Bioprocess engineering principles

SOLUTIONS MANUAL Bioprocess Engineering Principles Pauline M. Doran

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SOLUTIONS MANUAL Bioprocess Engineering Principles

Pauline M. Doran University of New South Wales, Sydney, Australia

ISBN 0 7334 15474 © Pauline M. Doran 1997

Table of Contents Solutions Page

Chapter 2

Introduction to Engineering Calculations

1

Chapter 3

Presentation and Analysis ofData

9

Chapter 4

Material Balances

17

ChapterS

Energy Balances

41

Chapter 6

Unsteady-State Material and Energy Balances

54

Chapter 7

Fluid Flow and Mixing

76

Chapter 8

Heat Transfer

86

Chapter 9

Mass Transfer

'98

Chapter 10

Unit Operations

106

Chapter 11 Homogeneous Reactions

122

Chapter 12 Heterogeneous Reactions

139

Chapter 13 Reactor Engineering

151

NOTE All equations, tables, figures, page numbers, etc., mentioned in this manual refer to the textbook,

Bioprocess Engineering Principles.

Introduction to Engineering Calculations 2.1

Unit conversion

(a)

From Table A.9 (Appendix A): 1 cP::::: 1O~3 kg m- I k 1 1 m= lOOcrn Therefore: 1.5

x

10-6 cP ::::: 1.5 x 10-6 cP

,1 10-

3 g k ;-1 t

s-ll.ll~~mI = 1.5 x 10-11 kg s-1 cm- 1

Answer: 1.5 x 10- 11 kg s-1 em- t

(b)

From Table A.S (Appendix A): 1 bp (British)::::: 42.41 Btu min-I Therefore:

Answer: 5.17 Btu min- 1 (e)

From Table A.S (Appendix. A): 1 mmHg:= 1.316 x 10-3 attn From Table A.I (Appendix A): 1 ft = 0.3048 m From Table A.7 (Appendix A): 11 atm 9.604 x 10-2 Btu From Table A8 (Appendix A): 1 Btu min-I::::: 2.391 x to- 2 metric horsepower Im=lOOcm 11= lOOOcm3 Ih=60min

=

Therefore: 670mmHgft3

3

3 atml.19.604X 1O-2Btul.I掳.3048m 1 mmHg llatm 1 it

= 670 mmHg ft3 .11.316X 10-

1

I I= ,

2 .391 x 10-2 metric horsepower I I h. . -60 1 1Btu minmm 1

3

1 100 em 1

1m

=

From Table A.7 (Appendix A): 1 Btu 0.2520 kcal From Table A.3 (Appendix A): Ilb = 453.6 g Therefore: 1

345 Btu Ib- = 345 BtulbAnswer: 0.192 kcal g-1

2.2

Unit conversion

Case 1 Convert to units of kg, m, s.

From Table AJ (Appendix A), lIb = 0.4536 kg

1

11 lOOOcm3

I.

956 x 10-4 metric . horsepower h

Answer: 9.56 x 10-4 metric horsepower h (d)

1

.1 o.2i~~Call路14;3~: g I = O.192kcal g-l

Solutions: Chapter 2

2 From Table A.2 (Appendix A): 1 tt3 =: 2.832 x 10-2 m 3 From Table A.9 (Appendix A): 1 cP::: 10-3 kg m- l s·l 1 rn= tOOcm= lOOOmm Therefore, using Eq. (2.1):

n(3cms-l.l~n(251bfC3I0.4536kgl.1 Ift ~ l00cmU lIb 2.832 x 10 2 m 3U _ 3

Dup Re::: - p -

(2mm.1

=

1m l000mmU

-6 P

10

I

10

c .

-3

kgm 1 cP

1

-11 s

- 2.4 x 10

7

Answer: 2.4 x 107

ease 2 Convert to units of kg, m, s.

From Table Al (Appendix A): 1 in.::: 2.54 x 10-2 m From Table A.9 (Appendix A): 1 Ibm ft-I h- 1 ::: 4.134 x 10-4 kg m-t s·t

Ih=3600s Therefore, usingEq. (2.1):

13~sl

= 1.5 x 104

Answer: 1.5 x 104

2.3

Dimensionless groups and property data

From the Chemical Engineers' Handbook, the diffusivity of oxygen in water at 2S"C is 2.5 x 10- 5 cm2 s·l. Assuming this is the same at 28"C, !lJ= 2.5 x 10-5 cm1 s·t, Also, from the Chemical Engineers' Handbook, the density of water at 28"e is PL ::: 0.9962652 g cm-3, and the viscosity of water at 28"C is JlL::: 0.87 cPo The density of oxygen at 28°C and 1 atm pressure can be calculated using the ideal gas law. As molar density is the same as n,V, from Eq. (2.32):

Temperature in the ideal gas equation is absolute temperature; therefore, from Eq. (2.24):

T = (28 + 273.15) K = 301.15 K From Table 2.5, R "" 82.057 cm3 atm K-I gmol~l. Substituting parameter values into the density equation gives: Pa ""

L

RT

""

latm

(82.057cm3 almK:""1 gmol 1)(301.15K)

"" 4.05 x 10-5 gmolcm-3

From the atomie weights in Table B.l (Appendix B), the molecular weight of oxygen is 32.0. Converting the result for Pa to mass tenns: Pa "" 4.05 x lO-5gmolem-

3

·1 i~~~ll

3 3 "" l.30XlO- gcm-

From Table A.9 (Appendix A): 1 eP "" 10- 2 g cm~l s-l; from Table A.l (Appendix A): 1 ft "" 0.3048 m "" 30.48 cm.

The parameter values and conversion factors, together with Db "" 2 mm "" 0.2 em, can now be used to calculate the dimensionless groups in the equation for the Sherwood number.

3

Solutions,' Chapter 2

0,87eP. S_J1.L_

-1 s-11 1'0-2 gem leP

-349

I) -

c - PrJ) - (0.9962652g cn,-3)(2.5 x 10 5em2 s

Therefore:

From the equation for Sh:

_Sh:lJ_(ll.21(2.5xl0-5cm2s-I)_I·A 10-3 -I 02 -,<tUX ems

kL - - - -

D1J

.

em

Answer: lAO x 10~3 em s·1

2.4

Mass and weight

From the definition of density on p 16, mass is equal to volume multiplied by density, Therefore:

From p 16, weight is-the force with wrncha body is attracted to the centre of the earth by gravity. According to Newton's law(p 15), this force is equal to the mass of the body multiplied by the gravitational acceleration. (a) From pIS, at sea level and 45° latitude. gravitational acceleration g = 32.174 it s-2. Therefore:

Weight:::: 624 Ibm (32.174 ft s-2) = 2.008x 104 Ibm it s·Z Converting these units to lbj' from Eq. (2.16), 1lbf= 32,174 Ibm it s·2; therefore:

4 2 Weight = 2.008 x 10 lbmfts- ,

llbf

2 .= 6241b

f

32.174Ibm ft s

Answer: 624lbI When g :::: 32.174 ft s·2, the number of lb mass is equal to the number of lb force, (b) From Table A.I (Appendix A): 1 m= 3.281 ft. Using the same procedure as in (a):

Weight = 624 Ibm (9.76 m s-Z .13.2~ft~ = 1.998 x 104 Ibm fts- Z Converting to Ibf

4 2 llbf Weight = 1.998 x 10 lbmfts- . -----''-~I 32.1741b m ft s-2

= 6211bf

Answer: 621lbf

2.5

Dimensionless numbers

First, evaluate the units of the groups cCP J.ll k) and (D G1p):

. (e

p ")

Umtsof - - :::: k

I (Bmlb- '1'1) Ibh- I ft- I BIUh-1 ft-2('Pft-1tl 1

2

UnilsOf(DG) = (ftllbh- ft- = 1 J1.

Ibh1ft1

= 1

Solutions: Chapter 2

4

Therefore, these groups are dimensionless. For the equation to be dimensionally homogeneous, (hICp G) must also be dimensionless; the units of h must therefore cancel the units of Cp G.

Unitsofh "'" unitsofCpG "'" (Btulb- 1 "p-l)(1bh- 1 fr 2) "'" Btu "F-l h- 1 ft- 2 The dimensions of h can be deduced from its units. From Table A.7 (Appendix A), Btu is a unit of energy with dimensions "'" L2Ml2. OF is a unit of temperature which, from Table 2.1, has the dimensional symbol E>. h is a unit of time with dimension"'" T; ft is a unit of length with dimension"'" L. Therefore: Dimensions of h = L2M'l2 a-I 1'"1 L-2 = MT" 3e- 1

Answer: Units = Btu opi h- 1 ft-2; dimensions =Ml3e-1

2.6

Dimensional homogeneity and Cc

From Table A8 (Appendix A), dimensions of P = L2MT"3 Dimensions of g = LT-z Dimensions of p =ML路3 Dimensions of Di "'" L From p 11, the dimensions oirotational speed, Nj =T-!; from p 15, the dimensions of gc= 1. Therefore:

As Np is a dimensionless number, equation (i) is not dimensionally homogeneous and therefore cannot be correct.

Equation (ii) is dimensionally homogeneous and therefore likely to be correct

Answer: (li)

2.7

Molar units

From the atomic weights in Table Rl (Appendix B), the molecular weight of NaOH is 40.0. (a) FromEq. (2.19):

lb moles NaOH

=

20.0 lb = 050 lbmol 4O.0lblbmol 1

Answer: 050 lbmol (b) From Table A.3 (Appendix A): lIb

=453.6 g.

20.01b

Therefore:

453.6gl = 9072 g = 20.0Ib. 1-lib

From Eq. (2.18): gram moles NaOH

=

9072g 4O.0g gmol-

I

= 227 grool

Answer: 227 gmol (c)

From p 16, 1 kgmol::::: 1000 gmot. Therefore, from (b):

.-----

5

Solutions: Chapter 2

kg molesNaOR

1 kgmol = 227 gmol. 11000 gmol I = 0.227 kgmol

Answer: 0.227 kgmol

2.8

Density and specific gravity

(a)

From p 16, the density of water at 4掳C can be taken as exactly 1 g cm-3. Therefore, for a substance with specific gravity L5129i~, the density at 20掳C is 1.5129 g cm*3, (I)

lkg=l000g 1 m: 100cm Therefore:

Answer: 1512.9 kg m- 3 (il)

From the atomic weights in Table 8.1 (Appendix B), the molecular weight of nitric acid (RN03) is 63.0. In 1 cm3 RNO" from Eq. (2.18)0 gram moles :

LS129g

1 : 0.0240 gmot

63.0ggmor Therefore, the molar density is 0.0240 gmol cm-3 . From the definition of specific volume on p 16: Molar specific volume :

1 1 d . : .,-_ _,,-1-:--:::;- : 41.67cm3 gmor 1 mo ar enslty 0.0240 gmol em 3

Answer: 41.67 cm3 gmol-l (b) (I)

From p 16, as density is defined as the mass per unit volume, the mass flow rate is equal to the volumetric flow rate multiplied by the density:

Answer: 80 g min- l (ii) From the atomic weights in Table B.l (Appendix B), the molecular weight of carbon tetrachloride, CC14, is 153.8.

Using the mass flow rate from (a): Molar flow rate : 80 g min-

l

.l :5~0~1 =

0.52gmolmin- 1

Answer: 0.52 grool min- I

2.9

Molecular weight

From p 17, the composition of air is close to 21 % oxygen and 79% nitrogen. For gases at low pressures, this means 21 mol% 02 and 79 mol% NZ. Therefore, in 1 gmol air, there are 0.21 gmot Oz and 0.79 gmol NZ From the atomic weights in Table B.l (Appendix B), the molecular weights of Oz and NZ are 3Z.0 and 28.0, respectively. The molecular weight of air is equal to the number of grams in 1 gmol: 1 gmolair

Answer. 28.8

= 0.21 gmOlOz路1 ;~~ll + 0.79 gmol NZ 路1 ;~;ll = 28.8g

6

2.10

Solutions.' Chapter 2

Mole fraction

The molecular weights can be obtained from Table B.7 (Appendix B): water 18.0; ethanol 46.1; methanol 32.0; glycerol 92.1; acetic acid 60.1; benzaldehyde 106.1. In 100 g solution, there are 30 g water, 25 g ethanol. 15 g methanol, 12 g glycerol, 10 g acetic acid, 8 g benzaldehyde, and no other components. Therefore:

Moles water = 30 g _I Moles ethanol:: 25

~r:;~ll = 1.67 gruol

g.1 ~~~ll :: 054 gmol

Molesmethanol:: 15 g _I

.

Molesglycerol

;~~ll = 0.47 gmol

= 12g. IIgmaII 92.1g = O.13gmol

Moles acetic acid

= 10g.1 ~~~II

Moles benzaldehyde: 8 g _I

= 0.17 gmol

:~o~ I = 0.08 gmal

The total number of moles is 1.67 + 0.54 + 0.47 + 0.13 + 0.17 + 0.08:: 3.06 gmal. From Eq. (2.20):

Mole fraction water = Mole fraction ethanol

;:~~ = 0.55

= ~:~ = 0.18

Mole fraction methanol =

~:~~

Mole fraction glycerol =

~:~ = 0.04

Mole fraction acetic acid

= 0.15

= ~:~ : : : 0.06

Mole fraction benzaldehyde :::::

~::

::::: 0.03

Answer: 0.55 water; 0.18 ethanol; 0.15 methanol; 0.04 glycerol; 0.06 acetic acid; 0.03 benzaldehyde

2.11

Temperature scales

From Eq. (2.27\ -40 ::::: 1.8 T(0C) + 32 Tee) = -40

From Eq. (2.25), T (OR) ::::: -40 + 459.67 T(째R) = 420 From Eq. (2.24) and the result for T ("C); T(K) = -40+273.15 T(K) = 233

2.12

Pressure scales

(a)

Assume that the atmospheric pressure is 14.7 psi. From Eq. (2.28): Absolute pressure ::::: 15 psi + 14.7 psi::::: 29,7 psi

7

Solutions: Chapter 2

From Table A.5 (Appendix A): 1 psi = 6.805 x Hy2 atm. Therefore: 2 . 16.805 x 10- atm 1 Absolutepressure = 29.7psl.

1 psi = 2.02atm

Answer: 29.7 psi; 2.02 atm (b) From p 19, vacuum pressure is the pressure below atmospheric. If the atmospheric pressure is 14.7 psi:

Absolutepressure = 14.7psi-3psi = IL7psi Answer: 11.7 psi

2.13

Stoichiometry and incomplete reaction

(a) The molecular weights are calculated from TableB.l (Appendix B): penicillin 334.;4; glucose 180.2. The maximum theoretical yield from the stoichiometric equation is 1 gruol of penicillin for every 1.67 gruol of glucose.

'This is equivalent to 334.4 g penicillin per 1.67 x 180.2 = 300.9 g glucose. or 1.1 g g-l.

=

=

Answer: 1.1 g g~1 (b)

The maximum theoretical yield in (a) is obtained when all the glucose consumed is directed into penicillin production according to the stoichiometric equation. If only 6% of the glucose is used in this way, the actual yield of penicillin from glucose is much lower, at 334.4 g penicillin per (300.9 x 100/6) g glucose, or 0.067 g g~l.

Answer: 0.067 g g-1 (c)

From the atomic weights in Table B.I (Appendix B), the molecular weight of phenylacetic acid is 136.2.

(I) The only possible limiting substrates are glucose and phenylacetic acid. Using a basis of II medium, if (50 - 5.5) = 44.5 g glucose are consumed but only 6% is available for penicillin synthesis, the mass of glucose used in the penicillin reaction is 44.5 x 6/ 100 2.67 g. 'This is equivalent to 2.67 g/180.2 g gmol-1 1.48 K 10- 2 gmol glucose available for penicillin synthesis. At the same time, 4 g or 4 g/136.2 g gmol-1 2.94 x 10~2 gruol phenylacetic acid is available which, according to the stoichiometric equation, requires 1.67 x 2.94 x 10- 2 4.91 x 10-2 gruol glucose for complete reaction. As the gmol glucose required is greater than the gmol glucose available after growth and maintenance activities, glucose is the limiting substrate.

=

=

=

=

Answer: Glucose (il)

Of the 44.5 g I-I glucose consumed, 24% or 10.7 g I-I is used for growth. In a H~)-litre tank. the total mass of glucose consumed for growth is therefore 1070 g or 1.07 kg.

Answer: 1.07 kg (iii)

From (i), 1.48 x 10- 2 gmol glucose is used in the penicillin reaction per litre. According to the stoichiometry, this produces 1.48 x 10-2/1.67 = 8.86 x 1O~3 gmol penicillin per litre. Therefore, in a l00-litre tank, 0.886 gmol or 0.886 gmol x 334.4 g gmol-1 296 g penicillin are formed.

=

Answer: 296 g

(iv) IT, from (i), 1.48 x 10-2 gmol [1 glucose is used in the penicillin reaction, 1.48 x 10-2/1.67 = 8.86 x 10- 3 gmoll-l phenylacetic acid must also be used. This is equivalent to 8.86 x 10- 3 gmoll-l x 136.2 g gmol-1 = 1.21 g t I phenylacetic acid. As 4 g I-I are provided, (4 - 1.21) = 2.79 g I~I phenylacetic acid must remain.

Answer: 2.79 g I-I

Solutions: Chapter 2

8

2.14

Stoichiometry, yield and the ideal gas law

(a)

Adding up the numbers of C, H, 0 and N atoms on both sides of the equation shows that the equation is balanced.

Answer: Yes (b) The molecular weights are calculated from Table B.I (Appendix B).

Cells: 91.5 Hexadecane: 226.4 From the stoichiometry, as 1 gmol of hexadecane is required to produce 1.65 gmol of cells, the maximum yield is 1.65 gmol x 91.5 g gmol-l =: 151 g cells per 226A g hexadecane, or 0.67 g g-I, Answer: 0.67 g g-1 (e)

From the atomic weights in Table RI (Appendix B), the molecular weight of oxygen is 32.0. From the stoichiometry, 16.28 gmol of oxygen is required to produce 1.65 gmal of cells which, from (b), is equal to 151 g cells.

The maximum yield is therefore 151 g cells per (16.28 groal x 32.0 g groot-I) =: 521 g oxygen, or 0.29 g g-1.

Answer: 0.29 g g-1 (d) Production of 2.5 kg cells is equivalent to 2500 g =: 2500 g/91.5 g gmoI-l = 27.3 gmol cells. The minimum amounts of substrates are required when 100% of the hexadecane is converted according to the stoichiometric equation.

(I)

From the stoichiometry, production of 27.3 gmol cells requires 27.3/1.65 3736 g = 3.74 kg hexadecane.

=16.5 gmol =16.5 gmol x 226.4 g gmol-I =

Answer: 3.74 kg (il) From the answer in (d)(i), the concentration ofhexadecane required is 3.74 kg in 3 m3, or 1.25 kg m- 3.

Answer: 1.25 kg m- 3 (ill) According to the stoichiometric equation, production of 27.3 gmol cells requires 27.3 x 16.28/1.65 269.4 gmol oxygen. As air at low pressure contains close to 21 mol% oxygen (p 11), the total moles of air required is 269.410.21 = 1282.9 gmot The volume of air required can be calculated using the ideal gas law. From Eq. (2.32):

=

V = nRT p Temperature in the ideal gas equation is absolute temperature; from Eq. (2.24):

T = (20 +273.15) K = 293.15K From Table 2.5, R "" 82.057 cm3 atm K-I gmol-I. Substituting these values into the equation for V gives:

v= Answer: 31 m 3

1-.!.."'...1

3 1 3 = 31 3 (1282.9gmolj(82.057cm atmK- gmol-l)(293.15Kj 1 atm . l00cm m

Presentation and Analysis of Data 3.1

c~

Combination of errors

""

0.25 mol m- 3 ±4%"= 0.25 ±O.OlOmol m- 3

CAL = 0.183 mol m- 3 ± 4% =: O.183±O.OO73molm-3 OTR = O.Qll mol m- 3 s-1 ± 5% For subtraction, absolute errors are added. TherefOre:

C~ -CAL"'" (O.25-0.183)±{O.OlO+0.0073)molm-3 "'" O.067±O.0173molm-3

=:

0.067 molm-3 ± 25.8%

For division, relative errors are added. Therefore: kLa

=:

o.OllmOlm-3~-1

±{25.8 + 5)% =O.16s-1 ±31%

= O.16± 0.05 s-l

O.Q67molm

Answer~

31 %. This example illustrates how a combination of small measurement errors can result in a relatively large uncertainty in the final result.

3.2

Mean and standard deviation

(a) The best estimate is the mean,

X.

FromEq. (3.1):

x = 5.15+5.45+5.50+5.35 = 5.36 Answer: 5.36 (b) Calculate the standard deviation from Eq. (3.2):

(5.15 - 5.36)' + (5.45 - 5.36)' + (5.50- 5.36)' + (5.35 - 5.36)' = 0.15 4-1 Answer. The standard deviation is 0.15. Note that standard error, which can be calculated from the standard deviation, is a more direct indication of the precision ofa mean. (c)

x = 5.15 +2 5.45 = 5.30 Standard deviation is not appropriate for expressing the accuracy of a mean evaluated using only two samples. In this case the maximum error, Le. the difference between the mean and either of the two measured values, might be used instead. The maximum error in this example is (5.30 - 5.15) 0.15.

=

Answer. 5.30; an indication of the accuracy is ± 0.15 (d)

x= 5.15+5.45+5.50+5.35+5.15+5.45+5.50+5.35 = 5.36

Solutions: Chapter 3

10

2 (S.lS -S.36f + 2(S.4S-S.36)~ + ~ (S.50-S.36f +2(S.3S -S.36)2 = 0.14

Answer: The best estimate of optimal pH is unchanged at 5.36, but the standard deviation is slightly lower at 0.14.

This example illustrates that although the standard deviation decreases as the number ofmeasurements is increased, (j is not strongly dependent on n. The best way to improve the reliability of the mean is to ensure that the individual measurements are as intrinsically accurate as possible, rather thanrepeat the measurement many times.

3.3

Linear and non-linear models

(a)

Xl = 1; Yl =10 X2=8;Y2=0.5 A straight line plot of y versus x on linear coordinates means that the data can be represented using Eq. (3.6). From

Eqs (3.7) and (3.8), A = (Yz-Y1) = 0.5-10 =-136 (x2

8-1

Xl)

.

B = YI-Axi = 10-(-1.36)1 = 11.4

Answer: y = -1.36 x + 11.4 (h) Xl = 3.2; Yl 14.5 );2 = 8.9; Y2 = 38.5 A straight line plot of y versus x Ih on linear coordinates means that the data can be represented using the equation:

=

Y=Axlh+B with A and B given by the equations:

A =

YrY1 = 38.5-14.S = 201 112_ Xl112 89Ih_32'k . x2 . ,

B = YI-Axi h = 14.5-20.1(3.2 112) = -21.5 Answer:y=20.1x'k -21.5 (0) Xl=5;Yl=6

X2= l;Y2=3 A straight line plot of Ity versus xl on linear coordinates means that the data can be represented using the equation:

lly = Ax2 +B with A and B given by the equations: A =

lin - Ity} 2

2

x2-xl

B = 1/y1

=

1/3 -lI6 2

-3

2 =-6.9xlO

1 -5

-Ax; = 1/6-(-Mx 1O-3)<S2) = 0.34

Answer: l/y = -6.9 x 10-3 xl + 0.34 (d)

Xl=0.5;Yl=25 x2 550; Y2 2600

=

=

A straight line plot of y versus x on log-log coordinates means that the data can be represented using Eq, (3.10).

From Eqs (3.13) and (3.14),

11

Solutions: Chapter 3

A

= (lnY2- ln Yl) = ln2600-ln25 = 0.663 (lnx2-lnxI)

lnB

ln550-1nO.5

= InYI-Alnxl = ln25-(0.663}ln0.5 = 3.678 B = e3.678 = 39.6

Answer: Y= 39.6 x O.663 (eJ Xl

X2

= 1.5; YI = 2.5 = 10; Y2 = 0.036

A straight line' plot of y,versus From Eqs (3.17) and (3.18):

X

on semi·log coordinates means that the data can be represented using Eq. (3.15).

A

= (lnY2- ln YI) = lnO.036-ln2.5 = -0.50 (x2 - xl)

10 - 1.5

In B = In Y1 - A Xl = In 2.5 - (-o.50J 1.5 = 1.666

B = el.666 = 5.29 Answer: Y= 5.29 e-o.sOx

3.4

Linear curve fitting

(aJ The results determined using Eqs (3.1) and (3.2) are listed below.

Sucrose concentration (g l~l)

Mean peak area

Standard deviation

6.0 12.0 18.0

56.84 112.82 170.63 232.74 302.04

1.21 2.06

24.0 30.0

2.54 1.80 2.21

(bJ 35

--.

30

.9 25

•• 0

~• g 8

20 15

~g

10

"'

5 0

0

50

100

150

200

Peak area

250

300

350

Solutions: Chapter 3

12

(e)

The linear least squares fit of the data is: y ::: 0,098 x

+ 0,83

Answer: y "'" 0.098 x + 0.83, where y is sucrose concentration in g 1-1 and x is peak area. (d)

For x::: 209,86, the equation in (c) gives a sucrose concentration oi2IA g I-I. Answer: 21.4 g I-I

3.5

Non-linear model: calcnlation of parameters

(a) The proposed model equation has the general form of Eq. (3.15); therefore, if the model is suitable, a plot of a versus lIT on semi-logarithmic coordinates will give a straight line. As T in the equation is absolute temperature, "'c must first be converted to degrees Kelvin using Eq. (2.24). The data are listed and plotted below.

Temperature (0C)

Temperature (K)

IIT(K-l)

Relative mutation frequency. a

15 20 25 30 35

288.15 293.15 298.15 303.15 308.15

3.47 x 10-3

4.4 x 10- 15 2.0x 10- 14 8.6x 10.. 14 3.5 x Jer I3

3.41 x lO w3 3.35 x leT3 3.30 x lfr·3 3.25 x 10-3

lAx }0"'12

10- 11 r---~--.,...--~----r--~---.

[" •"

fc

.Q

10- 12

10- 13

lii

"E ~

=

-lllJ:

10. 14

10-'5 L._~~_--'__~__....l._ _~_ _....J 3,2 x 10-3

3.4x 10-3

3.3x 10-3

3.5 x 10-3

1ITemperature (K-1)

As the data give a straight line on semi~logarithmiccoordinates, the model can be considered to fit the data well. (b) The equation for the straight line in (a) is:

y

=:

9.66 x 1024 e-26,12lx

where y is relative mutation frequency and x is reciprocal temperature in units of K~l. For dimensional homogeneity the exponent must be dimensionless (p 12), so that -26,121 has units of K, and EIR in the model equation is equal to 26,121 K From Table 2.5, R =: 8.3144 J gmol-l K·1; therefore: E

=:

(26,121 K) (8.3144 J gmol~l K*l)

=:

217,180.4 J gmol-1

=:

217.2 kJ gmot 1

13

Solutions: Chapter 3 Answer: 217.2 kJ gmol-l (c) From the equation in (b) for the straight line,

ao is equal to 9.66 x 10Z4.

Answer: 9.66 X 1024

3.6

Linear regression: distribution of residuals

(aJ

--

16

••

, 14 !'!l c

••i'!

12

•g

10

C 0 0

~

~

••m •~

.i1

•

8

4 2

• •• •

•• •

6

•

0 0

12345

6

Decrease in medium conductivity (mS cm-1)

The linear least squares fit of the data is:

y

=:

1.58 + 2.10x

where yis increase in biomass concentration in g r 1 and xis decrease in medium conductivity in mS cm~l. (b) The residuals are calculated as the difference between the measured values for increase in biomass concentration and the values for y obtained from the equation in (a).

Decrease in medium conductivity (mS

Residual

o

-1.58

0.12 0.31 0.41 0.82 1.03

0.57 -!l.23 0.36 1.20 1.36 1.28 0041 0.19 -0.46 -!l.50 -!l.73 -1.69 -1.99

lAO 1.91 2.11 2.42

2.44 2.74 2.91 3.53 4.39 5.21 5.24

5.55

-1.00 1.48 0.02 1.37

Solutions: Chapter 3

14

These results are plotted below.

3.-_-,._ _.,-_-,_ _.,..._ _,-_-, 2 1

·2 ·3 ' - _ - ' -_ _-'-_---'_ _......_ _' - _ - '

o

1

2

3

4

5

6

Decrease in medium conductivity (mS em- 1)

The residuals are not randomly distributed: they are mainly positive at low values of decrease in medium conductivity, then negative, then positive again. Therefore, the straight line fit of the data cannot be considered a very

good one.

3.7

Discriminating between rival models

(aJ The results are plotted using linear coordinates below.

0.11 ~

'"

0.10

os'

0.09

.§.

.~

~

.""

• •

0.08 0.07

'E

,1l.

•

0.06

~

0.05

~

::J

0.04 0.00

•

•

0.02 0.04 0.06 0.08 Gas superficial velocity, uG (m s-1)

0.10

The data are reasonably well fitted using a linear model. The linear least squares equation for the straight line fit is:

y

= 0.054 + 0.466 x

where y is liquid superficial velocity in m s-l andx is gas superficial velocity in m s·l. The sum of the squares of the residuals between the measured values for liquid superficial velocity and the values for y obtained from the above equation is 8.4 x 10-5,

15

Solutions: Chapter 3

(b) The proposed power law equation has the same form as Eq. (3.10). Therefore, if the power law model is suitable. the data should give a straight line when plotted on log-log coordinates.

0.01

om

0.1

Gas superficial velocity, uG (m s·1)

The data are reasonably well fitted using a power law- modeL The equation for the straight line in the plot is:

y = 0.199 x O.309 where y is liquid superficial velocity in m s·l andx is gas superficial velocity in m s·l. The sum of the squares of the residuals between the measured values for liquid superficial velocity and the values obtained from the above equation is 4.2 x 10-5.

«) The non·linear model is better because the sum of squares of the residuals is smaller.

3.8

Non-linear model: calculation of parameters

(a), (b) The proposed model equation has the same form as Eq; (3.15). Therefore, if the model is suitable, the data should give a straight line when plotted on semi-logarithmic coordinates.

10' . , . - - . , . . . - . . . . , - - , - - - - , - - - - , , - -. . .- - - .

103

101

100 '--_.1-_-'-_-'-_-'-_-L_----'-_---'

o

5

10

15

20

Time (min)

25

30

35

Solutions: Chapter 3

16 (c), (d)

The equation for the straight line in the figure is: y

=.

2.13 x 104 e-O.353x

where y is the number of viable cells and x is time in min. As the exponent must be dimensionless to preserve dimensional homogeneity (p 12), the dimensions of kd are T-t; therefore k(! =. 0353 min-I, The dimensions of No are the same as N, i.e, No is dimensionless and equal to 2.13 x 104.

Answer: kd =. 0.353 min-I. No =. 2.13 x 104; the dimensions of ka. are T"l, No is dimensionless

Material Balances 4.1

Cell concentration using membranes

L (i)

Assemble Flow sheet

Buffer solution in 80 kg min- 1 Cell concentrate 6% bacteria

(li)

I

I

I

I

I

Hollow-fibre membranes

Buffer solution out

I

I

I

I

I

Fennentation broth 350 kg min- 1 1% bacteria 99% water

System boundary

The system boundary is shown on the flow sheet. (iii) Reaction equation

No reaction occurs.

2.

Analyse

(i) Assumptions - steady state

-no leaks - only water passes across the membrane (ii) Extra data No extra data are required. (ill) Basis 1 min, or 350 kg fennentation broth (iv)

Compounds involved in reaction

No compounds are involved in reaction. (v) Mass-balance equation As there is no reaction, the appropriate mass-balance equation is Eq. (4.3):

mass in = mass out 3.

Calculate

(i) Calculation table The calculation table below shows all given quantities in kg. The total mass of cell concentrate is denoted c; the total mass of buffer out is denoted B. The columns for water refer to water originating in the fermentation broth.

18

Solutions: Chapter 4

Out

In Stream

Fermentation broth

Water 346.5

Buffer solution in Cell concentrate

0

Bacteria 3.5

Buffer 0

0

80

Total 350

Water

Bacteria

Buffer

Total

? ?

O.06C 0

0 80

B

?

0.06C

80

C+B

Out Buffer

Total

80

Buffer solution

C

out Total

346.5

3.5

80

430

(li) Mass-balance calculations Bacteria balance

3.5 kg bacteria in "'" 0.06 C kg bacteria out C

= 58.3 kg

Total mass balance

430 kg total mass in = (C + B) kg total mass out

Using the result for C:

B "'" 371.7 kg Water balance

346.5 kg water in = water out

Water out = 346.5 kg These calculations allow completion of the mass~balance table with all quantities in kg.

In

Stream

Fennentation

Water 346.5

broth Buffer solution in

0

Bacteria 3.5

814fer 0

0

80

Total 350

Water

Bacteria

80

Cell concentrate

54.8

Buffer solution out

291.7

3.50 0

0 80

58.3 371.7

346.5

3.50

80

430

346.5

Total (iii)

3.5

80

430

Check the results

All columns and rows of the completed table add up correctly.

4.

Finalise

(a)

After rounding to three significant figures, the total flow rate of buffer solution out of the annulus is 372 kg min-I.

Answer: 372 kg min,t (b)

The total flow rate of cell concentrate from the membrane tubes is 58.3 kg min-I. Answer. 58.3 kg min路 I

Solutions: Chapter 4

19

4.2

Membrane reactor

t.

Assemble Flow sheet

(i)

system boundary~

- -- - --

Feed 40 kg h- 1

10% glucose 90% water

I

I

I

I

I Solvent 40 kg h- 1

- 1

Membrane system

Aqueous residue 0.2% glucose 0.5% ethanol

I

I

I

I

I

Product

System boundary The system boundary is shown on the flow sheet.

(li)

(iii)

Reaction equation

2.

Analyse Assumptions - steady state -no leaks - yeast cells do not grow or dislodge from the membrane - no evaporation - all C02 produced leaves in the off~gas - no side reactions (li) Extra data Molecular weights (Table B.t, Appendix B): glucose = 180.2 ethanol =. 46.1 (i)

CO, =44.0 Basis 1 h, or 40 kg feed solution (iv) Compounds involved in reaction Glucose, ethanol and carbon dioxide are involved in the reaction. (v) Mass~balanceequations For glucose, ethanol and carbon dioxide, the appropriate mass-balance equation is Eq. (4.2): (iii)

mass in + mass generated

=.

mass out + mass consumed

For water, solvent and total mass, the appropriate mass-balance equation is Eq. (4.3): mass in 3.

=.

mass out

Calculate Calculation table The calculation table below shows all given quantities in kg. The total mass of aqueous residue is denoted R; the total mass of product out is denoted p; the total mass of carbon dioxide out is denoted G.

(i)

20

Solutions: Chapter 4

Stream

------Feed Solvent

Glucose Ethanol

4 0

0 0

In CO, Solvent H,O Total 40 0 0 36 0 0 40 40

Aqueous

Out

Glucose Ethotwl CO, Solvent H2 O Total 0_002R 0.005 R 0

0

?

R

0 0

residue Product

Off-gas Total

0

4

0

40

36

80

?

0

0 ?

?

0

0 0

P G

0.002 R ?

?

?

?

R+P

(ii) Mass-balance calculations Solvent balance

Solvent is a tie component.

40 kg solvent in ::::: solvent out

Solvent out ::::: 40 kg Water balance Water is a tie component.

36 kg water in ::::: water out Water out::::: 36 kg As water appears on the Out side of the table only in the aqueous residue stream: 0.002 R + 0.005 R + 36 kg ::::: R

R = 36.254 kg

Therefore, the residual glucose in the aqueous residue stream.::::: 0.002 R ::::: 0.073 kg; the ethanol in the aqueous residue stream::::: 0.005 R::::: 0.181 kg.

Glucose balance

4 kg glucose in + 0 kg glucose generated ::::: 0.073 kg glucose out + glucose consumed Glucose consumed ::::: 3.927 kg

Converting the glucose consumed to molar terms: 1 kgmOll ::::: O.0218kgmol 3.927 kg glucose ::::: 3.927kg. 180.2kg

1

From the reaction stoichiometry, conversion of this amount of glucose generates 2 X 0.0218 = 0.0436 kgmol ethanol and 2 x 0.0218 = 0.0436 kgmol COZ. Converting these molar quantities to mass: 0.0436kgmolethanol

= 0.0436kgmOI路1 iZ:~~11 = 2.010kg

0.0436kgmo1CO z = OJM36kgmOLI

~:~ll

= 1.92kg

C02 balance

okg COZ in + 1.92 kg C02 generated = C02 out + 0 kg C02 consumed C02 out = 1.92 kg = G Ethanol balance

o kg ethanol in + 2.010 kg ethanol generated Ethanol out

= ethanol out + 0 kg ethanol consumed

= 2.010 kg

Ethanol leaves the system only in the product and aqueous residue streams. Therefore:

21

Solutions: Chapter 4

Ethanol out in the product stream = (2.010 - 0.181) kg = 1.829 kg As the product stream consists of ethanol and solvent only:

P = (1.829 + 40) kg = 41.829 kg These calculations allow completion of the mass~balance table with all quantities in kg.

In

Stream

Feed Solvent Aqueous residue Product Off-gas Total

Out

Glucose Ethanol CO, Solvent H,O Total 40 4 0 0 0 36 40 0 40 0 0 0

Glucose Ethanol

CO2 Solvent H,O Total

-

0.073

0.181

0

0 0

1.829 0

0.073

2.010

4

0

0

40

36

80

0

36

36.254

0 1.92

40 0

0 0

41.829路 1.92

1.92

40

36

80.00

(iii) Check the results All columns and rows of the completed table add up correctly.

4.

Finalise

(0)

1.829 kg ethanol are contained in 41.829 kg of product stream. The ethanol concentration is therefore 1.829/41.829 x 100%:::: 4.4%.

Answer: 4.4% (h)

The mass flow rate ofCOz is 1.92 kg b- 1.

Answer: 1.92 kg h~l

4.3

Ethanol distillation

1. (i)

Assemble Flow sheet

System

Feed 50,000 kg h-l 10%ethanal 90%watar

boUnda'l'\. r

-t---,

I I I I

I I I I I I I I I

Distillation column

I

I I I I

-

-

-+- -

-

Distillate 5,000 kg h- 1 45% ethanol 55% water

Bottoms

SolutilJns: Chapter 4

22 (li)

System boundary

The system boundary is shown on the flow sheet. (ill) Reaction equation No reaction occurs.

2.

Analyse

(i) Assumptions - steady state

-no leaks (ll)

Extra data

No extra data are required. (ill) Basis 1 h, or 50,000 kg feed (Iv) Compounds involved in reaction

No compounds are involved in reaction.

Mass짜balance equation

(v)

As there is no reaction, the appropriate mass-balance equation is Eq. (4.3): mass in ::::: mass out

3.

Calculate

(i) Calculation table The calculation table shows all given quantities in kg.

In

Stream

Ethanol Feed

5,000

Water 45,000

Out Total 50,000

Distillate Bottoms

Total

5,000

45,000

50,000

Ethanol

Water

Total

2,250 ?

2,750 ?

5,000 ?

?

?

?

(li) Mass-balance calculations Total mass balance 50,000 kg total mass in ::::: total mass out

Total mass out ::::: 50,000 kg Therefore, from the total column on the Out side of the table: Bottoms out::::: (50,000 - 5,000) kg ::::: 45,000 kg Ethanol balance 5,000 kg ethanol in ::::: ethanol out

Ethanol out ::::: 5,000 kg From the ethanol column of the Out side of the table: Ethanol out in the bottoms

:=

(5,000..,... 2,250) kg

:=

2,750 kg

Water balance 45,000 kg water in Water out

:=

:=

water out

45,000 kg

From the water column of the Out side of the table: Water out in the bottoms == (45,000 - 2,750) kg == 42,250 kg These calculations allow completion of the mass~balance table with all quantities in kg.

Solutions: Chapter4

23

In

Stream

Out

Ethanol

Water

Total

Feed Distillate Bottoms

5,000

45,000

50,000

Total

5,000

45,000

50,000

Ethanol

Water

Total

2,250 2,750

2,750 42,250

5,000 45,000

5,000

45,000

50,000

(ill) Check the results All columns and rows of the completed table add up correctly.

4.

Finalise

(a) The bottoms contains 2,750 kg ctha,")l and 42,250 kg water in a total of 45,000 kg. Therefore, the composition is 2,750/45,000 x 100% = 6.1 % ethanol. and 42,250/45,000 x 100% = 93.9% water.

Answer: 6.1% ethanol, 93.9% water (b) Directly from the table, the rate of alcohol loss in the bottoms is 2,750 kg h- I .

