Bio Process Engineering Principles [Solutions Manual] - P. Doran (1997) WW

SOLUTIONS MANUALBioprocess Engineering Principles

Pauline M. Doran

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SOLUTIONS MANUALBioprocess 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 11 m= lOOcrnTherefore:

1.5x10-6 cP ::::: 1.5 x 10-6 cP ,1 10-

3ktg;-1 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-ITherefore:

Answer: 5.17 Btu min-1

(e)From Table A.S (Appendix. A): 1 mmHg:= 1.316 x 10-3 attnFrom Table A.I (Appendix A): 1 ft = 0.3048 mFrom Table A.7 (Appendix A): 11 atm =9.604 x 10-2 BtuFrom Table A8 (Appendix A): 1 Btu min-I::::: 2.391 x to-2 metric horsepowerIm=lOOcm11= lOOOcm3Ih=60minTherefore:

670mmHgft3 =670 mmHg ft3 .11.316X 10-3 atml.19.604X 1O-2Btul.I°.3048m 1

31 100 em 1

3 1 11 I.1 mmHg llatm 1 it 1 m lOOOcm3

1

2.391 x 10-2

metric horsepowerI I I h I 956 10-4 . h h1 . -60. = , x metric orsepower

1Btu min- mm

Answer: 9.56 x 10-4 metric horsepower h

(d)From Table A.7 (Appendix A): 1 Btu =0.2520 kcalFrom Table A.3 (Appendix A): Ilb = 453.6 gTherefore:

345 Btu Ib-1 = 345 Btulb-

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

Answer: 0.192 kcal g-1

2.2 Unit conversionCase 1Convert to units of kg, m, s.From Table AJ (Appendix A), lIb = 0.4536 kg

2 Solutions: Chapter 2

From Table A.2 (Appendix A): 1 tt3 =: 2.832 x 10-2 m3

From Table A.9 (Appendix A): 1 cP::: 10-3 kg m-l s·l1 rn= tOOcm= lOOOmmTherefore, using Eq. (2.1):

(2mm.1 1m n(3cms-l.l~n(251bfC3I0.4536kgl.1 Ift3

~Dup l000mmU l00cmU lIb 2.832 x 10 2m3U _ 7

Re::: -p- = I -3 1 -11 - 2.4 x 1010-6 P 10 kgm s

c . 1cP

Answer: 2.4 x 107

ease 2Convert to units of kg, m, s.From Table Al (Appendix A): 1 in.::: 2.54 x 10-2 mFrom Table A.9 (Appendix A): 1 Ibm ft-I h-1 ::: 4.134 x 10-4 kg m-t s·tIh=3600sTherefore, usingEq. (2.1):

= 1.5 x 104

13~sl

Answer: 1.5 x 104

2.3 Dimensionless groups and property dataFrom the Chemical Engineers' Handbook, the diffusivity of oxygen in water at 2S"C is 2.5 x 10-5 cm2 s·l. Assumingthis is the same at 28"C, !lJ= 2.5 x 10-5 cm1 s·t, Also, from the Chemical Engineers' Handbook, the density of waterat 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°Cand 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 "" latm "" 4.05 x 10-5 gmolcm-3RT (82.057cm3 almK:""1 gmol 1)(301.15K)

From the atomie weights in Table B.l (Appendix B), the molecular weight of oxygen is 32.0. Converting the resultfor Pa to mass tenns:

Pa "" 4.05 x lO-5gmolem-3 ·1 i~~~ll "" l.30XlO-

3gcm-

3

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 thedimensionless groups in the equation for the Sherwood number.

Solutions,' Chapter 2

1'0-2 -1 -110,87eP. gem sS_J1.L_ leP -349c - PrJ) - (0.9962652g cn,-3)(2.5 x 10 5em2s I) -

Therefore:

From the equation for Sh:

k_Sh:lJ_(ll.21(2.5xl0-5cm2s-I)_I·A 10-3 -I

L---- 02 -,<tUX emsD1J . em

Answer: lAO x 10~3 em s·1

2.4 Mass and weightFrom the definition of density on p 16, mass is equal to volume multiplied by density, Therefore:

3

From p 16, weight is-the force with wrncha body is attracted to the centre of the earth by gravity. According toNewton'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 llbfWeight = 2.008 x 10 lbmfts- , 2 .= 6241bf32.174Ibm 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 llbfWeight = 1.998 x 10 lbmfts- . -----''-~I = 6211bf

32.1741bm ft s-2

Answer: 621lbf

2.5 Dimensionless numbersFirst, evaluate the units of the groups cCP J.llk) and (D G1p):

. (ep") (Bmlb-I '1'1) Ibh-I ft-IUmtsof -- :::: = 1k BIUh-1 ft-2('Pft-1tl

UnilsOf(DG) = (ftllbh-1ft-

2= 1

J1. Ibh1ft1


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

Unitsofh "'" unitsofCpG "'" (Btulb-1 "p-l)(1bh-1fr2) "'" 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 withdimensions "'" L2Ml2. OF is a unit of temperature which, from Table 2.1, has the dimensional symbol E>. h is a unitof time with dimension"'" T; ft is a unit of length with dimension"'" L. Therefore:

Dimensions ofh = 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 CcFrom Table A8 (Appendix A), dimensions of P = L2MT"3Dimensions of g = LT-zDimensions of p =ML·3Dimensions ofDi "'" LFrom 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 unitsFrom 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 lbmol4O.0lblbmol 1

Answer: 050 lbmol

(b)From Table A.3 (Appendix A): lIb =453.6 g. Therefore:

1

453.6gl20.01b =20.0Ib. -lib =9072 g

From Eq. (2.18):

9072ggram moles NaOH = I =227 grool

4O.0g gmol-

Answer: 227 gmol

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

.-----

Solutions: Chapter 2

Answer: 0.227 kgmol

11 kgmol Ikg molesNaOR =227 gmol. 1000 gmol =0.227 kgmol

5

2.8 Density and specific gravity(a)From p 16, the density of water at 4°C can be taken as exactly 1 g cm-3. Therefore, for a substance with specificgravity L5129i~, the density at 20°C is 1.5129 g cm*3,(I)lkg=l000g1 m: 100cmTherefore:

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

LS129ggram moles : 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 d1

. : .,-__,,-1-:--:::;- : 41.67cm3 gmor1mo 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 ratemultiplied 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 weightFrom p 17, the composition of air is close to 21 % oxygen and 79% nitrogen. For gases at low pressures, this means21 mol% 02 and 79 mol% NZ. Therefore, in 1 gmol air, there are 0.21 gmot Oz and 0.79 gmol NZ From the atomicweights in Table B.l (Appendix B), the molecular weights of Oz and NZ are 3Z.0 and 28.0, respectively. Themolecular weight of air is equal to the number of grams in 1 gmol:

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

Answer. 28.8

6 Solutions.' Chapter 2

2.10 Mole fractionThe 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 gmethanol, 12 g glycerol, 10 g acetic acid, 8 g benzaldehyde, and no other components. Therefore:

Moles water = 30 g _I ~r:;~ll = 1.67 gruol

Moles ethanol:: 25 g.1 ~~~ll :: 054 gmol

Molesmethanol:: 15 g _I ;~~ll =0.47 gmol

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

Moles acetic acid = 10g.1 ~~~II = 0.17 gmol

Moles benzaldehyde: 8 g _I :~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 = ;:~~ = 0.55

Mole fraction ethanol = ~:~ =0.18

Mole fraction methanol = ~:~~ = 0.15

Mole fraction glycerol = ~:~ = 0.04

Mole fraction acetic acid = ~:~ ::::: 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 scalesFrom Eq. (2.27\

-40 ::::: 1.8 T(0C) + 32Tee) = -40

From Eq. (2.25),

T (OR) ::::: -40 + 459.67T(°R) = 420

From Eq. (2.24) and the result for T ("C);

T(K) = -40+273.15T(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


From Table A.5 (Appendix A): 1 psi = 6.805 x Hy2 atm. Therefore:

. 16.805 x 10-2 atm 1Absolutepressure = 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

7

2.13 Stoichiometry and incomplete reaction(a)The molecular weights are calculated from TableB.l (Appendix B): penicillin =334.;4; glucose = 180.2. Themaximum 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 productionaccording to the stoichiometric equation. If only 6% of the glucose is used in this way, the actual yield of penicillinfrom 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 thepenicillin 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 glucoseavailable 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 isavailable which, according to the stoichiometric equation, requires 1.67 x 2.94 x 10-2 =4.91 x 10-2 gruol glucose forcomplete reaction. As the gmol glucose required is greater than the gmol glucose available after growth andmaintenance 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 glucoseconsumed 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, thisproduces 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 0.886gmol 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-lphenylacetic 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


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: 91.5Hexadecane: 226.4From the stoichiometry, as 1 gmol of hexadecane is required to produce 1.65 gmol of cells, the maximum yield is 1.65gmol 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 thestoichiometry, 16.28 gmol of oxygen is required to produce 1.65 gmal ofcells 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 amountsof 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 =16.5 gmol =16.5 gmol x 226.4 g gmol-I =3736 g = 3.74 kg hexadecane.

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 gmoloxygen. 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 ofair required can be calculated using the ideal gas law. From Eq. (2.32):

V = nRTp

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= (1282.9gmolj(82.057cm3 atmK-1 gmol-l)(293.15Kj 1-.!.."'...13 = 31 31 atm . l00cm m

Answer: 31 m 3

Presentation and Analysis of Data

3.1 Combination of errorsc~ "" 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-lO.Q67molm

Answer~ 31%. This example illustrates how a combination of small measurement errors can result in a relativelylarge 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)'4-1

= 0.15

Answer. The standard deviation is 0.15. Note that standard error, which can be calculated from the standarddeviation, is a more direct indication of the precision ofa mean.

(c)

x = 5.15 +5.45 =5.302

Standard deviation is not appropriate for expressing the accuracy of a mean evaluated using only two samples. In thiscase the maximum error, Le. the difference between the mean and either of the two measured values, might be usedinstead. 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


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 individualmeasurements are as intrinsically accurate as possible, rather thanrepeat the measurement many times.

3.3 Linear and non-linear models(a)Xl =1; Yl =10X2=8;Y2=0.5A straight line plot of y versus x on linear coordinates means that the data can be represented using Eq. (3.6). FromEqs (3.7) and (3.8),

A = (Yz-Y1) = 0.5-10 =-136(x2 Xl) 8-1 .

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 = 38.5A straight line plot of y versus xIh 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 = 201112_ 112 89Ih_32'k .x2 Xl . ,

B = YI-Axih = 14.5-20.1(3.2112) = -21.5

'kAnswer:y=20.1x -21.5

(0)Xl=5;Yl=6X2= l;Y2=3A 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} 1/3 -lI6 -3

2 2 = 2 2 =-6.9xlOx2-xl 1 -5

B = 1/y1 -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=25x2 =550; Y2 =2600A 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),


Answer: Y= 39.6 xO.663

A = (lnY2- ln Yl) =ln2600-ln25 =0.663(lnx2-lnxI) ln550-1nO.5

lnB =InYI-Alnxl =ln25-(0.663}ln0.5 = 3.678

B = e3.678 = 39.6

11

(eJXl = 1.5; YI = 2.5X2 = 10; Y2 = 0.036A straight line' plot of y,versus X on semi·log coordinates means that the data can be represented using Eq. (3.15).From Eqs (3.17) and (3.18):

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

Sucrose concentration (g l~l)

6.012.018.024.030.0

(bJ

Mean peak area

56.84112.82170.63232.74302.04

Standard deviation

1.212.062.541.802.21

35

30--..9 250••~

20

•g 158

~ 10g

"' 5

00 50 100 150 200 250 300 350

Peak area


(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 versuslIT on semi-logarithmic coordinates will give a straight line. As T in the equation is absolute temperature, "'c mustfirst 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

15 288.15 3.47 x 10-3 4.4 x 10-15

20 293.15 3.41 x lOw3 2.0x 10-14

25 298.15 3.35 x leT3 8.6x 10..14

30 303.15 3.30 x lfr·3 3.5 x JerI335 308.15 3.25 x 10-3 lAx }0"'12

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

"[ 10-12

•"fc

10-13.Q

lii"E~= 10.14-lllJ:

3.5 x 10-33.3x 10-3 3.4x 10-3

1ITemperature (K-1)

10-'5L._~~_--'__~__....l.__~__....J

3,2 x 10-3

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 homogeneitythe exponent must be dimensionless (p 12), so that -26,121 has units of K, and EIR in the model equation is equal to26,121 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 gmot1


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

13

12345 6Decrease in medium conductivity (mS cm-1)

16--, 14!'!lc•• 12i'!C 10•g00 8

~ 6~•• 4m• •~ 2.i1

00

•• • • •• •

•

• •

The linear least squares fit of the data is:

y =: 1.58 + 2.10x

where yis increase in biomass concentration in g r1 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 andthe values for y obtained from the equation in (a).

Decrease in medium conductivity (mS

o0.120.310.410.821.03lAO1.912.112.422.442.742.913.534.395.215.245.55

Residual

-1.580.57

-!l.230.361.201.361.2800410.19

-0.46-!l.50-!l.73-1.69-1.99-1.00

1.480.021.37


These results are plotted below.

3.-_-,.__.,-_-,__.,...__,-_-,

2

1

·2

61 2 3 4 5Decrease in medium conductivity (mS em-1)

·3 '-_-'-__-'-_---'__......__'-_-'

o

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

3.7 Discriminating between rival models(aJThe results are plotted using linear coordinates below.

••

••

0.11

~

'"0.10

.§.os' 0.09

.~

~0.08

" 0.07."'E1l. 0.06,•~~ 0.05::J

0.040.00 0.02 0.04 0.06 0.08 0.10

Gas superficial velocity, uG (m s-1)

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 theresiduals between the measured values for liquid superficial velocity and the values for y obtained from the aboveequation is 8.4 x 10-5,

Solutions: Chapter 3 15

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

0.01om

Gas superficial velocity, uG (m s·1)0.1

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 theresiduals between the measured values for liquid superficial velocity and the values obtained from the above equationis 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 shouldgive a straight line when plotted on semi-logarithmic coordinates.

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

103

101

35302515 20Time (min)

100 '--_.1-_-'-_-'-_-'-_-L_----'-_---'o 5 10

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 preservedimensional homogeneity (p 12), the dimensions of kd are T-t; therefore k(! =. 0353 min-I, The dimensions of No arethe 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 membranesL Assemble(i) Flow sheet

Buffer solution in80 kg min-1

Cell concentrate6% bacteria

I II I

I Hollow-fibre membranes II I

I I

Buffer solution out

Fennentation broth350 kg min-11% bacteria99% water

(li) System boundaryThe system boundary is shown on the flow sheet.(iii) Reaction equationNo reaction occurs.

2. Analyse(i) Assumptions- steady state-no leaks- only water passes across the membrane(ii) Extra dataNo extra data are required.(ill) Basis1 min, or 350 kg fennentation broth(iv) Compounds involved in reactionNo compounds are involved in reaction.(v) Mass-balance equationAs there is no reaction, the appropriate mass-balance equation is Eq. (4.3):

mass in = mass out

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


In OutStream

Water Bacteria Buffer Total Water Bacteria Buffer TotalFermentation 346.5 3.5 0 350

brothBuffer solution in 0 0 80 80Cell concentrate ? O.06C 0 CBuffer solution ? 0 80 Bout

Total 346.5 3.5 80 430 ? 0.06C 80 C+B

(li) Mass-balance calculationsBacteria balance

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

C = 58.3 kg

Total mass balance430 kg total mass in = (C +B) kg total mass out

Using the result for C:B "'" 371.7 kg

Water balance346.5 kg water in = water out

Water out = 346.5 kg

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

Stream In OutWater Bacteria 814fer Total Water Bacteria Buffer Total

Fennentation 346.5 3.5 0 350broth

Buffer solution in 0 0 80 80Cell concentrate 54.8 3.50 0 58.3Buffer solution 291.7 0 80 371.7out

Total 346.5 3.5 80 430 346.5 3.50 80 430

(iii) Check the resultsAll 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· I


4.2 Membrane reactort. Assemble(i) Flow sheet

Product

Aqueous residue0.2% glucose0.5% ethanol

ystem boundary~

- -- - -- - 1I II I

I Membrane system II I

I I

s

Solvent40 kg h-1

Feed40 kg h-1

10% glucose90% water

(li) System boundaryThe system boundary is shown on the flow sheet.(iii) Reaction equation

2. Analyse(i) 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 dataMolecular weights (Table B.t, Appendix B): glucose = 180.2

ethanol =. 46.1CO, =44.0

(iii) Basis1 h, or 40 kg feed solution(iv) Compounds involved in reactionGlucose, ethanol and carbon dioxide are involved in the reaction.(v) Mass~balanceequationsFor glucose, ethanol and carbon dioxide, the appropriate mass-balance equation is Eq. (4.2):

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 =. mass out

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


Stream In OutGlucose Ethanol CO, Solvent H,O Total Glucose Ethotwl CO, Solvent H2O Total-------

Feed 4 0 0 0 36 40Solvent 0 0 0 40 0 40Aqueous 0_002R 0.005 R 0 0 ? Rresidue

Product 0 ? 0 ? 0 POff-gas 0 0 ? 0 0 G

Total 4 0 0 40 36 80 0.002 R ? ? ? ? R+P

(ii) Mass-balance calculationsSolvent balanceSolvent is a tie component.

40 kg solvent in ::::: solvent out

Solvent out ::::: 40 kg

Water balanceWater 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 residuestream::::: 0.005 R::::: 0.181 kg.

Glucose balance4 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:

11 kgmOll3.927 kg glucose ::::: 3.927kg. 180.2kg ::::: O.0218kgmol

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

0.0436kgmolethanol =0.0436kgmOI·1 iZ:~~11 = 2.010kg

0.0436kgmo1COz = 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 balanceokg ethanol in + 2.010 kg ethanol generated = ethanol out + 0 kg ethanol consumed

Ethanol out =2.010 kg

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


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


Stream In OutGlucose Ethanol CO, Solvent H,O Total Glucose Ethanol CO2 Solvent H,O Total

Feed 4 0 0 0 36 40Solvent 0 0 0 40 0 40Aqueous - 0.073 0.181 0 0 36 36.254

residueProduct 0 1.829 0 40 0 41.829·Off-gas 0 0 1.92 0 0 1.92

Total 4 0 0 40 36 80 0.073 2.010 1.92 40 36 80.00

(iii) Check the resultsAll 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 x100%:::: 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 distillation1. Assemble(i) Flow sheet

Bottoms

Distillate5,000 kg h-1

45% ethanol55% water

System boUnda'l'\.r -t---,I II II II

Distillation II column I

kg h-lthanal I Iatar I I

I II I- - -+- - -

Feed50,00010%e90%w

22

(li) System boundaryThe system boundary is shown on the flow sheet.(ill) Reaction equationNo reaction occurs.

2. Analyse(i) Assumptions- steady state-no leaks(ll) Extra dataNo extra data are required.(ill) Basis1 h, or 50,000 kg feed(Iv) Compounds involved in reactionNo compounds are involved in reaction.(v) Mass¥balance equationAs there is no reaction, the appropriate mass-balance equation is Eq. (4.3):

mass in ::::: mass out

3. Calculate(i) Calculation tableThe calculation table shows all given quantities in kg.

SolutilJns: Chapter 4

Stream

FeedDistillateBottoms

Total

Ethanol5,000

5,000

Water45,000

45,000

InTotal50,000

50,000

OutEthanol Water Total

2,250 2,750 5,000? ? ?

? ? ?

(li) Mass-balance calculationsTotal 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 balance5,000 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 balance45,000 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


Solutions: Chapter4 23

Stream

FeedDistillateBottoms

Total

Ethanol5,000

5,000

Water45,000

45,000

In OutTotal Ethanol Water Total50,000

2,250 2,750 5,0002,750 42,250 45,000

50,000 5,000 45,000 50,000

(ill) Check the resultsAll 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 is2,750/45,000 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 egg1. Assemble(i) Flow sheet

Off-gas

Enzyme reactor

--t- -

System boundary

r\.II1

Egg slurry -----L.o-l3000 kg h·1 I2% glucose20% water I

78% egg solids I

-------,1I

I--.L1---_- ProductI 0.2% glucose

1I

Inlet air18 kg h-1 02

(n) System boundaryThe 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


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

02=32.0NZ::::: 28.0H20 =IS.0gluconic acid::::: 196.2HZOz::::: 34.0

Composition of air (p 17): 21% 02, 79% Nz by volume(ill) Basis1 h, or 3000 kg egg slurry(iv) Compounds involved in reactionGlucose, 02, water, gluconic acid and HzOz are involved in the reaction.(v) Mass-balance equationsFor glucose, 02, water, gluconic acid and HZOz,the appropriate mass~balanceequation 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 tableThe mass of NZ accompanying 18 kg 02 in air can be calculated from the known composition of air. Converting 18kg 02 to molar units:

1

1 kgmOllISkg02 = ISkg02 · 32.0kg =0.563kgmo102

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

12S.0kg I2,118kgmo1N2 = 2.118kgmoIN2· lkgmol =59.30kg N2

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


59.30 kg NZ in ::::: NZ out

NZ out::::: 59,30 kg

Glucose balance60 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:

25

1

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

Glucoseconsumed ::::: (0.333-1.11 x 10-5 p) kgmol

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

(0.333 -1.11 x 10-5 p) kgmo102 = (0.333-1.11 x 10-5 p) kgm01.1 ;~~~11 = (10.656- 3.552x 10-4 p) kg 02

02 balam.:e18 kg Oz in +·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») 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 balance3077.3 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 10-4P + 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

26

Water balance


Gluconic acid generated = 0.300 kgmOI.! i9:~~ I= 58.86 kg

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

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

Water out =594.6 kg

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

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 OutGlucose Water Egg solids 02 N2 Gluconic acid H2O, Total

Egg slurry -All-Off-gas 0 0 0 8.41 59.30 0 0 67.71Product 6.02 594.6 2340 0 0 58.86 10.20 3009.6

Total 6.02 594.6 2340 8.41 59.30 58.86 10.20 3077.3

(iii) Check the resultsAll 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 reactionmust be determined. From the mass-balance table for streams in:

I1 kg mOll

Moles glucose = 60 kg. 180.2 kg = 0.333 kgmol

11kgmOIl

Moleswater = 600 kg . 18.0kg = 33.3kgmol

11kgmOIl

Moles 02 = 18 kg. 32.0 kg = 0.563 kgmol

As the substrates are required in the molar stoichiometric ratio of 1: 1: 1 and glucose is available in the smallest molarquantity, 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 tocompletion, from Eq, (2.34);

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

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

Answer: 9900% excess water; 69% excess 02


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

11kgmOll8.41 kg 02 = 8.4lkg02 · 32.0kg =0.263kgmo102

As the number of kgmol N2 was determined in the preliminary calculations to be 2.118, the total number of moles ofoff-gas is (0.263 + 2.118) =2.381 kgmoJ. Therefore, the composition of the off-gas is 0.263/2.381 =0.11 02, and2.118/2.381 = 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 kgglucose, 594.6 kg water, 2340 kg egg solids, 58.86 kg gluconic acid and 10.20 kg H202. Therefore, the compositionis:

6.02 0002 13009,6 =. g ucose

594.63009.6 =0.198 water

2340 .3009.6 = 0.778 egg solids

58.86 0020 I . ·d3009.6 =. g UCONC act

10.203009.6 =0.003 H20 ,

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

4.5 Azeotropic distillation1. Assemble(0 Flow sheet,...-----------------..,

--1---

--j-System boundary"r-11

Feed -----r-~95% ethanol I

5% water I1

Benzene feed ------'-1-ol1

1

1

(ii) System boundaryThe system boundary is shown on the flow sheet

Distillationtower

.,1

1

1

11

1

11

1

Overhead74.1% benzene18.5% ethanol7.4"1" H20

Product100% ethanol

28

(iii) Reaction equationNo reaction occurs.

