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SOLUTIONS MANUAL
Bioprocess Engineering Principles
Pauline M. Doran
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SOLUTIONS MANUAL
Bioprocess Engineering Principles
Pauline M. Doran
University of New South Wales, Sydney, Australia
ISBN 0 7334 15474
© Pauline M. Doran 1997
Table of Contents
Solutions
Page
Chapter 2 Introduction to Engineering Calculations 1
Chapter 3 Presentation and Analysis ofData 9
Chapter 4 Material Balances 17
ChapterS Energy Balances 41
Chapter 6 Unsteady-State Material and Energy Balances 54
Chapter 7 Fluid Flow and Mixing 76
Chapter 8 Heat Transfer 86
Chapter 9 Mass Transfer '98
Chapter 10 Unit Operations 106
Chapter 11 Homogeneous Reactions 122
Chapter 12 Heterogeneous Reactions 139
Chapter 13 Reactor Engineering 151
NOTE
All equations, tables, figures, page numbers, etc., mentioned in this manual refer to the textbook,
Bioprocess Engineering Principles.
Introduction to Engineering Calculations
2.1 Unit conversion
(a)
From Table A.9 (Appendix A): 1 cP::::: 1O~3 kg m-I k 1
1 m= lOOcrn
Therefore:
1.5x10-6 cP ::::: 1.5 x 10-6 cP ,1 10- 3 k
t
g;-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-I
Therefore:
Answer: 5.17 Btu min-1
(e)
From Table A.S (Appendix. A): 1 mmHg:= 1.316 x 10-3 attn
From Table A.I (Appendix A): 1 ft = 0.3048 m
From Table A.7 (Appendix A): 11 atm =9.604 x 10-2 Btu
From Table A8 (Appendix A): 1 Btu min-I::::: 2.391 x to-2 metric horsepower
Im=lOOcm
11= lOOOcm3
Ih=60min
Therefore:
670mmHgft3 =670 mmHg ft3 .11.316X 10-3 atml.19.604X 1O-2Btul.I°.3048m 13 1 100 em 13 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 h
1 . -60. = , x metric orsepower1Btu min- mm
Answer: 9.56 x 10-4 metric horsepower h
(d)
From Table A.7 (Appendix A): 1 Btu =0.2520 kcal
From Table A.3 (Appendix A): Ilb = 453.6 g
Therefore:
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 conversion
Case 1
Convert 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·l
1 rn= tOOcm= lOOOmm
Therefore, 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 10
10-6 P 10 kgm sc . 1cP
Answer: 2.4 x 107
ease 2
Convert to units of kg, m, s.
From Table Al (Appendix A): 1 in.::: 2.54 x 10-2 m
From Table A.9 (Appendix A): 1 Ibm ft-I h-1 ::: 4.134 x 10-4 kg m-t s·t
Ih=3600s
Therefore, usingEq. (2.1):
= 1.5 x 104
13~sl
Answer: 1.5 x 104
2.3 Dimensionless groups and property data
From the Chemical Engineers' Handbook, the diffusivity of oxygen in water at 2S"C is 2.5 x 10-5 cm2 s·l. Assuming
this is the same at 28"C, !lJ= 2.5 x 10-5 cm1 s·t, Also, from the Chemical Engineers' Handbook, the density of water
at 28"e is PL ::: 0.9962652 g cm-3, and the viscosity of water at 28"C is JlL::: 0.87 cPo The density of oxygen at 28°C
and 1 atm pressure can be calculated using the ideal gas law. As molar density is the same as n,V, from Eq. (2.32):
Temperature in the ideal gas equation is absolute temperature; therefore, from Eq. (2.24):
T = (28 + 273.15) K = 301.15 K
From Table 2.5, R "" 82.057 cm3 atm K-I gmol~l. Substituting parameter values into the density equation gives:
Pa "" L "" latm "" 4.05 x 10-5 gmolcm-3
RT (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 result
for Pa to mass tenns:
Pa "" 4.05 x lO-5gmolem-3 ·1 i~~~ll "" l.30XlO-3 gcm-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 the
dimensionless groups in the equation for the Sherwood number.
Solutions,' Chapter 2
1'0-2 -1 -110,87eP. gem sS_J1.L_ leP -349
c - 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 -IL---- 02 -,<tUX emsD1J . em
Answer: lAO x 10~3 em s·1
2.4 Mass and weight
From the definition of density on p 16, mass is equal to volume multiplied by density, Therefore:
3
From p 16, weight is-the force with wrncha body is attracted to the centre of the earth by gravity. According to
Newton's law(p 15), this force is equal to the mass of the body multiplied by the gravitational acceleration.
(a)
From pIS, at sea level and 45° latitude. gravitational acceleration g = 32.174 it s-2. Therefore:
Weight:::: 624 Ibm (32.174 ft s-2) = 2.008x 104 Ibm it s·Z
Converting these units to lbj' from Eq. (2.16), 1lbf= 32,174 Ibm it s·2; therefore:
4 2 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 numbers
First, evaluate the units of the groups cCP J.llk) and (D G1p):
. (ep") (Bmlb-I '1'1) Ibh-I ft-IUmtsof -- :::: = 1
k BIUh-1 ft-2('Pft-1tl
UnilsOf(DG) = (ftllbh-1ft-2 = 1
J1. Ibh1ft1
4 Solutions: Chapter 2
Therefore, these groups are dimensionless. For the equation to be dimensionally homogeneous, (hICp G) must also be
dimensionless; the units of h must therefore cancel the units of Cp G.
Unitsofh "'" unitsofCpG "'" (Btulb-1 "p-l)(1bh-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 with
dimensions "'" L2Ml2. OF is a unit of temperature which, from Table 2.1, has the dimensional symbol E>. h is a unit
of time with dimension"'" T; ft is a unit of length with dimension"'" L. Therefore:
Dimensions 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 Cc
From Table A8 (Appendix A), dimensions of P = L2MT"3
Dimensions of g = LT-z
Dimensions of p =ML·3
Dimensions ofDi "'" L
From p 11, the dimensions oirotational speed, Nj =T-!; from p 15, the dimensions of gc= 1. Therefore:
As Np is a dimensionless number, equation (i) is not dimensionally homogeneous and therefore cannot be correct.
Equation (ii) is dimensionally homogeneous and therefore likely to be correct
Answer: (li)
2.7 Molar units
From the atomic weights in Table Rl (Appendix B), the molecular weight of NaOH is 40.0.
(a)
FromEq. (2.19):
lb moles NaOH = 20.0 lb = 050 lbmol
4O.0lblbmol 1
Answer: 050 lbmol
(b)
From Table A.3 (Appendix A): lIb =453.6 g. 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
1
1 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 specific
gravity L5129i~, the density at 20°C is 1.5129 g cm*3,
(I)
lkg=l000g
1 m: 100cm
Therefore:
Answer: 1512.9 kg m-3
(il)
From the atomic weights in Table 8.1 (Appendix B), the molecular weight of nitric acid (RN03) is 63.0. In 1 cm3
RNO" from Eq. (2.18)0
LS129ggram moles : 1 : 0.0240gmot
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 gmor1
mo ar enslty 0.0240 gmol em 3
Answer: 41.67 cm3 gmol-l
(b)
(I)
From p 16, as density is defined as the mass per unit volume, the mass flow rate is equal to the volumetric flow rate
multiplied by the density:
Answer: 80 g min- l
(ii)
From the atomic weights in Table B.l (Appendix B), the molecular weight of carbon tetrachloride, CC14, is 153.8.
Using the mass flow rate from (a):
Molar flow rate : 80 g min-l .l :5~0~1 =0.52gmolmin-1
Answer: 0.52 grool min- I
2.9 Molecular weight
From p 17, the composition of air is close to 21 % oxygen and 79% nitrogen. For gases at low pressures, this means
21 mol% 02 and 79 mol% NZ. Therefore, in 1 gmol air, there are 0.21 gmot Oz and 0.79 gmol NZ From the atomic
weights in Table B.l (Appendix B), the molecular weights of Oz and NZ are 3Z.0 and 28.0, respectively. The
molecular weight of air is equal to the number of grams in 1 gmol:
1gmolair =0.21 gmOlOz·1 ;~~ll + 0.79 gmol NZ ·1 ;~;ll =28.8g
Answer. 28.8
6 Solutions.' Chapter 2
2.10 Mole fraction
The molecular weights can be obtained from Table B.7 (Appendix B): water 18.0; ethanol 46.1; methanol 32.0;
glycerol 92.1; acetic acid 60.1; benzaldehyde 106.1. In 100 g solution, there are 30 g water, 25 g ethanol. 15 g
methanol, 12 g glycerol, 10 g acetic acid, 8 g benzaldehyde, and no other components. Therefore:
Moles water = 30 g _I ~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 scales
From Eq. (2.27\
-40 ::::: 1.8 T(0C) + 32
Tee) = -40
From Eq. (2.25),
T (OR) ::::: -40 + 459.67
T(°R) = 420
From Eq. (2.24) and the result for T ("C);
T(K) = -40+273.15
T(K) = 233
2.12 Pressure scales
(a)
Assume that the atmospheric pressure is 14.7 psi. From Eq. (2.28):
Absolute pressure ::::: 15 psi + 14.7 psi::::: 29,7 psi
Solutions: Chapter 2
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. The
maximum theoretical yield from the stoichiometric equation is 1 gruol of penicillin for every 1.67 gruol of glucose.
