Solutions Manual for Basic Principles and Calculations in Chemical Engineering, 9th edition By David M. Himmelblau, James B. Riggs

Solutions Manual for Basic Principles and Calculations in Chemical Engineering, 9th edition By David M. Himmelblau, James B.
Riggs




Solutions Manual for Basic Principles and Calculations in Chemical Engineering, 9th edition By David M. Himmelblau, James B.
Riggs

,Solutions Manual for Basic Principles and Calculations in Chemical Engineering, 9th edition By David M.
Himmelblau, James B. Riggs


Chapter 2 Solutions

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

2.2 Unit Conversions

2.2.1
2.5 mi 1610 m
(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
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) = 1.50 m3 m-2 s-1
s ft 2 gal 0.30482 m2

816.67o R 5 K
(i) 357o F+459.67=816.67oR = 454 K
9o R
Solutions Manual for Basic Principles and Calculations in Chemical Engineering, 9th edition By David M.
Himmelblau, James B. Riggs

,Solutions Manual for Basic Principles and Calculations in Chemical Engineering, 9th edition By David M.
Himmelblau, James B. Riggs

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
1106 Btu 2.93 10-4 kW h
(d) = 293 kW
h Btu
-3
7.5 kg m 2.2046 lbm 16 oz
(e) -3 = 9.3 oz bushel-1
m 28.3776 bushel kg lbm
15.367 lb ft 2 lb s 2 1.28510-3 Btu
(f) m f = 6.1374 10−4 Btu
2
s 32.174 lbm ft lbf ft
0.2433 kg N s 2 Pa m 2 1.41410-4 psi
(g) = 3.440 10−5 psi
ms2 kg m N Pa
10.1 A V W kW
(h) = 0.0101 kW
A V 1000 W

32.17 ft 36002 s 2 0.3048 m 1000 mm
(i) 2 2
= 1.2711011 mm h-2
s 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 lb f ft h
m



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
= 2.68 106 AU
s h d 7
ly 1610 m 9.2910 mi

2.2.6
8.6 ly 2.99792 108 m 3600 s 24 h 365 d pc = 2.6 pc
s h d ly 3.08671016
mi
Solutions Manual for Basic Principles and Calculations in Chemical Engineering, 9th edition By David M.
Himmelblau, James B. Riggs

, Solutions Manual for Basic Principles and Calculations in Chemical Engineering, 9th edition By David M.
Himmelblau, James B. Riggs


2.2.7
400,000 bbl 42 gal 3.875 l 850 g lbm d
= 5.075  106 lb m h-1
d bbl gal l 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
297 s in 1000 cm 3 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
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
kg mol K 101.3 kP a m3 2.2046 lbmol 9 R lbmol R
2.2.12
3500 ft2 3 in ft 7.481 gal
V = Ah = = 6546 gal
12 in ft3

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 
=

b. Basis: 1 ft3/s




Solutions Manual for Basic Principles and Calculations in Chemical Engineering, 9th edition By David M.
Himmelblau, James B. Riggs