ADVANCED THERMODYNAMICS ENGINEERING phần 3 pdf

ADVANCED THERMODYNAMICS ENGINEERING phần 3 pdf

ADVANCED THERMODYNAMICS ENGINEERING phần 3 pdf

... 0.4 ÷ 1.2 = 0 .33 3 m 3 , and the total volume V = V W + V A = 0.001 + 0 .33 3 = 0 .33 4 m 3 . The volume of the coffee cup V W =0.001 m 3 , and it contains internal energy U W = 32 2.2 kJ. The ... i.e., s (g,240 .3) = 83. 5 + 142.51 = 226.01 kJ kmole –1 K –1 (point G in Figure 33 ). Therefore, s (g,240 .3) = 226 kJ kmole –1 K –1 , and h(g, 240 .3) = 131 32 + 20.058 = 33 ,190...
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ADVANCED THERMODYNAMICS ENGINEERING phần 9 pdf

ADVANCED THERMODYNAMICS ENGINEERING phần 9 pdf

... 39 4.2 919. 13 0.1612 37 .7 23 0. 232 53 Oxygen O 2 32 154.6 50.5 0.288 0.0 734 54.4 90.2 13. 9 2 13. 3 17 .39 0.0221 1 .38 017 0. 031 82 Ozone O 3 48 261 53. 7 0.220 0.0889 80.5 161 .3 232 .9 60.56 0. 035 0 3. 69922 ... 469.6 33 .7 0.262 0 .30 4 0.251 1 43. 4 30 9.0 11 63. 3 36 0.2 419. 03 0.1004 19.08 23 0.14482 Propane C 3 H 8 44.1 36 9.8 42.5 0.281 0.2 03 0.1 53 85.5 2...
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ADVANCED THERMODYNAMICS ENGINEERING phần 1 ppt

ADVANCED THERMODYNAMICS ENGINEERING phần 1 ppt

... volume m 3 ft 3 35 .31 5 V volume m 3 gallon 264.2 V velocity m s –1 ft s –1 3. 281 v specific volume (mass basis) m 3 kg –1 ft 3 lb m –1 16.018 v specific volume (mole basis) m 3 kmole –1 ft 3 lbmole –1 16.018 W ... x 3 , i.e., Z = Z(x 1 ,x 2 ,x 3 ). ( 23) The total differential of Z is dZ = ∂Z/∂x 1 dx 1 + ∂Z/∂x 2 dx 2 + (∂Z/∂x 3 ) dx 3 . (24) Since dZ is exact ∂Z...
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ADVANCED THERMODYNAMICS ENGINEERING phần 2 ppsx

ADVANCED THERMODYNAMICS ENGINEERING phần 2 ppsx

... chap- ters. F. APPENDIX 1. Air Composition Species Mole % Mass % Ar 0. 934 1.288 CO 2 0. 033 0.050 N 2 78.084 75.521 O 2 20.946 23. 139 Rare gases 0.0 03 0.002 Molecular Weight: 28.96 kg kmole –1 . 2. Proof of ... Applying Eq. (12) for all processes, i.e., 1–2, 2 3, 3 4, , and 8–1, δδQQ Q Q WW W W ∫∫ ===+++ +++ 12 23 81 12 23 81 L L , so that δW ∫ = 0 30 0 – 2000 + 0 + 0 –200 + 0 + 4000...
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ADVANCED THERMODYNAMICS ENGINEERING phần 4 potx

ADVANCED THERMODYNAMICS ENGINEERING phần 4 potx

... 1 93. 5 + 100 × 31 .4 = –48 ,33 3 kJ kmole –1 . Therefore, φ ∞ = 0.6 × (–46981) + 0.4 × (–4 833 3) = –47521.8 kJ kmole –1 , and w ch = –524 73. 8 – (–47521.8) = –4952 kJ kmole –1 . w u, max, ∞ = 33 ,221.6+ ... – 34 22 = 837 kJ kg –1 . The heat rejected in the condenser q 23 – w 23 = h 3 – h 2 , i.e., q 23 = q out = 192 – 2585 = – 239 3 kJ kg –1 . Likewise, in the pump q 34...
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ADVANCED THERMODYNAMICS ENGINEERING phần 5 potx