Answer: 2,750 kg h- I

4.4

Removal of glucose from dried egg

1. (i)

Assemble Flow sheet Off-gas

System boundary

r\. -------, I I 1 -----L.o-l 3000 kg h路1 I Egg slurry

2% glucose 20% water 78% egg solids

I I

1 I Enzyme reactor

--t- Inlet air 18 kg h-1 02

(n) System boundary The system boundary is shown on the flow sheet (ill) Reaction equation

2.

Analyse

(i) Assumptions - steady state -no leaks - air and off-gas are dry - gases are at low pressure so vol% = mol%

- H202 remains in the liquid phase

I--.L 1 - - - _ - Product I

1 I

0.2% glucose

24

Solutions: Chapter 4

(ti) Extra data Molecular weights (Table B.I, A.ppendix B): glucose::::: 180.2

02=32.0 NZ::::: 28.0

H20

= IS.0

gluconic acid::::: 196.2 HZOz::::: 34.0

Composition of air (p 17): 21% 02, 79% Nz by volume (ill) Basis 1 h, or 3000 kg egg slurry (iv) Compounds involved in reaction Glucose, 02, water, gluconic acid and HzOz are involved in the reaction. (v) Mass-balance equations For glucose, 02, water, gluconic acid and HZOz,the appropriate mass~balance equation is Eq. (4.2): mass in + mass generated ::::: mass out + mass consumed For egg solids. Nz and total mass, the appropriate mass-balance equation is Eq, (4.3):

mass in = mass out 3.

Calculate

(i) Calculation table The mass of NZ accompanying 18 kg 02 in air can be calculated from the known composition of air. Converting 18 kg 02 to molar units:

ISkg02

kgmOll = ISkg02 路 1 132.0kg = 0.563kgmo102

Therefore, 79/21 x 0563 kgmol = 2.118 kgmol NZ enter in the air stream. Converting this to mass units:

2S.0kg I = 59.30kg N2 2,118kgmo1N2 = 2.118kgmoIN2路 lkgmol

1

The calculation tables below show all known quantities in kg. The total mass of off-gas is denoted G; the total mass of product is denoted P. The In side of the mass-balance table is complete,

Solutions: Chapter 4

25 59.30 kg NZ in ::::: NZ out

NZ out::::: 59,30 kg Glucose balance 60 kg glucose in + 0 kg glucose generated ::::: 0.002 P kg glucose out + glucose consumed Glucose consumed ::::: (60 - 0.002 P) kg Converting the glucose consumed to molar tenus:

1 kgmOll (60 -0.002 P) Glucoseconsumed ::::: (60 - 0.002 P) kg. 180.2 kg::::: 180.2 kgmol 1

Glucoseconsumed ::::: (0.333-1.11 x 10-5

p) kgmol

From the reaction stoichiometry, conversion of this amount of glucose路 requires the same number of kgmol OZ. Converting this molar quantity to mass of Oz:

5 (0.333 -1.11 x 10-

p) kgmo102

= (0.333-1.11 x 10-5 p) kgm01.1

;~~~11

= (10.656- 3.552x 10-4 p) kg 02

02 balam.:e 18 kg Oz in +路0 kg Ozgenerated :::::{)Z out + (10.656 - 3.552 x 10-4 P) kg Oz consumed 02 out

= (18 -

(10.656 - 3.552 x

10"" P禄) kg

02; out::::: (7.344 + 3.552 x 10-4 P) kg Adding this mass of Oz to the mass of NZ in the off~gas:

G ::::: 59.30 + (7.344 + 3.55Z x 10-4 P) kg G

= (66.64 + 3.552 x 104 P) kg

Total mass balance 3077.3 kg total mass in ::::: (G + P) kg total mass out Substituting the expression for G into the total mass balance: 3077.3 kg ::::: (66.64 + 3552 x 10-4 P + P) kg

301O.7kg = I.0004Pkg P

= 3009.6kg

Therefore, from the oxygen balance: G ::::: (66.64 + 3.552 x 10-4 x 3009.6) kg G = 67.71 kg and: Oz out::::: (7.344 + 3.552

x 10,4 x 3009.6) kg

Oz out::::: 8.41 kg

The mass of glucose out is 0.002 x 3009.6::::: 6.02 kg. The moles of glucose consumed is: Glucose consumed::::: (0.333 - 1.11 x 10-5 x 3009.6) kgmol ::::: 0.300 kgmol Therefore, from stoichiometry and the molecular weights: Water consumed ::::: 0.300kgm01.1

:~:~il::::: 5.40kg

Solutions: Chapter 4

26

Gluconic acid generated = 0.300 kgmOI.!

i9:~~ I = 58.86 kg

HZOz generated = 0.300 kgmOl.ll ~oll = 10.20kg

Water balance 600 kg water in + 0 kg water generated = water out + 5.40 kg water consumed Water out

=594.6 kg

Gluconic add balance kg gluconic add in + 58.86kgglllcomc acid generated ::: gluconic acid out +0 kg gluconic acid consumed

o

Gluconic acid out ::: 58.86 kg

o kg HZOZ in + 10.20 kg HZOZ generated = HzOZ out + 0 kg HZOZ consumed HZOZ out

= 10.20 kg

These calculations allow completion of the Out side of the_mass~balancetab1e.withall,quantities-mJcg.. _ Stream

Out Glucose

Egg slurry

AllOff-gas

Water

Egg solids

02

N2

Gluconic acid

H2O,

Total

-

Product

0 6.02

0 594.6

0 2340

8.41 0

59.30 0

0 58.86

0 10.20

67.71 3009.6

Total

6.02

594.6

2340

8.41

59.30

58.86

10.20

3077.3

(iii)

Check the results

All columns and rows of the completed table add up correctly to within round-off error.

4.

Finalise

(a) To determine which is the limiting substrate, the number ofmoles available of each substrate involved in the reaction must be determined. From the mass-balance table for streams in: 1 kg mOll Moles glucose = 60 kg. 180.2 kg = 0.333 kgmol

I

1kgmOIl Moleswater = 600 kg . 18.0kg = 33.3kgmol

1

1kgmOIl Moles 02 = 18 kg. 32.0 kg = 0.563 kgmol

1

As the substrates are required in the molar stoichiometric ratio of 1: 1: 1 and glucose is available in the smallest molar quantity, the extent of the reaction must be limited by glucose.

Answer: Glucose (b) Water and 02 are available in excess. As only 0.333 kgmol of each will be used if the reaction proceeds to completion, from Eq, (2.34);

% excess water =

(33.3 - 0.333) kgmol 0.333 kgmol x 100% = 9900%

- 0.333) kgmol 100m _ 69% % excess 0 2 -- (0.563 0.333 kgmol x "10 -

Answer: 9900% excess water; 69% excess 02

Solutions: Chapter 4

27

(c) From the completed mass~balance table. the reactor off-gas contains 8.41 kg Oz and 59.30 kg N2. As gas compositions are normally expressed in molar or volumetric terms (p 17), these mass values must be converted to moles:

8.41 kg 02

kgmOll = 8.4lkg02 路 132.0kg 1

= 0.263kgmo102

As the number of kgmol N2 was determined in the preliminary calculations to be 2.118, the total number of moles of off-gas is (0.263 + 2.118) 2.381 kgmoJ. Therefore, the composition of the off-gas is 0.263/2.381 0.11 02, and 2.118/2.381 = 0.89 N,.

=

Answer: 0.11

=

Oz, 0.89 Nz

(d) From the completed mass-balance table, the product stream has a total mass of 3009.6 kg and contains 6.02 kg glucose, 594.6 kg water, 2340 kg egg solids, 58.86 kg gluconic acid and 10.20 kg H202. Therefore, the composition is:

6.02 0002 1 3009,6 =. g ucose 594.6 3009.6

= 0.198 water

2340 .

3009.6 = 0.778 egg solids 58.86 =.

0020 gIUCONC . act路d 3009.6 10.20 3009.6

= 0.003 H20 ,

Answer. 0.002 glucose. 0.198 water. 0.778 egg solids. 0.020 gluconic acid, 0.003 H202

4.5

Azeotropic distillation

1.

Assemble Flow sheet

(0

,...-----------------.., "rSystem boundary

--j-

1 1 Feed

95% ethanol 5% water

1

-----r-~

(ii)

I

Distillation tower

------'-1-ol

1 1 1

1

1

1

1

1

1

System boundary

7.4"1" H20

1

I

The system boundary is shown on the flow sheet

Overhead 74.1% benzene 18.5% ethanol

1

1 Benzene feed

.,

--1---

Product 100% ethanol

Solutions: Chapter 4

28

(iii) Reaction equation No reaction occurs.

2.

Analyse

(i) Assumptions - steady state

-no leaks (li) Extra data lOOOcm3 =11 l000g=lkg (iii) Basis

2501 ethanol product Compounds involved in reaction No compounds are involved in reaction. (v) Mass~balanceequation As there is no reaction, the appropriate mass-balance equation is Eq. (4.3): (iv)

mass in "'" mass out 3.

Calculate (i) Calculation table As all quantities in mass-balance calculations must be masses (rather-than vo!umes), 250 l-absolute ethanol must first be converted to mass. From the deflnition of density on p 16, mass is equal to volume multiplied by density: 3 2501absoluteethanol ::::: 2501 x 0.785 gcm-

.ll~~m31.ll~gl

"'"

196.25kg

The calculation table shows all given quantities in kg. The total mass of feed in is denoted F; the total mass of benzene feed in is denoted B; the total mass of overhead out is denoted v.

In

Stream Ethanol

Water

Feed Benzene feed Product Overhead

0.95 F

0.05F

Total

0.95F

o

o

Out

Benzene

o B

0.05F

Ethanol

Total F B

B

F+B

Water

F

0.741 V

0.074 V

(196.25 + 0.185 V) kg ethanol out

= (206.58 + 0.195 V) kg

Benzene balance B kg benzene in

=.

0.741 Vkg benzene out

B

=.

0.741 V

Total mass balance (F + B) kg total mass in =. (196.25 + V) kg total mass out

Substituting for F and B from the ethanol and benzene balances: (206.58 + 0.195 V + 0.741 V) kg 10.33

=.

= (196.25 + V) kg

0.064 V

V=. 161.4kg

Using this result in the ethanol and benzene balances gives: F = 238.1 kg

196.25 V

196.25 +

Mass-balance calculations =.

0 0.741 V

Ethanol balance 0.95 F kg ethanol in

Total

196.25 0 0.185 V 0.074 V

0.185 V (ii)

Benzene

196.25 + V

Solutions: Chapter 4

29

B = 119.6kg These calculations allow completion of the mass-balance table with all quantities in kg.

In

Stream Ethanol 226.2

F,ed

0

Benzene feed Product Overhead

Total

Water 11.9

0

226.2

11.9

Benzene

Total 238.1 119.6

0 119.6

119.6

357.7

Out Benzene

Ethanol

Water

Total

196.25 29.9

0

0

11.9

119.6

196.25 161.4

226.2

11.9

119.6

357.7

(ill) Check the results All columns audrows of the completed table add up correctly to within round-off error.

4.

Finalise Prom the completed mass-balance table, the mass of benzene required is 119.6 kg. Using the definition of density on p 16, volume is equal to mass divided by density: 1000g 11 3 1 "'" 1371 119.6 kg-3' 1l k li .

O.872gcm g l000cm

Volumeofbenzene"'"

Answer: 137 titres

4.6

Culture of plant roots

1.

Assemble Flow sheet

(i)

Off-gas 47 1m-as 02 15 litres CO2

System boundary )..

----l Medium 1425 9 3% glucose 1.75% ammonia 95.25% water

1

-----'-1_-I 1

1

I I

I Air-driven reactor

1

I

I 1 1

1

1

1 1

L

_

J

Air 22 cm 3 milr 1 10r 10 d 25째C, 1 atm

(n) System boundary The system boundary is shown on the flow sheet.

Drained liquid 1110g 0.063% glucose 1.7% ammonia

30 (iii)

Solutions: Chapter 4 Reaction equation

From Table B.2 (Appendix B), the molecular formula for glucose is C6H1206. The reaction equation is based on the general stoichiometric equation for aerobic growth, Eq. (4.4): C6H1206+aOZ+bNH3 ~ cCHaOpNo+dCOz+eHzO

2.

Analyse

(i) Assumptions - steady state -no leaks - air and off~gas are dry - all the COz produced leaves in the off~gas - gases are at low pressure so vol% = mol% (li) Extra data 11= lOOOcm3 Molecular weights (Table B.1, Appendix B): glucose = 180.2 02 = 32.0 NZ = 28.0 NH3 = 17.0 cOz 44.0 HzO= 18.0 Composition of air (p 17): 21% OZ, 79% NZ by volume Ideal gas constant (Table 2.5): R = 82.057 cm3 atm K~l gmol~ I (iii) Basis 10 d. or 1425 g nutrient medium (iv) Compounds involved in reaction Glucose, OZ' NH3. biomass, COZ and HZO are involved in the reaction. (v) Mass-balance equations For glucose. 02, NH3, biomass, C02 and HzO, the appropriate mass-balance equation is Eq. (4.2):

=

mass in + mass generated = mass out + mass consumed For NZ and total mass, the appropriate mass-balance equation is Eq. (4.3): mass in = mass out

3.

Calculate

(i) Calculation table Over 10 d. the volume of air sparged into the fennenter is:

Volumeofairin =

22cm3min-Ix10d.16~~nl.12:dhl

= 3.168x105 cm3

Converting tbis gas volume to moles using the ideal gas law, Eq. (2.32), with the temperature converted from <>C to degrees Kelvin using Eq. (2.24): Moles of air in = n = p V =

RT

5 3 = 12.95gmol 1 atm.(3.168 x 10 cm ) 3 1 1 82.057cm atmK- gmor (25 +273.15) K

From the known composition of air, the moles ofOZ in the incoming air is 0.21 X 12.95 = 2.72 gmol, and the moles of NZ is 0.79 x 12.95 = 10.23 gmo!. Converting these values to masses: MassofOZin = 2.72gmOI.I MassofNZin

{~;ll

= 87.04g

= 1O.23gmOI.I~~;11 =

286.4g

The total mass of air in is therefore (87.04 + 286.4) g = 373.44 g.

The gas volumes in the off-gas must also be converted to masses. First, convert the volumes of 02 and COZ to moles using the ideal gas law. Eq. (2.32), with the temperature converted from 째c to degrees Kelvin using Eq. (2.24):

31

Solutions: Chapter 4

.

11000em31 1 atm(47litres).

11

V

Mo1esofO z out::: n:::

L::: RT

82.057cm3 atmK- 1 gmor 1 (25+273.15) K

( . )I

1atm 15litres. 1000 11em'

V

Moles ofCOZ out

=n =L = RT

I

82.057em'.tmK Igmoll(25+273.15)K

:::

1.92gmol

= O.613gmol

Calculate the corresponding masses: MassofOz out

= 1.92 gmOl.! ~:~ll = 61.44g

MassofCOzout:::

0.613gmOl.l~~;11 = 26.97g

The calculation tables below show all known quantities in g. The In side of the mass~balance table is complete. The total mass of off~gas out is denoted G; the total biomassbarvested is denoted R As the ratio of biomass fresh weight to dry weight is 14:1, dry biomass comprises l!IS ::: 0.0667 of the total biomass. Because this problem requires an integral mass balance, the biomass remaining in the fermenter after 10 d culture must also be included in the table even though it is not contained in any of the streams flowing into or out of the vesseL

Using this result and adding up the row for the off*gas in the out table: G

= 61.44 + 286.4 + 26.97 = 374.81 g

Total mass balance

1798.44gtotalmassin::: (1110+G+B)gtotalmassout Using the result for G:

32

Solutions: Chapter 4

B

=:

313.63 g

Therefore, the dry biomass produced is 0.0667 x 313.63 313.63 = 292.71 g.

=20.92 g; the mass of water in the biomass is 0.9333 X

These calculations allow completion of the Out side of the mass路balance table with all quantities in g.

Further mass-balance calculations allow evaluation of the masses of components consumed 'or generated in the reaction.

Glucose balance 42.75 g glucose in + 0 g glucose generated = 0.699 g glucose out + glucose consumed

Glucose consumed = 42.05 g 02 balance 87.04 g 02 in + 0 g 02 generated = 61.44 g 02 out + 02 consumed

02; consumed NH3 balance 24.94 g NH3 in + 0 g NH3 generated

= 25.60 g

= 18,87 g NH3 out + NH3 consumed

NH3 consumed = 6.07 g C02 balance

og C02 in + C02 generated

= 26.97 g C02 out + 0 g C02 consumed

C02 generated

= 26.97 g

H20 balance 1357.31 g H20 in + H20 generated = (1090.43 + 0.9333 B) g H20 out + 0 g H20 consumed

Substituting the value for B from the total mass balance: H20 generated = 25.83 g (iii) Check the results All columns and row~ of the completed mass-balance table add up correctly.

4.

Finalise

(a)

Rounding to three significant figures from the completed mass~balance table, the mass of dry roots produced is 20.9 g.

Answer: 20.9 g (b) To determine the stoichiometry, the calculated masses of components consumed or generated in the reaction must be converted to molar quantities:

33

Solutions: Chapter 4

Moles of glucose consumed Moles of

= 42.05 g.1 :8:0~1 = 0.233 gmol

oz consumed = 25.60 g .1 ;~~11 = 0.800 gmol

MolesofNH3consumed = 6.07 g MolesofCOzgenerated=

.1 ~~~ll

= 0.357gmol

26.97g.I~~I! =

0.6l3gmol

MolesofHzO generated = 25.83 g .Ill~~ll = 1.435 gmol Moles ofbiomass generated =

~0:2 g

10mass

The moles of biomass generated is not yet known explicitly because the molecular formula for the dry biomass is unknown. The above molar quantities can be used as coefficients in the reaction equation:

20.92 0.233 Coli1206 + 0.800 0, + 0.357 NH3 -+ MW b'

lomass

CH"OoN + 0.613 CO, + 1.435 H,O P'" S

Dividing each coefficient by 0.233 to obtain the stoichiometry per gmol glucose:

89.79 CoH1206+3.430,+1.53NH3 -+ MWb'lOmass CH"OoNS+2.63CO,+6.16H,O fJ The values of a, /3 and 8 and the molecular formula for the biomass can obtained using elemental balances.

Cbalance:6

89.79 = MWb' +2.63 lomass

Therefore:

89:79

MWblomass

= 3.37

This result can be used in the remaining elemental balances for completion of the stoichiometric equation.

Hbalance: 12 + 3 x 1.53 = 3.37 a+ 2 x 6.16 --) a = 1.27 Obalance:6+2x3.43 3.37/3+2x2.63+6.l6 --) /3 = 0.43 3.37 8 --) 8 0.45 N balance: 1.53

=

=

=

A11SWer: 1he complete stoichiometric equation is:

The chemical formula for the dry roots is CHI .Z700.43N0,4S. (cj

Converting to moles the mass quantities of glucose, 0z and NH3 available for reaction on the In side of the massbalance table: Moles of glucose in

= 42.75g.1:8~o~1 = O.24gmol

Moles of 0z in = 87路04g.1 MoiesofNH3in =

;~~11 = 2.72gmol

24.93g.111'~~-~11

= 1.47gmol

From the stoichiometric equation, reaction of 0.24 gmol glucose requires 0.24 x 3.43 = 0.82 gmol 0z and 0.24 x 1.53 = 0.37 gmol NH3' As the molar quantities of Oz and NH3 available for reaction are in excess of these values, glucose must be the limiting substrate.

Answer: Glucose

34

Solutions: Chapter 4

(d)

The mass of glucose consumed is 42.05 g; the mass of dry biomass produced is 20.92 g. Therefore, the biomass yield from glucose is 20.92/42.05 = 0.50 g g-1 dry weight.

Answer: 0.50 g g-1 dry weight

4.7

Oxygen requirement for growth on glycerol

From TableB.2 (Appendix B), the molecular formula for glycerol is C3Hg03. From Table 4.3, the chemical formula for Klebsiella aerogenes can be taken as CH 1. 7S00.43NO.22. Substituting these formulae into the general stoichiometric equation for growth. Eq. (4.4), gives:

C 3Hg03 +a02 +b NH3 --t cCHI.7.s00:4JNO.22+dCOZ +eHZO From Table B:8 (Appendix B), the molecular weight of glycerol is"92.1. The biomass formula weight calculated from the atomic weights in Table B.l (Appendix B) is 23.74. Taking into account the 8% ash:

Biomassmolecularweight =

zg;:

= 25.8

The value of the stoichiometric coefficient c can be detennined from:the yield

Yxs =0.4 g g-1 and Eq. (4.12):

l

c = Yxs(MWsubstrate) = 0.40 g g-l (92.lggmOr ) = 1.43gmolgmol-1 MW cells 25.8 g grool 1 From Table B.2 (Appendix B), the degree of reduction of glycerol relative to NH3 is reduction of the biomass relative to NH3 is:

rs = 4.67.

The degree of

1B = 1x4+1.75xl-0.43x2-0.22x3 = 4.23 which is also listed in Table 4.3. The theoretical oxygen demand can be determined from Eq. (4.16) with/= 0; from Eq. (4.13). w = 3 for glycerol:

a = 1/4 (w ~ - c 1B) = 1/4 (3 x 4.67 - 1.43 x 4.23) = 1.99 Therefore. 1.99 gmo! oxygen are required per grool glycerol. From the atomic weights in Table B.l (Appendix B), the molecular weight of oxygen is 32.0. Converting a to mass terms:

a = 1.99gmolgmol-1 =

L99gmOlgmOI-I.I::~II.I~~~11 =

0.69gg-

1

Answer: 0.69 g oxygen is required per g glycerol conswned

4.8

Product yield In anaerobic digestion

From Eq. (4.13), the stoichiometric equation for anaerobic growth and product fonnation by methane bacteria can be written as:

0I3COOH+ b NH3

--j.

c CH1.400.4ONo.20+ deo2 + e H20 + /0I4

From Table B.8 (Appendix B), the molecular weight of acetic acid is 60.1. From Table B.l (Appendix B), the molecular weight of CO2 is 44.0. The value of the stoichiometric coefficient d can be determined based on Eq. (4.14) with carbon dioxide as the product and the yield YpS = 0.67 kg kg- 1 = 0.67 g g-l:

d =

Yps(MWsubstral~ = MWC02

I 1 0.67gg- (60.1ggmor j = 0.915 gmol gmol-l 44.0ggmol I

The other coefficients can be determined using this result and elemental balances.

Cbalance:2 = c+d+/= c+0.915+f--j.f= L085-c Hbalance:4+3b = 1.4c+2e+4f balance: 2 = 0.40c+2d+e = OAOc+2xO.915+e = OAOc+L83+e --j. e = O.17-0.40c N balance: b = 0.20 c

째

35

Solutions: Chapter 4 Substituting the expressions fort, e and b from the C, 0 and N balances, respectively, into the H balance:

4+3x0.20c

= l.4c+2x(0.17-0.40c)+4x(l.085-c) 4 c = 0.680 C :

0.170

Substituting this value for c into the expressions for the other coefficients gives b = 0,034, e = 0.102 and/: 0.915.

The yield of methane is therefore 0.915 gmol per gmol acetic acid

=

The maximum possible methane yield can be calculated using Eq. (4.20). From Eq. (4.13), w 2 for acetic acid and) 1 for methane. From TableB.2 (Appendix B); the degree of reduction of acetic acid relative to NH3 is it 4.00, and the degree of reduction of methane relative to NH3 is lP = 8.00. Substituting these values into Eq. (4.20) gives:

=

.

WYs

fmax = )IP =

=

:i ÂŤ1.00)

-1 1(8.00) = 1.0gmolgmol

The actual methane yield of 0.915 gmol gmol-l therefore represents 91:S%of the theoretical maximum.

Answer: 91.5% of the theoretical maximum

4.9

Stoichiometry of single-cell protein synthesis

(aJ From Table B.2 (Appendix B), the .molecular formula for glucose isC6H1206. If all carbon in the substrate is converted into biomass, production of carbon dioxide is zero. Therefore, from Eq. (4.4), the stoichiometric equation for anaerobic growth of Cellulomonas is: C~1206 + b NH 3 -+ c CH1.5iPO.~O.16 + e H20

The stoichiometric coefficients can be determined using elemental balances.

C balance: 6 = c Hbalance:12+3b: 1.56c+2e o balance: 6 0.54 c + e N balance: b : 0.16 c Substituting the value for c from the C balance ,into the and N balances gives e = 2.76 and b = 0.96. respectively.

The yield of biomass from substrate in molar terms is therefore 6 gmol gmol-l.

=

°

Using the atomic weights in Table B.l (Appendix B), the molecular weight of glucose is 180.2. The biomass formula weight calculated from Table B.l (Appendix B) is 24.46. Taking into account the 5% ash: Biomass molecularweight =

2;9~6

= 25.75

Therefore, in mass terms, the molar biomass yield of 6 gmol gmol-l= (6 x 25.75) g biomass per 180.2 g substrate: 0.86 g g-l.

The maximum possible biomass yield is calculated using Eq. (4.19). From Eq. (4.13), W = 6 for glucose. From Table B.2 (Appendix B). the degree ofreduclion of glucose relative to NH3 isrs = 4.00. The degree of reduction of the biomass relative to NH3 is:

JB =

I x4+ 1.56 x 1-0.54x2-0.16x3

= 4.

00

Substituting these values into Eq. (4.19): croax =

wYs

1i3 =

6(4.00) -1 4.00 = 6.0 gmol gmol

The theoretical maximum biomass yield is therefore the same as the actual biomass yield.

Solutions: Chapter 4

36

Answer: The biomass yield from substrate of 0.86 g g-t is 100% of the theoretical maximum. When there is no product formo.tion and no oxygen for electron transfer, all the available electrons from the substrate must go to the

biomass. (b) (I)

From Table B.2 (Appendix B), the molecular formula for methanol is CH40 and the degree of reduction relative to NH3 is 1S =6.00. The degree ofreduction of Methylophilus methylotrophus biomass relative to NH3 is: _ lx4+1.68xl-0.36x2-0.22x3 -430

JBFrom Eq. (4.13), W

""

1

-.

1 for methanol. Substituting'values into Eq. (4.19);

emu""

W JS Jh" ""

1 (6.00) 4.30

""

140 .

1 1-1 gIno gmo

From Table B.g (Appendix B). the molecular weight of methanol is 32.0. The biomass formula weight calculated from the atomic weights in Table B.I (Appendix B) is 22.55. With 6% ash:

Biomassmolecularweight =

~:: = 23.99

In mass terms, the maximum possible molar biomass yield of lAO gmol gmol路l is equal to (1.40 x 23.99) g biomass per 32,0 g substrate = 1.05 g g-l,

Answer: 1he maximum possible biomass yield from methanol is 1.05 g g-}, In terms of Catoms,.the biomass yield is 1.40 grool grool-l as both biomass and substrate have 1 C atom each. In comparison, the C*atQm biomass yield from glucose in (a) is 1 grool grool路}. The main reason for the increased yield in (b) is the high degree of reduction of methanol compared with glucose.

(il) The actual yield of biomass from methanol is c = OA2 x lAO gmol gmol路} = 0.59. The ox.ygen demand can be determined from Eq. (4.16) if biomass remains the only major product so thatf= O. Using the parameter values determined in (b) (i):

Therefore, 0.87 gruol ox.ygen is required r:.r gmol methanol. As the molecular weights of methanol and oxygen are the same, the oxygen demand is 0.87 g g. methanol.

Answer. 0.87 g oxygen is required per g methanol consumed

4.10

Ethanol production by yeast and bacteria

(a) From Table B.2 (Appendix B), the molecular formula for glucose is C6H1206 and the molecular formula for ethanol is C2H60. From Eq> (4.13), the stoichiometric equation for anaerobic growth and product formation is:

CJI}P6+bNH3 -+ cCHl.sO05NO.2+dC02+eH20+fC2H60 From Table B.8 (Appendix B), the molecular weight of ethanol is 46.1. Using the atomic weights in Table B.l (Appendix B), the molecular weight of glucose is 180.2 and the biomass molecular weight is 24.6. The values of the stoichiometric coefficients c can be determined from Eq. (4.12) and the yields Yxs -= O.ll g g.1 for yeast and Yxs 0.05 g g.1 for bacteria.

=

t. - Yxs(MWsubstrare) _ O.ll gg-' (180.2 g gmOr') - 081 1 1-' oryeas c MW II 1 - . gmo gmo ce s 24.6 g gmol

F

.

Yxs(MWsubstrate) For bactena, c = MW II ce s

1

(I80.2ggmor ) = 0.05 g g-' , = 037 . gmo1gmo1-' 24.6 g gmol-

The other coefficients can be determined using elemental balances.

37

Solutions: Chapter 4 Yeast Cbalance:6 = c+d+2j= 0.81+d+2j~ d = 5.19-2j Hbalance: l2+3b = 1.8c+2e+6j= 1.8xO.81+2e+6j~ 1O.54+3b = 2e+6j o balance: 6 = 0.sc+2d+e+f= 0.5xO.81+2d+e+f~5.595 = 2d+e+j Nbalance:b = 0.2e = O.2xO.81 = 0.16 Substituting the expression for b from the N balance into the H balances gives:

10.54+3xO.16 = 2e+6f

e=5.51-3f Substituting this and the expression for d" from the C balance into the 0 balance gives: ....â&#x20AC;˘.. \

5.595 =2x(5.19-2j)+(5.51-3j)+f 6f

= 10.295

f= 1.72 Therefore, for yeast, the yield of ethanol from glucose is 1.72 grool grool,l.

Bacteria C balance: 6 = c+d+2f= 0.37+d+2f~ d = 5.63-2f Hbalance:12+3b = 1.8c+2e+6f= 1.8x037+2e+6f~ 11.33+3b = 2e+6f o balance: 6 = 0.5c+2d+e+f= O.sxO.37+2d+e+f~5.815 = 2d+e+!

N balance: b = 0.2 c = 0.2 x 0.37 = 0.074 Using the same solution procedure as for yeast, substituting the expression for b from the N balance into the H balances gives:

11.33 +3xO.074

= 2e+6j

e=5.78-3f Substituting this and the expression for d from the C balance into the 0 balance gives:

5.815 = 2x(5.63-2j)+(5.78-3j)+f 6f= 11.225

f

= 1.87

Therefore, for bacteria. the yield of ethanol from glucose is 1.87 grool grool'}.

Answer~ 1.72 grool grool"} for yeast; 1.87 grool grool,l for bacteria

(b)

=

=

The maximum possible ethanol yield can be calculated using Eq. (4.20). From Eq. (4.13), w 6 for glucose andj 2 for ethanol. From Table B.2(Appendix B), the degree of reduction of glucose relative toNH3 is rs 4.00, and the degree of reduction of ethanol relative to NH3 is 6.00. Using these values in Eq. (4.20) gives:

Âť=

f max =

W JS

6 (4.00)

1P = 2(6.00) = 2.0gmolgmor

=

1

There:ore, the actu~ ethanol yield of 1.72 ~Ol gmol'} for yeast represe.nts 86%. of the theoretical maximum; for bactena, the actual yield of 1.87 grool grool' represents 94% of the theoretical maxunum.

Answer: 86% of the theoretical maximum for yeast; 94% of the theoretical maximum for bacteria

4.11

Detecting unknown prodncts

From Table B.2 (Appendix B), the molecular formula for glucose is C6H 1206. Assuming that no products other than biomass are formed, from Eq. (4.4), the stoichiometric equation for growth is~

Solutions: Chapter 4

38

C~1206+a02+bNH3 ~ cCHl.7~0.56NO.l7+dC02+eH:f)

Using the atomic weights in Table B.l (Appendix B), the molecular weight of glucose is 180.2. the molecular weight of oxygen is 32.0, and the biomass molecular weight is 25.16. The value of the stoichiometric coefficient c can be determined from the yield Yxs

c

=

=0.37 g g-l andEq. (4.12):

Yxs(MWsubstrat~ MW

II ce s

=

0.37gg- 1(180.2ggmor 1 25.16ggmor

j

-1

' = 2.65gmolgmol

Therefore, 2.65 groat cells are produced per gmol glucose consumed. Converting the oxygen demand to a molar basis:

O.88g02125.16gcellslllgmOlOZI 0.88 g 02 per g cells =~. 1 g ce!J.s . 19mol cells . 32.0 g 02

= 0.69 gmol 02 groot-1 cells

Combining this with the result for c, the observed oxygen demand a is: a=

0.69 gmol O2 (2.65 gmat cens) 1 gIno! cells

1 gmol glucose

=183 .

From Table B.2 (Appendix B), the degree of reduction of glucose relative to NH31srS'='4;00. The degree of reduction of the biomass relative to NH3 is: _ lx4+1.79xl-0.56x2-0.17x3 -416

- .

~-

If no products are formed other than biomass, the theoretical oxygen demand can be determined from Eq. (4.16) with!

0; from Eq. (4.13), w = 6 for glucose:

=

a = 1/4(w1!l-clll) = 1/4 (6x 4.00-2.65 x 4.16) = 3.24

As the theoretical oxygen demand is significantly higher than that observed, formation of other products acting as electron acceptors is likely to have occurred in the culture.

Answer: Yes

4.12

Medium formulation

Using the atomic weights'in Table B.l (Appendix B), the molecular weight of (N14)zS04 is 132,1 and the biomass molecular weight is 26.16. Using a basis of 1 litre, production of 25 g cells corresponds to 25/26,16 = 0.956 grool cells. As each gmolcells contains O.25gmol N, (0.956x0.25) = 0.239gmol Nare-neededfrom the'medium for biomass synthesis, As (NH4)zS04 is the sole N source and each gmol (Nl4)zS04contains 2gmol N, 0.239/2= 0.120 gmol (N14)zS04 is required. Multiplying this by the molecular weight, 0.120 x 132.-1 = 15.9 g (N14)zS04 are required. The minimum concentration of (N}4)zS04 is therefore 15.9 g 1*1.

Answer: 15.9 g I-I

4.13

Oxygen demand for production of recombinant protein

(a) Recombinant protein can be considered as a product of cell cuItureeven though it is not excreted from the cells; assume that recombinant protein is synthesised in addition to the normal E.- coli biomass composition. From Table 4.3, the chemical formula for E. coli can be taken as CH1.7700.49N0.24; from Table B.2 (Appendix B), the molecular formula for glucose is C6H1206' Substituting these formulae into the general stoichiometric equation for growth and product formation, Eq. (4.13), gives: C~1206 +a02 +bNH3

-lo

cCHL7,o0.4~O.24+dC02 +e H20+fCHl.s.sO0.3INO,2S

Using the atomic weights in Table B.l (Appendix B), the molecular weight of glucose is 180.2, the biomass molecular weight is 25.00, and the recombinant protein molecular weight is 22.03.

Solutions: Chapter 4

39

Assume that the biomass yield refers to the cell mass without recombinant protein. The value of the stoichiometric coefficient c can be determined from the yield Yxs = 0.48 g g-l and Eq, (4,12): c=

Yxs(MWsubs"ate) MWcells

=

0.48gg-1 (180.2g gmol-l) 346 I I-I =. gmogmo 25,OOggmoll

The value of the stoichiometric coefficientfcan be determined from the yield Yps

=0.20 x 0.48 =0.096 g g-l and Eq,

(4.14)'

f

1 _ 079 I I-I _- Yps(MWsubstrate) __ 0.096gg- (180.2ggmol-l) -,gmogmo MW product 22.03 g gmol 1

The ammonia requirement can be, detemPneq. using an elemental balance for N. N balance: b = 0.24 c + 0,25 f .'

Substituting the above valoesforc and/into the N balance gives b = 0,24 x 3.46 + 0.25 x 0,79 = 1.03,

Answer: 1.03 gmol grool-l glucose (h)

The oxygen demand can be determined using an .electron balance. From Table B,2 (Appendix B), the degree of reduction of glucose relative to NH3 is -ns 4.00; from Table 4.3," the degree of reduction 'Of E. coli relative to NH3 is JB 4.07. The degree of reduction of the recombinant protein relative to NH3 is:

=

=

_ lx4+1.55xl-0.31x2-0.25x3 _ 418

- .

~-

=

From Eq. (4.13), w 6 for glucose andj theoretical oxygen demand gives:

=1 for recombinant protein.

Substituting these values into Eq. (4.16) for the

a = 1/4(W~-C1tc-fjJP) = 1/4 (6 x 4.00-3.46x 4.07- 0.79 x 1 x4.18) = 1.65 Answer: 1.65 gmol per gmol glucose (e)

Iff in the stoichiometric equation is zero but c remains equal to 3.46, the ammonia requirement can be determined using an elemental balance for N as follows: N balance: b = 0.24 c = 0.24 x 3.46 = 0.83 Therefore, in wild-type E. coli the ammonia requirement is reduced from 1.03 to 0.83 grool gmol-1 glucose, a decrease of 19%, Eq. (4.16) for the oxygen requirement becomes:

Therefore, the oxygen demand is increased from 1.65 to 2.48 gmol gmol-l glucose, a rise of 50%.

Answer. The ammonia and oxygen requirements for wild-type E. coli are 0.83 grool and 2.48 gmol per gmol glucose, respectively. These values represent a 19% reduction and a 50% increase, respectively, compared with the genetically engineered strain.

4.14

Effect of growth on oxygen demand

The stoichiometric equation for acetic acid production using cell culture must include terms for growth. Based on Eq. (4.13), the stoichiometric equation for growth and product formation is:

C2H4>+a02+bNH3 ~ cCHLSOo.sNo.2+dC02+eH20+jC2f402 From Table 8.8 (Appendix B), the molecular weights of ethanol and acetic acid are 46.1 and 60.1, respectively. From the atomic weights in Table B.1 (Appendix B), the biomass molecular weight is 24.63. The value of the stoichiometric coefficient c can be determined from the yield Yxs 0.14 g g~l and Eq. (4.12):

=

c =

_Yx",s"(MW"","S::U,,bs::"_at-,-e) MW cells

=

0.14 g g-1 (46.1 g gmol-1) -1 "'" 0.26groolgmol 24.63 g gmol 1

40

Solutions: Chapter 4

The value of the stoichiometric coefficient!can be determined from the yield Yps = 0,92 g g-1 and Eq. (4.14):

f

=

Yps(MWsubstrate) MW odu pr ct

=.

O.92gg- t (46.1ggmol-l) 1 6O.1ggmor

=.

0.71 gmol gmol

-I

From Table B.2 (Appendix B), the degree of reduction of ethanol relative to NH3 is rs =. 6.00, and the degree of reduction of acetic acid relative to NH3 is 'no =. 4.00. The degree of reduction of the biomass relative to NH3 is :

_ lx4+1.8xl-0.5x2-0.2x3 _ 420

1ll-

From Eq. (4.13), W = 2 for ethanol and j theoretical oxygen demand gives: a=. 1/4(w/S-C1B-fj1P)

1

=.

-.

2 for acetic acid. Substituting these values into Eq. (4.16) for the

= 1/4(2x6.00-0.26x4.20-0.71x2x4.00)

=.

1.31

Therefore, with growth, 1.31 groat oxygen are required per gmol glucose consumed, compared with 1 gmoloxygen per gmol glucose without groWth. Therefore, with growth, the oxygen demand for acetic acid production is increased by31%.

Answer. The oxygen demand is increased by 31 %.

Energy Balances 5.1

Sensible energy cbange

(a) From Table B.5 (Appendix B), Cp for m~cresol between 25"'C andlOO"C is 0.551 calg- I oC-t.Thespecific enthalpy change calculated using Eq. (SIB) is:

M

= Cp aT = 0.551 cal g-IoC l (100 - 25)OC !ill : : : 41.3 cal g-t

Answer. 41.3 cal g-1 (b) From Table B.5 (Appendix B). Cp for ethylene glycol betweenlO"C and 20"C can be taken as 0.569 cal g-1 "C- 1.

