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probl9_29

# probl9_29 - then q41...

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parameters r k R T1 p1 15 1.4 0.287 310 100 Note: for each column below, the formula follow the template of Ex. 9.2 After anchoring, they are copied across columns so: v1 = 0.8897 and cv = 0.7175 rc = 1.5 1.7 1.9 2.1 2.3 2.5 T2 = T1 r^(k-1) = 915.79 915.79 915.79 915.79 915.79 915.79 therefore w12 = -434.66 -434.66 -434.66 -434.66 -434.66 -434.66 and p2 = 4431.27 4431.27 4431.27 4431.27 4431.27 4431.27 T3 = T2*rc = 1373.69 1556.85 1740.01 1923.17 2106.33 2289.49 then w23 = 262.83 262.83 262.83 262.83 262.83 262.83 and q23 = 591.37 722.79 854.21 985.62 1117.04 1248.46 T4 = T3/(r/rc)^(k-1) = 546.88 651.61 761.41 875.93 994.90 1118.09 and so w34 = 593.24 649.51 702.15 751.40 797.45 840.48
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Unformatted text preview: then q41 =-169.96-245.11-323.88-406.05-491.42-579.81 efficiency: 0.71 0.66 0.62 0.59 0.56 0.54 check: sum(q) = 421.42 477.68 530.32 579.57 625.62 668.65 sum(w) = 421.42 477.68 530.32 579.57 625.62 668.65 k, the ratio of specific heats, decreases slowly with increasin temperature. Modify it (1.38, 1.35. ..) and see the changes More accurate calculations would have a separate k for each process, or even better use the gas tables. Problem 9.29, cold air standard 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 efficiency:...
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