ADVANCED THERMODYNAMICS ENGINEERING phần 5 potx

... + 120) + 1 /3 = 0 .31 22 × cos ( 134 .29) + 0 .33 3 = –0.2180 + 0 .33 3 = 0.11 53. Z 3 = 2α 0.5 cos(φ /3 + 240) + 1 /3 = 0 .31 22 × cos(254.29) + 0 .33 333 = –0.084 53 + 0 .33 33 = 0.2488 The spreadsheet software ... = 0. 732 9, i.e., φ = 42.87. Z 1 = 2α 0.5 cos(φ /3) + 1 /3 = 2 × 0.02 437 0.5 cos(42.87 /3) + 1 /3 = 0 .31 22 × 0.9691 + 1 /3 = 0 .30 25 + 0 .33 3= 0. 635 9. Z...
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ADVANCED THERMODYNAMICS ENGINEERING phần 6 ppt

ADVANCED THERMODYNAMICS ENGINEERING phần 6 ppt

... 126.2 30 4.15 38 5.0 T ref (K) 2 73. 16 233 .14 233 .14 233 .15 64.1 43 216.55 233 .15 P ref (bar) 0.0061 13 1.0495 0.512615 .7177 .12 53 5.178 .6417 h ref (kJ kg –1 ) 0.01 0.0 0.0 150 .3 301.45 0.0 u ref ... CF 3 CH 2 F NH 3 N 2 CO 2 W m 18.015 86.476 102. 03 66.05 17. 03 28.0 13 44.01 120.92 P c (bar) 220.9 49.775 40.67 45.20 112.8 33 .9 73. 9 41.2 T c (K) 647 .3 369.15...
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ADVANCED THERMODYNAMICS ENGINEERING phần 7 ppt

ADVANCED THERMODYNAMICS ENGINEERING phần 7 ppt

... = (–9 530 .5 – (–9776.2))÷5 93 = 0.414 kJ kmole -1 K -1 , while for a change from D to the metastable vapor state K, σ = (–9 530 .5 – (–9 638 .8)) ÷5 93 = 0.1 83 k kmole -1 K - . -4000 -35 00 -30 00 -2500 -2000 -1500 -1000 -500 0 500 1000 0 ... the liquid molar volume to be 0.018 m 3 kmole -1 . Solution The pressure P β =P sat exp (v α (P α –P sat )/RT) = 133 exp (0.018 × (1– 133...
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ADVANCED THERMODYNAMICS ENGINEERING phần 8 doc

ADVANCED THERMODYNAMICS ENGINEERING phần 8 doc

... basis. Solution For water, g H2O( l ) (31 3 K,10 bar) = g HOg o 2 () (31 3 K) + 8 .31 4 31 3×ln ( f HO 2 ()l (31 3 K, 10 bar)/1) = g HOg 2 () (31 3 K, 1 bar) + 8 .31 4 31 3×ln ( f HO 2 ()l (31 3 K, 10 bar)/1), where (A) f HO 2 ()l (T,P)/ ... (F) Now, g HOg o 2 () (31 3 K, 1 bar) = ( h HOg o 2 () – T s HOg o 2 () ) 31 3 K, 1 bar = –24 132 1 – 31 3 × 190 .33 = 30 0894 kJ kmole –1 ....
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ADVANCED THERMODYNAMICS ENGINEERING phần 10 docx

ADVANCED THERMODYNAMICS ENGINEERING phần 10 docx

... 1.796 3. 2 83 3.516 3. 652 3. 744 3. 746 3. 671 3. 505 1.02 1.627 3. 039 3. 442 3. 595 3. 705 3. 718 3. 647 3. 484 1.05 1 .35 9 2. 034 3. 03 3 .39 8 3. 5 83 3. 632 3. 575 3. 42 1.1 1.12 1.487 2.2 03 2.965 3. 3 53 3.484 3. 4 53 ... 3. 788 3. 957 0.98 3. 264 3. 247 3. 268 3. 318 3. 412 3. 569 3. 701 3. 875 0.99 3. 0 93 3.082 3. 126 3. 195 3. 306 3. 477...
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