The specific enthalpy change calculated using Eq. (5,13) is:

M = Cp aT = 0.569 cal g-l °C·l (10- 20)OC

!ih :::: -5.69 cal g-1

Answer: -5.69 cal g~l (c) Prom Table B.6(Appendix B), Cp for succinic acid between 15"C-and 120°C is given by the expression 0.248 + 0.00153 T. where Tis in "c and Cp is in cal got °C~l. The sensible energy change is best determined from the integral of this equation between the limits T:::: 15"C and T:::: 1200C:

6h

'''''' CpdT = J,,,,,"c (0.248+ 0.(01531) dT calg-l = J.lS"C 1S"C 6h =

(0248T+0~153r2)[ calg-l M

= 36.9 cal g.l

(d) From Table B3 (Appendix B), the heat capacity of air between 65"C and 150"C is given by the equation:

Cp :::::: 28.94 + 0.4147 x W· 2 T + 0.3191 x lOw 5 y'2 _ 1.965 x lOw9 T3 where Cp is heat capacity in J gmol-l °C l and Tis temperature in °C. The sensible energy change can be determined by evaluating the integral of this expression between the limits T 150°C and T 65°C:

=

M

=

= J,'~ Cp dT = J,~~ (28.94 + 0.4147x 10-2 T+ 0.3191 x 10-5 f2 - 1.965 x 10-9 rJ) dT J gmor1 15O"C

l5O"C

Souaions: Chapter 5

42

t:.h = -2500.9 J gmol~l "" -2.50 kJ gmat- l

Answer: -2.50kJ gmal-]

5.2

Heat of vaporisation

-The latent heat of vaporisation of water at BOC is obtained from Table C.l (Appendix C). Taking the average of the values at nee and 34°C~ Ah v = 2423.55 kJ kg~l at 33°C From Eq. (5.16); Mf = M!>h v = 20 gb- I (2423.55 kJ kg-Ij .11~gl Mf = 48.5 kJ h- I

Answer: 48.5 kJ h- l

5.3

Steam tables

(aJ

The heat of vaporisation of water at 85"C is obtained from Table C.I (Appendix C). Taking the average of the values at 84"C and 86"C, !:J.hy = 2296.05 kJ kg-! at 85°C. Answer: 2296.05 kJ kg- l (bJ From Table C.l (Appendix C), the enthalpy of liquid water at IOce relative to the triple point is 42.0 kJ kg- t , The enthalpy of liquid water at 3S"C relative to the triple point can be estimated as the average of the values in Table Col for 34<>C and 36°C = 146.55 k:J kg-t. Using the relationship on p 89, the enthalpy of water at 35°C relative to 1000e is therefore (146.55 - 42.0) kJ kg-I"" 104.55 kJ kg-I.

Answer: 104.55 kJ kg-I (cJ -The enthalpy of saturated water vapour at 40°C relative to the triple point can be read directly from Table C.I (Appendix C) as 2574.4 kJ kg-I.

Answer: 2574.4 kJ kg- l (dJ The enthalpy of superheated steam at 2.5 atm and 275°C relative to the triple point can be obtained from Table C.3 (Appendix C). To convert the pressure to kPa, from Table A.S (Appendix A), 1 atm "" 1.013 x lOS Pa. Therefore:

2.5 atm

= 2.5 atm .11.Ql;:~5pal.lli:;al=253.3kPa

From Table C.3, the enthalpy at 100 kPa and 275°C is 3024 kJ kg-I; the enthalpy at 500 kPa and 275"'C is 3013 kJ kif 1. Interpolating between these values gives an enthalpy of 3019.8 kJ kg- l at 253.3 kPa.

Answer: 3019.8 kJkg- 1

5.4

Pre-heating nutrient medium

1.

Assemble

(i) Units kg, h, kJ, °C (ii) FkJw sheet

43

Solutions: Chapter 5

System boundar; \ .

Q (loss)

r Medium in

..l.-_~

Heating vessel

=0.22 kW

1 I I 1--'----_ Medium out 3250 kg h-1

3250 kg h-1

15°C

44°C

-Q (from condensing steam)

(ill) System boundary The system boundary is shown on the flow sbeet.

2.

Analyse (i) Assumptions - steady state -no leaks - system is homogenous - condensate temperature is 150°C - no shaft work (li)

Basis

1 h, or 3250 kg medium in (ill) Reference state H = 0 for water (steam) at its triple point H 0 for medium at 15°C Extra data (iv) C medium 0.9 cal g-l 0(;-1 0.9 kca1 kg-I °C- 1 1 v waterat 150°C = 2113.1 kJ kg- (fable C.2, Appendix C) 1 kcal = 4.187 x 103 J (fable A.7, Appendix A) 1 W = 1 J s-1 (fable A.S, Appendix A); therefore, 1 kW =·1 kJ s~l

=

6h

(v)

=

=

Mass balance

The mass balance is already complete. (vi) Energy-balance equation At steady state, Eq. (5.9) applies: :E(Mh) - :E(Mh)- Q+ W, = 0 input

output

"""""

"""""

3.

Calculate ldmtify ta"" in the 'nagy-balan" ,quatian Ws =O. There are two components for the beat term, Q: Qross and Q from the condensing steam. With symbol MD = medium. the energy-balance equation becomes: (i)

(M h)MD in - (M h)MD out - Q- Qloss

=0

(M h)MD in = 0 (reference state)

(M h)MD out at 44°C is calculated as a sensible energy change from H = 0 at 15°C using Eq. (5.13): (M h)MDuut

Converting to kJ:

= M Cp aT = 3250 kg (0.9 keal kg-l 0c- l )(44 -15loe = 8.483 x 104 keal

Solutions: Chapter 5

44

(M h)MD out

=

4 8.483 x 10 kcaL

14.187X103J111kJ 1 kcal '1000 1 I"" 3.55 x 105 kJ

The rate of heat loss is 0.22 kW. Converting to k:J hoi: O.22kW

= O.22kW. [lkJ,-1113600'1 1kW · l h =792kJh-1

Therefore, on the basis of 1 h, {hess "" 792 kJ. Substituting values into the energy~balance equation gives:

0-3.55 x 105 kJ-Q-792kJ "" 0

Q = -3.56 x 105

kJ

Q has a negative value which is consistent with the sign conventions outlined on pp 87-88: heat must be supplied to

the system from the surroundings. This heat is provided as the latent heat of vaporisation as saturated steam at 150<>C condenses. The enthalpy change from this change ofphase is calculated using Eq. (5.16) and must be equal to -Q. 3.56xloSkJ "" Msteam!JJl v

= Msteam(21l3,lkJkg- 1)

Mstearn "" 168 kg

4.

Finalise

Answer: 168 kg

5.5

Production of glutamic acid

(a) L Assemble (i) Units kg, b, kJ; °C (ii) Flow sheet Off-gas

System boundary

,~ - -- -

-

--

I I Feed 2000 kg h- 1 4% glucose 96% water

25°C

I

I I

/

r--t-

Q (cooling)

Inlet gas 100,000 litres min- 1 1 atm, 15"C 88% air 12%NHs

(ill)

System boundary

The system boundary is shown on the flow sheet.

I

Reactor (25,000 lltres)

-

1 I I

--

I I I

.

Product 0.5% glucose

45

Solutions: Chapter 5

(iv)

Reaction equation

2.

Analyse (i) Assumptions - steady state -no leaks - system is homogenous - solutions are ideal - inlet air and off-gas are dry - all excess NH3 is dissolved in the aqueous phase - all COz produced leaves in the off-gas - negligible sensible heat change - no evaporation - no shaft work (ii) Basis I h, or 2000 kg feed (ill) Reference state H = 0 for water at its triple point H = 0 for feed at 25째C (iv) Extra data Molecular'weights (Table B.t, Appendix B): glucose = 180.2 NH3 = 17.0 02=32.0 Nz = 28.0 glutamic acid = 147.1

C02 =44.0 H20 = 18.0 air = 28.8 (see Problem 2.9, Chapter 2) Ideal gas constant (fable 2.5): R = 0.0820571 atm K-l gmol-l Composition of air (p 17): 21% Oz, 79% Nz by volume Heats of combustion (Table B.8, Appendix B): Ah~ glucose = -2805.0kJ gmol-l Ah~NH3 =-382.6kJ gmol- 1

Ah~ glutamic acid =-2244.1 kJ gmor 1 (v) Compounds involved in reaction Glucose, ,ammonia, oxygen, glutamic acid, carbon dioxide and water are involved in the reaction. (vi) Mass-balance equations For,glucose, ammonia, oxygen; glutamic acid. carbon dioxide and water, the appropriate mass~balanceequation,is Eq. (4.2),

mass in + mass generated = mass out +.mass consumed For Nz and total mass, the appropriate mass-balance equation is Eq. (4.3): mass in = mass out (vii) Energy-balance equation The modified energy-balance equation, Eq. (526), applies:

-MIrxn-MyAhv-Q+ Ws 3.

=0

Calculate

(i) Mass balance As gas compositions are normally given in Yol%, 88,000 litres air and 12,000 litres NH3 enter the reactor every min.

On a basis ofl h, the volume of air in is 88,000 x 60 = 5.28 x 106 litres; the volume ofNE3 in is 12,000 x 60 = 7.20 x 105 litres. Using the known composition of air, the volume of Oz in = 0.21 x 5.28 x 106 = 1.11 x 106 Utres; the volume ofNZ in = 0.79 x 5.28 x 106 = 4.17 x l06litres. Converting these gas volumes to moles using the ideal gas law, Eq. (2.32), with the temperature converted from 째C to degrees Kelvin using Eq. (2.24):

Solutions: Chapter 5

46

Moles of 02 in = pV =

1 atm(l.l1 x 1Q6t) = 4.69xI04gmol 0.0820571alm r' gmor 1 (15 + 273.15) K

RT

路 _pV -_ laJln(4.17XI0'1) -176 10' gmoI MI o esofN 2m - - . x RT 0.0820'71aJlnK 19moll(I'+273.I5)K

MolesofNH in=pV = 3

RT

latm(7.20xt051) =3.05x10 4 g.mol 0.0820571aJlnK 'gmol-'(15+273.I5)K .

These molar quantities can now be converted to masses using the molecular weights: 4 Mass of02 in = 4.69 x 10

~Ol路l i~~II.ll~gl = 1500.8kg

5

~OLI i~~II.ll~gl = 4928.0kg

Mass ofN2 in = L76x 10

4 MassofNH3 in = 3.05 x 10 gmol.!

;~;II.ll~gl = 5185 kg

Therefore, the total mass of inlet gas is 1500.8 + 4928.0 + 518.5 = 6947.3 kg.

The calculation tables below show all known quantities in kg. The total mass of off~gas is denoted G; the total mass of product is denoted P.

In Gluconic acid C02

Stream

Glucose Peed Inlet gas Off-gas

o

80

o

518.5

02

o

o

1500.8

4928.0

o

0 0

o

HZO 1920

o

Total 2000

6947.3

Product 518.5 1500.8 4928.0 0 -=__~=",-_-",="-_==,--,,--

Total _ _ 80

-==,-_

0 1920 -,,,-_ _--,,=_ _ 8947.3

Out

Stream

Glucose

NHS

掳z

NZ

Gluconic acid COz

HZO

Total

Product

0 0.005P

0 ?

?

0

?

0

0 ?

?

0

0 ?

G P

Total

0.005P

?

?

?

?

?

?

G+P

Peed Inlet gas Off~gas

N2 balance N2 is a tie component.

4928.0 kg N2 in = N2 out N2 out = 4928.0 kg

Glucose balance 80 kg glucose in + 0 kg glucose generated = 0.005 P kg glucose out + glucose consumed Glucose consumed :::: (80 - 0.005 P) kg Converting the glucose consumed to molar terms:

Olucoseconsumed = (80-0.005 P) kg 路1

:8~:~ 1= (0.444-2.775x 10-' P)kgmol

From the reaction stoichiometry, conversion of this number of kgmol glucose requires the same number of kgmol NH3 and 1.5 x the number of kgmo1 02. Converting these molar quantities to masses:

47

Solutions: Chapter 5

NH, consumed

=(0.444-2.775 x 10-5 P)kgmol =(0.444-2.775 x 10-5 p) kgm01.1 ;~~~ll NH3 consumed = (7.548-4.718 x 10-4 P) kg

02 consumed

= 1.5 x (0.444-2.775 x 10-5 P)kgmol = 1.5X(0.444-2.775X 10-5 p)kgm01.j ;~:~11 02 consumed

= (21.312-1.332 x 10-3 P) kg

Similarly, expressions for the masses of glutamic acid, C02 and water generated can be determined ,from the stoichiometry:

Glutamicacidgenerated = (0.444-2.775 x 10-5 P)kgmol = (0.444-2.775X W-5 p)kgm01.1 ~4:~~~ Glutamic acid generated

I

= (65.312 -4.082 x 10-3 P) kg

CO2 generated = (0.444-2.775 x 10-5 P)kgmol = (0.444-2.775 x 10-5 p)kgm01.1

~:~ll

C02 generated = (19.536 - 1.221 x 10- 3 P) kg

Water generated

= 3 x (0.444-2.775 x 10-5 P)kgmol =3 x (0.444-2.775 x 10-5 p)kgm01.1 ;~:~11 Water generated

= (23.976 -

1.499 x 10-3 P) kg

02 balance 1500.8 kg02 in + 0 kg 02 generated

= 02 out + (21.312 -

1.332 x 10-3 P) kg 02 consumed

02 out = (1500.8 - (21.312 - 1.332 x 10-' P)) kg 02 out

= (1479.5 + 1.332 x 10-3 P) kg

C02 balance

okg C02 in + (19.536 -

1.221

x 10-3 P) kg C02 generated

C02 out = (19.536 - 1.221

= C02 out + 0 kg C02 consumed

x 10-3 P) kg

Adding together ,the masses ofN2, 02 and C02 out gives the total mass of off-gas,O:

a

= 4928.0 kg N2 + (1479.5 + 1.332 x 10- 3 P) kg 02 + (19.536 - 1.221 x 10-3 P) kg C02 G = (6427.0+ 1.11 x 10-4 P) kg

Total mass balance 8947.3 kg total mass in

= (0 + P) kg total mass out

Substituting the above expression for a into the total mass balance: 8947.3kg = (6427.0+ 1.11 x 10-4 P+ P)kg 2520.3 kg = 1.000 P kg

P = 2520.3 kg Therefore, from the total mass balance:

G

= (8947.3 - 2520.3) kg G = 6427.0kg

Substituting the result for P into the glucose, 02 and C02 balances gives:

Solutions: Chapter 5

48

Glucose consumed = (80 - 0.005 P) kg = 67.40 kg Glucose out

= 0.005 P kg =

12.60 kg

Ozconsumed = (2L312-1.332xlO*3p) kg = 17.95kg 02 out = (14795 + 1.332 x 10-3 P) kg = 1482.9 kg C02 out

= C02 generated = (19.536 -

1.221 x 10-3 P) kg

=:

16.46 kg

Using the result for P to evaluate the masses of the other reactants and products involved in the reaction: NH3consumed = (7.548-4.718 x 104 P) kg

= 6.36 kg

Glutamic acid generated = (65.312..;,. 4.082 x 10.3 P) kg :::: 55.02 kg

Water generated = (23.976 - 1.499 x 10*3 P) kg = 20.20 kg These results can be used directly in the energy balance for evaluation of the cooling requirements. However, completion of the mass balance allows the calculations to be checked.

NH3 balance

518.5 kg NH3 in + 0 kg NB3 generated

= NH3 out + 6.36 kg NB3 consumed

NH3 out =512.14kg

Water balance 1920 kg water in + 20.20 kg water generated = water out + 0 kg water consumed Water out::: 1940.20 kg Glutamic acid balance kg glutamic acid in

o

+ 55.02 kg glutamic acid generated ::: glutamic acid out + 0 kg glutamic acid consumed Glutamic acid out::: 55.02 kg

The Out side of the mass-balance table can now be completed with all quantities in kg.

Out

Stream

掳z

NZ

Gluconic acid CO2

0 512.14

1482.9 0

4928.0 0

0 55.02

512.14

1482.9

4928.0

55.02

Glucose

NHl

Inlet gas Off-gas Product

0 12.60

Total

12.60

Feed

HZO

Total

16.46 0

0 1940.20

6427.4 2520

16.46

1940.20

8947.3

All columns and rows of the completed table add up correctly to within round-off error. (ii) Ws

Energy balance Therefore, the energy*balance equation becomes:

=0; Mv =O.

-AHrxn-Q::: 0 The heat of reaction is evaluated using Eq. (5.20). As the heat of combustion ofCOZ and H20 is zero:

rxn = (nM~)G+(nM~)A -(nM~)GA where G ::: glucose, A ::: NH3 and GA = glutamic add. The nin this equation are the moles of reactant or product Mf

involved in the reaction. Converting the masses of reactants and products consumed or generated to moles: Glucoseconsumed::: 67.40kg ::: 67AOkg 路ll~gl路1 :8~o~

I::: 374gmol

49

Solutions: Chapter 5

As NH3 and glutamic acid are involved in the reaction in stoichiometric quantities. 374 gmol NH3 are consumed and 374 gmol glutamic acid are produced. Substituting these quantities into the heat of reaction equation gives:

MI"" : 374 gmol (-2805.0kJ gmol-lj + 374 gmol(-382.6kJ gmOl-lj_ 374 gmol(-2244.1 kJ gmol-lj M:lrxn

= -353 x lOS kJ

Substituting this result into the energy~balance equation:

3.53 x loS kJ -Q : 0

Q : 3.53 x loS kJ From the sign conventions outlined on W 87-88. Q positive indicates that,heat must be removed from the system.

Answer: 353 x 105 kJ h- 1 (b)

If cooling were not provided, the heat of reaction would be absorbed as sensible heat by the streams passing through the reactor. For a rough calculation of the effect of this heat on the temperature of the reactor, assume that the 353 x 105 kJ h- 1 is absorbed by 2000 kg h- 1 aqueous medium and 6947-3 kg l:r 1 gas. ,Assume that the heat capacity of the

=

aqueous medium is close to that of water:: 75.4 J gmol-l °C-l (Table B-3, Appendix B) 75.4 kJ kgmol-l °C-l, and that the heat capacity of the gas stream is equal to that of air :: approx. 29 J gmol-l 0C- 1 (Table B.3, Appendix B):: 29 kJ kgmol-l °C-l. From Eq. (5.12):

6T:

MI

(M Cplliquid + (M Cpl.., 5 1 353 x 10 kJ h-

T ::

