ADVANCED THERMODYNAMICS ENGINEERING phần 6 ppt

ADVANCED THERMODYNAMICS ENGINEERING phần 6 ppt

ADVANCED THERMODYNAMICS ENGINEERING phần 6 ppt

... NH 3 N 2 CO 2 W m 18.015 86. 4 76 102.03 66 .05 17.03 28.013 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) 64 7.3 369 .15 374.30 3 86. 44 405.5 1 26. 2 304.15 385.0 T ref (K) 273. 16 233.14 ... 524.534 169 .59 A o 30.54 22.54 16. 778 17.229 29.75 28.58 29.1519 26. 765 † B o 0.01030 0.1141077 0.2 865 0.4757 .025 .00377 –.001573 0.17594 † C o 0.0 1301 96....
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ADVANCED THERMODYNAMICS ENGINEERING phần 1 ppt

ADVANCED THERMODYNAMICS ENGINEERING phần 1 ppt

... references and index. ISBN 0-8493-2553 -6 (alk. paper) 1. Thermodynamics. I. Puri, Ishwar Kanwar, 1959- II. Title. III. Series. TJ 265 .A55 2001 62 1.402 ′ 1—dc21 200103 562 4 1.First Law 2.Adiabatic ... taught several courses at Texas A&M including Advanced Thermodynamics, Combustion Science and Engineering, Conduction at the graduate level and Thermodynamics, Heat Transfe...
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ADVANCED THERMODYNAMICS ENGINEERING phần 7 ppt

ADVANCED THERMODYNAMICS ENGINEERING phần 7 ppt

... (0.0 866 4/(0.4275 R)(a m /b m )) 2/3 , and P c ´ = 0.0 866 4 RT c ´/b m , i.e., a m = (0 .6 × 142 .6 0.5 + 0.4 × 220 1/2 ) 2 = 171 .6 bar k 1/2 m 6 kmole -2 , b m = (0 .6 × 0.0211 + 0.4 × 0.0 462 ) ... m 3 /kmol e P, bars 567 593 61 5 T 3 = 61 5 K T 1 = 59 3 T 2 = 56 7 A NS F G V apo r Liqui d Vapor Spinodal Cur v Liquid Spinodal Cur v B M D Figure 19: Spinodal curves at...
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ADVANCED THERMODYNAMICS ENGINEERING phần 2 ppsx

ADVANCED THERMODYNAMICS ENGINEERING phần 2 ppsx

... Eq. (16a) by dZ m (dV P /dt) dz = δW – (P o dV + mg dz), where the boundary work δW = P b dV. Therefore, δW = m (dV P /dt) dz + P o dV + mg dz. (16b) Using the relation dz = V P dt in Eq. (16a), ... c vo /R). (G) ∴ V 2 /V 1 = (1+ 0.7 × 100 ÷ (0.287 × 200)) ÷ (1+ 0.7 ÷ 0.287) = 0 .65 , and V 2 = 0 .65 × 0.1 = 0. 065 m 3 . Remarks The potential energy of the weight is converted into thermal e...
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ADVANCED THERMODYNAMICS ENGINEERING phần 3 pdf

ADVANCED THERMODYNAMICS ENGINEERING phần 3 pdf

... R ln(4÷1) = 2 16. 6 kJ kmole –1 K –1 . Likewise, s N 2 (1200K, 6. 6 bar) = s N 2 0 (1200 K) – R ln (6. 6÷1) = 279.3 – 8.314 × ln (6. 6÷1) = 263 .6 kJ kmole –1 K –1 , and s N 2 (1000K, 6 bar) = 269 .2 – ... × ln 6 = 254.3 kJ kmole –1 K –1 . Using Eqs. (C) and (D) S 1 = 0.04 × 2 16. 6 + 0. 06 × 254.3 = 23.92 kJ K –1 , S 2 = 0.04 × 221.8 + 0. 06 × 263 .6 = 24 .69 kJ K –1 , an...
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ADVANCED THERMODYNAMICS ENGINEERING phần 4 potx

