A homeowner with a large house has recently installed a heat pump with the working medium R134a. The process can be cons

[answerhappygod]

A homeowner with a large house has recently installed a heat
pump with the working medium R134a. The process can be considered
as an ideal steam compression cooling cycle with two exceptions. In
the compression step there are some losses and after the
condensation the liquid is cooled to 48 ° C. Determine the
COPHP of the system if the evaporator pressure is
320 k Pa and the condenser pressure is 1.6 M Pa. The temperature in
the compressor outlet is 90 ° C.
Also determine the compressor efficiency and draw the process in a
T-s diagram and in an enthalpy - pressure (P-h) diagram. Both with
saturation lines.
I have attached the calculations but donate understand where the
values for h and s are coming from. Could someone please clarify
this?

A Homeowner With A Large House Has Recently Installed A Heat Pump With The Working Medium R134a The Process Can Be Cons 1 (149.67 KiB) Viewed 17 times

T 3 (48°c) COP HP 1,6 MPa = 320kPa 4 25 20 (100°c) 1QH) |WINI h₁ = hg, 320kPa = h3 =h ¢,48°C • 320kPa = 251,9 kg / kg = 120,4 -~ haa (90°C, 1.6 hPa) = 316,5 kg/kg COPHP 316,5-120,4 3,04 316,5 - 251,9 has -h₁ Ок hea - h1 3 ५ S mal (haa - h3) in wint (haa - ha) Kond Foring are za QL 5₁ = 0,9303 kg (kesk) ку 3165 has via Sos = S₁ vid givet P 1
h2s = 286 курку 286 251,9 ђк 316,5 257,9 = 0,528 P 25 20 ду в ку/ку биси сорнР = 3,0, 9 к= 5300