Solution_7.pdf - L An actual heat pump cycle with R-134a as...

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Unformatted text preview: L. An actual heat pump cycle with R-134a as the refrigerant is considered. The isentropic efficiency of t e compressor, the rate of heat supplied to the heated room, the COP of the heat pump, and the COP and the rate of heat supplied to the heated room if this heat pump operated on the ideal vapor-compression cycle between the same pressure limits are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) The properties of refrigerant-134a are (Tables A-11 through A-13) P2 = 800 kPa T2 = 55°C T 3 = [email protected] kPa = 29-06°C P3 = 750 kPa T 3 = (29.06 — 3)°C h4 = 113 = 87.91 kJ/kg T [email protected] kPa = —10.09°C S P1 = 200 kPa h] = 247.87 kJ/kg T1 = (—10.09 + 4)°c s1 = 0.9506 kJ/kg P2 = 800 kPa 52 =S1 }h2 = 291.76 kJ/kg }h3 = 87.91kJ/kg }hZS = 277.26 The isentropic efficiency of the compressor is hZS — hl _ 277.26 — 247.87 — = 0.670 It; — 11.1 291.76 -— 247.87 ’70: (b) The rate of heat supplied to the room is QH = rh(h2 —h3) = (0.018 kg/s)(291.76 —87.91)kJ/kg = 3.67 kW (c) The power input and the COP are Win = rh(h2 — h1) = (0.018 kg/s)(291.76 — 247.87)kJ/kg = 0.790 kW COP = Q.” :31 = 4.64 W- 0.790 m (d) The ideal vapor—compression cycle analysis of the cycle is as follows: 51 = [email protected] 200 m = 0.9377 kJ/kg.K P2 = 800 MPa }h2 = 273.25 kJ/kg S2 = S] h—h . — . 2 3=27325 9547 :6.” h2 —h1 273.25—244.46 COP = QH = mmz — h3) = (0.018 kg/s)(273.25 — 95.47)kJ/kg = 3.20 kW ‘2 . A geothermal heat pump is considered. The degrees of subcooling done on the refrigerant in the condenser, the mass flow rate of the refrigerant, the heating load, the COP of the heat pump, the minimum power input are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) From the refrigerant-134a tables (Tables A-11 through A-13) T4 =20°C P4 =572.1kPa x4 = 0.23 }h4 = 121.24 kJ/kg h3 = h4 P1 = 572.1 kPa }h1 = 261.59 kJ/kg x1 = 1(sat. vap.) s1 = 0.9223 kJ/kg P2 = 1400 kPa h2 = 280.00 kJ/kg S 2 Z 51 From the steam tables (Table A-4) hW1 = hf @50°c = 209.34 kJ/kg hw2 = hf @ 4m = 167.53 kJ/kg The saturation temperature at the condenser pressure of 1400 kPa and the actual temperature at the condenser outlet are [email protected] kPa = 52.40°C P3 =1400 kPa r3 = 48.59°C (from EES) h3 = 121.24 M Then, the degrees of subcooling is ATsubcool = Tsat —T3 = 52.40 — 48.59 = 3.81 °C (b) The rate of heat absorbed from the geothermal water in the evaporator is Q L = n'tw (hw1 — hwz) = (0.065 kg/s)(209.34 —167.53)kJ/kg = 2.718 kW This heat is absorbed by the refrigerant in the evaporator QL _ flfl— = 0.01936 kg/s h1 —h4 ' (261.59 —121.24)kJ/kg (c) The power input to the compressor, the heating load and the COP are W“ = th (h2 — h1)+ Q6111 = (0.01936 kg/s)(280.00 — 261.59)kJ/kg = 0.6564 kW 1 Q',, = ritR (h2 —h3) = (0.01936 kg/s)(280.00—121.24)kJ/kg = 3.074 kW COP = Q.” = M = 4.68 W. 0.6564 kW m (d) The reversible COP of the cycle is 1 1 —- = ———— = 12 92 1— TL /TH 1— (25 + 273) /(50 + 273) The corresponding minimum power input is W _ QH _3.074kW "”"m COPrev 12.92 MR: COPrev : = 0.238 kW . ‘ A gas refrigeration cycle with helium as the working fluid is considered. The minimum temperature in the cycle, the COP, and the mass flow rate of the helium are to be determined. Assumptions 1 Steady operating conditions exist. 2 Helium is an ideal gas with constant specific heats. 3 Kinetic and potential energy changes are negligible. Properties The properties of helium are cp = 5.1926 kJ/kg'K and k = 1.667 (Table A-2). T Analysis (0) From the isentropic relations, P (k—1)/k 0.667/1.667 TZS 2 T1 [7:] = (263K)(3) = 408.2K 50°C P (k—1)/k 1 0.667/1.667 40°C T45 = T3 —4 = (323K{—] = 208.1K P3 3 and _ h3—h4 _ T3—T4 hS—h4s T3_T4s =h2s —l11 =T2s _Tl h2_hl T2_T1 ——> T4 = T3 — UT (T3 — T43 )= 323 — (0.80)(323 — 208.1) = 231.1 K = Tmin 771 —> T2 = T1 + (T25 — T1)/ 27C = 263 + (408.2 — 263)/(0.80) = 444.5 K 77C (b) The COP of this gas refrigeration cycle is determined from COPR : 61L = 4L wnet,in wcomp,in — wturb,out _ hl —h4 (hz -la)-(h3 -h4) _ T1 _ T4 (T2 -T1)-(T3 -T4) 263 — 231.1 2 —__=0. 56 (444.5 — 263)— (323 — 231.1) 3 (c) The mass flow rate of helium is determined from rh = Qrefrig = Qrefrig : Qrefrig : 18 kJ/5 20.109 kg/S qL h1 —h4 cp(T1—T4) (5.1926 kJ/kg-KX263—231.1)K An ideal gas refrigeration cycle with air as the working fluid is considered. The minimum pressure ratio for this system to operate properly is to be determined. Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant specific heats. 3 Kinetic and potential energy changes are negligible. Properties The properties of air at room temperature are cp = 1.005 kJ/kgK and k = 1.4 (Table A-2a). T Analysis An energy balance on process 4-1 gives qRefrig = 6,. (T1 -T4) 20°C - -5°C T4 = T1 — “em = 268 K ——20kJ—/kg—— = 248.1 K cp 1.005 kJ/kg - K The minimum temperature at the turbine inlet would be the same as that to which the heat is rejected. That is, T3 = 293 K Then the minimum pressure ratio is determined from the isentropic relation to be £_ T_3k/(k-1)_ 293K 1.4/0.4—179 P4 T4 248.1K ' ...
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