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

SOLUTIONS MANUAL Bioprocess Engineering Principles Pauline M. Doran

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SOLUTIONS MANUAL Bioprocess Engineering Principles

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

ISBN 0 7334 15474 © Pauline M. Doran 1997

Table of Contents Solutions Page

Chapter 2

Introduction to Engineering Calculations

1

Chapter 3

Presentation and Analysis ofData

9

Chapter 4

Material Balances

17

ChapterS

Energy Balances

41

Chapter 6

Unsteady-State Material and Energy Balances

54

Chapter 7

Fluid Flow and Mixing

76

Chapter 8

Heat Transfer

86

Chapter 9

Mass Transfer

'98

Chapter 10

Unit Operations

106

Chapter 11 Homogeneous Reactions

122

Chapter 12 Heterogeneous Reactions

139

Chapter 13 Reactor Engineering

151

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

Bioprocess Engineering Principles.

Introduction to Engineering Calculations 2.1

Unit conversion

(a)

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

x

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

,1 10-

3 g k ;-1 t

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

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

(b)

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

Answer: 5.17 Btu min- 1 (e)

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

=

Therefore: 670mmHgft3

3

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

= 670 mmHg ft3 .11.316X 10-

1

I I= ,

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

3

1 100 em 1

1m

=

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

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

2.2

Unit conversion

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

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

1

11 lOOOcm3

I.

956 x 10-4 metric . horsepower h

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

1

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

Solutions: Chapter 2

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

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

Dup Re::: - p -

(2mm.1

=

1m l000mmU

-6 P

10

I

10

c .

-3

kgm 1 cP

1

-11 s

- 2.4 x 10

7

Answer: 2.4 x 107

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

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

13~sl

= 1.5 x 104

Answer: 1.5 x 104

2.3

Dimensionless groups and property data

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

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

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

L

RT

""

latm

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

"" 4.05 x 10-5 gmolcm-3

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

3

·1 i~~~ll

3 3 "" l.30XlO- gcm-

From Table A.9 (Appendix A): 1 eP "" 10- 2 g cm~l s-l; from Table A.l (Appendix A): 1 ft "" 0.3048 m "" 30.48 cm. The parameter values and conversion factors, together with Db "" 2 mm "" 0.2 em, can now be used to calculate the dimensionless groups in the equation for the Sherwood number.

3

Solutions,' Chapter 2

0,87eP. S_J1.L_

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

-349

I) -

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

Therefore:

From the equation for Sh:

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

0 ' - - - - - -.......- - - - - - ' - - - - - - - ' 0.00

0.Q1

0.02

0.03

s (mol 1-1) The equation for the straight line in the plot is slv = 1.70 + 445 s,wheres has units ofmoll-} and Slv bas units aiulin, Therefore, from Eq. (11.39), I/vmax = 445 mOlll min, so that Vmax = 2.25 x 10"3 moII-} min-I. Also from Eq. (11.39), Ktn/ vmax 1.70. Multiplying this value by the result for Vmax. K m = 3.83 x 10-3 moll- t ,

=

Answer: Vmax =2.25 X 10-3 mol 1-1 min'I; K m = 3.83 x 10-3 mol 1-1

11.5

Effect of temperature on hydrolysis of starch

(a)

The activation energy is determined from the Arrhenius equation, Eq. (11.21), with k equal to the initial fate of glucose production. According to the Arrhenius equation, a plot of k versus liT on semi-logarithmic coordinates should give a straight line.. 'The parameter values are listed and plotted below; Tis converted to degrees Kelvin using

Eq. (2.24). T('C)

T(K)

liT (K'I)

Rate, k (mmel ro w3 s' I)

20 30 40 60

293.15 303.15 313.15 333.15

3,41 x 10.3 3.30 x 10-3

0.31 0.66 1.20 6.33

3.19 x Icy3 3.00 x Icy3

Solutions: Chapter II

125

"1

10

." E

§.

"g 'g

•

1

I

~

~

'5

~

0.1 2.9

3.1

3.0

3.2

1/Temperature

3.3 X

3.4

3.5

103 (K 1) o

=

The equation for the straight line in the plot is k 1.87 x lO lD c 73OO/ T, where k has units ofmmol m-3 5"1 and T has units ofK. Therefore, from Eq. (11.21), EIR = 7300K From Table 2.5,R = 8.3144 J K~1 gmol- l = 8.3144 x 10-3 kJ Kw l gmol-l; therefore, E= 7300 K x 8.3144 x 10-3 kJ K-I gmol-l = 60.7 kJ grool-l.

Answer: 60.7 kJ gmol- l . (b) Converting 55C to degrees Kelvin using Eq. (2.24), T= 55 + 273.15 equation for k obtained in (a):

k = 1.87 x 1010 e-73001T Similarly, for T= 25"C

=328.15 K.

Substituting this value into the

= 1.87 x 1010 e-73001328.15 = 4.08 romol m-3 s·1

=25 + 273.15 = 298.15 K:

k = 1.87 x 1010 e-73OO/T = 1.87 x WiD e-73001298.15 = 0.43 mmol mw3 s·1 Therefore, the rate at 55"C is 4.08/0.43 = 9.5 times faster than at 25"C.

