# Answer in Mechanical Engineering for SID #211714

A refrigerator using R-134a is located in a 20 0 C room. Consider the cycle to be ideal, except that the

compressor is neither adiabatic nor reversible. Saturated vapor at -20 0 C enters the compressor, and the

R-134a exits the compressor at 50 0 C. The condenser temperature is 40 0 C. The mass flow rate of

refrigerant around the cycle is 0.2kg/s and the c.o.p is measured and found to be 2.3. Find the power

input to the compressor and the rate of entropy generation in the compressor process.

“T = 20u00b0C”

“T_{svp, i} = -20u00b0C”

“T_{svp, e} = 50u00b0C”

“T_{con.} = 40u00b0C”

mass flow rate = 0.2kg/s

c.o.p = 2.3

P2 = P3 = Psat 40°C = 1017 kPa

h4 = h3 = 256.54 kJ/kg

s2 = 1.7472; h2 = 430.87; sq = 1.7395; h1 = 386.08 = qL/wC

wC =qL/(h1-h4) = (386.08 – 256.54)/2.3 = 56.32

WC = mwC = 11.26 kW C.V.

Compressor h1 +wC+q=h2>> gin=mmh2 – h1 – WC = 430.87 386.08- 56.32 = -11.53 i.e. a heat loss s 1 + dQ/T + sgen = s2 s gen = s2 – s1 – q / To = 1.7472-1.7395+ (11.53/293.15) = 0.047.. Sgen = m sgen = 0.2 0.047 = 0.0094 kW

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