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Solution Manual For Basic Principles and Calculations in Chemical Engineering, 9th Edition by David M. Himmelblau, All Chapter 2-11 $19.99   Add to cart

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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

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  • October 19, 2024
  • 502
  • 2024/2025
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Chapter 2 Solutions

2.1 Units of Measure


2.1.1
(a) N/mm or nm (nanometer)
R
(b) °C/M/s
(c) 100 kPa
U
(d) 273.15 K
(e) 1.50m, 45 kg
(f) 250°C
E
(g) J/s
(h) 250 N
S S
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
I
O
system were applied on the surface of the moon, the measurement would be much
smaller.
N
2.2 Unit Conversions

2.2.1
N
2.5 mi 1610 m
(a) 3
= 4.0 × 10 m or 4.0 km
O
mi
52.66 Btu 1.055 kJ
(b) = 55.56 kJ
C
Btu
5.00 hp 0.7457 kW
(c) = 3.73 kW
D
hp
-3 3
50. gal 3.875 × 10 m min
E
−3 3 -1
(d) = 3.2 × 10 m s
min gal 60 s
M
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
R
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
U
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
S
h Btu
-3
7.5 kg m 2.2046 lbm 16 oz
S
-1
(e) -3
= 9.3 oz bushel
m 28.3776 bushel kg lb m
I
2 2
15.367 lb m ft lb f s1.285×10-3 Btu −4
(f) = 6.1374 × 10 Btu
O
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)
N
−5
= 3.440 × 10 psi
ms
2
kg m N Pa
10.1 A V W kW
N
(h) = 0.0101 kW
AV 1000 W
2 2
O
32.17 ft 3600 s 0.3048 m 1000 mm
(i) 11
= 1.271 × 10 mm h
-2

s
2
h
2
ft m
C
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
D
2.2.3 Fractional reduction in miles driven = 1000/13500 = 0.07407
E
Gasoline saved = 0.07407(136.8) = 10.13 billion gallons

2.2.4
M
2000 Btu kW h $0.14 24 h 30 d -1
= $18, 000 mo
h 11.2 Btu kW h d 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
3
= 1.8 gal min
297 s in 1000 cm 3.875 l min
3
R
2.2.9
12km / 3.875 L/ mi
U
= 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
S
L $1 CAD gal
2.2.11
S
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
I
2.2.12
O
3500 ft/ 2 3 in
/ ft
/ 7.481 gal
V Ah
= = = 6546 gal
12 in ft/ 3
N
/

2.2.13 Basis: 1 mi3
N
a.
1mi 3  5280 ft  3  12 in  3  2. 54 cm  3  1 m  3
O
 1 mi   1 ft   1 in   100 cm 
=
C
b. Basis: 1 ft3/s
E D
M

, 2.2.14

a.



b.
R
c.
E U
S S
2.2.15 a. Basis: 60.0 mile/hr
60.0 mile 5280 ft 1 hr ft
I
= 88
O
hr 1 mile 3600 sec sec

b. Basis: 50.0 lbm/(in)2
N
50.0/lb m 454 g 1 kg 1(in) 2 (100 cm) 2 kg
= 3.52 × 104 2
N
2 2 2
(in) 1 lb 1000 g (2.54 cm) (1 m) m
O
c. Basis: 6.20 cm/(hr)2
6.20 cm 1 m 109 nm 1(hr) 2 nm
= 4.79 2
C
2 2
(hr) 100 cm 1 m (3600 sec) sec
D
2.2.16
14.91 kW
E
M
, not enough power even at 100% efficiency; 68 kW = 91.2 hp.

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