One kg water at 500oC and 1 kg saturated water vapor both at 200 kPa are mixed i

Question

One kg water at 500oC and 1 kg saturated water vapor both at 200 kPa are mixed in a constant pressure and adiabatic process. Find the final temperature and the entropy generation for the process.

Explanation / Answer

m2 - mA – mB = 0 Energy Eq.5.11: m2u2 - mAuA – mBuB = –1W2 Entropy Eq.8.14: m2s2 - mAsA – mBsB = ? dQ/T + 1S2 gen Process: P = Constant => 1W2 = ? PdV = P(V2 - V1) Q = 0 Substitute the work term into the energy equation and rearrange to get m2u2 + P2V2 = m2h2 = mAuA + mBuB+ PV1 = mAhA + mBhB where the last rewrite used PV1 = PVA + PVB. State A1: Table B.1.3 hA= 3487.03 kJ/kg, sA= 8.5132 kJ/kg K State B1: Table B.1.2 hB = 2706.63 kJ/kg, sB= 7.1271 kJ/kg K Energy equation gives: h2 = mA m2 hA + mB m2 hB = 12 3487.03 + 12 2706.63 = 3096.83 State 2: P2, h2 = 3096.83 kJ/kg => s2 = 7.9328 kJ/kg K; T2 = 312.2°C With the zero heat transfer we have 1S2 gen = m2s2 - mAsA – mBsB = 2 × 7.9328 – 1 × 8.5132 – 1 × 7.1271 = 0.225 kJ/K

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