Handbook of Mechanical Engineering Calculations ar Episode 3 Part 2 ppsx
Handbook of Mechanical Engineering Calculations P1
... before, ϭ (21 .5 ϫ 10 6 )/[(1195.7 Ϫ 33 2) ϩ 0.05 (33 8.5 Ϫ 33 2) ] ϭ 25 ,115.7 lb /h (11,0 43 kg /h). Then, the assumed duty for the economizer, Q a ϭ (25 ,115.7)(1.05) (33 2 Ϫ 198.5) ϭ 3. 52 MM Btu /h (1. 03 MW). 6. ... Based on 36 0 Њ F (1 82. 2 Њ C) water leaving the economizer, Q a ϭ 3. 52 MM Btu /h (1. 03 MW). Solving for t g 2 as before ϭ 38 2 Ϫ [ (3. 52 ϫ 10 6...
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Handbook of Mechanical Engineering Calculations P2
... rate, gal /min ϭ 950 (25 ,000)/ [500(94.6 Ϫ 85)] ϭ 4950 gal / min (31 2 .3 L /s). Then A ϭ 0.8 12( 4950) ϭ 4 020 ft 2 (37 3.5 m 2 ), and loading ϭ 25 ,000/ 4 020 ϭ 6. 23 lb/ ft 2 (30 .4 kg/ m 2 ). Check the correction ... min) ϭ 21 7 .2 in 3 /min (35 59 .3 cm 3 /min). Since 1 U.S. gal ϭ 23 1 in 3 (3. 8 L), the gal /min flow at a 1 ft /s (0 .3 m/ s) velocity ϭ 21 7 .2/...
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Tài liệu Handbook of Mechanical Engineering Calculations P3 pdf
... above, or 30 3.85 ϩ 58.5 ϭ 36 2 .35 ft 3 (10 .26 1 m 3 ). By using the procedure in step 5, the percent CO 2 , wet basis ϭ 53. 6 /36 2 .35 ϭ 14.8 percent. The percent CO 2 , dry basis ϭ 53. 8/ (36 2 .35 Ϫ 17.6) ϭ 15.6 ... and the fuel. 5. Compute the CO 2 content of the flue gas The CO 2 , wet basis ϭ 31 .5/ 23 3 .2 ϭ 0. 135 , or 13. 5 percent. The CO 2 , dry basi...
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Tài liệu Handbook of Mechanical Engineering Calculations P4 pptx
... diameter is D ϭ (2) (10 .25 )( 13) / (10 .25 ϩ 13) ϭ 11.45 ft (3. 5 m) ϭ 137 .3 in (34 8.7 cm). Duct friction is then p s ϭ [0. 03( 9)/ 137 .3 1 .24 ]( 127 5/ 1000) 1.84 ϭ 0.000 934 inH 2 O (0. 23 Pa). Velocity ... 76 .3 (22 .4) 101.6 (29 .8) Exhaust gas flow, lb/ h (kg / h) 1 52, 000 (69,008) 1 53, 140 (69, 526 ) 154 ,33 0 (70,066) 155,570 (70, 629 ) Exit gas temperature, Њ F( Њ C...
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Tài liệu Handbook of Mechanical Engineering Calculations P5 pdf
... Thus, ( 32 7.81 Ϫ 23 8 ) Ϫ ( 32 7.81 Ϫ 31 8) t ϭ m ln [ 32 7.81 Ϫ 23 8 / ( 32 7.81 Ϫ 31 8)] ϭ 36 .5 Њ F (20 .3 Њ C) The average film temperature t f for any closed heater is then t ϭ t Ϫ 0.8t fs m ϭ 32 7.81 Ϫ 29 .2 ϭ 29 8.6 Њ F ... (35 21 k? 20 00 psia ( 13, 780 kPa) 430 °F (22 1°C) 409 (9 53 kJ/kg) 35 0°F (177°C) 32 4.4 (756 kJ/kg) 450 psia (31 01 kPa) 1000°F ( 538 °C) 1 521...