ADVANCED THERMODYNAMICS ENGINEERING phần 4 potx

... (–0.1 46) = 0. 062 11 kJ for every kJ of heat pumped into the house. Since the ambient temperature is 0ºC, 1 2 13 76 i =789 W sh =837 1 56 3 4 21 56 I = 379 Pump 158 Condenser Turbine 6 Boiler 9 Figure ... 2 565 .3, s g (25C) = 8.558, with R= 8.314/18.02=0. 461 kJ/kgk, s g1 =s g sat (25C) - Rln(p v1 /P v sat (T 1 ))= 8 .66 72 kJ/kg K, s g2 = 8.3531 kJ/kg K, h f,3 = 188.45 kJ kg...
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ADVANCED THERMODYNAMICS ENGINEERING phần 5 potx

ADVANCED THERMODYNAMICS ENGINEERING phần 5 potx

... 0.125 -0.02 0.0778 0.0778 0.0 866 4 0.0 866 4 0.125 c´ 0 0.145 0.1878 26 0.1878 26 0.0 866 4 0.0 866 4 0 d´ 0 0.375 -0.03223 -0.03223 0 0 0 n 11000.500 f(ω), H 2 O 0.8732 36 1.00 062 9 Note that Z c is required ... rearranging, –s Res (T R ,v R ´)/R = (s o (T,v) – s(T,v))/R = ln(1– (0.0 866 4/v R ´)) – (2. 467 0/T R 3/2 ) (ln(1+ (0.0 866 4/v R ´))) (54) We observed from Example 16 that...
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ADVANCED THERMODYNAMICS ENGINEERING phần 8 doc

ADVANCED THERMODYNAMICS ENGINEERING phần 8 doc

... 0.205 269 .3 282.52 –444302 H 2 O(g) 5 –215849 0.1022 0.2 56 232.7 244.05 –288 561 O 2 3.25 22707 0. 066 5 0. 166 243 .6 258.51 –543 26 N 2 36. 66 21 460 0.7495 1.874 228.2 222.99 –44973 Applying the entropy ... K –1 . ν k h X k p k =X k P ˆ s s k o ( h – T o ˆ s ) k Reactants C 4 H 10 1 –1 261 48 0.021 0.0527 310.2 334.7 –225887 O 2 9.75 0 0.2057 0.5141 205.1 210.7 62 781 N 2 36. 66...
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ADVANCED THERMODYNAMICS ENGINEERING phần 9 pdf

ADVANCED THERMODYNAMICS ENGINEERING phần 9 pdf

... C 8 H 8 104.2 64 7 .6 39.99 0. 261 0.3518 0.234 242.5 418.3 105.1 351.4 788 .63 0.1 166 30.582 0. 168 3 Sulfur dioxide SO 2 64 . 06 430.8 78.8 0. 268 0.122 0.2 56 290.0 263 .2 115.5 385.4 144.45 0.0394 6. 867 98 0.0 568 2 Sulfur ... SO 3 80. 06 490.9 82.07 0.2 56 0.1271 0.422 290.0 317.9 24 .6 502.5 192.24 0.0431 8. 562 6 0. 062 16 Sulfuric acid H 2 SO 4 98.08 925 40 .66 0.159 0.3 1...
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ADVANCED THERMODYNAMICS ENGINEERING phần 10 docx

ADVANCED THERMODYNAMICS ENGINEERING phần 10 docx

... 62 8.07 18. 36 450.09 96. 92 2.443 56 -8 86. 9372 63 0 68 3 .63 19.84 457.78 92.84 2. 460 48 - 866 .4724 64 0 64 9.22 20 .64 465 .50 88.99 2.477 16 -9 36. 162 4 65 0 65 9.84 21. 86 473.25 85.34 2.49 364 - 961 .0 26 660 67 0.47 ... 3.445 16 -3531.77 1520 166 0.23 63 6.5 1223.87 6. 854 3. 461 20 - 360 0.794 1540 168 4.51 67 2.8 1242.43 6. 569 3.47712 - 367 0.2548 1 560 1708.82...
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