'This is equivalent to 334.4 g penicillin per 1.67 x 180.2 = 300.9 g glucose. or 1.1 g g-l.
Answer: 1.1 g g~1
(b)
The maximum theoretical yield in (a) is obtained when all the glucose consumed is directed into penicillin production
according to the stoichiometric equation. If only 6% of the glucose is used in this way, the actual yield of penicillin
from glucose is much lower, at 334.4 g penicillin per (300.9 x 100/6) g glucose, or 0.067 g g~l.
Answer: 0.067 g g-1
(c)
From the atomic weights in Table B.I (Appendix B), the molecular weight of phenylacetic acid is 136.2.
(I)
The only possible limiting substrates are glucose and phenylacetic acid. Using a basis of II medium, if (50 - 5.5) =
44.5 g glucose are consumed but only 6% is available for penicillin synthesis, the mass of glucose used in the
penicillin reaction is 44.5 x 6/100 =2.67 g. 'This is equivalent to 2.67 g/180.2 g gmol-1 =1.48 K 10-2 gmol glucose
available for penicillin synthesis. At the same time, 4 g or 4 g/136.2 g gmol-1 =2.94 x 10~2 gruol phenylacetic acid is
available which, according to the stoichiometric equation, requires 1.67 x 2.94 x 10-2 =4.91 x 10-2 gruol glucose for
complete reaction. As the gmol glucose required is greater than the gmol glucose available after growth and
maintenance activities, glucose is the limiting substrate.
Answer: Glucose
(il)
Of the 44.5 g I-I glucose consumed, 24% or 10.7 g I-I is used for growth. In a H~)-litre tank. the total mass of glucose
consumed for growth is therefore 1070 g or 1.07 kg.
Answer: 1.07 kg
(iii)
From (i), 1.48 x 10-2 gmol glucose is used in the penicillin reaction per litre. According to the stoichiometry, this
produces 1.48 x 10-2/1.67 = 8.86 x 1O~3 gmol penicillin per litre. Therefore, in a l00-litre tank, 0.886 gmol or 0.886
gmol x 334.4 g gmol-1 =296 g penicillin are formed.
Answer: 296 g
(iv)
IT, from (i), 1.48 x 10-2 gmol [1 glucose is used in the penicillin reaction, 1.48 x 10-2/1.67 = 8.86 x 10-3 gmoll-l
phenylacetic acid must also be used. This is equivalent to 8.86 x 10-3 gmoll-l x 136.2 g gmol-1 = 1.21 g t I
phenylacetic acid. As 4 g I-I are provided, (4 - 1.21) = 2.79 g I~I phenylacetic acid must remain.
Answer: 2.79 g I-I
8 Solutions: Chapter 2
2.14 Stoichiometry, yield and the ideal gas law
(a)
Adding up the numbers of C, H, 0 and N atoms on both sides of the equation shows that the equation is balanced.
Answer: Yes
(b)
The molecular weights are calculated from Table B.I (Appendix B).
Cells: 91.5
Hexadecane: 226.4
From the stoichiometry, as 1 gmol of hexadecane is required to produce 1.65 gmol of cells, the maximum yield is 1.65
gmol x 91.5 g gmol-l =: 151 g cells per 226A g hexadecane, or 0.67 g g-I,
Answer: 0.67 g g-1
(e)
From the atomic weights in Table RI (Appendix B), the molecular weight of oxygen is 32.0. From the
stoichiometry, 16.28 gmol of oxygen is required to produce 1.65 gmal 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 amounts
of substrates are required when 100% of the hexadecane is converted according to the stoichiometric equation.
(I)
From the stoichiometry, production of 27.3 gmol cells requires 27.3/1.65 =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 gmol
oxygen. As air at low pressure contains close to 21 mol% oxygen (p 11), the total moles of air required is 269.410.21
= 1282.9 gmot The volume ofair required can be calculated using the ideal gas law. From Eq. (2.32):
V = nRT
p
Temperature in the ideal gas equation is absolute temperature; from Eq. (2.24):
T = (20 +273.15) K = 293.15K
From Table 2.5, R "" 82.057 cm3 atm K-I gmol-I. Substituting these values into the equation for V gives:
v= (1282.9gmolj(82.057cm3 atmK-1 gmol-l)(293.15Kj 1-.!.."'...13 = 31 3
1 atm . l00cm m
Answer: 31 m 3
Presentation and Analysis of Data
3.1 Combination of errors
c~ "" 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 areadded. Therefore:
kLa =: o.OllmOlm-3~-1 ±{25.8 + 5)% =O.16s-1±31% = O.16± 0.05 s-l
O.Q67molm
Answer~ 31%. This example illustrates how a combination of small measurement errors can result in a relatively
large uncertainty in the final result.
3.2 Mean and standard deviation
(a)
The best estimate is the mean, X. FromEq. (3.1):
x = 5.15+5.45+5.50+5.35 = 5.36
Answer: 5.36
(b)
Calculate the standard deviation from Eq. (3.2):
(5.15 - 5.36)' + (5.45 - 5.36)' + (5.50- 5.36)' + (5.35 - 5.36)'
4-1 = 0.15
Answer. The standard deviation is 0.15. Note that standard error, which can be calculated from the standard
deviation, is a more direct indication of the precision ofa mean.
(c)
x = 5.15 +5.45 =5.30
2
Standard deviation is not appropriate for expressing the accuracy of a mean evaluated using only two samples. In this
case the maximum error, Le. the difference between the mean and either of the two measured values, might be used
instead. The maximum error in this example is (5.30 - 5.15) =0.15.
Answer. 5.30; an indication of the accuracy is ± 0.15
(d)
x= 5.15+5.45+5.50+5.35+5.15+5.45+5.50+5.35 = 5.36
10 Solutions: Chapter 3
2 (S.lS -S.36f + 2(S.4S-S.36)~ + ~ (S.50-S.36f +2(S.3S -S.36)2 = 0.14
Answer: The best estimate of optimal pH is unchanged at 5.36, but the standard deviation is slightly lower at 0.14.
This example illustrates that although the standard deviation decreases as the number ofmeasurements is increased,
(j is not strongly dependent on n. The best way to improve the reliability of the mean is to ensure that the individual
measurements are as intrinsically accurate as possible, rather thanrepeat the measurement many times.
3.3 Linear and non-linear models
(a)
Xl =1; Yl =10
X2=8;Y2=0.5
A straight line plot of y versus x on linear coordinates means that the data can be represented using Eq. (3.6). From
Eqs (3.7) and (3.8),
A = (Yz-Y1) = 0.5-10 =-136
(x2 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 = 38.5
A 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 = 201
112_ 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=6
X2= l;Y2=3
A straight line plot of Ity versus xl on linear coordinates means that the data can be represented using the equation:
lly = Ax2 +B
with A and B given by the equations:
A =
lin - Ity} 1/3 -lI6 -3
2 2 = 2 2 =-6.9xlO
x2-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=25
x2 =550; Y2 =2600
A straight line plot of y versus x on log-log coordinates means that the data can be represented using Eq, (3.10).
From Eqs (3.13) and (3.14),
Solutions: Chapter 3
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
(eJ
Xl = 1.5; YI = 2.5
X2 = 10; Y2 = 0.036
A 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
(aJ
The results determined using Eqs (3.1) and (3.2) are listed below.
Sucrose concentration (g l~l)
6.0
12.0
18.0
24.0
30.0
(bJ
Mean peak area
56.84
112.82
170.63
232.74
302.04
Standard deviation
1.21
2.06
2.54
1.80
2.21
35
30
-
-
.
.9 25
0
••~ 20
•g 158
~ 10g
"' 5
0
0 50 100 150 200 250 300 350
Peak area
12 Solutions: Chapter 3
(e)
The linear least squares fit of the data is:
y ::: 0,098 x + 0,83
Answer: y "'" 0.098 x + 0.83, where y is sucrose concentration in g 1-1 and x is peak area.
