Question

An air conditioner using refrigerant-134a as the working fluid and operating on the ideal vapor-compression refrigeration cycle is to maintain a space at 22°C while operating its condenser at 1000 kPa. Determine the COP of the system when a temperature difference of 2°C is allowed for the transfer of heat in the evaporator.

Answer 1

Answer:

note:

solution is attached due to error in mathematical equation. please find the attachment

Answer 2

Answer:

COP = 13.31

Explanation:

We have an allowed temperature difference of 2°C, thus, let's make use of temperature of 20°C in the evaporator.

Now, looking at table A-11 i have attached and looking at temperature of 20°C, we will see that the enthalpy(h1) = 261.59 Kj/Kg

While the enthropy(s1) = 0.92234 Kj/KgK

Now, the enthalpy at the second state will be gotten from the given condenser pressure under the condition s2 = s1.

Thus, looking at table A-13 which i have attached, direct 20°C is not there, so when we interpolate between the enthalpy values at 15.71°C and 21.55°C, we get an enthalpy of  273.18 Kj/Kg.

Now, the enthalpy at the third and fourth states is again obtained from interpolation between values at temperatures of 18.73 and 21.55 of the saturated liquid value in table A-12 i have attached.

Thus, h3=h4 = 107.34 Kj/kg

Formula for COP = QL/w = (h1- h4) / (h2 - h1)

COP = (261.59 - 107.34)/( 273.18 - 261.59) = 13.31