1 (2000 kg h-1)(75.4kJ kgmor oc-1j.1

:: 23"C

~~~~ll +(6947.3 kg h- 1)(29 kJkgmor1 oc-1j .1

;~~~ll

As a temperature rise of 23°C in the reactor would not be well tolerated by most commercial organisms, provision of adequate cooling for this reaction is an important consideration. Assuming that the usual temperature for the reaction is 25°C, the temperature without cooling woulJ,i increase to (25 + 23)OC 48°C.

=

5.6

Bacterial production of alginate

Alginate production at a rate of 5 kg h~ 1 requires:

5 kgh-1

: 1.25kgOZh-1

4kgkg 1 Converting this quantity to gInol using the molecular weight of 02 1 0zrequired:: 1.25kgh-

=32.0 (Table B.I, Appendix B):

0: 1.1 ~~~ll

.1 1

g

:: 39.06gmolh-

1

The heat of reaction for aerobic metabolism is approximately -460 kJ gmol- 1 Oz (p 100). Therefore, the heat of reaction for alginate production is:

'This result can be used in the modified energy-balance equation, Eq. (5.26): -6Hrxn -Mv tih v -Q+ Ws :: 0

with Ws :: 1.5 kW andMv s·1 and:

:=

0 (no evaporation). From Table A.8 (Appendix A), 1 W:: 1 J s-1; therefore 1 kW:: 1 kJ

Solutions: Chapter 5

50

1.80x 104 kJh-1_O_ Q+ 1.5 kJ 8- 1 路13~s 1 :: 0

Q = 2.3 x 10" kJ h- I From the sign conventions outlined on pp 87-88, Q positive indicates that heat must be removed from the system.

Answer: 2.3 x 104kJh路 1

5.7

Acid fermentation

From Tables B.S andB,l(Appendix B), the molecular fannulae and molecular weights are: sucrose = C12H2Z011: MW:: 342.3

propionicacid=C3H60 Z: MW =74.1" acetic acid = Czf40Z: MW =60.1 butyric acid=C4HgOZ: MW = 88.1 lactic acid C3H603: MW =90.1 From p 75. the biomass molecular formula can be taken..asthe,average, (Appendix B). the molecular weight of the biomass is 25.9.

=

CH1.80o.sNo.z'. From the end of Table E.8

The reaction equation can be obtained by modifying Eq. (4.13) for anaerobic growth andproo.uct-fonnation: C 12HZZ0 11 +bNH3 -+ cCHl.gOo.sNo.2 +dCOZ +eHZO+!I C3~02 +h C2H 40 2 +Jj C4H ,02 +14 C,H60 3 The biomass yield from substrate Yxs = 0.12 g gol, This value can be used to determine the stoichiometric coefficient c using Eq. (4.l2): _ Yxs (MW substraJe)) _ 0.12 gg-l (342.3) _ MW cells 25.9 - 1.59

c -

The coefficientsh ,/2./3 and/4 can be determined similarly using the product yields and Eq. (4.14):

11

h

0040 g g-I (342.3) Yps (MW substrate)) = MWpropionicacid = 74.1 = 1.85

= Yps(MWsubstrate)) = 0.20gg- 1 (342.3) = 114 MW acetic acid Yps (MW substrate))

Jj = MW butyric acid

60.1

=

.

0.05 g g-I (342.3) = 0. 19 88.1

_ Yps (MW substrate)) _ 0.034 g g-l (342.3) _ MWI achC . acl路d 90..I - 0.13

/4 -

Of the remaining coefficients b, d and e, because CO2 and H20 do not figure in heat of reaction calculations as their heat of combustion = 0, only b need be determined. nus can be done using an elemental balance on N. N balance: b = 0.2 c = 0.2 x 1.59 = 0.32.

To calculate the heat of reaction, the heats of combustion of the reactants and.products are required from Table E.8 (Appendix B): M~ sucrose=-5644.9kJ gulOr l

M~NH3 =-382.6kJ gmol-l

M~biomass=-552kJ gmor 1 M~ propionic acid = -l527.3kJ gmol-l M~ acetic acid=-874.2kJ gulOr l

M~ butyric acid =-2183.6kJ gmor 1 M~ lactic acid = -1368.3 kJ gmol-l

51

Solutions.' Chapter 5 The heat of reaction is determined using Eq. (5.20). As the heat of combustion of C02 and H20 is zero:

Mlrxn = (n 6.h~)s +(nah~)A -(n M~)B -(n M~)PA -(nM~)AA -(nah~)BA -(nM~tA =

=

=

=

where S sucrose, A = NH3, B = biomass, PA propionic acid, AA acetic acid, BA butyric acid, and LA = lactic acid. Using a basis of I gmol sucrose, the n in this equation are the stoichiometric coefficients, Substituting values:

Iili""

= 1 gmol(-5644.9kJ gmol-l) +0.32gmol (-382.6kJ gmol-l)_ 1.59 gmol(-552kJ gmol-l) - 1.85 gmol(-1527.3 kJ gmol-1)-1.l4 gmol (-874.2 kJ gmol-l) -0.19 gmol(-2183.6kJ gmol-l) -0.13 gmol(-1368.3 kJ gmol-l) t:Jirxn

= -474.8 kJ

This M rxn was determined on the basis of I gmol sucrose. For 30 kg sucrose consumed over a period of 10 d:

M rxn

=30kg.

llkg .

IOOOgll,gmOll( 342.3g -474.8 kJ gmol-I)

=-4.16x 104 kJ

The cooling requirements are determined using the modified energy~balanceequation,Eq. (5.26). For no evaporation and no shaft work, M v = Ws = 0, so that: Q = -Mfrxn

= 4.16 x 104 kJ

From the sign conventions outlined on pp 87-88. Q positive means that heat must be removed from the system.

Answer:4.16x 104 kJ

5.8

Ethanol fermentation

From Table B2 (Appendix B), the molecular formula for glucose is C6H1206 and the molecular fonnula for ethanol is C2H60. From Eq. (4.13), the stoichiometric equation under anaerobic conditions is: C6H1206+bNH3 ~ CCH1.7.s00.5SNO.lS+dC02+eH20+fC2H60 From Table B.8 (Appendix B), the molecular weight of ethanol is 46,1. From the atomic weights in Table B.I (Appendix B), the molecular weight of glucose is 180.2 and the biomass fonnula weight is 25.58. Taking into account the 8% ash: Biomassmolecularweight = 2;':::

= 27.80

The value of the stoichiometric coefficient/can be determined from the yield Yps= 0.45-g g-1 and Eq. (4.14):

f =

Yps(MWsubstrate) MW product

=

0.45gg- 1 (180.2ggmol-l) -1 = 1.76gmolgmol 46.1 g gmol I

The other coefficients can be determined using elemental balances.

Cbalance:6 c+d+2f= e+d+2x1.76 -+ c 2.48-d Hbalance: 12+3b 1.75c+2e+6f= 1.75c+2e+6x1.76 ~ 1.44+3b 1.75c+2e o balance: 6 = 0.58c+2d+e+f= 0.58c+2d+e+1x1.76 ~ 4.24 = 0.58c+2d+e Nbalance: b = 0.18e Substituting the expression for c from the C balance into the N balance:

=

=

=

b

=

= 0.18 (2.48-d) = 0.45-0.18d

Substituting this and the results from the C and N balances ioto the H balance: L44+ 3 (0.45-0.18d) = 1.75 (2.48-d) + 2e 1.21d-L55

= 2e

e = 0.61 d-0.78

Solutions: Chapter 5

52 Substituting the expressions for c, b and e into the 0 balance:

4.24

= 0.58 (2.48 - d) + 2 d + (0.61 d - 0.78) 3.58 = 2.03 d d = 1.76

Substituting this value· for d into the expressions for the other coefficients gives c "" 0.72, b = 0.13 and e "" 0.29. The

completed stoichiometric equation is therefore: C6H1206+0.13NH3 ~ O.72CH u sOO.SSNO. 18+ 1.76 CO2 + O.29HZO + 1.76CZH60 Using a basis of 1 h, 0.4 kg ethanol are produced. Converting this to moles:

Molesethanolproduced= OAkg .11~gl.1

~~11

"

8.68gmol

From stoichiometry:

1 Moles glucose consumed "" 8.68gmolx 1.76

= 4.93gmol

Moles NH3 consumed "" 8.68 gmol x ~:~~ "" 0.64 gmal Moles biomass produced "" 8.68 gmol x ~:;~ "" 3.55 gmat

The heats of combustion from Table B.8 (Appendix B) are: 8h~ glucose: -Z805.0kJ gmor 1

8h~NH3 =-382.6kJ gmol-l tJh~ ethanol"" -1366.8 k:J gmar l From p 101, the heat of combustion of yeast can be taken as -21.2 kJ g-1. The heat of reaction is determined using Eq. (5.20). As the heat of combustion of C02 and H20 is zero:

Mlrxn

=

= (nAh~)G+(nAh~)A -(nah~)B-(nAh~)E

=

where G glucose, A = NH3' B biomass and E = ethanol. The n in this equation are the actual moles of reactants and products consumed or produced. Substituting values gives:

/lH= = 4.93 gmol (-2805.0kJgmor1) + O.64gmol (-382.6kJ gmol-1)_3.55 gmol(c21.2kJ g-l)·1 ~7:o~

I

- 8.68 gmol (-1366.8 kJ gmol-l)

= -1175 kJ Using the modified energy~balance equation, Eq. (5.26), with M v =0 and Ws =0: Mlrx,n

Q=-Mlrx,n= 117.5kJ From the sign conventions outlined on ,pp87-88, Q positive means that heat must be removed from the system; in this case, 117.5 kJ h- I is used to raise the temperature' of 25 I h- I water from lOoC. The sensible heat change of the water can be calculated from Eq. (5.12). Using a value of 1 kg I-I for the density of water, Cp for water = 75.4 J gmol-I °C*I (Table B-3, Appendix B), and the molecular weight of water 18.0 (Table B.l, Appendix B):

=

lJ.T

= ~_ Mep

=

117.5 kJ h-

1

The final temperature of the water is therefore lOoe + ll.2°C = 2L2°C

Answer: 21.2°C

_ 112"<:

(25Ih-1 Ilkg~(754J 1-1oC- 1 11gmOllllOOOgll~fi . . 11 U . gmo . 18.0g· lkg . lOOOJU

.

Solutions: Chapter 5

5.9

53

Production of bakers' yeast

From Table B,8 (Appendix B), the molecular formula for sucrose is C12HZ2011' Therefore, from Eq. (4.4), the stoichiometric equation for aerobic cell growth is: CIZH22011 +a02+bNH3

-?

cCHl.8300.5sNo.17+dCOZ+eHZO

Using the atomic weights in Table R1 (Appendix B), the molecular weight of sucrose is 342.3 and the biomass formula weight is 25.04. Taking into account the 5% ash: Biomass molecularweight

= ~~ : 26.36

The degree of reduction of the biomass relative to NH3 is:

ni:::h ~,X4+J.83X1-0.55X2-0.17x3: . ....

1

4.22

..

The degree of reduction of sucrose relative to NH3 is:

rs _-

12x4+22x1-11x2 _ 400

- .

The value of the stoichiometric coefficient c can be determined from the yield Yxs: 0.5 g g~l and Eq. (4.12): C

= Yxs(MWsubstrate) = 0.5gg-1 (342.3ggmol-1) = 6.49gmolgmol-l MW cells

26.36 g gmol 1

The oxygen requirements can be determined from Eq. (4.16) with/: 0; from Eq (4.13), w: 12 for sucrose:

a

= 1/4 (w)S-cJ1!) = 1/4 (12x 4.00-6.49 x 4.22) = 5.15

Therefore, 5.15 gmol Oz are required for each 6.49 gmol biomass produced; this corresponds to 0.79 gmol Oz per gmol biomass. Converting this oxygen demand to mass terms using the molecular weight of oxygen: 32.0 from Table B.I (Appendix B):

1_

. _ 0.79 gmolOz 132.0 g Ozlll gmol biomass -1 0.79 gmol 0Zper gmol blOmasS - 1 gmol biomass' 1 gmol0 . 26.36 g biomass - 0.96 g g z The specific growth rate represents a rate of growth of 0.45 g biomass produced per g biomass per h. As 0.96 g 02 are required per g biomass produced, the specific rate of Oz consumption is 0.45 x 0.96: 0.43 g Oz.r,r g biomass per h. When the biomass concentration is 10 g 1~1 in a 50,000 litre fermenter, the mass of cells is 10 g 1- x 50/}00 1: 5 x 105 g. Therefore, the total rate of 02 consumption is 0.43 g g-1 h- 1 x 5 x 105 g: 2.15 x loS g Oz h路 1. Converting this to moles: 5 Rate of 0 1 consumption : 2.15x 10 gh-1 .l

;~~ll :

6.72x 10 3 gmolh- 1

From p 100, the heat of reaction for aerobic growth is approximately -460 kJ gmol-l 02 consumed. Therefore:

The rate of heat removal from the fermenter is determined using the modified energy路balance equation, Eq. (5.26),

withWs:Mv:O:

From the sign conventions outlined on pp 87-88, Q positive confrrms that heat is removed from the system.

Answer: 3.09 X 106 kJ h- 1

Unsteady-State Material and Energy Balances 6.1

Dilution of sewage

(i)

Flew sheet and system boundary

These are shown in the figure below,

System

,

boUndary\

Water F::::: 40,000 I h路 1

r I I I I I I I

-

"

- 1---

-

1

-

V

CA

I I I I I I I

Outlet stream

F=40,OOO I h- 1 CA

(Ii) Define variables V volume of material in the tank; F ::::: volumetric flow rate into and out of the vessel::::: 40,000 1 h路 I ; CA :::::

=

concentration of suspended solids (ill) Assumptions -no leaks -tank is well mixed; therefore CA in the outlet stream::::: CA inside the tank - density of the solids is the same as that of water"'" 1 kg I-I (iv) Boundary conditions At t 0, V::::: Vo ::: 440,000 litres + the volume of the solids. The initial mass of solids in the tank is 10,000 kg; therefore. if the density is 1 kg I-I, Vo::::: (440,000 + 10,000) 1 450,000 L Att::::: 0, CA::::: CAO:

=

=

kg 0022k 1-1 CAO::::: 10,000 450,ooof:::::' g (v)

Total mass balance As the volumetric flow rates of material in and out are the same and the density of the inlet and outlet streams are assumed to be equal, the volume of material in the tank is constant (see Example 6.2) so that V = Vo at all times.

Solids mass balance (vi) The general unsteady-state mass-balance equation is Eq. (6.5). As there is no reaction, RO RC = 0; in this problem, Mi 0, Mo F CA and M= V CA. Substituting into Eq. (6.5) gives:

=

=

=

d(VCA)

dt As V is constant, it can be taken outside of the differential:

= -FCA

55

Solutions: Chapter 6

As F is also constant, the differential equation contains only two variables, CA and t. Separating variables gives: dCA -F -=-dt CA V Integrating:

f

J-FV

=

dCA

CA

dt

Using integration rules (D'.27) 楼td(D.24) from Appendix D and combining the constants of integration:

-F

InCA = -yl+K From the initial condition for CA, at t = 0, In CAO = K, Substituting this value of K into the equation gives:

-F

InCA = -yt+InCAO CA

InCAO

C~

-F

=-1

V

= CAOe (-FIV; t

Substituting the known values for CAo. F and V: CA = 0.022 e-O路089 t

where CA has units of kg I-I and t has units of b. From this equation, at t= 5 h, CA

=0,014 kg I-I,

Answer: 0.014 kg I-I

6.2

Production of fish-protein concentrate

(i) System The system is the whole gutted fish placed in the batch drier at time zero. During drying, the mass of the system decreases as water is removed. (li) Assumptions No additional assumptions are required, (iii) Boundary conditions At t = 0, the mass of water in the fish in the drier is equal to M() (iv) Water mass balance The general unsteady-state mass-balance equation is Eq. (6.5), As there is no reaction, RO = RC = O. No water is added during drying; the~fore Mi = 0, At any time during drying, as the rate of water removal Me is proportional to the moisture contentM, Me k M where k is a constant, Substituting into Eq, (6.5) gives:

=

dM

Cit

=-kM

where M and t are the only variables. Separating variables and integrating:

Using integration rules (D.27) and (0,24) from Appendix. D and combining the constants of integration:

lnM=-kt+K At I"'" 0, M = Mo; therefore. K"'" In Mo- Substituting this value of K into the equation gives:

Solutions: Chapter 6

56 InM = -kt+1nMo

M In M

o

= -let

At t= 20 min, M = 0.5 MO. Substituting these values into the equation: In

O.5M

-:MOo = -k (20 min)

In 0.5 = -k (20 min) k = 0.0347 min~l

Therefore: M In Mo

= -0.0347 t

where t has units of min. When 95% of the water has been removed, M =0.05 Mo_ Therefore: In

O.05MO M = -0.0347 t

o

In 0.05 t

:=

-0.0347 t

= 86.3 min 86.3min.I601~n! = =:

l.44h

Answer. 1.44 h

6.3

Contamination of vegetable oil

Flow sheet and system boundary These are shown in the figure below. (i)

System boUndary\

Inlet stream Cod-liver oil drum

F; P

- -

r I I I I I I I

L

\.

;

-

- -

I I I I I I I

V p </

- - -

Oil tank

- - -

1

-

Outlet stream Fo

p

</

J

(li) Define variables V = volume of oil in the tank; Fi "'" volumetric flow rate of cod~liver oil into the oil tank; F 0 volumetric flow rate of mixed oil out of the oil tank; ¢ = mass fraction of vegetable oil; p"'" density of vegetable and cod-liver oils (ill) Assumptions - no leaks - oil tank: is well mixed; therefore ¢ in the outlet stream ¢ inside the tank - densities of each oil and the oil mixture are the same

=

=

57

Solutions: Chapter 6 (iv) Boundary conditions At t= 0 when cod-liver oil first enters the oil tank, V = Vo = 60 litres. At t= 0,4>= ¢o = 1. (a)

Total mass balance The general unsteady~state mass--balance equation is Eq. (6.5). As there is no reaction, RO = RC = O. The t2.ta1 mass flow rate into the oil tank is equal to the volumetric flow rate multiplied by the density of the cod-liver oil: Mi Fi p.

Similarly, Mo = Fo p. The total mass of oil in the tank is equal to the volume of oil multiplied by its density: M = V p.

Substituting these expressions into Eq. (6.5) gives:

=

d(VP)=Fp_Fp dt ' 0 As P is constant it can be taken outside of the differential and cancelled:

dV

Pdt

= (Fi-FdJp

dV

dt = (Fi-FdJ The differential equation contains only two variables, V and t. Separating variables and integrating:

I

dV

=

I

(Fi-FdJdt

Using integration rule (0.24) from Appendix D and combining the constants of integration:

From the initial condition for V, at t = 0, K = Vo. Substituting this value of K into the equation gives:

V= (£i-Fo)t+ Vo

=

=

The time between 8 p.m. and 9 a.m. the next morning is 13 h. From the above equation for V,- for t 13 h, Fi 7.5 1 h- 1 ,Fo =4.81h- 1 and Vo=60l,V""'95.11. As this volume is less than the tank capacity of 1001, the tank will not overflow. '

Answer: No (b) Total mass balance As the volumetric flow rates of oils into and out of the tank are the same and the density of the inlet and outlet streams are assumed to be equal, the volume of oil in the tank is constant (see Example 6.2)-s0 that V = Vo;:::; 60 1 at all times.

Vegetable oil mass balance The general unsteady~state mass-balance equation is Eq. (6.5). As there is no reaction, RG "'" RC O. No vegetable oil enters the tank; therefore Mi "'" O. The mass flow rate of vegetable oil out is Mo "'" F o P 4>. The mass of vegetable oil in the tank at any time is M"", V p 4>. Substituting these expressions into Eq. (6.5) gives:

=

d(V;:¢)

= -Fop¢

As both V and p are constants, they can be taken outside of the differential and p can be cancelled:

VP~ = -Fop¢ V~

= -Fo ¢

As F0 is also constant, the differential equation contains only two variables, 4> and t. Separating variables and integrating:

Solutions: Chapter 6

58

Using integration rules (D.27) and (D.24) from Appendix D and combining the constants of integration: -Fo In垄::::: yt+K From the initial condition for 41, at t

=0, In <Po::::: K.. Substituting this value of K into the equation gives: ln~

-Fo

= y'+lnib ~

-Fo

lnib=y' ~ =

In this problem. F0

:::::

ib e(-F,II'J'

4.8 I h-l. Substituting the known values for </Jo and V:

II : : :

1 e....(l路080 t

is dimensionless and t has units of h. The time between 8 p.m. and midnight is 4 h. From the above 4'= 0.73. Therefore, at midnight, the composition of oil in the tank is 73% vegetable oil and 27% cod-liver oil. where

<P

equation, at t= 4 h,

Answer. 73% vegetable oil, 27% cod-liver oil

6.4

Batch growth of bacteria

The general unsteady-state mass-balance equation is Eq. (6.5). For a batch culture, Mj ::::: Mo : : : O. For a mass balance on cells. assuming there is no loss of cells from the system, e.g. by lysis, RC::::: O. The rate of generation of cells RG is proportional to the concentration of cells present: RO = J.t x V where jJ is a constant, xis the cell concentration and路 V is the culture volume. The total mass of cells M is equal to the culture volume V multiplied by the cell concentration x: M = V x, Substituting these results into Eq. (6.5) gives: d(Vx)

= J.txV

dt Assuming that V is constant throughout the batch culture, it can be taken outside of the differential:

dx

V dt

dx dt

= J.tx V

= J.tx

The differential equation contains only two variables, x and t. Separating variables and integrating:

-dxx = I'dt

Using integration rules (0.27) and (0.24) from Appendix. D and combining the constants of integration:

lnx = j1.t+K

=

Assume an initial condition: at t 0 at the beginning of ex.ponential growth, x Substituting this value of K into the equation gives:

=xo.

Therefore, at t

=0, In Xo =K.

59

Solutions: Chapter 6

lnx =

j.tt+lnxo

x In- = XO

j1,t

When t = 45 min, x = 2 xo. Substituting these values into the equation:

2XO I n - = ,u(45 min) XO ln2 = j.t

j.t

(45 min)

= 0.0154 min'l

Therefore:

In''::' =

0.0154t

x =

eO.Ol54 t

XO

XQ

where thas units of min. For t= 12h = 12 x 60= no min: x=6.54x 104 XQ

After 12 h, the cell concentration is 6.54 x 104 times the cell concentration at the beginning of exponential growth. If there is a lag phase, the cell concentration at the beginning of exponential growth is usually very close to that at inoculation; therefore, the cell concentration is about 6.54 x 104 times the inoculum level.

Answer: 6.54 X 104 times the inoculum level

6.S

Radioactive decay

The general unsteady~state m.ass~ba1ance equation is Eq. (6.5). In this problem, M j = Mo = O. For a mass balance on isotope, RO O. The rate of isotope decay RC is proportional to the concentration of isotope present: RC -/q C V, where kl is a constant, C is the isotope concentration and V is the solution volume. The total mass of isotope M is equal to the solution volume V multiplied by the isotope concentration C: M = V C. Substituting these results into Eq. (6.5) gives:

=

=

d(:CJ =-k,CV If we assume that the density of the solution is constant during isotope decay, V is constant and can be taken outside of the differential and cancelled: dC V(if =-k 1 CV dC (if

= -k1 C

The differential equation contains only two variables, C and t, Separating variables and integrating:

dC

C

= -k 1 dt

Jdg = f

-k) dt

Using integration rules (D27) and (D.24) from Appendix D and combining the constants of integration:

lnC=-k1 t + K

60

Solutions: Chapter 6

Assume an initial condition: at t "'" 0, C "'" CO, Therefore, at t "'" 0, In Co = K, Substituting this value of K into the equation: InC = -ktt+lnCo

C In-=-k i t

Co

(a)

When t= !h, the half-life of the isotope; C = 0.5

Co.

Substituting these values into the equation:

0.5 Co I n - - = -kIth Co In 0.5 = -ki !h

_ -tn0.5 th - - -

k,

Using mathematical rule (D. 10) in Appendix D, -In 0.5

=In 1/0.5 =In 2.

Therefore:

In2

th

= kl

Answer. Q.E.D.

(b) From the equation derived in (a):

For t:h

=14.3 days: kl

= 4.85 x 10-2 d- I

Substituting this value of kl into the general equation for isotope concentration: In

go = -4.85 x 10-2

t

where thas units of days. ForC=O.Ol CO:

In

0.01 Co Co

In 0.01 t

Answer. 95 days

-2

= -4.85 x 10

= -4.85 x 10.2 t = 95 days

t

61

Solutions: Chapter 6

6.6

Continuous fermentation

(i)

Flow sheet and system boundary

'These are shown for a continuous fermenter in the figure below,

System

Feed F

xI =0

"

(li)

....

boUndary\

"

- - r- - - -

r I I I I I I I

V x

•

1

I I I I I I I

Product F

x

•

Define variables

=volume of broth in the fennenter; F = volumetric flow rate into and out of the vessel; x = concentration of cells; s = concentration of substrate; Xi. =concentration of cells in the feed; Si =concentration of substrate in the feed V

(iii)

Assumptions

-no leaks - fermenter is well mixed; therefore x and s in the outlet stream =x and s, respectively, inside the fermenter - density of the fermentation broth is the same as that of the feed Boundary condition (iv) Att=O,x=XO· (v) Total mass balance As the volumetric flow rates into and out of the fermenter are the same and the density of the inlet and outlet streams are assumed equal, the volume of broth in the fermenter is constant (see Example 6.2) 8..Iid equal to Vat all times. (aj Cell mass balance

The general unsteady-state mass-balance equation is Eq, (6.5). As no cells enter in the feed stream, Mi =O. The mass flow rate of cells out Mo = F x. The rate of cell generation RG = rx V = kl x V. Assuming that cell lysis is negligible, RC = O. The mass of cells in the fermenter M is equal to V x. Substituting these terms into Eq. (6.5) gives: d(Vx) -_.= -Fx+k1xV

d'

As V is constant, it can be taken outside of the differential.

dx V dt = -Fx+k1xV Dividing through by V and grouping terms: dx F - = x(kt--) d'

V

F

dx

Answer: dt = x (k I - V) (b)

At steady state, dxldt = O. Therefore, from the equation derived in (a), at steady state kl must be equal to FIV.

Answer: kl

=f.<iv

62

Solutions: Chapter 6

(e)

As F, V and kl are constants, the differential equation derived in (a) contains only two variables, x and t. Separating variables and integrating:

<Ix F -;- = (k'-l1)dt

J~ = f(k,-~)dt Using integration rules (D.27) and (D.24) from Appendix D and combining the constants of integration:

F

lux:::: (k1-V)t+K

From the initial condition for x,at t= 0, In Xo = K. Substituting this value of K into the equation gives:

F

Inx:::: (k1-y)t+1nXO â&#x20AC;˘

F

'0

V

10- :::: (kl--)t F

x=.xoe(kl--)t ,v F (k --) , Answer: x:::: XOe 1 V (d)

Substituting the parameter values into the equation derived in (c} -1

22001h-1

x:::: (0.5gl-1)e(0.33h - 10,0001)/

x :::: 0.5 eO.IlI where x has units of g 1-1 and thas units ofh. For x:::: 4.0 g I-I;

4 = 0.5 eO.II t inS = 0.11 t t::::

18,9h

Answer: 18.9 h (e)

Substrate mass balance

The general unsteady-state mass-balance equation is Eq. (6.5). For substrate, the mass flow rates in and out are Mj = F si and Mo = F s. The rate of substrate generation RG =0; the rate of substrate consumption RC =IS V = k2 x V.The mass of substrate in the fermenter M is equal to V s. Substituting these terms into Eq. (6.5) gives: d(Vs) ( i t = Fs i -Fs-k 2 xV

As V is constant, it can be taken outside of the differential:

Dividing through by V and grouping tenus:

Substituting the expression for x from (c):

63

Solutions: Chapter 6

In this equation, F, V, kl and X{) are constants and there are only two variables, s and t. However, the variables cannot be easily separated as in the previous problems, making algebraic solution difficult.

&

Answer. dt

= VF (si -s)-k1X{)e ~l-~t V

(f) An equation for &/dt in terms of x was dt;:rived in (e) as:

,

At steady state, &/dt =O. Therefore:

F V (si -s) = k 2 x Vk Z si- s = F'x

Answer. s

6.7

Vk

Z = si -Tx

Fed-batch fermentation

Flow sheet and system boundary These are shown for a fed-batch fermenter in the figure below. (i)

System bOUndary\

Feed F s;

p

-

r I I I I I I I L

\.

-

'"- -

1

I I I I

V

x s p

1 I I

J

(ll) Define variables V = volume of broth in the fermenter; F = volumetric flow rate into the fermenter; x = concentration of cells in the fermenter; s = concentration of substrate in the fermenter; Si = concentration of substrate in the feed; p = density of the

feed and fermentation broth (iii) Assumptions -no leaks - fermenter is well mixed - density of the fermentation broth is the same as that of the feed

64

Solutions: Chapter 6

(iv) Boundary condition Att=O, V=Vo. (a)

Total mass balance The general unsteady-state mass-balance equation is Eq. (6.5). As total mass cannot be generated or consumed, RO = RC =O. No mass leaves the fennenter; therefore Mo =O. The mass flow rate in Mj = p F. The total mass in the fermenter M is equal to V p. Substituting these terms into Eq. (6.5) gives: d(VP)=pF dt As P is constant, it can be taken outside of the differential and cancelled:

dV Pdi

= pF dV = F dt

As F is constant, the differential equation contains only two variables, V and t. Separating variables and integrating:

dV = Fdt

Using integration rule (D.24) from Appendix D and combining the constants of integration:

v= From the initial condition for V, at t

=0, Vo =K.

Ft+K

Substituting this value of K into the equation gives: v= Ft+Vo

Answer: V=Ft+Vo (b) Substrate mass balance The general unsteady-state mass-balance equation is Eq. (6.5). The mass flow rates of substrate in and out are Mj F si and Mo = o. The rate of substrate generation RO 0; the rate of substrate consumption RC rs V kl S V. The mass of substrate in the fermenter M is equal to V s. Substituting these terms into Eq. (6.5) gives:

=

=

=

=

d(V,) -at =Fsi-klsV

As neither V nor S is constant, both must be kept in the differential as a product. Expanding the differential using the product rule (D.22) from Appendix D:

Using the equation dV1dt

=F derived in (a) :

Grouping terms gives:

<is

Answer: dt

F s = V(Si-S)-kt

65

Solutions: Chapter 6

6.8

Plug-flow reactor

(i) Flow sheet and system boundary These are shown for a plug-flow reactor in the figure below.

System bOUndary,

...

~_ _!..I

Fee~--"""-II ~

:

- - - " ' - - -_ _~

-=-.,.....,.:.. z

I

S-.......- ~rOduct

~Ml

J

L

(ii) Define variables u := fluid linear velocity; z distance along the reactor; CA reactant in the feed stream; A = reactor cross-sectional area (iii) Assumptions -no leaks -plug flow

=

:=

concentration of reactant; CAi

= concentration of

(a)

Reactant balance Consider the system to be a small section of the reactor located between z and z + &. The general unsteady-state mass-balance equation is Eq. (6.5). The rate ofentry of reactant into the system is:

Mj

= CAuAl

k

where means that the parameter values are those at distance which reactant leaves the system is:

Mo =

CAuAI

z

z from the front of the reactor. Similarly, the rate at

,+<\,

Reactant is not generated; therefore RO = O. The rate of consumption of reactant is given by the equation:

Rc=rcV:=rc A & where A ilz is the volume of the system and rc is the volumetric rate of reaction. At steady state there is no accumulation in the system and dM/dt = O. Substituting these expressions into Eq. (6.5) gives:

0= CAuAlz-CAuAlz+hz-rcA&; As A is constant and does notdepend on z, it can be cancelled from each of the terms:

0= CAulz-CAulz+&-rcAz Dividing through by A.<;:

Taking the limit as Llz approaches zero and applying the definition of the derivative (0.13) in Appendix D: -d(eA u)

0= '"--az--rC

or d(CA u)

dz

= -rC

As the fluid velocity is constant throughout the reactor, u can be taken outside of the differential:

Solutions: Chapter 6

66

dCA

Answer: u liZ

"" -rC

(b)

Answer: at Z"" O. CA =CAi (e)

If the reaction is first-order,

As

rc =: kl CA where kt is the first-order rate constant.

The differential equation becomes:

u and kl are constants, the differential equation contains only two variables, CA and z. Separating variables and

integrating: dCA CA

= -k) dz U

f f dCA "" CA

-k) dz U

Using integration rules (D.2?) and (D.24) from Appendix D and combining the constants of integration:

InCA

-k)

= -z+K U

From the initial condition in (b), at z = 0, In CAi =: K. Substituting this value of K into the equation gives:

CA -kl 1n-=-z CN

CA

=:

U

CAi e(-k,lu) z

(d) The equation derived in (c) is directly analogous to the equation for change. in reactant concentration in a batch reactor. As z "" u t where t is the time taken for the fluid to travel distance z. the above equation can be written as: CA "" CAie""*lt

which is the same as the equation for reactant concentration in a batch reactor where CAi is the concentration at time

zero.

Answer: Essentially identical

6.9

Boiling water

The system is the beaker containing water. (i) Assumptions - no evaporation - water is well mixed - no shaft work - heat capacity is independent of temperature - beat losses are negligible - the density of water is constant between 18"C and lOO"C

67

Solutions: Chapter 6 (ii) Extra data Density of water = 1 kg I-I Cp water = 75.4 J gmol-1 "'C-l (fable B.3, Appendix B) = 75.4 kJ kgmol-l "'C-l Molecular weight of water (fable B.l, Appendix B) 18.0 1 W = 1 J s-l (fable A.8, Appendix A); therefore, 1 kW = 1 kJ s·l (iii) Boundary conditions Att=O,T=TO=18°C

=

(a) The general unsteady-state energy-balance equation is Eq. (6.10). For a batch system, Mi = Mo =0; also Ws = O. Energy is accumulated by the system in the fonn of sensible beat only; therefore:

dE <IT (jJ=MCpdt where M is the mass of water in the beaker and T is its temperature. Substituting these expressions into Eq. (6.10) gives:

dT MC =-Q Pdt dT

-

Answer: M Cp dt = -Q (b) If Q, Cp and M are constant, T and t are the only variables in the differential equation. Separating variables and integrating:

Using integration rule (D,24) from Appendix D and combining the constants of integration:

T= From the initial condition for T, at t= 0, To

=K.

-Q

MCp

t+K

Substituting this value of K into the equation gives:

-12

T= MC t+To p

Using the density of water = 1 kg terms:

t- 1, the mass of 2 litres of water M= 2 kg. Converting the Cp for water to mass

1 1 Cp = 75.4kJ kgmor "'C- .1 T

=the boiling temperature of water =100°C; t =11 min. lOO"C

=

-Q

= 4.189 kJ kg- 1 "'e- I

Substituting the parameter values into the equation for T:

2kg 4.189kJkg

1=

Q = -62.45kJmin-

(

~::~11

1

-62.45kJmin-

"'e-

1

I) (11 min) + 18°C

·116~snl·ll ~;11 = -L04kW

From the sign conventions outlined on pp 87-88, the negative value for

Answer: L04 kW

Q confirms that heat is added to the system.

68

6.10

Solutions: Chapter 6

Healing glycerol solution

The system is the stirred tank containing the solution of glycerol in water. (i) Assumptions -no leaks

- no evaporation - tank is well mixed - no shaft work - heat capacities are independent of temperature between 15<>C and 90"C - ideal solution - system is adiabatic; therefore heat losses are negligible (ii) Extra data

Cp glycerol = 0.576 cal g-t 째C- l (Table B.s, Appendix B) = 0.576 keal kg-I "C- 1 Cp water = 75.4 J gmol-l 째C- l (Table B.3, Appendix B) = 75.4 kJ kgmol-l 째C- 1 Molecular weight of water (Table B.I, Appendix B) "" 18.0 1 kcal "" 4.187 x 103 J (Table A.7, Appendix A) = 4.187 k1 1 W =1 J s-1 (Table A.S, Appendix A); therefore, 1 kW = 1 kJ 5- 1 (iii) Boundary conditions Att=O,T=To=15"C (a)

=

The general unsteady-state energy-balance equation is Eq. (6.10). For a batch system, Mj =Mo 0; also Ws = O. Energy is accumulated by the system in the form of sensible heat only; therefore:

dE=MC dT dt Pdt

where M is the mass of glycerol solution in the tank and T is its temperature, Substituting these expressions into Eq, (6.10) gives:

M has two components, glycerol and water, which have different heat capacities. Therefore, this equation can be written:

where Mw is the mass of water in the tank, MO is the mass of glycerol, CpW is the heat capacity of water, and CpO is the heat capacity of glyceroL

<IT

-

Answero(MWCpW+MOCpG)"dI =-Q (b) If

ii,

Cpw, CpO, MW and variables and integrating:

Mo are constant, T and t are the only variables in the differential equation. Separating

Using integration rule (0.24) from Appendix D and combining the constants of integration:

T=

-0.

t+K

(MwCpW+MGCpG)

From the initial condition for T, at t= 0, To:: K. Substituting this value of K into the equation gives:

69

Solutions: Chapter 6

Answer: T

-12

= (MwCpw+MOCpG) t+ To

(c) The mass of glycerol in the tank MO = 4S kg; the mass of water Mw terms:

Cpw = 7SAkJkgmor l

<)c-l',111~~~11

=55 kg.

Converting the Cp for water to mass

= 4.189 kJ kg-1 <)C- 1

Converting the Cp for glycerol to kJ: CpO

= 0.576kCalkg-loc-l.14;~:1 = 2.412kJkg-l oc-l

The rate of heat input to the system is 0.88 x 2.5 kW "'" 2.2 kW "'" 2,2 kJ s-l.From the sign conventions outlined on pp 87-88, Q must be negative as heat is added to the system; therefore, Q"'" -2.2 kJ s-l. Substituting the parameter values into the equation for Twith To 15<)C and T= 90<)C:

=

9O"C "'"

1 2.2kJst+ 15"C 55kg(4.189kJkg loC l)+45kg(2.412kJkg loe 1) t

= 1.l6x104 s = 1.l6X104S'13~sl "'" 3.2h

Answer: 3.2 h

6.11

Heating molasses

(i) Flow sheet and system boundary These are shown in the figure below.

System boundary~

- - -

-- -

I I Molasses solution in 1020 kg h- 1

20'C

-

1 I I

I

I

Heating tank

I I I

-

/ - --

M T

- - -

I I I

Molasses solution out 1020 kg h- 1 T

- -

-Q (from steam)

(ii) Assumptions - no leaks - no evaporation - tank is well mixed; therefore the temperature of the molasses solution out is the same as in the tank - no shaft work - beat capacity is independent of temperature - negligible heat losses

70

Solutions: Chapter 6

- condensate from the steam leaves at saturation conditions Reference state Tref"" 20oe; H"" 0 for molasses solution at 200 e (iv) Extra data

(iii)

Cp "" 0.85 kcal kg~l °C- 1 1 psi = 6.895 x 103 Pa (fable A.S, Appendix A) = 6.895 kPa Converting 40 psi to kPa:

40psi

= 40PSi.16.8;;~al

= 275.8kPa

The temperature of saturated steam at 275.8 kPa interpolated from Table C.2 (Appendix C) "" l30.7°e. (v) Boundary conditions " Att=O, T=To=20°C (vi) 'Total mass balance As the mass flow rates into and out of the tank are the same, the mass of molasses solution in the tank M is constant

and equal to 5000 kg at all times. (a)

The general unsteady-state energy-balance equation is Eq~ (6.10). From the reference state, hi = 0; The value of h o relative to the reference state is equal to the sensible heat-absorbed by the molasses'solution between Tref and the exit temperature T. From Eq. (5.13): ho

Ws = O.

= ~ = Cp(T-Tref)

The rate at which the molasses solution is heated is given by the equation:

Q = -U A (Tsteam -1) From the sign conventions outlined on pp 87-88, Q must be negative as heat is added to the system. Energy is accumulated in the form of sensible heat only; therefore: dE=MC dT <It P <It

Substituting these expressions into Eq. (6.10) gives:

dT

-

M Cp dt = -Mo Cp (T - Tref) + U A (Tsteam -

n

After rearranging, the differential equation is: dT = UATsteam+MoCpTref _(fJoCp+UA)T dt MCp MCp

dT

Answer: dt

(b) As U, A, Tsteam, Mo. Cp • Tref and M are all constant, Tand t are the only variables in the differential equation derived in (a). Substituting the known numerical values for the parameters:

dT _ 190kcal m-2 .-1 0C- (1.5 m2) (130.7'C) + 1020 kg .-1 (0.85 kCalkg-1 oC- ) (20"C) ' ' <It 5000kg(0.85kcalkg lo C 1) _(1020 kg .-1 (0.85 kcal kg- 1 oC- ) + 190 kcalm-2 .-1 oC' (l.5 m 2)) T ' 5000kg(0.85kcalkg loC 1) dT dt

