Question

A reverse Joule cycle is working with a pressure ratio of 5. If the temperatures at the inlet of compressor and expander are ${27}^{\circ }C$ and ${57}^{\circ }C$ respectively and the efficiencies of compressor and turbine are 0.8 or $80\%$, then the COP of the system is

• A. 0.5535
• B. 1.7128
• C. 1.9326
• D. 2.3439

Answer 1

The correct option is A 0.5535


Given: $r_p=5,\gamma =1.4,{\eta }_C={\eta }_T=0.8$

$T_1=300\text{K},T_3=330\text{K}$

$T_{2s}=T_1\times (r_p{)}^{\frac{\gamma -1}{\gamma }}$

$T_{2s}=475.1459\text{K}$

Now,
${\eta }_C=\frac{T_{2s}-T_1}{T_2-T_1}$

$T_2=\frac{T_{2s}-T_1}{{\eta }_C}+T_1$

$T_2=518.9324\text{K}$

$T_{4s}=\frac{T_5}{(r_p{)}^{\frac{\gamma -1}{\gamma }}}$

$T_{4s}=208.3571\text{K}$

Now, ${\eta }_T=\frac{T_3-T_4}{T_3-T_{4s}}$

$T_4=T_3-{\eta }_{\gamma }(T_3-T_{4s})$

$T_4=232.6857\text{K}$

COP $=\frac{Q_{14}}{W_C-W_T}=\frac{c_p(T_1-T_4)}{c_p(T_2-T_1)-c_p(T_3-T_4)}$


$=\frac{300-232.6857}{(518.93-300)-(330-232.6857)}=0.5535$