(d)
For x::: 209,86, the equation in (c) gives a sucrose concentration oi2IA g I-I.
Answer: 21.4 g I-I
3.5 Non-linear model: calcnlation of parameters
(a)
The proposed model equation has the general form of Eq. (3.15); therefore, if the model is suitable, a plot of a versus
lIT on semi-logarithmic coordinates will give a straight line. As T in the equation is absolute temperature, "'c must
first be converted to degrees Kelvin using Eq. (2.24). The data are listed and plotted below.
Temperature (0C) Temperature (K) IIT(K-l) Relative mutation frequency. a
15 288.15 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 JerI3
35 308.15 3.25 x 10-3 lAx }0"'12
10-11 r---~--.,...--~----r--~---.
"[ 10-12
•
"f
c 10-13.Q
lii
"E
~
= 10.14
-ll
lJ:
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 homogeneity
the exponent must be dimensionless (p 12), so that -26,121 has units of K, and EIR in the model equation is equal to
26,121 K From Table 2.5, R =: 8.3144 J gmol-l K·1; therefore:
E =: (26,121 K) (8.3144 J gmol~l K*l) =: 217,180.4 J gmol-1 =: 217.2 kJ gmot1
Solutions: Chapter 3
Answer: 217.2 kJ gmol-l
(c)
From the equation in (b) for the straight line, ao is equal to 9.66 x 10Z4.
Answer: 9.66 X 1024
3.6 Linear regression: distribution of residuals
(aJ
13
12345 6
Decrease in medium conductivity (mS cm-1)
16
-
-
, 14
!'!l
c
•• 12i'!
C 10•g
0
0 8~ 6~
•• 4m
• •~ 2
.i1
0
0
••
•
• ••
•
•
• •
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 and
the values for y obtained from the equation in (a).
Decrease in medium conductivity (mS
o
0.12
0.31
0.41
0.82
1.03
lAO
1.91
2.11
2.42
2.44
2.74
2.91
3.53
4.39
5.21
5.24
5.55
Residual
-1.58
0.57
-!l.23
0.36
1.20
1.36
1.28
0041
0.19
-0.46
-!l.50
-!l.73
-1.69
-1.99
-1.00
1.48
0.02
1.37
14 Solutions: Chapter 3
These results are plotted below.
3.-_-,.__.,-_-,__.,...__,-_-,
2
1
·2
61 2 3 4 5
Decrease in medium conductivity (mS em-1)
·3 '-_-'-__-'-_---'__......__'-_-'
o
The residuals are not randomly distributed: they are mainly positive at low values of decrease in medium
conductivity, then negative, then positive again. Therefore, the straight line fit of the data cannot be considered a very
good one.
3.7 Discriminating between rival models
(aJ
The results are plotted using linear coordinates below.
••
•
•
0.11
~
'"
0.10
.§.
os' 0.09
.~
~ 0.08
"
0.07
."
'E
1l. 0.06,
•~
~ 0.05
::J
0.04
0.00 0.02 0.04 0.06 0.08 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 the
residuals between the measured values for liquid superficial velocity and the values for y obtained from the above
equation is 8.4 x 10-5,
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. the
data should give a straight line when plotted on log-log coordinates.0.01
om
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 the
residuals between the measured values for liquid superficial velocity and the values obtained from the above equation
is 4.2 x 10-5.
«)
The non·linear model is better because the sum of squares of the residuals is smaller.
3.8 Non-linear model: calculation of parameters
(a), (b)
The proposed model equation has the same form as Eq; (3.15). Therefore, if the model is suitable, the data should
give a straight line when plotted on semi-logarithmic coordinates.
10' .,.--.,...-....,--,----,----,,--...---.
103
101
35302515 20
Time (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
Solutions: Chapter 3
where y is the number of viable cells and x is time in min. As the exponent must be dimensionless to preserve
dimensional homogeneity (p 12), the dimensions of kd are T-t; therefore k(! =. 0353 min-I, The dimensions of No are
the same as N, i.e, No is dimensionless and equal to 2.13 x 104.
Answer: kd =. 0.353 min-I. No =. 2.13 x 104; the dimensions of ka. are T"l, No is dimensionless
Material Balances
4.1 Cell concentration using membranes
L Assemble
(i) Flow sheet
Buffer solution in
80 kg min-1
Cell concentrate
6% bacteria
I I
I I
I Hollow-fibre membranes I
I I
I I
Buffer solution out
Fennentation broth
350 kg min-1
1% bacteria
99% water
(li) System boundary
The system boundary is shown on the flow sheet.
(iii) Reaction equation
No reaction occurs.
2. Analyse
(i) Assumptions
- steady state
-no leaks
- only water passes across the membrane
(ii) Extra data
No extra data are required.
(ill) Basis
1 min, or 350 kg fennentation broth
(iv) Compounds involved in reaction
No compounds are involved in reaction.
(v) Mass-balance equation
As there is no reaction, the appropriate mass-balance equation is Eq. (4.3):
mass in = mass out
3. Calculate
(i) Calculation table
The calculation table below shows all given quantities in kg. The total mass of cell concentrate is denoted c; the total
mass of buffer out is denoted B. The columns for water refer to water originating in the fermentation broth.
18 Solutions: Chapter 4
In Out
Stream
Water Bacteria Buffer Total Water Bacteria Buffer Total
Fermentation 346.5 3.5 0 350
broth
Buffer solution in 0 0 80 80
Cell concentrate ? O.06C 0 C
Buffer solution ? 0 80 B
out
Total 346.5 3.5 80 430 ? 0.06C 80 C+B
(li) Mass-balance calculations
Bacteria balance
3.5 kg bacteria in "'" 0.06 C kg bacteria out
C = 58.3 kg
Total mass balance
430 kg total mass in = (C +B) kg total mass out
Using the result for C:
B "'" 371.7 kg
Water balance
346.5 kg water in = water out
Water out = 346.5 kg
These calculations allow completion of the mass~balance table with all quantities in kg.
Stream In Out
Water Bacteria 814fer Total Water Bacteria Buffer Total
Fennentation 346.5 3.5 0 350
broth
Buffer solution in 0 0 80 80
Cell concentrate 54.8 3.50 0 58.3
Buffer solution 291.7 0 80 371.7
out
Total 346.5 3.5 80 430 346.5 3.50 80 430
(iii) Check the results
All columns and rows of the completed table add up correctly.
4. Finalise
(a)
After rounding to three significant figures, the total flow rate of buffer solution out of the annulus is 372 kg min-I.
Answer: 372 kg min,t
(b)
The total flow rate of cell concentrate from the membrane tubes is 58.3 kg min-I.
Answer. 58.3 kg min· I
Solutions: Chapter 4 19
4.2 Membrane reactor
t. Assemble
(i) Flow sheet
Product
Aqueous residue
0.2% glucose
0.5% ethanol
ystem boundary~
- -- - --
- 1
I I
I I
I Membrane system I
I I
I I
s
Solvent
40 kg h-1
Feed
40 kg h-1
10% glucose
90% water
(li) System boundary
The 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 data
Molecular weights (Table B.t, Appendix B): glucose = 180.2
ethanol =. 46.1
CO, =44.0
(iii) Basis
1 h, or 40 kg feed solution
(iv) Compounds involved in reaction
Glucose, ethanol and carbon dioxide are involved in the reaction.
(v) Mass~balanceequations
For glucose, ethanol and carbon dioxide, the appropriate mass-balance equation is Eq. (4.2):
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 table
The calculation table below shows all given quantities in kg. The total mass of aqueous residue is denoted R; the total
mass of product out is denoted p; the total mass of carbon dioxide out is denoted G.
20 Solutions: Chapter 4
Stream In Out
Glucose Ethanol CO, Solvent H,O Total Glucose Ethotwl CO, Solvent H2O Total
-------
Feed 4 0 0 0 36 40
Solvent 0 0 0 40 0 40
Aqueous 0_002R 0.005 R 0 0 ? R
residue
Product 0 ? 0 ? 0 P
Off-gas 0 0 ? 0 0 G
Total 4 0 0 40 36 80 0.002 R ? ? ? ? R+P
(ii) Mass-balance calculations
Solvent balance
Solvent is a tie component.
40 kg solvent in ::::: solvent out
Solvent out ::::: 40 kg
Water balance
Water is a tie component.
36 kg water in ::::: water out
Water out::::: 36 kg
As water appears on the Out side of the table only in the aqueous residue stream:
0.002 R + 0.005 R + 36 kg ::::: R
R = 36.254 kg
Therefore, the residual glucose in the aqueous residue stream.::::: 0.002 R ::::: 0.073 kg; the ethanol in the aqueous residue
stream::::: 0.005 R::::: 0.181 kg.
