Handbook of Mechanical Engineering Calculations ar Episode 2 Part 6 potx
Handbook of Mechanical Engineering Calculations P1
... isentropic ϭ (c p )(T 2 Ϫ T 1 )/ (c p ) and solve for the temperature after isentropic compression. Solv-(T Ϫ T ), T , 2 Ј 12 Ј ing, ϭ 0. 82 ϭ 0 .24 (7 72. 7 Ϫ 520 )/0 .24 ϭ 828 .4 Њ R, or 368 Њ F. ThisT (T Ϫ 520 ) 2 Ј 2 Ј is ... given, ϭ (1.4 Ϫ 1) / 1.4 T ϭ 520 [( 56/ 14)] 2 7 72. 7 Њ R, or 7 72. 7 Ϫ 520 ϭ 25 2.7 Њ F ( 122 .6 Њ C). (a) Here we have isentropic compressi...
Ngày tải lên: 09/11/2013, 04:15
Handbook of Mechanical Engineering Calculations P2
... 0. 62 2 lb (0 .28 kg) of water per pound (kg) of dry air. Since the water content of the air is a function of the partial pressures, (0. 62 2 ) (0.5 067 ) /[ (2 ϫ 0.5 067 )] ϭ 0 .67 3 lb of water per lb of ... x ϭ 0.8 12( 555/ 500) ϭ 0.90; e x ϭ 2. 7183 0.90 ϭ 2. 46; t 2 ϭ 101.1 Ϫ (101.1 Ϫ 85/ 2. 46) ϭ 94 .6 Њ F (34.8 Њ C). Check ⌬ t ϭ t s Ϫ t 2 ϭ 101.1 Ϫ 94 .6 ϭ 6. 5...
Ngày tải lên: 09/11/2013, 04:15
Tài liệu Handbook of Mechanical Engineering Calculations P3 pdf
... burned ϭ 12. 39 /22 .1 ϭ 0. 560 lb (0 .25 2 kg). This, plus the weight of N 2 in the fuel, step 1, is 0. 560 ϩ 0.0 025 ϭ 0.5 62 5 lb (0 .25 3 kg) of N 2 in the products of combustion. Next, find the total weight of ... weights are: for C ϭ 1 ϫ MC ϭ 1 ϫ 12 ϭ 12; O 2 ϭ 1 .2 ϫ MO 2 ϭ 1 .2 ϫ 32 ϭ 38.4; N 2 ϭ (1 .2 ϫ 3.78)MN 2 ϭ (1 .2 ϫ 3.78 ϫ 28 ) ϭ 127 . For the...
Ngày tải lên: 17/12/2013, 11:15
Tài liệu Handbook of Mechanical Engineering Calculations P4 pptx
... 1 52, 000 (69 ,008) 153,140 (69 , 5 26 ) 154,330 (70, 066 ) 155,570 (70, 62 9 ) Exit gas temperature, Њ F( Њ C) 319 (159) 28 5 (141) 27 3 (134) 26 9 (1 32) Fuel fired, MM Btu / h LHV basis (MW) 0 (0) 24 .50 (7 .2) ... 110, 467 ft 3 /min ( 52. 1 m 3 /s), with an outlet velocity of 4400 ft /min (22 .4 m /s), an outlet velocity pressure of 1 .21 0 inH 2 O (0.30 kPa), a speed of 122...
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Tài liệu Handbook of Mechanical Engineering Calculations P5 pdf
... (2. 5 cm). The number of tubes per ft 2 of tube sheet ϭ 166 /(pitch) 2 , or 166 /1 2 ϭ 166 tubes per ft 2 (17 86. 8 per m 2 ). Since the heater has 1 12 tubes, the area of the tube sheet ϭ 1 12/ 166 ϭ 0 .67 5 ... tube sheet ϭ 1 12/ 166 ϭ 0 .67 5 ft 2 ,or97in 2 ( 62 5 .8 cm 2 ). The inside diameter of the heater shell ϭ (tube sheet area, in 2 /0.7854) 0.5 ϭ (97/...
