Question

A refrigerator based on ideal vapour compression cycle operates between the temperature limits of $-{20}^{\circ }C$ and ${40}^{\circ }C$. The refrigerant enters the condenser as saturated vapour and leaves as saturated liquid. The enthalpy and entropy values for saturated liquid and vapour at these temperatures are given in table below.

$\begin{array}{ccccc}T& h_f& h_g& s_f& s_g\\ {(}^{\circ }C)& \left(kJ/kg\right)& \left(kJ/kg\right)& \left(kJ/kgK\right)& \left(kJ/kgK\right)\\ -20& 20& 180& 0.07& 0.7366\\ 40& 80& 200& 0.3& 0.67\end{array}$

If refrigerant circulation rate is $0.025kg/s$, the refrigeration effect is equal to

• A. $2.1kW$
• B. $2.5kW$
• C. $3.0kW$
• D. $4.0kW$

Answer 1

The correct option is A $2.1kW$

$m=0.025kg/s$

At $T_c={40}^{\circ }C$

From given table, we get

$h_2=h_g=200kJ/kg$

$s_2=s_g=0.67kj/kgK$

$h_3=h_f=80kJ/kg=h_4$



At $T_e=-{20}^{\circ }C$

From given table, we get

$s_f=0.07kJ/kgK$

$s_g=0.7366kJ/kgK$

$h_f=20kJ/kg,h_g=180kJ/kg$

$s_1=s_2=s_f+x_1(s_g-s_f)$

$0.67=0.07+x_1(0.7366-0.07)$

or $x_1=0.9$

Specific enthalpy at point 1,
$h_1=h_f+x_1(h_g-h_f)$

$=20+0.9(180-20)$

$=164kJ/kg$

Refrigeration effect,
$RE=m(h_1-h_4)$

$=0.025(164-80)=2.1kW$