A Kansas farmer heats his house with a heat pump using R134a, generating a tem p

Question

A Kansas farmer heats his house with a heat pump using R134a, generating a temperature of 50 deg C at the exit of the (isentropic) compressor. The evaporator is buried at a depth of 2 meters and has a temperature of 15 degrees C. The evaporator exit, which is the compressor entrance, is saturated vapor. Assume that the condenser exit is liquid, whose temperature is 7 degrees below saturated liquid. Assume a throttling valve is downstream of the condenser. Determine the coefficient of performance.

<hint> : After compression, you know T and s. If you need the pressure, you should be able to determine it from these

Explanation / Answer

State 1 = Compressor inlet / Evaporator outlet

State 2 = Compressor outlet / Condensor inlet

State 3 = Condensor outlet / Throttle valve inlet

State 4 = Throttle valve outlet / Evaporator inlet

From R-134a properties,

At T1 = 15 deg C and x1 = 1 (sat. vapor) we get h1 = 259 kJ / kg and s1 = 924 J / kg-K

At T2 = 50 deg C and s2 = s1 = 924 J / kg-K, we get h2 = 278 kJ/kg, P2 = 1225 kPa

At P3 = P2 = 1225 kPa we get T_sat = 47.1 deg C

T3 = T_sat - 7 = 47.1 - 7 = 39.9 deg C

At T3 = 39.9 deg C and P3 = 1225 kPa we get, h3 = 108 kJ / kg

In throttling, h4 = h3 = 108 kJ / kg

COP = (h2 - h3) / (h2 - h1)

= (278 - 108) / (278 - 259)

= 8.947

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