Glucose balance
4 kg glucose in + 0 kg glucose generated ::::: 0.073 kg glucose out +glucose consumed
Glucose consumed ::::: 3.927 kg
Converting the glucose consumed to molar terms:
1
1 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 ethanol
and 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 balance
okg 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:
Solutions: Chapter 4 21
Ethanol out in the product stream = (2.010 - 0.181) kg = 1.829 kg
As the product stream consists of ethanol and solvent only:
P = (1.829 + 40) kg = 41.829 kg
These calculations allow completion of the mass~balance table with all quantities in kg.
Stream In Out
Glucose Ethanol CO, Solvent H,O Total Glucose Ethanol CO2 Solvent H,O Total
Feed 4 0 0 0 36 40
Solvent 0 0 0 40 0 40
Aqueous - 0.073 0.181 0 0 36 36.254
residue
Product 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 results
All columns and rows of the completed table add up correctly.
4. Finalise
(0)
1.829 kg ethanol are contained in 41.829 kg of product stream. The ethanol concentration is therefore 1.829/41.829 x
100%:::: 4.4%.
Answer: 4.4%
(h)
The mass flow rate ofCOz is 1.92 kg b-1.
Answer: 1.92 kg h~l
4.3 Ethanol distillation
1. Assemble
(i) Flow sheet
Bottoms
Distillate
5,000 kg h-1
45% ethanol
55% water
System boUnda'l'\.
r -t---,
I II I
I I
I Distillation I
I column I
kg h-l
thanal I I
atar I I
I I
I I
- -
-+- - -
Feed
50,000
10%e
90%w
22
(li) System boundary
The system boundary is shown on the flow sheet.
(ill) Reaction equation
No reaction occurs.
2. Analyse
(i) Assumptions
- steady state
-no leaks
(ll) Extra data
No extra data are required.
(ill) Basis
1 h, or 50,000 kg feed
(Iv) Compounds involved in reaction
No compounds are involved in reaction.
(v) Mass¥balance equation
As there is no reaction, the appropriate mass-balance equation is Eq. (4.3):
mass in ::::: mass out
3. Calculate
(i) Calculation table
The calculation table shows all given quantities in kg.
SolutilJns: Chapter 4
Stream
Feed
Distillate
Bottoms
Total
Ethanol
5,000
5,000
Water
45,000
45,000
In
Total
50,000
50,000
Out
Ethanol Water Total
2,250 2,750 5,000
? ? ?
? ? ?
(li) Mass-balance calculations
Total mass balance
50,000 kg total mass in ::::: total mass out
Total mass out ::::: 50,000 kg
Therefore, from the total column on the Out side of the table:
Bottoms out::::: (50,000 - 5,000) kg ::::: 45,000 kg
Ethanol balance
5,000 kg ethanol in ::::: ethanol out
Ethanol out ::::: 5,000 kg
From the ethanol column of the Out side of the table:
Ethanol out in the bottoms := (5,000..,... 2,250) kg := 2,750 kg
Water balance
45,000 kg water in := water out
Water out := 45,000 kg
From the water column of the Out side of the table:
Water out in the bottoms == (45,000 - 2,750) kg == 42,250 kg
These calculations allow completion of the mass~balance table with all quantities in kg.
Solutions: Chapter4 23
Stream
Feed
Distillate
Bottoms
Total
Ethanol
5,000
5,000
Water
45,000
45,000
In Out
Total Ethanol Water Total
50,000
2,250 2,750 5,000
2,750 42,250 45,000
50,000 5,000 45,000 50,000
(ill) Check the results
All columns and rows of the completed table add up correctly.
4. Finalise
(a)
The bottoms contains 2,750 kg ctha,")l and 42,250 kg water in a total of 45,000 kg. Therefore, the composition is
2,750/45,000 x 100% = 6.1 % ethanol. and 42,250/45,000 x 100% = 93.9% water.
Answer: 6.1% ethanol, 93.9% water
(b)
Directly from the table, the rate of alcohol loss in the bottoms is 2,750 kg h-I.
Answer: 2,750 kg h-I
4.4 Removal of glucose from dried egg
1. Assemble
(i) Flow sheet
Off-gas
Enzyme reactor
--t- -
System boundary
r\.
I
I
1Egg slurry -----L.o-l
3000 kg h·1 I
2% glucose
20% water I
78% egg solids I
-------,
1
I
I--.L1---_- Product
I 0.2% glucose
1
I
Inlet air
18 kg h-1 02
(n) System boundary
The system boundary is shown on the flow sheet
(ill) Reaction equation
2. Analyse
(i) Assumptions
- steady state
-no leaks
- air and off-gas are dry
- gases are at low pressure so vol% = mol%
- H202 remains in the liquid phase
24 Solutions: Chapter 4
(ti) Extra data
Molecular weights (Table B.I, A.ppendix B): glucose::::: 180.2
02=32.0
NZ::::: 28.0
H20 =IS.0
gluconic acid::::: 196.2
HZOz::::: 34.0
Composition of air (p 17): 21% 02, 79% Nz by volume
(ill) Basis
1 h, or 3000 kg egg slurry
(iv) Compounds involved in reaction
Glucose, 02, water, gluconic acid and HzOz are involved in the reaction.
(v) Mass-balance equations
For glucose, 02, water, gluconic acid and HZOz,the appropriate mass~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 table
The mass of NZ accompanying 18 kg 02 in air can be calculated from the known composition of air. Converting 18
kg 02 to molar units:
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:
1
2S.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 mass
of product is denoted P. The In side of the mass-balance table is complete,
Solutions: Chapter 4
59.30 kg NZ in ::::: NZ out
NZ out::::: 59,30 kg
Glucose balance
60 kg glucose in + 0 kg glucose generated ::::: 0.002 P kg glucose out + glucose consumed
Glucose consumed ::::: (60 - 0.002 P) kg
Converting the glucose consumed to molar tenus:
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.:e
18 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 balance
3077.3 kg total mass in ::::: (G + P) kg total mass out
Substituting the expression for G into the total mass balance:
3077.3 kg ::::: (66.64 +3552 x 10-4P + 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
Solutions: Chapter 4
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 balance
okg 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 Out
Glucose Water Egg solids 02 N2 Gluconic acid H2O, Total
Egg slurry -
All-
Off-gas 0 0 0 8.41 59.30 0 0 67.71
Product 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 results
All columns and rows of the completed table add up correctly to within round-off error.
4. Finalise
(a)
To determine which is the limiting substrate, the number ofmoles available of each substrate involved in the reaction
must be determined. From the mass-balance table for streams in:
I1 kg mOllMoles glucose = 60 kg. 180.2 kg = 0.333 kgmol
1
1kgmOIlMoleswater = 600 kg . 18.0kg = 33.3kgmol
1
1kgmOIlMoles 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 molar
quantity, the extent of the reaction must be limited by glucose.
Answer: Glucose
(b)
Water and 02 are available in excess. As only 0.333 kgmol of each will be used if the reaction proceeds to
completion, from Eq, (2.34);
(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
Solutions: Chapter 4 27
(c)
From the completed mass~balance table. the reactor off-gas contains 8.41 kg Oz and 59.30 kg N2. As gas
compositions are normallyexpressed in molar or volumetric terms (p 17), these mass values must be converted to
moles:
1
1kgmOll8.41 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 of
off-gas is (0.263 + 2.118) =2.381 kgmoJ. Therefore, the composition of the off-gas is 0.263/2.381 =0.11 02, and
2.118/2.381 = 0.89 N,.
Answer: 0.11 Oz, 0.89 Nz
(d)
From the completed mass-balance table, the product stream has a total mass of 3009.6 kg and contains 6.02 kg
glucose, 594.6 kg water, 2340 kg egg solids, 58.86 kg gluconic acid and 10.20 kg H202. Therefore, the composition
is:
6.02 0002 13009,6 =. g ucose
594.6
3009.6 =0.198 water
2340 .
3009.6 = 0.778 egg solids
58.86 0020 I . ·d3009.6 =. g UCONC act
10.20
3009.6 =0.003 H20 ,
Answer. 0.002 glucose. 0.198 water. 0.778 egg solids. 0.020 gluconic acid, 0.003 H202
4.5 Azeotropic distillation
1. Assemble
(0 Flow sheet
,...-----------------..,
--1---
--j-System boundary"r-
1
1
Feed -----r-~
95% ethanol I
5% water I
1
Benzene feed ------'-1-ol
1
1
1
(ii) System boundary
The system boundary is shown on the flow sheet
Distillation
tower
.,
1
1
1
1
1
1
1
1
1
Overhead
74.1% benzene
18.5% ethanol
7.4"1" H20
Product
100% ethanol
28
(iii) Reaction equation
No reaction occurs.
