Handbook of Mechanical Engineering Calculations ar Episode 3 Part 2 ppsx

Handbook of Mechanical Engineering Calculations P1

... before,ϭ (21 .5ϫ106)/[(1195.7Ϫ 33 2) ϩ0.05 (33 8.5Ϫ 33 2) ]ϭ 25 ,115.7 lb /h (11,0 43 kg /h). Then, the assumed duty for the economizer,Qaϭ (25 ,115.7)(1.05) (33 2 Ϫ198.5)ϭ 3. 52 MM Btu /h (1. 03 MW).6. ... Based on 36 0ЊF (1 82. 2ЊC) water leavingthe economizer, Qaϭ 3. 52 MM Btu /h (1. 03 MW). Solving for tg 2 as beforeϭ 38 2 Ϫ[ (3. 52 ϫ106)/(165,000)(0.9)(0 .2 53) ]ϭ 38 8Ϫ85ϭ 30 3ЊF (150.6ЊC). ... Substituting,(TϪT ), T 3 Ј 2 Ј 3 Ј 20 0ϭ0 .24 Solving,ϭ1661 .3 ЊR; 120 1 .3 ЊF (649.6ЊC).(TϪ 828 ). T 3 Ј 3 ЈDownloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com)Copyright...

Handbook of Mechanical Engineering Calculations P2

... rate, gal /minϭ950 (25 ,000)/ [500(94.6Ϫ85)]ϭ4950 gal / min (31 2 .3 L /s). Then Aϭ0.8 12( 4950)ϭ4 020 ft 2 (37 3.5 m 2 ), andloadingϭ 25 ,000/ 4 020 ϭ6. 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. galϭ 23 1 in 3 (3. 8 L), the gal /min ﬂow at a 1 ft /s (0 .3 m/ s) velocityϭ 21 7 .2/ 23 1 ϭ0.94 gal/ min (0.059 L/s).Downloaded from Digital Engineering ... requiredarea is computed from Aϭ(kL/V)(gpm)ϭ0.8 12( 4800)ϭ 39 00 ft 2 (36 2 .3 m 2 ).Then loadingϭS/Aϭ 25 ,000/ 39 00ϭ6.4 lb/ ft 2 (31 .2 kg/ m 2 ).Since the loading is less than 8 lb /ft 2 (39 .1...

Tài liệu Handbook of Mechanical Engineering Calculations P3 pdf

... above, or 30 3.85ϩ58.5ϭ 36 2 .35 ft 3 (10 .26 1 m 3 ).By using the procedure in step 5, the percent CO 2 , wet basisϭ 53. 6 /36 2 .35 ϭ14.8 percent. The percent CO 2 , dry basisϭ 53. 8/ (36 2 .35 Ϫ17.6)ϭ15.6 ... and the fuel.5. Compute the CO 2 content of the ﬂue gasThe CO 2 , wet basisϭ 31 .5/ 23 3 .2 ϭ0. 135 , or 13. 5 percent. The CO 2 , dry basisϭ 31 .5/( 23 3 .2 Ϫ 28 .6Ϫ 35 .9)ϭ0.187, or 18.7 percent.6. ... per-centage, or (6. 527 )(0 .20 ) (35 9/ 28 .95) (2. 15)ϭ 34 .8 ft 3 (0.985 m 3 ). Add this to thevolume of the products of combustion found in step 4, or 23 3 .2 ϩ 34 .8ϭ 26 8.0ft 3 (7.587 m 3 ).By using the...

Tài liệu Handbook of Mechanical Engineering Calculations P4 pptx

... diameter is Dϭ (2) (10 .25 )( 13) / (10 .25 ϩ 13) ϭ11.45 ft (3. 5 m)ϭ 137 .3 in (34 8.7 cm). Duct friction is then psϭ[0. 03( 9)/ 137 .3 1 .24 ]( 127 5/ 1000)1.84ϭ0.000 934 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,ЊF(ЊC) 31 9 (159) 28 5 (141) 2 73 ( 134 ) 26 9 ( 1 32 )Fuel ... 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 2r/ min, and an input hp of 25 5.5 bhp...

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Tài liệu Handbook of Mechanical Engineering Calculations P5 pdf

... Thus, ( 32 7.81Ϫ 23 8 )Ϫ ( 32 7.81Ϫ 31 8)tϭmln [ 32 7.81Ϫ 23 8 / ( 32 7.81Ϫ 31 8)]ϭ 36 .5ЊF (20 .3 ЊC)The average film temperature tffor any closed heater is thentϭtϪ0.8tfs mϭ 32 7.81Ϫ 29 .2 ϭ 29 8.6ЊF ... (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 Btu/lb (35 44 kJ/kg) 500°F (26 0°C) 20 0 psia ( 137 8 kPa) 126 9 ... hr 2, 000 psia 35 0°FH5 = 32 4.4H7 = 449.4Heater1,000,000 lb/hr (454,000 kg/hr) 1800 psia ( 12, 4 02 kPa) 1050°F (565°C)500 psia (34 45 kPa) 740°F (39 3°C) 137 9 .3 Btu/lb ( 32 14 kJ/kg) 1511 .3 Btu/lb...

