A commercial refrigerator with refrigerant-134a as the

Step 1 of 4

(a)

Obtain the enthalpy of the refrigerant, R134a entering into the condenser at 1.2 MP1 and

\(50^{\circ} \mathrm{C}\) from the table A-13.

\(h_{R_{1}}=278.27 \mathrm{~kJ} / \mathrm{kg}\)

Obtain the enthalpy of the saturated liquid refrigerant. R134a at 1.2 MP3 from the table A-12.

\(h_{\text {sat } @ 1.2 \mathrm{MPa}}=117.7 \mathrm{~kJ} / \mathrm{kg}\)

Calculate the enthalpy of the refrigerant, R134a at 1.2 MPa that is sub-cooled by \(50^{\circ} \mathrm{C}\) as

follows,

\(h_{R_{2}}=h_{\text {sat } @ 1.2 \mathrm{MPa}}-C_{P_{. \pi}} \Delta T\)

Here, \(h_{\text {sat } @ 1.2 \mathrm{MPa}}\) is the enthalpy of the saturated liquid refrigerant, R134a at 1.2 MPa. \(C_{P_{, R}}\) is

the specific heat capacity of the liquid refrigerant. R134a and \(\Delta T\) is the degree of sub-cooling done.

Substitute \(117.7 \mathrm{~kJ} / \mathrm{kg} \text { for } h_{\text {sat } @ 1.2} \mathrm{MPa}, 5^{\circ} \mathrm{C} \text { for } \Delta T\) and \(1.4944 \mathrm{~kJ} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C} \text { for } C_{P_{\cdot, k}} \text {. }\)

\(\begin{aligned}
h_{R_{2}} & =117.7-1.4944 \times 5 \\
& =110.228 \mathrm{~kJ} / \mathrm{kg}
\end{aligned}\)