2. Analyse
(i) Assumptions
- steady state
-no leaks
(li) Extra data
lOOOcm3 =11
l000g=lkg
(iii) Basis
2501 ethanol product
(iv) Compounds involved in reaction
No compounds are involved in reaction.
(v) Mass~balanceequation
As there is no reaction, the appropriate mass-balance equation is Eq. (4.3):
mass in "'" mass out
Solutions: Chapter 4
3. Calculate
(i) Calculation table
As all quantities in mass-balance calculations must be masses (rather-than vo!umes), 250 l-absolute ethanol must first
be converted to mass. From the deflnition of density on p 16, mass is equal to volume multiplied by density:
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 of
benzene feed in is denoted B; the total mass of overhead out is denoted v.
Stream
Feed
Benzene feed
Product
Overhead
Total
Ethanol
0.95 F
o
0.95F
Water
0.05F
o
0.05F
In
Benzene
o
B
B
Total
F
B
F+B
Out
Ethanol Water Benzene Total
196.25 0 0 196.25
0.185 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 calculations
Ethanol balance
0.95 F kg ethanol in =. (196.25 + 0.185 V) kg ethanol out
F = (206.58 + 0.195 V) kg
Benzene balance
B kg benzene in =. 0.741 Vkg benzene out
B =. 0.741 V
Total mass balance
(F +B) kg total mass in =. (196.25 + V) kg total mass out
Substituting for F and B from the ethanol and benzene balances:
(206.58 + 0.195 V + 0.741 V) kg = (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
Solutions: Chapter 4 29
B = 119.6kg
These calculations allow completion of the mass-balance table with all quantities in kg.
Stream In Out
Ethanol Water Benzene Total Ethanol Water Benzene Total
F,ed 226.2 11.9 0 238.1
Benzene feed 0 0 119.6 119.6
Product 196.25 0 0 196.25
Overhead 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 results
All columns audrows of the completed table add up correctly to within round-off error.
4. Finalise
Prom the completed mass-balance table, the mass of benzene required is 119.6 kg. Using the definition of density on
p 16, volume is equal to mass divided by density:
119.6 kg 11000g li 11 1Volumeofbenzene"'" -3' lk . 3 "'" 1371
O.872gcm g l000cm
Answer: 137 titres
4.6 Culture of plant roots
1. Assemble
(i) Flow sheet
Off-gas
47 1m-as 02
15 litres CO2
Drained liquid
1110g
0.063% glucose
1.7% ammonia
1
J
Air-driven
reactor
----l
1
1
I
1
I
1
1
System boundary )..
Medium -----'-1_-I
1425 9 1
3% glucose
1.75% ammonia I
95.25% water I
1
1
I
1L _
Air
22 cm3 milr1 10r 10 d
25°C, 1 atm
(n) System boundary
The system boundary is shown on the flow sheet.
Moles of air in = n = p V =RT
30 Solutions: Chapter 4
(iii) Reaction equation
From Table B.2 (Appendix B), the molecular formula for glucose is C6H1206. The reaction equation is based on the
general stoichiometric equation for aerobic growth, Eq. (4.4):
C6H1206+aOZ+bNH3 ~ cCHaOpNo+dCOz+eHzO
2. Analyse
(i) Assumptions
- steady state
-no leaks
- air and off~gas are dry
- all the COz produced leaves in the off~gas
- gases are at low pressure so vol% = mol%
(li) Extra data
11= lOOOcm3
Molecular weights (Table B.1, Appendix B): glucose = 180.2
02 = 32.0
NZ = 28.0
NH3 = 17.0
cOz =44.0
HzO= 18.0
Composition of air (p 17): 21% OZ, 79% NZ by volume
Ideal gas constant (Table 2.5): R = 82.057 cm3 atm K~l gmol~I
(iii) Basis
10 d. or 1425 g nutrient medium
(iv) Compounds involved in reaction
Glucose, OZ' NH3. biomass, COZ and HZO are involved in the reaction.
(v) Mass-balance equations
For glucose. 02, NH3, biomass, C02 and HzO, the appropriate mass-balance equation is Eq. (4.2):
mass in + mass generated = mass out + mass consumed
For NZ and total mass, the appropriate mass-balance equation is Eq. (4.3):
mass in = mass out
3. Calculate
(i) Calculation table
Over 10 d. the volume of air sparged into the fennenter is:
Volumeofairin = 22cm3min-Ix10d.16~~nl.12:dhl = 3.168x105 cm3
Converting tbis gas volume to moles using the ideal gas law, Eq. (2.32), with the temperature converted from <>C to
degrees Kelvin using Eq. (2.24):
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 of
NZ 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 moles
using the ideal gas law. Eq. (2.32), with the temperature converted from °c to degrees Kelvin using Eq. (2.24):
Solutions: Chapter 4
. 11000em31V 1atm(47litres). 11
Mo1esofOz out::: n::: L::: ::: 1.92gmolR T 82.057cm3atmK-1gmor1(25+273.15) K
( . )I1000 em'IV 1atm 15litres. 11
Moles 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. The
total mass of off~gas out is denoted G; the total biomassbarvested is denoted R As the ratio of biomass fresh weight
to dry weight is 14:1, dry biomass comprises l!IS ::: 0.0667 of the total biomass. Because this problem requires an
integral mass balance, the biomass remaining in the fermenter after 10 d culture must also be included in the table
even though it is not contained in any of the streams flowing into or out of the vesseL
Using this result and adding up the row for the off*gas in the out table:
G =61.44 + 286.4 + 26.97 = 374.81 g
Total mass balance
1798.44gtotalmassin::: (1110+G+B)gtotalmassout
Using the result for G:
32
B =: 313.63 g
Solutions: Chapter 4
Therefore, the dry biomass produced is 0.0667 x 313.63 =20.92 g; the mass of water in the biomass is 0.9333 X
313.63 = 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 evaluationof the masses of components consumed 'or generated in the
reaction.
Glucose balance
42.75 g glucose in + 0 g glucose generated = 0.699 g glucose out + glucose consumed
Glucose consumed = 42.05 g
02 balance
87.04 g 02 in +0 g 02 generated = 61.44 g 02 out + 02 consumed
02; consumed = 25.60 g
NH3 balance
24.94 g NH3 in +0 g NH3 generated = 18,87 g NH3 out + NH3 consumed
NH3 consumed = 6.07 g
C02 balance
og C02 in + C02 generated = 26.97 g C02 out + 0 g C02 consumed
C02 generated =26.97 g
H20 balance
1357.31 g H20 in +H20 generated = (1090.43 + 0.9333 B) g H20 out + 0 g H20 consumed
Substituting the value for B from the total mass balance:
H20 generated = 25.83 g
(iii) Check the results
All columns and row~ of the completed mass-balance table add up correctly.
4. Finalise
(a)
Rounding to three significant figures from the completed mass~balance table, the mass of dry roots produced is 20.9 g.
Answer: 20.9 g
(b)
To determine the stoichiometry, the calculated masses of components consumed or generated in the reaction must be
converted to molar quantities:
Solutions: Chapter 4
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 g
10mass
33
The moles of biomass generated is not yet known explicitly because the molecular formula for the dry biomass is
unknown. The above molar quantities can be used as coefficients in the reaction equation:
20.920.233 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:
89.79
CoH1206+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' +2.63lomass
Therefore:
89:79 = 3.37
MWblomass
This result can be used in the remaining elemental balances for completion of the stoichiometric equation.
Hbalance: 12 + 3 x 1.53 = 3.37 a+ 2 x 6.16 --) a = 1.27
Obalance:6+2x3.43 = 3.37/3+2x2.63+6.l6 --) /3 = 0.43
N 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.
(cj
Converting to moles the mass quantities of glucose, 0z and NH3 available for reaction on the In side of the mass-
balance 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, glucose
must be the limiting substrate.
Answer: Glucose
34 Solutions: Chapter 4
(d)
The mass of glucose consumed is 42.05 g; the mass ofdry biomass produced is 20.92 g. Therefore, the biomass yield
from glucose is 20.92/42.05 = 0.50 g g-1 dry weight.