2. Analyse(i) Assumptions- steady state-no leaks(li) Extra datalOOOcm3 =11l000g=lkg(iii) Basis2501 ethanol product(iv) Compounds involved in reactionNo compounds are involved in reaction.(v) Mass~balanceequationAs there is no reaction, the appropriate mass-balance equation is Eq. (4.3):

mass in "'" mass out


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

2501absoluteethanol ::::: 2501 x 0.785 gcm-3 .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 ofbenzene feed in is denoted B; the total mass of overhead out is denoted v.

Stream

FeedBenzene feedProductOverhead

Total

Ethanol0.95 Fo

0.95F

Water0.05Fo

0.05F

InBenzeneoB

B

TotalFB

F+B

OutEthanol Water Benzene Total

196.25 0 0 196.250.185 V 0.074 V 0.741 V V

196.25 + 0.074 V 0.741 V 196.25 +0.185 V V

(ii) Mass-balance calculationsEthanol balance

0.95 F kg ethanol in =. (196.25 + 0.185 V) kg ethanol out

F = (206.58 + 0.195 V) kg

Benzene balanceB 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 = (196.25 +V) kg

10.33 =. 0.064 V

V=. 161.4kg

Using this result in the ethanol and benzene balances gives:

F = 238.1 kg


B = 119.6kg

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

Stream In OutEthanol Water Benzene Total Ethanol Water Benzene Total

F,ed 226.2 11.9 0 238.1Benzene feed 0 0 119.6 119.6Product 196.25 0 0 196.25Overhead 29.9 11.9 119.6 161.4

Total 226.2 11.9 119.6 357.7 226.2 11.9 119.6 357.7

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

4. FinaliseProm the completed mass-balance table, the mass of benzene required is 119.6 kg. Using the definition of density onp 16, volume is equal to mass divided by density:

119.6 kg 11000g li 11 1Volumeofbenzene"'" -3' lk . 3 "'" 1371O.872gcm g l000cm

Answer: 137 titres

4.6 Culture of plant roots1. Assemble(i) Flow sheet

Off-gas47 1m-as 02

15 litres CO2

Drained liquid1110g0.063% glucose1.7% ammonia

1

J

Air-drivenreactor

----l

1

1

I1

I11

System boundary )..

Medium -----'-1_-I1425 9 1

3% glucose1.75% ammonia I

95.25% water I

1

1

I1L _

Air22 cm3 milr1 10r 10 d

25°C, 1 atm

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

Moles of air in = n = p V =RT


(iii) Reaction equationFrom Table B.2 (Appendix B), the molecular formula for glucose is C6H1206. The reaction equation is based on thegeneral 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 data11= lOOOcm3

Molecular weights (Table B.1, Appendix B): glucose = 180.202 = 32.0NZ = 28.0NH3 = 17.0cOz =44.0HzO= 18.0

Composition of air (p 17): 21% OZ, 79% NZ by volumeIdeal gas constant (Table 2.5): R = 82.057 cm3 atm K~l gmol~I(iii) Basis10 d. or 1425 g nutrient medium(iv) Compounds involved in reactionGlucose, OZ' NH3. biomass, COZ and HZO are involved in the reaction.(v) Mass-balance equationsFor 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 tableOver 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 todegrees Kelvin using Eq. (2.24):

1 atm.(3.168 x 105

cm3) = 12.95gmol

82.057cm3 atmK-1gmor1(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 ofNZ is 0.79 x 12.95 = 10.23 gmo!. Converting these values to masses:

MassofOZin = 2.72gmOI.I {~;ll = 87.04g

MassofNZin = 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 molesusing the ideal gas law. Eq. (2.32), with the temperature converted from °c to degrees Kelvin using Eq. (2.24):


. 11000em31V 1atm(47litres). 11Mo1esofOz out::: n::: L::: ::: 1.92gmol

R T 82.057cm3atmK-1gmor1(25+273.15) K

( . )I1000 em'IV 1atm 15litres. 11Moles ofCOZ out =n =L = =O.613gmol

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

Calculate the corresponding masses:

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

MassofCOzout::: 0.613gmOl.l~~;11= 26.97g

31

The calculation tables below show all known quantities in g. The In side of the mass~balance table is complete. Thetotal mass of off~gas out is denoted G; the total biomassbarvested is denoted R As the ratio of biomass fresh weightto dry weight is 14:1, dry biomass comprises l!IS ::: 0.0667 of the total biomass. Because this problem requires anintegral mass balance, the biomass remaining in the fermenter after 10 d culture must also be included in the tableeven 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 balance1798.44gtotalmassin::: (1110+G+B)gtotalmassout

Using the result for G:

32

B =: 313.63 g


Therefore, the dry biomass produced is 0.0667 x 313.63 =20.92 g; the mass of water in the biomass is 0.9333 X313.63 = 292.71 g.

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

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

Glucose balance42.75 g glucose in + 0 g glucose generated = 0.699 g glucose out + glucose consumed

Glucose consumed = 42.05 g

02 balance87.04 g 02 in +0 g 02 generated = 61.44 g 02 out + 02 consumed

02; consumed = 25.60 g

NH3 balance24.94 g NH3 in +0 g NH3 generated = 18,87 g NH3 out + NH3 consumed

NH3 consumed = 6.07 g

C02 balanceog C02 in + C02 generated = 26.97 g C02 out + 0 g C02 consumed

C02 generated =26.97 g

H20 balance1357.31 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 resultsAll 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 beconverted to molar quantities:


Moles ofglucose consumed =42.05 g.1 :8:0~1 =0.233 gmol

Moles ofoz consumed =25.60 g .1 ;~~11 =0.800 gmol

MolesofNH3consumed = 6.07 g .1 ~~~ll = 0.357gmol

MolesofCOzgenerated= 26.97g.I~~I!= 0.6l3gmol

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

Moles ofbiomass generated = ~0:2 g10mass

33

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

20.920.233 Coli1206 + 0.800 0, +0.357 NH3 -+ MW b' CH"OoNS + 0.613 CO, + 1.435 H,Olomass P'"

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

89.79CoH1206+3.430,+1.53NH3 -+ MWb' CH"OoNS+2.63CO,+6.16H,OlOmass fJ

The values of a, /3 and 8 and the molecular formula for the biomass can obtained using elemental balances.89.79

Cbalance:6 = MWb' +2.63lomassTherefore:

89:79 = 3.37MWblomass

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

A11SWer: 1he complete stoichiometric equation is:

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

(cjConverting 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 = 87·04g.1 ;~~11 =2.72gmol

MoiesofNH3in = 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, glucosemust be the limiting substrate.

Answer: Glucose


(d)The mass of glucose consumed is 42.05 g; the mass ofdry biomass produced is 20.92 g. Therefore, the biomass yieldfrom 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 glycerolFrom TableB.2 (Appendix B), the molecular formula for glycerol is C3Hg03. From Table 4.3, the chemical formulafor Klebsiella aerogenes can be taken as CH1. 7S00.43NO.22. Substituting these formulae into the generalstoichiometric equation for growth. Eq. (4.4), gives:

C3Hg03 +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 fromthe 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 ccan be detennined from:the yield Yxs =0.4 g g-1 and Eq. (4.12):

c = Yxs(MWsubstrate) = 0.40 g g-l (92.lggmOrl) = 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 rs = 4.67. The degree ofreduction of the biomass relative to NH3 is:

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; fromEq. (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 digestionFrom Eq. (4.13), the stoichiometric equation for anaerobic growth and product fonnation by methane bacteria can bewritten as:

0I3COOH+ bNH3 --j. c CH1.400.4ONo.20+ deo2 + eH20 +/0I4

From Table B.8 (Appendix B), the molecular weight of acetic acid is 60.1. From Table B.l (Appendix B), themolecular 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~ = 0.67gg-I(60.1ggmor

1j = 0.915 gmol gmol-lMWC02 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-cHbalance:4+3b = 1.4c+2e+4f°balance: 2 = 0.40c+2d+e = OAOc+2xO.915+e = OAOc+L83+e --j. e = O.17-0.40cN balance: b = 0.20 c


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

35

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 :i «1.00) -1fmax = )IP = 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(aJFrom Table B.2 (Appendix B), the .molecular formula for glucose isC6H1206. If all carbon in the substrate isconverted into biomass, production of carbon dioxide is zero. Therefore, from Eq. (4.4), the stoichiometric equationfor anaerobic growth of Cellulomonas is:

C~1206 +b NH3 -+ c CH1.5iPO.~O.16+ e H20

The stoichiometric coefficients can be determined using elemental balances.C balance: 6 = cHbalance:12+3b: 1.56c+2eo balance: 6 =0.54 c + eN balance: b : 0.16 cSubstituting 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 formulaweight 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 TableB.2 (Appendix B). the degree ofreduclion of glucose relative to NH3 isrs = 4.00. The degree of reduction of thebiomass relative to NH3 is:

I x4+ 1.56 x 1-0.54x2-0.16x3 00JB = = 4.

Substituting these values into Eq. (4.19):

wYs 6(4.00) -1croax = 1i3 = 4.00 = 6.0 gmol gmol

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


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

(b)(I)From Table B.2 (Appendix B), the molecular formula for methanol is CH40 and the degree of reduction relative toNH3 is 1S =6.00. The degree ofreduction of Methylophilus methylotrophus biomass relative to NH3 is:

_ lx4+1.68xl-0.36x2-0.22x3 -430JB- 1 -.

From Eq. (4.13), W "" 1 for methanol. Substituting'values into Eq. (4.19);

W JS 1 (6.00) 140 1 1-1emu"" Jh" "" 4.30 "" . gIno gmo

From Table B.g (Appendix B). the molecular weight of methanol is 32.0. The biomass formula weight calculatedfrom 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·l is equal to (1.40 x 23.99) g biomassper 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 is1.40 grool grool-l as both biomass and substrate have 1 C atom each. In comparison, the C*atQm biomass yield fromglucose in (a) is 1 grool grool·}. The main reason for the increased yield in (b) is the high degree of reduction ofmethanol compared with glucose.

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

Therefore, 0.87 gruol ox.ygen is required r:.r gmol methanol. As the molecular weights of methanol and oxygen arethe 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 ethanolis 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 thestoichiometric 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.

F 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 b . Yxs(MWsubstrate) 0.05 g g-' (I80.2ggmor1) 037 1 1-'or actena, c = MW II = ,= . gmo gmo

ce s 24.6 g gmol-

The other coefficients can be determined using elemental balances.


YeastCbalance:6 = c+d+2j= 0.81+d+2j~d = 5.19-2jHbalance: l2+3b = 1.8c+2e+6j= 1.8xO.81+2e+6j~1O.54+3b = 2e+6jo balance: 6 = 0.sc+2d+e+f= 0.5xO.81+2d+e+f~5.595 = 2d+e+jNbalance:b = 0.2e = O.2xO.81 = 0.16Substituting 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:\ ....•.. "

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.

37

BacteriaC balance: 6 = c+d+2f= 0.37+d+2f~d = 5.63-2fHbalance:12+3b = 1.8c+2e+6f= 1.8x037+2e+6f~11.33+3b = 2e+6fo 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 = 0.074Using the same solution procedure as for yeast, substituting the expression for b from the N balance into the Hbalances 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 =2for ethanol. From Table B.2(Appendix B), the degree of reduction of glucose relative toNH3 is rs =4.00, and thedegree of reduction of ethanol relative to NH3 is »= 6.00. Using these values in Eq. (4.20) gives:

W JS 6 (4.00) 1fmax = 1P = 2(6.00) = 2.0gmolgmor

There:ore, the actu~ ethanol yield of 1.72 ~Ol gmol'} for yeast represe.nts 86%. of the theoretical maximum; forbactena, 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 prodnctsFrom Table B.2 (Appendix B), the molecular formula for glucose is C6H1206. Assuming that no products other thanbiomass are formed, from Eq. (4.4), the stoichiometric equation for growth is~


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 weightof oxygen is 32.0, and the biomass molecular weight is 25.16. The value of the stoichiometric coefficient c can bedetermined from the yield Yxs =0.37 g g-l andEq. (4.12):

Yxs(MWsubstrat~ 0.37gg-1(180.2ggmor'j -1

c = MW II = 1 = 2.65gmolgmolce s 25.16ggmor

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

O.88g02125.16gcellslllgmOlOZI -10.88 g 02 per g cells =~. 1g ce!J.s . 19mol cells . 32.0 g 02 =0.69 gmol 02 groot cells

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

0.69 gmol O2 (2.65 gmat cens)a= =1831gIno! cells 1gmol glucose .

From Table B.2 (Appendix B), the degree of reduction of glucose relative to NH31srS'='4;00. The degree ofreduction 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 aselectron acceptors is likely to have occurred in the culture.

Answer: Yes

4.12 Medium formulationUsing the atomic weights'in Table B.l (Appendix B), the molecular weight of (N14)zS04 is 132,1 and the biomassmolecular weight is 26.16. Using a basis of 1 litre, production of 25 g cells corresponds to 25/26,16 = 0.956 groolcells. As each gmolcells contains O.25gmol N, (0.956x0.25) = 0.239gmol Nare-neededfrom the'medium forbiomass synthesis, As (NH4)zS04 is the sole N source and each gmol (Nl4)zS04contains 2gmol N, 0.239/2= 0.120gmol (N14)zS04 is required. Multiplying this by the molecular weight, 0.120 x 132.-1 = 15.9 g (N14)zS04 arerequired. 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 Table4.3, the chemical formula for E. coli can be taken as CH1.7700.49N0.24; from Table B.2 (Appendix B), the molecularformula for glucose is C6H1206' Substituting these formulae into the general stoichiometric equation for growth andproduct 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 molecularweight is 25.00, and the recombinant protein molecular weight is 22.03.


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

Yxs(MWsubs"ate) 0.48gg-1(180.2g gmol-l) 346 I I-Ic= = =. gmogmo

MWcells 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- Yps(MWsubstrate) _ 0.096gg-1(180.2ggmol-l) _ 079 I I-I_ _ -,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 ofreduction of glucose relative to NH3 is -ns =4.00; from Table 4.3," the degree of reduction 'Of E. coli relative to NH3 isJB =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 =1 for recombinant protein. Substituting these values into Eq. (4.16) for thetheoretical oxygen demand gives:

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 determinedusing an elemental balance for N as follows:N balance: b = 0.24 c = 0.24 x 3.46 = 0.83Therefore, in wild-type E. coli the ammonia requirement is reduced from 1.03 to 0.83 grool gmol-1 glucose, adecrease 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 geneticallyengineered strain.

4.14 Effect of growth on oxygen demandThe 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 ofethanol and acetic acid are 46.1 and 60.1, respectively. Fromthe atomic weights in Table B.1 (Appendix B), the biomass molecular weight is 24.63. The value of thestoichiometric coefficient c can be determined from the yield Yxs =0.14 g g~l and Eq. (4.12):

_Yx",s"(MW"","S::U,,bs::"_at-,-e) 0.14 g g-1 (46.1 g gmol-1) -1c = = "'" 0.26groolgmol

MW cells 24.63 g gmol 1


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

Yps(MWsubstrate) O.92gg-t (46.1ggmol-l) -If = MW odu =. 1 =. 0.71 gmol gmol

pr ct 6O.1ggmor

From Table B.2 (Appendix B), the degree of reduction of ethanol relative to NH3 is rs =. 6.00, and the degree ofreduction 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 _ 4201ll- 1 -.

From Eq. (4.13), W = 2 for ethanol and j =. 2 for acetic acid. Substituting these values into Eq. (4.16) for thetheoretical oxygen demand gives:

a=. 1/4(w/S-C1B-fj1P) = 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 gmoloxygenper gmol glucose without groWth. Therefore, with growth, the oxygen demand for acetic acid production is increasedby31%.

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 enthalpychange 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 integralof this equation between the limits T:::: 15"C and T:::: 1200C:

J.'''''' J,,,,,"c

6h = CpdT = (0.248+ 0.(01531) dT calg-llS"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 lOw5 y'2 _ 1.965 x lOw9 T3

where Cp is heat capacity in J gmol-l °Cl and Tis temperature in °C. The sensible energy change can be determinedby evaluating the integral of this expression between the limits T =150°C and T =65°C:

J,'~ J,~~

M = Cp dT = (28.94 + 0.4147x 10-2 T+ 0.3191 x 10-5 f2 - 1.965 x 10-9 rJ) dT J gmor115O"C l5O"C

42 Souaions: Chapter 5

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 thevalues at nee and 34°C~ Ahv =2423.55 kJ kg~l at 33°C From Eq. (5.16);

Mf = M!>hv = 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(aJThe heat of vaporisation of water at 85"C is obtained from Table C.I (Appendix C). Taking the average of the valuesat 84"C and 86"C, !:J.hy = 2296.05 kJ kg-! at 85°C.

Answer: 2296.05 kJ kg-l

(bJFrom Table C.l (Appendix C), the enthalpy of liquid water at IOce relative to the triple point is 42.0 kJ kg- t , Theenthalpy of liquid water at 3S"C relative to the triple point can be estimated as the average of the values in Table Colfor 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 istherefore (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

(dJThe 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 kJkif1. 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 medium1. Assemble(i) Unitskg, h, kJ, °C(ii) FkJw sheet


System boundar;\.

r

Q (loss) =0.22 kW

1II

Medium in ..l.-_~

3250 kg h-1

15°C

Heating vessel 1--'----_ Medium out3250 kg h-1

44°C

-Q (from condensing steam)

(ill) System boundaryThe 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) Basis1 h, or 3250 kg medium in(ill) Reference stateH = 0 for water (steam) at its triple pointH =0 for medium at 15°C(iv) Extra dataC medium =0.9 cal g-l 0(;-1 =0.9 kca1 kg-I °C-16hv waterat 150°C = 2113.1 kJ kg-1 (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(v) Mass balanceThe mass balance is already complete.(vi) Energy-balance equationAt steady state, Eq. (5.9) applies:

:E(Mh) - :E(Mh)- Q+ W, = 0input output

""""" """""3. Calculate(i) ldmtify ta"" in the 'nagy-balan" ,quatianWs =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:

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

Converting to kJ:

44

4 14.187X103J111kJ I 5(M h)MD out = 8.483 x 10 kcaL 1 kcal '1000 1 "" 3.55 x 10 kJ

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

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


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 kJQ has a negative value which is consistent with the sign conventions outlined on pp 87-88: heat must be supplied tothe system from the surroundings. This heat is provided as the latent heat of vaporisation as saturated steam at 150<>Ccondenses. The enthalpy change from this change ofphase is calculated using Eq. (5.16) and must be equal to -Q.

3.56xloSkJ "" Msteam!JJlv =Msteam(21l3,lkJkg-1)

Mstearn "" 168 kg

4. Finalise

Answer: 168 kg

5.5 Production of glutamic acid(a)L Assemble(i) Unitskg, b, kJ; °C(ii) Flow sheet

Off-gas

System boundary

,~ - -- - - -- 1I II I

FeedI Reactor I

(25,000 lltres)Product

2000 kg h-1 I I 0.5% glucose4% glucose

I / I96% water25°C r--t- I

- -- .

Q (cooling)Inlet gas

100,000 litres min-11 atm, 15"C

88% air12%NHs

(ill) System boundaryThe system boundary is shown on the flow sheet.


(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) BasisI h, or 2000 kg feed(ill) Reference stateH = 0 for water at its triple pointH = 0 for feed at 25°C(iv) Extra dataMolecular'weights (Table B.t, Appendix B): glucose = 180.2

NH3 = 17.002=32.0Nz = 28.0glutamic acid = 147.1C02 =44.0H20 = 18.0air = 28.8 (see Problem 2.9, Chapter 2)

Ideal gas constant (fable 2.5): R = 0.0820571 atm K-l gmol-lComposition of air (p 17): 21% Oz, 79% Nz by volumeHeats 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 gmor1

45

(v) Compounds involved in reactionGlucose, ,ammonia, oxygen, glutamic acid, carbon dioxide and water are involved in the reaction.(vi) Mass-balance equationsFor,glucose, ammonia, oxygen; glutamic acid. carbon dioxide and water, the appropriate mass~balanceequation,isEq.(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 equationThe modified energy-balance equation, Eq. (526), applies:

-MIrxn-MyAhv-Q+ Ws = 0

3. Calculate(i) Mass balanceAs 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 = 7.20x 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; thevolume ofNZ in = 0.79 x 5.28 x 106 = 4.17 x l06litres. Converting these gas volumes to moles using the ideal gaslaw, Eq. (2.32), with the temperature converted from °C to degrees Kelvin using Eq. (2.24):


Moles of02 in = pV = 1atm(l.l1 x 1Q6t) = 4.69xI04gmolR T 0.0820571almr' gmor1(15 + 273.15) K

MI fN · _pV _ laJln(4.17XI0'1) -176 10' Io eso 2m - - - - . x gmoRT 0.0820'71aJlnK 19moll(I'+273.I5)K

MolesofNH in=pV = latm(7.20xt051) =3.05x104 g.mol3 RT 0.0820571aJlnK 'gmol-'(15+273.I5)K .

