11-13An ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The rate of heat removal from the refrigerated space, the power input to the compressor, the rate of heat rejection to the environment, and the COP are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a ) In an ideal vapor-compression refrigeration cycle, the compression process is isentropic, the refrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenser as saturated liquid at the condenser pressure. From the refrigerant tables (Tables A-12 and A-13),()()throttlingkJ/kg 82.88kJ/kg82.88liquid sat.MP a 7.0C 95.34kJ/kg 50.273MP a 7.0K kJ/kg 94779.0kJ/kg97.236 vapor sat.kP a 12034MPa 7.0 @ 3322122kPa 120 @ 1kPa 120 @ 11=≅==⎭⎬⎫=︒==⎭⎬⎫==⋅====⎭⎬⎫=h h h h P T h s s P s s h h P f g gThen the rate of heat removal from therefrigerated space and the power input to the compressor are determined fromand()()()()()()kW 1.83kW 7.41=-=-==-=-=kJ/kg 236.97273.50kg/s 0.05kJ/kg 82.8897.236kg/s 0.0512in41h h m W h h m Q L(b ) The rate of heat rejection to the environment is determined fromkW 9.23=+=+=83.141.7inW Q Q L H (c ) The COP of the refrigerator is determined from its definition,4.06===kW 1.83kW7.41COP inR W Q L11-15An ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The throttling valve in the cycle is replaced by an isentropic turbine. The percentage increase in the COP and in the rate of heat removal from the refrigerated space due to this replacement are to be determined.sAssumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis If the throttling valve in the previous problem is replaced by an isentropic turbine, we would have s 4s = s 3 = s f @ 0.7 MPa = 0.33230 kJ/kg·K and the enthalpy at the turbine exit would be()()()kJ/kg58.8248.2142802.049.222802.085503.009275.033230.0kPa120 @44kPa120 @34=+=+==-=⎪⎪⎭⎫⎝⎛-=fgs f s fg fs h x h h s s s xThen,()()()kW 7.72kJ/kg 82.58236.97kg/s 0.0541=-=-=s Lh h m Q and23.4kW 1.83kW 7.72COP inR ===W Q LThen the percentage increase in Qand COP becomes4.2%4.2%=-=∆==-=∆=06.406.423.4COP COP COP in Increase 41.741.772.7in Increase R R R L L L Q Q Q11-23A vapor-compression refrigeration cycle with refrigerant-134a as the working fluid isconsidered. The amount of cooling, the work input, and the COP are to bedetermined. Also, the same parameters are to be determined if the cycle operated on the ideal vapor-compression refrigeration cycle between the same temperature limits. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) The expansion process through the expansionsvalve is isenthalpic: h 4 = h 3. Then,kJ/kg 159.3=-=-=19.24349.40241h h q L kJ/kg 8.21019.24300.45432=-=-=h h q H kJ/kg 51.51=-=-=49.40200.45412in h h w3.093===kJ/kg51.51kJ/kg3.159COP in w q L (c) Ideal vapor-compression refrigeration cycle solution:kJ/kg 149.2=-=-=80.24904.39941h h q LkJ/kg 190.980.24971.44032=-=-=h h q H kJ/kg 41.67=-=-=04.39971.44012in h h w3.582===kJ/kg67.41kJ/kg 2.149COP in w q LDiscussion In the ideal operation, the refrigeration load decreases by 6.3% and the work input by 19.1% while the COP increases by 15.8%.11-75A regenerative gas refrigeration cycle using air as the working fluid is considered. The effectiveness of the regenerator, the rate of heat removal from the refrigerated space, the COP of the cycle, and the refrigeration load and the COP if this system operated on the simple gas refrigeration cycle are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats.Properties The properties of air at room temperature are c p = 1.005 kJ/kg·K and k = 1.4 (Table A-2).Analysis (a ) From the isentropic relations, ()()()K 4.4325K 2.2734.1/4.0k/1k 1212==⎪⎪⎭⎫ ⎝⎛=-PP T T sK5.4722.2732.2734.43280.02212121212=−→−--=--=--=T T T T TT h h h h s s C ηThe temperature at state 4 can be determined by solvingsthe following two equations simultaneously:()4.1/4.04k/1k 454551⎪⎭⎫ ⎝⎛=⎪⎪⎭⎫ ⎝⎛=-T PP T T sss T T T T h h h h 54454542.19385.0--=→--=ηUsing EES, we obtain T 4 = 281.3 K.An energy balance on the regenerator may be written asor,()()K3.2463.2812.3082.273431661436143=+-=+-=-=-−→−-=-T T T T T T T T T T c m T T c m p pThe effectiveness of the regenerator is0.434=--=--=--=3.2462.3083.2812.30863436343regen T T T T h h h h ε (b ) The refrigeration load iskW 21.36=-=-=K )2.19346.3kJ/kg.K)(2 5kg/s)(1.00 4.0()(56T T c m Q p L(c ) The turbine and compressor powers and the COP of the cycle are kW 13.80K )2.27372.5kJ/kg.K)(4 5kg/s)(1.00 4.0()(12in C,=-=-=T T c mW p kW 43.35kJ/kg )2.19381.3kJ/kg.K)(2 5kg/s)(1.00 4.0()(54out T,=-=-=T T c m W p0.478=-=-==43.3513.8036.21COP outT,in C,innet,W W Q WQ LL0︒C35︒-80︒(d ) The simple gas refrigeration cycle analysis is as follows:()()K 6.19451K 2.30814.1/4.0k/1k 34=⎪⎭⎫⎝⎛=⎪⎭⎫ ⎝⎛=-r T T sK6.2116.1942.3082.30885.0444343=−→−--=−→−--=T T T T T T s T η kW24.74=-=-=kJ/kg )6.21173.2kJ/kg.K)(2 5kg/s)(1.00 4.0()(41T T c m Q p L[]kW32.41kJ/kg )6.211(308.2)2.273(472.5kJ/kg.K)5kg/s)(1.00 4.0()()(4312in net,=---=---=T T c m T T c mW p p 0.599===32.4174.24COP innet,WQ Ls35︒。