Answer: 0.50 g g-1 dry weight
4.7 Oxygen requirement for growth on glycerol
From TableB.2 (Appendix B), the molecular formula for glycerol is C3Hg03. From Table 4.3, the chemical formula
for Klebsiella aerogenes can be taken as CH1. 7S00.43NO.22. Substituting these formulae into the general
stoichiometric 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 from
the atomic weights in Table B.l (Appendix B) is 23.74. Taking into account the 8% ash:
Biomassmolecularweight = zg;: = 25.8
The value of the stoichiometric coefficient 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 of
reduction 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; from
Eq. (4.13). w = 3 for glycerol:
a = 1/4 (w ~ - c 1B) = 1/4 (3 x 4.67 - 1.43 x 4.23) = 1.99
Therefore. 1.99 gmo! oxygen are required per grool glycerol. From the atomic weights in Table B.l (Appendix B),
the molecular weight of oxygen is 32.0. Converting a to mass terms:
a = 1.99gmolgmol-1 = L99gmOlgmOI-I.I::~II.I~~~11 = 0.69gg-1
Answer: 0.69 g oxygen is required per g glycerol conswned
4.8 Product yield In anaerobic digestion
From Eq. (4.13), the stoichiometric equation for anaerobic growth and product fonnation by methane bacteria can be
written as:
0I3COOH+ 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), the
molecular weight of CO2 is 44.0. The value of the stoichiometric coefficient d can be determined based on Eq. (4.14)
with carbon dioxide as the product and the yield YpS = 0.67 kg kg-1 = 0.67 g g-l:
d = Yps(MWsubstral~ = 0.67gg-I (60.1ggmor1j = 0.915 gmol gmol-l
MWC02 44.0ggmol I
The other coefficients can be determined using this result and elemental balances.
Cbalance:2 = c+d+/= c+0.915+f--j.f= L085-c
Hbalance:4+3b = 1.4c+2e+4f
°balance: 2 = 0.40c+2d+e = OAOc+2xO.915+e = OAOc+L83+e --j. e = O.17-0.40c
N balance: b = 0.20 c
Solutions: Chapter 4
Substituting the expressions fort, e and b from the C, 0 and N balances, respectively, into the H balance:
4+3x0.20c = l.4c+2x(0.17-0.40c)+4x(l.085-c)
4 c = 0.680
C : 0.170
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) -1
fmax = )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
(aJ
From Table B.2 (Appendix B), the .molecular formula for glucose isC6H1206. If all carbon in the substrate is
converted into biomass, production of carbon dioxide is zero. Therefore, from Eq. (4.4), the stoichiometric equation
for anaerobic growth of Cellulomonas is:
C~1206 +b NH3 -+ c CH1.5iPO.~O.16+ e H20
The stoichiometric coefficients can be determined using elemental balances.
C balance: 6 = c
Hbalance:12+3b: 1.56c+2e
o balance: 6 =0.54 c + e
N balance: b : 0.16 c
Substituting the value for c from the C balance ,into the °and N balances gives e = 2.76 and b = 0.96. respectively.
The yield of biomass from substrate in molar terms is therefore 6 gmol gmol-l.
Using the atomic weights in Table B.l (Appendix B), the molecular weight of glucose is 180.2. The biomass formula
weight calculated from Table B.l (Appendix B) is 24.46. Taking into account the 5% ash:
Biomass molecularweight = 2;9~6 = 25.75
Therefore, in mass terms, the molar biomass yield of 6 gmol gmol-l= (6 x 25.75) g biomass per 180.2 g substrate:
0.86 g g-l.
The maximum possible biomass yield is calculated using Eq. (4.19). From Eq. (4.13), W = 6 for glucose. From Table
B.2 (Appendix B). the degree ofreduclion of glucose relative to NH3 isrs = 4.00. The degree of reduction of the
biomass relative to NH3is:
I x4+ 1.56 x 1-0.54x2-0.16x3 00JB = = 4.
Substituting these values into Eq. (4.19):
wYs 6(4.00) -1
croax = 1i3 = 4.00 = 6.0 gmol gmol
The theoretical maximum biomass yield is therefore the same as the actual biomass yield.
36 Solutions: Chapter 4
Answer: The biomass yield from substrate of 0.86 g g-t is 100% of the theoretical maximum. When there is no
product formo.tion and no oxygen for electron transfer, all the available electrons from the substrate must go to the
biomass.
(b)
(I)
From Table B.2 (Appendix B), the molecular formula for methanol is CH40 and the degree of reduction relative to
NH3 is 1S =6.00. The degree ofreduction of Methylophilus methylotrophus biomass relative to NH3 is:
_ lx4+1.68xl-0.36x2-0.22x3 -430
JB- 1 -.
From Eq. (4.13), W "" 1 for methanol. Substituting'values into Eq. (4.19);
W JS 1 (6.00) 140 1 1-1
emu"" Jh" "" 4.30 "" . gIno gmo
From Table B.g (Appendix B). the molecular weight of methanol is 32.0. The biomass formula weight calculated
from the atomic weights in Table B.I (Appendix B) is 22.55. With 6% ash:
Biomassmolecularweight = ~:: =23.99
In mass terms, the maximum possible molar biomass yield of lAO gmol gmol·l is equal to (1.40 x 23.99) g biomass
per 32,0 g substrate = 1.05 g g-l,
Answer: 1he maximum possible biomass yield from methanol is 1.05 g g-}, In terms of Catoms,.the biomass yield is
1.40 grool grool-l as both biomass and substrate have 1 C atom each. In comparison, the C*atQm biomass yield from
glucose in (a) is 1 grool grool·}. The main reason for the increased yield in (b) is the high degree of reduction of
methanol compared with glucose.
(il)
The actual yield ofbiomass from methanol is c = OA2 x lAO gmol gmol·} = 0.59. The ox.ygen demand can be
determined from Eq. (4.16) if biomass remains the only major product so thatf= O. Using the parameter values
determined in (b) (i):
Therefore, 0.87 gruol ox.ygen is required r:.r gmol methanol. As the molecular weights of methanol and oxygen are
the same, the oxygen demand is 0.87 g g. methanol.
Answer. 0.87 g oxygen is required per g methanol consumed
4.10 Ethanol production by yeast and bacteria
(a)
From Table B.2 (Appendix B), the molecular formula for glucose is C6H1206 and the molecular formula for ethanol
is C2H60. From Eq> (4.13), the stoichiometric equation for anaerobic growth and product formation is:
CJI}P6+bNH3 -+ cCHl.sO05NO.2+dC02+eH20+fC2H60
From Table B.8 (Appendix B), the molecular weight of ethanol is 46.1. Using the atomic weights in Table B.l
(Appendix B), the molecular weight of glucose is 180.2 and the biomass molecular weight is 24.6. The values of the
stoichiometric coefficients c can be determined from Eq. (4.12) and the yields Yxs -= O.ll g g.1 for yeast and Yxs =
0.05 g g.1 for bacteria.
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-' (I80.2ggmor
1) 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.
Solutions: Chapter 4
Yeast
Cbalance:6 = c+d+2j= 0.81+d+2j~d = 5.19-2j
Hbalance: l2+3b = 1.8c+2e+6j= 1.8xO.81+2e+6j~1O.54+3b = 2e+6j
o balance: 6 = 0.sc+2d+e+f= 0.5xO.81+2d+e+f~5.595 = 2d+e+j
Nbalance:b = 0.2e = O.2xO.81 = 0.16
Substituting the expression for b from the N balance into the H balances gives:
10.54+3xO.16 = 2e+6f
e=5.51-3f
Substituting this and the expression for d from the C balance into the 0 balance gives:
\ ....•.. "
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
Bacteria
C balance: 6 = c+d+2f= 0.37+d+2f~d = 5.63-2f
Hbalance:12+3b = 1.8c+2e+6f= 1.8x037+2e+6f~11.33+3b = 2e+6f
o balance: 6 = 0.5c+2d+e+f= O.sxO.37+2d+e+f~5.815 = 2d+e+!
N balance: b = 0.2 c = 0.2 x 0.37 = 0.074
Using the same solution procedure as for yeast, substituting the expression for b from the N balance into the H
balances gives:
11.33 +3xO.074 = 2e+6j
e=5.78-3f
Substituting this and the expression for d from the C balance into the 0 balance gives:
5.815 = 2x(5.63-2j)+(5.78-3j)+f
6f= 11.225
f = 1.87
Therefore, for bacteria. the yield of ethanol from glucose is 1.87 grool grool'}.
Answer~ 1.72 grool grool"} for yeast; 1.87 grool grool,l for bacteria
(b)
The maximum possible ethanol yield can be calculated using Eq. (4.20). From Eq. (4.13), w =6 for glucose andj =2
for ethanol. From Table B.2(Appendix B), the degree of reduction of glucose relative toNH3 is rs =4.00, and the
degree of reduction of ethanol relative to NH3 is »= 6.00. Using these values in Eq. (4.20) gives:
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; for
bactena, the actual yield of 1.87 grool grool' represents 94% of the theoretical maxunum.
