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(Solved): please explain all steps and write clear the answer. thanks  6-58 Refrigerant-134a enters the e ...



please explain all steps and write clear the answer. thanks 

6-58 Refrigerant-134a enters the evaporator coils placed at the back of the freezer section of a household refrigerator at \(
6-58 Refrigerant-134a enters the evaporator coils placed at the back of the freezer section of a household refrigerator at \( 100 \mathrm{kPa} \) with a quality of 20 percent and leaves at \( 100 \mathrm{kPa} \) and \( -26^{\circ} \mathrm{C} \). If the compressor consumes \( 600 \mathrm{~W} \) of power and the COP of the refrigerator is \( 1.2 \), determine \( (a) \) the mass flow rate of the refrigerant and \( (b) \) the rate of heat rejected to the kitchen air. Answers: (a) \( 0.00414 \mathrm{~kg} / \mathrm{s} \), (b) \( 1320 \mathrm{~W} \)


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From table of R-134a at P = 100kPa - hf=17.28kJkg hfg=217.16kJkg Now enthalpe at state 1 which is entry of evaporator is - h4=hf+x4hfg h4=17.28+0.2×21
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