Question

A refrigerator operates between $120kPa$ and $800kPa$ in an ideal vapor compression cycle with $R-134a$ as the refrigerant. The refrigerant enters the compressor as saturated vapor and leaves the condenser as saturated liquid. The mass flow rate of the refrigerant is $0.2kg/s.$ Properties for $R-134a$ are as follows:

$\begin{array}{cccccc}& & & \text{Saturated R- 134a}& & \\ P\left(k{P}_a\right)& T{(}^{\circ }C)& h_f\left(kJ/kg\right)& h_s\left(kJ/kg\right)& s_f\left(kJ/kgK\right)& s_g\left(kJ/kgK\right)\\ 120& -22.32& 22.5& 237& 0.093& 0.95\\ 800& 31.31& 95.5& 267.3& 0.354& 0.918\\ & & & \text{Superheated R- 134a}& & \\ P\left(k{P}_a\right)& & T{(}^{\circ }C)& h\left(kJ/kg\right)& & s\left(kJ/kgK\right)\\ 800& & 40& 276.45& & 0.95\end{array}$

The rate at which heat is extracted, in $kJ/s$ from the refrigerated space is

• A. $28.3$
• B. $42.9$
• C. $34.4$
• D. $14.6$

Answer 1

The correct option is A $28.3$

From given $R-134a$ tables ,
we get
$h_1=237kJ/kg$
$h_2=276.45kJ/kg$
$h_3=h_4=95.5kJ/kg$


Mass flow rate: $m=0.2kg/s$

Rate of heat extracted:
$Q_{4-1}=m(h_1-h_4)=0.2(237-95.5)$

$=28.3kW$