= 12.84-0.271 T

where T has units °C and t has units h. Separating variables and integrating:

71

Solutions: Chapter 6

= eIt

dT

12.84-0.271 T

f

dT

-felt

12.84-0.271 T -

Using integration rules (D.28) and (D.24) from Appendix D and combining the constants of integration: -1 0.271 In (12.84-0.2711)

= t+ K

From the initial condition for T, at t = 0: -I = 0.271 In (12.84-0.271 To) Applying the numerical value of To =20째C, K =-7.395. Substituting this value for K into the equation gives:

K

-I 0.271In(l2.84-0.2711)+7.395 -I Answer: 0.271 In (12.84- 0.2711) + 7.395 =

=t

t

(c)

Values of t corresponding to various temperatures in the tank can be calculated from the equation derived in (b).

Time, t

Temperature, T

CC)

(h)

20

0.00

25

0.74 1.68 2.93 4.84 9.01 11.0 15.8 21.5

30 35 40 45

46 47 47.3

The results are shown in the figure below.

50

r----,----,----,,----,-----,

20 ..-_ _-'-_ _..L_ _-'-_ _-'-_ _- '

o

5

10

15 Time (h)

20

25

Solutions: Chapter 6

72 (d)

From the equation derived in (b), as the logarithm of zero and negative numbers is not defined (p 413), the theoretical maximum temperature that can be achieved in the tank occurs when 12.84 - 0.271 T= 0; Le. when T= 47.4°C.

(e) 'The temperature cbanges constantly with time; therefore, strictly speaking, there is no steady state. For practical purposes, however, the temperature approaches a constant value after about 16 h.

Answer. About 16 b (f) From the calculation table in (c), the temperature reaches 40°C after 4.84 h.

Answer: 4.84 h-

6.12

Pre-heating culture medium

The system is the glass fermenter containing nutrient medium. (i) Assumptions - no evaporation

-

fennenter is well mixed no shaft work heat capacities are independent of temperature between 15°C and 36°C beat losses are negligible

(li) Extra data Cp glass vessel"" 0.20 cal g-l 0C- l "" 0.20 keal kg-I oct Cp medium"" 0,92 cal g-l °C l "" 0.92 keal kg-loCi

1 W"'" 1.433 x 10-2 keal min-I (Table A.S, Appendix A)

(ill) Boundary conditions Att=O,T=To=15°C (iv) Energy balance The general unsteady-state energy·balance equation is Eq. (6.10). For a batch system., Mi = Mo = 0; also Ws =0.

Energy is accumulated by the system in the form of sensible heat only; therefore:

dE dT dt =MCpdt M has two components, the glass vessel and the medium, which have different heat capacities. Therefore, this equation can be written:

where My is the mass of the glass vessel, MM is the mass of the medium, Cpy is the heat capacity of the vessel, and CpM is the heat capacity of the medium. Substituting into Eq. (6.10) gives:

dT (MyCpy+MMCpM) dt =-Q As (i, Cpy, CpM, My and MM are constant, T and t are the only variables in the differential equation. Separating variables and integrating:

dT=

-Q

dt

(My Cpy + MM CpM)

f dT-f -

-Q

(My Cpy+MM CpM)

dt

Using integration rule (D.24) from Appendix D and combining the constants of integration:

73

Solutions: Chapter 6

-Q

T= From the initial condition for T, at t= 0, To

(MV CpV + MM CpM)

=K.

t+K

Substituting this value of K into the equation gives:

T=

-Q

(Mv CpV + MM CpM)

t+To

The rate of heat input to the system is 450 W. Converting this to kcaI min-l:

Q = -450W

=-450W.!1.433X lO;~kcalmin-l! =-6A5kcalmin-t •

Co

From the sign conventions outlined on pp 87-88, Q must be negative as heat is added to the system. Substituting this' and the other parameter values into the equation for Twith To = 15¢C and T= 36¢C:

36'C

=

l 6.45 kcal mint+ 150C 12.75 kg (0.20 kcal kg-I 'C- I) + 7.5 kg (O.92kcal kg-I 'C- I) t

= 30.8 min

Answer: 30.8 min

6.13

Water heater

The system is the tank containing the water. (i) Assumptions - no leaks - no evaporation - tank is well mixed - no shaft work - heat capacity is independent of temperature - condensate from the steam leaves at saturation conditions (li) Extra data Cp water 75.4 J gmol~1 oC-L(Table B.3, Appendix B) 75.4 kJ kgmol-t 0C- 1 Molecular weight of water (Table R.l, Appendix B) 18.0 1 kcal = 4.187 x 103 J (fable A.7, Appendix A) 4.187 kJ (iii) Boundary conditions Att=O, T=To =24"C (iv) Energy balam:e The general unsteady-state energy-balance equation is Bq. (6.10). For a batch system, Mi=Mo =O;a1soWs =O. Energy is accumulated by the system in the form of sensible heat only; therefore:

=

=

=

dE

dt

=

dT

= MCPdt

where Mis the mass of water in the tank and Tis its temperature. Substituting these expressions into Eq. (6.10) gives:

dT

-

MCpdi=-Q (aj There are two components to

Q: the rate of heating from the steam, and the rate of heat loss to the surrounding air: Q = UtAt (T-Tair)-UZAz(Tsteam -1)

This equation reflects the sign conventions outlined on pp 87-88: the term for the heat loss to the atmosphere is positive to indicate heat removal from the system, while the term for heat input from the steam is negative.

Substituting into Eq. (6.10) gives:

Solutions: Chapter 6

74

MC dT Pdt

dT

Cit

UZA2(Tsteam-1) - VtA I (T-Tair)

=

MCp

As Vt. A 1> Uz. AZ, T steam, Yair, Cp and M are all constant, T and t are the only variables in the differential equation.

Substituting the known parameter values:

dT

=

220 kcal m-2 h- I 0c- l (03 m 2)(130-1)째C _ 25 kcal m-2 h- I 0C-l (O.9m 2) (T - 20Y'C

n

( - 1 "'C-1 . 1118.0kg kgmOllllkCai . 4.187k:JU

dt

lOOOkg 75.4kJkgmol dT

dt = 9.026-0.088 T

where T has units of"C and t has units of h. Separating variables and integrating: dT

9.026 - 0.088 T

f 9.026~.088T

=dt

=

f

dt

Using integration rules (0.28) and (D.24) from Appendix D and combining the constants of integration:

0~81n(9.026-0.0881) = '+K From the initial condition for T. at t= 0: -I

K = 0.088 In (9.026-0.088 ToJ

Applying the numerical value of To =24"C, K = -21.97. Substituting this value for K into the equation gives: -1

0.088 In (9.026-0.0881) + 21.97 = , From this equation, for T = 80"C, t =14.2 h.

Answer: 14.2 h (b)

If beat losses can be neglected,

Qhas only one component and Eq. (6.10) becomes: dT MCp <It

dT

Cit

=

UZAZ (Tsteam -1) MC p

Substituting the known numerical values: dT

dt

220kcalm-2 h- 1 0C-l (0.3m 2)(130-n"C

= -1-000-'k=g7(7'=S=.4"'kJ=-k-'gm=-OI--I"'o-c--"'1'11'i~';;;=~=k0:i;TI'I"',/':/i::8~::C~"-;;:~ dT dt

= 8.576 - 0.066 T

where Thas units of "C and t has units ofh. Separating variables and integrating:

75

Solutions: Chapter 6

dT

= <It

8.576- 0.066 T

f

8.576

~.066 T = f <It

Using integration rules (0.28) and (D.24) from Appendix D and combining the constants of integration: -1

0.066 In (8.576-0.0661)

= t+K

From the initial condition for T, at t= 0: K

-1

= 0.066 In (8.576-0.066 ToJ

Applying the numerical value of To =24째C, K =-29.47. Substituting this value for K into the equation gives: -1 0.066 In (8.576-0.0661) + 29.47

=t

From this equation, for T = 80째C, t "'" 11.4 h, Therefore, from (a), the time saved is (14,2..:,- 11.4) h"'" 2.& h, which corresponds to 20% of the time required when heat losses occur.

Answer:

2.8 h, or 20% of the time required when heat losses occur

Fluid Flow and Mixing 7.1

Rheology of fermentation broth

(a)

The rheogram is obtained by plotting shear stress againSt shear rate.

700

...---r---,---,,--,.----r---,

600 600

400 300

200 100

ol-_...L_......l._---''-_-'-_....L_--' o

200

400

600

800

1000

1200

Shear rate (s路1) (b)

The rheogram in (a) is similar to those for pseudoplastic and Casson plastic fluids in Figure 7.7. It is not clear from 0 or not; therefore. both non-Newtonian models are worth the rheogram whether the fluid exhibits a yield stress at checking. From the equation in Figure 7.7 for pseudoplastic fluids, a plot of shear stress versus shear rate on log-log

r=

coordinates would be expected to give a straight line. This plot is shown below.

100

1000

Shear rate (s-l)

10000

Solutions: Chapter 7

77

'1:=

11.43 ;-0587. Therefore, the flow behaviour index n = The equation for the straight line in the log-log plot is 0.587 and the consistency index K =. 11.43 dyn sit cm- 2. The sum of squares of the residuals for this data fit is' 1137.

From the equation in Figure 7.7 for Casson plastic fluids, a plot of the square root of shear stress versus the square root of shear rate on linear coordinates would be expected to give a straight line. This plot is shown below.

The equation for the straight line in the plot is liZ =. 5.93 + 0.648 ytl2.. From the equation in Figure 7.7, this means that K p = 0.648 dyn 11z Sllz cm- l and the yield stress 't{) = 35.2 dyn cm- 2. The sum of squares of the residuals for this data fit is 8986.

Comparison of the residuals from the two models suggests that the equation for a pseudoplastic fluid is the better fit Answer. Pseudoplastic fluid: n = 0.587; K = 11.43 dyn sn cm- 2 (e)

The apparent viscosity for a pseudoplastic fluid is given by Eq. (7.8) and can be calculated using the parameter values determined in (b).

(I)

Answer. 3.7 dyn s cm- 2 (Ii)

J1a

= Kyn-l = 11.43dynsO.587 cm-2 (200 S_I)O.587-1 = 1.3dynscm-2

Answer: 1.3 dyn s cm-2

7.2

Rheology of yeast suspensions

From Eq. (7.8), for pseudoplastic fluids, a plot of apparent viscosity versus shear rate on log-log coordinates can be expected to give a straight line. Log-log plots for the different cell concentrations are shown below.

78

Solutions: Chapter 7

10 Cen concentrations: .1.5% 03% D 6%

0::-

.£.

••t

L

• 10.5%

I

• t3-

~

"

1 1

10

100

Shear rate (8- 1)

100 Cell concentration: 12%

0::-

.£.

t

••E

e

!l

~

10 1

10

100

Shear rate (sol)

,

1000 Cell concentration: 18%

0::-

.£. ~

8

.~ 100 ~ E

....

...

I

10 1

10 Shear rate (s·l)

100

Solutions: Chapter 7

79

Cell concentration: 21%

•

f

I

10 Shear rate (5- 1)

100

The equations and parameter values for the straight lines in each plot are listed below.

Cell concentration (%)

Flow behaviour index, n

Equation

I', = 1.5

1.5

Consistency index, K (cp sn-l)

r·0 = 1.5

1

1.5

·0

1',=2.0r =2.0 Pa = 2.91 r-:O,050

1

2.0

0.95

2.9

10.5

I', = 5.38

<>.07.

0.92

5.4

12

0.61

18

Jl.a = 50.1 -:0.395 J.la = 162 7-:0·307

21

Jla = 833 7-:0.251

0.75

3 6

r..

r

50

0.69

K and n are plotted as a function of cell concentration below.

1.4,--,.---..--....--"..---..... index, n • Flow behaviour

o

c

~. -" 5

! !

£

Consistency, index, K

1.2

1.0

0.8 10 0.6 0.4 '--_ _..J....._ _.....J.._ _- ' ' -_ _- ' -_ _.....Il

o

5

10 15 Cell concentration ('Yo)

20

25

Solutions: Chapter 7

80

The cell broth is Newtonian up to a cell concentration of about 2%, then becomes pseudoplastic. The flow behaviour index continues to decrease until a cell concentration of about 12% is reached. The consistency index rises !broughout the culture with increasing cell concentration.

7.3

Impeller viscometer

(a) If the rheology can be described using a power~law model, a plot of shear stress versus shear rate on log-log coordinates can be expected to give a straight line. Values of shear stress and shear rate can be determined from torque and stirrer speed data using Eqs (7.11) and (7.12) with k =10,1 and Dj, "'" 4 em "'" 4 x 1<r2 m. The results are

listed below and plotted below.

Stirrer speed (s·l)

Torque (Nm)

Shear stress (N m- 2)

Shear rate (gw 1)

0.185 0.163 0.126 0.111

3.57 x 3.45 x 10-6 3.31 x 10-6 3.20 x 10.6

0.0559 0.0540 0.0518 0.0501

1.89 1.66 1.29 1.13

0.07 0.06

i

0

0.05

~

11~

e

"m"

0.04

~

<J)

0.03'1

..L...

1.5

-'

2

Shear rate (s-1)

The equation for the straight line on the log-log plot is 1'=O,049jr°·20. From Eq. (7.7), this means that the flow behaviour index n is 0.20 and the consistency index K is 0.049 N sn m- 2.

Answer: Yes: n = 0.20; K = 0,049 N s1l m-2 (b) The impeller Reynolds number for pseudoplastic fluids is defined in Eq. (7.23). If Rei at the highest stirrer speed is in the laminar range, conditions at the other stirrer speeds are also laminar. Therefore, for Ni = 0.185 s·1 and using the unit conversion factor 1 N 1 kg m s·2 (Table A.4, Appendix A):

=

N'~ D' P

Rei

(0.185,-1)2-0.20 (O.04m)' l000kg m-3

= ~1 ~ = -->=="(r'---"==-"IC"':""-"'-:"O,,, = 10,0 10.20.20-1 0.049NsO.20m-2. lk~~S-

This value for Rei is at the upper limit of the laminar range (p 137); therefore, we can conclude that flow at this and lower stirrer speeds is laminar.

Answer: Flow is laminar.

Solutions: Chapter 7

81

(oj

From Table 7.4, k = 30 for a helical-ribbon impeller. From Figure 7.25, flow with this type of impeller is laminar up to about Rei = 103. These conditions allow use of higher stirrer speeds within the laminar region than is possible using a turbine impeller. Solving for Ni from Eq. (7.23) and substituting values of n and Kfor the fluid from (a), the maximwn stirrer speed at Rej = 30 can be calculated as:

N?--n

= Rei ~1 K

1

!if..".20 =

vf P

-21

103 (30o.zo-1) (0.049 N sO.20 m

1 kgms-2 ~ 1N

II

(0.04m)21000kgm 3

I

Nl- 8= 2.02s-1.8 Ni = 1.48 s-l

r=

Therefore, from Eq. (7.11), as 30Ni for a helical-ribbon impeller, shear rates up to 30 x 1.48 = 44 s-1 can be used.

This is a considerable improvement on the restricted range-of.up to about 1.9 s-l as determined in (a) for the upper limit of laminar flow with a Rushton turbine impeller.

Answer: The shear rate range can be extended from 1.9 s-1 with a Rushton turbine impeller to 44 s-1 with a helicalribbon impeller.

7.4

Particle suspension and gas dispersion

The Zwietering equation is an equation in numerics (p 12); therefore, the parameter values used in the equation must have the units specified. V p =10 J.llll =10 xlo-6 m; g =9.8 m s~2 (p 16); Vj =30 em =0.3 m; PL =density of water = 1000 kg m- 3 . The dynamic viscosity of water J4... at 20"C is about 1 cP (p 133); from Table A.9 (Appendix A), this is the same as J4... = 1O~3 kg m- I s~I. Using the definition of .kinematic viscosity from p 133: VL

f.lL 10-3 kgm- 1 8-1 =-= =10-u m2 s-1 PI. lOOOkgm-3

The density of the cells PP is 1.04 g cm-3. Converting to kg m- 3 ; -3 Iloocml3 . gcm.

Pp -_ 104 m . I~I_ lOOOg - lO4Okg m-3 Substituting values in the correct units into the Zwietering equation:

This minimum stirrer speed for suspension of the cells can be compared with the minimum stirrer speed for dispersion of air bubbles. Taking the average minimum tip speed of 2.0 m s-1 for bubble dispersion: N 1

= tipspeed = 2.0ms- I = 2.1 s-1 1t Vi

1t (0.3

m)

The stirrer speed for air dispersion is 3.7 times higher than for ceIl suspension. As power in the turbulent regime is proportional to the stirrer speed cubed, about 50 times more power is required to disperse the air bubbles than to keep the cells in suspension.

Answer: Bubble dispersion requires significantly more power than cell suspension.

Solutions: Chapter 7

82

7.s

Scale-up of mixing system

(aJ The Reynolds number for a Newtonian fluid in a stirred vessel is defined in Eq. (7.2). Using this equation to evaluate the maximum viscosity jJ. under turbulent conditions with the density of water p= 1000 kg m- 3:

I.t

=

D~ P 1 I

.1

(SOOmin- 1 160miDll(5 cm)2

N

Rei

=

S

U

2

) 1000 kg m-3 .11001m1 em

= 33. x 10-3 kgm-1-1 s

104

From Table A.9 (Appendix A), 1 .kg m- I s路1 = 103 cP; therefore:

-1 s -1.

33 10-3kg m J.l=.x

3 10 cP l l = '33 cP

1kgm- s-

Answer: 3.3 cP

(bJ The tip speed in the laboratory equipment is: Tipspeed = 7tNj Di

= 1拢 (800min-1 -I 16:

sD

1)(5 em .11~~ml) = 2.1 m

1

8-

In the large-scale vessel. Di =15 x 5 em =75 em. If the tip speed is kept the same after scale-up: M.

tip speed _ _ 1tDj

_

'" _

2.1 ill s-1 1m

( 1t

75cm.l lOocm

l

_ 089 -1

) _ .

s

Using this value of Ni to calculate the maximum viscosity as in (a):

N:,.D,~p O.89S-I((75Cm)2路1100Im 1)lOOOkg m-3 2

JJ.=

Rei

=

.em

10

=O,05kgm- l s- 1

Converting to cP:

= 50cP Answer: Scale-up increases the maximum viscosity for turbulent conditions to 50 cP.

7.6 (aJ Dj (1)

Effect of viscosity on power requirements

=1 m;p= 1 gcm-3 =1000 kg m-3.

The viscosity Of water at 20"C is about 1 cP (p 133); from Table A.9 (Appendix A), this is the same as lO-3 kg m- 1 s-l. Substituting parameter values into Eq.(7.2): NiDrp Rei

=

Ji

=

(90min-I路116~sn~(lm)21000kgm-3 lO

3

11

kgm- C

=

6 1.5 x 10

From Figure 7.24 for a Rushton turbine, this value of Rei corresponds to turbulent flow and The power required is evaluated using Eq. (720): P =

Np can be taken as 5.8,

NppNf Dr = 5.8 (lOOOkg m-3)(90 min- I .II:sn~3 (l m)5 = 1.96x 104 kg m2 s-3

83

Solutions: Chapter 7 From Table A8 (Appendix A), 1 W =I kg m 2 s-3; therefore: P = L96xl04w

= 19.6kW

Answer: 19.6 kW (ti)

For a viscosity of 100 x 10-3 kg m- 1 s-1 =0.1 kg m- 1 s-l, Rei is: NiDf p

Ret' =

(90min-

1

.11~n~(1 m)21000kgm-3

=

#

1

0.1 kgm- s-

1

4 = 1.5 x 10

From Figure 7.24 for a Rushton turbine, this value of Rei is still within theturbulent regime so that Np isagain5.8.

Therefore, the power required is the same as that calculated in (i): P =19.6 kW. Answer: 19.6 kW (iii)

For a viscosity of 104 x 10-3 kg m- 1 s-l = 10 kg m- 1 s-l, RCi is: NiD[ p ~=

J.l

(90 min=

1

.11=~(l m)2 1000kgm-3 IOkgm1s1

=1~

From Figure 7.24 for a Rushton turbine, this value of Rei is within the transition regime. Np read from Figure 7,24 is, about 3.5. The power required is evaluated using Eq. (7.18):

P = NppNf Dr = 3.5 {tOOOkg m-3) (90 min- 1

.! 16~n~3

(l m)5 = 1.18 x 104 kg m2 s-3

From Table A8 (Appendix A), 1 W = 1 kg m2 s-3; therefore: P = L18xl04W = 11.8kW

Answer: 11.8 kW (b)

From Figure 7.24 for a Rushton turbine, turbulence with Np =5.8 is achieved at a minimum Reynolds number of about 104. For a viscosity of 1000 x 10-3 kg m- 1 s-1 = 1kg m- 1 s-l, the stirrer speed required can be detennined from Eq. (7.2),

Using this result in Eq. (7.20):

P= NppNI D; = 5.8 (1000 kg m-3j(1O s-lj3 (I m)S = 5.80x 106 kg m2 s-3 From Table A8 (Appendix A), 1 W = 1 kg m2 s-3; therefore: P = 5.80 x 106 W = 5.80 x 1()3 kW

Answer: 5.80 x 103 kW

84

7.7

Solutions: Chapter 7

Electrical power required for mixing

Di =- 7 em =- 0.07 m; p =- 1000 kg m~3. The viscosity of water at 20°C is about I cP (p 133); from Table A.9 (Appendix A), this is the same as 10.3 kg m- 1 8. 1, Substituting parameter values into Eq. (7.2):

Rei =

2 NiDi P

--'-cC:-C- = J.l.

(900' -1 11minn(007 )2 1OOOk -3 nun. 60s U . m gm 4 3 I I = - 7Ax 10 10- kgm s

From Figure 7.24 for a Rushton turbine, this value of Rei corresponds to turbulent flow and N~ can be taken as 5.8.

The power required is evaluated using Eq. (7.20): p

= NpPNf Dr = 5.8 (1000 kg m-3) (900min-1 .11~~3 (O.07m)5

= 32.9kgm2 s-3

From Table A,S (Appendix A), 1 W =- 1 kg m2 8-3; therefore: P = 32.9W This value is considerably lower that the electrical power consumed by the stirrer motor.' Much of the remainder of the electrical power is converted into heat within the motor bousing.

Answer: 32.9 W; a significant fraction of the electrical power is dissipated as beat within the motor housing

7.8

Mixing time with aeradon

Di = 0,67 m; p = density of water = 1000 kg m- 3. From TableA.9 (Appendix A). I cP = 10.3 kg mol s-l; therefore. Jl = 4 cP = 4 x 10,3 kg m~I s'l. For a cylindrical tank of diameter Dr= 2 m and height H= 2m. the volume Vis:

V = 7t{;Y H = 7t{2;t 2m = 6.28m

3

(aJ If the maximum specific power consumption is 1.5 kW m'3. the maximum power Pis: P = (L5kWm-3)6.28m3 = 9.42kW From Table A.8 (Appendix A), I W = I kg m 2 s'3; therefore 1 kW = 1000 kg m2 s-3. ConvertingP to kg m 2 s·3:

P = 9.42kW =

9A2kw.ll°oo~-:2s-31 = 9.42 x 103 kg m 2 C 3

Assume for now that the fluid flow is turbulent and Np = 5.8. The stirrer speed can be evaluated using Eq. (7.20):

N:= ,

P

NppDi

=~42xI03kgm2s-3

5.S (1000 kg m-3) (O.67m)5

=120s-3 .

Ni = 2.29 sol = 137 rpm

Check that this stirrer speed provides turbulent mixing conditions by evaluating the Reynolds number. Rei for Newtonian fluids in a stirred vessel is defined in Eq. (7.2): Rei = NiDf P = (2.29s-l)(O.67m)21000kgm-3 = 2.6x 105 j.l

4 x 10 3 kg m- I s 1

From Figure 7.24 for a Rushton turbine. this value of Rei is well into the turbulent regime; therefore the value for ~ assumed above is valid.

For high Rei> the mixing time can be calculated using Eq. (7.16):

85

Solutions: Chapter 7

= 1.54 V = 1.54(6.28m') Nj (O.67m)3 2.29S-1

t

D;

m

= 14s

Answer: The maximum allowable stirrer speed is 2.29 s-1 or 137 rpm; the mixing time is 14 s. (b)

The ungassed power number was 5.8; therefore, the power number with gassing (Np)g =0.5 x 5.8 = 2.9. The stirrer speed which delivers the maximum power P= 9.42 x 103 kg m2 s-3 can be evaluated using Eq. (7.18):

Nt =

P

5

=

(Np) pD; g

3 2 3 9,42x10 kgm s- 241 -3 2.9(lOOOkgm-'j(O.67m)5 . s

Nj = 2.89 s路1 = 173 rpm

The mixing time evaluated using Eq. (7.16) is: t m

= !541' = DfN

1

1.54(6.28m') (O.67m)3 2 .89s-1

= II s

Answer: The maximum allowable stirrer speed is 2.89 s-1 or 173 rpm; the mixing time is 11 s.

Heat Transfer 8.1

Rate of condnction

(a) B::::: 15 em "" 0.15 m; liT"", qoo -.80) ::::: 620"C. which is equal to 620 K as temperature differences are the same on the Celsius and Kelvin scales (p 18). The rate of heat conduction can be calculated using Eq. (8.10):

l

1

2

Q = kA !'J = (0.3 Wrn- K- ) Urn (620K) = 1860W = 1.86kW B

a.15m

Answer: 1.86 kW (b)

In this case there are two thermal resistances in series. Their magnitudes are calculated using Eq. (8.15). For the firebrick:

For the asbestos, B2 ::::: 4 em ::::: 0.04 m, so that: RZ::::: 82:::::

a.04m

(o,tWm lK 1)1.5m2

k2A

::::: O.27KW-1

~ Therefore, the total wall resistance RW= RI + RZ ::::: (0.33 + 0.27) K W~ I ::::: 0.60 K W~l, For thennal resistances in series, the rate of heat conduction is calculated using Eq. (8.14):

Q=

I1T =

Rw

620K = 1033W= 1.03kW O.60KVI

Answer: 1.03 kW

8.2

OveraU heat-transfer coefliclent

The overall heaNransfer coefficient is calculated using Eq. (8.24) with hfh::::: 830 W m~2 K-I, hh::::: 1.2 kW m路 2 K-I, B =6mm::::: 0.006 m, k= 19 Wm,l K-l, he"" 1.7 kWm w2 K-l and hfc= O.

1 lIB 1 1 - = -+-+-+-+U

1. = U

1

1 830Wm 2r-

+

1

hfh

hh

1.2kWm-2r-1.ll~:1

k

+

he

hfc

O.OO6m

19Wm IK-

i, = 2.94 x 10-3 W-l m2 K U= 340Wm路2 K-I

1

+

1

1.7kwm-2K'"1.11~:1

+0

87

Solutions: Chapter 8

8.3

Effect of cooling-coH length on coolant requirements

(a)

The steady~state energy~balance equation for the cooling~coil is Eq. (8.32):

Q = McC",(T",-Tct) = (0.5kgs- I)4.18kJkg- l oc- l (15-S)"C = 14.6kJs-1 Answer. 14.6 kJ sol (b)

The mean temperature difference between the fermentation fluid and the cooling water is calculated from Eq. (8.35): /iTA:

2 Tp-(Tct+Tco) (2 x 35)OC-(S + 151'C 2 : 2 : 23.5OC

Answer. 23.5"C (c) U A is evaluated using Eq. (8.19) with /iT: /iTA: UA

=

Q

ATA

kJ s-l = 14.6 23.5"<:

(d)

UA': 1.5UA: 1.5xO.62kJs- I "C-): 0.93kJs- I0C- 1 Answer: 0.93 kJ s'} °C·l (c)

Applying Eq. (8.19) to determine /iT: /iTA for the new coil:

!iTA :

-iL :

U A'

1 14.6 kJ s1 0.93 kJ s-l °C

= 15.7"<:

The new cooling water outlet temperature Tco is determined using Eq. (8.35):

Answer: 30.6"C (f)

From Eq. (8.32), for the new coil with

M -

Q: 14.6kJ s-l, Tci = goC and Teo =30.6"C:

-

Q

-I

-

14.6kJ, _ 015k -I c - C",(Tco-Tct) - 4.18kJkg-IoC I(30.6- Sl'C - . gs

Therefore, installation of the new coil allows a 70% reduction in cooling-water requirements; Answer: 70%

8.4

Calculation of heat~transferarea in fermenter design

=

=

=

=

=

Ni: 80 rpm 80'60 1.33 s·l. Cp culture fluid 4.2 kJ kg· l °c l : 4.2 x 1()3 J kg'} "c l . Cpc Cp water 75.4 J gmol'} "C- I (Table B.3, Appendix B) 75.4 kJ kgmol-l "C- l . Converting Cpc ta a mass basis using the molecular weight af water 18.0 (Table B.I. Appendix B):

=

Cpc

=

= 75.4 kJ kgmol-l "C- l

: 75.4kJkgmor l

"c-I.1 ~~~ll

;:: 4.19kJkg- 1 "C- l

10.3 N s m· 2; from Table A.9 (Appendix A), I N sm·2 =} kg m· l s-l; therefore Jlb = 10-3 kgm- l s·l. ktb = 0.6 W mol "e l ; from Table A.8 (Appendix A), 1 W = 1 J s·l; therefore k fb 0.6 J 8"1 m- l "C-l. B: 6 mm;:: 0.006 m.

,llb;::

=

Solutions: Chapter 8

88

From p 97, as the heat of reaction for an exothermic reaction is negative, Aiirxn =-2500kW. From the modified energy-balance equation, Eq. (8.33), when evaporation and shaft work can be neglected. Q=-Mrxn =2500 kW =2.5 x 106 W, According to the sign conventions on pp 88-89, Q positive is consistent with heat being removed from the system. From Table A.8 (Appendix A), 1 W

=1 J 8. 1; therefore Q= 2.5 x 106 J s·1 =2500 kJ s-I,

(a) The fermenter-side heat-transfer coefficient can be evaluated using the empirical correlation, Eq. ·(8A5). The dimensionless nwnbers in this equation are Rei, Pr and Nu. Rei is given by Eq. (8.39):

Pr is given by Eq. (8AO):

From Eq. (8.45):

Nu

= 0.87 Re?,62 pr033(::fI4 = O.87(3.84X 106)°.62 (7.0)0.33 (1)0.14= 2.0x 104

From the definition of Nu in Eq. (8.37): Nukfb

h=--= D

2.0x 104 (0.6 Wm- l °C-1) 5m

= 2.4 X 103 Wm-2 0C; I

Therefore, as the fennenter fluid is the hot fluid, hh = 2.4 x 103 W m-2 °e 1.

Answer: 204 x 103 W m-Z oC- 1 (b) The overall heat-transfer coefficient in the absence of fouling layers can be calculated using Eq. (8.23). Assuming that the heat-transfer coefficient for the cooling water h c can be neglected (p 185):

1- = .!.+~ = V

hh

k

+

1 2AxI03Wm 20(;"1

U = L40x

O.OO6m

= 7.17xlO-4W-1mZoC

20Wm-1OC 1

103 wm-Zoe 1 =

1.40kWm-20e l

From Eqs (8.21) and (8.22), the beat-transfer resistance due to the pipe wall is B/k A; from Eq. (8.20), the total resistance to beat transfer is l/Vk Therefore, the proportion of tbetotalresistance,duetothe-pipe wallis:

BfkA = BU = O.OO6m (L40 x 103 Wm-2 oc-1j = 042 l/VA k 20Wm 10C-1 .

Answer: lAO kW m- 2 °C- 1. The pipe wall contributes 42% of the total resistance to heat transfer. (c) The energy-balance equation for the cooling water, Eq. (8.32), relates the cooling water flow rate, the inlet and outlet temperatures and the rate of heat transfer. From this equation, an expression for the outlet cooling-water temperature is:

89

Solutions: Chapter8

where Tco has units of °C and Me has units of kg b· 1. An expression for the mean temperature difference between the fermentation fluid and the cooling water can be determined from Eq. (8.35); _ 2 Tp-(Tci + Teo} _ (lx30)OC-(1O+Tco}OC _ 6O"C-10"C-Tco _ Teo ATA 2 2 2 - 25- 2 where !!TA and Teo have units of °C. Substituting this expression into Eq. (8.19) for evaluation of A with AT= ATA and the value of Ufrom (b);

=~ =

A

U!1TA

2500kW 1.40kWm-Zoc-1(25_

T;'fe

= 1.8x 103 25-

T~

where A has units of mZand .Teo has units of °C. The above equations for Teo and A allow evaluation of these parameters as a function of Me' The results for several values of Me between 1.2 x loS kg b· 1 and 2 x 106 kg b· 1 are listed and plotted below.

M,(kgb- I )

Tco (0C)

A (m 2)

1.2 x loS 2.0xloS 3.0x loS 4.0x loS 8.0x loS 1.0 x 106 2.0 x J(j6

27.9 20.8 17.2 15.4 12.7 12.2 ILl

163 123 1I0 104 97 95 93

30,.....---..,.,.....---.,..-----.,..-.----,200 41

• Temperature o Area

o'o

J-

-'-

5

10

Cooling-water flow rate,

...J''--_ _--' 50 15

Me

20

x 1Q-5 (kg h· 1)

(d) From the,equations developed in (c), at a cooling.water flow rate of 5 x loS kg h*l, Teo 14.3°C and A = 101 .tn.2.

The area A of a cylindrical cooling·coil is equal to 2 1t r L, where r is the cylinder radius and L is its length. For a radius of 5 em 0.05 m:

=

=

A

L

101 m 2

= 21tr = 21t(0.05m)

= 321 m

Answer: 321 m. This is a long cooling·coil, representing a considerable expeme whenfabricatedfrom stainless steel.

90

Solutions: Chapter 8

8.5

Effect of fouling on heat-transfer resistance

(a) Cpc Cp water = 75.4 J groat-! 0C"l (Table B.3, Appendix B):o 75.4 kJ kgmo}路l 掳C~l, Converting to mass terms using the molecular weight of water = 18.0 (Table B.I. Appendix B):

=

Cpc

= 75.4 kJ kgmorl 'C-l = 75,4 kJ kgmorl

'c-ll i~~11 =

l I 4,19kJ kg- 'C-

From Eq. (8.32), the rate of heat removal to the cooling water before cleaning is:

= 20 kg s-I (4,19kJkg- 1 'C- I) (2S-12)'C = 1340,SkJs-1

Q = McCpc(Tco-Tci)

The meantemperatute difference between the fennentation fluid and the cooling water can be calculated from Eq. (835),

LiTA

=

2 Tp-(Tci + Tco) 2

=

(2 x 37)'C-(12 + 2S)'C 2

= 17'C

The area A ofthe cylindrical cooling~coi1 is equal to 2 1t r L, where r is the cylinder radius and L is its length, For r=

6cm = O.06mandL= 150m: A

= 2 1t (0.06 m) (150 m)

:0

56.5 m 2

=

Evaluating the overall heat-transfer coefficient U from Eq. (8.19) with aT aTA: -

1

- 1340.8kJs- -140kJ -1 -Z'C-1 - -Q- U sm ALiTA (565m2)I7'C '

(b)

If the fermentation temperature is maintained the same after cleaning the cooling-coil, the rates of heat removal Q before and after cleaning must be equal. The new outlet cooling~water temperature can be calculated from Eq. (8.32):

+ 120C = 36.6OC

1340.8kJ s-1

13kgs-I(4,19kJkg-l,C I)

(e) .The mean temperature difference between the fermentation fluid and the cooling water after cleaning is detennined from Eq, (835),

1J.TA

=

2 TF-(Tci + Too) 2

(2 x 37)'C -(12 + 36,6)'C

= _.

2

= 12.7OC

The beat-transfer area is the same as before cleaning: A = 56.5 m2 . Evaluating Ufrom Eq. (8.19) with AT= tiTA:

U

-

=~ = ALiTA

,

1340.8kJs-

(565 m2) 12,7'C

= 1.87kJs-1m-20c-1

The overall beat-transfer resistance is given by Eq. (8.20):

RT = _1_

UA

=

I

L87kJs-'m 2'C-l (S65m 2)

= 9.46 X 10..,3.k]""1 s 0C

This is the beat-transfer resistance after cleaning the cooling coil. 'The resistance before cleaning is calculated from Eg. (8.20) using the value of U from (a):

91

Solutions: Chapter 8 The resistance due to the fouling deposits is equal to the difference between these two overall resistances 10-2 _ 9.46 x 10"3) = 3.14 x 10-3 kJ- I s 0c. Therefore, fouling contributes: 3.14x 1O-3 kr 1 soC ==~=:T'-:':: L26x 10 2 kri soC

=:

(1.26 x

= 0.25

of the total resistance before cleaning.

Answer. 0.25

8.6

Pre-beating of nutrient medium

=

Cph =: Cpc =: Cp water =: 75.4 J gmol"l °C l (Table 8,3, Appendix B) 75.4 kJkgmot l °C l . Converting to mass terms using the molecular weight of water = IS.O(Table B.I, Appendix B):

Cph

= Cpc = 75.4kJkgmorl 0c-1

=:

75.4kJkgmorl OC-1

.1 ;;~~ll

= 4.19kJkg- 1 °C-1

P medium =: p water =: 1000 kg m"3. The viscosity of water at 20°C is about 1 cP (p 133); from Table A.9 (Appendix A), this is the same as 10"3 kgm- I s-l; therefore,,u medium =:,u water =: H}"'"3 kg m-I s-l. kfb medium = 0.54 W m- 1 °C-1; from Table A.S (Appendix A), this is the same as 0.541 5""1 m~l 0C- 1.;tberefore, kfb medium =: 0.54 x 10-3 kJ s-l m- 1 °C l . From Table S.l, kfb water at 303 K= 0.62 Wm- 1 °C-l;from Table AS-{A-ppendix A), this is the same as 0.62 J s-l mol °C-1; therefore, ktb water =0.62 x 10-3 kJ 5- 1 ar 1 0C- I . From Eq. (2.24), 303 K =: approx. 30°C, which is close enough to the conditions in the heat exchanger for evaluation of thermal conductivity. k pipe wall = 50 W mol °C-I; from Table A.8 (Appendix A), this is the same as 50 J s-l mol °C-l; therefore, k = 50 x 10..3 kJ s-l m- l °C l . B =: 5 mID =: 0.005 m. (al From the definition of density (P-16), the mass flow rate of the medium is equal to the volumetric flow rate multiplied by the density:

The steady-state energy-balance equation for medium in the tubes of the heat exchanger is Eq. (8.32); therefore:

Q = M,Cpc(T,o-Td) = 13.9kg,-1 (4.19kJkg- 1 'C1)(28-1O)OC = 1048.3kJ,-1 Answer. 1048.3 kJ s-I (b)

The heat-transfer coefficient for the medium in the tubes of the heat exchanger can be calculated using the empirical correlation, Eq. (8.42). The parameters in this equation are Re, Pr and Nu. Re is given by Eq. (8.38). The linear velocity of the fluid u is equal to the volumetric fIowrate per tube divided bythecross.sectional area of the tube. The cross-sectional area ofa cylindrical tube =: 1£,.z where ris the tube radius = 25 cm =: 0.025m. Therefore:

u

=:

50

volumetricflowrate

=:

numberoftube,(n?)

m

3 h-1

I~I

. 3600s

=:

0.236ms-1

(30) n (0.025 ml'

=5 cm =0.05 m: 3 Re = Dup = 0.05m(0.236m,-I)IOOOkgm- = 1.18 x 104

Substituting parameter values into Eq. (8.38) with the tube diameter D

JIb

1O-3 kgm Is 1

This value of Re is within the range of validity of Eq. (8.42). Pr is given by Eq. (8.40):

_ Cpi'b _ 4.19kJkg- 1 "C 1 (10-3 kgm-I ,-I) Pr____ =78 k tb

0.54x1O-3kJs-1m-1oC I

This value of Pris also within the range of validity ofEq. (S.42). From Eq. (8.42):

.

Solutions: Chapter 8

92

Nu = 0.023 ReO.S PrOA = 0.023 (1.18 X 104)°.8 (7.8)0,4

= 94.6

From the definition of Nu in Eq. (8,37): _ Nukfb _ 94.6{O.54xlO-3 .kJs-1 m-1 oc-1) _ 102kJ -1 -2 "C-1 h D O.05m -. s m As the medium in the tubes is the cold fluid in this heat exchange system,

he =1.02 kJ s·l m-2 0C- l ,

The heat-transfer coefficient for the water flowing in the shell of the heat exchanger can be calculated using the empirical correlation, Eq. (8.44). The parameters in this equation are C, Remax, Prand Nu. As the tubes of the heat exchanger are arranged in line, from p 183. C 0.26. Remax is given by Eq. (8.38) with D equal to the outer tube diameterdetennined as the sum of the inner tube. diameter and the pipe wall thicknesses: D = 0.05 m + 2 x 0.005 m 0.06 m. Substituting parameter values into Eq. (8.38):

=

=

= Dup = 0.06m{O.lSms-1)tOOOkgm-3 = 9.00 x 103

Re max

10-3 kg m- l s-1

fib

As this value is > 6 x UP, Remax is within the range of validity of Eq. (8.44). Prior theshell·side water is given by

Eq. (8.40):

Therefore. from Eq. (8.44): Nu

= CRe~Z,Jr°.33 = O.26(9.00X 103}O.6 (6.8)°.33 = 115

From the definition of Nu in Eq. (8.37) with D the outside tube diameter:

h

Nukfb

=-D- =

115 (0.62 x 10-3 kJ s-l m- l OC- 1)

O.06m

= 1.19kJs-

1

2

m- "C-

1

As water is the hot fluid in this heat exchange system, hb = 1.19 kJ 5- 1 m- 2 "C-l.

(e)

The overall heat-transfer coefficient without fouling factors is calculated using Eq. (8.23);

~ = 1.92kJ1 sm 2 °C U

= 0.52 kJ s·l m~2 0C- 1

(d)

The outlet temperature of the water from the shell is detennined from Eq. (8.31):

~

- ~.

.Lho - .Lbl -

-

.....

Q

MhCph

-- 60~ ~

1048.3kJs-

-

3x

I

= 30~

104kgh-l.13~sl{4.19kJkg-loc-l)

~

The fluid flow directions and the inlet and outlet temperatures for a single~pass countercurrent shell-and-tube heat exchanger are represented graphically below.

Solutions: Chapter 8

93

j60 C 0

Tei

/

=- 10"C

Teo

/'

=- 28<>C

Heat exchanger Tho=-30"C

The temperature differences at the two ends of the exchanger are fl.Tl =- (30 - to) ::::: 20"C and /1T2 ::::: (60 - 28) ::::: 32"C. Substituting these values into Eq. (8.34) for the log-mean temperature difference: <IT. = LlTZ-LlTI

L

In(LlT2I<lTd

r

= (32- 20 e = 25.5OC In (32120)

Answer: 25.5"C (e) The heat-transfer area is determined from Eq. (8.19) with AT::::: /1TV A ::::: ~ :::::

U LlTL

I

1048.3k1s-

0.52kJ s-I m-2 oc- I (25.5OC)

::::: 79m2

Answer: 79 m 2 (f) The total area A of the tubes in a shall-and-tube heat exchanger is equal to 2 1t r L N, where r is the tube radius, L is the length of the tubes and N is the number of tubes. For r::::: 2.5 cm::::: 0.025 m and N::::: 30:

A 79m2 L::::: 21trN::::: 21t(0.025m)30 ::::: 16.8m From these results, LID::::: 16.8/0.05 ::::: 336, where D is the tube diameter. As this value is > 60, application of Eq. (8.42) used.to detennine the tube-side heat-transfer coefficient is valid.

Answer: 16.8 m

8.7

Suitability of an existing cooling-coil

Ni =- 50 rpm::::: 50/60::::: 0.83 s-l. Cph::::: Cpc::::: Cp water::::: 75.4 J gmol-l °C-l (Table B.3, Appendix B)::::: 75.4 k1 kgmol·l °C- 1. Converting to mass tenns using the molecular weight of water::::: 18.0 (Table B.l, Appendix B); Cph =- Cpc =-75.4kJkgmor l °C- 1 =- 75.4kJkgmor 1 "'c- l

.! ~::~1! : :

4.19kJkg-1 "'C-l

kfb fennentation fluid =-ktb water. From Table 8.1, kfb water at 303 K =- 0.62 W urI "'C-l; from Table A.8 (Appendix A), this is the same as 0.62 J s-l m- l oCl; therefore, kfb::::: 0.62 x 10-3 kJ sol m- 1 °C 1. From Eq. (2.24), 303 K =approx. 30°C, which is close enough to the conditions in the fennentation system for evaluation of thennal conductivity. p fennentation fluid::::: p water =- 103 kg m'3. # fennentation fluid =- # water. The viscosity of water at 20"'C is about 1 cP (p 133); from Table A.9 (Appendix A), this is the same as to- 3 kg m- l sol. Assume that this value applies under the conditions in this system, and that the fluid viscosity at the wall J1w is equal to the bulk viscosity J.I.b.

The fennenter-side heat-transfer coefficient can be evaluated using the empirical correlation, Eq. (8.45). The dimensionless numbers in this equation are Rei, Pr and Nu. Rei is given by Eq. (8.39):

