝑒𝑐𝑡𝑖𝑜𝑛

𝑉𝑒𝑛𝑡 𝑉𝑎𝑙𝑣𝑒: ●𝐹𝑖𝑎𝑙𝑠 𝑜𝑝𝑒𝑛. ●𝐹𝑖𝑎𝑙𝑠 𝑐𝑙𝑜𝑠𝑒.

●𝑀𝑒𝑐ℎ𝑎𝑛𝑖𝑐𝑎𝑙 𝑓𝑎𝑖𝑙𝑢𝑟𝑒. ●𝐻𝑢𝑚𝑎𝑛 𝑒𝑟𝑟𝑜𝑟.

𝑅𝑎𝑟𝑒

●𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑒.

●𝑅𝑒𝑔𝑢𝑙𝑎𝑟 𝑖𝑛𝑠𝑝𝑒𝑐𝑡𝑖𝑜𝑛.

𝑅𝑒𝑙𝑒𝑖𝑓 𝑉𝑎𝑙𝑣𝑒 ●𝐹𝑖𝑎𝑙𝑠 𝑜𝑝𝑒𝑛. ●𝐹𝑖𝑎𝑙𝑠 𝑐𝑙𝑜𝑠𝑒.

●𝑀𝑒𝑐ℎ𝑎𝑛𝑖𝑐𝑎𝑙 𝑓𝑎𝑖𝑙𝑢𝑟𝑒. ●𝐻𝑢𝑚𝑎𝑛 𝑒𝑟𝑟𝑜𝑟.

𝑅𝑎𝑟𝑒

●𝑂𝑣𝑒𝑟𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑖𝑛 𝐻𝑒𝑎𝑡 𝑒𝑥𝑐ℎ𝑎𝑛𝑔𝑒𝑟.

●𝑅𝑒𝑔𝑢𝑙𝑎𝑟 𝑖𝑛𝑠𝑝𝑒𝑐𝑡𝑖𝑜𝑛 𝑎𝑛𝑑 𝑚𝑎𝑖𝑛𝑡𝑒𝑛𝑎𝑛𝑐𝑒.

𝐷𝑟𝑎𝑖𝑛 𝑉𝑎𝑙𝑣𝑒 ●𝐹𝑖𝑎𝑙𝑠 𝑜𝑝𝑒𝑛. ●𝐹𝑖𝑎𝑙𝑠 𝑐𝑙𝑜𝑠𝑒.

●𝑀𝑒𝑐ℎ𝑎𝑛𝑖𝑐𝑎𝑙 𝑓𝑎𝑖𝑙𝑢𝑟𝑒. ●𝐵𝑙𝑜𝑐𝑘𝑎𝑔𝑒.

𝑅𝑎𝑟𝑒

●𝐿𝑜𝑠𝑠 𝑜𝑓 𝑝𝑟𝑜𝑑𝑢𝑐𝑡. ●𝑅𝑒𝑔𝑢𝑙𝑎𝑟 𝑖𝑛𝑠𝑝𝑒𝑐𝑡𝑖𝑜𝑛 𝑎𝑛𝑑 ●𝑈𝑛𝑎𝑏𝑙𝑒 𝑡𝑜 𝑑𝑟𝑎𝑖𝑛 𝑚𝑎𝑖𝑛𝑡𝑒𝑛𝑎𝑛𝑐𝑒. 𝑡ℎ𝑒 𝑒𝑥𝑐ℎ𝑎𝑛𝑔𝑒𝑟 𝑜𝑟 𝑑𝑟𝑎𝑖𝑛 𝑖𝑡.

𝑇𝑢𝑏𝑒 𝑆ℎ𝑒𝑒𝑡

●𝑉𝑎𝑏𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑡𝑢𝑏𝑒𝑠.

𝐶𝑜𝑚𝑚𝑜𝑛

●𝐼𝑛𝑛𝑒𝑟 𝑤𝑒𝑙𝑑𝑖𝑛𝑔 𝑐𝑟𝑎𝑐𝑘

●𝐴𝑣𝑜𝑖𝑑 𝑠𝑢𝑑𝑑𝑒𝑛 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛

𝑅𝑎𝑟𝑒

● 𝑆ℎ𝑢𝑡𝑑𝑜𝑤𝑛. ● 𝑁𝑜 ℎ𝑒𝑎𝑡 𝑒𝑥𝑐ℎ𝑎𝑛𝑔𝑒.

● 𝑊𝑜𝑟𝑘 𝑡ℎ𝑒 𝑒𝑞𝑢𝑖𝑝𝑚𝑒𝑛𝑡 𝑖𝑛 𝑎 𝑠𝑎𝑓𝑒 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛. ● 𝑅𝑒𝑔𝑢𝑙𝑎𝑟 𝑖𝑛𝑠𝑝𝑒𝑐𝑡𝑖𝑜𝑛. ● 𝑅𝑒𝑔𝑢𝑙𝑎𝑟 𝑡𝑒𝑠𝑡 𝑓𝑜𝑟 𝑒𝑚𝑒𝑟𝑔𝑒𝑛𝑐𝑦 𝑠ℎ𝑢𝑡𝑑𝑜𝑤𝑛.

𝐸𝑞𝑢𝑖𝑏𝑚𝑒𝑛𝑡 𝐹𝑎𝑖𝑙𝑢𝑟𝑒 ● 𝐴𝑏𝑛𝑜𝑟𝑚𝑎𝑙 𝑜𝑝𝑒𝑟𝑎𝑡𝑖𝑛𝑔 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛. ● 𝑃𝑟𝑜𝑏𝑙𝑒𝑚𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑑𝑒𝑠𝑖𝑔𝑛.

22 | P a g e

Critical Review Design The properties that are used in the all calculation were provided by Hysys. The design includes: the calculation of the duty, the estimation of the log mean temperature, specify the inner and outer diameter, calculation of the shell side diameter, specify the baffle spacing, calculation of the shell and tube sides heat transfer coefficient, the pressure drop for both sides, brief hazard, startup and shutdown, and the pressure relief system.

References Coulson, J.M, Richardson, J.F. 1983. Chemical EngineeringDesign.6th ed. University College of Swansea: Robert Maxwell, M.C. Coulson, J. M, J. F Richardson, J. R Backhurst, and J. H Harker. 1999. Chemical Engineering. 1st ed. Oxford: Butterworth-Heinemann. Incropera, Frank P, and DeWitt,David P. 1981. Fundamentals Of Heat Transfer. 4th ed. New York: Wiley. Kakac, S.,& Liu, H.1998. Heat exchanger: Selection, rating, and theraml design. Boca Raton, FL: CRC press. Kuppan, T. 2000. Heat Exchanger Design Handbook. 1st ed. New York: Marcel Dekker. Perry, Robert H, Don W Green, and James O Maloney. 2008. Perry's Chemical Engineers' Handbook. 8th ed. London: McGraw-Hill. Towler, Gavin P, and R. K Sinnott. 2008. Chemical Engineering Design. 5th ed. Amsterdam: Elsevier/Butterworth-Heinemann.

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