Question

through the home central air conditioning as a 1000 kpa. Any heat transfer from 1. Saturated R 134a vapor enters the uninsulated compresor of an wystem atutes the flow rate of Rwaya and the electrical power input is 5 kul. The exit compressor is to the ambient environment at 10. a Determine the rate of en tropy production for the process in kulis. b) It the change in enthalpy for the isentropic compression is 29.5 usleg, Cactual and isentropic compression process in a T-s diagram compresa te is the

Answer 1

1000 kPa 652 2 25 1000 kPa 25 2 P 77 42 sat 3.5bal S h 1-25= Isentropic Compression 1^2= Actual Compression. Sat. Vapow at 4 is taking. The number seems like 4. Wir 5 kW (on compressor 0.1 kg/3 m que at From P-h chart, sat. Vapout at 46 5= 1.73 kilig-k h = 400 kJ/kg We know s=5.25 (Isentropic process has = 428.5 rolig has -h,= 28.5 kJ/kg ons at P = 1000 kPa & 65°c h₂ = 447.5 ki/ng S2=1.8 kJ/kg-k 10

2 We have Energy Q W ( balance equation by neglecting KIE & Pot. Energy, mi (ha-h.) W = n5kW (on the system Q-(-5)=0 =0.1(447-5-400 Q: -0.25 kW Means heat rejected or. heat loss to surrounding Now Rate of Entropy Produced Algen As sys t To= 27C in (52-5.) + = 300k heat 0.1 (18-1-73) + gained from 0.0078 KW/k DS surround For Schrounding 0.25 300 ( i3 has - hi hathi I sentropic Efficie Piciency v V = 28.5 0.6 = 160% ] 447.5 - 400 Thank you. All the Best. Small Rindly provide hence values tolerence on final answer taken from chart. Thank you. ave