Solutions: ChapterS

94

Pr is given by Eq. (8.40):

From Eq. (8.45): Nu = 0.87 Re?,62 Pr0.33

(::::)""14 = 0.87(8.3 x 1O~0,62 (6.8P·33 (1)0,14 = 7,7 x 103

From the definition of Nu in Eg. (8.37) with D::::: the tank diameter:= 3 m: Nu k 7.7 X 103 (0.62 x 10-3 kJ s-1 m- I oc- l ) h =~ = ,;", 1.59kJ 8- 1 m-2 °C- 1 D 3m

Therefore. as the fermentation fluid is the bot fluid, hh i:::: 1.59 .kJ s·l m- 2 oe·!. The overall heat-transfer coefficient in the absence of fouling layers is .calculatedusing Eq. (8.23). Assuniingthat the beat-transfer coefficient for the cooling water he and the tube wall resistance O'kean be neglected (p 185), from Eq. (8.23), U:= klJ = 1.59 kJ s-1 m- 2oC- I. To maintain thefennentation temperature constant, tbecooling-coil must be capable of removing heat from the ,fermenter attbe rate at which it isprodu.ced. From p100,the heatof reaction for aerobic cultures is -460 kJ pergmol ~ygen consumed, Therefore. if the maximum oxygen demand = 90 gmol m- 3 b- l and the fennenter volume = 20

From the modified energy-balance equation, Eq, (8.33), when evaporation and sbaft work can be neglected, Q= -AHrxn = 230 kJ s·l. According to the sign conventions on pp 88-89, Qpositive is consistent with beat being removed from the system, From the definition of density(p 16), the mass flow rate of-the- cooling water is equal to the volumetric flow rate multiplied by the density:

Xl,

=

20m3h-I(103kgm-3)'13i~sl

= 5.56kgs- 1

From Eq. (8.32), the outlet cooling-water temperature is:

Tco =

Q

---- + Tei = Mc Cpc

230kJs-I

( )- + 12"C = 21.9"C 5.56 kg s-1 4.19 kJ kg-I oc;l

The mean temperature difference between the fermentation fluid and the cooling water can be determined from Eq. (8.35): tli'A =

2 Tp-(Tci + T,o) (2 x 28)'C-(12 +21.9)"C 2 = 2 = 11.1"C

Substituting the results into Eq. (8.19) for evaluation of A with aT= !lTA:

The area A of a cylindrical cooling-coil is equal to 2 1t r L, where r is the cylinder radius and L is its length. For r = 7.5/2 = 3.75 em = 0.0375 m: A L=--= 21tr

13m2

_ =55m 21t(0.0375m)

95

Solutions: Chapter 8

As the required length of cooling-coil is considerably longer than the 45 m available in the offered fermenter, the second·hand fermenter without modification is unsuitable for the proposed culture.

Answer: No

8.8

Optimum stirring speed for removal of heat from viscous broth

J.tb = 10,000 cPo From Table A.9 (Appendix A), 1 cP = 10-3 kg m- I s-l; therefore, JJb = 10,000 x 1O~3 .kg m- I s~I = 10 kgm- I sol. kfu = 2 W m- l °C- l . From Table A.8 (Appendix A), this is the same as 2 J gol mol °C- l ; therefore,ktb = 2 J s -1 m- 1 °C = 2 x 1O~3 kJ s-l mol °C- 1.

(a). (b) and (e)

The fermenter-side beat-transfer coefficient can be evaluated using the empirical correlation, Eq. (8.45). The dimensionless numbers in this equation are Rei, Pr and Nu. The dependence of Rejon stirrer speed is given by Eq. (8.39),

where Ni has units s-l. Pr is given by Eq. (8.40): C.LIb

2kJkg-Ioc-l(lOkgm-ls-I)

ktb

2 x 10 3 k1 C l m- l 0C-l

Pr=-P-=

4

=10

From Eq. (8.45), assuming that the viscosity at the wall is equal to the viscosity of the bulk-fluid: Nu = 0.87 Rep·62

p,O.33(:::t 14 = 0.87 (60.8 N;)0.62 (104)".33 (1)0.14 = 232AP,·62

From the definition of Nu in Eq. (8.37) with D = the tank diameter = 2.3 m: Nukfb

h

= -V =

232NP·62(2XlO-3kJs-lm-loc-l) 203m

.11.62

=0.20Ni

-1

kJs

-2

m

0

-1

C

Therefore, as the fermentation broth is the bot fluid: hh = 0.20N?·62 kJs -l m-2oC-1

Assuming that the heat~transfer coefficient for the cooling water he and £he tube wall resistance Blk can be neglected (p 185), from Eq. (8.23), U = hh = 0.20N?·62 kJs -I m-2 °C- 1

Substituting the known parameters into Eq. (8.19) with liT = liTA = WOC:

Q = UA6TA =

0,20AP,·62 kJ ,-l m -2'C 1 (14m 2jzO"C = 56.0AP,·62 kJ ,-1

The power dissipated from the stirrer, W s ' is equal to the power P calculated using Eq. (7.18), In this system, the value of Np with gassing (Npg) is 40% lower or 0.6 x the value of Np read from Figure 7.24. Therefore:

W,

= P = NpgPN;

Dr = 0.6Np(103 kgm-31N; (0.78m)' = 173NpN; kgm2 ,-3

where the value of Np depends onRej, and Nj has units of s~l. From Table A.8 (Appendix A), 1 J s·l = I kg m 2 s·3; therefore, 1 kJ s~l = 103 kg m2 s·3 and:

Ws =

173NpN[ kg m2 s-3

.1

kJS-~

-31 = 0.173NpNf kJ s-l

1 lOOOkgm s

From the modified steady-state energy-balance equation, Eq. (8.33), assuming evaporation is negligible:

Solutions: Chapter 8

96

Therefore:

-Mrxn

=

Q- WS = (S6.0,vp·62_0.173NPNf)kJS-l

Values of Q, W and -tJirxn calculated using the equations derived above are listed below as a function of Ni. S Values of Np as a function of Rei were obtained from the original reference (I.H. Rushton, E.W. Costich and H.I. Everett, 1950, Power characteristics of mixing impellers. Chern. Eng. Prog.46. 467-476) for more accurate interpolation of the power curve in Figure 7.24,

Nj

(s~l)

0.5 0.8 l.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 10.0

The values of

Rei

Q(1<1 ,.1)

Np

30.4 48.6 60.8 122 182 243 304 365 426 486 608

36.4 48.8 56.0 86.1 111 132 152 170 187 203 233

4.5 4.1 4.0 3.7 3.7 3.6 3.6 3.8 3.8 4.0 4.0

Ws

-Mirxn (kJ s·l)

(kJ s-l)

0.10 0.36 0.69 5.12 17.3 39.9 77.9

142 225 354 692

36.3 48.4 55.3 80.9 93.4 92.4 74.0 28.1 -38 -151 -459

ii, Ws and -tilirxn are plotted as a function of Nj below.

00 a Ws • -AHrxn

5

10

Stirrerspeed, Nj (s·1) (d)

The rate of removal of metabolic heat (the -Aiirxn component of

'h

reaches a tnaltimurn at aroundNj

=3 8. 1,

Answer: About 3 s·1 (e)

From p 100, the heat of reaction for aerobic cultures is -460.kJ per groal oxygen consumed. Therefore, if the rate of oxygen consumption per g cells is 6 rnmol g.I h· I = 6 x IO~3 gmol g.I h· I, when the stirrer speed is 3 s·l and 6Hrxn

=-93AkJs· I :

Solutions: Chdpter 8

Biomass =

97

1 -93.4k:Js=L22XIIYg I 6 10-' gmolg-I h- .I'6~sl (-460kJ gmol-I)

x

The cell concentration is equal to the biomass divided by the fermenter volume: .

CeII concentration =

5

L22xl0 g 22 I-I 3 11000 11 = 1 . g 10m . --,1m

Answer: 12.2 g l~l (I)

From the graph, as the stirrer speed is raised, the overall heat-transfer coefficient increases as beat transfer through the boundary layer in the fermentation broth is improved. Therefore, the rate at which heat can be removed from the system Qis increased. However, the rate at which heat is dissipated by the stirrer W s also increases with stirrer speed.. When the curves for Qand W intersect, the entire heat-transfer capacity of the fennenter cooling system is s being used just to remove the heat from stirring; the system in unable to remove any of the heat generated from reaction. Accordingly, at high stirrer speeds, the system bas limited capacity to handle exothermic reactions.

Mass Transfer 9.1

Rate-controlling processes in fennentation

Converting the units of the maximum specific ox.ygen uptake rate using the molecular weigbt of oxygen =32.0 (Table B.I, Appendix B):

444

I I gmol 1. 11gmol 32.0g II~I qo = 5 mmo1g-I b- . 1lOOOmmol . 3600s = '

x 10-5 g g-1-1 s

At a cell density of 40 g 1~1, the maximum oxygen requirement is:

qox

= 4.44XIO-5gg-I,-I(40grl)rl~lll~gl = 1.78 x 10-3 kg m-3 ,-I

The rate of oxygen transfer is given by Eq. (9.37); NAis maximum when CAL "" 0:

kLaC~ ""

O.15s- 1 (ax 1O-3 kgm-3) "" 1.20 x 10-3 kg m-3 s-l

As the maximum oxygen demand of the culture is greater than the maximum oxygen transfer rate in the fennenter, the system will be limited by mass transfer.

Answer: Limited by mass transfer.

9.2

leLa required to maintain critical oxygen concentration

Converting the units of the oxygen uptake rate using the molecular weight of oxygen"" 32.0 (Table B.I. Appendix B):

gmol

11~IIIOOOIII~1 1m3 ' 36008 =

1 I 1 1. 119mol 32.0g . l000g' qox = 80 mmo11- h- . 1lOOOmmol

711 . x 10-4 kgm-3-1 s

Converting the units of the critical oxygen concentration:

IlgmOI I

lI...'!LI

g 1 = t0004mM= 0004 Ce路 n ' , m m11o , l000mmol ' 1-,2.0 19mol . l000g' 1100011-128 1m3 - . x 10-4 kgm-3 (aJ From Table 9.2. the solubility of oxygen in water at 30"C under 1 atm air pressure is 8,05 x 10-3 kg m路 3 . If the solubility in medium is 10% lower;

C~ = 0.9 x (8.05 x 10-3) kg m-3 = 7.25x 10-3 kgm-3 To maintain the oxygen concentration in the medium at the critical level, from Eq. (9.41):

(kLa). = ent

qox

(C~ -Cerit)

=

7.11XlO-4 kgm -3 s-1 (7.25X 1O-3kgm-3 _1.28X 10-4 kg m- 3)

= O.10 s-1

Answer: 0.10 s-1 (bJ From Table 9.1, the solubility of oxygen in water at 30"C under 1 atln oxygen pressure is 3.84 Therefore:

x 10"2 kg m-3.

Solutions: Chapter 9

99

From Eq. (9.41),

= . . qox

(kLa) .

=

(CAL -Cerlt)

ent

7.llxlO-4 kgm-3 s-1

= 0.021s-1

(0.0346kgm-3-1.28XI0-4kgm-3)

Answer: 0.021 s-l

Single-point !<La determination using tbe oxygen-balance metbod

9.3

(a) The oxygen transfer rate for kLa determination by the oxygen~balance method is given as Eq. (9.48). From p 17, the partial pressure of oxygen in the inlet air at 1 atm is (0.21 x 1 atm) = 0.21 atm; assuming that the exit gas leaves the fermenter at the fermentation conditions (l atm pressure and 28°C), the partial pressure of oxygen in the exit gas is (0.201 x 1 atm) 0.201 atm. Using R 0.000082057 m3 atm K-1 gmol- 1 from Table 2.5 and converting the temperatures to degrees Kelvin using Eq. (2.24):

=

=

N = _ I [(FgPAG) _(FgPAG) 1 A

RVL

TiT

0

[(200lmin-l·I~10.2Iatm) (1891min-I·I~10.20Iatm)1

1 NA = 0.OOOO82057m3 attn K-1 gmol 1(200 I)

(20+ 273.15) K

NA

= 0.0174gmolm-3 s· 1

-

(28 + 273.15)K

Answer: 0,0174 gmol m~3 s~l (b)

From Eq. (9,37) with the units of NA from (a) converted to mass terms using the molecular weight of oxygen = 32.0 (Table B.t, Appendix B):

Answer: 0.15 s-l (e)

=

If the measured exit gas composition of 20.1% 02 is an overestimation, the actual value is (l/Ll x 20,1)% 18.3% 02. Therefore, the partial pressure of oxygen in the exit gas is (0.183 x 1 atm) = 0.183 atm. From Eq, (9.48):

NA = _ I [(FgPAG) _(FgPAG) 1 RVL TiT 0

_

[(200lmin-l·I~10.2Iatm) (189Imin-I·I~10.183atm)1

1

NA - 0.OOOO82057m3 atmK-1 gmorl (2001)

(20+273.15)K

NA

= 0,0289 gmol m- 3 s-1

-

(28 + 273.15)K

Therefore, from Eq. (9.37) with NA converted to mass units:

kLa

=...

NA

(CAL - CAL)

=

II...I.!>LI

1 -3 s-I . 132.08 0.0289 groom 1 01'1000

g 3 3 3 kg (7.8 x 10- kg m- -0.52x 7.8 x 10m-3) gIn

= 0,25s

_ 1

The kr.,a value obtained in (b) using the incorrectly calibrated oxygen analyser is 60% of the actual leLa value; the error is therefore 40%.

Solutions: Chapter 9

100

Answer: 40%. This calculation illustrates the sensitivity of the oxygen balance method to the accuracy ofall of the measured parameters used in Eq. (9.48). This sensitivity arises from the subtraction of two similar numbers for the moles of oxygen in and out of the system. When similar errors in both Fg terms are taken into consideration. the error in the final kIft value can be very large,

9.4

leLa measurement

(a)

From p 16, g "" 9.8 m 8.2 . Assuming the density of the culture broth to be the same as warer, p "" 1000 kg m- 3. The static pressure at the sparger is: Ps = P g h "" 1000 kg m- 3 (9.8 m s·2) (3.5 m) "" 3.43 x t()4 kg mwl s·l ,From Table A.S (Appendix A), 1 kg m- l s·2 = 9.869 x 10"6 aim; therefore: Ps -_ 343 . x

104 kg m-1 8-1

-_ 343 . x

104 kgm-1 s-1 . 19.869Xlo-6atml_ 1 I - 034 . atm 1 kgm- s-

Ps is the pressure due to_the head of liquid above the sparger; oThctotal pressure at the'llparger iSh + atmospheric pressure "" 1.34 atm.

Answer: 1.34 aim (b)

From Tables B.8 and B.l (Appendix B), the molecular formulae for glucose and sucrose and the molecular weights of the medium components are: glucose (C6H1206) = 180.2, sucrose (C12H22011) = 342.3, CaC03 = 100.1, (NH4)2S04 = 132.1, Na2HP04 = 142.0 and KH2P04 = 136.1.

The parameter values for application in Eq. (9.45) are listed below. Values of Hi and Kj are taken from Table 95.

Medium component

Hi or Kj(m 3 mo1·l)

Glucose

0.119 x 10.3

Sucrose

0.149 x 10.3

Zi

GiL or CJL (mol m· 3)

20g1-111mOllfoooll 180.2g 1m3 = 11l 1mol 11100011 8.5g1-I . 342.3g· 1m3 =24.8

I

-0.303 x 10.3

2

1.3g1-1.11~~~~I'111~11 =

C032-

0.485 x 1<r3

2

l.3gl-I.ll~~~I'II~311 = 13.0

NR,+

-0.720 x 1<r3

1

2x13 . gr

S04 2-

0.453 x 10.3

2

11 1m3 1= 9.8 1.3g1-I . lImo] 132.1g·

-0.550 x 10-3

1

1mol 11100011 2xO.09g1-I . 142.0g· 1m3 = 1.3

0.485 x 10.3

2

0.09g1-1.11~;~~I'111~'1 = 0.63

-0596 x 10-3

1

0,12g1-I . 11mol 136.1g' 1 1 m 3 1 = 0.88

1.037x 10-3

1

1 1 1 m3 1 = 0.88 0.12 g 1-I , 11mol 136.1 g'

Ca2+

Na+

HP042K+

H2P04-

Substituting these values into Eq. (9.45) gives:

1

13.0

10001 1molg' I 1 1m3 1 = 197 . 132.1 .

I

10001

I

I 10001 10001

101

Solutions: Chapter 9

10g1O(C~) = O.5~HiZ?CiL + ~KjCJL CAL j

t

10-') (2)213.0+(0.485 x 10-') (2)2 13.0 + (-o.720x 10-') (1)2 19.7 +(0.453 x 10-') (2)2 9.8 +] (-o.550X 10-') (1)2 1.3 + (0.485 X 10-') (2)2 0.63 + (-0.596 x 1O-')cl)2 0 .88 + (1.037 x 10-') (1)2 0.88 + [(0.119X 10-') 111 + (0.149 x 10-') 24.8J

_ 05 [(-0.303 x -.

•

• CAL = 0.95CAIJ) Solutes in the medium reduce the oxygen solubility by about 5%. From Table 9.2, the solubility of oxygen in water at 35°C and 1 attn air pressure is 7.52 x 10-3 kg m~3. Using this value for C~, C~ = 0,95 x 7,52 x 10-3 kg m*3 = 7,14x 1O~3kgm~3.

Answer:. 7,14 x 10-3 kg m-3 (e)

From Henry's law, Eq, (9.43), the solubility of a gas is directly proportional to the total pressure. From (a), the total pressure at the bottom of the tank is 1.34 attn rather than 1 attn; therefore:

C~

= L34(7.14xlO-3 kgm-3} = 957xlO-3 kgm-3

Answer: 957 x 10-3 kg m- 3 (d) The logarithmic~meanconcentration difference is given by Eq. (9.53). From (b), the solubility of oxygen at the top or outlet end of the vessel is 7.14 x 10-3 kg m~3; the dissolved oxygen concentration at this location is 50% air saturation or 0.50 x 7.14 x 10-3 = 3.57 x 10- 3 kg m- 3. From (c), the solubility of oxygen at the bottom or inlet end of the vessel is 9.57 x 10-3 kg m~3; the dissolved oxygen concentration at this location is 65% air saturation or 0.65 x 9.57 x 1O~3 6.22 x 10*3 kg m~3. Substituting these values into Eq. (9.53):

=

(c'" AL

-CAL)

=

(C~-CAL)o-(C~-CAL)i

L

[(c:u.-CALlo In

(C~L -CALli

= {7.14-3.57)XIO-3 kgm-3 -(9.57-6.22)xlO-3 kgm-3 10[(7.14-3.57) X 10 'kgm-'j {9.57-6.22)xIO 3 kgm 3

(C~ -CALt = 3A6x 10-3 kgm-3 Answer: 3.46 x 10-3 kg m- 3 (e)

The oxygen transfer rate for !La determination by the oxygen-balance method is given in Eq. (9A8). Assume that the volumetric flow rates of gas into and out of the fermenter are measured under ambient pressure, Le. at I attn. From p 17, the partial pressure of oxygen in the inlet air at 1 attn is (0.21 x I attn) = 0.21 attn; the partial pressure of oxygen in the outlet gas is (0.2015 x 1 aim) = 0.2015 atm. Assuming that the exit gas leaves the fermenter at the fermentation temperature (35°C), using R = 0.000082057 m3 attn K-l gmol-l from Table 2.5 and converting the temperatures to degrees Kelvin using Eq. (2.24):

Solutions: Chapter 9

102

Y~lo.2latm) (4071S-1·1~10.2015atm)1

1 3 I [(25 m minNA - 0.OOOO8Z057m3atmK 1gmor1(ZOm3) (25+273.15)K _

NA

=-

(35 + 273.I5)K

-

0.0167 gInol m· 3 s~l

Answer: 0.0167 gmol m- 3 s·l (f) ..

• From Eq. (9.37) with (CAL - CAL) = (CAL - CAL)L from (d) and units of NA converted to mass terms using the molecular weight of oxygen = 32.0 (fable B.t, Appendix B):

kLa

=-

'"

NA

=-

3Z 0g 00167 . Illkg . gIno1m-3 s-1 - 1lOf . 1000 I g

gIll

_

=- 0.15 s 1

3.46 x 10 3kgm 3

(CAL -CALt Answer: 0.15 s-1

(g) The maximum cell concentration supported by the fennenter can be calculated using Eq. (9.40) with C~equal to the average solubility of oxygen between the top and bottom of the fermenter:

_ kLa(c~) _ max x -

qo

0I5s-I[C·14;9.s7)XIO-3]kgm-31~1

_

-1

-74 l_lh_lllgmOII132.0gll~II~I-19g1 . mmo g . lOOOmmol . 1gmol . looog . 36008

Answer: 19 g I-I

9.5 Dynamic kLa measurement kLa is determined using Eq. (9.52) and the method described in Section 9.10.2. Take!} =- 10 s and CALI =- 43.5% air saturation. The steady-state dissolved oxygen tension CAL = 73.5%. Calculated values of

In

(.,~p.:::,,-=-;;~p.~"'1 )

and (12 - (1) are listed and plotted below.

0.00 OAI

0.80 1.6

2.3 3.0 4.1

o 5 10 20 30 40 60

Solutions: Chapter 9

103

â&#x20AC;˘

4

1

fq,a is equal to the slope of the straight line in the plot"'" 0.069 s-l.

Answer: 0.069 s~l

9.6 Measurement of kLa as a function of stirrer speed: the oxygen-balance metbod of Mukhopadhyay and Ghose From p 17, the mole fraction of oxygen in the inlet air CYAoh "'" 0210. Convert the data for CAL to units of gmol m- 3 using the molecular weight of oxygen"'" 32 (Table B.l, Appendix B) and assuming that the density of the fermentation broth is the same as that of water = 1000 kg m~3;

1

ppm =

1: 109

(lOookgm-J) .11O~XJgl.1 g .;~Oll g

= 3.125x 10-2

gmol m-J

Values of CAL and fCYAO)i - CYAO)oJ for each stirrer speed are listed below.

Agitator speed

Fermentation time (h)

300 rpm

0 4 5 6 7 8 9 10 11

500 rpm

700 rpm

CAL [(YAGJi - (YAG)o] (gmol m¡ 3)

CAL [(YAG); -(YAG)o) (gmol m- 3)

[(YAG); - (YAG)o] CAL (gmol m- 3)

0.184

0

0.184 0.175

0 0.001

0.184 0.178

0 0.001

0.166

0.001

0.147 0.128 0.106 0.106 0.109

0.002 0.003 0.004 0.004 0.003

0.163 0.153 0.138 0.125 0.125 0.131

0.002 0.003 0.004 0.005 0.005 0.004

0.169 0.159 0.147 0.128 0.128 0.131

0.002 0.003 0.004 0.006 0.006 0.005

The tabulated values of CAL and [(y AO)i - (y AO)oJ for 0-10 b are plotted below. As CAL increases at 11 b after being constant at 9 and 10 b, the final values in the table for each stirrer speed most likely reflect the decline phase of culture growth and are therefore not included in the lqp analysis.

104

Soluticns: Chapter 9

0.20.---,,.---,,.----,----,---,-----.

• 300 rpm o 500 rpm • 700 rpm

f

i

0.15

<o~

0.10 L-_.....L_ _....J.._ _.l-_-''-_.....L_ _...J 0,000

0.001

0.002

0.005

0.004

0.003

0.006

(a)

The slopes of the straight lines in the plot for 300, 500 and 700 rpm are -19.58, -12.04 and -9.68 groot m-3, respectively. These values can be used to determine,qa using R = 0.000082057 m3 atm gmol-l K~l from Table 2.5 and converting the temperature to degrees Kelvin using Eg. (224): ' -1

- 3 atm (31 nun

-J1TF.

kLa =

.

liminfi 66SV

_ -2.02 gmol m-3 8-1

RTVLX~ope = O,OOOO82057m3atmgmor1r1(29+273.15)K{31)xs}ope -

stope

For 300 rpm:

For SOOrpm:

At 700 rpm' k La =

-2.02gmolm-3s-1 -9.68 gmol m-3

1

= 0.21 s-

Answer: 0.10 s·l at 300 rpm; 0.17 s·l at 500 rpm; 0.21 s·1 at 700 rpm (b)

The intercepts of the straight lines in the plot for 300, 500 and 700 rpm are 0.185, 0.186 and 0.187 gmat m-3, respectively. Taking the average, C~ is 0.186 gmol m- 3. Converting to mass terms using the molecular weight of oxygen = 32.0 (Table B.t, Appendix B):

C~ = 0.l86gmolm-3 = 0.186 gmol m-3 .1 i~~ll·11~gl Answer. 5.95 x 1<r3 kg m~3 (c)

Equating Eqs (9.39) and (9P6.2),

= 5.95 x 10-3 kgm-3

Solutions: Chapter 9

105

qo x

qox

=

PTFg

= R T II. [(YAG)i - (YAGlo]

3atm(3Imin-lr~~[(YAG)i-(YAG)oj O'{)OOO82057m3atmgmolIrI(29+273.15)K(31)

__

= 2.02[(YAG)-(YAG) jgmolm 3, I I

0

The maximum rate of oxygen uptake occurs when [(yAO:>i - (yAO)oJ is maximum. At 300 rpm. the maximum value of [(YAo)i - (YAeVo] is 0.004; therefore, .(qO x)max = 2.02 x 0.004 = 8.1 x 10- 3 gmol nr 3 s-l. At 500 rpm, the maximum value of [(YAo)i - (YAO)o] is 0.005; therefore, (qoxhnax = 2.02x 0.005 = 1.0 x 1O~2 gmol m-3 s-l. At 700 rpm, the maximum value of [(YAo)i - (YAo)oJ is 0.006; therefore. (qO x)max = 2.02 x 0.006=1.2 x 10-2 gmoJ m- 3 s-1.

Answer: 8.1 x 10-3 gmoJ m-3 s路l at 300 rpm; 1.0 x 10-2 gmol m-3 8- 1 at 500 rpm; 1.2 x 10- 2 grool m路 3 s路l at 700 rpm

Unit Operations 10.1

Bacterial filtration

(aJ

According to Eqs (10.12)-(10.14), a plot of fl Vf versus Vf should yield a straight,line for determination of the filtration parameters. The data after converting the units to s and m3 are listed and plotted below.

Time, t (8)

120 180 360 600 900

1200

Filtrate volume, Vf(m 3)

tlVr(s rn*3)

0.0108 0.0l2l 0.0180 0.0218 0.0284 0.0320

1.11 x 104 1.49 x 104 2.00 x 104 2.75 x 104 3.17x 104 3.75 x 104

4.0

1 ., .!!.

3.0

•

-

0

x ....1:::.- 2.0

•

~ ~

g ~~

•

1.0

~

Il:

0.0 0.00

0.01

0.02

0.03

0.04

Filtrate volume, ~ (m3)

The slope of the 'Straight line in the plot KI = 1.18 x 106 s m- 6 ; the intercept K2 = -363 s m o3 . From Eq. (10.13), using the conversion factors 1 mmHg = 1.333 x 102 kg m- I s·2 (Table AS, Appendix A) and 1 cP = 10- 3 kg m-I g-l (Table A.9, Appendix A), the specific cake resistance a is:

2(0.25m2)'36OmmHg.I1.333\1~:-1

s-'j(1.l8x 106 sm-6)

= ----,c=c---':i-~=-=""-:-----.l:-:-----:-­ 4.00p.110 3kgm Is Ij(22g1-111°OO1j.I~~ lcP 1m3 lOOOglJ

a

= 8.0 x lO lO m kg-!

The intercept K2 is a relatively small negative number, which may practically be taken as zero. Therefore, from

Solutions: Chapter 10

107

Eq. (10.14), the filter medium resistance rm is zero.

Answer. a = 8.0 x 1010 m kg-I; rm is effectively zero. (h)

tl yr = 7.5 x 10-3 min 1~1. From(a), K2 =0; therefore, from Eq. (10.12): 3

1

_ t/Vr = 7.5xl0- minl- 1100011'160' 1 = 113 10' -6 K 1Vf 40001' 1m3 'lmin . x sm

The filter area is obtained fromEq. (10.13) using the result for a from (a):

,

I'tac

4.0cP

.1 10-

3 \g:;:-I '-'I&·ox 10

1O

1 1 mkg- (22 g l-

A = 2Kt6p = 2(1.13X 10' 'm-6) 360mmHg.

ff;;;¥I·I~I)

I1.333Xi~:-1 I

4

= 649m

,-'

A = 25.5m2

10.2

FUlratlon ofmyceUaI suspensions

(a)

After converting units using the conversion factor 1 m3 = 106 mI, the data for tlyfand Yr are listed and plotted below.

Pressure drop (mmHg) 100

250

350

10.5 10.5 10.5 10-5

2.2 x 106 3.5 x 106 4.5 x 106 5.8 x 106

1O~5

6.7 x 8.1 x 9.2 x l.0 x

10.5 10.5 10.5

750

7.0 x lOS 9.3x1OS 1.4 x 1()6 1.7 x 1()6 2.1 x 1()6 2.4 x 1()6 2.& x 1()6 3.1 x 1()6 3.5 x 106

5.0 x lOS &.0 x lOS 1.1 x 106 1.4 x 1()6 1.7xl()6 2.0 x 1()6 2.3 x 1()6 25 x 1()6 2.& x 1()6

550

750

IfYr(S m· 3)

Vf(m 3) pelleted 1.0 x 1.5 x 2.0 x 2.5 x 3.0x 3.5 x 4.0x 4.5 x 5.0 x

550

106 106 106 107

H,.5

12 x 1.7 x 2.5 x 3.0x 3.7 x 4.3 x 4.& x 5.3 x 6.0 x

1()6 1()6

106 106 1()6 106 1()6 1()6 1()6

9.0 x 1.3 x 1.& x 2.4 x 2.9 x 3.4 x 3.9 x 4.3 x 4.8 x

lOS 1()6 1()6 1()6

106 1()6 106

106 106

Pressure drop (mmHg) 100 Vf

1.0 x 1.5 x 2.0 x 2.5 x 3.0 x 3.5 x

250

350

t/Vr(s m· 3)

filamentous 1(T5 1(T5 10".5 1(T5 1(T5 10-5

4.0 x 10"5 4.5 x 10"5 5.0x lOwS

3.6 x 1()6 55 x 1()6 72x106 9.0 x 106 1J x 107 1.3 x 107

2.2 x 3.1 x 4.3 x 5.3 x 65 x 7.5 x 8.5 x 9.6 x

1()6 1()6 1()6 1()6 I1J6 106 106 106

1.7 x 1()6 2.7 x 1()6 3.6 x 1()6 4.4 x 1()6 52 x I1J6 6.IXI1J6 7.1 x I1J6 7.& x 11J6 &.& X11J6

1.3 x 2.1 x 2.7 x 3.4 x

106 106

106 106

4.0 x 106 4.7 x 106

5.6x 106 6.2 x 1()6 6.& x 1()6

1.1 x 1.7 x 2.3 x 2.& x 3.3 x

106

1()6 1()6 1()6 I1J6 4.0 x 106 45 x 1()6 5.1 x 106 5.7 x 106

Solutions: Chapter 10

108

12 Pressure drops:

f

10

o 250mmHg .350mmHg

$

~

o

550mmHg .. 750mmHg

8

6

~

x

-1"'-

PaUeted

• 100mmHg

6

~

•

~

~g

~I

2 0 2 3 Filtrate volume,

0

•

5

6

V; x 105 (m3)

,.

Filamentous

Pressure drops:

12

..

• 100mmHg o 250mmHg

10

o

x

8

f $

.350mmHg

550mmHg .. 750mmHg

0

_1:::.,0

E ~

~g F

i

6

• 2 0 0

1

2

3

Filtrate volume,

V;

•

6

6

x 105 (m3)

From Eq. (10.12), the slopes of the straight lines in the plots are equal to Kl for each pressure drop. 'These values are listed below.

Pressure drop (romHg) 350

100

250

550

750

2.26 x 1011

L20x lOll

9.93

x lO lD

7.07 x 1010

5.77 x 10 10

3.73 x lOll

2.13 X 1011

1.75 x 1011

1.38 X lOll

Ll4 x 1011

Pelleted Slope::: Kl (s m· 6)

Filamentous Slope:::: KI (8 rn·6)

=

From Eq. (10.13), using the conversion factors 1 mmHg 1.333 x 102 kg m- I 8.2 (Table A.S, Appendix A) and 1 cP ::::: 1O~3 kg m- I s·1 (Table A.9, Appendix A), the specific cake resistance aofthe pelleted cells is:

Solutions: Chapter 10

109

2(1.8Cm2 .11C:O~mI2)2 /;p.1 L333\1~:-1 s-ZjKI

2AZ .6pK

3k;;-l I

a = -~,.....:.I = -"-,--0'='-.::;,"-\,,-'---+'7',",--'--, = 2.47 x 1O-5 /;pKI mkg- I 1.4cP

Jl{c·

.1 10-

l . 1~~~1'11::~1

s- I(0.25 gml-

where tip has units ofmmHg andKl has units of s m-6. Similarly for the filamentous cells: 2A2 _

a-

K /;p

Jl{c

1_

-

2(1.8cmZ.I~P)2 /;p·11.333xlOZkgm-ls-ZjKl 1.4cP.

loocml I10-3kf; 1 lit s-

ImmHg

6I 1.Il~mU:

0.lgml-

1~5A.K

-617

)- .

.ll::gl

x

v

~

k-l

Im g

The results for specific cake resistance obtained after substituting the values for tJ.p and K 1 into the equations are

listed below.

250

558 x 108

7.41 x

loB

2.30 x 109

3.29 x 10'

Pelleted a(mkg~l)

Pressure drop (mmHg) 350

100

550

750

8.58 x 108

9.60 x 108

1.07 x 109

3.78 x 109

4.68 xla'

5.28

Filamentous

a(mkg- I )

X109

(b)

From Eq. (10.2), the compressibility can be obtained by plotting the specific cake resistance a versus lip on log-log coordinates. Using the conversion factor 1 mmHg:::: 1.333 x 102 kg m- 1 s·Z (Table A.5, Appendix A) to convert the units of lip, the data are listed and plotted below.

1.33 x 1()4

Pressure drop (kg m- l s-Z) 3.33 x 1()4 4.67 x 1()4 7.33 x 1()4

1.00 x lOS

5.58 x loB

7.41 x 108

8.58 X 108

9.60 x 108

l.07 X 109

2.30x109

3.29 X 109

3.78 X 109

4.68 x 10'

5.28 x 109

Pelleted

arm kg-I) Filamentous

a(mkg- I )

_ 100 . -

,,-

!if

S

~--~-_~-~_~...,

o Pelfeted • Filamentous

6

x

"

Pressure drop, Apx 10'4 (kg m·1 s·2)

10

Solutions: Chapter 10

110

From Eq. (ID.2), the equations for the two lines in tJ:e figure have the form, a= a' 1¥,s. For the pelleted suspension. a= 2.60 x 107 ApO.323; for the filamentous suspenSIon, a= 4.42 x 107 Ap0.415, where lip has umts kg m· l s·2, a has

units m kg-I, and a' bas units kg-(l +s) m 1+3 sZs. Therefore. the compressibility is 0.323 for the pelleted suspension and 0.415 for the fIlamentous suspension.

Answer: 0.323 for the pelleted suspension; OA15 for the filamentous suspension (0) The filtration equation for a compressible filter cake is Eq. (10.11) with a= a' ti[l. Assuming that the filter medium resistance rm is negligible:

Solving for (61')$-1: 2

(Ap)J'-l :: _I ~

v3f I'f a' c

Substituting the parameter values for the filamentous suspension using consistent units, including the results ex':: 4.42 x 107 kg..(l+s) m 1+s s2f and s= 0.415 from (b):

(Ap)0.415-1 ::

Ih .

Ih1 13600' 2 x

(20m 3)

IAcP

.1 10- 3 klgc~-l ,-114.42X 107 kg-{I+O.415)m l +O.415 ,2x 0.415 ( 0.1 g ml-1 .

6 10 3 1 mm]

'I~I) Iooog

{1ip)-O.585 :: 6.54 x 10-4 kg-o.585 mO.585 s1.170

= 2.78 x loS kg mol s·2 Converting units using the conversion factor I mmHg = 1.333 x 102 kg mol s·2 (fable A.5, Appendix A): !¥J = 2.78 x 105 kgm- 1 8-2 = 2.78x 105 kg m- l 8-2 .1 I mmHg I = 2086mmHg Ap

1.333 x 102 kg m I s-2

Answer: 2086 mmHg. An assumption associated with this answer is that the filter cake characteristics measured at pressures between 100 and 750 mmHg apply at the much higher pressure of2086 mmHg.

10.3

Rotary-drum vacuum mtration

(a) For negligible filter medium resistance, the filtration equation for a compressible fdter cake is Eq. (lO.lI) with rm and a= c£ !¥Js:

t Vf

=0

=

Solving for fJf a' c:

2

J1{a'c

t 2A =----

Vi t>p>-1

Substituting the parameter values for the laboratory filter and using the conversion factor I psi s·2 (fable A.5, Appendix A):

=6.895 x 103 kg mol

Solutions: Chapter 10

lI..,tXc ""

r,

Answer: 3.67 x

111

2 1m 2 21 lOOcm 1) 2 ( Scm.

. -1-' 60s 23.5mm.

1 nun I

(Sooml. ~ )2 ( 12 si.16.89sxI03k~m-l 106 m1 P 1 pst

e21~0.57-1

"" 367x 105 kg0.43 m-2.43 s0.l4 .

105 kg0.43 m-2.43 sO. 14

(h) The cycle time is lIN hours per revolution.

(el

As 30% of the rotating filter cloth is submerged at any time, each cm2 of cloth is submerged for 0.3 x the cycle time "" O.3/N h.

(d) The volume filtered per revolution is 20 m3 h~1 x liN h "" 20INm3.

(el Eq. (10.11) with rm "" 0 and a",,ÂŤ !J.ps is:

lit a' c ;ips-I 2 V, 2A Applying this equation to a single revolution of the filter, t "" O.3IN h and Vf"" 20lN m3 . Substituting the filter cake parameter values from (a), evaluating the filter area A "" 2 1t r Wwhere r is the drum radius"" 0.75 m and W is the drumwidth "" 1.2 m, and using the conversion factor 1 psi = 6.89S-x 103 kg m- I s-2 (Table AS, Appendix A);

I I

0.3 h 3600s N'lh

-'-'-~';--c:.=.

20 m3 N

-

3kg -1 _2~0.57-1 367 105kg0.43 -2.430.14 45 . . 95 xlO m s .x m s . pSt. Ii (20) 3

(

-

168

2(2n(0:75m)(1.2ml)'

ps

N

m

where N is the number of revolutions per hour. Calculating both sides of the equation gives: 54sm-3 "" 134~.9 sm-3

N = 24.9rph Answer: 24.9 revolutions per hour (I) The ftltration time is increased from O.3IN h to O,SIN h. Substituting this value for t into the left-hand-side of the equation in (e):

h O.5 N .

13600 Ih '1 _ 1343.9

20 _m3 N

-3 - -,;;-sm

Solutio1lS: Chapter 10

112

N

= 14.9 rph

Answer; 14.9 revolutions per hour

10.4

Centrifugation of yeast

(a) Convert the parameter values to units of kg, m, s:

p

P

Pr

=:

3 3 1.06gcm- = L06gcm-

3 0.997 gene

=:

1 Q = SOOlh- =

:=

'lli.:gl.ll~:r = l060kgm-3

3 O.997gcm-

.Ili.:g1路ll~~m

3 1

= 997kgm-3

500lh-l'!l~JbJ~sl = l.39xlO-4 m3 s-1 =

From Table A.9 (Appendix A), INs m- 2 = 1 kg mol g-1; 'therefore J.I. 1.36 x 10- 3 kgm- I 8- 1. D p = 5 MlD= 5 x 10-6 m. From Eq. (10.15) with g =: 9.8 m s-2 (p 16), the sedimentation velocity is:

u g

= Pp -PfD 2g = 18p

P

3 (l060-997)kgm-

(SXI0-6m)29.8ms-Z_=6.31XIO-7ms-l

18 (1.36 x 10 3 kgm 1s-l)

Substituting this result into Eq. (10.18):

I: =

JL

4

3

1.39x W .. m s-1 = 110m2 2 (6.31 x 10 7 ms-l)

=:

lUg Answer: 110 m 2 (b)

From the relationship between specific gravity and density (p 16), as the density of water is close to 1.0000 g em-3 (p 16),

3

Pp Dp

1kg 1. 11m 100 em 1 = 20. gem-3 = 20. gem-31. l000g

=0.1 nun = 0.1 x 10-3 m.

=:

2000kgm-3

From Eq. (10.15), the sedimentation velocity for the quartz particles is:

3 2 _ _ Pp -Pf (2000-997) kg m)2 98 (01 10-3 m ug _ -D g _ ( ) . x . ms-2 -_ 402 . x 10-3 ms -1 3 18,u p 18 1.36x 10- kgm 1 s:-1 From Eq. (10.18), the sigma factor is:

r = -lL = lUg From the result in (a),

L39x 1O-4 m 3 s-1

r for the yeast cells is 110/0.017 =6470 times that for the quartz particles.

Answer: By a factor of 6470.

10.5

= 0.017m2

Z(4.02xlo-3ms-l)

Centrifugation offood particles

Convert the parameter values to units of kg, m, s:

Solutions: Chapter 10

113

3

3

Pr = 1.00gcm- = 1.00gcm-

·ll~gl·ll~~mr = l000kgm-3

From Table A.9 (Appendix A), I Pa s = I kg m~l s~l; therefore /.l = 1.25 x 10"3 kg m~1 8"1. D p = 10--2 mm = 10--5 m; b = 70 cm = 0.70 m; r= 11.5 cm = 0.115 m. As one revolution = 2ft radians where radians in a non~dimensional unit (p 11), converting the centrifuge speed to radians s~1: (J)

From Eq. (10.22) withg

= IO,OOOrpm = 2ft (lO,OOO) min-I

.!

1::-1 = 1.047 x 103 s-l

=9.8 m s~2 (p 16), the sigma factor for the tubular~bowl centrifuge is:

" 1: = 21t~J:!: = 21t ( 1.047x1O3 s_1)2 0.70m(0.115m)2 = 6.506 x 103 m2 g 9.8ms 2

From Eq. (10.15),

Pp -P'D2 ug __ g _ _ 18,u

p

3

)2 98 _ .

1307 x 10-6 ms -1 ((l030-1000)1Qlm- ) (10-5 m . ms-2 18 1.25 x Io-3 kgm-1 s-1

From Eq. (10.18),

Answer. 0.017 m 3 s~l

10.6

Scale-up of disc-stack ceutrifuge

=

The sigma factor for the pilot~scale disc~stack bowl centrifuge 1:1 is calculated using Eq. (10.20) withg 9.8 m s·2 (p 16), and (J) converted to rad s·l using the conversion factor I revolution = 2 1t radians where radians is a non~ dimensional unit (p 11). For the pilot-scale device, '2 = 5 cm = 0.05 m and rl = I cm = m:

om

:!: = 2noJ-(N-ll(,3_,1j = 1 3gIanO 2 1

1 . ~2 (25-1) 21t (21tx3000min- l . ~

1 60s 3(9.8ms 2)clan35')

(0.05m)3_(0.0Im)3) = 89.6m2

From Eq. (10.19), the sigma factor·:rz for the bigger centrifuge is therefore:

I :!: = Q2:!:I = 801min- (89.6m2) = 200m2 2

3.5 1OOn-1

Q1

From Eq. (10.20), for the bigger centrifuge with 1"2 = 75 cm = 0.075 m and rt = 2.35 cm = 0.0235 m:

oJ-

=

3glanO:!:,

2 n(N - IlH -

,1)

=

2 3 (9.8 m s-2)clan 45') 2048m = 4 34x 10' s-2 2 n (55 -1) (0.075 m)3 - (0.0235m)3) .

_ 659 r ad s-1 .

(J)-

II

Answer. 6290 rpm

10.7

CeutrifUgatiou of yeast and ceO debris

Combining Eqs (10.15) and (10.18),

and

II 1-

revolution 60 s - 6290 rpm 21trad . lmin

Solutions: Chapter 10

114

where subscript 1 refers to the centrifugation conditions for the yeast suspension, and subscript 2 refers to the centrifugation conditions for the cell debris. As the same centrifuge operated at the same speed is used in both applications, ,1;1 =1:2 and:

Assuming that the particle and fluid densities are the same in both applications, cancelling terms gives: ~1 Ql !'2 Q2 - 2 - =-2D pl Dp2

or

Answer. 0.0671 min-I

10.8

Cen disruption

The relationship between pressure and protein release for a Manton-Gaulin homogeniser is given by Eq. (10.23). At constant pressure, a semi-lagplat ofRm/(RnrR) versusN should yield a straight line, so thatthe value of k pa can be determined from the slope.

The data for % protein release represent values of RIR m x 100. These data can be converted to Rm/(Rm_R) as

follows: Rm Rm-R

100

100

= loo-RIRm x 100 = 100-% protein release

The results are listed and plotted below.

Pressure drop (kgjcm- 2) 200

300

3 4

5 6

500

550

Rm

N 1 2

400

Rm-R 1.05 1.10 1.16 1.22 1.28 1.35

1.16 1.31 1.50 1.75 1.90 2.22

1.30 1.67 2.11 2.99 3.70 4.88

1.56 2.41 4.00 5.71 8.70 11.5

1.72 2.94

6.13 8.70 18.2

Solutions: Chapter 10

liS

30...--,---,----,---,--,--....---,

10

â&#x20AC;˘ o â&#x20AC;˘ o ..

200 kg f an-2 300kg f an"2 400 kg f an"2 500 kg,an'2 550 kg,an'2

2

1

3 4 Number of passes, N

5

6

7

The values of k pa obtained from the slopes of the straight lines for each pressure are listed below.

200

300

0.050

0.129

0.267

500

550

0.406

0.580

A log-log plot of these values versus p can be expected to give a straight line with the value of a obtained from the slope.

'l,.

0.1

'"

0.01 L100

---._ _~ _ ~ ~ _ ~ ~ ~ _ ' 1000 Pressure, p (kgfcm- 2 )

As the equation to the straight line in the plot is k pa = 1.71 x 10-7 p2.37, a = 2.37. Therefore, for this system, Eq. (10.23) becomes:

where p bas units ofkgjcm- 2.

116

Solutions: Chapter 10

(a) For 80% protein release, Rm/(Rm_R) "" 5.0. Rearranging the above empirical equation for the homogeniser and substituting p "" 460 kgfcnc 2 gives:

N=

In(R=~R)

ln5.0 = 4.6 = 1.71 x 10-7 p2.37 1.71 x 10-7 (460)2.37

Therefore, 80% protein release is achieved within 5 passes through the homogeniser.

Answer: 5 (b)

For 70% protein release, Rm/(Rm_R) "" 3.33. Rearranging the empirical equation for the homogeniser and substituting

N=2 gives:

In(R=~R)

_

p2.37 = -"-'''--,fln3.33 "" 3.S2x 106 7 7 L71xlO- N-1.71xlO- (2) p

:=

578 kgjcm路 2

Answer: 578 kgjcm路 2.An assumption associated with this answer is that the homogenisation characteristics measured at pressures between 200 and 550 kgfcm路2 apply at the higher pressure of578 klJfcm路2,

10.9

Enzyme purification using two-phase aqueous partitioning

(a)

From Eq. (10.24), for K"", 3.5, the product partitions into the upper phase. Eq. (10.27) withK:= 3.5 and Yu "" 0.8 is:

Vu

0.8 =

v.

I

Vu + 3.5 Rearranging gives: 0,8 Vu + 0,23 VI = Vu 0,23 VI

= 0,2 Vu

v,

YJ = LI5

Answer: US (b)

The mass of enzyme in the two phases must be equal to the mass of enzyme in the original homogenate, The mass~ balance equation is:

From Eq. (10,24), for K = 3.5, CAu = 3.5 CAl. If VJ = 100 I, from (a) for 80% recovery, Vu = 115 l. Substituting these values into the mass~balance equation with Vo = 150 I and CAD = 3.2 u ml- J = 3.2 x 103 u 1-1:

where CAl has units ofu 1~1. Solving the equation fOTCM 502.5 CAl = 4.8 x loS CAl = 955 u 1~1

Solutions: Chapter 10

=

117

As CAu 3.5 CAl> CAn = 3.34 the upper phase is:

x 103 u I~I. From Eq. (10.29), the concentration factor for product partitioning into

Answer: 1.04

10.10 Recovery of viral particles

=

From Eq. (10.24), for K,= 10..2â&#x20AC;˘ CAn m¡2 CAl and the product partitions into the lower phase. If51 culture volume is added to 2 I polymer solution and, after phase separation, the volume of the lower phase is 1 1, the volume of the upper phase must be (5 + 2) - 1 6 L

=

(a) The yield of virus in the lower phase is given by Eq. (10.28) with VI

Vr

=11, Vu =61 and K = 10-2:

11

Y, = VuK + V, = 61(10-2)+ 11 = 0.94 Answer: 0.94 (b) The mass of viral particles in the two phases must be equal to the mass of viral particles in the original culture broth.

The mass..balance equation is:

(e) From Eq. (10.24):

Rearranging the mass-balance equation in (b) and substituting for CAI gives: CAu

