Consider a Carnot Refrigeration cycle with a working fluid of R-134a operating b

Question

Consider a Carnot Refrigeration cycle with a working fluid of R-134a operating between temperatures of -10?C and 20?C. The compressor in the cycle requires 500 W of power.

a)  What is the maximum coefficient of performance?

b)  What is the rate of heat that is removed from the refrigerated space?

c)  What is the rate of heat that is rejected from the cycle?

d)  The R-134a enters the evaporator as a saturated mixture with a quality of 0.1. If the R-134a leaves the evaporator as a saturated mixture with a quality of 0.9 what is the flow rate?

Explanation / Answer

For refrigerant R134a

at -10 C

hf1 = 186.7kJkg,hg1=392.66 kJ/kg

sf1=0.9506, sg1=1.7334

at 20 C

hf2=227.47kJ/kg, hg2=409.75kJ/kg

sf2=1.0962, sg2= 1.7180

Now

h1=hg1=392.66,

s2=sg2

Now

since s1<sg2

so point 1 will inside the vpour dome and let x be the quality at inlet of compressor

so, s1=s2

      sf1+x(sg1-sf1)=s2g

      0.9506+x(1.7334-0.9506)=1.7180

on solving we get x= 0.98

so h1=hf1+x(hg1-hf1)=186.7+0.98(392.66-186.7)=388.54

h2=h2g=409.75

h3=hf2=227.47

h4=h3=227.47

Win=0.5kW

so mass flow rate(m)=Win/(h2-h1)=0.5/(409.75-388.54)=0.0235 kg/s

(a) Maximum COP=T2/(T1-T2)= (-10+273)/30=8.76

(b) rate of heat removal =m(h1-h4)= 0.0235(388.54-227.47)=3.797kW

(c) heat rejected in condenser= m(h2-h3)=0.0235(409.75-227.47)=4.283kW

(d) for this case

h4= hf+0.1(hg-hf)=186.7+0.1(392.66-186.7)=207.3kJ/kg

h1=hf+0.9(hg-hf)=186.7+0.9(392.66-186.7)=372.06 kJ/kg

so flow rate = heat removal rate/(h1-h4)=3.797/(372.06-207.3)=0.0230 kg/s

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