Answer: The reaction rateat55"C is 4.08 mmol m- 3 s·t or 9.5 times faster than the rate of 0.43 rnmol ro· 3 s·l at25"C. (e)

From Table 2.5, the value of R in the appropriate units.is 1.9872 cal K-l gmol- 1, At 55"C = 328.15 K: k

d

= 2.25 x 1027 e-4I,630IRT = 2.25x 1027 e-4l,630/(1.9872 x 328.15) = 0.42h-l

Therefore, from Eq. (11.45), the half-life of the enzyme at 55"C is: t = In2 = In2 = 1.65h h kd 0.42b 1 At 25"C

=298.15 K: kd = 2.25x 1027 e-41,63O!RT = 2.25x 1027 e-4I,630;(1.9872 x 298. 15)

and the enzyme half-life is: th - In2 _ In2 - kd - 6.87x lO-4 h -l

= l009h

= 6.87x lO-4 h-1

126

Solutions: Chapter II

Although the reaction rate is 9.5 times faster at 55 C than at 25°C, the rate of deactivation is 0.42/(6.87 x 104) =: 611 times greater. Therefore, unless there are other considerations. 25°C would probably be the more practical temperature for processing operations. Q

Answer: The enzyme half·life at 25°C is 1009 h or 611 times longer than the practical operating temperature is probably 2jl'C.

n.6

half~life

of 1.65 h at 55°C. The more

Enzyme reaction and deactivation

K m = 5 roM = 5 x 10-3 gmol t 1 = 5 x Hy6 gmol m- 3. The concentration offat is reduced from 45 gmol m~3 to 0.2 x 45 gmol m- 3 = 9,0 groat m-3: As this concentration range is well above the value of .&tn. s» K m and. from p 269. v "" vmax ' As v= -ds1dt. combining Eqs (11.35) and (11.44) gives:

-::: = VrnaxQ e-kd t In this equation. s and t are the only variables. Separating variables and integrating:

J-- 0 is all that is required to ensure maximum reaction rate.

Answer: Negligible

12.6

Immobllised.enzyme reaction kinetics

(3)

R =0.8 nun = 0.8 x 10-3 m. As external boundary~layers have been eliminated. CAs::::: CAb =0.85 kg m-3 and fie =1. The value of Pas defined on p 313 is: 3

jJ::::: Km ::::: 35kgmCAs

::::: 4.12

0.85kgm 3

From Figures 12.10-12.12, this value of p means that the reaction kinetics can be considered effectively f:trst-order. Evaluating the observable Thiele modulus tP from the equation in Table 12.4 for spherical geometry:

4>

= (li)2 3

From Figure 12.11. at

Answer: 0.12

3

3

1

rA,obs = (0.8 x 10- m)2 1.25 x 10- kg 5- m-3 ::::: 8 0 2 3 . .PAe CAs 3 1.3 x 1O-1l m s-1 (0.85 kg m- )

= 8.0. flil =0.12.

Therefore. from Eq. (12.46). TJT

=11i 11e::::: 0.12 x I ::::: 0.12.

Solutions: Chapter 12

147

(b) From the definition of the effectiveness factor. Eq. (12.26): 3 L25xtO- .kgs-I m-3 =OOlO4k -I -3 0.12 . gs m Por first~order kinetics. r~s

=kl CAs; therefore: k

l

=

• rAJ;

3

= O.OI04kg s-1 m-

= 0.0122s-1

0.85kgm-3

CAs

Answer: 0.0122 s-I (e) The value of 0.0122 s·1 for kt corresponds to an enzyme loading of 0.1 Ilmol got. The Thiele modulus 4't can be evaluated as a function of enzyme loading usingothe equation ·for first-order reaction and spherical geometry from Table 12.2. with kt·directly proportional to the' enzyme 10ad.ing,Por¢>I < 1O,.the-intemal effectiveness factor l1il is determined using the equation in Table 12.3 and the' definition·of coth x; for tf1I > 10. from Eq. (12.30) 1]u II¢!. For each value of kJ, r~s kt CAs. and rA,obs can be determined from these results and the,definition of the effectiveness factor in Eq. (12.26); Calculated values of these'parameters for several different enzyme loadings are listed below.

=

=

Enzyme loading (llJllol g~ 1)

kl (,-I)

¢I

'lit

r'"A, (kg s-t m-3)

rA,obs (kg s-t m- 3)

0.01 0.05 0.10 0.20 050 0.80 1.0 1.3

0.0012 0.0061 0.0122 0.0244 0.0610 0.0976 0.122 0.159 0.183 0.220 0.244

2.6 5.8 8.2 11.6 18.3 23.1 25.8 295 31.6 34.7 365

0.34 0.16 0.12 0.086 0.055 0.043 0.039 0.034 0.032 0.029 0.027

0.0010 0.0052 0.0104 0.021 0.052 0.083 0.104 0.135 0.156 0.187 0.207

3.40x lO4 8.32 x lO4 1.25 x Ily3 1.81 X lO-3 2.86 x lO~3 3.57 x 10~3 4.06 x lO~3 4.59 X 10-3 4,99 x lO-3 5.42 x lO-3 5.59 x to- 3

1.5 1.8 2.0

The resultsfor 11i1 and rA.obs are plotted below as a function of enzyme loading.

0.4,----,,----,----,-----, 6 5

r¥-

I! =w~

0.3

'A,obs

4

3

02

1"

" ~ '0

x ~