CAuVu+CAIVi CAO

=

Vo

=

CAuVu+KVI

The concentration factor for product partitioning into the lower phase is given by Eq. (10.29). Substituting for CAl and CAO in this equation gives:

(d) Using the equation for

4: derived in (c) with Vo = 51, VI = I 1, Vu = 61 and K = 10,2: 51

Answer: 4.7

118

Solutions: Chapter 10

10.11 Gel chromatography scale-up (a)

The elution volume for the toxoid is lower than for the impurity. Therefore, as the toxoid stays in the column for the shorter time. it must be the larger molecule.

Answer: Toxoid (b) The internal pore volume in the gel in the laboratory reactor can be calculated using Eq. (10,42):

!'i = aW, = IOgoll~gl(0003Sm3kg-l)oll~:~1 = 3Sml As V 0= 23 ml, the partition coefficients for the toxoid and impurity can be detennined using the measured elution volumes and Eq. (10,41);

°171

Taxoid Kp

= (V,-V; Vol = (29-23)m! = 35ml

Impun°tyKp

o) = (45-23)m! = 0629 = (V,-V Vi 35m1 .

,

.

,

Answer: Toxoid =- 0.171; impurity =- 0.629 (c) Let subscripts 1 and 2 denote the small and large columns, respectively. The total volume of the laboratory columnaf inner diameter Del 15 em 0.015 m and height HI::::' 0.4 m is:

=

=

VTl ::::.

The total volume oitbe

1t

(2DI)2 HI::::' C

1&

(OoOISm)2 2 OAm::::.

large~scale column of diameter D c2::::'

VTI

=-

1t

7.07 x 10-5 m3

0.5 m and height HZ =- 0.6 m is:

3 ~ O.6m =- 0.118m (TD02)2 HZ =- 1t (osm)2

If the void fraction in the large column is the same as in the small column, Vo2 in the large column is: 3

23m!.

V02

= rVol

TI

VTI =

~ 6

10 5 m!3 ( 3) 3 0.118m = 0.0384m

7.07 x 10

m

If the pore volume fraction is also the same:

ViZ

= r!'itT1 VTI =

3 35m!. 1: 10 m!

( 3 ) = 0.0584m3

5 3 0.118m 7.07 x 10- m

If the large·scale column is operated with the same packing and flow conditions, the partition coefficients can be assumed to be the same-as those in the laboratory column. Therefore, from Eq. (10.41), for toxoid in the large column with the value of K p from (b): Ve2 = Kp Va + Vo2 = 0.171 (0.0584 m 3) + 0.0384 m 3 = 0.0484 m3

Similarly for the impurity: Ve2 = K p Vi2 + Vo2 = 0.629 (0.0584 m3) + 0.0384 m 3 = 0.0751 m3

Answer: Toxoid

=0.0484 m 3 ; impurity =0.0751 m3

Solutions: Chapter 10

119

(d) The liquid flow rate is scaled up in proportion to the column cross-sectional area. As the cross-sectional area "" 1&

(DC/2)2, the volumetric flow rate v2 in the large column is:

D2)2 ( n 20

v2 "" VI

n(D;I)

1m 0.5 m 2 3 (D)2 "" 14mlmm. -1 . 131 (0015 ) "" 0.0156m

c2 2 "" VI -D cl

-6-'

10 ml

.

m

,-1 IDlD

,

Answer: 0.0156 m 3 min-I (e)

The retention time t'R is equal to the elution volume divided by the volumetric flow rate. For the toxoid in the large column, using the results from (c) and (d): Ve2

IR=-= v2

0.0484 m 3 = 3.1 min 0.0156m3 min 1

Answer: 3.1 min

10.12 Protein separation nsing chromatography (a)

Selected values of u and H are listed and plotted below.

Linear liquid velocity, u

HETP,H

(m s-l)

(m)

0,1 x 104 0.2 x 104 0.3 x 10-4 0.5 x 10-4 0,8 x 10-4 1.0 x 10-4 1.2 x 10-4 1.5 x 104 1.8 x 10.4 2.0 x 10.4

2.72 x 10.4 1.87 x 10-4 1.69 x 10-4 1.72 x 10-4 2.02 x 10-4 2.27 x 10-4 2.54 x 10-4 2,95 x 10-4 3.38 x 10.4 3.67 x 10-4

4,----,--...,---,---,----,

1 '--_---'_ _- ' -_ _- ' -_ _.L._--' 0.0

0.5 1.0 1.5 2.0 Unear rrquid velcoity, u x 1()4 (m $"1)

2.5

120

Solutions: Chapter 10

(h) From the graph in (a), the minimum HETP is around 1.7 X H;4 m. This can be confirmed by differentiating the equation for H and solving for dH/du = 0:

dB --A - = -+8 = 0 du

.2

Therefore:

The value of H corresponding to this u is:

A 2xlO"""m2 ,-1 ( ) H=-+Bu+C= 5 1 +1.5s 3.65x1O-5 ms- 1 +5,7x1O-5 m= 1.67xl0-4 m u 3.65x1O ms

Answer: The minimum HETP is 1.67 X 10-4 m, obtained at a liquid velocity of 3.65 X 10-5 m s~1. (c) The column diameter Dc = 25 em = 0.25 m~ The --volumetric flow ,.rate v is 0.311 min"1, ,The-linear ,flow-rate u is obtained by dividing the volumetric flow rate by the column cross-sectional area:

u =

v

K(~'r

=

I mini

'm3 0311 . -I I60s . 110001 I . mm.

K(02gmr

_ = L05x 1O-4 ms 1

Substituting this result into the equation for H:

A 2x 1O"""m2 ,-1 ( ) H=-+Bu+C= -4 1+1.5s 1.05xl0-4 ms-l +5.7xlO-5 m= 2.34xlO-4 m u 1.05xlO ms The capacity factors for the A and B chains are kA= 0.85 andkn = 1.05, respectively. From Eq. (l0.38):

o= kkB

A

= 1.05 = 1.235 0.85

Substituting these parameter values into Eq. (10.51) with L = 1 m for the larger column and k2 = k]J:

R N,

=!

fT(O-I)(~)=\/ 1m 0 kB+l 4 2.34xI0-4

4Vli

(1.235-1)( 1.05 )=159 m

1.235

1.05+1

.

From p 248, as RN> 1.5. the two peaks are completely separated.

Answer: Yes (d)

'.Por L = 0.7 m for the smaller column, the value of H corresponding to RN= 1.5 or virtual complete separation can be detennined using Eq. 00.51):

.fIl =! .fL(O-I)(~) =! .fiifID(1.235-1)( 4RN

8

kB+I

41.5

1.235

1.05 ) = 0.0136Jiil

1.05+1

H= 1.85xlO-4 m The linear liquid velocity at which this value of H is obtained can be determined by rearranging the expression for H

as a function of u and solving for u:

121

Solutions: Chapter 10

The solution to this quadratic equation is:

u

=

-{C -H) ±\!<C _H)' -4B A

2B

H-C±,!<C-H)2_ 4BA

=

2B

Substituting the parameter values:

u

= (1.85X 104

-5.7 x 1O-5)m± u

V

(5.7 x 10-5 m-1.85x 104 m)' -4 (1.5 s)(2X 10-9 m2 s-l) 2 (1.5 s)

= 4.27 x 1O,S ms- l ±2.21 x 10-5 ms- I

Therefore: u

= 6.48 x 10-5 m s·l

or

u

= 2.06xlO-S ms· 1

1be maximum flow rate for complete separation-is-u =·6.48wlO-5 m s-l. 'Betweenu = 2.06 x IO-S m s·l and u = 6.48 x Io-S m s·I,H < 1.85 x 104 m SO tbatRN> 1.5 and-eomplete:separationismaintained. The volumetric flow rate v is equal to u multiplied by the column cross--sectional area: v

= U1t(~cr = {6.48XlO-S ms-1}1t(o.2gmt = 3.18xW"·Q m 3 s-1

Converting units:

10001

60s 1 = 0191 ·-1 v _ - 3. 18 x 10-6 m3 s-I . 1 1m3 1 ' 1lmin . nun

Answer: 0.191 min-I

Homogeneous Reactions 11.1

Reaction eqnilibrium

From p 258. take the standard conditions to be 25°C and 1 attn pressure. The temperature is converted to degrees Kelvin using Eq. (224); from Table 2.5, R =8.3144 J K-l gmol-l = 8.3144x 10-3 kJ K-I gmol-l. (a) FromEq. (11.3):

inK

-t1GO =~ =

RT

1

14.1klmol= 5.69 8.3144xlO-3kJr1gmoll(25+273.15K)

K = 295 The large magnitude of K indicates that the reaction can be considered .irreversible.

Answer: 295; irreversible (b)

FromEq. (11.3): _AGO

In K

=~ = RT

1

-32kJ mor = -1.29 8.3144xlO 3 kJK Igmoll(25+273.15K) K

= 0.275

The relatively small magnitude of K indicates that the reaction is reversible.

Answer: 0.275; reversible

11.2

Eqnilibrium yicld

(a)

From Eq. (11.2) with G6P and GIP representing glucose 6-phosphate and glucose I-phosphate, respectively: K =

CGtiP at equilibrium 0,038 M = = 19 C01patequilibrium O.OO2M

Answer: 19 (b)

From Eq. (11.9) and the reaction stoichiometry:

..

Theorettcal yIeld =

molesG6Pfonned 1mol -1 mo = I mol mol mo1es GlP us ed to ÂŁorm G6P = -1-1

Answer:: I mol mol~l

(c)

From p 260 and using a basis of I litre: Gross ield = molesG6Pfo~ = 0,038m?1 = 0.95 mol mol-1 Y moles GIP supplied 0.04 mol Answer: 0.95 mol mol~l

123

Solutions: Chapter 11

11.3

Reaction rate

(a)

Tenns used to express reaction rate are outlined on p 261.

(I)

Answer: The volumetric productivity is unaffected by change in volume. (ij)

Answer: The specific productivity is unaffected by change in volume. (iii)

Answer. The total productivity is doubled if the fermenter volume is doubled, (b)

Answer. The volumetric productivity is doubled, the specific productivity is unaffected. and the total productivity is doubled. (e) (I) RA

=100 kg d~l; rA =0.8 g I-I h- I , V

From Eq. (1 1.16):

= RA = rA

I

= 52081 08gl-lb-ll~:III~gl lookgd-

Answer: 5208 I (ij)

FromEq, (11.67):

11.4

Enzyme kinetics

Evaluate the enzyme kinetic parameters by plotting slv versus s as a Langmuir plot(p 272), The values are listed and plotted below, s(moll- 1)

Slv (min)

0.0250

12.89

0.0227

11.88

0.0184 0.0135

9.95 750 7.02 5.00 3.93 2.62

0.0125

0.00730 0.00460

0.00204

Solutions: Chapter 11

124

15,----,-----,.----...,

10

!

-I> 5

0 ' - - - - - -.......- - - - - - ' - - - - - - - ' 0.00

0.Q1

0.02

0.03

s (mol 1-1) The equation for the straight line in the plot is slv = 1.70 + 445 s,wheres has units ofmoll-} and Slv bas units aiulin, Therefore, from Eq. (11.39), I/vmax = 445 mOlll min, so that Vmax = 2.25 x 10"3 moII-} min-I. Also from Eq. (11.39), Ktn/ vmax 1.70. Multiplying this value by the result for Vmax. K m = 3.83 x 10-3 moll- t ,

=

Answer: Vmax =2.25 X 10-3 mol 1-1 min'I; K m = 3.83 x 10-3 mol 1-1

11.5

Effect of temperature on hydrolysis of starch

(a)

The activation energy is determined from the Arrhenius equation, Eq. (11.21), with k equal to the initial fate of glucose production. According to the Arrhenius equation, a plot of k versus liT on semi-logarithmic coordinates should give a straight line.. 'The parameter values are listed and plotted below; Tis converted to degrees Kelvin using

Eq. (2.24).

T('C)

T(K)

liT (K'I)

Rate, k (mmel ro w3 s' I)

20 30 40 60

293.15 303.15 313.15 333.15

3,41 x 10.3 3.30 x 10-3

0.31 0.66 1.20 6.33

3.19 x Icy3 3.00 x Icy3

Solutions: Chapter II

125

"1

10

."

E

§.

"g 'g

•

1

I

~

~

'5

~

0.1 2.9

3.1

3.0

3.2

1/Temperature

3.3 X

3.4

3.5

103 (K 1) o

=

The equation for the straight line in the plot is k 1.87 x lO lD c 73OO/ T, where k has units ofmmol m-3 5"1 and T has units ofK. Therefore, from Eq. (11.21), EIR = 7300K From Table 2.5,R = 8.3144 J K~1 gmol- l = 8.3144 x 10-3 kJ Kw l gmol-l; therefore, E= 7300 K x 8.3144 x 10-3 kJ K-I gmol-l = 60.7 kJ grool-l.

Answer: 60.7 kJ gmol- l . (b) Converting 55<>C to degrees Kelvin using Eq. (2.24), T= 55 + 273.15 equation for k obtained in (a):

k = 1.87 x 1010 e-73001T Similarly, for T= 25"C

=328.15 K.

Substituting this value into the

= 1.87 x 1010 e-73001328.15 = 4.08 romol m-3 s·1

=25 + 273.15 = 298.15 K:

k = 1.87 x 1010 e-73OO/T = 1.87 x WiD e-73001298.15 = 0.43 mmol mw3 s·1 Therefore, the rate at 55"C is 4.08/0.43 = 9.5 times faster than at 25"C.

Answer: The reaction rateat55"C is 4.08 mmol m- 3 s·t or 9.5 times faster than the rate of 0.43 rnmol ro· 3 s·l at25"C. (e)

From Table 2.5, the value of R in the appropriate units.is 1.9872 cal K-l gmol- 1, At 55"C = 328.15 K: k

d

= 2.25 x 1027 e-4I,630IRT = 2.25x 1027 e-4l,630/(1.9872 x 328.15) = 0.42h-l

Therefore, from Eq. (11.45), the half-life of the enzyme at 55"C is: t = In2 = In2 = 1.65h h kd 0.42b 1 At 25"C

=298.15 K: kd = 2.25x 1027 e-41,63O!RT = 2.25x 1027 e-4I,630;(1.9872 x 298. 15)

and the enzyme half-life is: th - In2 _ In2 - kd - 6.87x lO-4 h -l

= l009h

= 6.87x lO-4 h-1

126

Solutions: Chapter II

Although the reaction rate is 9.5 times faster at 55 C than at 25°C, the rate of deactivation is 0.42/(6.87 x 104) =: 611 times greater. Therefore, unless there are other considerations. 25°C would probably be the more practical temperature for processing operations.

Q

Answer: The enzyme half·life at 25°C is 1009 h or 611 times longer than the practical operating temperature is probably 2jl'C.

n.6

half~life

of 1.65 h at 55°C. The more

Enzyme reaction and deactivation

K m = 5 roM = 5 x 10-3 gmol t 1 = 5 x Hy6 gmol m- 3. The concentration offat is reduced from 45 gmol m~3 to 0.2 x 45 gmol m- 3 = 9,0 groat m-3: As this concentration range is well above the value of .&tn. s» K m and. from p 269. v "" vmax ' As v= -ds1dt. combining Eqs (11.35) and (11.44) gives:

-::: = VrnaxQ e-kd t In this equation. s and t are the only variables. Separating variables and integrating:

J--<Is = f vmaxOe-kdtdt Applying calculus rules (D.24) and (0.17) from AppendixD and combining the oonstants-ofintegration:

-" = -v"""" e-kdt+ K kd

The initial condition is: at t

=0, s= sO.

Therefore: -Vmaxfl

-$0=

id+ K

Substituting this expression for K into the equation for -s gives: -k t VmaxO -" = --vmaxO - - e ' +---'0 kd k

d

S

= $ 0vmoxO - - - ( 1-e-<d I) kd

At the beginning of the reaction, SO = 45 gmol m-3. Converting the units of VrnaxO:

10001

60 -1. 1lOOOnunol' 1 gmol 11 1 m 11 '1 42 1 -3 ·-1 vmaxO = 007 . nuno11-1 s-1 = 007 . mmo11-1 s 3 . min = . gmo m nun From Eq. (11.45):

_ln2 _ In2 -0087 mm ·-1 k d-----.--. th

8nun

Substituting parameter values into the equation for $, when 80% of the fat is hydrolysed: 0.2 x 45 gmolm-3

= 45 gmol m-3 _ 4.2 gmol m-3 ~in-l (1_e-(l·087 t) 0.087 min

where t has units of min. Grouping tenus:

= 48.276 e-O·087 t e-o·087 t = 0.254 -0.087 t = -1.369

12.276

Solutions: Chapter 11

127

=-

t

15.7 min

Answer:: 15.7 min

11.7

Growth parameters for recombinant E. coli

(a)

The average rate-equal area construction is used to determine growth rates from the concentration data. The data and calculations are tabulated below.

Time, t(h)

x (kg m-3)

<!.x (kg m-3)

<it (h)

<!.xl<it (kg m-3 h- I )

0.0

0.20

0.01

0.33

0.03

0.33

0.21

0.17

0.06

0.22

om

0.5

0.10

0.25

0.40

0.75

0.32

0.15

0.25

0.60

1.0

0.47

0.53

0.50

1.06

1.5

1.00

1.10

0.50

2.20

2.0

2.10

2.32

0.50

4.64

2.5

4.42

2.48

0.30

8.27

2.8

6.9

2.50

0.20

12.50

3.0

9.4

1.50

0.10

15.00

3.1

10.9

3.2

11.6

3.5

11.7

3.7

11.6

0.70

0.10

7.00

0.10

0.30

0.33

-0.10

0.20

-0.50

The values of luIfit are plotted as a function of time below according to the method described on p 263.

16

,

,

,

,

,

,

-

14

rf

12 10

if

8 6

1/

4

.--r-

2

-

/

/

I:".

o -2 0.0

,

,

,

,

,

,

0.5

1.0

1.5

2.0

2.5

3.0

Time (h)

:l 3.5

4.0

128

Solutions: Chapter 11

Results for <ix/dt read from the average rate-equal area curve and the calculated specific growth rates are listed below.

Time, t(h)

o

0.0 0.33 0.5 0.75 1.0 1.5 2.0

o

0.05 0.20 0.45 0.70 1.5 3.3 6.5 10.4 14.5 11.0 3.5

2.5 2.8 3.0 3.1 32

024 0.91 1.41 1.49 1.50 1.57

1.47 1.51 1.54 1.01 0.30

-0.03 -0.07

-0.3 -0.8

3.5 3.7

Values of the-specific growth rate,u are plotted as a function of time below.

2.0

i:.

1.5

~

1.0

..

1'"

0.5

t'" / 0.0

•

-0.5 0

•

1

2

3

4

TIme (h)

(b) As expected for most batch cell cultures, growth occurs at around the maximum specific growth rate for most of the culture period. Taking the average of the values of J1. between times 0,75 hand 3 h• .Ltmax = 1.50 ± 0.05 h- l , where 0.05 is the standard deviation.

Answer: 1.50 h- I (c)

,

From Eq. (11.49), the observed biomass yield Yxs at any point in time is equal to the ratio of the observed growth and substrate consumption rates. The average rate-equal area construction is used to determine the substrate consumption rates rs = -ds'dt from the concentration data, as shown below.

Solutions: Chapter 11

129

Time, t(h)

,(kg m-3)

0.0

25.0

0.33 0.5

24.8 24.8

0.75

24.6

1.0

LI.Y (kg m-3)

Lv (h)

-0.2 0.0

0.33 0.17

0.0

-0.2 -0.3

0.25 0.25

0.80 1.20

24.3

-1.0

0.50

2.00

1.5

23.3

-2.6

0.50

5.20

2.0

20.7

-5.0

0.50

10.0

2.5

15.7

-5.5

2.8

10.2

-5.0

0.30 0.20

18.3 25.0

3.0

5.2

-3.55

0.10

35.5

3.1

-1.45

0.10

14.5

3.2

1.65 0.2

-0.2

0.30

0.67

3.5

0.0

0.0

0.20

0.0

3.7

0.0

0.61

The values of-As/tit are plotted as a function of time below according to the method described on p 263.

,

40

·

35 30 _

·

l-

J

25

1,

1

20

11~

15

~

10

I-

rf I

l-

·

1/ /

5

Y

.t>..

•

o 0.0

·

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

Time (h)

Results for -<is/ dt read from the average rate-equal area curve are listed below, together with the instantaneous biomass yield coefficients calculated using Eq. (11.49) and the values of rx from (a).

Solutions: Chapter 11

130

Time. t(h) 0.0 0.33 0.5 0.75 1.0 1.5 2.0 2.5 2.8 3.0 3.1 3.2 3.5 3.7

0.'

o

0.5 0.6 1.0 1.6 3.0

0.10 0.33 0.45 0.44 0.50 0.45 0.47 0.47

7.'

13.9 22.2 33.0 27.5 7.0 0.4

0."

0.40 0.50 -{J.75

o

, Values of the observed biomass yield from substrate Yxs are plotted as a function oftime below.

• 0

"x

D.•

~ ,

D.•

E

0.4

;;

~

0

e-~

-

'll~

.'" '" 0.3 0"

~

E

0.2

~

0.1

~

.il0

0.0

J 0

-

, 1

,

,

2

3

•

Time (h)

,

During the exponential growth phase between 0.75 b and 3 h when /.l = ,umax, Yxs is approximately constant with an average value of 0.46 ± 0.02 kg kg-I, where 0.02 is the standard deviation.

Answer:OA6 kg kg-I; ~s is approximately constant during exponential growth

U.S

Growth parameters for hairy roots

(a) The mid-point slope method is used to determine rates from the concentration data. The growth data are listed and plotted below according to the method described on pp 264-265. Values of [(x)t+e- (x)t_el are read from the graph.

Time, ted) 0

5 10 15

Biomass concentration, x (g 1-1)

0.64 1.95

4.21 5.5.

e

[(X),+e- (x),-e1 (g 1,1)

5 5 5 5

3.0 3.6 3.8

131

Solutions: Chapter 11 6.98 9.50 10.3 12.0

20 25 30 35 40 45 50 55

3.7 3.4 2.6 1.9 1.2 0.8 0.6

5 5 5 5 5 5 5 5

12.7 13.1 13.5 13.7

16 1.

f

"'"

~

~

12 10

~

j 0 0

~

~

8 6

â&#x20AC;˘

â&#x20AC;˘

2 0 0

10

20

30 Time (d)

40

50

60

The growth rate dxldt is determined using the central-difference formula. Eq. (11.23). These results and values of the specific growth rate J.I. are listed and plotted below as a function of time.

Time, tId)

o 5 10 15 20

25 30 35 40 45 50

0.30 0.36 0.38 0.37 0.34 0.26 0.19 0.12

O.os 0.06

55 Values of the specific growth rate J.I. are plotted below,

0.154 0.086 0.069 0.053 0.036 0.025 0.016 0.009 0.006 0.004

132

Solutions: Chapter 11

-

~

0.15

:!?"-

{

,e ~

0,10

"'" l"'l

;e

0.05

'"

0.00

L~,-~,-~,-::::b!:=L----J

o

10

ro

M

~

W

00

Time (d)

-Answer: Near the beginning of the culture. (bl

The mid~point slope method is used to determine the rate of substrate uptake as a function of time, The sugar concentration data are plotted below according tathe method described onpp 264-265.

35

30

-

25

~

.$ c 0

~

20

~ ~ c

8

,

15

,il,"

'"

I I I I I I I I

10

5

<I I I I I I

0 0

10

20

30

40

Time (d)

Values of[(s)t+e- (s}t-el read from the graph are listed in the table below.

50

60

Solutions: Chapter 11

133

Sugar concentration, s (g 1- 1)

Time, t (d)

e

o

30.0

5

5

5

30 35 40 45 50

27.4 23.6 21.0 18.4 14.8 13.3 9.7 8.0 6.8 5.7

55

5.1

10 15 20

25

-6.2 -6.1

5 5 5 5 5 5 5 5 5 5

-5.7 -5.4

-5.2 -4.6 -4.4 -3.9 -2.7 -1.9

The rate of substrate uptake -ds1dt is determined using the central-difference formula, Eq. (l1.23). These results and values of the specific rate of substrate uptake qs are listed and plotted as a function of time below.

Time, t (d)

o 5

0.62 0.61 0.57 0.54 0.52

10

15 20 25 30 35

0.318 0.145 0.103

0.077 0.055 0.045 0.037 0.031 0.021 0.014

0.46 0.44

40

0.39

45

0.27

50

0.19

55

~

8'

i ! â&#x20AC;˘" l\!

Jl

0.35 0.30 0.25 0.20

"

0.15

2

0.10

0

7;

e

t '"

0.05

0.00 0

10

20

30

40

50

60

TIme (d) (c)

The instantaneous biomass yield coefficient is calculated using Eq. (11.49) and the values of rx and (b). The results are listed and plotted below.

rs from (a) and

134

Solutions: Chapter 11

Time, t(h)

o 5

0.48

10

0.59 0.67 0.69 0.65 0.57

15 20 25 30

35 40 45 50 55

0.43

0.31 0.30 0.32

_ 0

0.8

1!i e jg

0.7

"x

0.6

~ E

,g

·m-•

3iZ~

'"

0.5 0.4

.S- '"

E

.Q

0.3

'"

0.2

~

•

0.1

<3

0.0

~

0

10

20

30 Time (d)

40

50

60

Y

Answer: xs varies during the culture period within the range 0.30-0.69 g g~l dry weight.

11.9

Ethanol fennentation by yeast and bacteria

(al From Table B.8 (Appendix B), the molecular formulae for glucose and ethanol are C6H1206 and C2H{jO, respectively. The reaction equation for fermentation of glucose to ethanol without cell growth is:

From the stoichiometry, the maximum theoretical yield of ethanol from glucose is 2 gmol gmol~l. From Table B.1

(Appendix B) the mol,ecular weights are: glucose:: 180.2; ethanol:: 46,1, Therefore:

Maximumtheoreticalyield =

;gmo:e~anol = gmogucose

2 gmolethanol

,I ~.l~ll = gm

11802g1 1 gmol glucose. 1 ~ol

Answer: 0.51 g g-1 (h)

In the absence of growth. fJ. in Eq. (11.83) is zero and the equation reduces to:

_

0.51gg 1

Solutions: Chapter 11

135

, YpS

mp =-

ms

Therefore, for Yr.s equal to the maximum theoretical yield: mp

= 051 ms

=

=

Foi S, cerevisiae, ms 0,18 kg kg-I h- I ; for Z mobiUs, mS 22 kg kg-I h- l . Applying these parameter values in the above equation gives mp =0.092 h- l for S. cerevisiae and mp =1.12 h- l for Z rrwbilis.

Answer: 0.092 h- I for S. cerevisiae; 1.12 h- l forZ mobilis

(cJ During batch culture at usual glucose concentrations, J1. = J1.rnax for most of the culture period (p 279). Eq. (11.83) can therefore be written as:

,

Yps

=

_y,,;PXc:I'm~,,,,,-+_m-,,p

Jlmax

--+ms Yxs

Substituting the parameter values for S. cerevisiae, including the result for mp from (b):

, _ 3.9kgkg- I (0.4h- I)+0.092h-I _ 043k k- I Yops1 -.gg 0.4 h + 0.18kg kg-I h-I O.llkgkg 1 Similarly for Z 11Wbilis:

ips--

I I I 7.7 kg kg- (O.3h- ) + 1.12h- _ 048k kg-I 1 -. g O.3h+ 2.2 kg kg- I h- I O.06kgkg 1

Answer: 0.43 kg k:g"l for S. cerevisiae; 0.48 kg k:g"l for Z mobilis (dj

Using the results from (8) and (c), for S. cerevisiae: Efficiency

=

, Yps

maximum theoretical yield

= 0.43 kg kg-I = 084 0.51 g g-1

.

For Z mobilis:

, Efficiency = Yps maximum theoretical yield

Answer: 0.84 for S. cerevisiae; 0.94 for Z mobilis (ej

From Eq. (11.70) withf.l = J.!max, for S. cerevisiae:

For Z. mobilis:

Answer: 1.65 h- 1 forS. cerevisiae; 3.43 h- l forZ mobilis

= 0.48kgkg-l = 094 0.51 g g-1

.

136

Solutions: Chapter 11

(f) In Eq (11.70), the growth~associated term is Ypx ,umax, the nonwgrowthwassociated term is mp_ Therefore, using the result from (e), the proportion of ethanol production from growth for S, cerevisiae is: 3.9 kg kg-I (0.4h- l ) Ypxl'max+mp - 3.9kgkg(OAh- 1)+0.092h 1 = 0.94 Ypxl'max

_

so that the proportion from non~growth metabolism is 0.06. For Z mobilis: Ypxl'max Ypxl'max+mp and the proportion from substantial for Z. mobilis.

non~growth

=

I 7.7 kg kg-I (0.3 b- ) 7.7kgkg-I(0.3h-I)+L12h 1

metabolism is 0.33.

= 0.67

Non·growth~associated ethanol

production is more

Answer: For S. cerevisiae, 0.94 of the ethanol production is growth~~ociated and 0.06 is non-growth·associated. For Z mobilis, 0.67 is growth~associatedand 0.33 is non-growth~associated. Z. mobilis produces a more substantial proportion of its ethanol in non~growth·associated metabolism. (g) Fromp261;volumetric productivity is equal to specific productivity multiplied byceU concentration. From (e), as the specific ethanol productivity qp for Z mobilis is 3.43/1.65 = 2.1 times that of S. cerevisiae, to achieve the same volumetric productivity, the concentration of yeast must be 2.1 times that of bacteria.

Answer: 2.1 times the concentration of bacteria (h) From p 261, total productivity is equal to specific productivity multiplied by cell concentration and fermenter volume.

At zero growth. ,u in Eq. (11.70) is zero and q~ = mp, Therefore, from (b), the s~cific productivity qp for S. cerevisiae is 0.092 h· l , and for Z. mobilis, LI2 h' , As the value for Z. mobilis is LI /0.092 = 12.2 times that of S. cerevisiae, to achieve the same total productivity at the same cell concentration, the fermenter volume for the yeast culture must be 12.2 times that for the bacteria.

Answer: 12.2 times the volume for the bacterial culture

(0 From Eq. (11.81) with,u= #max, for S. cerevisiae:

-J- = _1_ + _m_,_ = Yxs

YXS

Pmax

l 1 + 0.18 kg kg- h- l 0.11 kg kg-l O.4h-l ,

Yxs = 0.105 kg kg

= 9.54 kg kg-l

-I

For Z mobilis:

~=_l_+_m_,_= Yxs Yxs J4nax

I 0.06 kg kg- l ,

Yx,

1

+ 2.2 kg kg- h-

l

= 24.0 kg kg-l

O.3h 1

= 0.042 kg kg-I

Answer: 0.105 kg kg.l for S. cerevisiae; 0.042 kg kg- l for2. mobilis. As Z mobilis produces less biomass per mass of substrate consumed and per mass of ethanol produced than S, cerevisiae. biomass disposal is less of a problem with the bacteria.

OJ The ethanol yield from substrate is 12% higher using Z mobilis than S. cerevisiae. the specific productivity is 2.1 times higher so that a smaller and cheaper fermentation vessel is required to achieve the same rate of ethanol production. and Z mobilis produces less than half the amount of biomass generated by S. cerevisiae per mass of glucose consumed. All of these factors mean that Z. mobilis perfonns better than S.· cerevisiae for ethanol production.

Solutions: Chapter 11

137

However, other aspects of the cultures and the ethanol production industry also need to be considered. Z mobilis requires a higher pH (5.0) for growth than S. cerevisiae (3.5-4.0), and is therefore more susceptible to contamination.

Z mobilis also does not grow well on molasses, a common substrate material for fermentations, because of the high salt content. The biomass produced in ethanol fermentation by yeast is often sold for use in animal feeds, whereas application of bacteria for this purpose is not as well accepted in the industry. These are some of the reasons why Z mobilis has not been widely adopted for industrial ethanol production, despite its superior ethanol production characteristics.

11.10 Plasmid loss during culture maintenance FromEq. (11.65), 1

a = f.r = 0.033h- = 1.32 J.l+

O.025h 1

The number of generations of plasmid-containing cells over the 28..day period is, from Eq. (11.66): I

_ ,,", _ (0.025h- )28d路12::I_ nln2 _24.2 ln2

If the fraction of cells containing plasmid is F= 0.66, fromEq. (11.64): 066 =

.

l-a-p = l-L32-p = 0.32+p = 0.32+p 1_1.32_(224.2(1.32+p Il)p l_a_2 n(a+p-lJ p 0.32+ (2 24.2 (0.32+Pl)p 0.32+214.38p(224.2")

Multiplying through by the denominator: 0.211 + 141.49p(2 24 .2") =0.32+p

141.49p(224.2 p)-0.109_p = 0 1bis equation can be solved for p by trial and error. Values of the left-band-side of the equation for various pare listed below, starting withp = 0.001 as the first guess.

141.49 p (2 24.2p) -O.I09-p

p

0.001 0.002 0.0007 0.00076 0.00077 0.000766

0.0339 0.1816 -0.0095 -0.0008 0.0006 0.000017

As the value of the expression wben p = 0.000766 is sufficiently close to zero, the solution can be taken as p = 0.000766.

Answer: 0.000766

11.11 Medinm sterilisation (a), (h) aad (c) The specific death constant kd is evaluated using Eq. (11.46) with A = 1036.2 s-1 and Ed = 283 k:J gmolw1. From Table 2.5, R 8.3144 J K-l grool-1 8.3144 x 10-3 kJ K-l gmol-1, Converting the temperatures to degrees Kelvin using Eq. (2.24), 80"C (80 + 273.15) 353.15 K, 121"C (121 + 273.15) 394.15 K, and 140"C (140 + 273.15) 413.15 K. Therefore, at 8()'>C:

=

=

=

=

=

=

=

1 3 kd = A e--EalR T = 10362 s-l e-283 kJ gmor /(8.3144x 10- kJK""1 gmo1-1 x 353.15K) = 2.20x 10--6 s-1

=

Solutions: Chapter 11

138 At 121째C: k

d

=.

A e-ErJ/RT

= 1036.2 s- 1 e-283 kJ gmorl/(8.3144x 10-3 kJ~l gmor1x 394.l5K)

=.

0.04978-1

At 141째C:

The relationship between the number of viable cells and time is given by Eq. (11.87). Converting the units of the

initial concentration of contaminants xo: XC)

= 108 cells 1-1

=.

108ceusl-l'111~11

=.

10

11

3 cellsm-

Therefore. per m 3, No =. 1011; N =. 10.3 . Substituting values into Eq. (11.87) gives:

10-3

1011 e-/cd t

=.

10-14 -32.24

=. =.

e-kdt -kd t

32.24

t=--

kd

Therefore, at 8We: t =. 32.24 =.

kd

t =. 32.24

kd

t

3224 2.20 x 10-6 8-1

=

kd

=.

3600s

_II min I = 10.8 min

32.24

0.04978-1

= 32.24

.I~I = 4070h

32.24

60s

2.638- 1

Answer: 4070 b at 80oe; 10.8 min at 121째C; 12.3 s at 140째C

= 12.3 s

Heterogeneous Reactious 12.1

Diffusion and reaction in a waste treatment lagoon

(a) A substrate shell mass~balance is perfonned around a thin slice of sludge of area A and thickness Llz located distance

z

from the bottom of the lagoon, as shown in the diagram below.

Wastewater

Iu

Siudga

-----------------------------1------z Bottom of lagoon

//// Substrate diffuses into the shell across the upper boundary at

z+.dz. and diffuses out across the lower boundary at z.

Substrate consumption within the shell follows firstMorder kinetics. Following the procedure on p 301, from Pick's law, the rate of substrate input by diffusion is:

where Vse is the effective diffusivity of substrate in the sludge. Similarly, the rate of substrate output by diffusion is:

The rate of substrate consumption in the shell is kl sA Az.Substituting these expressions into the mass-balance equation, Eq. (4.1), with the rate of substrate generation = 0, at steady state:

Assuming that the substrate diffusivity and sludge area do not vary with distance ;::::

!t'soA(:!,.",-:l)-k1SA'" =

0

Cancelling A and dividing through by tlz" gives:

Taking the limit as A;::: --+ 0 and invoking the definition of the derivative as in Eq. (D. 12) in Appendix D:

~e:(:)-klS = 0

Solutions: Chapter 12

140

or

d 2s Answer: 2JSe d? -k 1 S

:::::

0

(b)

At z::::: L, s::::: Sb. At Z "" O. we assume that the substrate concentration profile reaches a minimum so that ds/dz Answer: At z::::: L, s::::: Jb; at z:::: O. cIs/dz::::: O. (c)

(I) If s:::: N efJz, using differentiation rule Eq. (0.17) in Appendix D:

:

= pNeP'

Wld

Substituting the expressions for s and d2s/dz2 into the differential equation in (a):

!D"p2 N ePZ-k j NeI" = 0 Dividing through by N ePz and solving for p2;

or

(H)

If s::::: N ePZ + M e-Pz, substituting the values for p from (I) above:

s;:; NezVk,/!Dse

+ Me-1:Jk,/Vse

Differentiating this equation using differentiation rule Eq. (D.17) in Appendix D:

:

= N Jkll!lJs. ezJk11flJ" - M Jk 11!D" e-zJk11flJ"

Applying the boundary condition at z::::: 0:

Cancelling the square root terms gives: N=M

Answer.N=M (iii) Substituting N for M in the equation for $ in (6):

=O.

Solutions: Chapter 12

141

, = Applying the boundary condition at z

N(ezJk,l"s. +e-<Jk""s.)

=L: 'b =

N(,LJk, 12lo, +e-LJk,l"s.)

Solving for N:

(Iv) From the definition of cosh x, the numerator in the equation for s in (iii) is equal to 2 cosh (z the denominator is equal to2 cosh (L kl/psc)' Therefore:

J

Jk1/.q.)

- 'b2Cosh(LJkl/.q.)

, _

2 cosh (Z

or

, _ cosh(zJk 11.q.) 'b - cosh(LJk 1/.q,l Answer: QE.D. (d) Taking the derivative of s using the equation for s in (c) (iv);

Evaluating the derivative at z

=L:

Substituting this result into the expression for r A,obs:

r

(L Jk]I:lJ,

sinh - ~ A-~ Jk I Se ~o 1 !OSe COSh(L

A,obs -

)

Jk 1I!OSeSe )

Jkl/.27s )- Similarly, e

Solutions.' Chapter 12

142 (e)

The internal effectiveness factor is defined in Eq. (12.26). For first-order reaction kinetics at substrate concentration Sb everywhere in the sludge, the rate of reaction within the entire sludge volume is:

Substituting this and the expression for rA,obs from (d) into Eq. (12.26) for fIrSt-order kinetics:

rA,obs

1]il

= -.-

'A'

=

Cancelling and grouping terms and applying the definition of tanh x gives:

17i1

=

tanh(L

JkIffD,,)

LJ k1ffDs,

Therefore, applying the definition of 1'1;

tanh¢j

~il

= -¢j-

Answer: Q.E.D, (f)

From the definition of tanh x;

For L = 2 em, the values of ¢I. tanh <ftI and Tlil evaluated for the three sets of conditions are listed below.

Condition

¢I

(1)

0.50 2.0 10

(2) (3)

1]il

0.46 0.96 1.0

0.92 0.48 0.10

The substrate concentration profiles are calculated using the equation for s as a ,function of z from (c) (iii) and applying the definition of 1'1:

The results for Sb =10-5 grool em-] at various values of z are listed and plotted below.

Distance, z (em)

Condition (1)

0.0 0.3 0.7 1.0 1.2

8.87 x 10-<' 8.89x 10-6 9.00 x 10-6 9.15 x 10--6 9.27 x 10-6 9.50 x 10-6 9.78 x 10-6 1.00 x 10-5

I.S 1.8 2.0

Substrate concentration, S (gmo1 em- 3) Condition (2) 2.66 x 10-6 2.78 x 10--6 3.34 x 10-6 4.10 x 10--6 4.81 x 10-6 6.25 x 10-6 8.26 x 10-6 1.00 x 10-5

Condition (3) 9.08 x 10- 10 2.14x 10-9 1.50 x 10- 8 6.74 x 10-8

1.83 x 10-7 8.21 x 10-7 3.68 x 10-<' 1.00 x 10-5

Solutions: Chapter 12

143

.,

1.2

E 0

"

~

-•

1.0

Condition (1)

'& 0.8

4; = 0.50;

1111 = 0.92

x

f

0.6 0.4

g

~ I!,

'"

Condition (2)

0.2 Condition(3) ;1=10;1111=0.10

0.0 0.0

0.5

1.0

1.5

2.0

Distance from the bottom of the lagoon, z (cm)

As tPi increases, 1Jil decreases. the concentration profile becomes steeper, and the minimum suhstrateconcentl'ation in the sludge is reduced.

12.2

Oxygen prome in immobilised-enzyme catalyst

(aJ From p 269, as CAs is O.5mM/0.015:tnM= 33 times the value of Km• as a:frrst approximation we can consider the kinetics to be effectively zero order with ko vmax. Converting the units of ~ to a per volume of gel basis:

=

k o = Vmax

= 0.12 mol s-1 kg- l (0.012kgm-3) .113;~11.11~gl = 4.61 x to-5 kgs- l m-3

Converting CAs to units ofkg m- 3: CAs

III

-1 11000 = 0.5mM = 05x to-3 gmoH. 1m3

III I

g . looog kg = 0.Ol6kgm-3 . 132 gmol

For zero-order reaction. the equation used to determine the substrate concentration inside the beads depends on whether CA remains > 0 throughout the particle. The maximum particle radius for which this occurs can be calculated using Eq. (12.17);

= Therefore. the maximum particle diameter for CA >0 everywhere is 4.2 nun.· Because the immobilised-enzyme beads are smaller than this, CA > 0 and the oxygen concentration profile can be calculated using the equation for zero-order reaction and spherical geometry in Table 12.1. Values for CA asa function of r areJisted and plotted below.

Radius, r (m)

Oxygen concentration, CA (kg m-3)

2.0x

l.60x

1.7 x 10-3 1.5 x 10-3

1.19 x 10-2

1.2 x 1(J3 1.0 x 1(J3 0.8 x 10-3 05 x 10-3 0.2 x 10-3

0.0

9.60 x 10-3 6.63 x 10-3 5.02 x 10-3 3.71 x 10-3 2.28 x 10- 3 L51 x 10-3 1.37 x 10-3

Solutions: Chapter 12

144

20

f

0>

"- 15 M 0

x

<J

"

~

~ § 0

c

~

0

10

5

O'-_ _--''--_ _- l .

0.0

--'-

1.5

0.5

-'

2.0

As the minimum value ofCA at·the centre, of the bead is still about 2.9 timesKm• the assumption of zero·order kinetics is reasonable. (b) As CA > 0 everywhere within the bead. for zero-order reaction tbismeans that the entire bead volume is active.

Answer: 1.0 (0)

FOf zero·order reaction, the maximum conversion rate occurs when the oxygen concentration is greater than zero everywhere in the particle. The largest bead size for this to occur was calculated in (a) is 4.2 rom.

Answer: 4.2 rom

12.3

Effect of oxygen transfer on recombinant cells

(a)

Converting the units of ko to mass: l

3

k o = 10-3 mols- m-

.113~~II.ll~gl = 3.2x 10-5 kg s-l m-3

The maximum particle radius for which oxygen concentration inside the beads remains greater than zero is calculated usingEq. (12,17):

= Therefore, the maximum particle diameter for aerobic conditions is 2 x 1 A5 = 2.9 mm.

Answer: 2.9 rom (b) The particle radius is 1.45/2 "'" 0.725 rom "'" 7.25 x 10-4 m. As this radius is less than Rmax determined in (a), oxygen is present everywhere in the particle. Therefore, as the kinetics are zero-order, T1i "'" 1. From Eq. (12.26) with 1Ji "'" I:

rA,obs = r~s "'" k o "'" 3.2 x 10-5 kg s-1 m-3 Substituting parameter values into the equation for the observable Thiele modulus tP in Table 12.4 for spherical geometry:

145

Solutions: Chapter /2

Using this value, the minimum intraparticle oxygen concentration can be calculated from the equation in Table 12.5 for spherical geometry and f/) < 0.667: CA,min

= CAs(I-~tJÂť) = 8XIO-3kgm-3(1_~(0.167Âť) = 6.0xlO-3 kgm-3

Answer: 6.0 X 10-3 kg m- 3 (c) If the cell density is reduced bya factor of 5, ko is reduced to 1/5 its previous value, Therefore, ko = 115 (3.2 x 10-5 kg s-1 m- 3) = 6.4 x 10-6 kg s-l m- 3 . From Eq, (12,17):

1) 3 3 6 LxI 4 0 9 m2 s8xlO kgm =3,24xlO-3 m=3.24mm 6.4 x 10-6 kg s 1 m 3 Therefore, the maximum particle diameter for aerobic conditions is 2 x 3.24 = 6.5 mm.

Answer: 6.5 rom diameter

12.4

Ammonia oxidation by Immobillsed ceUs

(a)

R = 1.5 mm = 1.5 x 10-3 m. Calculating the observable modulus Dfor spherical geometry from Table 12.6:

3 3 = 0.031 rA,obs = 1.5xlO- m 2.2 x 1O-5 kgs-l mS 6xlO- ms 1(6xIO-3kgm-3} 3 kSCAb 3

l2=!i

From Eqs (12.43) and (12.44),

As CAs"'" CAb, external mass-transfer effects are insignificant.

Answer: Insignificant; the surface oxygen concentration is only 3% lower than in the bulk medium. (b)

From Table 12,7 for zero-order oxygen uptake kinetics, 1Jeo =1. The internal effectiveness factor l1io can be determined from Figure 12.11 as a function of the observable Thiele modulus CPo Evaluating tP from the equation in Table 12.4 for spherical geometry using the result for CAJ> from (a):

R)2 rA,obs (I.5XlO-3 m)2 2.2XlO-5 kgs- 1 m-3 ( 'vAe CAs = fJ)='3 3 1.9XlO-9m2s-1(0.97X6xI03kgm3)=0.s0 From Figure 12.11, attP=0.50,1Jio= I. Using Eq,(12.46), 11T= 1Jio l1eo= I X I = I.