These molar quantities can now be converted to masses using the molecular weights:

Mass of02 in = 4.69 x 104 ~Ol·l i~~II.ll~gl = 1500.8kg

Mass ofN2 in = L76x 105 ~OLI i~~II.ll~gl = 4928.0kg

MassofNH3 in = 3.05 x 104

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 massof product is denoted P.

20006947.3

Total

1920o

HZOInGluconic acid C02o 0o 0

o4928.0

o1500.8

02o

518.580o

80 518.5 1500.8 4928.0 0 0 1920 8947.3

OutGlucose NHS °z NZ Gluconic acid COz HZO Total

0 0 ? ? 0 ? 0 G0.005P ? 0 0 ? 0 ? P

0.005P ? ? ? ? ? ? G+P

GlucoseStream

PeedInlet gasOff-gasProduct

Total __-=__~=",-_-",="-_==,--,,-- -,,,-__--,,=__-==,-_Stream

PeedInlet gasOff~gas

Product

Total

N2 balanceN2 is a tie component.

4928.0 kg N2 in = N2 out

N2 out = 4928.0 kg

Glucose balance80 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 ·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 kgmolNH3 and 1.5 x the number of kgmo1 02. Converting these molar quantities to masses:


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

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

47

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

Glutamicacidgenerated = (0.444-2.775 x 10-5 P)kgmol = (0.444-2.775X W-5p)kgm01.1 ~4:~~~ IGlutamic acid generated = (65.312 -4.082 x 10-3 P) kg

CO2 generated = (0.444-2.775 x 10-5P)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 10-5P)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 balance1500.8 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 balanceokg C02 in + (19.536 - 1.221 x 10-3 P) kg C02 generated = C02 out + 0 kg C02 consumed

C02 out = (19.536 - 1.221 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 10-3P) kg 02 + (19.536 - 1.221 x 10-3 P) kg C02

G = (6427.0+ 1.11 x 10-4P) kg

Total mass balance8947.3 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:

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 10.3P) 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 balance518.5 kg NH3 in + 0 kg NB3 generated = NH3 out + 6.36 kg NB3 consumed

NH3 out =512.14kg

Water balance1920 kg water in + 20.20 kg water generated = water out + 0 kg water consumed

Water out::: 1940.20 kg

Glutamic acid balanceokg glutamic acid in +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.

Stream OutGlucose NHl °z NZ Gluconic acid CO2 HZO Total

FeedInlet gasOff-gas 0 0 1482.9 4928.0 0 16.46 0 6427.4Product 12.60 512.14 0 0 55.02 0 1940.20 2520

Total 12.60 512.14 1482.9 4928.0 55.02 16.46 1940.20 8947.3

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

(ii) Energy balanceWs =0; Mv =O. Therefore, the energy*balance equation becomes:

-AHrxn-Q::: 0

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

Mfrxn = (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 productinvolved in the reaction. Converting the masses of reactants and products consumed or generated to moles:

Glucoseconsumed::: 67.40kg ::: 67AOkg ·ll~gl·1 :8~o~I::: 374gmol


: 1.25kgOZh-1

As NH3 and glutamic acid are involved in the reaction in stoichiometric quantities. 374 gmol NH3 are consumed and374 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 throughthe reactor. For a rough calculation of the effect of this heat on the temperature of the reactor, assume that the 353 x105 kJ h-1 is absorbed by 2000 kg h-1 aqueous medium and 6947-3 kg l:r1 gas. ,Assume that the heat capacity of theaqueous 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, andthat 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):: 29kJ kgmol-l °C-l. From Eq. (5.12):

6T: MI(M Cplliquid +(M Cpl..,

T :: 353 x 105

kJ h-1

:: 23"C

(2000 kg h-1)(75.4kJ kgmor1oc-1j.1 ~~~~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 ofadequate cooling for this reaction is an important consideration. Assuming that the usual temperature for the reactionis 25°C, the temperature without cooling woulJ,i increase to (25 + 23)OC =48°C.

5.6 Bacterial production of alginateAlginate production at a rate of 5 kg h~1 requires:

5 kgh-1

4kgkg 1

Converting this quantity to gInol using the molecular weight of 02 =32.0 (Table B.I, Appendix B):

0zrequired:: 1.25kgh-1 .1 10:g 1.1 ~~~ll :: 39.06gmolh-1

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

'This result can be used in the modified energy-balance equation, Eq. (5.26):

-6Hrxn -Mv tihv -Q+ Ws :: 0

with Ws :: 1.5 kW andMv := 0 (no evaporation). From Table A.8 (Appendix A), 1 W:: 1 J s-1; therefore 1 kW:: 1 kJs·1 and:

50

1.80x 104 kJh-1_O_ Q+ 1.5 kJ 8-1·13~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·1

5.7 Acid fermentationFrom Tables B.S andB,l(Appendix B), the molecular fannulae and molecular weights are:sucrose = C12H2Z011: MW:: 342.3propionicacid=C3H60 Z: MW =74.1" -acetic acid = Czf40Z: MW =60.1butyric acid=C4HgOZ: MW = 88.1lactic acid =C3H603: MW =90.1From p 75. the biomass molecular formula can be taken..asthe,average, CH1.80o.sNo.z'. From the end of Table E.8(Appendix B). the molecular weight of the biomass is 25.9.

The reaction equation can be obtained by modifying Eq. (4.13) for anaerobic growth andproo.uct-fonnation:

C 12HZZ011 +bNH3 -+ cCHl.gOo.sNo.2 +dCOZ +eHZO+!I C3~02

+h C2H40 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 coefficientc using Eq. (4.l2):

_ Yxs (MW substraJe)) _ 0.12 gg-l (342.3) _c - MW cells - 25.9 - 1.59

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

Yps (MW substrate)) 0040 g g-I (342.3)11 = MWpropionicacid = 74.1 = 1.85

= Yps(MWsubstrate)) = 0.20gg-1 (342.3) = 114h MW acetic acid 60.1 .

Yps (MW substrate)) 0.05 g g-I (342.3)Jj = = = 0 19

MW butyric acid 88.1 .

_ Yps (MW substrate)) _ 0.034 g g-l (342.3) _/4 - MWI . ·d - 90 I - 0.13achC acl ..

Of the remaining coefficients b, d and e, because CO2 and H20 do not figure in heat of reaction calculations as theirheat 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 ofcombustion of the reactants and.products are required from Table E.8(Appendix B):

M~ sucrose=-5644.9kJ gulOrl

M~NH3 =-382.6kJ gmol-l

M~biomass=-552kJ gmor1

M~ propionic acid = -l527.3kJ gmol-l

M~ acetic acid=-874.2kJ gulOrl

M~ butyric acid =-2183.6kJ gmor1

M~ lactic acid = -1368.3 kJ gmol-l

Solutions.' Chapter 5 51

The heat of reaction is determined using Eq. (5.20). As the heat ofcombustion 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 = lacticacid. 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:

lIOOOgll,gmOll( -I) 4M rxn =30kg. lkg . 342.3g -474.8 kJ gmol =-4.16x 10 kJ

The cooling requirements are determined using the modified energy~balanceequation,Eq. (5.26). For no evaporationand no shaft work, Mv = Ws = 0, so that:

Q = -Mfrxn =4.16 x 104 kJ

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

Answer:4.16x 104 kJ

5.8 Ethanol fermentationFrom Table B2 (Appendix B), the molecular formula for glucose is C6H1206 and the molecular fonnula for ethanolis 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 intoaccount 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):

Yps(MWsubstrate) 0.45gg-1(180.2ggmol-l) -1f = = - = 1.76gmolgmol

MW product 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-dHbalance: 12+3b = 1.75c+2e+6f= 1.75c+2e+6x1.76 ~ 1.44+3b = 1.75c+2eo balance: 6 = 0.58c+2d+e+f= 0.58c+2d+e+1x1.76 ~ 4.24 = 0.58c+2d+eNbalance: b = 0.18eSubstituting 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

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. Thecompleted stoichiometric equation is therefore:

C6H1206+0.13NH3 ~ O.72CHu 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:

1Moles 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 gmor1

8h~NH3 =-382.6kJ gmol-l

tJh~ ethanol"" -1366.8 k:J gmarl

From p 101, the heat of combustion of yeast can be taken as -21.2 kJ g-1. The heat of reaction is determined usingEq. (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 reactantsand 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)

Mlrx,n =-1175 kJ

Using the modified energy~balanceequation, Eq. (5.26), with Mv =0 and Ws =0:

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 thiscase, 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 watercan 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 =~_ = 117.5 kJ h-1

_ 112"<:

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

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

Answer: 21.2°C


5.9 Production of bakers' yeastFrom Table B,8 (Appendix B), the molecular formula for sucrose is C12HZ2011' Therefore, from Eq. (4.4), thestoichiometric 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 biomassformula 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: 4.22. .... 1 ..

The degree of reduction of sucrose relative to NH3 is:

_ 12x4+22x1-11x2 _ 400rs - - .

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-lMW 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 pergmol biomass. Converting this oxygen demand to mass terms using the molecular weight of oxygen: 32.0 fromTable B.I (Appendix B):

. _ 0.79 gmolOz 132.0 g Ozlll gmol biomass 1_ -10.79 gmol 0Zper gmol blOmasS - 1gmol biomass' 1 gmol0

z. 26.36 g biomass - 0.96 g g

The specific growth rate represents a rate of growth of 0.45 g biomass produced per g biomass per h. As 0.96 g 02are 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 perh. 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 x105 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·1. Convertingthis to moles:

Rate of0 1 consumption : 2.15x 105 gh-1 .l ;~~ll : 6.72x 103 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·balance equation, Eq. (5.26),withWs:Mv:O:

From the sign conventions outlined on pp 87-88, Qpositive 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 boundaryThese are shown in the figure below,

WaterF::::: 40,000 I h·1

System boUndary\ , "- - 1--- - 1rI II - II II II II V II

CA I

Outlet streamF=40,OOO Ih-1CA

(Ii) Define variablesV =volume of material in the tank; F ::::: volumetric flow rate into and out of the vessel::::: 40,000 1 h· 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 conditionsAt 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:

C 10,000 kg 0022k 1-1AO::::: 450,ooof:::::' g

(v) Total mass balanceAs the volumetric flow rates of material in and out are the same and the density of the inlet and outlet streams areassumed to be equal, the volume of material in the tank is constant (see Example 6.2) so that V = Vo at all times.(vi) Solids mass balanceThe 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 = -FCA

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


As F is also constant, the differential equation contains only two variables, CA and t. Separating variables gives:

dCA -F-=-dtCA V

Integrating:

fdCA = J-F dtCA V

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

-FInCA = -yl+K

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

-FInCA = -yt+InCAO

CA -FIn- =-1

CAO V

C C (-FIV; t~ = AOe

Substituting the known values for CAo. F and V:

CA = 0.022 e-O·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

55

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

dMCit =-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:

56

InM = -kt+1nMo

MIn M

o= -let

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

O.5MoIn -:MO =-k (20 min)

In 0.5 = -k (20 min)

k = 0.0347 min~l

Therefore:

MIn Mo =-0.0347 t

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

O.05MOIn M

o= -0.0347 t

In 0.05 := -0.0347 t

t =86.3 min =: 86.3min.I601~n!= l.44h

Answer. 1.44 h

6.3 Contamination of vegetable oil(i) Flow sheet and system boundaryThese are shown in the figure below.


System boUndary\ \. ;

- - - - - 1rI II I

Inlet stream I IOutlet stream

Cod-liver oil drum F; I I FoP I I p

V </I p II </ I

L Oil tankJ- - - - - - -

(li) Define variablesV = volume of oil in the tank; Fi "'" volumetric flow rate of cod~liveroil into the oil tank; F0 =volumetric flow rate ofmixed 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


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

57

(a)Total mass balanceThe general unsteady~statemass--balance equation is Eq. (6.5). As there is no reaction, RO = RC = O. The t2.ta1 massflow 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_Fpdt ' 0

As P is constant it can be taken outside of the differential and cancelled:

dVPdt = (Fi-FdJp

dVdt = (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 1h-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 notoverflow. '

Answer: No

(b)Total mass balanceAs the volumetric flow rates of oils into and out of the tank are the same and the density of the inlet and outlet streamsare 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 balanceThe general unsteady~statemass-balance equation is Eq. (6.5). As there is no reaction, RG "'" RC =O. No vegetable oilenters the tank; therefore Mi "'" O. The mass flow rate of vegetable oil out is Mo "'" Fo P 4>. The mass of vegetable oilin 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 andintegrating:


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

-FoIn¢::::: yt+K

From the initial condition for 41, at t =0, In <Po::::: K.. Substituting this value of K into the equation gives:

-Foln~ = y'+lnib

~ -Folnib=y'

~ = ib e(-F,II'J'

In this problem. F0 ::::: 4.8 I h-l. Substituting the known values for </Jo and V:

II ::::: 1 e....(l·080 t

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

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

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

d(Vx) = J.txVdt

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

dxV dt = J.tx V

dxdt =J.tx

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

dx- = I'dtx

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 =xo. Therefore, at t =0, In Xo =K.Substituting this value ofK into the equation gives:


lnx = j.tt+lnxo

xIn- = j1,t

XO

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

2XOIn- = ,u(45 min)

XO

ln2 = j.t (45 min)

j.t = 0.0154 min'l

Therefore:

In''::' = 0.0154tXO

x = XQ eO.Ol54 t

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

x=6.54x 104 XQ

59

After 12 h, the cell concentration is 6.54 x 104 times the cell concentration at the beginning of exponential growth. Ifthere is a lag phase, the cell concentration at the beginning of exponential growth is usually very close to that atinoculation; 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 decayThe general unsteady~state m.ass~ba1ance equation is Eq. (6.5). In this problem, M j = Mo = O. For a mass balance onisotope, 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 isequal 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 outsideof the differential and cancelled:

dCV(if =-k1CV

dC(if = -k1 C

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

dCC = -k1dt

Jdg =f -k) dt

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

lnC=-k1 t + K


Assume an initial condition: at t "'" 0, C "'" CO, Therefore, at t "'" 0, In Co = K, Substituting this value of K into theequation:

InC = -ktt+lnCo

CIn-=-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 CoIn-- = -kIth

Co

In 0.5 = -ki !h

t_ -tn0.5

h---k,

Using mathematical rule (D. 10) in Appendix D, -In 0.5 =In 1/0.5 =In 2. Therefore:

In2th =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 ofdays. ForC=O.Ol CO:

0.01 Co -2In Co = -4.85 x 10 t

In 0.01 = -4.85 x 10.2 t

t = 95 days

Answer. 95 days


6.6 Continuous fermentation(i) Flow sheet and system boundary'These are shown for a continuous fermenter in the figure below,

61

Sy

FeedF

xI =0

"

stem boUndary\ .... "- - r- - - - 1rI II II II II

VI

I x II • I

ProductFx

•

(li) Define variablesV =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(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(iv) Boundary conditionAtt=O,x=XO·(v) Total mass balanceAs the volumetric flow rates into and out of the fermenter are the same and the density of the inlet and outlet streamsare assumed equal, the volume of broth in the fermenter is constant (see Example 6.2) 8..Iid equal to Vat all times.

(ajCell mass balanceThe general unsteady-state mass-balance equation is Eq, (6.5). As no cells enter in the feed stream, Mi =O. The massflow 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+k1xVd'

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

dxV dt = -Fx+k1xV

Dividing through by V and grouping terms:

dx F- = x(kt--)d' V

dx FAnswer: 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


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

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

J~ = f(k,-~)dt


Flux:::: (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:

FInx:::: (k1-y)t+1nXO

• F10- :::: (kl--)t

'0 V

F(kl--)tx=.xoe ,v

F(k1--) ,

Answer: x:::: XOe 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 balanceThe general unsteady-state mass-balance equation is Eq. (6.5). For substrate, the mass flow rates in and out areMj =F si and Mo = F s. The rate of substrate generation RG =0; the rate of substrate consumption RC =IS V = k2 xV.The mass of substrate in the fermenter M is equal to V s. Substituting these terms into Eq. (6.5) gives:

d(Vs)(it = Fsi -Fs-k2xV

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):


In this equation, F, V, kl and X{) are constants and there are only two variables, s and t. However, the variables cannotbe easily separated as in the previous problems, making algebraic solution difficult.

& F ~l-~tAnswer. dt =V (si -s)-k1X{)e V

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

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

FV (si -s) =k2 x

Vk Zsi- s = F'x

Vk ZAnswer. s = si -Tx

6.7 Fed-batch fermentation(i) Flow sheet and system boundaryThese are shown for a fed-batch fermenter in the figure below.

1

IIII1II

L J

stem bOUndary\ \. '"- - - -r

II -III V

xI s

Ip

s;

p

Sy

FeedF

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

(iv) Boundary conditionAtt=O, V=Vo.


(a)Total mass balanceThe 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 thefermenter M is equal to V p. Substituting these terms into Eq. (6.5) gives:

d(VP)=pFdt

As P is constant, it can be taken outside of the differential and cancelled:

dVPdi =pF

dV =Fdt

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 = Ft+K

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

v= Ft+Vo

Answer: V=Ft+Vo

(b)Substrate mass balanceThe general unsteady-state mass-balance equation is Eq. (6.5). The mass flow rates of substrate in and out areMj =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 theproduct rule (D.22) from Appendix D:

Using the equation dV1dt =F derived in (a) :

Grouping terms gives:

<is FAnswer: dt =V(Si-S)-kt s


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

System bOUndary, ...

~__!..I ---"'---__~

Fee~--"""-II : -=-.,.....,.:.. S-.......- ~rOduct~ I z ~Ml

L J

65

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

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

Mj = CAuAlz

where kmeans that the parameter values are those at distance z from the front of the reactor. Similarly, the rate atwhich reactant leaves the system is:

Mo = CAuAI,+<\,

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 noaccumulation 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:

66

dCAAnswer: uliZ "" -rC

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


(e)If the reaction is first-order, rc =: kl CA where kt is the first-order rate constant. The differential equation becomes:

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

dCA = -k) dzCA U

fdCA "" f-k) dzCA U

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

-k)InCA =-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 -kl1n-=-z

CN U

CA =: CAi e(-k,lu) z

(d)The equation derived in (c) is directly analogous to the equation for change. in reactant concentration in a batchreactor. 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 timezero.

Answer: Essentially identical

6.9 Boiling waterThe 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


(ii) Extra dataDensity of water = 1 kg I-ICp water = 75.4 J gmol-1 "'C-l (fable B.3, Appendix B) = 75.4 kJ kgmol-l "'C-lMolecular weight of water (fable B.l, Appendix B) =18.01 W = 1 J s-l (fable A.8, Appendix A); therefore, 1 kW = 1 kJ s·l(iii) Boundary conditionsAtt=O,T=TO=18°C

67

(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 -MCPdt =-Q

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 andintegrating:

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

T= -Q t+KMCp

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

-12T= MC t+To

p

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

Cp = 75.4kJ kgmor1 "'C-1 .1 ~::~11 = 4.189 kJ kg-1 "'e-I

T =the boiling temperature of water =100°C; t =11 min. Substituting the parameter values into the equation for T:

lOO"C = ( -Q 1 I) (11 min) + 18°C2kg 4.189kJkg "'e-

Q = -62.45kJmin-1= -62.45kJmin-

1·116~snl·ll ~;11 = -L04kW

From the sign conventions outlined on pp 87-88, the negative value for Qconfirms that heat is added to the system.

Answer: L04 kW

68

6.10 Healing glycerol solutionThe 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 dataCp glycerol = 0.576 cal g-t °C-l (Table B.s, Appendix B) = 0.576 keal kg-I "C-1

Cp water = 75.4 J gmol-l °C-l (Table B.3, Appendix B) = 75.4 kJ kgmol-l °C-1

Molecular weight of water (Table B.I, Appendix B) "" 18.01 kcal "" 4.187 x 103 J (Table A.7, Appendix A) = 4.187 k11 W =1 J s-1 (Table A.S, Appendix A); therefore, 1 kW = 1kJ 5- 1

(iii) Boundary conditionsAtt=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 ofsensible heat only; therefore:

dE=MC dTdt 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 bewritten:

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 isthe heat capacity of glyceroL

<IT -Answero(MWCpW+MOCpG)"dI =-Q

(b)

If ii, Cpw, CpO, MW and Mo are constant, T and t are the only variables in the differential equation. Separatingvariables and integrating:

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:


-12Answer: T = (MwCpw+MOCpG) t+ To

(c)The mass of glycerol in the tank MO = 4S kg; the mass of water Mw =55 kg. Converting the Cp for water to massterms:

Cpw = 7SAkJkgmorl <)c-l',111~~~11 = 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 pp87-88, Q must be negative as heat is added to the system; therefore, Q"'" -2.2 kJ s-l. Substituting the parametervalues into the equation for Twith To =15<)C and T= 90<)C:

9O"C "'" 2.2kJs-1

t+ 15"C55kg(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 boundaryThese are shown in the figure below.