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Tài liệu Handbook of Mechanical Engineering Calculations P6 pdf
... for 9 percent CO 2 in the exhaust gases, 0.09 ϭ (CO 2 )/(CO 2 ϩ N 2 ) ϭ 0. 065 4/(0. 065 4 ϩ N 2 ). Solving for N 2 , we find N 2 ϭ 0 .66 1 mole. The weight of N 2 therefore ϭ 0 .66 1 ϫ 28 ϭ 18.5 lb (8.399 ... cylinder)(number of cylinders)(rpm /2) /60 s/min. Substituting, engine displacement ϭ 0.04 92( 6) ( 120 0/ 2) /60 ϭ 2. 9 52 ft 3 /s (0.0084 m 3 /s). Then, the...
Ngày tải lên: 17/12/2013, 11:15
Tài liệu Handbook of Mechanical Engineering Calculations P7 docx
... lb/ h (63 ,63 6 kg/h) ϭ aflowof (140,000)(0. 120 2) ϭ 16, 840 gal/ h (63 ,739.4 L / h), or 16, 840 /60 ϭ 28 1 gal/ min (17.7 L/ s). Then the liquid velocity v ϭ gpm /2. 448d 2 ϭ 28 1 /2. 448(4. 0 26 ) 2 ϭ 7.1 ft/s ... velocity head v 2 /2g ϭ (7.1) 2 /2( 32. 2) ϭ 0.784 ft (0 .24 m). Using k values from Fig. 11 in this section, h e ϭ k v 2 /2g ϭ (0.5)(0.784) ϭ 0.3 92 ft (0. 12 m)...
Ngày tải lên: 17/12/2013, 11:15
Tài liệu Handbook of Mechanical Engineering Calculations P8 doc
... size, known/new Pressure factor, P 2/ 1 5.0 92 1 /2 0.03 86 3/ 2 2.58 2/ 3 0.150 4/ 3 1.919 3/4 0 .27 1 6/ 4 2. 67 4 4 /6 0.1399 8/ 6 1.933 6/ 8 0 . 26 7 10/ 8 1. 7 26 8/10 0.335 12/ 10 1.5 42 10/ 12 0. 421 When computing such ... diameter of pipe, in (cm). From a table of pipe properties, d ϭ 22 . 62 6 in (574.7 mm). Also, k ϭ z/S ϭ 75/0.85 ϭ 88 .2 cSt. Then R ϭ 31 62...
Ngày tải lên: 17/12/2013, 11:15
Tài liệu Handbook of Mechanical Engineering Calculations P9 pdf
... air draws in 12, 000 ft 3 /min (339 .6 m 3 /min) of air at a pressure of 14 lb/in 2 (abs) ( 96. 46 kPa) and a temperature of 60 Њ F (15 .6 Њ C). Downloaded from Digital Engineering Library @ McGraw-Hill ... d ϭ 0 .23 6( 100) 0.5 (0.0) 0 .25 / (21 4.7) 0 .25 ϭ 0 .60 in (1. 52 cm). Related Calculations. Rupture disks are used in a variety of applications— process, chemical...
Ngày tải lên: 17/12/2013, 11:15
Tài liệu Handbook of Mechanical Engineering Calculations P11 doc
... 3 1 ⁄ 2 3 1 ⁄ 2 Product of factors: Tubeside 0.535 0.514 0.514 0.513 Shellside 0. 424 0. 367 0 .29 6 0 .29 3 Tube wall 0.0 52 0.0 42 0.0 42 0.0 42 Fouling 0 .25 0 0 .20 4 0 .20 4 0 .20 2 Total sum of products 1 . 26 1 1. 127 1.0 56 1.050 Tubeside ⌬ P, ... 0.95 ϫ 1.015 66 .5 30 ϫ 27 .5 ϫ 0.5 M ϭ 2. 92 ϫ 1.370 ϫ 73 .28 ϫ 0.00080 32 ϭ 0 .23 5 Tube wall, Eq. (11): ⌬ T 1 3. 42...
Ngày tải lên: 17/12/2013, 11:15