Answer to Question #211714 in Mechanical Engineering for SID

Question

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.

Answer 1

$T = 20°C$

$T_{svp, i} = -20°C$

$T_{svp, e} = 50°C$

$T_{con.} = 40°C$

mass flow rate = 0.2kg/s

c.o.p = 2.3

P₂ = P₃ = P_sat 40°C = 1017 kPa

h₄ = h₃ = 256.54 kJ/kg

s₂ = 1.7472; h₂ = 430.87; s_q = 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