86kjkgbecausethecompressorisisentropics1

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Unformatted text preview: .05 kg/s. Show the cycle on a T‐s diagram with respect to saturation lines. Determine (a) the rate of heat removal from the refrigerated space and the power input to the compressor, (b) the rate of heat rejection to the environment, and (c) the coefficient of performance (COP). Answers: (a) 7 kW, 2 kW; (b) 9 kW; (c) 4 Solution The ideal vapor‐compression refrigeration cycle is shown below on a T‐s diagram. T P2 = 0.7 MPa qH 2 win P1 = 0.12 MPa 3 1 4 qL s a) We will find qL and win first. From Table A‐11 (saturated pressure table), s1 = sg1 = 0.9354 kJ/(kg K) and h1 = hg1 = 233.86 kJ/kg. Because the compressor is isentropic, s1 = s2. Interpolating in Table A‐12 (superheated r‐134a) with s2 to find h2 gives h2 = 270.22 kJ/kg. From Table A‐11, h3 = hf3 = 86.78 kJ/kg. For a throttling device, h3 = h4 = 86.78 kJ/kg. Therefore, QL mh1 h4 0.05233 .86 86.78 7.354 kW and Win mh2 h1 0.05270 .22 233 .86 1.818 kW . b) The rate of heat removal from the compressor is Q...
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