Question

An air refrigeration system operating in open air cycle is required to produce 25 ton refrigeration with a cooler pressure of 12 bar and the refrigerator pressure of 1 bar. The temperature of air entering the compressor is at $0^{\circ }C$ and entering the expander is ${25}^{\circ }C$. Assume the isentropic efficiency for compressor to be 90% and polytropic expansion for expander to be n = 1.35. The COP of refrigeration system is (Take $\gamma$ for air = 1.4 and $c_p$ for air = 1.005 kJ/kgK) .
(Correct upto 3 decimal places)

• A. 0.739

Answer 1

The correct option is A 0.739

$\frac{P_2}{P_1}=12$

$T_3={25}^{\circ }C=298K$

$T_1=0^{\circ }C=273K$

$\frac{T_2}{T_1}={\left(\frac{P_2}{P_1}\right)}^{\frac{\gamma -1}{\gamma }}=(12{)}^{0.2857}$

$T_2=555.245K$

$\frac{T_4}{T_3}={\left(\frac{P_4}{P_3}\right)}^{\frac{n-1}{n}}=\frac{1}{\left(12\right)\frac{1.35-1}{0.35}}$

$T_4=156.47K$

${\eta }_c=\frac{T_2-T_1}{T_2-T_1}$

$a=\frac{555.245-273}{T_2^{\prime }-273}$

$T_2^{\prime }=586.60K$

So, $W=c_p(T_2^{\prime }-T_1)$

$W_c=1.005(586.60-273)=315.168kJ/kg$

$W_T=\frac{n}{(n-1)}R(T_3-T_4)=\frac{1.35}{0.35}\times 0.287\times (298-156.47)=156.673kJ/kg$

$W_{net}=W_c-W_T=158.494kJ/kg$

$COP=\frac{Q_2}{W_{net}}=\frac{c_p(T_1-T_4)}{158.494}=\frac{1.005\times (273-156.47)}{158.494}=0.739$