Question

Consider a Carnot refrigeration cycle executed in a closed system in the saturated liquid-vapor mixture region using 0.909 kg of refrigerant-12 as the working fluid. It is known that the maximum absolute temperature in the cycle is 1.25 times the minimum absolute temperature, and the net work input to the cycle is 25 kl. If the refrigerant changes from saturated liquid to saturated vapor during the heat rejection process, determine the minimum pressure in the cycle, in kPa.

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Answer 1

Son: The data given for carnot refrigeration cycle are mzo Tmin - The cop of & COP = car onot TL = TH-Ti nax = 1.25. Work input. W= 25 KI Refrigeration cycle is Tmin e t- Tmax - Tmm - Tmax T max- Temin cop Tmin e Tras-1 top-1.25 -1 = 0.25, :. (cop=25= 4) also, cop= RE TRE = 4x 25 = 160 KT) R.E Cop= E = 4x 25 = 100 kJ W .R - The T-S and Pr-y diggram for Carnot refrigeration cycle co PV=C 0 OVOU PV=c pvc - > - V~ from p-v diagram it is observed that and also obsered that, T2 = 73=Tmax emin is at state I Ti Tha Tmin) - (for R-12, 8=1.14) (7) _ 6.154 0 = ( Thaime lei14- 12 = ( Tmax ) - 1 = ( 1025 ) 119-1 Tihan] V2 = 4.923 -

- compressor work for cannot refrigeration cycle as, d a n W.R- P2 : V2-PIVI - MR C T2-T - o. gog, 0.0688. (T2-TDX103 1.14 - -(for R-12, R = 0.0608 Kol ng k) Dit is 100 25X103 = 0.909 X6-0688x18 (T2-TD 1:14-1 - 2 - = 15.9 x - 2) . - Tmar- Tmin= 55.965- (3) But Tomar Tmn ):25 PRC ont bils Sao TDA Tminx1.25- Tmin= 55.965 0.25 Tmin= 55.965 Tmin= 223.9kZ: Tmax = 279.825k/

i. Properties of R-12 at Tmin=223.9k v=0.3831 m3/kg, but mrt 2, N=0.3482 m)

,, PV = m РТ. Р. = oqoyxo. 06 88 х 10 х 223 Ф• 3482 JP, – Pm) = 4o. 21ч ЕР° |