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Solutions for Basic Principles and Calculations in Chemical Engineering, 9th edition by David M. Himmelblau, All Chapters

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Solutions for Basic Principles and Calculations in Chemical Engineering, 9th edition by David M. Himmelblau, All Chapters

Institution
Chemical Engineering, 9th Edition
Course
Chemical Engineering, 9th edition











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Institution
Chemical Engineering, 9th edition
Course
Chemical Engineering, 9th edition

Document information

Uploaded on
May 15, 2025
Number of pages
502
Written in
2024/2025
Type
Exam (elaborations)
Contains
Questions & answers

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UYTREW



Chapter 2 Solutions ALL CHAPTERS

2.1 Units of Measure


2.1.1
(a) N/mm or nm (nanometer)
(b) °C/M/s
(c) 100 kPa
(d) 273.15 K
(e) 1.50m, 45 kg
(f) 250°C
ST
(g) J/s
(h) 250 N


2.1.2 a. The device measures equivalent masses when it is balanced (both pans aligned
horizontally).
b. The compressed spring measures the force under the influence of gravity. If this
U
system were applied on the surface of the moon, the measurement would be much
smaller.
D
2.2 Unit Conversions

2.2.1
2.5 mi 1610 m
YL
(a) = 4.0 103 m or 4.0 km
mi
52.66 Btu 1.055 kJ
(b) = 55.56 kJ
Btu
5.00 hp 0.7457 kW
(c) = 3.73 kW
AB
hp
50. gal 3.875 10-3 m3 min
(d) = 3.2 10−3 m3 s-1
min gal 60 s

22.5 lbf 6.958 kPa
(e) = 157 kPa
in 2 lb f /in 2
45.6 slug 32.174 lbm 0.4536 kg
(f) = 665 kg s-1
s slug lbm
17.0 hp h 745.7 J 3600 s
(g) = 4.56 106 J=4.56 103 kJ=4.56 MJ
hp s h

35.9 gal 3.875 10-3 m3 ft 2
(h) 2 = 1.50 m3 m-2 s-1
s ft 2 gal 0.3048 m2

816.67o R 5 K
(i) 357o F+459.67=816.67oR = 454 K
9o R




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, UYTREW


7.6 lbm 0.4536 kg ft 3
(j) = 120 kg m-3
ft 3 lbm 0.02832 m3



2.2.2
6.000 ft 0.3048 m 106m
(a) = 1.829 106 m
ft m

100 km 3.937 104 in h s
(b) = 1.094 10−3 in s-1
h km 3600 s 10 s
6


500K 9o R
(c) = 900o R 900o R-459.67=440oF
5K
ST
1106 Btu 2.93 10-4 kW h
(d) = 293 kW
h Btu
-3
7.5 kg 2.2046 lbm 16 oz = 9.3 oz bushel-1
m
(e) kg
m
-3
28.3776 bushel lb
m
2 2
15.367 lb m ft lb f s 1.28510-3 Btu
(f) = 6.1374 10−4 Btu
U
2
s 32.174 lbm ft lb f ft
0.2433 kg N s 2 Pa m 2 1.41410-4 psi
(g) = 3.440 10−5 psi
D
ms 2 kg m N Pa
10.1 A V W kW
(h) = 0.0101 kW
A V 1000 W
YL
32.17 ft 36002 s 2 0.3048 m 1000 mm
(i) 2
= 1.2711011 mm h-2
s
2
h ft m
0.779 lb lb s 2
ft 3.76610-7 kW h 3600 s
(j) m f = 3.28 10−5 kW
s
3 32.174 lb f lbf ft h
m
AB
2.2.3 Fractional reduction in miles driven = 1000/13500 = 0.07407
Gasoline saved = 0.07407(136.8) = 10.13 billion gallons

2.2.4
2000 Btu kW h $0.14 24 h 30 d
= $18, 000 mo-1
h 11.2 Btu kW h d mo

2.2.5
4.24 ly 2.99792 108 m 3600 s 24 h 365 d mi AU
7 = 2.68 106 AU
s h d ly 1610 m 9.2910 mi

2.2.6
8.6 ly 2.99792 108 m 3600 s 24 h 365 d pc
16
= 2.6 pc
s h d ly 3.086710 mi




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2.2.7 850 g
400,000 bbl 42 gal 3.875 l lbm d m
l = 5.075  106 lb h-1
d bbl gal 454.3 g 24 h
2.2.8
18 15  8 in 3 2.54 cm 3
3
l gal 60 s
= 1.8 gal min-1
3 3
297 s in 1000 cm 3.875 l min
2.2.9
12km  3.875 L mi
= 28.9 mi/gal
L gal 1.61 km
2.2.10
$1.29 CAD $0.79 USD 3.875 L
= $3.95 USD/gal
ST
L $1 CAD gal
2.2.11
8.314 kP a m

3
atm 35.31 ft3 kg mol 5 K atm ft3
= 0.7303
101.3 kP a 
3
kg mol K m 2.2046 lbmol 9 R lbmol R
2.2.12
U
2
3500 ft 3 in ft 7.481 gal
V = Ah = = 6546 gal
12 in ft3
D
2.2.13 a. Basis: 1 mi3
1mi3  5280 ft   12 in   2. 54 cm   1 m 
3 3 3 3


 1 mi   1 ft   1 in   100 cm 
YL
=

b. Basis: 1 ft3/s
AB



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, UYTREW



2.2.14

a.



b.


c.
ST

2.2.15 a. Basis: 60.0 mile/hr
60.0 mile 5280 ft 1 hr ft
= 88
U
hr 1 mile 3600 sec sec

b. Basis: 50.0 lbm/(in)2
D
50.0/lb 454 g 1 kg 1(in)2 (100 cm)2 kg
= 3.52  10
4
m
2 2 2
(in) 1 lb 1000 g (2.54 cm) (1 m) m2
YL
c. Basis: 6.20 cm/(hr)2
6.20 cm 1m 109nm 1(hr)2 nm
= 4.79
(hr)2 100 cm 1 m (3600 sec)2 sec2
AB
2.2.16
14.91 kW


, not enough power even at 100% efficiency; 68 kW = 91.2 hp.




YTREW

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