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* 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 hq through ho in the space provided below, and show all your work on your paper. • At the compressor inlet, T1 = 20°C and P1 = 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 W 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
fOne question but divided into sub connected questions

Answer 1

syn P=1&ring 3 و معنا -.P=0.2014 20°C -10.09 - 3.3 kw 4+ Stale( P,-0.2MNA, TI = 20°C) from pressure fasle, At p=0.2 Mina, Tear = -10.09°c Tr> Tsar i fence stale Ouperheated state from superheated farle, hi = 267.78 ls/ks Af state2 (P = 1.8 une To = 1050 c) from superheated farce h₂ 333.25 lglies At states (P=18mpe, a=0) from pressure lashes 142.22 hoahol 3.3 = n(haha) 3-3 (333.25 - 267.28) 0.0004 resp m ure fasce, you Heat removed sy Isrne condenser (Sort) = in (h₂ h₂) Sort At state 6 (P = 0.2 Nng, x=1) from pressure talle 2413 lug ho = hel 0-2M na -0.050yX(333.25-142-22) 9.6288 low Heat required by water heater me at hy = (mare) = (200X06001)x 109 x 42 x (55-20) 29400 les han hinho = hiz why ho Chicho) a hua 142022 – (267.78 - 2413) - hy = 115.74 chy 294ou lg huchs = 115-74 laste time = gaus 9.6288 (19) BO53.33 see 50.889 minutes (In Heat pump, as desired effect- Cory is ufilised

lismng Qook24 116 115.74 142.22 241.3 267.78 333.25 hikilig) An м ondenser = 0.0004 kofs q (a) power absorbed by water heater Heat removed by 9. 6288 kW 9.6288 2.917 COP WH = 3.3 Rate of heat transfer to evaporater from the air slowing pate of heat taken by evaporater (9) = M( ho-hs) 0.0504(241.3-115.74) = 6.328 10w 6.328 lo917 3.3 (6) (COP), Win 2 Max Upx 6.328 air Rate of theat transfer sy air = Mr G (AT) X AT = > max 1.0854 15 Mai Mass flow rate of =) Ma=0.4198 ks/