Exam (elaborations)

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

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Chemical Engineering

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

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Uploaded on December 31, 2024
Number of pages 503
Written in 2024/2025
Type Exam (elaborations)
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solutions

Institution Chemical Engineering
Course Chemical Engineering

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Basic Principles and Calculations in Chemical Engineering, 9th edition

SOLUTIONS
DR

Basic Principles and
E AM

Calculations in Chemical
Engineering, 9th edition
AC
HI

Authors:
David M. Himmelblau
VE

◊ ALL CHAPTERS COVERED
RS

◊ INSTANT PDF DOWNLOAD💯💯💯
??

◊ ORIGINAL FROM PUBLISHER
?

DREAMACHIVERS ©2025

, Chapter 2 Solutions

2.1 Units of Measure

2.1.1
DR
(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
E
(g) J/s
(h) 250 N
AM

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
system were applied on the surface of the moon, the measurement would be much
smaller.
AC

2.2 Unit Conversions

2.2.1
HI
2.5 mi 1610 m
(a) 3
= 4.0 × 10 m or 4.0 km
mi
52.66 Btu 1.055 kJ
(b) = 55.56 kJ
VE
Btu
5.00 hp 0.7457 kW
(c) = 3.73 kW
hp
-3 3
50. gal 3.875 × 10 m min
RS
−3 3 -1
(d) = 3.2 × 10 m s
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) = 665 kg s
-1

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

2.2.2
6.000 ft 0.3048 m 106 µ m
(a) 6
= 1.829 × 10 µ m
ft m
DR
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
E
6 -4
1 × 10 Btu 2.93 × 10 kW h
(d) = 293 kW
AM
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
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
AC
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
10.1 A V W kW
(h) = 0.0101 kW
AV 1000 W
HI
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
VE
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
RS
Gasoline saved = 0.07407(136.8) = 10.13 billion gallons

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

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
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
= 1.8 gal min
DR
3
297 s in 1000 cm 3.875 l min
3

2.2.9
12km / 3.875 L/ mi
= 28.9 mi/gal
L/ gal 1.61 km/
2.2.10
E
$1.29 CAD $0.79 USD 3.875 L
= $3.95 USD/gal
AM
L $1 CAD gal
2.2.11
3
8.314 kPa
/ m/ atm 35.31 ft 3 kgmol
/ 5 K/ atm ft 3
3
= 0.7303
kgmol
/ K/ 101.3 kPa
/ m
/ 2.2046 lbmol 9° R lbmol° R
2.2.12
AC
3500 ft/ 2 3 in
/ ft
/ 7.481 gal
V Ah
= = = 6546 gal
12 in
/ ft/ 3

2.2.13 a. Basis: 1 mi3
HI
1mi 3  5280 ft  3  12 in  3  2. 54 cm  3  1 m  3
 1 mi   1 ft   1 in   100 cm 
=
VE
b. Basis: 1 ft3/s
RS
??
?

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Exam (elaborations)

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

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