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Tài liệu Handbook of Mechanical Engineering Calculations P6 pdf

... gravity of the fuel. Hence, sϭ141.5( 131 .5ϩ 25 )ϭ0.904. Since 1 gal (3. 8 L) of water weighs 8 .33 lb (3. 8 kg)at 60ЊF (15.6ЊC), 1 gal (3. 8 L) of this oil weighs (0.904)(8 .33 )ϭ7. 529 lb (3. 39kg). ... actual charge drawn into the cylinderϭ(lb of air perlb of fuel)(specific volume of the air, ft 3 /lb)(fuel consumption, lb/h) /36 00 s/h. Or 24 .1( 13. 02) (28 ) /36 00ϭ 2. 44 ft 3 (0.69 m 3 ) per second.5. ... hp) [3. 5ft 3 /(min⅐hp)]ϭ 35 00 ft 3 /min (99.1 m 3 /min). 2. Compute the quantity of dust entering the engineEach 1000 ft 3 (28 .3 m 3 ) of air entering the engine contains 1.6 gr (1 03. 7 mg) of dust....

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Tài liệu Handbook of Mechanical Engineering Calculations P7 docx

... ( 73. 1 L/ s); Hpϭ(1.905) 2 (0 .33 5) 2 (35 0)ϭ1 42. 5 ft ( 43. 4 m); NPSHpϭ(1.905) 2 (0 .33 5) 2 (10)ϭ4.06ft (1 .24 m); Ppϭ(1.905)5(0 .33 5) 3 (55)ϭ51.8 hp (38 .6 kW). 3. Compute the specific ... velocity headv 2 /2gϭ(7.1) 2 /2 ( 32 .2) ϭ0.784 ft (0 .24 m).Using k values from Fig. 11 in this section, heϭkv 2 /2gϭ(0.5)(0.784)ϭ0 .39 2 ft (0. 12 m); hexϭ v 2 /2gϭ0.784 ft (0 .24 m).4. ... Values*CRE ValuesPHCRMean time betweenfailures, months0–6 6– 12 12 24 Ͼ 24 Aa1a2a3a4Ba2b1b2b3Ca3b2c1c2Da4b3c2d1*Chemical Engineering. TABLE 5Predictive MaintenanceFrequencies for Rotating...

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Tài liệu Handbook of Mechanical Engineering Calculations P8 doc

... Leϭ(10)(7.981/4. 026 )5ϭ 30 6 ft ( 93. 3m). Then the total equivalent length of 8-in (2 03. 2- mm) pipeϭsum of the equiv-alent lengthsϭ 42. 6ϩ 39 0ϩ 20 00ϩ 30 6ϭ 2 738 .6 ft ( 834 .7 m); or, by roundingoff, 27 40 ... 2. 58 2/ 3 0.1504/ 3 1.919 3/ 4 0 .27 16/ 4 2. 674 4/6 0. 139 98/ 6 1. 933 6/8 0 .26 710/ 8 1. 726 8/10 0 .33 5 12/ 10 1.5 42 10/ 12 0. 421 When computing such a listing, the actual inside diameter of the ... equivalentflowϭ(500)(5.6)ϭ 28 00 ft 3 /h (79 .3 m 3 /h), or 28 00 /36 00 s/hϭ0.778 ft 3 /s (0. 02 m 3 /s). Since 6-in (1 52. 4-mm), schedule 40 pipe has a cross-sectional area of 0 .20 06ft 2 (0. 02 m 2 ) internally,...

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Tài liệu Handbook of Mechanical Engineering Calculations P9 pdf

... table of air properties, Iϭ1000(0.074 93/ 0.7785)ϭ9 62. 49 ft 3 (27 .2 m 3 ). With an inlettemperature of 90ЊF ( 32 .2 ЊC), using the inside -of- the-building air intake, Iϭ1000(0.074 93/ 0.0 721 9)ϭ1 037 .95 ... gsϭ 32 .174lb⅐ft/lb⅐s 2 , a conversion factor. Substituting appropriate values gives Psϭ 2. 036 ( 12. 75)(60 .33 ) ( 32 .0/ [144ϫ 32 .174])ϭ10. 82 in Hg (36 .64 kPa). 3. Compute the head space volume ... 1 gram per 1000 ft 3 (28 .3 m 3 ), 720 0 grains of dirt will pass into the air compressor in 1 week’s oper-ation. With the higher level of 4 grains per 1000 ft 3 (28 .3 m 3 ), 28 ,800 grains, orover...

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Tài liệu Handbook of Mechanical Engineering Calculations P11 doc

... 66.5 30 ϫ 27 .5ϫ0.5Mϭ 2. 92 ϫ1 .37 0ϫ 73. 28 ϫ0.0008 0 32 ϭ0. 23 5 Tube wall, Eq. (11):⌬T 1 3. 422 ϫ945 0.098sϭ88ͫͬͫ ͬͫ ͬ⌬T 26 66.5 30 ϫ 27 .5ϫ0.5Mϭ 3. 385ϫ48. 63 ϫ0.000 23 7 6ϭ0. 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 42 0.0 42 0.0 42 Fouling 0 .25 00 .20 4 0 .20 4 0 .20 2Total sum of products ... ͬ1.85 3. 750.999 30 0.4 02 ϭ0.0 026 8ϫ0.9151ϫ1.0ϫ 2, 574ϭ6. ( 43. 4 kPa)Shellside, Eq. (22 ): 2 62. 4 3. 422 Ϫ4⌬Pϭ4ϫ10ϫ 30 ͫͬͫ ͬ0.00 727 27 .5ϫ0 .31 25 2 ϭ0.00004ϫ858 .3 ϫ0.1586ϫ 30 ϭ0.163...

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