Answer: 1 (c)

Using the results from (a) and (b), the minimum intraparticle oxygen concentration can be calculated from the equation in Table 12.5 for spherical geometry and tJ) < 0.667: CA,min = CAS(l-itP)

= O,97X6XlO-3kgm-3(1-~(0.50)) =

This oxygen concentration is greater than the critical level.

Answer: Yes

1.5 x 10-3 kg m-3

Solutions: Chapter 12

146

12.5

Microcarrier culture and external mass transfer

D p "'" 120 JUD "'" 120 x 10.6 m. The external mass-transfer coefficient can be determined using the equations on p 322 for free~moving_spberes. The Grashof number is calculated from Eq. (12.51) with g "'" 9.8 m s路2 from p 16 and the uni{conversion factor 1 N s m- 2 = 1 kg m- l s路l from Table A.9. Appendix A:

Gr= gD~PL(pp-PL)

= 9.8 m s-2 (I20X lO-6 mj' 103 kg m-3 (1.2 x Io'kgm-3 _1o'kgm-3) = 3.39

K

(10-3 N sm-2

.11 kg m- I ,-I h2 U

1 Nsm 2

Therefore, from Eq. (l2.S2).Rep =Gr/18:= 3.39'18 =0.188. The Schmidt number from Eq. (12.49) is: 10-3 Nsm-2 _II kgm$c

=

=

IlL

PL.vAL

1

8;11

IN.sm

103kgm-3(2.3xlO-9m2s

I)

= 435

Therefore, Rep Sc = 0.188 x 435 = 81.8. As this value is less than 104. the Sherwood number can be evaluated using Eq. (12.55): Sh =

J4 +

1.21(Rep Sc)"路67 =

v' 4 + 1.21 (81.8P路67

= 5.21

From the definition of the Sherwood number in Eq. (12.50): _ Sh.vAL _ 5.21 (2.3 x 10-9 m2 ,-1) _ 999 10-5 -I kS . x ros Dp 120xl0--6 m Using this value of ks to determine

n from the equation in Table 12.6 for spherical geometry: 120xlO-6 m

D = R TA,obs = 3 k s CAb

2

3 0.015 mol s-1 m-

3

9.99 x 10 'ms I (0.2molm-3j

::::: 0.015

From Eqs (12.43) and (12.44);

CA, = CAb(l-D) = CAbO-0.QJ5) = 0.985 CAb External mass*transfer effects are insignificant as CAs ,., CAb. Because respiration is zero-order and the cells are present only-on the surface of the beads. CAs> 0 is all that is required to ensure maximum reaction rate.

Answer: Negligible

12.6

Immobllised.enzyme reaction kinetics

(3)

R =0.8 nun = 0.8 x 10-3 m. As external boundary~layers have been eliminated. CAs::::: CAb =0.85 kg m-3 and fie =1.

The value of Pas defined on p 313 is: 3

jJ::::: Km ::::: 35kgmCAs

::::: 4.12

0.85kgm 3

From Figures 12.10-12.12, this value of p means that the reaction kinetics can be considered effectively f:trst-order.

Evaluating the observable Thiele modulus tP from the equation in Table 12.4 for spherical geometry:

4>

= (li)2 3

From Figure 12.11. at

Answer: 0.12

3

3

1

rA,obs = (0.8 x 10- m)2 1.25 x 10- kg 5- m-3 ::::: 8 0 2 3 . .PAe CAs 3 1.3 x 1O-1l m s-1 (0.85 kg m- )

<1>= 8.0. flil =0.12.

Therefore. from Eq. (12.46). TJT

=11i 11e::::: 0.12 x I ::::: 0.12.

Solutions: Chapter 12

147

(b) From the definition of the effectiveness factor. Eq. (12.26): 3 L25xtO- .kgs-I m-3 =OOlO4k -I -3 0.12 . gs m Por first~order kinetics. r~s

=kl CAs; therefore: k

l

=

• rAJ;

3

= O.OI04kg s-1 m-

= 0.0122s-1

0.85kgm-3

CAs

Answer: 0.0122 s-I (e) The value of 0.0122 s·1 for kt corresponds to an enzyme loading of 0.1 Ilmol got. The Thiele modulus 4't can be evaluated as a function of enzyme loading usingothe equation ·for first-order reaction and spherical geometry from Table 12.2. with kt·directly proportional to the' enzyme 10ad.ing,Por¢>I < 1O,.the-intemal effectiveness factor l1il is determined using the equation in Table 12.3 and the' definition·of coth x; for tf1I > 10. from Eq. (12.30) 1]u II¢!. For each value of kJ, r~s kt CAs. and rA,obs can be determined from these results and the,definition of the effectiveness factor in Eq. (12.26); Calculated values of these'parameters for several different enzyme loadings are listed below.

=

=

Enzyme loading (llJllol g~ 1)

kl (,-I)

¢I

'lit

r'"A, (kg s-t m-3)

rA,obs (kg s-t m- 3)

0.01 0.05 0.10 0.20 050 0.80 1.0 1.3

0.0012 0.0061 0.0122 0.0244 0.0610 0.0976 0.122 0.159 0.183 0.220 0.244

2.6 5.8 8.2 11.6 18.3 23.1 25.8 295 31.6 34.7 365

0.34 0.16 0.12 0.086 0.055 0.043 0.039 0.034 0.032 0.029 0.027

0.0010 0.0052 0.0104 0.021 0.052 0.083 0.104 0.135 0.156 0.187 0.207

3.40x lO4 8.32 x lO4 1.25 x Ily3 1.81 X lO-3 2.86 x lO~3 3.57 x 10~3 4.06 x lO~3 4.59 X 10-3 4,99 x lO-3 5.42 x lO-3 5.59 x to- 3

1.5 1.8 2.0

The resultsfor 11i1 and rA.obs are plotted below as a function of enzyme loading.

0.4,----,,----,----,-----, 6 5

r¥-

I! =w~

0.3

'A,obs

4

3

02

1"

" ~ '0

x ~

<!-

2 0.1

~

l! c

0

1

II ~ ~

0.0 ' -_ _-..J'-_ _- l .

0.0

0.5

-'-

1.0 Enzyme loading (1lffi01 g·1)

1.5

...J

0 2.0

0:

148

Solutions: Chapter 12

As the enzyme loading is increased from 0.01 ~ol got, the effectiveness factor drops significantly. Although the reaction rate continues to rise with increasing enzyme loading, at loadings above about 0.5 !J.mol g-t, less than 5% of the potential activity of the enzyme is being utilised. Further increases in enzyme loading therefore represent an effective waste of more than 95% of that enzyme,

12.7

Mass-transfer effects in plant cell culture

(a) Dp ::::: 1.5 nun =: 1.5 x 10-3 m. The particle Reynolds number is evaluated using Eq. (12.48) withPL =: density afwater ::::: 103 kg m- 3, andJiL =: viscosity afwater::::: 1 cP (p 133) = 10-3 kg mol 5. 1 (Table A.9. Appendix A):

Re = DpUpLPL = 1.5 x 10-3 m(0.83 x 1O-2 ms-l)(lO'kgm-3j = 12.5 PL

p

10 3 kgm 1 8-1

As this value is within the range 10 < Rep < 104, the external mass-transfer coefficient can be determined using Eq. (12.57) for spberical particles in a packed bed. Converting the diffusivity units to m 2 8. 1:

!lJ.

Ae

iJ)AL

::::: 9x 1O-Qcm 2 s- 1

= 9x 10-6 cm2 s-1 .I~p 100cml

=2 x iJ)Ae =2 x 9 x 10-10 m 2 s·1 =1.8 x 10-9 m 2 s-l.

= 9x 1O-10 m 2 s-1

The Schmidt number from Eq; (12.49) is:

From Eq. (12.57), the Sherwood number can be evaluated as:

5 SeO.33 = 0.95 (12.51"·5 (556)°·33 = 27.0 Sh = 0.95 ReO. p From the definition of the Sherwood number in Eq. (1250): k

= Sha/AL s

9

2

1

= 27.0(1.8 X 1O- m s- ) = 3.24xlO-5 ms- I 1.5 x 10 3 m

Dp

'This value of ks can be used to-determine the observable modulus for external mass-transfer D from the equation in

Table 12.6 for spherical geometry. As the specific gravity of the wet cells is 1. from p 16, 1 g wet cells occupies a volume of 1 cm3 and rA,obs = 0.28 mg cm-3 hoi. Converting these units to kg s~I m- 3: rA,obs

= 0.28mgcm-3. h-I

3

I

= 0.28 mg cm- h-

'11~6k~gl.13~sl;IIi'~m j3 = 7.78 X 10-5 kg s-I m-3

Substituting this and the other parameter values into the equation for D: 1.5 X 10-3 m 2

7.78x 10-5 kg s-l m-3

_ 0075

3.24XlO-5ms-I(8mgl-I)·I...!!'Lj·jlOOO11- .

1 m3 1Q6 mg From Eqs (12.43) and (12.44):

As CAs is close to CAb. external mass-transfer effects are present but small.

Answer: The effect is small; the surface oxygen concentration is 7.5% lower than in the bulk medium. (b)

Evaluating the observable Thiele modulus tP from the equation in Table 12.4 for spherical geometry Using the result for CAs from (a):

149

Solutions: Chapter 12

<1>=(li)2 3

1.5X rA,obs

=(

aJAeCAs

to-3m)2

1 3 7.78XlO'.." skgs- m-

2

-073

9X10-lOm2s-1(o.925X8mgl-I).!16kg !.!1~1!- .

10 mg 1m

3

From Figure 12.11 for zero-order reaction, at t1J = 0.73, 17io is very close to but slightly less than 1.0. Therefore, internal mass-transfer effects can be considered negligible.

Answer: Negligible (cJ CAs = 0.925 x 8 mg 1" I = 7.4 mg I-I. As 1Jia is only very slightly less than 1.0, it is likely that oxygen is exhausted just close to the centre of the clumps. Taking rA,obs to be essentially equal to the intrinsic zero-order rate constantko, the maximum particle radius for the oxygen concentration to remain greater than zero throughout the clump can be evaluated using Eq. (12.17):

_ -

(9xto-lOm2S-1)7.4mg'-1路1'6kQI'~11路

F!-

Rmax

= 7.2 x 104

1Omgl'

1m

7.78 x 10-5 kg s-1 m 3

m

= 0.72 mm

Therefore, the maximum particle diameter for oxygen through to the centre of the clump is 2 x 0.72 = 1.4 mm, which is only slightly less than the plant cell clump diameter of 1.5 mm.

Answer. The oxygen concentration falls from 7.4 mg I-I at the external surface to zero just near the centre of the clumps.

12.8

Respiration in mycelial pellets

(aJ R = 2.5 mm = 2.5 x 10-3 m. The presence of external boundary-layers can be checked by calculating the observable modulus for external mass-transfer, D. From Table 12.6 for spherical goomelry:

,Q=!!..

rA,obs 3 kSCAb

= 2.5x10-3 m

5 1 3 8.7XlO- kgs- m3.8x10 5 ms 1(8XlO-3 kgm-3)

=024 .

From Eqs (12.43) and (12.44): CAs = CAb(1-D) = CAb(1-0.24) = 0.76 CAb = O.76(8Xl(r3 kgm- 3) = 6.1xlO-3 kgm- 3 As. CAs is significantly less than CAb, external mass-transfer effects are present.

Answer: Yes (bJ As oxygen uptake is considered a zero-order reaction, for CAs> 0, l1eo = 1.

Answer: 1 (c)

In the absence of internal an;! external mass-transfer resistances, th~ reaction rate is r~b corresponding to CA = CAb throughout the pellets. As rAb is related to rA,obs by Eq. (12.45), r b can be determined if we know 1JTo. Evaluating the observable Thiele modulus <P using the equation in Table 12.4 tor spherical geometry and the result for CAs from (a):

150

Solutions: Chapter 12

From Figure 12.11 for zero-order reaction. at <P = 5,66,1'/io = 0.30. Therefore, from Eq. (12.46), as TJeo "'" 1 from (b), l7To "'" TJio TJeo "'" 0.30 x 1 = 0.30. Using Eq. (12.45): r

'* A,obs rAb""'--"'" ~ro

87 10-'k -1 -3 ,x gs m = 29 ,x 10'""kgs -1 00-3 0.30

Answer. 2,9 x 10'4 kg lSI 00,3, or more than three times the rate actually observed (d)

If external mass-transfer effects were eliminated, CAs = CAb = 8 x 10-3 kg 00,3, and the observed reaction rate would be greater than 8,7 x-1O- 5 kg g-l.m- 3, Under these conditions. an-expression for the observable Thiele modulus ep from the equation in Table 12.4 for spherical geometry is:

~ _(R)2

.... - -

3

3

_ (2.5 x 10- 00)2 r A,obs 4 ( ) = 4.96x 10 rA,obs :tJAeCAs 3 1.75xI0-9 m 2 s 1 8xlO-3kgm 3 r A,obs

-

where.rAObs has units of kgs- 1 00-3 .- Because the reaction is zero--order, r~s = r~b; therefore, from the result in (c), r~s = 2,9'x 10-4 kg g-1 00,3, Using Eq. (12.26): r A,obs

'ho

r A,obs

=-.= 2.9 x 10-4 kg C 1 00-3 rAs

tP and TJio are related by the curve in Figure 12.11 for spheres and zero-order reaction. The value of rA,obs can be determined trial-and-error using Figure 12.11, and the equations derived above, As a ftrst guess, take r A,obs = 2.0 x 10-4 kg S' 00,3. Depending on the difference between the values of 1]io obtained from the figure and from the equation with r ~s' adjust rA,obs as shown in the table below.

bl

11iO = r~Obs

rA.obs (kg s-I 00-3)

<1>

1Ji.o (from Figure 12.11)

9.92 4.96

0.19 0.34

rA, 2.0 x 10-4 1.0 x 10-4

0.69 0.34

Since the values for 11io in the last row are as close as practical, rA,obs "" 1.0 x 10-4 kg s-I 00-3. Therefore, compared with the observed reaction rate of'8.7 x 10-5 kg Sol 00- 3 in the presence of both 'internal and external mass-transfer resistances, eliminating the external boundary layers increases the reaction rate by about 15%.

Answer. 1.0 x 10-4 kg lSI

00- 3

Reactor Engineering 13.1

Economics of batch enzyme conversion

For 75% conversion, sf = 0.25 so- TIle batch reaction time for enzyme processes is evaluated using Eq, (13.10):

Km I So sO-sf tb = - - n-+-Vmax sf vmax

=

1.5 g 1-1

In

O.9gI 1 h 1

3g1-1

+3gt- 1-O.25X3g1-1 __ 4.81h

O.25X3g1 1

O.9gI 1 h 1

The operating cost is therefore: Operating cost = 4,81 h

-I ~:~l

$4800 day-l

= $962

The cost of downstream processing per kg product is: C

= 155 -0.33 X = 155 -0.33 (75) = $130.25 t<g-I

The mass of product formed is detennined from the mass of substrate consumed. which is equal to the change in substrate concentration multiplied by the volume of the reactor V: Mass of substrate consumed = (so-sf) V = (3-0.25x3)gl-1 (16001) = 3600g As 1.2 g product are fonned per g substrate consumed:

Mass of product fonned = 1.2 x 3600 g "'" 4320 g "'" 4.32 kg Therefore:

Downstream processing cost = $130.25 kg-I (4.32 kg) = $563 The revenue from sale of the product is: Revenue"", $750 kg- 1 (4.32 kg) = $3240 Therefore the cost benefit at 75% substrate conversion is: Cost benefit = revenue - operating cost-downstream processing cost = $3240- $962 - $563 = $1715 Carrying out the calculations for 90% conversion. the batch reaction time for Sf = 0.10 So is:

= 6.84h At 90% conversion, the operating cost is increased due to the longer reaction time~ Operatingcost = 6.84h.j

;:~I.$4800day-l = $1368

The cost of downstream processing per kg product is~ C = 155 - 0.33 X = 155 - 0.33 (90) = $125.30 t<g-I

The mass of substrate consumed is:

152

Solutions: Chapter 13 Mass of substrate consumed

=:

($0- Sf) V

=:

(3 - 0.10 x 3) g I-I (1600 1)

=:

4320 g

and the mass of product formed is: Mass of product formed = 1.2 x 4320g "'" 5184g

=:

5.18 kg

Therefore, the downstream processing cost is: Downstream processing cost

=:

Sl2S.30kg-l (5.18 kg)

=:

$649

The sales revenue is: Revenue

=:

$750 kg-! (5.18 kg) = $3885

Therefore, the cost benefit at 90% substrate conversion is: Cost benefit

=:

revenue - operating cost- downstream processing cost "'" $3885 -$1368 - $649

The gain per batch from increasing the conversion from 75%:to90% is therefore $1868 -" $1715

=:

=:

$1868

$153.

Answer: There is a gain of $153 per batch; representing a 9% increase in the cost benefit at 75% conversion.

13.2

Batch production of aspartic acid using cell-hound ~nzyme

(0) The initial concentration of substrate So =: 15% (w/v) =: 15 g per 100 ml =: 150 g 11. The final substrate concentration Sf = 0.15 So = 0.15 x 150 g 1-1 = 22.5 g 1-1, Calculating the deactivation rate constant at 32"C using Eq, (11,45):

= ~ I~I = 275 IO-3 h-1 k d = In2 th 1O,5d'24h . x For enzyme subject to deactivation, the batch reaction time is evaluated using Eq. (13.13);

tb =-lln[l-k Km In SfSo + SO-S,)] d ( vmaxo k Vmaxo d

=

l 1 1 In 1_2.75XIO-3h-l( 4.0gl- 10 lSOgl- +(150-22.5Jgr ) = 23.6h 2.75 x 10-3 h- 1 5.9 gl 1 h-1 22.5 g I-I 5.9 g I-I h- 1 -I

At 37"C, the deactivation rate constant is;

k

d

= In2 = In2 I~I = 126 1O-2 h-l t

h

2.3d路 24h

.

x

The batch reaction time is;

m In So + so-S,)] ( K tb = -lln[l-k d k vmaxo Vmaxo Sf

d

=

-I L26xlO 2 h-l

1 1 In l-l.26XIO-Zh-l( 4.0gl- In 150gl- + (150-22.5j gl-I) = 17.7h 8.5gl- 1 h- 1

22.5g1 1

8.5g11h 1

As the batch reaction time is lower at 37"C, 37"C is the recommended operating temperature.

Answer: 37"C (h) From Eq. (13.33), the total batch reaction time at 37"C is:

153

Solutions: Chapter 13 tT =- 1b+tdn =- 17.7h+28h =- 45.7h

Therefore. in one year. the number of batches carried out is:

365d!W! Numberofbatches =- 45.7 h per batch =- 192 In each batch, the mass of ammonium fumarate converted is 0.85 x 150 g r 1 =- 127.5 g 1-1 multiplied by the reactor volume V. Therefore. the mass of substrates converted is 1275 V g =- 0.1275 V kg, where V has units of litres. From the reaction stoichiometry, as the molecular weights of ammonium fumarate and aspartic acid are approximately equal, the mass of aspartic acid produced is also 0.1275 Vkg. After one year or 192 batches. the mass of aspartic acid produced is 0.1275 V x 192 =- 24.5 V kg. Using the conversion factor 1 tonne =- 103 kg (Table A.3, Appendix A), the target level of aspartic acid production each year is 5000 x 103 =- 5 x 1()6 kg. To teach this target level: 24.5 Vkg = 5 x 106 kg

V= 2.04 x 1051

= 204m3

Answer: 204 m 3

13.3

Prediction of batch cnllnre time

(aJ

The initial cell concentration Xo =- 12 g/loo I =- 0.12 g 1-1. Assume that stationary phase is reached when Sf =- O. The batch culture time can be determined using Eq. (l3.27):

I

l] =- 4.3h

gg- ( XS ] =- - -Il I n [0.575 tb =- - -I I n [Y l+-(so-sr) 1+ I lOgl-I -0 Pmax Xo 0.9 h0.12 g I Answer: 4.3 h (bJ

If only 70% of the substrate is consumed,

tb = _1_ln[1 + YXS Pmax Xo

Sf

=- 0.3 So =- 0.3 x 10 g I-I =- 3 g I-I. From Eq. (13.27):

('o-'d] = _1_1 1n[1 + 0.575 g g~I flO g I-I _ 3 grIl] = 3.9 h 0.9h-

0.12g'-

The biomass density at this time can be calculated from Eq. (13.19): 1 xf =- Xo ePmax iI =- 0.12 g.-I e(O.9h- x 3.9 h) =- 4.0 g I-I

Answer: 4.0 g 1-1

13.4

Fed-batch schednling

(aJ

The initial substrate concentration So =- 3% (w/v) =-3 g per looml =- 30 g I-I. The batch culture time to achieveS{=- 0 . is determined using Eq. (13.27):

tb = _1_ln[l+ YXS(so-'d] = I 1 1n[1+ 0.5 gg-I(30gl-I-Ol] = 13.3d Jlmax Xo 0.18d1.5g1-1 The biomass density at this time can be calculated using Eq. (13.19): xf =- xoePmuft. =- 1.5 g 1-1

i0.18a' x 13.3 d) =-

16.4g1-1

Answer: The batch culture time is 13.3 days; the final biomass concentration is 16.4 g 1. 1.

154

Solutions: Chapter 13

(b) The mass of cells at the start of fed-batch operation is equal to the fmal batch cell concentration multiplied by the

initial medium volume: Xo = xf V = 16.4 g 1,1 (100 I) = 1640 g

The fmal mass of cells after 40 d fed-batch culture can be determined using Eq. (13.50):

Answer, 4.04 kg (c)

The mass of cells produced in each reactor run is equal to the final biomass minus the biomass used for inoculation: Biomass produced per run

:=

4040 g - 1.5 g j-l (100 I) = 3890 g = 3.89 kg

By analogy with Eq. (13.33), the total reaction time is:

where lb is the batcbreaction time and lfb is the fed-batch operation time. Substituting parameter values using the result for 1b from (a): IT

= !b+tfb+tdn:= 133d+40d+ld = 54.3d

In one year, the number of runs carried out is: 275d Numberofruns = 54.3dperrun :: 5.06 The total biomass produced annually is equal to the biomass produced per run multiplied by the number of runs per

year: Biomass produced per year = 3.89 kg x 5.06

:=

19.7 kg

Answer: 19.7 kg

13.5

Fed-batch production of cheese starter cultnre

(a)

An expression for the liquid volume as a function of time during fed路batch reactor operation can be derived from an unsteady-state total mass balance as shown in the solution to Problem 6.7afrom Chapter 6. Using this expression:

Vo

=:

V-Ft

=:

40m3 _4m 3 h路 l (6 h) = 16m3

Answer: 16 m3 (b) From the definition of the dilution rate in Eq. (13.39), after 6 h cffed-batch operation when V"" 40 m 3 ; D

=!:.. = V

3 1 4m b40m3

=:

O,lOh-1

Substituting this value into Eq. (13.45) for the substrate concentration at quasi*steady state:

s=

DKS Iimax-D

I

3

= O. 10 h- (0.I5kgm- ) = 0.06kgm-3 0.35h I_ O. lOb 1

Answer: 0.06 kg ro路 3 (c) Taking maintenance substrate requirements into account, for qp:: 0, Eq. (13.43) becomes:

Solutions: Chapter 13

155

£Is dt = At quasi~steady state, dsld! '" 0, J1 '" D, ands«

D(S.-S)-(.J!....+ms)x 1 Yxs

Sj.

Therefore, the equation reduces to:

o = DSi-(~s +ms)x Solving for r.

1

x

3

Ds, O.10h- (SOkgm- ) =,,-'=__~T'--""=':"-.1.= 14.0kgm-3 _D_+ ms Yxs

O.lOh I +0.135kgkg- I h-1 0.23 kg kg 1

Answer: 14.0 kg m-3 (d)

After 6 h fed~batch operation, the mass of cells is:

x = xV =

14.0 kg m-3 (40 m3) = 560kg

At the start of fed-batch 0r.:ration when the liquid volume is 16 m3 , if operation is at quasi~steady state, the cell concentration", 14.0 kg m- and:

x=xV=

14.0 kgm- 3 (16 m3) = 224 kg

Therefore, the mass of cells produced during fed~batch operation is (560 kg - 224 kg) = 336 kg.

Answer: 336 kg

13.6

Continuous enzyme conversion in a fIXed-bed reactor

=

Convert the parameter values to units of kg, m, s. Km. =0.54 g 1~1 =0.54 kg m- 3. During the reactor operation, s 0.02 g 1~1 = 0.02 kg m~3;si = 0.42 g 1~1 = 0.42 kg m~3. R = 1 mm = 10- 3 m. The active enzyme concentration per unit volume of catalyst ea is: ea = lO-4 g 250cm3

=

lO-4 g 250cm3

.1~1.I!oocmI3 = 4xlO-4 kgm-3 looog

1m

The effective diffusivity of urea in the gel VAe is: Ij)

Ae

= 7xlO-6 cm2 s-1 =

7XI0-6cm2s-I'I~12 l00cm

= 7xlO- lO m 2 s-1

From Table B.l (Appendix B), the molecular weight of urea is 60.1 and the molecular weight of NH4+ is 18.0.

Therefore, from the stoichiometry, reaction of 60.1 g urea produces 2 x 18.0 =36.0 g Nf4+. Expressing the turnover number k2 in terms of urea: k2

= l1,OOOg NH1 (g enzyme)-l s-l = 11,000 g NH1 (g enzyme)-l s-l .1 60 .1

ore: 1 36.0gNH4

k2

= 1.84 x 104 g urea (g enzyme)"l s~l

= 1.84x 104 kg kg-I s~l

From Eq. (11.33), Vmax expressed on a'per volume gel basis is: Vmax = k2 ea = 1.84 x 104 kg kg~l s·l (4 x 10-4 kg m~3) = 7.36 kg m~3 s-l

As there are 250 em 3 gel per litre of liquid in the reactor, Vrnax expressed on a per volume liquid basis is:

156

Solutions: Chapter 13

1~131100011= . 1m3

3 vrnax =736k . gm-3 s_1(2S0Cm 11 ) . lOOcm

184k . gm-3-1 s

The rate of reaction can be determined after evaluating the effectiveness factor in the absence of external boundary layers. From the definition of f30n p 313 with CAs = s:

p=

3 Km = O.54kgm- "'" 27 s O.02kgm-3

From Figure 12.10, for this value of p the reaction kinetics can be considered first-order. Based on Eq. (11.36), the effective first-order rate constant kl is:

-3-1 I .84kgills "'" 3.418-1 3 0.54kgm-

_Calculating the Thiele modulus from the equation in Table 12.2 for first-order kinetics and spherical geometry:

As lPI > 10. fromEq. (12,30):

1

~il = <Pi /

I 23.3 = 0.043

From Eq. (12.46), as TIe = 1, 1]T= 0.043. The flow rate of urea solution iDto and out of the reactor can be detennined by evaluating the dilution rate D in the mass-balance equation, Eq. (13.54): D =

7V~_T,;-V:;m7""';-'--:K (Km + $) (si

$)

3 = -,0",.04=3"(I".8"4",kg"m"-,3-:o,-_IL)0",.0,,,2:.okg=mc.-_,, = 7.06 x 10-3 s-I (0,54+ 0.02) kg m-3 (0.42-0.02) kg m-3

From the definition of the dilution rate, Eq. (13.39): F = D V = 7.06 x 10-3 s-l (11) = 7.06 x 10-3 1 S'"l

In 30 min. the volume of urea solution treated is: Volume treated = 7.06

x

3 10- I s-I (30 min)

路ll~n

I

= 12.71

Answer: 12.7litres

13.7

Balch and conllnuous hiomass production

= 4% (w/v) = 4 gper 100ml =40 g 1-1 =40 kgm- 3 . Sf = s=0.02x40kg m-3 =0.8 kg m- 3. For the batch reactor. X() = 0.01% (w/v) = 0.01 g per 100 ml = 0.1 g I-I = 0.1 kg m- 3.

SO= Si

The batch culture time can be determined from Eq. (13.27):

1

1

1

1

' 3] = 11.6h

4I - 3 (40-0.8) kg mtb = " In [Y I +-('O-Sf) = lin [10 +. r-max XO O.44h O.lkgm

xs

gg

The biomass density at this time is obtained fromEq. (13.19):

Xf = Xoe"mu;, = 0.1 kgm-3 i0.44h-1 x I 1.6 h) = 16.5kgm-3 Calculating the mass of cells produced per batch:

x

= (xf-x,,) V = (16.5 - 0.1) kg m- 3 (1000 m3) = 1.64 x 104 kg

Solutions: Chapter 13

157

If the downtime between batches tdn is 20 h, from Eq. (13.33):

IT = 1b+tdn = 11.6h+20h = 31.6h Therefore. in one year, the number of batches carried out is:

_ 365dl楼*1 _

Numberofbatches - 31.6bperbatcb - 277 The total annual.biomass.productionfrom batch culture is therefore 1.64 x 104 kg x 277 = 4.54

x 106 kg.

For continuous reactor operation, the steady'-state cell concentration is given by Eq. (13.62): x = (Si-S)Yxs = (40-0.8)kgm- 3 (0.41gg- l ) = 16.1kgm-3

The dilution rate D corresponding to S = 0.8 kg m- 3 can be detennined using Eqs (13.57) and (lL6O):

1 3 M4h- {O.8kgm- j

D = p = I'm",,' = Ks+s

07 . mg r

1

= O.44h-1

08kgm-3 . 11000111~1 3' 6 +.

1m 10 mg

From the definition of the dilution rate. Eq. (13.39):

F = D V = O.44h- l (1000m3) = 440m3 h- l The rate of biomaSs production F x= 440 m3 h- I x 16.1 kgm~3 = 7084 kg b- 1. The number of days per year available for continuous reactor operation is (365 - 25) = 340 d; this corresponds to 340 d x 24 h d-} = 8160 h. Therefore. the total biomass produced per year is 7084 kgh-} x 8160 h =5.78 x 107 kg. This production level is 5.78 x 107/4.54 x 106 = 12.7 times the amount produced using batch culture.

Answer: The annual biomass production using continuous operation is 5.78 X 107 kg, which is 12.7 times the production of 4.54 x 106 kg from batch culture.

13.8

Reactor design for immobilised enzymes

So = Si = 10% (w/v) = 10 g per 100 ml = 100 g 1. 1 = 100 kg m- 3. Sf = S = 0.01 x 100 kg m-3 = 1 kg m- 3. Based on the -unsteady路state mass-balance equation derived in Example 6.1 in Chapter 6 for first~order reaction, the equation for the rate of change of substrate concentration in a batch reactor is:

d(V,) _ -Ie V (itIS where V is the reaction volume and k} is the reaction rate constant. As V can be considered constant in a batch reactor, this term can be taken outside of the differential and cancelled from both sides of the equation: ds

dt =-k 1 s

The differential equation contains only two variables, sand t. Separating variables and integrating: ds

,

- = -kl dt

Using integration rules (D.27) and (D.24) from Appendix D and combining the constants of integration: Ins=-klt+K The initial condition is: at t = 0, S = so. From the equation, therefore, In So = K. Substituting this value of K into the equation gives:

Solutions: Chapter 13

158 lns = -kl t+lnsQ

,

In-=-kt

f

'0

The batch culture time 1b is the time requited for the substrate concentration to reach Sf. -l

sf -10-

So

tb = - - = k1

n

1kgm-3 3

lOOkgm

O.8XlO-4Cl路13~sl

= 16.0h

If the downtime between batches too is 20 b, from Eq. (13.33):

tT = tb+tdn

= 16.0h+20h = 36h

Therefore, in one year, the number of batches carried out is: 365d'

lITl 24h

Numberofbatches = 36hperbatcb = 243 To treat 400 tonues penicillin G annually, using the unit conversion factor 1 tonne:= 103 kg (Table A.3, Appendix A):

Mass of penicillin G treated per batch =

her 400 fh,onn " num 0 atehesperyear

I=

3 g 400tonnes 路1_l0__k_ = o.ci--'-l"to"n"ne", 243

1.65 x 103 kg

As the concentration of penicillin G added to the reactor is tOO kg m- 3:

Reactor volume

= I.65xl0

3

k:g 100 kg m-3

= 165m3

The batch reactor volume required is 16.5 m3.

For a CSTR operated under steady state conditions, Fi ::::: Fo = F, V is constant, and ds/dt = O. Therefore, themass~ balance equation for frrst路oro.er reaction derived in Example 6.1 in Chapter 6 becomes:

0:= FSi-Fs-1q sV 0:= FIV(Si-S)-kl sV Solving for Fly.

F;

_ kt s _ O.8 X lO-4 s-1(1kgm-3) _ 808 10-7 - 1 _.X s si -s (100-1) kg m 3

V----

The flow rate of penicillin G into the CSTR is 400 tonnes per year. Using the unit conversion factor 1 tonne:= 103 kg (Table A.3, Appendix A) and the concentration of substrate in the feed stream Si := 100 kg m-3, the total volumetric flow rate of the feed stream F is:

F:=

I I

i lId Illh

3 1Yeai -1 10 kg . 1 365 d . 24 h . 3600 s 400 tonnes year . 1 tonne

Applying this with the above result for Fly:

100 kg m-3

I

:= 1.27x 10-4 m3 s-1

159

SolutWns: Chapter 13

1.27 x 1O-4 m3 s-I F v=-= FIV 8.08xlO 7 8-1

= 157m3

The CSTR reactor volume required is 157 m3 .

For the PFTR, if the density of enzyme beads is four times greater than in the other reactors. kl = 4 x 0.8 x 104 s路l = 3.2 x 10-4 s路l. By analogy with Eq. (13.83), the differential equation for change in substrate concentration with position in the reactor for first路order kinetics is: <Is

Uctz=-k1s

The differential equation contains only two variables, s and z. Separating variables and integrating: <Is -k, -=-dz S u

Using integration rules (D.27) and (0.24) from Appendix D and combining the<;onstants"of integration:

-k,

Ins = -z+K u

The boundary condition is: atz= 0,

S

= si. From the equation, therefore, In sf= K. Substituting this value of Kinto

the equation gives:

-k,

Ins = -z+lnsj u

At the end of the PFIR, z = L and S = Sf, therefore:

Applying the defInition of the reactor residence time .. from Eq. (13.85): sf

In- =-kl't' si

Rearranging and solving for 't':

Note that this is 1/4 the value obtained for the batch reaction time th. as expected from the analogous kinetic characteristics ofbatch and PFTR reactors and the 4 x higher value ofk] in the PFTR.

As calculated for the CSTR, F = 1.27 X 10-4 m3 8. 1. Therefore. from the definition of .. in Eq. (13.51): V

= -rF = 4.0 h.! 3~~ 8 j1.27X 10-4 m3 s-I = 1.83 m3

The PFrR reactor volume required is 1.83 m3 .

Answer. The smallest reactor volume is 1.83 m3 for a PFrR.

Solutions: Chapter 13

160

13.9

Two-stage cbemostat for secondary metabolite production

(a) Sj:::::

10 g 1~1 ::::: 10 kg m- 3 , The dilution rate, which is the same for both reactors, is calculated using Eq. (13.39):

D

= f. =

1m3 501h-1 1 I 10001

.

V

a.5m3

= O.lOh-'

The cell and substrate concentrations entering the second reactor are the same as those leaving the first reactor. The substrate concentration can be determined using Eq. (1358):

s

= .-!'.~ = O.lOb-' (l.Okgm-3) = S.Okgm-3 JJmax- D

O.12h-1-O,lOh- 1

When maintenance requirements are significant, the cell concentration is calculated using Eq. (13.61): x:::: D(sj -s)

.D. +ms

Yxs

= _O"'c.lO"b"-.'",(lcoO.-.cS".O",),,kg,.m=.â&#x20AC;˘ -3_ = 2.2 kgm-3 O.lOh-

1

+ 0.025 kg kg-1 h-1

O.5kgkg I

Answer: The cell concentration is 2,2 kg m-3; the substrate concentration is 5,0 kg ro- 3. (b)

As growth is negligible in the second reactor,-x ::::; Xi ::::: 2.2 kg m- 3. The substrate concentration is determined by rearranging Eq. (13.59)and solving for s with fJ::::: 0: qp

)

s::::: $i- ( Y +mS x ps

V

p

Substituting the parameter values with Si ::= 5.0 kg m- 3;

For the two reactors together: (Si- S) (lO-O.3l)kg m-3 Overallsubstrateconversion::= _ - x 100%::= 3 x 100% = 97% si lOkgm

Answer: 97% (cj

As product is not formed in the first reactor, Pi = 0 for the second reactor. The product concentration is determined by rearranging Eq. (13.64) and solving for p: p

= qpxV = O.16kgkg-1b-'(22kg~-3)0.5m3 = 3.Skgm-3

501h-l¡ll~11

Answer: 3.5 kg m- 3

13.10 Kinetic analysis of bioremediatlng bacteria uslug a cbemostat (a) From Eq. (13.92), Jlmax and Ks can be detennined from the slope and intercept of a plot of SID versus s. From the definition of the dilution rate in Eq. (13.39), values of D are evaluated from the experimental flow rates using V = 11 = 1000 ml. The measured substrate concentration at 50 ml h- l indicates that washout occurs at this flow rate; therefore, this result is not included in the kinetic analysis. The data are listed and plotted below.

Solutions: Chapter 13

161

Flow rate, F (ml h"l)

Dilution rate, D (h"l)

10 15 20 25 30 35 50

0.010 0.015 0.020 0.025 0.030 0.035

Substrate concentration, S (J,lM)

1740

17.4 25.1 39.8

1673 1990 1872 2313 2289 2000

46.8 69.4 80.1 100

O.OSO

SID OtM h)

2500

• •

•

2000

:2

~

• •

ore,

•

1500

1000

l-~_...L_~_L.~_...L_~_L._~....J

o

20

40 60 Substrate concentration,

80 S

100

(p.M)

The slope of the straight line in the plot is 10.48 h; the intercept is 1493 liM h. From Eq. (13.92), the slope = llPmax; therefore, J.tmax =1/10.48 h = 0.095 h"l. The intercept = KS/p:max;thereforeKg = 1493 J.l,M hx· 0.095 h"l = 142 J.l,M.

Answer: Pmax =0.095 h"l; Kg = 142 liM (b) The critical dilution rate Dcrit is determined using Eq. (13.66): 0.095h- 1 (100/lM) -1 Derit = KS+sj = 142Jl,M+lOOliM = 0.039h

tlmaxSj

Calculating the flow rate from Eq. (13.39) with V = 1000 mI, F = D V"", 0.039 h"l x 1000 ml = 39 ml h~I.

Answer: 39 ml h"l

13.11 Kinetic and yield parameters of an auxotropWc mutant From Eq. (13.92), Jlmax and Ks can be determined from the slope and intercept of a plot of SID versus s. From the definition of dilution rate in Eq. (13.39), values of D are evaluated from the experimental flow rates using V"", 2 l.

The relevant data are listed and plotted below.

Flow rate. F (I hoI)

Dilution rate, D (h- 1)

Substrate concentration, S (g 1"1)

SID (g I-I h)

1.0 1.4 1.6 1.7 1.8 1.9

o.so

0.010 0.038 0.071 0.066 0.095 0.477

0.020 0.054 0.089 0.078 0.106

0.70 0.80 0.85

0.90 0.95

0.502

Solutions: Chapter 13

162

0.6 0.6 0.4

-

:2

~

.$

0.3

~ICl

0.2 0.1 0.0 0.0

0.1

0.2 0.3 Substrate concentration,

0.4 S

0.5

(g 1-1)

'The slope of the straight line in the plot is 1.027 h;the intercept is 0.0119 g I-I h. From Eq. (13.92), the slope: Iftlmax; therefore, J.Lmax:::: 1/1.027 h "'" 0.97 hoi. The intercept:::: KS/J.I.max; therefore Ks =0.0119 g I-I h x 0.97 h- t = 0.012 g l-t, From Eq. (13.93), Yxs and ms can be determined from the slope and intercept of a plot of 1I~ versus lID, where Y~s is calculated using Eq. {13.94) '\II(ith Sj "" 10 g I~l. The relevantdata are listed and plotted

beISt,

Flow rate, F (l h- 1)

Dilution rate, D (hw l )

liD (h)

Y~s(g g-l)

Ill"

1.0 1.4 1.6 1.7 1.8 1.9

050 0.70 0.80 0.85 0.90 0.95

2.00 1.43 1.25 1.18 1.11 LOS

0.315 0.323 0.329 0.328 0.324 0.326

3.175 3.096 3.040 3.049 3.086 3.067

XS

3.20,---,---,-.,--,,--,---r--,

•

3.15

--'?:-X b

.$ _ m 3.10

• 3.05

3.00 0.8

1.0

• •• 1.2

1.4

1.6

YOilution rate (h)

1.8

2.0

2.2

(g g-l)

Solutions: Chapter 13

163

The scatter in the plot is typical for measured values of l/Y~, The slope of the straight line in the plot is 0.12 g g.l h- I ; the intercept is 2.93 g g-1. From Eq. (13.93), the slope =- ms; therefore ms =- 0.12 g g.l h- 1. The intercept =I/ yxs ; therefore Yxs =- 1/2 .93 h-I =- 0.34 g g-l.

Answer:,umax =- 0.97 h· 1; Ks =- 0,012 g 1. 1; rns =- 0.12 g g-1 h· l ; Yxs =- 0.34 g g.1

13.12 Continuous sterilisation From the definition of dilution rate in Eq. (13.39), the medium volumetric flow rate F =- D V =- 0.1 h- 1 x 15 m 3 =- 1,5 m 3 h· 1. The linear velocity u in the holding section of the steriliser is determined by dividing F by the pipe cross· sectional area A =- n: where r is the pipe radius. For r =- 6 cm =- 0.06 m:

,.z,

U

f. =-

=-

A

3 1 1.5m h1t (0.06 m)Z

=-

132.6mh-1

The value of the specific death constant is evaluated using Eq. (11.46) with R =- 8.3144 1. gruol·l ,K~l from Table 2.5, Ed =- 288.5 kJgmol-l =- 2.885 x loS J gruol·I, A =- 7.5x 1039 h· l , and the temperature convertedfromoC to degrees Kelvin using Eq. (2.24): kd

=A e-EdiR T == 7.5 x 1039 h-1 e-Z.885 x lOS J gmorlt[(8.3144J gmor1 K"1)(130+ Z73.15H(] = 313.1 h- 1

Within a period of 3 months "" 90 d, the number of cells Nl entering the steriliser is equal to the medium volumetric flow rate Fmultiplied by the cell concentration and the time: Nj

= 1.5 m 3 h-I ( 105 ml-1

1~:~)90d.12::1 = 3.24 x 10 14

Within the same 3-month period, the acCeptable number of cells remaining at the end of the sterilisation treatment is NZ =- L Therefore:

NZ Nt

=-

1 3.24x 10 14

=-

3.09 x 10-15

(a) For perfect plug flow .withno:axial dispersion, the sterilisation time can be determined using Eq. (13,97):

N1 In-

N2

thd

=-

kd =

1n 3.24XlO

14

1 313.1h 1

= O.107h

To allow the medium to remain for this period. of time in the.holding ,section,of:the steriliserpipe,.the length of pipe required is equal to the linear velocity of the medium u multiplied by t'hd:

L = u!hd =- 132.6mh-1 xOJ07h =- 14.2m

Answer: 14.2 m (h)

Calculating the Reynolds number for pipe flow using Eq, (7.1) with pipe diameter D =- 12 cm =- 0,12 m:

Dup 0,12 m(132.6m h-1) lOOOkgm-3 Re=---= =-3978 ,u 4kgm-1 h 1 The value of !J)vu Dcorresponding to this Re is found from Figure 13.40. Using the experimental curve as this gives a higher 2i than the theoretical curve and thus a more conservative design, 1Jzlu D "" 1.5. Therefore:

From Eq. (13.101), an expression for the Peelet number Pe is:

164

Solutions: Chapter 13

where L has units of m. Similarly, an expression for the Damkohler number Da from Eq. (13.102) is: k L

Da

=- du "'"

(313.1 h- 1) L

132.6mb-1

"'" 2.36ÂŁ

The design problem can be solved from this point using tria1~andwerror methods and Figure 13.41. As a first guess, try

L = 20 ID. The values for Pe and Da are evaluated using the equations determined above, and the corresponding value for N2/Nl read from Figure 13.41. Depending on how this value compares with the target of 3.09 x 10- 15 , the value

of L is adjusted until the results for N2/N} coincide. The calculations are shown in the table below.

L(m)

Pe

Da

20

110

18 19

99

47 42

105

45

(from Figure 13.41)

4x 1 x 10- 14 1.5 x to- 15

The last value of N2/NI is as close as practicable to 3.09 x 10- 15 considering the resolution of Figure 13.4l.

Therefore. the required length of pipe in the holding section is about 19 m. or 34% longer than that detennined for ideal plug flow.

Answer: About 19 m

(e) For L =- 14,2 m, from the equations developed in (b), Pe :::: 78 and Va:::: 34, From Figure 13.41, N2/Nl is about 5 x 10- 12; therefore, NI/N2 :::: 2 X 1011. As N2 :::: 1, N} :::: 2 x 1011, i.e. one contaminant enters the fermentetfor every 2 x lOll that enter the steriliser, For F:::: 1.5 m 3 h- l and an input contaminant concentration of lOS ml- l , the time required for 2 x 1011 contaminants to enter the steriliser is: 2 x 1011

Therefore, contaminants enter the fermenter at a rate of one every 80 min.

Answer: One contaminant enters the fermenter every 80 min.