Answer: 86% of the theoretical maximum for yeast; 94% of the theoretical maximum for bacteria
4.11 Detecting unknown prodncts
From Table B.2 (Appendix B), the molecular formula for glucose is C6H1206. Assuming that no products other than
biomass are formed, from Eq. (4.4), the stoichiometric equation for growth is~
38 Solutions: Chapter 4
C~1206+a02+bNH3 ~ cCHl.7~0.56NO.l7+dC02+eH:f)
Using the atomic weights in Table B.l (Appendix B), the molecular weight of glucose is 180.2. the molecular weight
of oxygen is 32.0, and the biomass molecular weight is 25.16. The value of the stoichiometric coefficient c can be
determined from the yield Yxs =0.37 g g-l andEq. (4.12):
Yxs(MWsubstrat~ 0.37gg-1(180.2ggmor
'
j -1
c = MW II = 1 = 2.65gmolgmol
ce s 25.16ggmor
Therefore, 2.65 groat cells are produced per gmol glucose consumed. Converting the oxygen demand to a molar
basis:
O.88g02125.16gcellslllgmOlOZI -1
0.88 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 of
reduction of the biomass relative to NH3 is:
_ lx4+1.79xl-0.56x2-0.17x3 -416
~- - .
If no products are formed other than biomass, the theoretical oxygen demand can be determined from Eq. (4.16) with!
=0; from Eq. (4.13), w = 6 for glucose:
a = 1/4(w1!l-clll) = 1/4 (6x 4.00-2.65 x 4.16) = 3.24
As the theoretical oxygen demand is significantly higher than that observed, formation of other products acting as
electron acceptors is likely to have occurred in the culture.
Answer: Yes
4.12 Medium formulation
Using the atomic weights'in Table B.l (Appendix B), the molecular weight of (N14)zS04 is 132,1 and the biomass
molecular weight is 26.16. Using a basis of 1 litre, production of 25 g cells corresponds to 25/26,16 = 0.956 grool
cells. As each gmolcells contains O.25gmol N, (0.956x0.25) = 0.239gmol Nare-neededfrom the'medium for
biomass synthesis, As (NH4)zS04 is the sole N source and each gmol (Nl4)zS04contains 2gmol N, 0.239/2= 0.120
gmol (N14)zS04 is required. Multiplying this by the molecular weight, 0.120 x 132.-1 = 15.9 g (N14)zS04 are
required. The minimum concentration of (N}4)zS04 is therefore 15.9 g 1*1.
Answer: 15.9 g I-I
4.13 Oxygen demand for production of recombinant protein
(a)
Recombinant protein can be considered as a product of cell cuItureeven though it is not excreted from the cells;
assume that recombinant protein is synthesised in addition to the normal E.- coli biomass composition. From Table
4.3, the chemical formula for E. coli can be taken as CH1.7700.49N0.24; from Table B.2 (Appendix B), the molecular
formula for glucose is C6H1206' Substituting these formulae into the general stoichiometric equation for growth and
productformation, Eq. (4.13), gives:
C~1206+a02 +bNH3 -lo cCHL7,o0.4~O.24+dC02 +e H20+fCHl.s.sO0.3INO,2S
Using the atomic weights in Table B.l (Appendix B), the molecular weight of glucose is 180.2, the biomass molecular
weight is 25.00, and the recombinant protein molecular weight is 22.03.
Solutions: Chapter 4 39
Assume that the biomass yield refers to the cell mass without recombinant protein. The value of the stoichiometric
coefficient c can be determined from the yield Yxs = 0.48 g g-l and Eq, (4,12):
Yxs(MWsubs"ate) 0.48gg-1(180.2g gmol-l) 346 I I-I
c= = =. 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 of
reduction of glucose relative to NH3 is -ns =4.00; from Table 4.3," the degree of reduction 'Of E. coli relative to NH3 is
JB =4.07. The degree of reduction of the recombinant protein relative to NH3 is:
_ lx4+1.55xl-0.31x2-0.25x3 _ 418
~- - .
From Eq. (4.13), w =6 for glucose andj =1 for recombinant protein. Substituting these values into Eq. (4.16) for the
theoretical 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 determined
using an elemental balance for N as follows:
N balance: b = 0.24 c = 0.24 x 3.46 = 0.83
Therefore, in wild-type E. coli the ammonia requirement is reduced from 1.03 to 0.83 grool gmol-1 glucose, a
decrease of 19%, Eq. (4.16) for the oxygen requirement becomes:
Therefore, the oxygen demand is increased from 1.65 to 2.48 gmol gmol-l glucose, a rise of 50%.
Answer. The ammonia and oxygen requirements for wild-type E. coli are 0.83 grool and 2.48 gmol per gmol glucose,
respectively. These values represent a 19% reduction and a 50% increase, respectively, compared with the genetically
engineered strain.
4.14 Effect of growth on oxygen demand
The stoichiometric equation for acetic acid production using cell culture must include terms for growth. Based on Eq.
(4.13), the stoichiometric equation for growth and product formation is:
C2H4>+a02+bNH3 ~ cCHLSOo.sNo.2+dC02+eH20+jC2f402
From Table 8.8 (Appendix B), the molecular weights ofethanol and acetic acid are 46.1 and 60.1, respectively. From
the atomic weights in Table B.1 (Appendix B), the biomass molecular weight is 24.63. The value of the
stoichiometric coefficient c can be determined from the yield Yxs =0.14 g g~l and Eq. (4.12):
_Yx",s"(MW"","S::U,,bs::"_at-,-e) 0.14 g g-1 (46.1 g gmol-1) -1
c = = "'" 0.26groolgmol
MW cells 24.63 g gmol 1
40 Solutions: Chapter 4
The value of the stoichiometric coefficient!can be determined from the yield Yps = 0,92 g g-1 and Eq. (4.14):
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 of
reduction of acetic acid relative to NH3 is 'no =. 4.00. The degree of reduction of the biomass relative to NH3 is :
_ lx4+1.8xl-0.5x2-0.2x3 _ 420
1ll- 1 -.
From Eq. (4.13), W = 2 for ethanol and j =. 2 for acetic acid. Substituting these values into Eq. (4.16) for the
theoretical 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 gmoloxygen
per gmol glucose without groWth. Therefore, with growth, the oxygen demand for acetic acid production is increased
by31%.
Answer. The oxygen demand is increased by 31%.
Energy Balances
5.1 Sensible energy cbange
(a)
From Table B.5 (Appendix B), Cp for m~cresol between 25"'C andlOO"C is 0.551 calg-I oC-t.Thespecific enthalpy
change calculated using Eq. (SIB) is:
M = Cp aT = 0.551 cal g-IoC l (100 - 25)OC
!ill :::::: 41.3 cal g-t
Answer. 41.3 cal g-1
(b)
From Table B.5 (Appendix B). Cp for ethylene glycol betweenlO"C and 20"C can be taken as 0.569 cal g-1 "C-1.
The specific enthalpy change calculated using Eq. (5,13) is:
M = Cp aT = 0.569 cal g-l °C·l (10- 20)OC
!ih :::: -5.69 cal g-1
Answer: -5.69 cal g~l
(c)
Prom Table B.6(Appendix B), Cp for succinic acid between 15"C-and 120°C is given by the expression 0.248 +
0.00153 T. where Tis in "c and Cp is in cal got °C~l. The sensible energy change is best determined from the integral
of this equation between the limits T:::: 15"C and T:::: 1200C:
J.'''''' J,,,,,"c6h = 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 determined
by 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 the
values 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
(aJ
The heat of vaporisation of water at 85"C is obtained from Table C.I (Appendix C). Taking the average of the values
at 84"C and 86"C, !:J.hy = 2296.05 kJ kg-! at 85°C.
Answer: 2296.05 kJ kg-l
(bJ
From Table C.l (Appendix C), the enthalpy of liquid water at IOce relative to the triple point is 42.0 kJ kg- t , The
enthalpy of liquid water at 3S"C relative to the triple point can be estimated as the average of the values in Table Col
for 34<>C and 36°C = 146.55 k:J kg-t. Using the relationship on p 89, the enthalpy of water at 35°C relative to 1000e is
therefore (146.55 - 42.0) kJ kg-I"" 104.55 kJ kg-I.
Answer: 104.55 kJ kg-I
(cJ
-The enthalpy of saturated water vapour at 40°C relative to the triple point can be read directly from Table C.I
(Appendix C) as 2574.4 kJ kg-I.
