Me 235 exam 2 solution november 22 2005 c borgnakke me

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Unformatted text preview: ME 235 EXAM 2, SOLUTION NOVEMBER 22, 2005 C. BORGNAKKE, ME 1) [30 Points] Take CV as the air out to the source surface. The first process is a constant volume process so 1W2 = 0 Energy Eq: m(u2 u1) = 1Q2 1W2 = 1Q2 => q = u2 - u1 Entropy Eq.: m(s2 s1) = 1Q2/Ts + 1S2 gen P2V2 = mRT2 Process & ideal gas: P1V1 = mRT1, P2 = P1 T2 /T1 = 100 1000 / 300 = 333.3 kPa From energy eq.: 1q2 = u2 - u1 = Cv o (T2 T1) = 0.717 (1000 300) = 501.9 kJ/kg 1000 s2 s1 = Cv o ln(T2 / T1) = 0.717 ln 300 = 0.8632 kJ/kgK From entropy eq.: 1s2 gen = s2 s1 1q2/Ts 501.9 = 0.8632 1500 = 0.5286 kJ/kgK [20 Points] The second process is isentropic, q = 0, so with constant heat capacity we get: From Eq.8.??: 1000 0.4 P2 = P1 (T2 /T1) = 100 ( 300 ) = 6762 kPa 1w2 = u1 - u2 = Cv o (T1 T2) = 0.717 (300 1000) = -501.9 kJ/kg R 1 = 1 - k (P2v2 - P1v1) = 1 - k (T2 T1) = same. -------------------------------------------------------------Solving using Table A.7 gives: 1q2 = u2 u1 = 759.19 214.36 = 544.83 kJ/kg s2 s1 = sT2 sT1 R ln (P2 / P1) 1000 = 8.13493 6.86926...
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This note was uploaded on 05/05/2008 for the course ME 235 taught by Professor Borgnakke during the Winter '07 term at University of Michigan.

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