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Tài liệu Handbook of Mechanical Engineering Calculations P6 pdf
... gravity of the fuel. Hence, s ϭ 141.5( 131 .5 ϩ 25 ) ϭ 0.904. Since 1 gal (3. 8 L) of water weighs 8 .33 lb (3. 8 kg) at 60 Њ F (15.6 Њ C), 1 gal (3. 8 L) of this oil weighs (0.904)(8 .33 ) ϭ 7. 529 lb (3. 39 kg). ... actual charge drawn into the cylinder ϭ (lb of air per lb of fuel)(specific volume of the air, ft 3 /lb)(fuel consumption, lb/h) /36 00 s/h. Or 24 .1( 13. 02) (28...
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Tài liệu Handbook of Mechanical Engineering Calculations P7 docx
... ( 73. 1 L/ s); H p ϭ (1.905) 2 (0 .33 5) 2 (35 0) ϭ 1 42. 5 ft ( 43. 4 m); NPSH p ϭ (1.905) 2 (0 .33 5) 2 (10) ϭ 4.06 ft (1 .24 m); P p ϭ (1.905) 5 (0 .33 5) 3 (55) ϭ 51.8 hp (38 .6 kW). 3. Compute the specific ... 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 .39 2 ft (0. 1...
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Tài liệu Handbook of Mechanical Engineering Calculations P8 doc
... L e ϭ (10)(7.981/4. 026 ) 5 ϭ 30 6 ft ( 93. 3 m). Then the total equivalent length of 8-in (2 03. 2- mm) pipe ϭ sum of the equiv- alent lengths ϭ 42. 6 ϩ 39 0 ϩ 20 00 ϩ 30 6 ϭ 2 738 .6 ft ( 834 .7 m); or, by rounding off, 27 40 ... 2. 58 2/ 3 0.150 4/ 3 1.919 3/ 4 0 .27 1 6/ 4 2. 674 4/6 0. 139 9 8/ 6 1. 933 6/8 0 .26 7 10/ 8 1. 726 8/10 0 .33 5 12/ 10 1.5 42 10/ 12 0...
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Tài liệu Handbook of Mechanical Engineering Calculations P9 pdf
... table of air properties, I ϭ 1000(0.074 93/ 0.7785) ϭ 9 62. 49 ft 3 (27 .2 m 3 ). With an inlet temperature of 90 Њ F ( 32 .2 Њ C), using the inside -of- the-building air intake, I ϭ 1000(0.074 93/ 0.0 721 9) ϭ 1 037 .95 ... g s ϭ 32 .174 lb ⅐ ft/lb ⅐ s 2 , a conversion factor. Substituting appropriate values gives P s ϭ 2. 036 ( 12. 75)(60 .33 ) ( 32 .0/ [144 ϫ 32 .174]) ϭ 1...
Ngày tải lên: 17/12/2013, 11:15
Tài liệu Handbook of Mechanical Engineering Calculations P11 doc
... 66.5 30 ϫ 27 .5 ϫ 0.5 M ϭ 2. 92 ϫ 1 .37 0 ϫ 73. 28 ϫ 0.0008 0 32 ϭ 0. 23 5 Tube wall, Eq. (11): ⌬ T 1 3. 422 ϫ 945 0.098 s ϭ 88 ͫͬͫ ͬͫ ͬ ⌬ T 26 66.5 30 ϫ 27 .5 ϫ 0.5 M ϭ 3. 385 ϫ 48. 63 ϫ 0.000 23 7 6 ϭ 0. 039 Fouling, ... in 5 5 3 1 ⁄ 2 3 1 ⁄ 2 Product of factors: Tubeside 0. 535 0.514 0.514 0.5 13 Shellside 0. 424 0 .36 7 0 .29 6 0 .2 93 Tube wall 0.0 52 0.0...
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