Manual De Respuestas

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 ContentsSolutions
Page

Chapter 2 Chapter 3 Chapter 4 ChapterS Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10

Introduction to Engineering Calculations Presentation and Analysis ofData Material Balances Energy Balances Unsteady-State Material and Energy Balances Fluid Flow and Mixing Heat Transfer Mass Transfer Unit Operations

1

9
17 41 54 76 86
'98

106 122

Chapter 11 HomogeneousReactions Chapter 12 Heterogeneous Reactions Chapter 13 Reactor Engineering

139
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
(a)

Unit conversion

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

x10-6 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 x10-2 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-

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

3

1 100 em 1

3

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

I I= ,

1m

1

11 lOOOcm3

I.

956 x 10-4 metric horsepower h .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- = 345 BtulbAnswer: 0.192 kcal g-1

1

1

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

2.2
Case 1

Unit conversion

Convert to units of kg, m, s.

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

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):

Solutions: Chapter 2

Dup
Re::: - p -

(2mm.1

=

1m l000mmU

n(3cms-l.l~n(251bfC3I0.4536kgl.1 x 10 2 3U _ Ift ~ l00cmU lIb 2.832 m
3
10
-6 P

I

c .

10

-3

kgm 1 cP

1

-11 s

- 2.4 x 10

7

Answer: 2.4 x 107

ease 2Convert to units of kg, m, s. From Table Al (Appendix A): 1 in.::: 2.54 x 10-2 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
Answer: 1.5 x 104

= 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. Assumingthis is the same at 28"C, !lJ= 2.5 x 10-5 cm1 s·t, Also, from the Chemical Engineers' Handbook, the density of 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 isabsolute 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

""

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

latm

"" 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...