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I need your help solving this thermodynamics problem please:

Question Completion Status: QUESTION 1 20 points Save Ans 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 ha through h6 in the space provided below, and show all your work on your paper. • At the compressor inlet, T1 = 20°C and P4 = 200 kPa. . At the compressor outlet, T2 = 105°C and P2 = 1.8 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 = 3.3 kW. evaporator section condenser section Click Save and Submit to save and submit. Click Save All Answers to save all answers. Save All Answers Save and Submit

Answer 1

solution: Ti z 20°C, P,= 200kPa T2 =105°c ,P2 = 1.8 MPa incomp 23.3 KW P-H chart the is for suplem given below: a 4 3 1.8 Mila 2 200 ila 5 6 h of Risha, from superheated rapour properties T, -20°C, hi= 270.2 kJ/kg P=200 ICPA دما T2 = 105°C P2 = 1.8 MPa 2 h2= 330.8k5/1g State 3 is saturated liquid at P = 1.8 MPa :. h3=ht3 = 144.1 KS / kg

P State 6 is sahurated rapons, = 200 kPa :. h6=hg6 = 244.5 kJ/kg = 3.3 KW (63): Compressor work Wcomp mcha-hi) im (330.8-270.2) = 3.3 m=0.0545 kg/s = 3.27 kg/min - (Aurs) There is an exchange of heat at adiabatic heat ex changer 76 4 3 Heat exchanger Energy balance equation for Heat exchanger, hi-h6= ha-ha L. 270.2 - 244.5 = 144,1-44 h4 = 118.3 kJ/kg Process 4-5 is an adiabatic thusaltling process huchs is :h5=118.3 KS/1g

Enthalpay at various stages are tabulated below: State Enthalpy (15/16) hi 270.2 h2 330.8 h3 144.1 hu 118.3 hs 118.3 ho 244.5 Wheater (04) Power absorbed by water heater incha-ha) Wheater = = 0.0545 (330.8-144.1) :) Wheater = 10.175 KW Aus) in Chaaha) 10-175 COPWN = il camp 3.3 COR WH=3.083 (Ans)

(105): Mwater 200 liters = 200 2001g AT-55-20 = 35 specific heat of water at constant pressure up2 4:182 15/12 :mcha-h3)t = mwater & Upx AT .(to time baken) Es mwater & Cp x AT mcha-ha) 2004.1828 35 10:175 t = 2877.05 s = 47.95 min (Aus) (06): Rate heat transfer to the evaporator, Wevap ncho-hs) Wevap 0.0545 ( 249.5-118.3) Werapa 6.872 KW COPAC 2 WEvap inicomp 6.872 3.3 COPAC = 3.572

(L07): AT ain 15°C Cpai = 1.00565/1g Heat by Energy loss by air passing through duct gained R134a through section .: mais x 1.005x15 = 6.872 : Mair =0.456 Igls s --(And