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
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Table of Contents Solutions Page
Chapter 2
Introduction to Engineering Calculations
1
Chapter 3
Presentation and Analysis of Data
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 $"1 1 m::: lOOcrn Therefore: 1.5
x
10-6 cP "" 1.5 x 10-6 cP
,1 10-
3 g k ;-1 1
s-ll.ll~~m I = 1.5 x 10-11 kg s-1 cm- 1
Answer: 1.5 x 10- 11 kg s-1 em- 1
(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 10"2 metric horsepower Im=lOOcm 11= lOOOcm3 Ih=60min Therefore: 670mmHgft3
= 670 mmHg ft3 .11.31~~ atmI19.604X 1O-2Btul.I°.3048m 13 llatm
1ft
2 .391 x 10-2 metric horsepower 1 1 I h 1 . -60 . 1 Btu min-1 mm 1
3
1 100 em 1
1m
(d)
From Table A.7 (Appendix A): 1 Btu =0.2520 keal From Table A.3 (Appendix A): lIb = 453.6 g
Therefore:
Answer: 0.192 kcal g-l
2.2
Unit conversion
Case 1 Convert to units of kg, m, s.
From Table A.3 (Appendix A), lIb = 0.4536 kg
.! o.2i~!Call.14;3~:
I
11 lOOOcm3
I.
. horsepower h = 956 , x 10-4 metric
Answer: 9.56 x 10-4 metric horsepower h
345 Btu Ib- 1 = 345 Btulb- 1
1
g = O.192kcal g-1
Solutions: Chapter 2
2 From Table A.2 (Appendix A): 1 ft3 ::: 2.832 x 10-2 m3 From Table A.9 (Appendix A): 1 cP::: 10-3 kg m- l s·l 1 rn= lOOcm= lOOOmm Therefore, using Eq. (2.1):
1m n(3cms-l.l~n(251bft-3I0.4536kgl.1 1ft ~ (2mm.1 l000mmU l00cmU lIb 2.832 x 10 2m3 U _ 3
Dup Re::: - p -
=
I
-11
- 2.4 x 10
-3 1 s 10-0 cP . 10 kgm 1 cP
7
Answer: 2.4 x 107
ease 2 Convert to units of kg, m, s. From Table Al (Appendix A): 1 in.::: 2.54 x 10-2 m From Table A.9 (Appendix A): 1 Ibm ft-I h- 1 ::: 4.134 x 10-4 kg mol s·l Ih=3600s Therefore, usingEq. (2.1):
13~sl
= 1.5 x 104
Answer: 1.5 x 104
2.3
Dimensionless groups and property data
From the Chemical Engineers' Handbook. the diffusivity of oxygen in water at 2S"C is 2.5 x 10- 5 cm2 s-l. Assuming this is the same at 28"C, !lJ= 2.5 x 10-5 cm2 s·l, Also, from the Chemical Engineers' Handbook, the density of water at 28"C is PL ::: 0.9962652 g cm-3, and the viscosity of water at 28"C is JlL::: 0.87 cP. 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 gruol-I. Substituting parameter values into the density equation gives: PG ""
L
""
RT
I atm
(82.057cm3 aImK:""1 gmol 1)(301.15K)
"" 4.05 x 10-5 gmolcm-3
From the atomic weights in Table B.l (Appendix B), the molecular weight of oxygen is 32.0. Converting the result for PG to mass tenns: PG "" 4.05 x
Io-5gmolcm-3 ·1 i~~!ll
3 3 "" L30xlO- gcm-
From Table A9 (Appendix A): 1 cP "" 10- 2 g cm- 1 s-l; from Table Al (Appendix A): 1 ft "" 0.3048 m = 30.48 cm. The parameter values and conversion factors, together with Db "" 2 mm "" 0.2 cm, can now be used to calculate the dimensionless groups in the equation for the Sherwood number.
