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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 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 k1g ;-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 1 100 em 13 1 11 I.

llatm 1ft 1m lOOOcm3

1

2.391 x 10-2

metric horsepower 1 1 I h 1 956 10-4 . h h 1 . -60 . = ,x metric orsepower

1 Btu min- mm

Answer: 9.56 x 10-4 metric horsepower h

(d) From Table A.7 (Appendix A): 1 Btu = 0.2520 keal From Table A.3 (Appendix A): lIb = 453.6 g Therefore:

345 Btu Ib-1 = 345 Btulb-1 .! o.2i~!Call.14;3~: g I = O.192kcal g-1

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

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):

Solutions: Chapter 2

(2mm.1 1m n(3cms-l.l~n(251bft-3I0.4536kgl.1 1ft3 ~

Dup l000mmU l00cmU lIb 2.832 x 10 2m3U _ 7 Re::: -p- = I -3 1 -11 - 2.4 x 10

10-0 P 10 kgm s c . 1 cP

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):

Answer: 1.5 x 104

2.3 Dimensionless groups and property data

= 1.5 x 104

13~sl

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 "" I atm "" 4.05 x 10-5 gmolcm-3 RT (82.057cm3 aImK:""1 gmol 1)(301.15K)

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 "" L30xlO-3gcm-3

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.

Solutions,' Chapter 2 3

1'0-2 -1 -11 0,87eP. gem s S_J1.L_ leP -349

c - Pc 11 - (0.9962652g em-3)(2.5 x 10 5 em2 s I) -

Therefore:

From the equation for Sh:

k _Sh11_(ll.21(2.5xIO-5cm2s-I)_I·A 10-3 -I

L---- 02 -.<tUX ems DlJ . em

Answer: lAO x 10~3 em s·1

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

From p 16, weight is the force with 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 llbf Weight = 2.008 x 10 lbmfts- , 2 .= 6241bf 32.174Ibm ft s

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):

Converting to lbf

Answer: 621lbf

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

4 2 llbf Weight = 1.998 x 10 Ibmfts- , -----''-~I = 6211bj

32.I741bm ft s-2

2.5 Dimensionless numbers First, evaluate the units of the groups cCP J.llk) and (D GJp):

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

UnitsOf(DG) = (ftllbh-1ft-

2 = 1

J1. Ibh1ft1

4 Solutions: Chapter 2

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 fr2) "'" 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. Therefore:

1

453.6gl 20.01b = 20.0Ib. -lib = 9072 g

From Eq. (2.18):

9072g gram moles NaOH = I = 227 gmol

4O.0g gmol-

Answer: 227 gmol

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

. ----

Solutions: Chapter 2 5

11 kgmol I kg molesNaOR = 227 gmol. 1000 gmol = 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)

1.5129 g gram moles : 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 d1 .: 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 = 0.21 gm0102·1 ;~~ll + 0.79 gmol NZ .! ;~!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 .1 ~io~l! = 1.67 gmol

Moles ethanol = 25 g.1 ~~~ll = 0.54 gmol

Moles methanol = 15 g .! ;io~ll = 0.47 gmol

. IlgmOl1 Molesglycerol = 12g. 92.1g = O.13gmol

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

Moles benzaldehyde= 8 g .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 = !:~~ = 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\

From Eq. (2.25),

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

2.12 Pressure scales (a)

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

T (OR) = -40 + 459.67 T(°R) = 420

T(K) = -40+273.15 T(K) = 233

Assume that the atmospheric pressure is 14.7 psi. From Eq. (2.28):

Absolute pressure = 15 psi + 14.7 psi = 29.7 psi


From Table A.5 (Appendix A): 1 psi = 6.805 x 10--2 atm. Therefore:

. 16.805 x 10-2 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)

7

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

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 = 16.5 gmol = 16.5 gmol x 226.4 g gmol-l = 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 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= (1282.9gmo1)(82.057cm3atmK-1 gmol-l)(293.15K) 1~13 = 31 3 1 atm . lOOcm m

Answer: 31 m3

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 = 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 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 of a 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


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 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 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_ lh 89Ih_32'k . x2 Xl . .

B = YI-Axjh = 14.5-20.1 (3.21h) = -21.5

'k Answer:y=20.1x -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} 1/3 -116 -3

2 2 = 2 2 =-6.9xlO x2-xl 1 -5

B = 1/y1 -Ax; = 1/6-(-Mx 1O-3)<S2) = 0.34

Answer: I/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),


Answer: Y = 39.6 x O.663

(eJ Xl = L5;YI =2.5 X2 = 10; Y2 = 0.036

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

InB = InYI-Alnxl = In 25-(0.663) In 0.5 = 3.678

B = e3.678 = 39.6

11

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);

Answer: Y = 5.29 e-O.50x

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

3.4 Linear curve fitting (aJ The results determined using Eqs (3.1) and (3.2) are listed below,

Sucrose concentration (g 1~1)

6.0 12.0 18.0 24.0 30.0

(bJ

-. .9 0 •• ~ • g 8

~ g

"'

35

30

25

20

15

10

5

0 0 50 100

Mean peak area

56.84 112.82 170.63 232.74 302.04

150 200 Peak area

250 300

Standard deviation

1.21 2.06 2.54 1.80 2.21

350


(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

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

20 293.15 3.41 x 10-3 2.Qx 10-14

25 298.15 3.35 x leT3 g,6x 10-14

30 303.15 3.30 x lfr·3 3.5 x Hr 13 35 308.15 3.25 x 10-3 lAx 10"'12

" [ 10-12

• " f c

10-13 .2 lii " E ~ = 10.14 -!l c:

10-15 L._~~_--' __ ~ __ ....I. __ ~ __ ....J

3,2 x 10-3 3.3x 10-3 3,4x 10-3 3.5 x 10-3

1 IT emperature (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


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 •• 12 i'! C 10 • g 0 0 8

~ 6 ~ •• m

4

• ~ .i1

2

0

•• •

0 1

•

2

• • •

3

•

•

4 5 Decrease in medium conductivity (mS cm-1)

The linear least squares fit of the data is:

y "'" 1.58 + 2.10x

• •

6

where y is increase in biomass concentration in g rl and xis decrease in medium conductivity in mS cm-I .

(bJ

13

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 OAI 0.19

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

1.48 0.02 1.37


These results are plotted below.

3,-__ -. ____ ,-__ -, ____ ,-____ ,-__ -,

2

1

·2

o 1 2 3 4 5 6 Decrease in medium conductivity (mS cm·1)

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

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

0.11

~

" 0.10

.§. oS' 0.09

.~ • • ~ 0.08

" 0.07 ." 'E • 1l. 0,06 , • • ~ ~ 0.05 ::J

0.04 0.00 0.02 0.04 0.06 0.06 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,

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.

1

-" E -5' .i .il ~ 0.1 :ffi .0 t • ~ " • 1 ::J

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.

104r----r---,----~--_r--_,r_--._--_.

103

101

1~L---L---~--~--~---L--~--~ o 5 10 15 20 25 30 35

Time (min)

solutions doran

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