Question

available ur tank V=0.2 m reversible, adiabatic compressor R22 Thank = -20°C Vii = 0.05 m I Wc Tomo = -10°C m=0.1 kg/s for t = 10 min Pcout = 18 bar Problem sketch A tank with volume V = 0.2 m2 contains refrigerant R22 at Ttank = -20°C, as shown in the problem sketch. Initially the tank contains V1,1 = 0.05 m2 of liquid R22 and the rest of the tank contains vapor R22. A reversible, adiabatic compressor operating at steady state pulls only saturated vapor from the tank and compresses it to Pc,out 18 bar. The mass flow rate through the compressor is m = 0.1 kg/s. The compressor operates for t = 10 min. During this process, the tank is maintained at a constant temperature by heat transfer from the surroundings at Tamb = -10°C. The R22 in the tank remains in a two-phase state throughout the process. Use these tables for R22 properties. show answers a. Determine the temperature of the R22 leaving the compressor. b. Determine the power required by the compressor. C. What is the initial quality of the R22 in the tank? d. What is the quality of the R22 that remains in the tank? e. What is the heat transfer to the tank required to maintain its temperature? f. Is this process reversible? If not, explain how entropy is generated by the process and calculate the amount of entropy generated by the process.

Answer 1

given: 1+ thitial conditinak tank v="2m3 tank I final condition states of tank Trank T=-26c Wc = -20°c Uf = .05m3 V = = 62-105=115m3 Tamba10c #m=0.1Kg/S t=lomin Prout = 18 bar 20 mfting from R-22 table! at Tsat =-200 Uf = .00074 m3 1kg Vo = 109 268 m3lkg mass of •05 me= Vf liquid = 67.567 kg VE 106074 mass of Mg = g = -15 = 1.618 kg m, a my tong = 69.185kg vapeur Vg · 09268 mg --02339 initial quality of R22 to tank y = 1777-02339 22041 = Mi = 177+6233x220.1 = 182.12 Sq = 1.783 kJ lkgK h2 = 39.701 KG 1kg since compressor is reversible adiabatic re. Kentropic Sz=53 T 1.783 = 1.60+ 1.069 In Temp of R-22 leaving comp. T=348.769 K h3 = 417.1 + 1.669x 348-769-319.4) n3 449.16 4 K5 1kg Power = = 1X ( 448.174-3971) P= 5.106 kW his 273+46 71 in (ha-ha) Am Mont = mxt = 1X 10x60 = 60 kg Volume of Tank - .2 mass left = 67.56771.618-60 V=2 =40277 me 9.85kg 9.15 021775.00074 + ul.09 268-.00074) X = 2287 quality of R22 semaining in tank

from energy balance? mu, to m'ultam hz mil - mass heft in tank Assumption! mass out uch mi hith=minit Amhz hi'= 177+ 22878 220.1 = 227.33 krig om = e 69. 195X18212+9 = 9.105x22733 +60x 3971 Heat Transfer Q = 13314 KJ As =-(s +_+ im's!+0m2) Tamid AS – 9.185XP0123+60X1-783-69.18 5X0-9342-13314 263 as = 1.9402 kJ IK S, -935+. 0239.869=9342 hene since as is Positive Si = .91354.2287X18695 given process is foreversible process. Si=1.1123 Ilkok PLSS RATE THE SOLNTION"