System boundary~

- - - -- - - 1I II I

Molasses solution in IHeating tank

IMolasses solution out

1020 kg h-1 IM I 1020 kg h-1

20'C/

TI T II I- - -- - - - - --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

- condensate from the steam leaves at saturation conditions(iii) Reference stateTref"" 20oe; H"" 0 for molasses solution at 200 e(iv) Extra dataCp "" 0.85 kcal kg~l °C-1

1 psi = 6.895 x 103 Pa (fable A.S, Appendix A) = 6.895 kPaConverting 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 balanceAs the mass flow rates into and out of the tank are the same, the mass of molasses solution in the tank M is constantand 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 horelative to the reference state is equal to the sensible heat-absorbed by the molasses'solution between Tref and the exittemperature T. From Eq. (5.13):

ho =~ =Cp(T-Tref)

Ws = O. The rate at which the molasses solution is heated is given by the equation:

Q =-UA (Tsteam -1)

From the sign conventions outlined on pp 87-88, Q must be negative as heat is added to the system. Energy isaccumulated in the form ofsensible 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)Tdt MCp MCp

dTAnswer: dt

(b)As U, A, Tsteam, Mo. Cp • Tref and M are all constant, Tand t are the only variables in the differential equation derivedin (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-1oC-') (20"C)

<It - 5000kg(0.85kcalkg loC 1)

_(1020 kg .-1 (0.85 kcal kg-1oC-')+ 190 kcalm-2 .-1 oC' (l.5 m2)) T

5000kg(0.85kcalkg loC 1)

dTdt = 12.84-0.271 T

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


dT = eIt12.84-0.271 T

f dT -felt12.84-0.271 T -


-10.271 In (12.84-0.2711) = t+ K

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

-IK = 0.271 In (12.84-0.271 To)

Applying the numerical value of To =20°C, K =-7.395. Substituting this value for K into the equation gives:

-I0.271In(l2.84-0.2711)+7.395 = t

-IAnswer: 0.271 In (12.84- 0.2711) +7.395 = t

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

Temperature, TCC)

Time, t(h)

202530354045464747.3

0.000.741.682.934.849.01

11.015.821.5

The results are shown in the figure below.

50 r----,----,----,,----,-----,

252010 15

Time (h)

5

20 ..-__-'-__..L__-'-__-'-__-'

o


(d)From the equation derived in (b), as the logarithm of zero and negative numbers is not defined (p 413), the theoreticalmaximum 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 practicalpurposes, 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 mediumThe 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 dataCp glass vessel"" 0.20 cal g-l 0C-l "" 0.20 keal kg-I octCp medium"" 0,92 cal g-l °Cl "" 0.92 keal kg-loCi1 W"'" 1.433 x 10-2 keal min-I (Table A.S, Appendix A)(ill) Boundary conditionsAtt=O,T=To=15°C(iv) Energy balanceThe 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 dTdt =MCpdt

M has two components, the glass vessel and the medium, which have different heat capacities. Therefore, thisequation 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, andCpM 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. Separatingvariables and integrating:

dT= -Q dt(My Cpy + MM CpM)

fdT-f -Q dt- (My Cpy+MM CpM)

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


T= -Q t+K(MV CpV +MM CpM)

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

T= -Q t+To(Mv CpV +MM CpM)

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

73

36'C =

From the sign conventions outlined on pp 87-88, Qmust 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:

6.45 kcal min-l t+ 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 heaterThe 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 dataCp 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) =18.01 kcal = 4.187 x 103 J (fable A.7, Appendix A) =4.187 kJ(iii) Boundary conditionsAtt=O, T=To =24"C(iv) Energy balam:eThe 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 dTdt =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

(ajThere 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 ispositive to indicate heat removal from the system, while the term for heat input from the steam is negative.Substituting into Eq. (6.10) gives:

74

MC dTPdt

dTCit =

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

MCp


As Vt. A 1> Uz. AZ, Tsteam, 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 m2)(130-1)°C _ 25 kcal m-2 h-I 0C-l (O.9m2) (T - 20Y'C=

dt (-1 -1 11 kgmOllllkCai nlOOOkg 75.4kJkgmol "'C . 18.0kg . 4.187k:JU

dTdt = 9.026-0.088 T

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

dT =dt9.026 - 0.088 T

f 9.026~.088T = 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:

-IK = 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:

-10.088 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:

dTMCp <It

dT UZAZ (Tsteam-1)

Cit = MCp

Substituting the known numerical values:

dT 220kcalm-2 h-1 0C-l (0.3m2)(130-n"C

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

dTdt =8.576 - 0.066 T

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


dT = <It8.576- 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:

-10.066 In (8.576-0.0661) = t+K

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

-1K = 0.066 In (8.576-0.066 ToJ

Applying the numerical value of To =24°C, K =-29.47. Substituting this value for K into the equation gives:

-10.066 In (8.576-0.0661) +29.47 = t

75

From this equation, for T =80°C, t "'" 11.4 h, Therefore, from (a), the time saved is (14,2..:,- 11.4) h"'" 2.& h, whichcorresponds 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---,

12001000400 600 800

Shear rate (s·1)

200ol-_...L_......l._---''-_-'-_....L_--'

o

300

600

400

600

200

100

(b)The rheogram in (a) is similar to those for pseudoplastic and Casson plastic fluids in Figure 7.7. It is not clear fromthe rheogram whether the fluid exhibits a yield stress at r= 0 or not; therefore. both non-Newtonian models are worthchecking. From the equation in Figure 7.7 for pseudoplastic fluids, a plot of shear stress versus shear rate on log-logcoordinates would be expected to give a straight line. This plot is shown below.

100 1000

Shear rate (s-l)

10000


The equation for the straight line in the log-log plot is '1:= 11.43 ;-0587. Therefore, the flow behaviour index n =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 squareroot 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 thatKp = 0.648 dyn11z Sllz cm- l and the yield stress 't{) = 35.2 dyn cm-2. The sum of squares of the residuals for this datafit 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 valuesdetermined 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 suspensionsFrom Eq. (7.8), for pseudoplastic fluids, a plot of apparent viscosity versus shear rate on log-log coordinates can beexpected to give a straight line. Log-log plots for the different cell concentrations are shown below.


10Cen concentrations:.1.5%03%

0::- D 6% L.£.

t • 10.5% •••I t3- "~

11 10 100

Shear rate (8-1)

100Cell concentration: 12%

0::-.£.

t••Ee!l~

101 10 100

Shear rate (sol)

1000 ,Cell concentration: 18%

0::-.£.~ ....8.~ 100 ~

E

I...

101 10 100

Shear rate (s·l)


Cell concentration: 21%

•fI

10Shear rate (5-1)

100

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

Cell concentration (%) Equation Flow behaviour index, n Consistency index, K(cp sn-l)

1.5

3

6

10.5

12

18

21

·0I', = 1.5 r = 1.5·01',=2.0r =2.0

Pa = 2.91 r-:O,050

I',= 5.38 r..<>.07.

Jl.a = 50.1 r-:0.395

J.la = 162 7-:0·307

Jla = 833 7-:0.251

1

1

0.95

0.92

0.61

0.69

0.75

1.5

2.0

2.9

5.4

50

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

10

c

.~-" 1.0

0.6

5

!! 0.8

£

1.4,--,.---..--....--"..---.....1.2

• Flow behaviour index, no Consistency, index, K

0.4 '--__..J.....__.....J..__-''-__-'-__.....Il

o 5 10 15 20 25Cell concentration ('Yo)


The cell broth is Newtonian up to a cell concentration of about 2%, then becomes pseudoplastic. The flow behaviourindex 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-logcoordinates can be expected to give a straight line. Values of shear stress and shear rate can be determined fromtorque 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 arelisted below and plotted below.

Stirrer speed (s·l) Torque (Nm) Shear stress (N m-2) Shear rate (gw 1)

0.185 3.57 x 0.0559 1.890.163 3.45 x 10-6 0.0540 1.660.126 3.31 x 10-6 0.0518 1.290.111 3.20 x 10.6 0.0501 1.13

0.07

0.06

i 0.050~

11-~e"" 0.04m~<J)

21.5Shear rate (s-1)

0.03'- ..L... -'

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 flowbehaviour 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). IfRei at the highest stirrer speed is inthe laminar range, conditions at the other stirrer speeds are also laminar. Therefore, for Ni = 0.185 s·1 and using theunit conversion factor 1 N =1 kg m s·2 (Table A.4, Appendix A):

N'~D' P (0.185,-1)2-0.20 (O.04m)' l000kg m-3

Rei = ~1 ~ =-->=="(r'---"==-"IC"':""-"'-:"O,,, = 10,010.20.20-1 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 andlower stirrer speeds is laminar.

Answer: Flow is laminar.


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

N?--n = Rei~1 K

1 vf P

103 (30o.zo-1) (0.049 N sO.20 m-21 1kgms-2~

!if..".20 = 1N III (0.04m)21000kgm 3

Nl-8= 2.02s-1.8

Ni = 1.48 s-l

Therefore, from Eq. (7.11), as r= 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 upperlimit 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 dispersionThe Zwietering equation is an equation in numerics (p 12); therefore, the parameter values used in the equation musthave the units specified. Vp =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 isthe same as J4... = 1O~3 kg m-I s~I. Using the definition of .kinematic viscosity from p 133:

f.lL 10-3 kgm-18-1VL =-= =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;

_ 104 -3 Iloocml3 I~I_ lO4Okg -3Pp - . gcm. m . lOOOg - m

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 dispersionof air bubbles. Taking the average minimum tip speed of 2.0 m s-1 for bubble dispersion:

N = tipspeed = 2.0ms-I =2.1 s-1

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 isproportional to the stirrer speed cubed, about 50 times more power is required to disperse the air bubbles than to keepthe cells in suspension.

Answer: Bubble dispersion requires significantly more power than cell suspension.


7.s Scale-up of mixing system(aJThe Reynolds number for a Newtonian fluid in a stirred vessel is defined in Eq. (7.2). Using this equation to evaluatethe maximum viscosity jJ. under turbulent conditions with the density of water p= 1000kg m-3:

N D~ P (SOOmin-1 .1 160miDll(5 cm)2 .11001m12) 1000 kg m-31 I S U em 33 10-3 k -1-1I.t = = = . x gm sRei 104

From Table A.9 (Appendix A), 1 .kg m-I s·1 =103 cP; therefore:

333 10-3k -1 -1 10 cP 33 PJ.l=.x gms. ll='c

1kgm- s-

Answer: 3.3 cP

(bJThe tip speed in the laboratory equipment is:

Tipspeed = 7tNj Di =1£ (800min-1

-I 16: sD 1)(5 em .11~~ml) = 2.1 m 8-1

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 _ 2.1 ill s-1 _ 089 -1'" _ _ ( ) _ . s1tDj 1m

1t 75cm.l lOocmlUsing this value of Ni to calculate the maximum viscosity as in (a):

N:,.D,~p O.89S-I((75Cm)2·1100Im 12)lOOOkg m-3JJ.= = .em =O,05kgm-l s-1

Rei 10

Converting to cP:

= 50cP

Answer: Scale-up increases the maximum viscosity for turbulent conditions to 50 cP.

7.6 Effect of viscosity on power requirements(aJDj =1 m;p= 1 gcm-3=1000 kg m-3.(1)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-1s-l. Substituting parameter values into Eq.(7.2):

NiDrp (90min-I·116~sn~(lm)21000kgm-3 6Rei = = 3 11 =1.5 x 10

Ji lO kgm- C

From Figure 7.24 for a Rushton turbine, this value of Rei corresponds to turbulent flow and Np can be taken as 5.8,The power required is evaluated using Eq. (720):

P = NppNf Dr =5.8 (lOOOkg m-3)(90 min-I .II:sn~3 (l m)5 = 1.96x 104 kg m2 s-3


From Table A8 (Appendix A), 1 W =I kg m2 s-3; therefore:

P = L96xl04w = 19.6kW

Answer: 19.6 kW

83

(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 (90min-1.11~n~(1 m)21000kgm-3Ret' = = 1 1 = 1.5 x 10

4# 0.1 kgm- s-

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-1s-l = 10 kg m-1 s-l, RCi is:

NiD[ p (90 min-1.11=~(l m)2 1000kgm-3

~= = =1~J.l IOkgm1s1

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 m2s-3

From Table A8 (Appendix A), 1 W = 1kg 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 ofabout 104. For a viscosity of 1000 x 10-3 kg m-1s-1 = 1kg m-1 s-l, the stirrer speed required can be detennined fromEq. (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 = 1kg m2 s-3; therefore:

P = 5.80 x 106 W = 5.80 x 1()3 kW

Answer: 5.80 x 103 kW


7.7 Electrical power required for mixingDi =- 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 10.3kg m-18.1, Substituting parameter values into Eq. (7.2):

2 (900' -1 11minn(007 )2 1OOOk -3NiDi P nun. 60s U . m gm 4Rei = --'-cC:-C- = 3 II=- 7Ax 10

J.l. 10- kgm s

From Figure 7.24 for a Rushton turbine, this value of Rei corresponds to turbulentflow 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 ofthe 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 aeradonDi = 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 = 6.28m3

(aJIf 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 m2 s'3; therefore 1 kW = 1000 kg m2 s-3. ConvertingP to kg m2 s·3:

P = 9.42kW = 9A2kw.ll°oo~-:2s-31 = 9.42 x 103 kg m2C 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 =~42xI03kgm2s-3 =120s-3, NppDi 5.S (1000kg m-3) (O.67m)5 .

Ni = 2.29 sol = 137 rpm

Check that this stirrer speed provides turbulent mixing conditions by evaluating the Reynolds number. Rei forNewtonian fluids in a stirred vessel is defined in Eq. (7.2):

Rei = NiDf P = (2.29s-l)(O.67m)21000kgm-3 = 2.6x 105j.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):


t = 1.54 V = 1.54(6.28m')m D; Nj (O.67m)3 2.29S-1

= 14s

85

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 stirrerspeed which delivers the maximum power P= 9.42 x 103 kg m2 s-3 can be evaluated using Eq. (7.18):

Nt = P 5 = 9,42x103

kgm2

s-3

- 241 -3(Np) pD; 2.9(lOOOkgm-'j(O.67m)5 - . s

g

Nj = 2.89 s·1 = 173 rpm

The mixing time evaluated using Eq. (7.16) is:

t =!541' = 1.54(6.28m')m DfN

1(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 theCelsius and Kelvin scales (p 18). The rate of heat conduction can be calculated using Eq. (8.10):

Q = kA !'J = (0.3 Wrn-l

K-1) Urn

2(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 thefirebrick:

For the asbestos, B2 ::::: 4 em ::::: 0.04 m, so that:

RZ::::: 82::::: a.04m ::::: O.27KW-1k2A (o,tWm lK 1)1.5m2

~ Therefore, the total wall resistance RW= RI + RZ ::::: (0.33 + 0.27) K W~ I ::::: 0.60 K W~l, For thennal resistances inseries, the rate of heat conduction is calculated using Eq. (8.14):

Q= I1T = 620K = 1033W= 1.03kWRw O.60KVI

Answer: 1.03 kW

8.2 OveraU heat-transfer coefliclentThe overall heaNransfer coefficient is calculated using Eq. (8.24) with hfh::::: 830 W m~2 K-I, hh::::: 1.2 kW m·2 K-I, B=6mm::::: 0.006 m, k= 19 Wm,l K-l, he"" 1.7 kWmw2 K-l and hfc= O.

1 lIB 1 1- = -+-+-+-+-U hfh hh k he hfc

1. = 1 + 1 + O.OO6m + 1 +0

U 830Wm 2r-1 1.2kWm-2r-1.ll~:1 19Wm IK-

1 1.7kwm-2K'"1.11~:1

i, = 2.94 x 10-3 W-l m2 K

U= 340Wm·2 K-I


8.3 Effect ofcooling-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

87

(b)The mean temperature difference between the fermentation fluid and the cooling water is calculated from Eq. (8.35):

2 Tp-(Tct+Tco) (2 x 35)OC-(S + 151'C/iTA: 2 : 2 : 23.5OC

Answer. 23.5"C

(c)U A is evaluated using Eq. (8.19) with /iT: /iTA:

UA = Q =ATA

(d)

14.6 kJ s-l

23.5"<:

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:

- 1!iTA : -iL : 14.6 kJ s-

UA' 0.93 kJ s-l °C1 = 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 Q: 14.6kJ s-l, Tci = goC and Teo =30.6"C:

- -IM - Q - 14.6kJ, _ 015k -Ic - 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 designNi: 80 rpm =80'60 =1.33 s·l. Cp culture fluid =4.2 kJ kg· l °cl : 4.2 x 1()3 J kg'} "c l . Cpc =Cp water =75.4 Jgmol'} "C- I (Table B.3, Appendix B) =75.4 kJ kgmol-l "C- l . Converting Cpc ta a mass basis using the molecularweight af water =18.0 (Table B.I. Appendix B):

Cpc =75.4 kJ kgmol-l "C-l : 75.4kJkgmorl "c-I.1 ~~~ll ;:: 4.19kJkg-1 "C-l

,llb;:: 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 = 0.6W 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.


From p 97, as the heat of reaction for an exothermic reaction is negative, Aiirxn =-2500kW. From the modifiedenergy-balance equation, Eq. (8.33), when evaporation and shaft work can be neglected. Q=-Mrxn =2500 kW =2.5x 106 W, According to the sign conventions on pp 88-89, Qpositive is consistent with heat being removed from thesystem. 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). Thedimensionless 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?,62pr033(::fI4 =O.87(3.84X 106)°.62 (7.0)0.33 (1)0.14= 2.0x 104

From the definition ofNu in Eq. (8.37):

Nukfb 2.0x 104 (0.6 Wm-l °C-1) 3 2 Ih=--= =2.4 X 10 Wm- 0C;

D 5m

Therefore, as the fennenter fluid is the hot fluid, hh = 2.4 x 103 W m-2 °e1.

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). Assumingthat the heat-transfer coefficient for the cooling water hc can be neglected (p 185):

1- =.!.+~ = 1 + O.OO6m =7.17xlO-4W-1mZoCV hh k 2AxI03Wm 20(;"1 20Wm-1OC 1

U = L40x 103 wm-Zoe1 = 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 totalresistance 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 = 042l/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 outlettemperatures and the rate of heat transfer. From this equation, an expression for the outlet cooling-water temperatureis:

Solutions: Chapter8 89

where Tco has units of °C and Me has units of kg b·1. An expression for the mean temperature difference between thefermentation 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 _ TeoATA - 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= ATAand the value of Ufrom (b);

A =~ = 2500kW = 1.8x 103

U!1TA 1.40kWm-Zoc-1(25_ T;'fe 25- T~

where A has units of mZand .Teo has units of °C. The above equations for Teo and A allow evaluation of theseparameters 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 arelisted and plotted below.

M,(kgb- I ) Tco (0C) A (m2)

1.2 x loS 27.9 1632.0xloS 20.8 1233.0x loS 17.2 1I04.0x loS 15.4 1048.0x loS 12.7 971.0 x 106 12.2 952.0 x J(j6 ILl 93

30,.....---..,.,.....---.,..-----.,..-.----,20041 • Temperature

o Area

o'- J- -'- ...J''--__--' 50o 5 10 15 20

Cooling-water flow rate, Me 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 aradius of 5 em =0.05 m:

A 101 m2

L = 21tr = 21t(0.05m) = 321 m

Answer: 321 m. This is a long cooling·coil, representing a considerable expeme whenfabricatedfrom stainless steel.


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}·l °C~l, Converting to mass termsusing 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 =4,19kJ kg-l 'C-I

From Eq. (8.32), the rate ofheat removal to the cooling water before cleaning is:

Q =McCpc(Tco-Tci) =20 kg s-I (4,19kJkg-1'C-I) (2S-12)'C = 1340,SkJs-1

The meantemperatute difference between the fennentation fluid and the cooling water can be calculated from Eq.(835),

2 Tp-(Tci + Tco) (2 x 37)'C-(12 + 2S)'CLiTA = = = 17'C2 2

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 m2

Evaluating the overall heat-transfer coefficient U from Eq. (8.19) with aT=aTA:

- 1U - Q - 1340.8kJs- -140kJ -1 -Z'C-1---- - sm

ALiTA (565m2)I7'C '

(b)If the fermentation temperature is maintained the same after cleaning the cooling-coil, the rates of heat removal Qbefore and after cleaning must be equal. The new outlet cooling~watertemperature can be calculated from Eq. (8.32):

1340.8kJ s-1 +120C = 36.6OC13kgs-I(4,19kJkg-l,C I)

(e).The mean temperature difference between the fermentation fluid and the cooling water after cleaning is detenninedfrom Eq, (835),

2 TF-(Tci + Too) (2 x 37)'C -(12 + 36,6)'C1J.TA = 2 =_. 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 =~ = 1340.8kJs- =1.87kJs-1m-20c-1

ALiTA (565 m2) 12,7'C

The overall beat-transfer resistance is given by Eq. (8.20):

RT = _1_ = I =9.46 X 10..,3.k]""1 s 0CUA L87kJs-'m 2'C-l (S65m2)

This is the beat-transfer resistance after cleaning the cooling coil. 'The resistance before cleaning is calculated fromEg. (8.20) using the value of U from (a):


The resistance due to the fouling deposits is equal to the difference between these two overall resistances =: (1.26 x10-2 _ 9.46 x 10"3) = 3.14 x 10-3 kJ- I s 0c. Therefore, fouling contributes:

3.14x 1O-3kr1 soC==~=:T'-:'::=0.25L26x 10 2kri soC

of the total resistance before cleaning.

Answer. 0.25

8.6 Pre-beating of nutrient mediumCph =: Cpc =: Cp water =: 75.4 J gmol"l °Cl (Table 8,3, Appendix B) =75.4 kJkgmot l °Cl . Converting to massterms 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 (AppendixA), 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 kJs-l m-1 °Cl . From Table S.l, kfb water at 303 K= 0.62 Wm-1 °C-l;from Table AS-{A-ppendix A), this is the sameas 0.62 J s-l mol °C-1; therefore, ktb water =0.62 x 10-3 kJ 5-1 ar1 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 = 50W 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

°Cl. B =: 5 mID =: 0.005 m.

(alFrom the definition ofdensity (P-16), the mass flow rate of the medium is equal to the volumetric flow rate multipliedby 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 empiricalcorrelation, Eq. (8.42). The parameters in this equation are Re, Pr and Nu. Re is given by Eq. (8.38). The linearvelocity of the fluid u is equal to the volumetric fIowrate per tube divided bythecross.sectional area of the tube. Thecross-sectional area ofa cylindrical tube =: 1£,.z where ris the tube radius = 25 cm =: 0.025m. Therefore:

50 3 h-1 I~Iu =: volumetricflowrate =: m . 3600s =: 0.236ms-1

numberoftube,(n?) (30) n (0.025 ml'

Substituting parameter values into Eq. (8.38) with the tube diameter D =5 cm =0.05 m:

Re =Dup =0.05m(0.236m,-I)IOOOkgm-3 =1.18 x 104JIb 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"C1(10-3 kgm-I ,-I)Pr____ =78

ktb 0.54x1O-3kJs-1m-1oC I .