3
Solutions,' Chapter 2
0,87eP. S_J1.L_
-2 gem-1 s
1'0
-11
leP
-349
c - Pc 11 - (0.9962652g em-3)(2.5 x 10 5 em2 s I) -
Therefore:
From the equation for Sh:
_Sh11_(ll.21(2.5xIO-5cm2s-I)_I·A 02 -.
kL - - - -
DlJ
.
em
10-3 ems-I
Answer: lAO x 10~3 em s·1
2.4
Mass and weight
From the definition of density on p 16, mass is equal to volume multiplied by density, Therefore:
From p 16, weight is the force with which a 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 lbf. from Eq. (2.16), I Ibj= 32,1741bm it s·2; therefore:
4 2 Weight = 2.008 x 10 lbmfts- ,
llbf 32.174Ibm ft s
2 .= 6241bf
Answer: 6241bj- 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 lbf
4 2 llbf Weight = 1.998 x 10 Ibmfts- , -----''-~I 32.I741b m ft s-2
= 6211bj
Answer: 621lbf
2.5
Dimensionless numbers
First, evaluate the units of the groups cCP J.ll k) and (D GJp):
. (e
")
Urutsof -p- :::: k
I (Btulb- '1'1) Ibh- I ft- I BIUh-1 ft-2('Pft-1tl 1
2
UnitsOf(DG) = (ftllbh- ft- = 1 J1.
Ibh1ft1
= 1
Solutions: Chapter 2
4
Therefore, these groups are dimensionless. For the equation to be dimensionally homogeneous, (hI Cp G) must also be dimensionless; the units of h must therefore cancel the units of Cp G. Unitsofh "'" unitsofCpG "'" (Btulb- 1 "p-l)(1bh- i fr 2) "'" Btu "F-l h- 1 ft- 2
The dimensions of h can be deduced from its units. From Table A.7 (Appendix A), Btu is a unit of energy with dimensions "'" L2Ml2. OF is a unit of temperature which, from Table 2.1, has the dimensional symbol E>. h is a unit of time with dimension"'" T; ft is a unit of length with dimension"", L. Therefore:
Dimensions of h = L2M'l2 e- I 1'"1 L-2 = MT"3€}-1
Answer: Units = Btu op-I h- I ft"2; dimensions =Ml3€}-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 of Di "'" L From p 11, the dimensions of rotational speed, Nj =T-1; 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 (U) is dimensionally homogeneous and therefore likely to be correct
Answer: (n)
2.7
Molar units
From the atomic weights in Table B.l (Appendix B), the molecular weight of NaOH is 40.0. (a)
FromEq. (2.19): lb moles NaOH =
20.01b = 0.50 lbmol 4O.0lblbmol 1
Answer: 0.50lbmol (b)
From Table A.3 (Appendix A): lIb
=453.6 g. 20.01b
Therefore: 453.6gl = 9072 g = 20.0Ib. 1-lib
From Eq. (2.18): gram moles NaOH
=
9072g
4O.0g gmol-
I
= 227 gmol
Answer: 227 gmol (c)
From p 16. 1 kgmol::::: 1000 gmol. Therefore. from (b):
.----
Solutions: Chapter 2
5
kg molesNaOR
1 kgmol = 227 gmol. 11000 gmol I = 0.227 kgmol
Answer: 0.227 kgmol
2.8
Density and specific gravity
(a)
From p 16, the density of water at 4°e can be taken as exactly 1 g cm·3. Therefore, for a substance with specific gravity 1.5129i~. the density at 200 e 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 B.1 (Appendix B), the molecular weight of nitric acid (RNO]) is 63.0. In 1 cm3 HN03, from Eq. (2.18) gram moles :
1.5129 g
1 : 0.0240 gmol
63.0ggmor Therefore, the molar density is 0.0240 gmol cm-3 . From the definition of specific volume on p 16: Molar specific volume :
1 1 d .: 1 : 41.67cm3 gmorl 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- 1 (ii) From the atomic weights in Table B,l (Appendix B), the molecular weight of carbon tetrachloride, CCI4, is 153.8. Using the mass flow rate from (a):
Molar flow rate : 80 g min-
1
.l :5~0~1 =
0.52gmolmin- l
Answer: 0.52 gmol min- l
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 gmol 02 and 0.79 gmol NZ From the atomic weights in Table B.l (Appendix B), the molecular weights of Oz and NZ are 3Z,0 and 28.0, respectively. The molecular weight of air is equal to the number of grams in 1 gmol: 1 gmolair
Answer: 28.8
.!