Answer: 2574.4 kJ kg- l
(dJ
The enthalpy of superheated steam at 2.5 atm and 275°C relative to the triple point can be obtained from Table C.3
(Appendix C). To convert the pressure to kPa, from Table A.S (Appendix A), 1 atm "" 1.013 x lOS Pa. Therefore:
2.5 atm =2.5 atm .11.Ql;:~5pal.lli:;al=253.3kPa
From Table C.3, the enthalpy at 100 kPa and 275°C is 3024 kJ kg-I; the enthalpy at 500 kPa and 275"'C is 3013 kJ
kif1. Interpolating between these values gives an enthalpy of 3019.8 kJ kg- l at 253.3 kPa.
Answer: 3019.8 kJkg-1
5.4 Pre-heating nutrient medium
1. Assemble
(i) Units
kg, h, kJ, °C
(ii) FkJw sheet
Solutions: Chapter 5 43
System boundar;\.
r
Q (loss) =0.22 kW
1
I
I
Medium in ..l.-_~
3250 kg h-1
15°C
Heating vessel 1--'----_ Medium out
3250 kg h-1
44°C
-Q (from condensing steam)
(ill) System boundary
The system boundary is shown on the flow sbeet.
2. Analyse
(i) Assumptions
- steady state
-no leaks
- system is homogenous
- condensate temperature is 150°C
- no shaft work
(li) Basis
1 h, or 3250 kg medium in
(ill)Reference state
H = 0 for water (steam) at its triple point
H =0 for medium at 15°C
(iv) Extra data
C 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 balance
The mass balance is already complete.
(vi) Energy-balance equation
At steady state, Eq. (5.9) applies:
:E(Mh) - :E(Mh)- Q+ W, = 0
input output
""""" """""
3. Calculate
(i) ldmtify ta"" in the 'nagy-balan" ,quatian
Ws =O. There are two components for the beat term, Q: Qross and Q from the condensing steam. With symbol MD =
medium. the energy-balance equation becomes:
(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
Solutions: Chapter 5
Therefore, on the basis of 1 h, {hess "" 792 kJ. Substituting values into the energy~balance equation gives:
0-3.55 x 105 kJ-Q-792kJ "" 0
Q = -3.56 x 105 kJ
Q has a negative value which is consistent with the sign conventions outlined on pp 87-88: heat must be supplied to
the system from the surroundings. This heat is provided as the latent heat of vaporisation as saturated steam at 150<>C
condenses. The enthalpy change from this change ofphase is calculated using Eq. (5.16) and must be equal to -Q.
3.56xloSkJ "" Msteam!JJlv =Msteam(21l3,lkJkg-1)
Mstearn "" 168 kg
4. Finalise
Answer: 168 kg
5.5 Production of glutamic acid
(a)
L Assemble
(i) Units
kg, b, kJ; °C
(ii) Flow sheet
Off-gas
System boundary,~
- -- - - -- 1
I I
I I
Feed I Reactor I(25,000 lltres) Product2000 kg h-1 I I 0.5% glucose
4% glucose I / I96% water
25°C r--t- I- -- .
Q (cooling)
Inlet gas
100,000 litres min-1
1 atm, 15"C
88% air
12%NHs
(ill) System boundary
The system boundary is shown on the flow sheet.
Solutions: Chapter 5
(iv) Reaction equation
2. Analyse
(i) Assumptions
- steady state
-no leaks
- system is homogenous
- solutions are ideal
- inlet air and off-gas are dry
- all excess NH3 is dissolved in the aqueous phase
- all COz produced leaves in the off-gas
- negligible sensible heat change
- no evaporation
- no shaft work
(ii) Basis
I h, or 2000 kg feed
(ill) Reference state
H = 0 for water at its triple point
H = 0 for feed at 25°C
(iv) Extra data
Molecular'weights (Table B.t, Appendix B): glucose = 180.2
NH3 = 17.0
02=32.0
Nz = 28.0
glutamic acid = 147.1
C02 =44.0
H20 = 18.0
air = 28.8 (see Problem 2.9, Chapter 2)
Ideal gas constant (fable 2.5): R = 0.0820571 atm K-l gmol-l
Composition of air (p 17): 21% Oz, 79% Nz by volume
Heats of combustion (Table B.8, Appendix B):
Ah~ glucose = -2805.0kJ gmol-l
Ah~NH3=-382.6kJ gmol-1
Ah~ glutamic acid =-2244.1 kJ gmor1
45
(v) Compounds involved in reaction
Glucose, ,ammonia, oxygen, glutamic acid, carbon dioxide and water are involved in the reaction.
(vi) Mass-balance equations
For,glucose, ammonia, oxygen; glutamic acid. carbon dioxide and water, the appropriate mass~balanceequation,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 equation
The modified energy-balance equation, Eq. (526), applies:
-MIrxn-MyAhv-Q+ Ws = 0
3. Calculate
(i) Mass balance
As gas compositions are normally given in Yol%, 88,000 litres air and 12,000 litres NH3 enter the reactor every min.
On a basis ofl h, the volume of air in is 88,000 x 60 = 5.28 x 106 litres; the volume ofNE3 in is 12,000 x 60 = 7.20
x 105 litres. Using the known composition of air, the volume of Oz in = 0.21 x 5.28 x 106 = 1.11 x 106 Utres; the
volume ofNZ in = 0.79 x 5.28 x 106 = 4.17 x l06litres. Converting these gas volumes to moles using the ideal gas
law, Eq. (2.32), with the temperature converted from °C to degrees Kelvin using Eq. (2.24):
46 Solutions: Chapter 5
Moles of02 in = pV = 1atm(l.l1 x 1Q6t) = 4.69xI04gmol
R T 0.0820571almr' gmor1(15 + 273.15) K
MI fN · _pV _ laJln(4.17XI0'1) -176 10' Io eso 2m - - - - . x gmo
RT 0.0820'71aJlnK 19moll(I'+273.I5)K
MolesofNH in=pV = latm(7.20xt051) =3.05x104 g.mol
3 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 10
4 ~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 mass
of product is denoted P.
2000
6947.3
Total
1920
o
HZO
In
Gluconic acid C02
o 0
o 0
o
4928.0
o
1500.8
02
o
518.5
80
o
80 518.5 1500.8 4928.0 0 0 1920 8947.3
Out
Glucose NHS °z NZ Gluconic acid COz HZO Total
0 0 ? ? 0 ? 0 G
0.005P ? 0 0 ? 0 ? P
0.005P ? ? ? ? ? ? G+P
Glucose
Stream
Peed
Inlet gas
Off-gas
Product
Total __-=__~=",-_-",="-_==,--,,-- -,,,-__--,,=__-==,-_
Stream
Peed
Inlet gas
Off~gas
Product
Total
N2 balance
N2 is a tie component.
4928.0 kg N2 in = N2 out
N2 out = 4928.0 kg
Glucose balance
80 kg glucose in + 0 kg glucose generated = 0.005 P kg glucose out + glucose consumed
Glucose consumed :::: (80 - 0.005 P) kg
Converting the glucose consumed to molar terms:
Olucoseconsumed = (80-0.005 P) kg ·1 :8~:~ 1= (0.444-2.775x 10-' P)kgmol
From the reaction stoichiometry, conversion of this number of kgmol glucose requires the same number of kgmol
NH3 and 1.5 x the number of kgmo1 02. Converting these molar quantities to masses:
Solutions: Chapter 5
NH, consumed =(0.444-2.775 x 10-5 P)kgmol =(0.444-2.775 x 10-5 p) kgm01.1 ;~~~ll
NH3 consumed = (7.548-4.718 x 10-4 P) kg
02 consumed =1.5 x (0.444-2.775 x 10-5P)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 the
stoichiometry:
Glutamicacidgenerated = (0.444-2.775 x 10-5 P)kgmol = (0.444-2.775X W-5p)kgm01.1 ~4:~~~ I
Glutamic 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 balance
1500.8 kg02 in + 0 kg 02 generated = 02 out + (21.312 - 1.332 x 10-3 P) kg 02 consumed
02 out = (1500.8 - (21.312 - 1.332 x 10-' P)) kg
02 out = (1479.5 + 1.332 x 10-3 P) kg
C02 balance
okg C02 in + (19.536 - 1.221 x 10-3 P) kg C02 generated = C02 out + 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 balance
8947.3 kg total mass in = (0 + P) kg total mass out
Substituting the above expression for a into the total mass balance:
8947.3kg = (6427.0+ 1.11 x 10-4 P+ P)kg
2520.3 kg = 1.000 P kg
P = 2520.3 kg
Therefore, from the total mass balance:
G = (8947.3 - 2520.3) kg
G = 6427.0kg
Substituting the result for P into the glucose, 02 and C02 balances gives:
48
Glucose consumed = (80 - 0.005 P) kg = 67.40 kg