Question

A vapor compression refrigeration system operates at steady state with refrigerant 134a as the working fluid. Superheated vapor enters the compressor at 10 lbf/in2 , 0 oF. The liquid leaving the condenser is at 180 lbf/in2 , 100 oF. There is no significant pressure drop in the evaporator or condenser. For compressor efficiency of 83% and refrigeration capacity of 6 tons, determine

(a) the compressor power input in horsepower, and

(b) the coefficient of performance.

A vapor compression refrigeration system operates at steady state with refrigerant 134a as the working fluid. Superheated vapor enters the compressor at 10 lbf/in, 0 °F. The liquid leaving the condenser is at 180 lbf/in?, 100 °F. There is no significant pressure drop in the evaporator or condenser. For compressor efficiency of 83% and refrigeration capacity of 6 tons, determine (a) the compressor power input in horsepower, and (b) the coefficient of performance.

Answer 1

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Solm Given, Super beated Vapoir enters the compressor al 10 dl line and or 10 lbfline = 10x 0.0689 - 0.689 bay 0°F = -17.7:18°C Hiquid refrigerant leaves the condensar at 180 Ilyf/in" and 100F 180 lbtind 180X0.0689 = 12:402 bay 100 F = 37.748Ć Refrigeration Table for R-134a Temp Enthalpy (kJ/kg) Liquid TVарочү Sp. Heat (K5 /kg) Bessure (bat) (°C) Entropy Liquid vapour MS2 (S) ch) (ha) Vapour Liquid Vapour (C) (09) 148.57 0.5122 374.16 0.7973 -40 1.7649 1.243 0.740 0.689 377.52 1.758 0.756 0.8436 161:10 380.45 0.8498 1.7519 1.2600. 771 11.901 265.42 422.091.21851.7094 .539 1-175 422.72 12.402 267.88 1.7087 1.5502 - 12. 527 268.49 1422.88 1.2279 1.7086 1.553 1.196 The under lined values are obtained by interpolation method. Use any standard refrigerant table. I used Kothandaraman Refrigerant Tables and Charts.

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FUSTELU 2 MILLI /9 P = 12.402 bar = 190 1 T3= 37.778€ . P = 0.689 bar = to 157 m2 T-17.78 S hi - hg at P-0.689 bar hi= hit Cg (T-I) = 374.52 +0.756(-17.778 - (-34.66) = 390.283 kJ/kg = 377.52 kg/kg Cg at P=0.689 bar = 0.756 KJ/xgK T = -17.778°C ( Given) = 255-222K T,' - Saturation temp corresponding - to P=0.689 bar = - 34.66°C = 238.34 K si's Sg at P=0.689 bar = 1.758 kJIngk S, = si'+ Cgen (1) 1.758 +0.75Gxln Windo255.202 238.34 = 1.81 kJ/gK Si = S2 = 1.81 KJ ligy ( For isentropic compression 1-2). Now, - Sg at P = 12.402 bar S2 = Sa't Cg In (172) - 1.7087 ksgk. =) S2 = 1.7087 +1.192 en/ (320.6 1.81 = 1.7087 + 1.192 en 12 320.6 Cg at P = 12.402 bay = 1.192 M3/gk. T,' = Auto Saturation temp corresponding to P-12.402 = 47.6°C. = 320-6K bar =) T2 = 349.03 K

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hy = h' + Cg (T2-T2') = 422.724 1.192 (349.03 – 320.6) P= 12. hoz bay = 422.72 KB Ing. Og at P= 12.402 bar = 1.192 kJ /kgK = 456.62 kJikg ha= ba't C, (T3-73) = 267-88 +1.5502 (47.6-37.718) = 283.106 kJ like ho's be at P= 12.402 bar = 267.88 KJ/kg. C at P = 12.402 bar = 1.5502 Koligk T3 = 31.778°c (Given) T3' = Saturation temp at - P= 12.402 bar = 47.6°C bg- h4 = 283. 106 Kiling lisenthalten Given , Refrigeration capacity, QR = 6 tons = 6x 3.5 KW = 21KW Now, {Iton = 3.5 kw} - QR = mp Chi-hu). =) mp = OR = 21 = 0.196 kg/ hi-h4 390.283 – 283.106 a) Compressor power input P= mie (ha-ha2 = 0.196 ( 456:62 – 390.283) 0-83 15.665KW = 15.665 hp 0.746 - a k . b) C-0.2 = ho-h4 = 390.283 - 283.106 = 1.6156 ha-hi 456.62 - 390.283

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