= 0.21 gm0102·1 ;~~ll + 0.79 gmol NZ ;~!ll = 28.8g
6
2.10
Solutions.' Chapter 2
Mole fraction
The molecular weights can be obtained from Table B.7 (Appendix B): water 18.0; ethanol 46.1; methanol 32.0; glycerol 92.1; acetic acid 60.1; benzaldehyde 106.1. In 100 g solution, there are 30 g water, 25 g ethanol. 15 g methanol, 12 g glycerol, 10 g acetic acid, 8 g benzaldehyde, and no other components. Therefore: Moles water = 30 g .1
~io~l! =
Moles ethanol = 25 g.1
~~~ll
Moles methanol = 15 g
1.67 gmol
= 0.54 gmol
.! ;io~ll
= 0.47 gmol
. Molesglycerol = 12g. IlgmOl1 92.1g = O.13gmol Moles acetic acid = 10g.1 Moles benzaldehyde= 8 g
~~~II
= 0.17 gmol
.1 :~o~ I = 0.08 gmol
The total number of moles is 1.67 + 0.54 + 0.47 + 0.13 + 0.17 + 0.08 = 3.06 gmol. From Eq. (2.20): Mole fraction water = Mole fraction ethanol
!:~~
= 0.55
= ~:~ = 0.18
Mole fraction methanol =
~:~~
Mole fraction glycerol =
~:~ = 0.04
Mole fraction acetic acid
= 0.15
= ~:~ = 0.06
Mole fraction benzaldehyde
= ~:: = 0.03
Answer: 0.55 water; 0.18 ethanol; 0,15 methanol; 0.04 glycerol; 0.06 acetic acid; 0.03 benzaldehyde
2.11
Temperature scales
From Eq. (2.27\ -40 = 1.8 T(0C) + 32 Tee) = -40
From Eq. (2.25), T (OR) = -40 + 459.67 T(°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
7
Solutions: Chapter 2
From Table A.5 (Appendix A): 1 psi = 6.805 x 10--2 atm. Therefore: 2 . 16.805 x 10- atm 1 Absolutepressure = 29.7psl. 1 psi = 2.02atm Answer: 29.7 psi; 2.02 atm (b) From p 19, vacuum pressure is the pressure below atmospheric. If the atmospheric pressure is 14.7 psi:
Absolutepressure = 14.7psi-3psi = 11.7psi Answer: 11.7 psi
2.13
Stoichiometry and incomplete reaction
(a) The molecular weights are calculated from Table B.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--1. Answer: 0.067 g g--1 (c)
From the atomic weights in Table B.l (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 11 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-l 1.48 K 10. 2 gmol glucose available for penicillin synthesis. At the same time, 4 g or 4 g/136.2 g gmol--l 2.94 x 10--2 gmol 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 1. 1 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 W· 3 gmol penicillin per litre. Therefore, in a lOO-litre tank:, 0.886 gmol or 0.886 gmol x 334.4 g gmol-l 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-l = 1.21 g t l 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 1. 1
Solutions: Chapter 2
8
2.14
Stoichiometry, yield and the ideal gas law
(a)
Adding up the numbers of C, H, 0 and N atoms on both sides of the equation shows that the equation is balanced. Answer: Yes (b) The molecular weights are calculated from Table B.l (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 (c)
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 gmol of cells which, from (b), is equal to 151 g cells. The maximum yield is therefore 151 g cells per (16.28 groal x 32.0 g grooI-I):: 521 g oxygen, or 0.29 g g-1.
Answer: 0.29 g g-1 (d) Production of 2.5 kg cells is equivalent to 2500 g:: 2500 g'91.5 g gmoI-l = 27.3 gmol cells. The minimum amounts of substrates are required when 100% of the hexadecane is converted according to the stoichiometric equation. (I)
From the stoichiometry, production of 27.3 gmol cells requires 27.3/1.65 3736 g = 3.74 kg hexadecane.
=16.5 gmol =16.5 gmol x 226.4 g gmol-l =
Answer: 3.74 kg (il) From the answer in (d)(i). the concentration ofhexadecane required is 3.74 kg in 3 m3, or 1.25 kg m- 3.
