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Solution Manual For Basic Principles and Calculations in Chemical Engineering, 9th Edition by David M. Himmelblau Chapter 2-11 $17.49   Add to cart

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

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

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  • May 3, 2024
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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 measure s 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 ( a) 3 2.5 mi 1610 m4.0 10 m or 4.0 km
mi= × ( b) 52.66 Btu 1.055 kJ55.56 kJ
Btu= ( c) 5.00 hp 0.7457 kW3.73 kW
hp= (d) -3 3
3 3 -1 50. gal 3.875 10 m min3.2 10 m s60 s min gal− ×= × ( e) f
22
f22.5 lb 6.958 kPa157 kPa
in lb /in= (f) -1 m
m45.6 slug 32.174 lb 0.4536 kg665 kg slb s slug= ( g) 63 17.0 hp h 745.7 J 3600 s4.56 10 J=4.56 10 kJ=4.56 MJh hp s= ×× ( h) -3 323 -2 -1
22 235.9 gal 3.875 10 m ft1.50 m m s0.3048 m s ft gal×= ( i) o
oo
o816.67 R
9R5 K357 F+459.67=816.67 R 454 K = ( j) 3-3 m
3 3
mft 7.6 lb 0.4536 kg120 kg m0.02832 m ft lb= 2.2.2 (a) 66 6.000 ft 0.3048 m 10 m1.829 10 mm ftµµ = × (b) 4
3 -1
6s 100 km 3.937 10 in h1.094 10 in s3600 s h km 10 sµ
µ− ×= × (c) o
o oo 9R
R R-500K900 900 459.67=440 F
5 K= (d) 6 -41 10 Btu 2.93 10 kW h293 kW
h Btu××= (e) -3
-1m
-3
mkg16 oz 7.5 kg m 2.2046 lb9.3 oz bushel
m 28.3776 bushel lb= (f) 22-34 mf
2
fm 15.367 lb ft lb s 1.285 10 Btu6.1374 10 Btulb ft s 32.174 lb ft− ×= × (g) 22 -45
20.2433 kg N s Pa m 1.414 10 psi3.440 10 psiN Pa m s kg m− ×= × (h) W10.1 A V W kW0.0101 kW1000 A V= (i) 22
11 -2
22ft32.17 ft 3600 s 0.3048 m 1000 mm1.271 10 mm hm sh= × (j) 2-75 mf
3
fm lb ft3600 s 0.779 lb ft lb s 3.766 10 kW h3.28 10 kW
s 32.174 lb f h− ×= × 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 -1
kW h2000 Btu kW h 24 h 30 d $0.14$18, 000 mo
h 11.2 Btu d mo= 2.2.5 8
6
7h4.24 ly 2.99792 10 m 24 h 365 d 3600 s mi AU2.68 10 AU1610 m 9.29 10 mi s d ly×= ×× 2.2.6 8
16h8.6 ly 2.99792 10 m 24 h 365 d 3600 s pc2.6 pc3.0867 10 mi s d ly×=× 2.2.7 6 -1 m
m 850 g 400,000 bbl 42 gal lb 3.875 l d5.075 10 lb hgal 24 h d bbl l454.3 g= × 2.2.8 3 33
-1
3 3gal 18 15 8 in 2.54 cm 60 s l 1.8 gal min
297 s in min 3.875 l 1000 cm××= 2.2.9 12km 3.875 L mi28.9 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 33 3
38.314 kPa m atm 35.31 ft kgmol 5 K atm ft0.7303
kgmol K 101.3 kPa m 2.2046 lbmol 9 R lbmol R/// /=
// ° ° / / 2.2.12 2
33500 ft 3 in ft 7.481 gal6546 gal
12 in ftV Ah// /= = =
/ / 2.2.13 a. Basis: 1 mi 3 1 mi 3 5280 ft 1 mi     3 12 in 1 ft     3 2 . 54 cm 1 in     3 1 m 100 cm     3 = b. Basis: 1 ft 3/s 2.2.14 a. b. c. 2.2.15 a. Bas is: 60.0 mile/hr 60.0 mile 5280 ft 1 hr ft = 88 hr 1 mile 3600 sec sec b. Basis: 50.0 lb m/(in)2 22
4 m
2 22 250.0/lb 454 g 1 kg 1(in) (100 cm) kg = 3.52 10(in) 1 lb 1000 g (2.54 cm) (1 m) m× c. Basis: 6.20 cm/(hr)
2 92
2 226.20 cm 1 m 10 nm 1(hr) nm = 4.79 (hr) 100 cm 1 m (3600 sec) sec 2.2.16 14.91 kW , not enough power even at 100% efficiency; 68 kW = 91.2 hp.

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