This value of Pris also within the range of validity ofEq. (S.42). From Eq. (8.42):


Nu = 0.023 ReO.SPrOA = 0.023 (1.18 X 104)°.8 (7.8)0,4 =94.6

From the definition ofNu in Eq. (8,37):

h_ Nukfb _ 94.6{O.54xlO-3 .kJs-1m-1oc-1) _ 102kJ -1 -2 "C-1- 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 theempirical correlation, Eq. (8.44). The parameters in this equation are C, Remax, Prand Nu. As the tubes of the heatexchanger are arranged in line, from p 183. C =0.26. Remax is given by Eq. (8.38) with D equal to the outer tubediameterdetennined 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):

Re =Dup = 0.06m{O.lSms-1)tOOOkgm-

3 =9.00 x 103max fib 10-3kg m-l s-1

As this value is > 6 x UP, Remax is within the range of validity of Eq. (8.44). Prior theshell·side water is given byEq. (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 ofNu in Eq. (8.37) with D the outside tube diameter:

Nukfb 115 (0.62 x 10-3 kJ s-l m-l OC-1) 1 2 1h =-- = = 1.19kJs- m- "C-

D O.06m

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);

~ =1.92kJ1sm2°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):

- I~ - ~. Q - 60~ 1048.3kJs- 30~.Lho - .Lbl - ..... - ~ - = ~

MhCph 3x 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 heatexchanger are represented graphically below.


j600C

Tei =- 10"C/ Teo =- 28<>C

/'Heat exchanger

Tho=-30"C

93

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 = (32- 20re = 25.5OCL In(LlT2I<lTd In (32120)

Answer: 25.5"C

(e)The heat-transfer area is determined from Eq. (8.19) with AT::::: /1TV

- IA :::::~ ::::: 1048.3k1s- ::::: 79m2

U LlTL 0.52kJ s-I m-2 oc-I (25.5OC)

Answer: 79 m2

(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 isthe 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-coilNi =- 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 k1kgmol·l °C-1. Converting to mass tenns using the molecular weight of water::::: 18.0 (Table B.l, Appendix B);

Cph =- Cpc =-75.4kJkgmorl °C-1 =- 75.4kJkgmor1"'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 (AppendixA), this is the same as 0.62 J s-l m- l oCl; therefore, kfb::::: 0.62 x 10-3 kJ sol m-1 °C1. From Eq. (2.24), 303 K =approx. 30°C, which is close enough to the conditions in the fennentation system for evaluation of thennalconductivity. p fennentation fluid::::: p water =- 103 kg m'3. # fennentation fluid =- # water. The viscosity of water at20"'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 valueapplies 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). Thedimensionless numbers in this equation are Rei, Pr and Nu. Rei is given by Eq. (8.39):

94

Pr is given by Eq. (8.40):

Solutions: ChapterS

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 ofNu in Eg. (8.37) with D::::: the tank diameter:= 3 m:

Nu k 7.7 X103 (0.62 x 10-3kJ s-1 m-I oc-l)h =~ = ,;", 1.59kJ 8-1m-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 inthe absence of fouling layers is .calculatedusing Eq. (8.23). Assuniingthat the beat-transfer coefficient for the coolingwater 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 beingremoved from the system, From the definition of density(p 16), the mass flow rate of-the- cooling water is equal tothe 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:

Q 230kJs-ITco = ---- + Tei = ( )- + 12"C = 21.9"C

Mc Cpc 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):

2 Tp-(Tci + T,o) (2 x 28)'C-(12 +21.9)"Ctli'A = 2 = 2 = 11.1"C

Substituting the results into Eq. (8.19) for evaluation ofA 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 13m2L=--= _ =55m

21tr 21t(0.0375m)


As the required length of cooling-coil is considerably longer than the 45 m available in the offered fermenter, thesecond·hand fermenter without modification is unsuitable for the proposed culture.

Answer: No

8.8 Optimum stirring speed for removal of heat from viscous brothJ.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 = 10kgm-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). Thedimensionless 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) 4Pr=-P-= =10

ktb 2 x 10 3 k1 C l m-l 0C-l

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 ofNu in Eq. (8.37) with D = the tank diameter = 2.3 m:

Nukfb 232NP·62(2XlO-3kJs-lm-loc-l) .11.62 -1 -20

-1h =-V = 203m =0.20Ni kJs m 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'C1(14m2jzO"C = 56.0AP,·62 kJ ,-1

The power dissipated from the stirrer, Ws' is equal to the power P calculated using Eq. (7.18), In this system, thevalue 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 m2 s·3;therefore, 1 kJ s~l = 103 kg m2 s·3 and:

Ws = 173NpN[ kgm2 s-3 .1 1 kJS-~ -31 = 0.173NpNf kJ s-llOOOkgm s

From the modified steady-state energy-balance equation, Eq. (8.33), assuming evaporation is negligible:


Therefore:

-Mrxn = Q- WS = (S6.0,vp·62_0.173NPNf)kJS-l

Values of Q, WS and -tJirxn calculated using the equations derived above are listed below as a function of Ni.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 accurateinterpolation of the power curve in Figure 7.24,

Nj (s~l) Rei Q(1<1 ,.1) Np Ws (kJ s-l) -Mirxn (kJ s·l)

0.5 30.4 36.4 4.5 0.10 36.30.8 48.6 48.8 4.1 0.36 48.4l.0 60.8 56.0 4.0 0.69 55.32.0 122 86.1 3.7 5.12 80.93.0 182 111 3.7 17.3 93.44.0 243 132 3.6 39.9 92.45.0 304 152 3.6 77.9 74.06.0 365 170 3.8 142 28.17.0 426 187 3.8 225 -388.0 486 203 4.0 354 -151

10.0 608 233 4.0 692 -459

The values of ii, Ws and -tilirxn are plotted as a function of Nj below.

105

00a Ws• -AHrxn

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 ofreaction for aerobic cultures is -460.kJ per groal oxygen consumed. Therefore, if the rate ofoxygen 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 97

Biomass = -93.4k:Js-1

=L22XIIYg

6x10-' gmolg-I h-I .I'6~sl (-460kJ gmol-I)

The cell concentration is equal to the biomass divided by the fermenter volume:

5II . L22xl0 g 22 I-I

Ce concentration = 3 11000 11 = 1 . g10m . --,-

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 theboundary layer in the fermentation broth is improved. Therefore, the rate at which heat can be removed from thesystem Qis increased. However, the rate at which heat is dissipated by the stirrer Ws also increases with stirrerspeed..When the curves for Qand Ws intersect, the entire heat-transfer capacity of the fennenter cooling system isbeing used just to remove the heat from stirring; the system in unable to remove any of the heat generated fromreaction. Accordingly, at high stirrer speeds, the system bas limited capacity to handle exothermic reactions.

Mass Transfer

9.1 Rate-controlling processes in fennentationConverting the units of the maximum specific ox.ygen uptake rate using the molecular weigbt of oxygen =32.0 (TableB.I, Appendix B):

= 5 1 -I b-I 1 I gmol 1132.0g II~I = 444 10-5 -1-1qo mmo g . lOOOmmol . 1gmol . 3600s ' x g g 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 10-3kg 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, thesystem will be limited by mass transfer.

Answer: Limited by mass transfer.

9.2 leLa required to maintain critical oxygen concentrationConverting the units of the oxygen uptake rate using the molecular weight of oxygen"" 32.0 (Table B.I. Appendix B):

= 80 11-1 h-I 1 1 gmol 1132.0g 11~IIIOOOIII~1 = 711 10-4 k -3-1qox mmo . lOOOmmol . 19mol . l000g' 1m3 ' 36008 . x gm s

Converting the units of the critical oxygen concentration:

C · = 0004mM= 0004 11-1 IlgmOI I 1-,2.0 g l I...'!LI 1100011-128 10-4 k -3ent' ,mmo, l000mmol ' 19mol . l000g' 1m3 - . x gm

(aJFrom 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·3. If thesolubility in medium is 10% lower;

C~ = 0.9 x (8.05 x 10-3) kg m-3 = 7.25x 10-3kgm-3

To maintain the oxygen concentration in the medium at the critical level, from Eq. (9.41):

(kLa). = qox = 7.11XlO-4 kgm-3 s-1 =O.10 s-1

ent (C~ -Cerit) (7.25X 1O-3kgm-3 _1.28X 10-4 kg m-3)

Answer: 0.10 s-1

(bJFrom Table 9.1, the solubility of oxygen in water at 30"C under 1 atln oxygen pressure is 3.84 x 10"2 kg m-3.Therefore:


From Eq. (9.41),

(kLa) . = ... qox = 7.llxlO-4 kgm-3 s-1 = 0.021s-1

ent (CAL -Cerlt) (0.0346kgm-3-1.28XI0-4kgm-3)

Answer: 0.021 s-l

99

9.3 Single-point !<La determination using tbe oxygen-balance metbod(a)The oxygen transfer rate for kLa determination by the oxygen~balance method is given as Eq. (9.48). From p 17, thepartial 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 thefermenter 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 thetemperatures to degrees Kelvin using Eq. (2.24):

N =_I [(FgPAG) _(FgPAG) 1A RVL TiT 0

1 [(200lmin-l·I~10.2Iatm) (1891min-I·I~10.20Iatm)1NA = 0.OOOO82057m3 attn K-1 gmol 1(200 I) (20+ 273.15) K - (28 +273.15)K

NA =0.0174gmolm-3 s·1

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):

N = _I [(FgPAG) _(FgPAG) 1A RVL TiT 0

_ 1 [(200lmin-l·I~10.2Iatm) (189Imin-I·I~10.183atm)1NA - 0.OOOO82057m3 atmK-1 gmorl (2001) (20+273.15)K - (28 +273.15)K

NA =0,0289 gmol m-3 s-1

Therefore, from Eq. (9.37) with NA converted to mass units:

0.0289 1 -3 -I 132.08 II...I.!>LINA groom s . 1 01'1000 _kLa =... = gIn g =0,25s 1

(CAL - CAL) (7.8 x 10-3 kg m-3 -0.52x 7.8 x 10-3 kg m-3)

The kr.,a value obtained in (b) using the incorrectly calibrated oxygen analyser is 60% of the actual leLa value; theerror is therefore 40%.


Answer: 40%. This calculation illustrates the sensitivity of the oxygen balance method to the accuracy ofall of themeasured parameters used in Eq. (9.48). This sensitivity arises from the subtraction of two similar numbers for themoles of oxygen in and out of the system. When similar errors in both Fg terms are taken into consideration. theerror 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. Thestatic pressure at the sparger is:

Ps = Pg 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:

_ 343 104 kg -1 -1 _ 343 104 k -1 -1 19.869Xlo-6atml_ 034Ps - . x m 8 - . x gm s . 1 I -. atm1kgm- s-

Ps is the pressure due to_the head of liquid above the sparger; oThctotal pressure at the'llparger iSh+ atmosphericpressure "" 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 ofthe 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(m3 mo1·l) Zi GiL or CJL (mol m·3)

Glucose 0.119 x 10.3 20g1-111mOllfoooll = 11l180.2g 1m3Sucrose 0.149 x 10.3 -I I1mol 111000118.5g1 . 342.3g· 1m3 =24.8

Ca2+ -0.303 x 10.3 2 1.3g1-1.11~~~~I'111~11 = 13.0

C032- 0.485 x 1<r3 2 l.3gl-I.ll~~~I'II~311 = 13.0

NR,+ -0.720 x 1<r3 1 2x13 r 1 I 1mol I 110001

1 = 197. g . 132.1 g' 1m3 .

S042- 0.453 x 10.3 2 -I lImo] 11 10001 11.3g1 . 132.1g· 1m3 = 9.8

Na+ -0.550 x 10-3 1 -I I 1mol 111000112xO.09g1 . 142.0g· 1m3 = 1.3

HP042- 0.485 x 10.3 2 0.09g1-1.11~;~~I'111~'1 = 0.63

K+ -0596 x 10-3 1 -I 11mol I110001

10,12g1 . 136.1g' 1 m3 = 0.88

H2P04- 1.037x 10-3 1 -I 11mol 1110001

10.12 g1 , 136.1 g' 1 m3 = 0.88

Substituting these values into Eq. (9.45) gives:


10g1O(C~) =O.5~HiZ?CiL +~KjCJLCAL t j

_ 05 [(-0.303 x 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 X10-') (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

• •CAL = 0.95CAIJ)

Solutes in the medium reduce the oxygen solubility by about 5%. From Table 9.2, the solubility of oxygen in water at35°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 totalpressure 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 ofoxygen at the top oroutlet end of the vessel is 7.14 x 10-3 kg m~3; the dissolved oxygen concentration at this location is 50% air saturationor 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 vesselis 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'" -CAL) = (C~-CAL)o-(C~-CAL)i = {7.14-3.57)XIO-3 kgm-3 -(9.57-6.22)xlO-3 kgm-3

AL L [(c:u.-CALlo 10[(7.14-3.57) X10 'kgm-'j

In (C~L -CALli {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 thevolumetric flow rates of gas into and out of the fermenter are measured under ambient pressure, Le. at I attn. From p17, 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 oxygenin the outlet gas is (0.2015 x 1 aim) = 0.2015 atm. Assuming that the exit gas leaves the fermenter at the fermentationtemperature (35°C), using R = 0.000082057 m3 attn K-l gmol-l from Table 2.5 and converting the temperatures todegrees Kelvin using Eq. (2.24):


_ I [(25 m3

min-1Y~lo.2latm) (4071S-1·1~10.2015atm)1

NA - 0.OOOO8Z057m3atmK 1gmor1(ZOm3) (25+273.15)K - (35 +273.I5)K

NA =- 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 themolecular weight of oxygen = 32.0 (fable B.t, Appendix B):

00167 1 -3 -1 13Z.0g Illkg INA . gIno m s - lOf . 1000 _

kLa =- '" =- gIll g =- 0.15 s 1(CAL -CALt 3.46 x 10 3kgm 3

Answer: 0.15 s-1

(g)The maximum cell concentration supported by the fennenter can be calculated using Eq. (9.40) with C~equalto theaverage solubility of oxygen between the top and bottom of the fermenter:

_kLa(c~) _ 0I5s-I[C·14;9.s7)XIO-3]kgm-31~1 _ -1

xmax

- qo -74 l_lh_lllgmOII132.0gll~II~I-19g1. mmo g . lOOOmmol . 1gmol . looog . 36008

Answer: 19 g I-I

9.5 Dynamic kLa measurementkLa is determined using Eq. (9.52) and the method described in Section 9.10.2. Take!} =- 10 s and CALI =- 43.5% airsaturation. The steady-state dissolved oxygen tension CAL = 73.5%. Calculated values of

In (.,~p.:::,,-=-;;~p.~"'1 )

and (12 - (1) are listed and plotted below.

0.00OAI0.801.62.33.04.1

o5

1020304060


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 ofMukhopadhyay and GhoseFrom 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 fermentationbroth is the same as that of water = 1000 kg m~3;

1 ppm = 1: (lOookgm-J) .11O~XJgl.1 ;~Oll = 3.125x 10-2 gmol m-J109 g.g

Values of CAL and fCYAO)i - CYAO)oJ for each stirrer speed are listed below.

Fermentation time Agitator speed

(h)300 rpm 500 rpm 700 rpm

CAL [(YAGJi - (YAG)o] CAL [(YAG); -(YAG)o) CAL [(YAG); - (YAG)o](gmol m·3) (gmol m-3) (gmol m-3)

0 0.184 0 0.184 0 0.184 04 0.175 0.001 0.178 0.0015 0.166 0.0016 0.163 0.002 0.169 0.0027 0.147 0.002 0.153 0.003 0.159 0.0038 0.128 0.003 0.138 0.004 0.147 0.0049 0.106 0.004 0.125 0.005 0.128 0.006

10 0.106 0.004 0.125 0.005 0.128 0.00611 0.109 0.003 0.131 0.004 0.131 0.005

The tabulated values of CAL and [(y AO)i - (yAO)oJ for 0-10 b are plotted below. As CAL increases at 11 b afterbeing constant at 9 and 10 b, the final values in the table for each stirrer speed most likely reflect the decline phase ofculture growth and are therefore not included in the lqp analysis.

104 Soluticns: Chapter 9

0.20.---,,.---,,.----,----,---,-----.• 300 rpmo 500 rpm• 700 rpm

fi 0.15

<o~

0.0060.0050.0040.0030.0020.001

0.10 L-_.....L__....J..__.l-_-''-_.....L__...J

0,000

(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 2.5and converting the temperature to degrees Kelvin using Eg. (224):

3 (31 ' -1 liminfi-J1TF. - atm nun . 66SV _ -2.02 gmol m-3 8-1

kLa = RTVLX~ope = O,OOOO82057m3atmgmor1r1(29+273.15)K{31)xs}ope - stope

For 300 rpm:

For SOOrpm:

At 700 rpm'

k-2.02gmolm-3s-1 1

La = = 0.21 s--9.68 gmol m-3

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 ofoxygen = 32.0 (Table B.t, Appendix B):

C~ = 0.l86gmolm-3 = 0.186 gmol m-3 .1 i~~ll·11~gl = 5.95 x 10-3 kgm-3

Answer. 5.95 x 1<r3 kg m~3

(c)Equating Eqs (9.39) and (9P6.2),


PTFgqo x = R T II. [(YAG)i - (YAGlo]

3atm(3Imin-lr~~[(YAG)i-(YAG)oj __qox = =2.02[(YAG)-(YAG) jgmolm 3, I

O'{)OOO82057m3atmgmolIrI(29+273.15)K(31) I 0

The maximum rate of oxygen uptake occurs when [(yAO:>i - (yAO)oJ is maximum. At 300 rpm. the maximum valueof [(YAo)i - (YAeVo] is 0.004; therefore, .(qO x)max = 2.02 x 0.004 = 8.1 x 10-3 gmol nr3 s-l. At 500 rpm, themaximum 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 700rpm, 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-3s-1.

Answer: 8.1 x 10-3 gmoJ m-3 s·l at 300 rpm; 1.0 x 10-2 gmol m-3 8-1 at 500 rpm; 1.2 x 10-2 grool m·3 s·l at 700 rpm

Unit Operations

10.1 Bacterial filtration(aJAccording to Eqs (10.12)-(10.14), a plot of flVf versus Vf should yield a straight,line for determination of thefiltration parameters. The data after converting the units to s and m3 are listed and plotted below.

Time, t (8) Filtrate volume, Vf(m3) tlVr(s rn*3)

120 0.0108 1.11 x 104180 0.0l2l 1.49 x 104360 0.0180 2.00 x 104600 0.0218 2.75 x 104900 0.0284 3.17x 104

1200 0.0320 3.75 x 104

4.0

1 •.!!. 3.0., •0-x....1:::.- 2.0

~~

~g

•~ 1.0~

Il:

0.00.00 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 mo3 . 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 106sm-6)= ----,c=c---':i-~=-=""-:-----.l:-:-----:-

4.00p.110 3kgm Is Ij(22g1-111°OO1j.I~~lcP 1m3 lOOOglJ

a = 8.0 x lOlO mkg-!

The intercept K2 is a relatively small negative number, which may practically be taken as zero. Therefore, from


Eq. (10.14), the filter medium resistance rm is zero.

Answer. a = 8.0 x 1010 m kg-I; rm is effectively zero.

(h)tlyr = 7.5 x 10-3 min 1~1. From(a), K2 =0; therefore, from Eq. (10.12):

K _ t/Vr = 7.5xl0-3

minl-1

1100011'160' 1 = 113 10' -61- Vf 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

1Omkg-

1(22 g l-

1ff;;;¥I·I~I)A = 2Kt6p = I I = 649m

4

2(1.13X 10' 'm-6) 360mmHg. 1.333Xi~:-1,-'

A = 25.5m2

107

10.2 FUlratlon ofmyceUaI suspensions(a)After converting units using the conversion factor 1m3 = 106 mI, the data for tlyfand Yr are listed and plotted below.

Pressure drop (mmHg)100 250 350

Vf(m3) pelleted IfYr(S m·3)

1.0 x 10.5 2.2 x 106 12 x 1()6 9.0 x lOS1.5 x 10.5 3.5 x 106 1.7 x 1()6 1.3 x 1()62.0 x 10.5 4.5 x 106 2.5 x 106 1.& x 1()62.5 x 10-5 5.8 x 106 3.0x 106 2.4 x 1()63.0x 1O~5 6.7 x 106 3.7 x 1()6 2.9 x 106

3.5 x 10.5 8.1 x 106 4.3 x 106 3.4 x 1()64.0x 10.5 9.2 x 106 4.& x 1()6 3.9 x 1064.5 x 10.5 l.0 x 107 5.3 x 1()6 4.3 x 1065.0 x H,.5 6.0 x 1()6 4.8 x 106

550

7.0 x lOS9.3x1OS1.4 x 1()61.7 x 1()62.1 x 1()62.4 x 1()62.& x 1()63.1 x 1()63.5 x 106

750

5.0 x lOS&.0 x lOS1.1 x 106

1.4 x 1()61.7xl()62.0 x 1()62.3 x 1()625 x 1()62.& x 1()6

Pressure drop (mmHg)100 250 350 550 750

Vf filamentous t/Vr(s m·3)

1.0 x 1(T5 3.6 x 1()6 2.2 x 1()6 1.7 x 1()6 1.3 x 106 1.1 x 1061.5 x 1(T5 55 x 1()6 3.1 x 1()6 2.7 x 1()6 2.1 x 106 1.7 x 1()62.0 x 10".5 72x106 4.3 x 1()6 3.6 x 1()6 2.7 x 106 2.3 x 1()62.5 x 1(T5 9.0 x 106 5.3 x 1()6 4.4 x 1()6 3.4 x 106 2.& x 1()63.0 x 1(T5 1J x 107 65 x I1J6 52 x I1J6 4.0 x 106 3.3 x I1J63.5 x 10-5 1.3 x 107 7.5 x 106 6.IXI1J6 4.7 x 106 4.0 x 1064.0 x 10"5 8.5 x 106 7.1 x I1J6 5.6x 106 45 x 1()64.5 x 10"5 9.6 x 106 7.& x 11J6 6.2 x 1()6 5.1 x 1065.0x lOwS &.& X11J6 6.& x 1()6 5.7 x 106


12Pressure drops: PaUeted

f 10 • 100mmHgo 250mmHg

$ .350mmHg~ 8 o 550mmHg6~ .. 750mmHgx

-1"'- 6

~~ •~g

~I 2

00 2 3 • 5 6

Filtrate volume, V; x 105 (m3)

,.Pressure drops: Filamentous

f 12 • 100mmHgo 250mmHg

$10 .350mmHg.. o 550mmHg0

x 8.. 750mmHg

_1:::.,-

0 6E~

~g •F i 2

00 1 2 3 • 6 6

Filtrate volume, V; 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 arelisted below.