Answer: 1.25 kg m- 3 (ill) According to the stoichiometric equation, production of 27.3 gmol cells requires 27.3 x 16.28/1.65 269.4 gmol oxygen. As air at low pressure contains close to 21 mol% oxygen (p 11), the total moles of air required is 269.410.21 = 1282.9 gmot The volume of air required can be calculated using the ideal gas law. From Eq. (2.32):
=
V = nRT p Temperature in the ideal gas equation is absolute temperature; from Eq. (2.24):
T = (20 +273.15) K = 293.15K From Table 2.5, R "" 82.057 cm3 atm K-I gmol-l. Substituting these values into the equation for V gives:
v= Answer: 31 m 3
3 1 (1282.9gmo1)(82.057cm atmK- gmol-l)(293.15K) 1~13 = 31 3 1 atm . lOOcm m
Presentation and Analysis of Data 3.1
c~
:::
Combination of errors 0.25 mol m- 3 ±4%"= 0.25 ±O.OlOmol m- 3
CAL = 0.183 mol m- 3 ± 4% =: O.183±O.OO73molm-3 OTR = Omi 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±O.05s-1
O.067molm 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)2 + (5.45 - 5.36)2 + (5.50- 5.36)2 + (5.35 - 5.36)2 = 0.15 4-1
Answer. The standard deviation is 0.15. Note that standard error, which can be calculated from the standard deviation, is a more direct indication of the precision of a mean. (c)
x = 5.15 +2 5.45 = 5.30 Standard deviation is not appropriate for expressing the accuracy of a mean evaluated using only two samples. In this case the maximum error, Le. the difference between the mean and either of the two measured values, might be used instead. The maximum error in this example is (5.30 - 5.15) 0.15.
=
Answer: 5.30; an indication of the accuracy is ± 0.15 (d)
x = 5.15+5.45+5.50+5.35+5.15+5.45+5.50+5.35 = 5.36
Solutions: Chapter 3
10
2 (S.lS -S.36f + 2(S.4S-S.36)~ + ~ (S.50-S.36f +2(S.3S -S.36)2 = 0.14
Answer: The best estimate of optimal pH is unchanged at 5.36, but the standard deviation is slightly lower at 0.14. This example illustrates that although the standard deviation decreases as the number of measurements 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 than repeat the measurement many times.
3.3
Linear and non-linear mndels
(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 = (Y2-Y1) = 0.5-10 =-136 (x2
8-1
xl)
.
B = YI-Axi = 10-(-1.36)1 = 11.4
Answer: y = -1.36 x + 11.4 (h) Xl = 3.2; Yl 14.5 X2 = 8.9; Y2 = 38.5 A straight line plot of y versus x Ih on linear coordinates means that the data can be represented using the equation:
=
y=Axlh+B with A and B given by the equations:
A =
YrY1 = 38.5-14.S = 201 Ih_ Xllh 89Ih_32'k . x2 . .
B = YI-Axjh = 14.5-20.1 (3.2 1h ) = -21.5 Answer:y=20.1x'k -21.5 (c)
Xl=5;YI=6 X2= 1;Y2=3 A straight line plot of Ity versus xl on linear coordinates means that the data can be represented using the equation:
Ily = Ax2 +B with A and B given by the equations: A =
lin - tty} 2
2
x2-xl
B = 1/y1
Answer: I/y
=
1/3 -116 2
-3
2 =-6.9xlO
1 -5
-Ax; = 1/6-(-Mx 1O-3)
=-6.9 x 10-3 xl + 0.34
(d)
Xl=0.5;Yl=25 x2 550; Y2 2600
=
=
A straight line plot of y versus x on log-log coordinates means that the data can be represented using Eq. (3.10).
From Eqs (3.13) and (3.14),
11
Solutions: Chapter 3
A
= (lnY2- ln Yt) = ln2600-ln25 = 0.663 (lnx2-lnxI)
InB
ln550-1nO.S
= InYI-Alnxl = In 25-(0.663) In 0.5 = 3.678 B = e3.678 = 39.6
Answer:
Y= 39.6 x O.663
(eJ Xl
X2
= L5;YI =2.5 = 10; Y2 = 0.036
A straight line' plot of y,versus From Eqs (3.17) and (3.18);
X
on semi-log coordinates means that the data can be represented using Eq. (3.15).
A
= (lnY2- 1n Yl) = InO.036-1n2.5 = -0.50 (x2 - xl)
10 - 1.5
InB = InYl-AX1 = 1n2.5-(-O.50J1.5 = 1.666
B = el.666 = 5.29 Answer:
Y= 5.29 e-O.50x
3.4
Linear curve fitting
(aJ The results determined using Eqs (3.1) and (3.2) are listed below,
Sucrose concentration (g 1~1)
Mean peak area
Standard deviation
6.0 12.0 18.0 24.0 30.0
56.84 112.82 170.63 232.74 302.04
1.21 2.06
2.54 1.80 2.21
(bJ 35
-.
