# ES_202_EES_Refrig_HW_Answers_Check - 1 equation page(for...

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ES 202 EES Project – Cycle Analysis Problem 1 Vapor-Compression Refrigeration Cycle A vapor-compression refrigeration cycle uses R-134a to provide cooling for air conditioning. The compressor has an efficiency of 85%. The mass flow rate of the R-134a is 6 kg/min. Using EES, complete the following table, then find the compressor work (kW), the refrigeration capacity (kW), and the COP R . Note: in every case, P 2s = P 2 = P 3 and P 1 = P 4 . Table 1. Refrigeration Cycle Properties State T (°C) P (kPa) h (kJ/kg) s (kJ/kg-K) other 1 4 Sat vapor 2s 1200 2 1200 3 1200 Sat liquid 4 Repeat the above exercises for the various values of T 1 and P 2 shown below in Table 2. Table 2. Refrigeration Cycle Results for Various T 1 and P 2 Values. T 1 (°C) P 2 (kPa) in W (kW) in Q (kW) COP R 4 1200 4 800 -10 1200 -10 800 Deliverables – Vapor Compression Refrigeration Cycle 5 tables – 4 versions of Table 1, plus Table 2 (you may, but need not, print these from EES)

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Unformatted text preview: 1 equation page (for the case with T 1 = 4 °C and P 2 = 1200 kPa) printed from EES 1 solutions page (for the case with T 1 = 4 °C and P 2 = 1200 kPa) printed from EES ES 202 EES Project – Cycle Analysis Correct answers to check your results: Vapor-Compression Refrigeration Cycle – Sample EES Results T1=4 [C] (given) p2=1200 [kPa] (given) x1=1 (given) x3=0 (given) η comp =0.85 T2=54.75 [C] T2s=50.71 [C] T3=46.29 [C] T4=4 [C] p1=337.9 [kPa] p3=1200 [kPa] p4=337.9 [kPa] h1=252.8 [kJ/kg] h2=283.7 [kJ/kg] h2s=279.1 [kJ/kg] h3=117.8 [kJ/kg] h4=117.8 [kJ/kg] s1=0.9293 [kJ/kg-K] s2=0.9435 [kJ/kg-K] s3=0.4244 [kJ/kg-K] s4=0.4422 [kJ/kg-K] x4=0.3095 powerin=3.097 [kW] qin=13.5 [kW] COP R =4.359...
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