Question

In the vapour compression cycle shown in the figure, the evaporating and condensing temperature are $260K$ and $310K$, respectively. the compressor takes in liquid-vapour mixture (state 1) and isentropically compresses it to a dry saturated vapour condition (state 2). the specific heat of the liquid refrigerant is $4.8kJ/kg-K$ may be treated as constant. The enthalpy of evaporation for the refrigerator at $310K$ is $1054kJ/kg$.


The difference between the enthalpies at state points $1$ and $0$ (in kJ/kg) is

Answer 1

Given data:
$T_e=260K$
$T_c=310K$
$c_{pl}=4.8kJ/kgK$
$h_{fg}=1054kJ/kg$ at $T_c=310K$


$\frac{h_1-h_0}{260}=s_1-s_0(\because s_1=s_2)$

$\frac{h_1-h_0}{260}=s_2-s_0....\left(i\right)$

$s_2-s_3=\frac{1054}{310}...\left(ii\right)$

$s_3-s_0=c_{liquid}\ln \frac{310}{260}$

$c_{liquid}=4.8kJ/kgK$

$(s_2-s_3)+(s_3-s_0)=s_2-s_0$

Adding equation $\left(ii\right)and\left(iii\right)$
$(s_2-s_3)+(s_3-s_0)=s_2-s_0$

$=4.8\ln \frac{310}{260}+\frac{1054}{310}$

$=4.2442kJ/kgK$

Therefore From equation (i)

$\frac{h_1-h_0}{260}=4.2442$

$h_1-h_0=1103.51kJ/kg$