Soln_Fall2008_HW4

Soln_Fall2008_HW4 - 1. Solution: C.V. Water in the piston...

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1. Solution: C.V. Water in the piston cylinder. Continuity: m2 = m1, Energy Eq. per unit mass: u2 - u1 = 1q2 - 1w2 Process: P = constant = P1, => 1 w 2 = ⌡⌠ 1 2 P dv = P1(v2 - v1) State 1: T1 , P1 => Table B.1.5 compressed solid, take as saturated solid. v1 = 1.09•10-3 m3/kg, u1 = -341.78 kJ/kg State 2: x = 1, P2 = P1 = 125 kPa due to process => Table B.1.2 v2 = vg(P2) = 1.3749 m3/kg, T2 = 106.0 ° C ; u2 = 2513.5 kJ/kg From the process equation 1w2 = P1(v2 -v1) = 125*(1.3749 -1.09•10-3) = 171.7kJ/kg From the energy equation 1q2 = u2 - u1 + 1w2 = 2513.5 - (-341.78) + 171.7 = 3026 kJ/kg 2 Solution: C.V.: Both rooms A and B in tank. Continuity Eq.: m2 = mA1 + mB1 ; Energy Eq.: m2 u2 - mA1 uA1 - mB1 uB1 = 1Q2 - 1W2 State 1A: (P, v) Table B.1.2, mA1 = VA/vA1 = 1.5/0.5 = 3 kg xA1 = (v – vf)/vfg = (0.5 - 0.001073)/ 0.60475 = 0.8250 uA1 = uf + xA1* ufg = 561.13 + 0.8250× 1982.43 = 2196.6 kJ/kg State 1B: Table B.1.3, vB1 = 0.4742, uB1 = 2881.12, VB = mB1vB1 =3.5 *0.4742= 1.66 m3 Process constant total volume: Vtot = VA + VB = 1.5+1.66=3.16 m3
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This note was uploaded on 09/29/2009 for the course ME 530.230 taught by Professor Katz during the Spring '09 term at Johns Hopkins.

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Soln_Fall2008_HW4 - 1. Solution: C.V. Water in the piston...

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