Question

An air refrigerator working on Bell Coleman cycle takes air into the compressor at $-5^{o}C$ and compresses polytropically with $P{V}^{1.35}=C$ upto $5$ bar and cools to ${25}^{o}C$ at the same pressure. It is further expanded isentropically in the expander to $1$ bar. The isentropic efficiency of expander is $90$%. The net work required is ____ $kJ/kg$. (Correct upto three decimal place)

• A. 54.265

Answer 1

The correct option is A 54.265

$\frac{T_2}{T_1}={\left(\frac{P_2}{P_1}\right)}^{n-1/n}$

$T_2=T_1(5{)}^{1.55-1/1.55}$

$T_1=-5+273=268K$

$T_2=268(5{)}^{0.55/1.55}$

$T_2=406.77K$

$\frac{T_3}{T_4}={\left(\frac{P_3}{P_4}\right)}^{\gamma -1/\gamma }$ $[{\gamma }_p=5,\gamma =1.4]$

$T_4=\frac{T_3}{({\gamma }_p{)}^{\gamma -1/\gamma }}$

$T_3=25+273=298K$

$T_4=\frac{298}{(5{)}^{0.4/1.4}}$

$T_4=188157K$

${\eta }_T=0.9=\frac{(T_3-T_4^{\prime })}{(T_3-T_4)}$

$T_4^{\prime }=199.141K$

$W_{net}=W_C-W_T$

$W_C=\frac{n}{(n-1)R(T_2-T_1)}=153.618kJ/kg$

$W_T=c_p(T_3-T_4^{\prime })$

$99.35kJ/kg$

$W_{net}=W_C-W_T=153.618-99.35=54.265kJ/kg$