Pressure drop (romHg)100 250 350 550 750

PelletedSlope::: Kl (s m·6) 2.26 x 1011 L20x lOll 9.93 x lOlD 7.07 x 1010 5.77 x 1010

FilamentousSlope:::: KI (8 rn·6) 3.73 x lOll 2.13 X 1011 1.75 x 1011 1.38 X lOll Ll4 x 1011

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:


o Pelfeted• Filamentous

2AZ .6pKI

2(1.8Cm2 .11C:O~mI2)2 /;p.1 L333\1~:-1 s-ZjKIa = -~,.....:. = -"-,--0'='-.::;,"-\,,-'---+'7',",--'--, = 2.47 x 1O-5 /;pKI mkg-I

Jl{c· 1.4cP .1 10-3k;;-l s-

II(0.25 gml-l. 1~~~1'11::~1

where tip has units ofmmHg andKl has units of s m-6. Similarly for the filamentous cells:

2A2 K 2(1.8cmZ.I~P)2 /;p·11.333xlOZkgm-ls-ZjKl_ /;p 1_ loocml ImmHg -617 1~5A.K k-l

a - - I 3 1 lit I 6 I )- . x v ~ Im gJl{c 1.4cP. 10- kf; s- 0.lgml-1. l~mU: .ll::gl

The results for specific cake resistance obtained after substituting the values for tJ.p and K1 into the equations arelisted below.

Pressure drop (mmHg)100 250 350 550 750

Pelleteda(mkg~l) 558 x 108 7.41 x loB 8.58 x 108 9.60 x 108 1.07 x 109

Filamentousa(mkg-I) 2.30 x 109 3.29 x 10' 3.78 x 109 4.68 xla' 5.28 X109

(b)From Eq. (10.2), the compressibility can be obtained by plotting the specific cake resistance a versus lip on log-logcoordinates. Using the conversion factor 1 mmHg:::: 1.333 x 102 kg m-1 s·Z (Table A.5, Appendix A) to convert theunits of lip, the data are listed and plotted below.

Pressure drop (kg m- l s-Z)1.33 x 1()4 3.33 x 1()4 4.67 x 1()4 7.33 x 1()4 1.00 x lOS

Pelletedarm kg-I) 5.58 x loB 7.41 x 108 8.58 X 108 9.60 x 108 l.07 X 109Filamentousa(mkg-I) 2.30x109 3.29 X 109 3.78 X 109 4.68 x 10' 5.28 x 109

_ 100 .- ~--~-_~-~_~...,

,,-!ifS6-x"

Pressure drop, Apx 10'4 (kg m·1 s·2)10


10

6

m] 'I~I)1m3 Iooog

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 hasunits m kg-I, and a' bas units kg-(l +s) m 1+3 sZs. Therefore. the compressibility is 0.323 for the pelleted suspensionand 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 mediumresistance rm is negligible:

Solving for (61')$-1:

2(Ap)J'-l :: _I~

v3 I'f a' cf

Substituting the parameter values for the filamentous suspension using consistent units, including the results ex':: 4.42x 107 kg..(l+s) m 1+s s2f and s= 0.415 from (b):

Ih 13600'1. I h(Ap)0.415-1 :: 2 x

(20m3)

IAcP .1 10-3 klgc~-l ,-114.42X 107kg-{I+O.415)m l+O.415 ,2x 0.415 ( 0.1 g ml-1 .

{1ip)-O.585 :: 6.54 x 10-4 kg-o.585 mO.585 s1.170

Ap = 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-18-2 =2.78x 105 kg m-l 8-2 .1 I mmHg I= 2086mmHg1.333 x 102 kg m I s-2

Answer: 2086 mmHg. An assumption associated with this answer is that the filter cake characteristics measured atpressures 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 =0and a= c£ !¥Js:

t=

Vf

Solving for fJf a' c:

t 2A2J1{a'c =----Vi t>p>-1

Substituting the parameter values for the laboratory filter and using the conversion factor I psi =6.895 x 103 kg mols·2 (fable A.5, Appendix A):


( )

2

. 60s 21 1m 12

23.5mm. -1-' 2 Scm. lOOcmlI..,tXc "" 1 nun I "" 367x 105 kg0.43 m-2.43 s0.l4

r, ( )2 ( ~0.57-1 .Sooml.~ 12 si.16.89sxI03k~m-l e21

106 m1 P 1pst

Answer: 3.67 x 105 kg0.43 m-2.43 sO. 14

(h)

The cycle time is lIN hours per revolution.

111

(elAs 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.

(elEq. (10.11) with rm "" 0 and a",,« !J.ps is:

lit a' c ;ips-I2 V,

2A

Applying this equation to a single revolution of the filter, t "" O.3IN h and Vf"" 20lN m3. Substituting the filter cakeparameter values from (a), evaluating the filter area A "" 2 1t r Wwhere r is the drum radius"" 0.75 m and W is thedrumwidth "" 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 ( 168 5 3kg -1 _2~0.57-10.3 h 3600s 367 105kg0.43 -2.430.14 45 .. 9 xlO m sN'lh .x m s . pSt. Ii (20)

-'-'-~';--c:.=. - ps m320 m3 - 2(2n(0:75m)(1.2ml)' NN

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 theequation in (e):

O.5 h 13600 '1N . Ih _ 1343.9 -320 - -,;;-sm_m3N

112

Answer; 14.9 revolutions per hour

10.4 Centrifugation of yeast(a)Convert the parameter values to units of kg, m, s:

N = 14.9 rph

Solutio1lS: Chapter 10

Pp=: 1.06gcm-3 = L06gcm-3 'lli.:gl.ll~:r = l060kgm-3

Pr =: 0.997 gene3

:= O.997gcm-3.Ili.:g1·ll~~m 1

3=997kgm-

3

Q = SOOlh-1 = 500lh-l'!l~JbJ~sl = l.39xlO-4 m3s-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. Dp = 5 MlD= 5 x10-6 m. From Eq. (10.15) with g =: 9.8 m s-2 (p 16), the sedimentation velocity is:

u = Pp -PfD2g = (l060-997)kgm-3

(SXI0-6m)29.8ms-Z_=6.31XIO-7ms-lg 18p P 18 (1.36 x 10 3 kgm 1s-l)

Substituting this result into Eq. (10.18):

I: = JL =: 1.39x W ..4 m3

s-1 = 110m2

lUg 2 (6.31 x 10 7ms-l)

Answer: 110 m2

(b)From the relationship between specific gravity and density (p 16), as the density of water is close to 1.0000 g em-3 (p16),

20 -3 20 -31 1kg 1 1100 em 13

2000k -3Pp = . gem = . gem . l000g . 1m =: gm

Dp =0.1 nun = 0.1 x 10-3 m. From Eq. (10.15), the sedimentation velocity for the quartz particles is:

_ Pp-Pf D2 _ (2000-997) kg m-3

(01 10-3 )2 98 -2 _ 402 10-3 -1ug _ -- g _ ( ) . x m. ms - . x ms18,u p 18 1.36x 10-3 kgm 1 s:-1

From Eq. (10.18), the sigma factor is:

r = -lL = L39x 1O-4m3 s-1 = 0.017m2

lUg Z(4.02xlo-3ms-l)

From the result in (a), r for the yeast cells is 110/0.017 =6470 times that for the quartz particles.

Answer: By a factor of 6470.

10.5 Centrifugation offood particlesConvert the parameter values to units of kg, m, s:


Pr = 1.00gcm-3 = 1.00gcm-

3 ·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. Dp = 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) = IO,OOOrpm = 2ft (lO,OOO) min-I .! 1::-1 = 1.047 x 103 s-l

From Eq. (10.22) withg =9.8 m s~2 (p 16), the sigma factor for the tubular~bowl centrifuge is:

" ( 3 _1)2 21: = 21t~J:!: = 21t 1.047x1O s 0.70m(0.115m) = 6.506 x 103m2g 9.8ms 2

From Eq. (10.15),

_ Pp -P'D2 _ (l030-1000)1Qlm-3 (10-5 )2 98 -2 _ 1307 10-6 -1ug _ -- g _ ( ) m. ms -. x ms18,u p 18 1.25 x Io-3 kgm-1 s-1

From Eq. (10.18),

Answer. 0.017 m3 s~l

10.6 Scale-up of disc-stack ceutrifugeThe sigma factor for the pilot~scale disc~stack bowl centrifuge 1:1 is calculated using Eq. (10.20) withg =9.8 m s·2 (p16), 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 = om m:

( 11 . ~221t 21tx3000min-l .~ (25-1)

:!: = 2noJ-(N-ll(,3_,1j = 60s (0.05m)3_(0.0Im)3) = 89.6m21 3gIanO 2 1 3(9.8ms 2)clan35')

From Eq. (10.19), the sigma factor·:rz for the bigger centrifuge is therefore:

:!: = Q2:!:I = 801min-I

(89.6m2) = 200m22 Q1 3.5 1OOn-1

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:!:, = 3 (9.8 m s-2)clan 45') 2048m2 = 4 34x 10' s-2

2 n(N - IlH - ,1) 2 n (55 -1) (0.075 m)3 - (0.0235m)3) .

_ 659 ad -1 II revolutionII 60 s 1- 6290(J)- r s . 21trad . lmin - rpm

Answer. 6290 rpm

10.7 CeutrifUgatiou of yeast and ceO debrisCombining Eqs (10.15) and (10.18),

and


where subscript 1 refers to the centrifugation conditions for the yeast suspension, and subscript 2 refers to thecentrifugation conditions for the cell debris. As the same centrifuge operated at the same speed is used in bothapplications, ,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- =-2-Dpl Dp2

or

Answer. 0.0671 min-I

10.8 Cen disruptionThe relationship between pressure and protein release for a Manton-Gaulin homogeniser is given by Eq. (10.23). Atconstant pressure, a semi-lagplat ofRm/(RnrR) versusN should yield a straight line, so thatthe value of k pa can bedetermined from the slope.

The data for % protein release represent values of RIRm x 100. These data can be converted to Rm/(Rm_R) as

follows:

Rm 100Rm-R = loo-RIR x 100 =

m

The results are listed and plotted below.

100100-% protein release

Pressure drop (kgjcm-2)200 300 400

NRm

Rm-R

1 1.05 1.16 1.302 1.10 1.31 1.673 1.16 1.50 2.114 1.22 1.75 2.995 1.28 1.90 3.706 1.35 2.22 4.88

500

1.562.414.005.718.70

11.5

550

1.722.946.138.70

18.2

Solutions: Chapter 10 liS

30...--,---,----,---,--,--....---,• 200 kgf an-2

o 300kgf an"2• 400 kgf an"2o 500 kg,an'2

.. 550 kg,an'2

762 3 4 5Number of passes, N

1

10

The values ofk pa obtained from the slopes of the straight lines for each pressure are listed below.

200 300 500 550

0.050 0.129 0.267 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 theslope.

'l,. 0.1

'"

0.01 L- ---.__~_~~_~~~_'

100 1000Pressure, 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.


ln5.0 = 4.61.71 x 10-7 (460)2.37

(a)For 80% protein release, Rm/(Rm_R) "" 5.0. Rearranging the above empirical equation for the homogeniser andsubstituting p "" 460 kgfcnc2 gives:

In(R=~R)N= =

1.71 x 10-7p2.37

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 = -"-'''--,f- ln3.33 "" 3.S2x 106

L71xlO-7N-1.71xlO-7(2)

p := 578 kgjcm·2

Answer: 578 kgjcm·2.An assumption associated with this answer is that the homogenisation characteristicsmeasured at pressures between 200 and 550 kgfcm·2 apply at the higherpressure of578 klJfcm·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:

Vu0.8 = v.

IVu + 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. Substitutingthese 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


As CAu =3.5 CAl> CAn = 3.34 x 103 u I~I. From Eq. (10.29), the concentration factor for product partitioning intothe upper phase is:

Answer: 1.04

10.10 Recovery of viral particlesFrom Eq. (10.24), for K,= 10..2•CAn =m·2 CAl and the product partitions into the lower phase. If51 culture volumeis added to 2 I polymer solution and, after phase separation, the volume of the lower phase is 1 1, the volume of theupper 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 =11, Vu =61 and K = 10-2:

Vr 11Y, = 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:

CAuCAuVu+CAIVi CAuVu+KVI

CAO = Vo =

The concentration factor for product partitioning into the lower phase is given by Eq. (10.29). Substituting for CAland 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


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 theshorter 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 V0= 23 ml, the partition coefficients for the toxoid and impurity can be detennined using the measured elutionvolumes and Eq. (10,41);

T idK = (V,- Vol = (29-23)m! =°171axo p V; 35ml .,

I °tyK = (V,-Vo) = (45-23)m! =0629mpun p 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 columnafinner diameter Del =15 em =0.015 m and height HI::::' 0.4 m is:

(DCI)2 (OoOISm)2 -5 3VTl ::::. 1t 2 HI::::' 1& 2 OAm::::. 7.07 x 10 m

The total volume oitbe large~scale column of diameter Dc2::::' 0.5 m and heightHZ =- 0.6 m is:

(D02)2 (osm)2 3VTI =- 1t T HZ =- 1t~ O.6m =- 0.118m

If the void fraction in the large column is the same as in the small column, Vo2 in the large column is:

323m!.~

Vol 106

m! (3) 3V02 =r VTI = 5 3 0.118m = 0.0384m

TI 7.07 x 10 m

If the pore volume fraction is also the same:

35m!. 1:3

!'it 10 m! (3) 3ViZ =r VTI = 5 3 0.118m =0.0584m

T1 7.07 x 10- m

If the large·scale column is operated with the same packing and flow conditions, the partition coefficients can beassumed to be the same-as those in the laboratory column. Therefore, from Eq. (10.41), for toxoid in the large columnwith the value of Kp from (b):

Ve2 = Kp Va + Vo2 = 0.171 (0.0584 m3) + 0.0384 m3 = 0.0484 m3

Similarly for the impurity:

Ve2 = Kp Vi2 + Vo2 = 0.629 (0.0584 m3) + 0.0384 m3 = 0.0751 m3

Answer: Toxoid =0.0484 m 3; impurity =0.0751 m3


(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:

(D02)2n- (D)2 131 22 c2 . -1 1m 0.5 m 3 ,-1

v2 "" VI 2 "" VI -D "" 14mlmm . -6-' (0015 ) "" 0.0156m IDlDn(D;I) cl 10 ml . m ,

Answer: 0.0156 m3 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 largecolumn, using the results from (c) and (d):

Ve2 0.0484 m3IR=-= = 3.1 min

v2 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(m s-l)

HETP,H(m)

0,1 x 104

0.2 x 104

0.3 x 10-40.5 x 10-40,8 x 10-41.0 x 10-41.2 x 10-41.5 x 1041.8 x 10.4

2.0 x 10.4

2.72 x 10.4

1.87 x 10-41.69 x 10-41.72 x 10-42.02 x 10-42.27 x 10-42.54 x 10-42,95 x 10-43.38 x 10.4

3.67 x 10-4

4,----,--...,---,---,----,

2.50.5 1.0 1.5 2.0

Unear rrquid velcoity, u x 1()4 (m $"1)

1 '--_---'__-'-__-'-__.L._--'0.0


(h)From the graph in (a), the minimum HETP is around 1.7 X H;4 m. This can be confirmed by differentiating theequation for H and solving for dH/du = 0:

dB --A- = -+8 = 0du .2

Therefore:

The value ofH 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= 1.67xl0-4m

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 isobtained by dividing the volumetric flow rate by the column cross-sectional area:

0311 . -I II mini 1'm3

Iv . mm. 60s . 10001 _u = = = L05x 1O-4 ms 1

K(~'r K(02gmrSubstituting 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= kB = 1.05 = 1.235kA 0.85

Substituting these parameter values into Eq. (10.51) with L = 1 m for the larger column and k2 = k]J:

R=! fT(O-I)(~)=\/ 1m (1.235-1)( 1.05 )=159N, 4Vli 0 kB+l 4 2.34xI0-4 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 ofH corresponding to RN= 1.5 or virtual complete separation can bedetennined using Eq. 00.51):

.fIl =! .fL(O-I)(~)=! .fiifID(1.235-1)( 1.05 ) = 0.0136Jiil4RN 8 kB+I 41.5 1.235 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 Has a function of u and solving for u:


The solution to this quadratic equation is:

-{C-H) ±\!<C_H)' -4B Au = 2B

H-C±,!<C-H)2_4BA= 2B

Substituting the parameter values:

u = (1.85X 104 -5.7 x 1O-5)m±V(5.7 x 10-5m-1.85x 104 m)' -4 (1.5 s)(2X 10-9 m2 s-l)2 (1.5 s)

u =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 = 6.48x Io-S m s·I,H < 1.85 x 104 m SO tbatRN> 1.5 and-eomplete:separationismaintained. The volumetric flow rate v isequal to u multiplied by the column cross--sectional area:

v =U1t(~cr ={6.48XlO-Sms-1}1t(o.2gmt = 3.18xW"·Q m3 s-1

Converting units:

_ 3 18 10-6 3 -I 110001

1 1 60s 1= 0191 ·-1v - . x m s . 1m3 ' lmin . nun

Answer: 0.191 min-I

Homogeneous Reactions

11.1 Reaction eqnilibriumFrom p 258. take the standard conditions to be 25°C and 1 attn pressure. The temperature is converted to degreesKelvin 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):

-t1GO 1inK =~ = 14.1klmol- = 5.69

RT 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 1In K =~ = -32kJ mor = -1.29

RT 8.3144xlO 3 kJK Igmoll(25+273.15K)

K =0.275

The relatively small magnitude ofK 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:

CGtiP at equilibrium 0,038 MK = = = 19

C01patequilibrium O.OO2M

Answer: 19

(b)From Eq. (11.9) and the reaction stoichiometry:

.. molesG6Pfonned 1mol -1Theorettcal yIeld = 1 GlP ed £ G6P = -1-1 = I mol molmo es us to orm mo

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-1Y moles GIP supplied 0.04 mol

Answer: 0.95 mol mol~l


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,

123

(b)Answer. The volumetric productivity is doubled, the specific productivity is unaffected. and the total productivity isdoubled.

(e)(I)RA =100 kg d~l; rA =0.8 g I-I h-I, From Eq. (1 1.16):

V =RA = lookgd-I =52081

rA 08gl-lb-ll~:III~gl

Answer: 5208 I

(ij)FromEq, (11.67):

11.4 Enzyme kineticsEvaluate the enzyme kinetic parameters by plotting slv versus s as a Langmuir plot(p 272), The values are listed andplotted below,

s(moll-1)

0.02500.02270.01840.01350.01250.007300.004600.00204

Slv (min)

12.8911.889.957507.025.003.932.62


15,----,-----,.----...,

10

!-I>

5

0.030.020.Q10'------.......------'-------'0.00

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. Km = 3.83 x 10-3 moll- t ,

Answer: Vmax =2.25 X 10-3 mol 1-1 min'I; Km = 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 ofglucose production. According to the Arrhenius equation, a plot of k versus liT on semi-logarithmic coordinatesshould give a straight line.. 'The parameter values are listed and plotted below; Tis converted to degrees Kelvin usingEq. (2.24).

T('C) T(K) liT (K'I) Rate, k (mmel row3 s' I)

20 293.15 3,41 x 10.3 0.3130 303.15 3.30 x 10-3 0.6640 313.15 3.19 x Icy3 1.2060 333.15 3.00 x Icy3 6.33

Solutions: Chapter II

- 10

"1."E§.

"g •'g 1

I~~'5

~ 0.12.9 3.0 3.1 3.2 3.3 3.4 3.5

1/Temperature X 103 (Ko 1)

125

The equation for the straight line in the plot is k =1.87 x lOlD c 73OO/T, where k has units ofmmol m-3 5"1 and T hasunits 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 kJKw 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 =328.15 K. Substituting this value into theequation for k obtained in (a):

k = 1.87 x 1010 e-73001T = 1.87 x 1010 e-73001328.15 =4.08 romol m-3 s·1

Similarly, for T= 25"C =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 = 1.65hh 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) =6.87x lO-4 h-1

and the enzyme half-life is:

th - In2 _ In2 = l009h- kd - 6.87x lO-4 h-l

126 Solutions: Chapter II

Although the reaction rate is 9.5 times faster at 55Q C than at 25°C, the rate of deactivation is 0.42/(6.87 x 104) =: 611times greater. Therefore, unless there are other considerations. 25°C would probably be the more practicaltemperature for processing operations.

