Question

The heat-pump system shown is designed to provide water-heating and space-cooling. The system includes a heat exchanger to subcool the R134a refrigerant at the outlet of the water heater while heating the refrigerant at the outlet of the evaporator. Given the following data, determine the enthalpy at each station, 1 through 6, using the R134a property tables (click here). Type your answers hy through h6 in the space provided below, and show all your work on your paper. · At the compressor inlet, T1 = 20°C and P1 = 320 kPa. • At the compressor outlet, T2 = 90°C and P2 = 2 MPa. · At the water heater outlet, the refrigerant is in a saturated-liquid state. · At the outlet of the evaporator, the refrigerant is in a saturated-vapor state. · The power supplied to the compressor is comp = 2.7 kW. evaporator section condenser section air flow Qcool P2, T2 AMAT water heater space cooling compressor (adiabatic) Qhw fan P5 = P6 insulated duct P6 = P1 sat. vapor 6) Wcomp throttle (adiabatic) heat exchanger (adiabatic) P1, T1 P3 = P2 sat.liquid P4 = P2 refrigerant R134a

Answer 1

compressor PG=P, 15= P6 Sat. Vapor throttle, (adiabatic) sat liquid P₂ = P2 Py = P2 At compressor inlel : Ti = 20°C, P = 320 kPa At the compressor outlet: T2 = 90°c and P2= 2 MPa From the property table of R134a : At T, = 20°c and P, = 320kPa h = 415.80 KJ kg At T2 = 90°c, P2= 2 MPa h2= 458.38 KJ kg Powere supplied to the compressor = 207 KW = m (ha hi m- 2.7 kW (458-38-415.80) K = 0.0634kg kg Mass flow rate of R134a = 0.0634 Kg so, Enthalpy at point m. hi = 0.0634 Kg x 415.80KT kg S Hi = 26.36 KW dermal stress

Х 458.38 KJ Enthalpy at polut @ = mh2 = 0.0634 u kg H₂ = 29.06 KW The H₂= mhz P2 = P₂ = 2 MPa and saturated liquid. (hg = Specific enthalpy from property lable) = 0.0634 kg x 299.95 KJ H₂ kg 19.02 KW As, Pu= P2 = 2 MPa and PG=P = 320kPa and saturated Vapour At point ②, Ho=m.ng = 0.0634 kg x 400.04 kJ R9 Ho = 25-36 kW we get Doint the enthalpy balance Across heat exchanger, Ho + H3 = Hit Hu 0 . Hy = Hot H₃ - Hi = (25-36 + 19:02 - 2636) KW Hy = 18.02 kW Hy = 18.02 kW Since thetting is bsoeuthalpic process: Hu= H5 18.02 kW Any

a Powere absorbed by the water heatere = H₂ - H₂ (29.06 - 19:02) kW 10.04 kW sus COP WH - Heat abrorbed by Water heaters Wondek done by compressor H₂ - H₂ 10.04 kW H₂ - Hi (29.06- 26.36) kW 3.72 = COPWH Aus m3 200 quantity of watere = 200 lit. .: Mass of watere - (200 x 10-3) m3 x 1000kg 1 kg AT= (550-20°c) = 35% Doing the energy balance across the water heatere: (time) x Heat absorbed 10.04 kWxt Quote e 200kg X 4.2 kJ x (55-20) kg.k t 200 X 4-2 x (55-20) kJ.ķ Rate of by watere = m.CPW AT - KJ 10.04 X Kys) kg.k хK + - 2928.287 sec It will take 2928.287 seconds. Ans.