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Solution Manual For Basic Principles and Calculations in Chemical Engineering, 9th Edition by David M. Himmelblau, All Chapter 2-11

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Solution Manual For Basic Principles and Calculations in Chemical Engineering, 9th Edition by David M. Himmelblau, All Chapter 2-11

Institution
Chemical Engineering
Course
Chemical Engineering











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Institution
Chemical Engineering
Course
Chemical Engineering

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Uploaded on
November 4, 2025
Number of pages
505
Written in
2025/2026
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Basic Principles and Calculations in Chemical
ST
Engineering – 9th Edition
UV

SOLUTION
IA

MANUAL
_A
PP

David M. Himmelblau
RO
VE
Comprehensive Solutions Manual for Instructors
and Students
D?
© David M. Himmelblau

All rights reserved. Reproduction or distribution without permission is prohibited.
??

©STUDYSTREAM

, Chapter 2 Solutions
ST
2.1 Units of Measure


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


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

2.2 Unit Conversions
PP
2.2.1
2.5 mi 1610 m
(a) 3
= 4.0 × 10 m or 4.0 km
mi
RO
52.66 Btu 1.055 kJ
(b) = 55.56 kJ
Btu
5.00 hp 0.7457 kW
(c) = 3.73 kW
hp
-3 3
50. gal 3.875 × 10 m min −3 3 -1
(d) = 3.2 × 10 m s
VE
min gal 60 s
22.5 lb f 6.958 kPa
(e) 2 2
= 157 kPa
in lb f /in
45.6 slug 32.174 lb m 0.4536 kg
(f) -1
D?
= 665 kg s
s slug lb m

17.0 hp h 745.7 J 3600 s
(g) 6 3
4.56 × 10 J=4.56 × 10 kJ=4.56 MJ
=
hp s h
-3 3
35.9 gal 3.875 × 10 m ft 2
(h) 3 -2 -1
= 1.50 m m s
??
s ft
2
gal 0.30482 m 2
o
816.67 R 5 K
(i) 357 o F+459.67=816.67 o R o
= 454 K
9 R

, 7.6 lb m 0.4536 kg ft 3 -3
(j) = 120 kg m
ft
3
lb m 0.02832 m3
ST
2.2.2
6.000 ft 0.3048 m 106 µ m
(a) 6
= 1.829 × 10 µ m
ft m
UV
4
100 km 3.937 × 10 in h s −3 -1
(b) = 1.094 × 10 in µ s
h km 3600 s 106 µ s
o
500K 9 R
(c) = 900 R
o o o
900 R-459.67=440 F
5K
IA
6 -4
1 × 10 Btu 2.93 × 10 kW h
(d) = 293 kW
h Btu
-3
7.5 kg m 2.2046 lbm 16 oz -1
(e) -3
= 9.3 oz bushel
m 28.3776 bushel kg lb m
_A
2 2
15.367 lb m ft lb f s1.285×10-3 Btu −4
(f) = 6.1374 × 10 Btu
s
2
32.174 lb m ft lb f ft
2
0.2433 kg N s Pa m 2 1.414×10-4 psi
(g) −5
= 3.440 × 10 psi
ms
2
kg m N Pa
PP
10.1 A V W kW
(h) = 0.0101 kW
AV 1000 W
2 2
32.17 ft 3600 s 0.3048 m 1000 mm
(i) 11
= 1.271 × 10 mm h
-2

s
2
h
2
ft m
RO
2
0.779 lb m ft 3.766×10-7 kW h 3600 s
lb f s
(j) 3
−5
= 3.28 × 10 kW
s 32.174 lb m f lbf ft h


2.2.3 Fractional reduction in miles driven = 1000/13500 = 0.07407
Gasoline saved = 0.07407(136.8) = 10.13 billion gallons
VE
2.2.4
2000 Btu kW h $0.14 24 h 30 d -1
= $18, 000 mo
h 11.2 Btu kW h d mo
D?
2.2.5
8
4.24 ly 2.99792 × 10 m 3600 s 24 h 365 d mi AU 6
7 = 2.68 × 10 AU
s h d ly 1610 m 9.29×10 mi
??
2.2.6
8
8.6 ly 2.99792 × 10 m 3600 s 24 h 365 d pc
= 2.6 pc
s h d ly 3.0867×1016 mi

, 2.2.7
400,000 bbl 42 gal 3.875 l 850 g lb m d 6 -1
= 5.075 × 10 lb m h
ST
d bbl gal l 454.3 g 24 h
2.2.8
3 3 3
18 × 15 × 8 in 2.54 cm l gal 60 s -1
3
= 1.8 gal min
297 s in 1000 cm 3.875 l min
3
UV
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
IA
= $3.95 USD/gal
L $1 CAD gal
2.2.11
3
8.314 kPa
/ m/ atm 35.31 ft 3 kgmol
/ 5 K/ atm ft 3
= 0.7303
_A
3
kgmol
/ K/ 101.3 kPa
/ m
/ 2.2046 lbmol 9° R lbmol° R
2.2.12
3500 ft/ 2 3 in
/ ft
/ 7.481 gal
V Ah
= = = 6546 gal
12 in
/ ft/ 3
PP
2.2.13 a. Basis: 1 mi3
1mi 3  5280 ft  3  12 in  3  2. 54 cm  3  1 m  3
 1 mi   1 ft   1 in   100 cm 
=
RO
b. Basis: 1 ft3/s
VE
D?
??

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