Answer: The enzyme half·life at 25°C is 1009 h or 611 times longer than the half~life of 1.65 h at 55°C. The morepractical operating temperature is probably 2jl'C.

n.6 Enzyme reaction and deactivationKm = 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 x45 gmol m-3 = 9,0 groat m-3: As this concentration range is well above the value of .&tn. s» Km 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+ Kkd

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:

-vmaxO -k t VmaxO-" =---e ' +---'0kd kd

vmoxO ( -< I)S = $0--- 1-e d

kd

At the beginning of the reaction, SO = 45 gmol m-3. Converting the units of VrnaxO:

007 11-1 -1 007 11-1 -1 1 1 gmol 11 10001 11 60 '1 42 1 -3 ·-1vmaxO = . nuno s = . mmo s. lOOOnunol' 1 m3 . min = . gmo m nun

From Eq. (11.45):

k _ln2 _ In2 -0087 ·-1d-----.--. mmth 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:

12.276 =48.276 e-O·087 t

e-o·087 t =0.254

-0.087 t =-1.369


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 andcalculations 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 0.030.33 0.21 om 0.17 0.060.5 0.22 0.10 0.25 0.400.75 0.32 0.15 0.25 0.601.0 0.47 0.53 0.50 1.061.5 1.00 1.10 0.50 2.202.0 2.10 2.32 0.50 4.642.5 4.42 2.48 0.30 8.272.8 6.9 2.50 0.20 12.503.0 9.4 1.50 0.10 15.003.1 10.9 0.70 0.10 7.003.2 11.6 0.10 0.30 0.333.5 11.7 -0.10 0.20 -0.503.7 11.6

The values of luIfit are plotted as a function of time below according to the method described on p 263.

4.03.53.01.5 2.0 2.5

Time (h)

1.00.5

, , , , , ,

-rf

if -

/1/

/

.--r-I:".

, , , , , , :l-20.0

o

2

4

6

8

12

14

10

16


Results for <ix/dt read from the average rate-equal area curve and the calculated specific growth rates are listed below.

Time, t(h)

0.00.330.50.751.01.52.02.52.83.03.1323.53.7

o0.050.200.450.701.53.36.5

10.414.511.03.5

-0.3-0.8

o0240.911.411.491.501.571.471.511.541.010.30

-0.03-0.07

Values of the-specific growth rate,u are plotted as a function of time below.

2.0

1.5-i:...~ 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 theculture period. Taking the average of the values of J1. between times 0,75 hand 3 h• .Ltmax = 1.50 ± 0.05 h- l , where0.05 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 andsubstrate consumption rates. The average rate-equal area construction is used to determine the substrate consumptionrates rs = -ds'dt from the concentration data, as shown below.


Time, t(h) ,(kg m-3) LI.Y (kg m-3) Lv (h)

0.0 25.0 -0.2 0.330.33 24.8 0.0 0.170.5 24.8 -0.2 0.250.75 24.6 -0.3 0.251.0 24.3 -1.0 0.501.5 23.3 -2.6 0.502.0 20.7 -5.0 0.502.5 15.7 -5.5 0.302.8 10.2 -5.0 0.203.0 5.2 -3.55 0.103.1 1.65 -1.45 0.103.2 0.2 -0.2 0.303.5 0.0 0.0 0.203.7 0.0

129

0.61

0.0

0.80

1.20

2.005.20

10.0

18.3

25.0

35.5

14.5

0.67

0.0

The values of-As/tit are plotted as a function of time below according to the method described on p 263.

4.03.53.01.5 2.0 2.5

Time (h)

1.00.5

,

·

l- ·

JI-

rf·

l- I·

1//

Y• .t>..o

0.0

5

30

40

10

35

_ 251,

1 20~

11~ 15

Results for -<is/dt read from the average rate-equal area curve are listed below, together with the instantaneousbiomass yield coefficients calculated using Eq. (11.49) and the values of rx from (a).

130

Time. t(h)

0.00.330.50.751.01.52.02.52.83.03.13.23.53.7

0.'0.50.61.01.63.07.'

13.922.233.027.5

7.00.4o


o0.100.330.450.440.500.450.470.47

0."0.400.50

-{J.75

,Values of the observed biomass yield from substrate Yxs are plotted as a function oftime below.

• 0 D.•"x~ D.•;;~,0 0.4 -Ee--~

'll~ 0.3.'" '"0"

~E 0.2~

~ 0.1 J.il0 0.0

0

,1

,2

Time (h)

,3

-

•,

During the exponential growth phase between 0.75 b and 3 h when /.l = ,umax, Yxs is approximately constant with anaverage value of0.46 ± 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 andplotted 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) Biomass concentration, x (g 1-1) e [(X),+e- (x),-e1 (g 1,1)

0 0.64 55 1.95 5 3.0

10 4.21 5 3.615 5.5. 5 3.8


20 6.98 5 3.725 9.50 5 3.430 10.3 5 2.635 12.0 5 1.940 12.7 5 1.245 13.1 5 0.850 13.5 5 0.655 13.7 5

16

1.

f 12

'"" 10~

~

~

j 8

600

~ •~

2

00

•

10 20 30Time (d)

40 50 60

The growth rate dxldt is determined using the central-difference formula. Eq. (11.23). These results and values of thespecific growth rate J.I. are listed and plotted below as a function of time.

Time, tId)

o5

10152025303540455055

0.300.360.380.370.340.260.190.12O.os0.06

0.1540.0860.0690.0530.0360.0250.0160.0090.0060.004

Values of the specific growth rate J.I. are plotted below,

132

-~ 0.15:!?-"-{~ 0,10,e'"";e"l'l 0.05

'"

0.00 L~,-~,-~,-::::b!:=L----Jo 10 ro M ~ W 00

Time (d)

-Answer: Near the beginning of the culture.


(blThe mid~point slope method is used to determine the rate of substrate uptake as a function of time, The sugarconcentration data are plotted below according tathe method described onpp 264-265.

35

30

- 25~

.$c0 20~

~~c 158 ," Iil,,

I

'" 10 <III II I

5 I II II I

00 10 20 30 40 50 60

Time (d)

Values of[(s)t+e- (s}t-el read from the graph are listed in the table below.


Time, t (d)

o5

10152025303540455055

Sugar concentration, s (g 1-1)

30.027.423.621.018.414.813.39.78.06.85.75.1

e

555555555555

133

-6.2-6.1-5.7-5.4-5.2-4.6-4.4-3.9-2.7-1.9

The rate of substrate uptake -ds1dt is determined using the central-difference formula, Eq. (l1.23). These results andvalues of the specific rate of substrate uptake qs are listed and plotted as a function of time below.

Time, t (d)

o5

10152025303540455055

0.620.610.570.540.520.460.440.390.270.19

0.3180.1450.1030.0770.0550.0450.0370.0310.0210.014

0.35-~ 0.308'i 0.25!"• 0.20l\!Jl

0.15"07;2 0.10e

t 0.05

'" 0.000 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 rs from (a) and(b). The results are listed and plotted below.


Time, t(h)

o5

10152025303540455055

0.480.590.670.690.650.570.430.310.300.32

_ 0 0.8"x1!i 0.7ejg 0.6~E 0.5,g3iZ~ 0.4·'".S- '"m- 0.3•E.Q

'" 0.2~

~ 0.1•<3 0.0

0 10 20 30 40 50 60Time (d)

Answer: Yxs 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(alFrom Table B.8 (Appendix B), the molecular formulae for glucose and ethanol are C6H1206 and C2H{jO,respectively. The reaction equation for fermentation ofglucose 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:

2 gmolethanol ,I ~.l~ll _Maximumtheoreticalyield = ;gmo:e~anol = gm =0.51gg 1

gmogucose 11802g11 gmol glucose. 1~ol

Answer: 0.51 g g-1

(h)In the absence ofgrowth. fJ. in Eq. (11.83) is zero and the equation reduces to:


, mpYpS =

ms

Therefore, for Yr.s equal to the maximum theoretical yield:

135

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 theabove 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

(cJDuring batch culture at usual glucose concentrations, J1. = J1.rnax for most of the culture period (p 279). Eq. (11.83) cantherefore be written as:

, _y,,;PXc:I'm~,,,,,-+_m-,,pYps = Jlmax

--+msYxs

Substituting the parameter values for S. cerevisiae, including the result for mp from (b):

Yo, _ 3.9kgkg-I (0.4h-I)+0.092h-I _ 043k k-Ips- 1 -.gg

0.4 h + 0.18kg kg-I h-IO.llkgkg 1

Similarly for Z 11Wbilis:

i - 7.7 kg kg-I

(O.3h-I) + 1.12h-I _ 048k kg-Ips- 1 -. g

O.3h- + 2.2 kg kg-I h-IO.06kgkg 1

Answer: 0.43 kg k:g"l for S. cerevisiae; 0.48 kg k:g"l for Z mobilis

(djUsing the results from (8) and (c), for S. cerevisiae:

,

Efficiency = Yps = 0.43 kg kg-I =084maximum theoretical yield 0.51 g g-1 .

For Z mobilis:,

Efficiency = Yps = 0.48kgkg-l =094maximum theoretical yield 0.51 g g-1 .

Answer: 0.84 for S. cerevisiae; 0.94 for Z mobilis

(ejFrom Eq. (11.70) withf.l = J.!max, for S. cerevisiae:

For Z. mobilis:

Answer: 1.65 h-1forS. cerevisiae; 3.43 h- l forZ mobilis


(f)In Eq (11.70), the growth~associatedterm is Ypx ,umax, the nonwgrowthwassociated term is mp_ Therefore, using theresult from (e), the proportion ofethanol production from growth for S, cerevisiae is:

Ypxl'max _ 3.9 kg kg-I (0.4h-l)

Ypxl'max+mp - 3.9kgkg(OAh-1)+0.092h 1 = 0.94

so that the proportion from non~growthmetabolism is 0.06. For Z mobilis:

Ypxl'max = 7.7 kg kg-I (0.3 b-I) =0.67

Ypxl'max+mp 7.7kgkg-I(0.3h-I)+L12h 1

and the proportion from non~growth metabolism is 0.33. Non·growth~associated ethanol production is moresubstantial for Z. mobilis.

Answer: For S. cerevisiae, 0.94 of the ethanol production is growth~~ociated and 0.06 is non-growth·associated. ForZ mobilis, 0.67 is growth~associatedand 0.33 is non-growth~associated. Z. mobilis produces a more substantialproportion of its ethanol in non~growth·associated metabolism.

(g)Fromp261;volumetric productivity is equal to specific productivity multiplied byceU concentration. From (e), asthe specific ethanol productivity qp for Z mobilis is 3.43/1.65 = 2.1 times that of S. cerevisiae, to achieve the samevolumetric 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 yeastculture must be 12.2 times that for the bacteria.

Answer: 12.2 times the volume for the bacterial culture

(0From Eq. (11.81) with,u= #max, for S. cerevisiae:

-J- = _1_ + _m_,_ = 1 +0.18 kg kg-l

h-l = 9.54 kg kg-l

Yxs YXS Pmax 0.11 kg kg-l O.4h-l

, -IYxs = 0.105 kg kg

For Z mobilis:

~=_l_+_m_,_= I + 2.2 kg kg-1h-

l =24.0 kg kg-lYxs Yxs J4nax 0.06 kg kg-l O.3h 1

, -IYx, =0.042 kg kg

Answer: 0.105 kg kg.l for S. cerevisiae; 0.042 kg kg- l for2. mobilis. As Z mobilis produces less biomass per massof substrate consumed and per mass of ethanol produced than S, cerevisiae. biomass disposal is less of a problem withthe bacteria.

OJThe ethanol yield from substrate is 12% higher using Z mobilis than S. cerevisiae. the specific productivity is 2.1times higher so that a smaller and cheaper fermentation vessel is required to achieve the same rate of ethanolproduction. and Z mobilis produces less than half the amount of biomass generated by S. cerevisiae per mass ofglucose consumed. All of these factors mean that Z. mobilis perfonns better than S.· cerevisiae for ethanol production.


However, other aspects of the cultures and the ethanol production industry also need to be considered. Z mobilisrequires 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 highsalt content. The biomass produced in ethanol fermentation by yeast is often sold for use in animal feeds, whereasapplication of bacteria for this purpose is not as well accepted in the industry. These are some of the reasons why Zmobilis has not been widely adopted for industrial ethanol production, despite its superior ethanol productioncharacteristics.

11.10 Plasmid loss during culture maintenanceFromEq. (11.65),

a = f.r = 0.033h-1 = 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):

_ ,,", _ (0.025h-I)28d·12::I_

n-ln2

- ln2 _24.2

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. l_a_2n(a+p-lJ p 1_1.32_(224.2(1.32+p Il)p 0.32+ (224.2(0.32+Pl)p 0.32+214.38p(224.2")

Multiplying through by the denominator:

0.211 + 141.49p(224.2") =0.32+p

141.49p(224.2p)-0.109_p = 0

1bis equation can be solved for p by trial and error.listed below, starting withp =0.001 as the first guess.

p

0.0010.0020.00070.000760.000770.000766

Values of the left-band-side of the equation for various pare

141.49 p (224.2p) -O.I09-p

0.03390.1816

-0.0095-0.0008

0.00060.000017

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 Table2.5, R =8.3144 J K-l grool-1 =8.3144 x 10-3 kJ K-l gmol-1, Converting the temperatures to degrees Kelvin usingEq. (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:

kd = A e--EalR T = 10362 s-l e-283 kJ gmor1/(8.3144x 10-3kJK""1 gmo1-1x 353.15K) = 2.20x 10--6 s-1


At 121°C:

kd

=. A e-ErJ/RT = 1036.2 s-1e-283 kJ gmorl/(8.3144x 10-3 kJ~l gmor1x 394.l5K) =. 0.04978-1

At 141°C:

The relationship between the number of viable cells and time is given by Eq. (11.87). Converting the units of theinitial concentration of contaminants xo:

XC) = 108 cells 1-1 =. 108ceusl-l'111~11 =. 1011 cellsm-3

Therefore. per m3, No =. 1011; N =. 10.3. Substituting values into Eq. (11.87) gives:

10-3 =. 1011 e-/cd t

10-14 =. e-kdt

-32.24 =. -kd t

32.24t=--

kd

Therefore, at 8We:

t =. 32.24 =. 3224 .I~I =4070hkd 2.20 x 10-6 8-1 3600s

t =. 32.24 = 32.24 _II min I= 10.8 minkd 0.04978-1 60s

t = 32.24 =. 32.24 = 12.3 skd 2.638-1

Answer: 4070 b at 80oe; 10.8 min at 121°C; 12.3 s at 140°C

Heterogeneous Reactious

12.1 Diffusion and reaction in a waste treatment lagoon(a)A substrate shell mass~balanceis perfonned around a thin slice of sludge of area A and thickness Llz located distance zfrom the bottom of the lagoon, as shown in the diagram below.

Wastewater

Siudga Iu---------------------- ---------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'slaw, 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-balanceequation, 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

140

or

d2sAnswer: 2JSe d? -k1S ::::: 0


(b)At z::::: L, s::::: Sb. At Z "" O. we assume that the substrate concentration profile reaches a minimum so that ds/dz =O.

Answer: At z::::: L, s::::: Jb; at z:::: O. cIs/dz::::: O.

(c)(I)Ifs:::: 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)Ifs::::: 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 Jk11!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):


, = N(ezJk,l"s. +e-<Jk""s.)Applying the boundary condition at z=L:

'b = N(,LJk,12lo, +e-LJk,l"s.)Solving for N:

141

(Iv)From the definition of cosh x, the numerator in the equation for s in (iii) is equal to 2 cosh (z Jkl/.27s )- Similarly,the denominator is equal to2 cosh (LJkl/psc)' Therefore: e

, _ 2cosh (Z Jk1/.q.)- 'b2Cosh(LJkl/.q.)

or

, _ cosh(zJk11.q.)'b - cosh(LJk1/.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 rA,obs:

sinh (L Jk]I:lJ, )r - ~ A-~ Jk I SeA,obs - Se ~o 1 !OSe COSh(LJk I )

1 !OSe

142 Solutions.' Chapter 12

(e)The internal effectiveness factor is defined in Eq. (12.26). For first-order reaction kinetics at substrate concentrationSb 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,obs1]il = -.- =

'A'

17i1 =

Cancelling and grouping terms and applying the definition of tanh x gives:

tanh(LJkIffD,,)LJk1ffDs,

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 evaluatedfor the three sets of conditions are listed below.

Condition

(1)(2)(3)

¢I

0.502.0

10

0.460.961.0

1]il

0.920.480.10

The substrate concentration profiles are calculated using the equation for s as a ,function of z from (c) (iii) andapplying the definition of 1'1:

The results for Sb =10-5 grool em-] at various values ofzare listed and plotted below.

Distance, z (em)

0.00.30.71.01.2I.S1.82.0

Condition (1)

8.87 x 10-<'8.89x 10-6

9.00 x 10-6

9.15 x 10--69.27 x 10-69.50 x 10-69.78 x 10-61.00 x 10-5

Substrate concentration, S (gmo1 em-3)Condition (2)

2.66 x 10-62.78 x 10--63.34 x 10-64.10 x 10--64.81 x 10-6

6.25 x 10-68.26 x 10-61.00 x 10-5

Condition (3)

9.08 x 10-10

2.14x 10-9

1.50 x 10-8

6.74 x 10-81.83 x 10-7

8.21 x 10-7

3.68 x 10-<'1.00 x 10-5

143

Condition(3) ;1=10;1111=0.10

4; = 0.50; 1111 = 0.92

Condition (1)

Condition (2)


., 1.2

E0

" 1.0

~'& 0.8-x•

f0.6

0.4g

~ 0.2

I!,'" 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 inthe sludge is reduced.

12.2 Oxygen prome in immobilised-enzyme catalyst(aJFrom p 269, as CAs is O.5mM/0.015:tnM= 33 times the value of Km• as a:frrst approximation we can consider thekinetics to be effectively zero order with ko =vmax. Converting the units of~ to a per volume of gel basis:

ko = 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:

-3 -1 11000III 32 g III kg I -3CAs =0.5mM =05x to gmoH. 1m3 . 1 gmol . looog = 0.Ol6kgm

For zero-order reaction. the equation used to determine the substrate concentration inside the beads depends onwhether CA remains > 0 throughout the particle. The maximum particle radius for which this occurs can becalculated using Eq. (12.17);

=

Therefore. the maximum particle diameter for CA >0 everywhere is 4.2 nun.· Because the immobilised-enzyme beadsare smaller than this, CA > 0 and the oxygen concentration profile can be calculated using the equation for zero-orderreaction 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)

2.0x1.7 x 10-3

1.5 x 10-3

1.2 x 1(J31.0 x 1(J3

0.8 x 10-3

05 x 10-3

0.2 x 10-3

0.0

l.60x1.19 x 10-29.60 x 10-3

6.63 x 10-3

5.02 x 10-3

3.71 x 10-32.28 x 10-3

L51 x 10-3

1.37 x 10-3


20

f0>

"-M 150-x<J

" 10~~§

50c

~0

2.01.50.5

O'-__--''--__-l. --'- -'

0.0

As the minimum value ofCA at·the centre, of the bead is still about 2.9 timesKm• the assumption of zero·orderkinetics 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 zeroeverywhere 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:

ko = 10-3 mols-l m-3 .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 calculatedusingEq. (12,17):

=

Therefore, the maximum particle diameter for aerobic conditions is 2 x 1A5 = 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), oxygenis 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 "'" ko "'" 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 sphericalgeometry:

Solutions: Chapter /2 145

Using this value, the minimum intraparticle oxygen concentration can be calculated from the equation in Table 12.5for spherical geometry and f/) < 0.667:

CA,min = CAs(I-~tJ») = 8XIO-3kgm-3(1_~(0.167») = 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 kgs-1 m-3) = 6.4 x 10-6 kg s-l m-3. From Eq, (12,17):

6 4 0 9 2 1) 3 3LxI m s8xlO kgm =3,24xlO-3 m=3.24mm

6.4 x 10-6 kg s 1m 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:

l2=!i rA,obs = 1.5xlO-3

m 2.2 x 1O-5 kgs-l m-3 = 0.0313 kSCAb 3 6xlO-S ms 1(6xIO-3kgm-3}

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 bedetermined from Figure 12.11 as a function of the observable Thiele modulus CPo Evaluating tP from the equation inTable 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-1m-3

fJ)='3 'vAe CAs = 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 theequation in Table 12.5 for spherical geometry and tJ) < 0.667:

CA,min = CAS(l-itP) =O,97X6XlO-3kgm-3(1-~(0.50)) = 1.5 x 10-3 kg m-3

This oxygen concentration is greater than the critical level.

Answer: Yes


12.5 Microcarrier culture and external mass transferDp "'" 120 JUD "'" 120 x 10.6 m. The external mass-transfer coefficient can be determined using the equations on p 322for free~moving_spberes. The Grashof number is calculated from Eq. (12.51) with g "'" 9.8 m s·2 from p 16 and theuni{conversion factor 1 N s m-2 = 1 kg m-l s·l from Table A.9. Appendix A:

Gr= gD~PL(pp-PL) =9.8 m s-2 (I20X lO-6 mj' 103kg 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

1 Nsm 2 UTherefore, 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-18;11

$c = IlL = IN.sm = 435PL.vAL 103kgm-3(2.3xlO-9m2s I)

Therefore, Rep Sc = 0.188 x 435 = 81.8. As this value is less than 104. the Sherwood number can be evaluated usingEq. (12.55):

Sh = J4 + 1.21(Rep Sc)"·67 = v'4 + 1.21 (81.8P·67 = 5.21

From the definition of the Sherwood number in Eq. (12.50):

k_ Sh.vAL _ 5.21 (2.3 x 10-9 m2 ,-1) _ 999 10-5 -I

S - - - . x rosDp 120xl0--6 m

Using this value of ks to determine n from the equation in Table 12.6 for spherical geometry:

D = R TA,obs =3 ks CAb

From Eqs (12.43) and (12.44);

120xlO-6 m

2 0.015 mol s-1 m-3

::::: 0.015

3 9.99 x 10 'ms I (0.2molm-3j

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 arepresent 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~layershave been eliminated. CAs::::: CAb =0.85 kg m-3 and fie =1.The value of Pas defined on p 313 is:

jJ::::: Km ::::: 35kgm-3

::::: 4.12CAs 0.85kgm 3

From Figures 12.10-12.12, this value of pmeans 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 rA,obs = (0.8 x 10-3

m)2 1.25 x 10-3

kg 5-1

m-3 ::::: 8 03 .PAe CAs 3 1.3 x 1O-1l m2 s-1 (0.85 kg m-3) .

From Figure 12.11. at <1>= 8.0. flil =0.12. Therefore. from Eq. (12.46). TJT =11i 11e::::: 0.12 x I ::::: 0.12.