30
.9 25
•• 0
~• g 8
20 15
~g
10
"'
5 0
0
50
100
150
200
Peak area
250
300
350
Solutions: Chapter 3
12
(e)
The linear least squares fit of the data is: y ::: 0,098 x
+ 0,83
Answer: y "'" 0.098 x + 0.83, where y is sucrose concentration in g 1-1 and x is peak area. (d)
For x::: 209,86, the equation in (c) gives a sucrose concentration of2IA g }-1, Answer: 21.4 g 1-1
3.5
Non-linear model: calculation of parameters
(a) The proposed model equation has the general form of Eq. (3.15); therefore, if the model is suitable, a plot of a versus lIT on semi-logarithmic coordinates will give a straight line. As T in the equation is absolute temperature, "'c must first be converted to degrees Kelvin using Eq. (2.24), The data are listed and plotted below.
Temperature (0C)
Temperature (K)
IIT(K-l)
Relative mutation frequency. a
15 20 25 30 35
288.15 293.15 298.15 303.15 308.15
3.47 x 10-3 3.41 x 10-3 3.35 x leT3 3.30 x lfr·3 3.25 x 10-3
4.4 x 10-15 2.Qx 10- 14 g,6x 10-14
[" •"
3.5 x Hr 13 lAx 10"'12
10- 12
f
c .2
10- 13
lii
"E ~
=
-!l c:
10. 14
10-15 L._~~_--'__~__....I._ _~_ _....J
3,2 x 10-3
3.3x 10-3
3,4x 10-3
3.5 x 10-3
1ITemperature (K-1)
As the data give a straight line on semi-logarithmic coordinates, 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·l; therefore: E :;; (26,121 K) (8.3144 J gmol~l K*l) :;; 217,180.4 J gmol-l = 217.2 kJ gmotl
13
Solutions: Chapter 3 Answer: 217.2 kJ gmol-l (cJ From the equation in (b) for the straight line,
ao is equal to 9.66 x 1()24.
Answer: 9.66 X 1024
3.6
Linear regression: distribution of residuals
(aJ
-
16
••
, 14 !'!l c
••i'!
12
•g
10
C 0 0
~
~
••m •~
.i1
•
8
•• •
6 4 2
• •• •
•
0 0
1 2 3 4 5 Decrease in medium conductivity (mS cm-1)
6
The linear least squares fit of the data is:
y "'" 1.58 + 2.10x where y is increase in biomass concentration in g rl and xis decrease in medium conductivity in mS cm- I .
(bJ The residuals are calculated as the difference between the measured values for increase in biomass concentration and the values for y obtained from the equation in (a). Decrease in medium conductivity (mS
Residual
o
-1.58
0.12 0.31 0.41 0.82 1.03 lAO 1.91 2.11 2.42 2.74 2.91 3.53
0.57 -!l.23 0.36 1.20 1.36 1.28 OAI 0.19 -0.46 -!l.50 -!l.73 -1.69 -1.99
4.39
-1.00
5.21 5.24
1.48 0.02 1.37
2.44
5.55
Solutions: Chapter 3
14
These results are plotted below.
3 , -__- .____, -__- ,____, -____, -__-, 2 1
·2
o
1 2 3 4 5 Decrease in medium conductivity (mS cm· 1)
6
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
0.04
0.02
0.06
0.06
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,
15
Solutions: Chapter 3
(b) The proposed power law equation has the same form as Eq. (3.10). Therefore, if the power law model is suitable. the data should give a straight line when plotted on log-log coordinates.
"-
1
E
5'
.i
.il
~ 0.1
:ffi .0 t
•
~
•"
1 ::J
0.01
om
0.1
Gas superficial velocity, uG (m s-1)
The data are reasonably well fitted using a power law- model. The equation for the straight line in the plot is:
y = 0.199 x O.309 where y is liquid superficial velocity in m s-l andx is gas superficial velocity in m s-l. The sum of the squares of the residuals between the measured values for liquid superficial velocity and the values obtained from the above equation is 4.2 x 10-5.
«) The non-linear model is better because the sum of squares of the residuals is smaller.
3.8
Non-linear model: calculation of parameters
(a), (b) The proposed model equation has the same form as Eq. (3.15). Therefore, if the model is suitable, the data should give a straight line when plotted on semi-logarithmic coordinates. 104r----r---,----~--_r--_,r_--._--_.
103
101
1~L---L---~--~--~---L--~--~
o
5
10
15
20
Time (min)
25
30
35
