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Obtain the enthalpy and specific volume at 180 \mathrm{kPa} for the refrigerant R-134a from the table A12 "Saturated refrigerant -134a-Pressure table".The value of enthalpy of saturated vapor of refrigerant \mathrm{R}-134 \mathrm{a} at 180 \mathrm{kPa} is h_{1}=242.86 \mathrm{~kJ} / \mathrm{kg}The value of specific volume of saturated vapor of refrigerant \mathrm{R}-134 \mathrm{a} at 180 \mathrm{kPa} is v_{1}=0.1104 \mathrm{~m}^{3} / \mathrm{kg} Calculate the mass flow rate of the refrigerator as follows:\dot{m}=\frac{\dot{V}_{1}}{v_{1}}Here, \dot{V}_{1} is the volume flow rate and v_{1} is the specific volume.Substitute 0.35 \mathrm{~m}^{3} / \mathrm{min}\left|\frac{1 \mathrm{~min}}{60 \mathrm{~s}}\right| for \dot{V}_{1} and 0.1104 \mathrm{~m}^{3} / \mathrm{kg} for v_{1}.\begin{aligned}\dot{m} & =\frac{(0.35 / 60)}{0.1104} \\& =0.0528 \mathrm{~kg} / \mathrm{s}\end{aligned}Therefore, the mass flow rate of the refrigerator is 0.0528 \mathrm{~kg} / \mathrm{s}. Calculate the enthalpy at exit of the compressor as follows:\begin{array}{l}\dot{W}_{i n}=\dot{m}\left(h_{2}-h_{1}\right) \\h_{2}=h_{1}+\frac{\dot{W}_{i n}}{\dot{m}}\end{array}Here, h_{2} is the enthalpy at the exit of the compressor, h_{1} is the enthalpy ate the inlet of the compressor, and \dot{W}_{i n} is the work input to the compressor.Substitute 242.86 \mathrm{~kJ} / \mathrm{kg} for h_{1}, 2.35 \mathrm{~kW} for \dot{W}_{i n}, and 0.0528 \mathrm{~kg} / \mathrm{s} for \dot{m}.\begin{aligned}h_{2} & =242.86+\frac{2.35}{0.0528} \\& =287.4 \mathrm{~kJ} / \mathrm{kg}\end{aligned} Since the enthalpy at the exit of the compressor is greater than the enthalpy of the saturated vapor of refrigerant \mathrm{R}-134 \mathrm{a} at 700 \mathrm{kPa} that shows that the condition of the refrigerant is superheated.From the table A-13 "Superheated refrigerant 134 \mathrm{a} " at 700 \mathrm{kPa} the enthalpy value of refrigerant at the exit of the compressor lies between the temperature limits T_{1}=40^{\circ} \mathrm{C} and T_{2}=50^{\circ} \mathrm{C}. The value of enthalpy of refrigerant 134 \mathrm{a} at 700 \mathrm{kPa}, T_{1}=40^{\circ} \mathrm{C} is h_{1}=278.57 \mathrm{~kJ} / \mathrm{kg}. The value of enthalpy of refrigerant 134 \mathrm{a} at 700 \mathrm{kPa}, T_{2}=50^{\circ} \mathrm{C} is h_{2}=288.53 \mathrm{~kJ} / \mathrm{kg}. Calculate the temperature of refrigerant R-134a at the exit of the compressor.Interpolate between the values h_{1}=278.57 \mathrm{~kJ} / \mathrm{kg}, T_{1}=40^{\circ} \mathrm{C} and h_{2}=288.53 \mathrm{~kJ} / \mathrm{kg},\begin{array}{l}T_{2}=50^{\circ} \mathrm{C} \\\frac{T-T_{1}}{T_{2}-T_{1}}=\frac{h-h_{1}}{h_{2}-h_{1}}\end{array}Substitute 40^{\circ} \mathrm{C} for T_{1}, 50^{\circ} \mathrm{C} for T_{2}, 278.57 \mathrm{~kJ} / \mathrm{kg} for h_{1}, 288.53 \mathrm{~kJ} / \mathrm{kg} for h_{2}, and 287.4 \mathrm{~kJ} / \mathrm{kg} for h\begin{array}{l}\frac{T_{2}-40}{50-40}=\frac{287.4-278.57}{288.53-278.57} \\\frac{T_{2}-40}{10}=0.8865 \\T_{2}=48.86^{\circ} \mathrm{C}\end{array}Therefore, temperature of refrigerant \mathrm{R}-134 \mathrm{a} at the exit of the compressor is 48.86^{\circ} \mathrm{C}.   P.S: If you are having any doubt, please comment here, I will surely reply you.Please ask questions seperately from next time. We will be provided very limited time for solving your query. Please understand and post the questions. Please provide your valuable feedback by a thumbsup. Your thumbsup is a result of our efforts. Your LIKES are very important for me so please LIKEu2026. Thanks in advance! Note: I've answered according to 's policy    ...