Question

Thermodynamics. No interpolation needed.

Problem #3. Refrigerant 134a is the working fluid for vapor-compression refrigeration cycle. The evaporator temperature is 8°C and the condenser pressure is 12 bar. Saturated vapor enters the compressor and superheated vapor enters the condenser at 60°C and exits the condenser as saturated liquid. For a refrigeration capacity of 8 tons or 2.816 x104 J/s determine the following: (1) The refrigerant mass flow rate in kg/s; (2) The compressor isentropic efficiency [Hint: Interpolation is required); (3) The compressor power in kW; (4) The system coefficient of performance; and (5) Sketch the Ts diagram.

Answer 1

Condenior Given:-) Eraporator temperature = 8°C condenser pressure = 12 but Superheated rapor enters the condenser = 60c Refrigeration capacity = 8 tous = 2.816810" Ils rapor compression Cycket at point 1 → T=8c from tables wi=255 kJ /kg S = 0.9273 KJ / kyk at point 2a → P2a = 12 bar, Mize = 278 5 noing constant entaropy Sza = Si= 0.9272 kJ/gk. at point 2 → P2 = 12 bar, T2 = 60c 12 = 2876 FTS at point 3 P=12 bar Hg = 1178 Ftlig constant entrally process til ₂ = hu=117.8 KI/B Erapor

1 mass flow rate. refafijeration capacity = m Che-ha) 2.816 X104 = (255-178) x103 m = 0.200 Fylkec compressor esenbiopic efficiency > The he-h2a - 255– 2485 255-289.6 he = 6792 4. » И — И. 9) COM DJu8km poo 02 = b (ha-k) = 0.204 (207.6-255) w = 7.06 kW 29 Coefficient of performance RC Cop= 2.816x104 wolk 706810? Cop =3.96