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The condenser in a refrigerator receives R-134a at 50°C, 700 kPa and it exits as saturated liquid at 25°C. The flow-rate is 0.1 kg/s and air flows in the condenser at ambient 15°C and leaving at 35°C. Find the heat exchanger second-law efficiency.

The condenser in a refrigerator receives R-134a at 50°C, 700 kPa and it exits as saturated liquid at 25°C. The flow-rate is 0.1 kg/s and air flows in the condenser at ambient 15°C and leaving at 35°C. Find the heat exchanger second-law efficiency. Correct Answer 0.77

m1h1 + mah3 = m1h2 + mah4 (here m is the mass flow-rate) ma = m1 × (h1 – h2)/(h4 – h3) = 0.1 × (436.89 – 234.59)/(1.004(35 – 15)) = 1.007 kg/s ψ1 – ψ2 = h1 – h2 – T0(s1 – s2) = 436.89 – 234.59 – 288.15(1.7919 – 1.1201) = 8.7208 kJ/kg ψ4 – ψ3 = h4 – h3 – T0(s4 – s3) = 1.004(35 – 15) – 288.15 × 1.004 × ln (308.15/288.15) = +0.666 kJ/kg η(II) = ma(ψ4 – ψ3)/m1(ψ1 – ψ2) = 1.007(0.666)/ = 0.77.

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