Bioprocess Engineering Principles — Solutions Manual (All 14 Chapters) With Latest Edition

Bioprocess Engineering Principles —
Solutions Manual




1

,Chapter 2
Introduction To Engineering Calculations

2.1 Unit Conversion
(A)
From Table A.9 (Appendix A): 1 Cp = 10–3 Kg M–1 S–
1
1 M = 100 Cm
Therefore:
6 6 103 kg m1 s1 1m 11 1 1
1.5  10 Cp = 1.5  10 Cp . .  1.5  10 Kg S Cm
1 cP 100 cm

ANSWER: 1.5 × 10–11 Kg S–1 Cm–1

(B)
From Table A.8 (Appendix A): 1 Hp (British) = 42.41 Btu Min–1
Therefore:
42.41 Btu min1
0.122 Hp = 0.122 Hp  5.17 Btu Min1
. 1 hp


ANSWER: 5.17 Btu Min–1

(C)
1 Min = 60 S
Rpm Means Revolutions Per Minute. As Revolutions Is A Non-Dimensional Quantity (Section 2.1.2), The
Units Of Rpm Are Min–1. Therefore:
1
10, 000 Min1  10, 000 Min1 .  167 S1
Min
60 S

ANSWER: 167 S–1

(D)
From Table A.8 (Appendix A): 1 W = 1 J S–1
From Table A.7 (Appendix A): 1 J = 9.869 × 10–3 L
Atm From Table A.1 (Appendix A): 1 M = 3.281 Ft
1 Min = 60 S
As Explained In Section 2.4.6, °C–1 Is The Same As K–1. Therefore:
2
1 J s1 9.869 103 l atm 60 s 1m
4335 W M °C  4335 W M °C .
2 1
. 2 1
. .
1W 1J 1 min 3.281 ft
 238.45 L Atm Min1ft2 K1
ANSWER: 238 L Atm Min–1 Ft–2 K–1

2

, Chapter 2 – Introduction to Engineering Calculations




2.2 Unit Conversion
(A)
From Table A.7 (Appendix A): 1 Btu = 0.2520 Kcal
From Table A.3 (Appendix A): 1 Lb = 453.6 G
Therefore:
0.2520 Kcal 1 Lb
345 Btu Lb1 = 345 Btu  0.192 Kcal G1
.
Lb1 . 1 Btu 453.6 G


ANSWER: 0.192 Kcal G–1

(B)
From Table A.5 (Appendix A): 1 Mmhg = 1.316 × 10–3 Atm
From Table A.1 (Appendix A): 1 Ft = 0.3048 M
From Table A.7 (Appendix A): 1 L Atm = 9.604 × 10–2 Btu
From Table A.8 (Appendix A): 1 Btu Min–1 = 2.391 × 10–2 Metric
Horsepower 1 M = 100 Cm
1 L = 1000 Cm3
1 H = 60 Min
Therefore:
0.3048 M 3 100 Cm 3
670 Mmhg Ft3  670 Mmhg Ft3 . 1.316 10 atm . 9.604 10 Btu .
3 2
. .
1 mmHg 1 l atm 1 Ft 1M
1l 2.391102 metric horsepower 1h
3
. 1
.
1000 cm 1 Btu min 60 min
 9.56 104 Metric Horsepower H
ANSWER: 9.56 × 10–4 Metric Horsepower H

(C)
From Table A.7 (Appendix A): 1 Kcal = 4.187 × 103 J
1 Kcal = 1000 Cal
1 Kg = 1000 G
As Explained In Section 2.4.6, °C–1 Is The Same As K–1. Therefore:
1 kcal 4.187 103 J 1000 g
0.554 Cal G1 °C1  0.554 Cal G1 °C1 . . .
1000 cal 1 kcal 1 kg
 2319.6 J Kg1 K1
ANSWER: 2320 J Kg–1 K–1

(D)
From Table A.2 (Appendix A): 1 M3 = 103 L


3

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