Answer: 0.12


(b)From the definition of the effectiveness factor. Eq. (12.26):

L25xtO-3

.kgs-I m-3 =OOlO4k -I -30.12 . gs m

Por first~orderkinetics. r~s =kl CAs; therefore:

•k

l= rAJ; = O.OI04kg s-1 m-

3 =0.0122s-1CAs 0.85kgm-3

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 beevaluated as a function of enzyme loading usingothe equation ·for first-order reaction and spherical geometry fromTable 12.2. with kt·directly proportional to the' enzyme 10ad.ing,Por¢>I < 1O,.the-intemal effectiveness factor l1il isdetermined using the equation in Table 12.3 and the' definition·of coth x; for tf1I > 10. from Eq. (12.30) 1]u =II¢!. Foreach value of kJ, r~s =kt CAs. and rA,obs can be determined from these results and the,definition of the effectivenessfactor 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'" (kg s-t m-3) rA,obs (kg s-t m-3)A,

0.01 0.0012 2.6 0.34 0.0010 3.40x lO40.05 0.0061 5.8 0.16 0.0052 8.32 x lO40.10 0.0122 8.2 0.12 0.0104 1.25 x Ily30.20 0.0244 11.6 0.086 0.021 1.81 X lO-3050 0.0610 18.3 0.055 0.052 2.86 x lO~3

0.80 0.0976 23.1 0.043 0.083 3.57 x 10~3

1.0 0.122 25.8 0.039 0.104 4.06 x lO~3

1.3 0.159 295 0.034 0.135 4.59 X 10-3

1.5 0.183 31.6 0.032 0.156 4,99 x lO-31.8 0.220 34.7 0.029 0.187 5.42 x lO-32.0 0.244 365 0.027 0.207 5.59 x to-3

The resultsfor 11i1 and rA.obs are plotted below as a function of enzyme loading.

6

5 1""4 ~'0-x

3 ~<!-

2 ~

l!c0

1 II~~

0:

02.0

'A,obs

0.5 1.0 1.5

Enzyme loading (1lffi01 g·1)

0.0 '-__-..J'-__-l. -'- ...J

0.0

r¥- 0.3

I! 02

=w~ 0.1

0.4,----,,----,----,-----,


As the enzyme loading is increased from 0.01 ~ol got, the effectiveness factor drops significantly. Although thereaction rate continues to rise with increasing enzyme loading, at loadings above about 0.5 !J.mol g-t, less than 5% ofthe potential activity of the enzyme is being utilised. Further increases in enzyme loading therefore represent aneffective waste ofmore 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 = 12.5p PL 10 3kgm 18-1

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 m2 8.1:

!lJ. ::::: 9x 1O-Qcm2s-1 = 9x 10-6 cm2 s-1.I~p =9x 1O-10m2 s-1Ae 100cml

iJ)AL =2 x iJ)Ae =2 x 9 x 10-10 m2 s·1 =1.8 x 10-9 m2 s-l. The Schmidt number from Eq; (12.49) is:

From Eq. (12.57), the Sherwood number can be evaluated as:

Sh = 0.95 ReO.5SeO.33 = 0.95 (12.51"·5 (556)°·33 = 27.0p

From the definition of the Sherwood number in Eq. (1250):

ks

= Sha/AL = 27.0(1.8 X 1O-9 m2

s-1) = 3.24xlO-5 ms-I

Dp 1.5 x 10 3m

'This value of ks can be used to-determine the observable modulus for external mass-transfer D from the equation inTable 12.6 for spherical geometry. As the specific gravity of the wet cells is 1. from p 16, 1 g wet cells occupies avolume 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 = 0.28 mg cm-3

h-I '11~6k~gl.13~sl;IIi'~m j3 = 7.78 X10-5 kg s-I m-3

Substituting this and the other parameter values into the equation for D:

1.5 X 10-3 m2 7.78x 10-5 kg s-l m-3 _ 0075

3.24XlO-5ms-I(8mgl-I)·I...!!'Lj·jlOOO11- .1Q6 mg 1 m3

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 resultfor CAs from (a):


(

1.5X to-3m)2<1>=(li)2 rA,obs = 2 7.78XlO'.."skgs-

1m-

3-073

3 aJAeCAs 3 9X10-lOm2s-1(o.925X8mgl-I).!16kg !.!1~1!- .10 mg 1m

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

(cJCAs = 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 exhaustedjust 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 beevaluated using Eq. (12.17):

F!-(9xto-lOm2S-1)7.4mg'-1·1'6kQI'~11·

_ 1Omgl' 1m

- 7.78 x 10-5 kg s-1 m 3

Rmax =7.2 x 104 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, whichis 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 theclumps.

12.8 Respiration in mycelial pellets(aJR = 2.5 mm = 2.5 x 10-3 m. The presence of external boundary-layers can be checked by calculating the observablemodulus for external mass-transfer, D. From Table 12.6 for spherical goomelry:

,Q=!!.. rA,obs =2.5x10-3

m 8.7XlO-5

kgs-1

m-3 =024

3 kSCAb 3.8x10 5 ms 1(8XlO-3 kgm-3) .

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

(bJAs 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 = CAbthroughout the pellets. As rAb is related to rA,obs by Eq. (12.45), r b can be determined if we know 1JTo. Evaluatingthe observable Thiele modulus <P using the equation in Table 12.4 tor spherical geometry and the result for CAs from(a):


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 87 10-'k -1 -3'* A,obs ,x gs m 29 10'""k -1 -3rAb""'--"'" = ,x gs 00~ro 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 wouldbe greater than 8,7 x-1O-5 kg g-l.m-3, Under these conditions. an-expression for the observable Thiele modulus epfrom the equation in Table 12.4 for spherical geometry is:

~ _(R)2 rA,obs _ (2.5 x 10-300)2 rA,obs 4.... - - - ()= 4.96x 10 rA,obs

3 :tJAeCAs 3 1.75xI0-9 m2 s 1 8xlO-3kgm 3

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 rA,obs'ho =-.- =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 bedetermined bl trial-and-error using Figure 12.11, and the equations derived above, As a ftrst guess, take rA,obs = 2.0x 10-4 kg S' 00,3. Depending on the difference between the values of 1]io obtained from the figure and from theequation with r~s' adjust rA,obs as shown in the table below.

rA.obs (kg s-I 00-3) 11iO = r~Obs <1> 1Ji.o (from Figure 12.11)rA,

2.0 x 10-4 0.69 9.92 0.191.0 x 10-4 0.34 4.96 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, comparedwith the observed reaction rate of'8.7 x 10-5 kg Sol 00-3 in the presence of both 'internal and external mass-transferresistances, 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 conversionFor 75% conversion, sf= 0.25 so- TIle batch reaction time for enzyme processes is evaluated using Eq, (13.10):

Km I So sO-sf 1.5 g 1-1 In 3g1-1 +3gt-1-O.25X3g1-1 __ 4.81htb = -- n-+-- =

Vmax sf vmax O.9gI 1h 1 O.25X3g1 1 O.9gI 1h 1

The operating cost is therefore:

Operatingcost = 4,81 h -I ~:~l $4800 day-l = $962

The cost ofdownstream 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 insubstrate 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 ofsubstrate consumed is:


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 =: $1868

The gain per batch from increasing the conversion from 75%:to90% is therefore $1868 -" $1715 =: $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 concentrationSf = 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):

k = In2 =~ I~I = 275 IO-3 h-1d th 1O,5d'24h . x

For enzyme subject to deactivation, the batch reaction time is evaluated using Eq. (13.13);

tb =-lln[l-kd ( Km In So + SO-S,)]kd vmaxo Sf Vmaxo

= -I In 1_2.75XIO-3h-l( 4.0gl-1

10 lSOgl-1

+(150-22.5Jgrl) = 23.6h

2.75 x 10-3 h-1 5.9 gl 1h-1 22.5 g I-I 5.9 g I-I h-1

At 37"C, the deactivation rate constant is;

k = In2 = In2 I~I = 126 1O-2 h-ld t

h2.3d· 24h . x

The batch reaction time is;

tb = -lln[l-kd ( Km In So + so-S,)]kd vmaxo Sf Vmaxo

= -I In l-l.26XIO-Zh-l( 4.0gl-1

In 150gl-1

+ (150-22.5j gl-I) = 17.7hL26xlO 2 h-l 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:


tT =- 1b+tdn =- 17.7h+28h =- 45.7h

153

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 r1 =- 127.5 g 1-1 multiplied by the reactorvolume V. Therefore. the mass of substrates converted is 1275 V g =- 0.1275 V kg, where V has units of litres. Fromthe reaction stoichiometry, as the molecular weights of ammonium fumarate and aspartic acid are approximatelyequal, the mass of aspartic acid produced is also 0.1275 Vkg. After one year or 192 batches. the mass of aspartic acidproduced is 0.1275 V x 192 =- 24.5 V kg. Using the conversion factor 1 tonne =- 103 kg (Table A.3, Appendix A), thetarget 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 m3

13.3 Prediction of batch cnllnre time(aJThe initial cell concentration Xo =- 12 g/loo I =- 0.12 g 1-1. Assume that stationary phase is reached when Sf =- O. Thebatch culture time can be determined using Eq. (l3.27):

I [YXS ] I [0.575 gg-I

( -I l]tb =- --In l+-(so-sr) =- --lIn 1+ I lOgl -0 =- 4.3hPmax Xo 0.9 h- 0.12 g I

Answer: 4.3 h

(bJIf only 70% of the substrate is consumed, Sf =- 0.3 So =- 0.3 x 10 g I-I =- 3 g I-I. From Eq. (13.27):

tb = _1_ln[1 + YXS ('o-'d] = _1_1

1n[1 + 0.575 g g~I flO g I-I _ 3grIl] = 3.9 hPmax Xo 0.9h- 0.12g'-

The biomass density at this time can be calculated from Eq. (13.19):

xf =- Xo ePmax iI =- 0.12 g.-I e(O.9h-1

x 3.9 h) =- 4.0 g I-I

Answer: 4.0 g 1-1

13.4 Fed-batch schednling(aJThe 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] = 13.3dJlmax Xo 0.18d- 1.5g1-1

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.


(b)The mass of cells at the start of fed-batch operation is equal to the fmal batch cell concentration multiplied by theinitial 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 ofcells 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 theresult for 1b from (a):

IT = !b+tfb+tdn:= 133d+40d+ld = 54.3d

In one year, the number of runs carried out is:

275dNumberofruns = 54.3dperrun :: 5.06

The total biomass produced annually is equal to the biomass produced per run multiplied by the number of runs peryear:

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·batch reactor operation can be derived from anunsteady-state total mass balance as shown in the solution to Problem 6.7afrom Chapter 6. Using this expression:

Vo =: V-Ft =: 40m3 _4m3 h·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 m3;

D =!:.. = 4m3

b-1

=: O,lOh-1V 40m3

Substituting this value into Eq. (13.45) for the substrate concentration at quasi*steady state:

s = DKS = O. 10 h-I

(0.I5kgm-3) = 0.06kgm-3

Iimax-D 0.35h I_ O.lOb 1

Answer: 0.06 kg ro·3

(c)Taking maintenance substrate requirements into account, for qp:: 0, Eq. (13.43) becomes:


£Is = D(S.-S)-(.J!....+ms)xdt 1 Yxs

At quasi~steadystate, dsld! '" 0, J1 '" D, ands« Sj. Therefore, the equation reduces to:

o = DSi-(~s +ms)x

Solving for r.

155

Ds, O.10h-1(SOkgm-3)x =,,-'- =__~T'--""=':"-.1.-

_D_+ ms O.lOh I +0.135kgkg-I h-1

Yxs 0.23 kg kg 1

Answer: 14.0 kg m-3

(d)After 6 h fed~batch operation, the mass of cells is:

= 14.0kgm-3

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 cellconcentration", 14.0 kg m- and:

x = x V = 14.0 kgm-3 (16 m3) = 224 kg

Therefore, the mass ofcells produced during fed~batch operation is (560 kg - 224 kg) = 336 kg.

Answer: 336 kg

13.6 Continuous enzyme conversion in a fIXed-bed reactorConvert 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 perunit volume of catalyst ea is:

ea = lO-4 g = lO-4 g .1~1.I!oocmI3 =4xlO-4 kgm-3250cm3 250cm3 looog 1m

The effective diffusivity of urea in the gel VAe is:

Ij) = 7xlO-6 cm2 s-1 = 7XI0-6cm2s-I'I~12 = 7xlO-lO m2 s-1Ae l00cm

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 turnovernumber 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: 136.0gNH4

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 em3 gel per litre of liquid in the reactor, Vrnax expressed on a per volume liquid basis is:


=736k -3 _1(2S0Cm3

) 1~131100011= 184k -3-1vrnax . gm s 11 . lOOcm . 1m3 . gm s

The rate of reaction can be determined after evaluating the effectiveness factor in the absence of external boundarylayers. From the definition of f30n p 313 with CAs = s:

p = Km = O.54kgm-3

"'" 27s 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), theeffective first-order rate constant kl is:

I 84k -3-1. gills "'" 3.418-1

0.54kgm-3

_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 I~il = <Pi / 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 detenninedby evaluating the dilution rate D in the mass-balance equation, Eq. (13.54):

D = 7V~_T,;-V:;m7""';-'--:K = -,0",.04=3"(I".8"4",kg"m"-,3-:o,-_IL)0",.0,,,2:.okg=mc.-_3,, = 7.06 x 10-3 s-I

(Km + $) (si $) (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 = 7.06x10-3 I s-I (30 min) ·ll~nI= 12.71

Answer: 12.7litres

13.7 Balch and conllnuous hiomass productionSO= Si = 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 batchreactor. X() = 0.01% (w/v) = 0.01 g per 100 ml = 0.1 g I-I = 0.1 kg m-3.

The batch culture time can be determined from Eq. (13.27):

1 [Yxs 1 1 [04I gg-1 ' 3]tb = " In I +-('O-Sf) = lin 1+. 3 (40-0.8) kg m- = 11.6h

r-max XO O.44h O.lkgm

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


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¥*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):

D = p = I'm",,' = M4h-1

{O.8kgm-3j = O.44h-1

Ks+s 07 r 1 11000111~1 08k -3. mg . 3' 6 +. gm1 m 10 mg

157

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 availablefor continuous reactor operation is (365 - 25) = 340 d; this corresponds to 340 d x 24 h d-} = 8160 h. Therefore. thetotal 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 x106 = 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 theproduction of 4.54 x 106 kg from batch culture.

13.8 Reactor design for immobilised enzymesSo = 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·state mass-balance equation derived in Example 6.1 in Chapter 6 for first~orderreaction, the equation for therate of change ofsubstrate concentration in a batch reactor is:

d(V,) _ -Ie V(it- IS

where V is the reaction volume and k} is the reaction rate constant. As V can be considered constant in a batchreactor, this term can be taken outside of the differential and cancelled from both sides of the equation:

dsdt =-k1 s

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

ds- = -kl dt,


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 theequation gives:

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.

sf -l 1kgm-3-10- n 3

So lOOkgmtb = -- = = 16.0h

k 1 O.8XlO-4Cl·13~sl

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:

l24hl365d' ITNumberofbatches = 36hperbatcb = 243


Reactor volume =

To treat 400 tonues penicillin G annually, using the unit conversion factor 1 tonne:= 103 kg (Table A.3, Appendix A):

400tonnes ·1_l0_3_k_g IMass of penicillin G treated per batch = her400fh,onn

h" = o.ci--'-l"to"n"ne", = 1.65 x 103kg

num 0 ate esperyear 243

As the concentration of penicillin G added to the reactor is tOO kg m-3:

I.65xl03

k:g = 165m3100 kg m-3

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·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 - 1V---- _.X ssi -s (100-1) kg m 3

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 volumetricflow rate of the feed stream F is:

-1 I103

kg I11Yeaii lId Illh I400 tonnes year . 1 tonne . 365 d . 24 h . 3600 s

F:= := 1.27x 10-4 m3 s-1100 kg m-3

Applying this with the above result for Fly:

SolutWns: Chapter 13

Fv=-=FIV

1.27 x 1O-4 m3 s-I

8.08xlO 7 8-1= 157m3

159

The CSTR reactor volume required is 157 m3.

For the PFTR, if the density ofenzyme beads is four times greater than in the other reactors. kl = 4 x 0.8 x 104 s·l =3.2 x 10-4 s·l. By analogy with Eq. (13.83), the differential equation for change in substrate concentration withposition in the reactor for first·order kinetics is:

<IsUctz=-k1s

The differential equation contains only two variables, s and z. Separating variables and integrating:

<Is -k,-=-dzS u

Using integration rules (D.27) and (0.24) from AppendixD 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 Kintothe equation gives:

-k,Ins = -z+lnsju

At the end of the PFIR, z= L and S = Sf, therefore:

Applying the defInition of the reactor residence time .. from Eq. (13.85):

sfIn- =-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 kineticcharacteristics 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.


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):

501h-1 11m3

ID =f. = . 10001 =O.lOh-'

V a.5m3

The cell and substrate concentrations entering the second reactor are the same as those leaving the first reactor. Thesubstrate 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) = _O"'c.lO"b"-.'",(lcoO.-.cS".O",),,kg,.m=.•-3_ = 2.2 kgm-3

.D. +ms O.lOh-1

+0.025 kg kg-1 h-1Yxs 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 byrearranging Eq. (13.59)and solving for s with fJ::::: 0:

(qp ) V

s::::: $i- Yps

+mS x p

Substituting the parameter values with Si ::= 5.0 kg m-3;

For the two reactors together:

(Si- S) (lO-O.3l)kg m-3Overallsubstrateconversion::= _-x 100%::= 3 x 100% = 97%

si lOkgm

Answer: 97%

(cjAs product is not formed in the first reactor, Pi = 0 for the second reactor. The product concentration is determined byrearranging Eq. (13.64) and solving for p:

p =qpxV =O.16kgkg-1b-'(22kg~-3)0.5m3 =3.Skgm-3

501h-l·ll~11Answer: 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 thedefinition 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.


Flow rate, F (ml h"l) Dilution rate, D (h"l) Substrate concentration, S (J,lM) SID OtM h)

10152025303550

0.0100.0150.0200.0250.0300.035O.OSO

17.425.139.846.869.480.1

100

1740167319901872231322892000

2500

• •2000 •

:2 •~ •ore, •

1500

10020 40 60 80

Substrate concentration, S (p.M)

1000 l-~_...L_~_L.~_...L_~_L._~....J

o

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):

tlmaxSj 0.095h-1 (100/lM) -1Derit = KS+sj = 142Jl,M+lOOliM = 0.039h

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 mutantFrom Eq. (13.92), Jlmax and Ks can be determined from the slope and intercept of a plot of SID versus s. From thedefinition 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)

1.01.41.61.71.81.9

o.so0.700.800.850.900.95

0.0100.0380.0710.0660.0950.477

0.0200.0540.0890.0780.1060.502

162

0.6

0.6

0.4:2-~ 0.3.$~ICl

0.2

0.1

0.00.0 0.1 0.2 0.3 0.4 0.5

Substrate concentration, S (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, whereY~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 (hwl) liD (h) Y~s(g g-l) Ill" (g g-l)XS

1.0 050 2.00 0.315 3.1751.4 0.70 1.43 0.323 3.0961.6 0.80 1.25 0.329 3.0401.7 0.85 1.18 0.328 3.0491.8 0.90 1.11 0.324 3.0861.9 0.95 LOS 0.326 3.067

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

•3.15

--b.$ 3.10_ m

'?:-X •- •3.05 • •3.00

0.8 1.0 1.2 1.4 1.6 1.8 2.0 2.2

YOilution rate (h)


= O.107h

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.lh-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 sterilisationFrom the definition of dilution rate in Eq. (13.39), the medium volumetric flow rate F =- D V =- 0.1 h-1 x 15 m3 =- 1,5m3 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: ,.z, where r is the pipe radius. For r =- 6 cm =- 0.06 m:

U =- f. =- 1.5m3

h-1

=- 132.6mh-1A 1t (0.06 m)Z

The value ofthe 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 degreesKelvin using Eq. (2.24):

kd =A e-EdiR T == 7.5 x 1039 h-1 e-Z.885 x lOS J gmorlt[(8.3144J gmor1K"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 volumetricflow rate Fmultiplied by the cell concentration and the time:

N j = 1.5 m3

h-I

( 105 ml-1 1~:~)90d.12::1=3.24 x 1014

Within the same 3-month period, the acCeptable number of cells remaining at the end of the sterilisation treatment isNZ =- L Therefore:

NZ 1 =- 3.09 x 10-15Nt =- 3.24x 1014

(a)For perfect plug flow .withno:axial dispersion, the sterilisation time can be determined using Eq. (13,97):

N1 14In- 1n 3.24XlO

N2 1thd =- kd = 313.1h 1

To allow the medium to remain for this period. of time in the.holding ,section,of:the steriliserpipe,.the length of piperequired is equal to the linear velocity of the medium u multiplied by t'hd:

L = u!hd =- 132.6mh-1xOJ07h =- 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-3Re=---= =-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 givesa 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:


where L has units of m. Similarly, an expression for the Damkohler number Da from Eq. (13.102) is:

kd L (313.1 h-1) LDa =- "'" "'" 2.36£

u 132.6mb-1

The design problem can be solved from this point using tria1~andwerror methods and Figure 13.41. As a first guess, tryL =20 ID. The values for Pe and Da are evaluated using the equations determined above, and the corresponding valuefor N2/Nl read from Figure 13.41. Depending on how this value compares with the target of 3.09 x 10-15, the valueof L is adjusted until the results for N2/N} coincide. The calculations are shown in the table below.

L(m)

201819

Pe

11099

105

Da

474245

(from Figure 13.41)

4x1 x 10-141.5 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 forideal 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 x10-12; therefore, NI/N2 :::: 2 X 1011. As N2 :::: 1, N} :::: 2 x 1011, i.e. one contaminant enters the fermentetfor every 2 xlOll that enter the steriliser, For F:::: 1.5 m3 h- l and an input contaminant concentration of lOS ml- l , the timerequired 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.

Bio Process Engineering Principles [Solutions Manual] - P. Doran (1997) WW

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