Explanation Sketch the T − s diagram for the vapor compression refrigeration cycle of heat pump as in Figure (1). Write the relation between specific enthalpies at inlet and exit of a throttling process. h 3 = h 4 (I) Here, the specific enthalpy at the inlet of throttling is h 3 and the specific enthalpy at the exit of throttling is h 4 . Write the formula to calculate the mass flow rate of refrigerant ( m ˙ ) from the heat rejected. m ˙ = Q ˙ H q H m ˙ = Q ˙ H h 2 − h 3 (II) Here, heating load is Q ˙ H , rate of heating load of the heat pump is q H , specific enthalpy of refrigerant at compressor exit is h 2 . And the specific enthalpy of refrigerant at condenser exit is h 3 . Write the formula to calculate the work input required ( W ˙ in ) to the vapor compression refrigeration cycle. W ˙ in = m ˙ ( h 2 − h 1 ) (III) Here, specific enthalpy at the compressor inlet is h 1 . Write the expression to determine the required electrical power ( W ˙ e ) without heat pump. W ˙ e = Q ˙ H (IV) Write the expression to calculate the electric power saved ( W ˙ saved ) . W ˙ saved = W ˙ e − W ˙ in (V) Conclusion: From the Table A-12E of “Saturated refrigerant R-134a: Pressure”, obtain the properties of refrigerant at compressor inlet pressure ( P 1 ) of 50 psia as follows: h 1 = h g = 108.83 Btu / lbm s 1 = s g = 0.22192 Btu / lbm ⋅ R Here, specific enthalpy of the saturated vapor is h g and the specific entropy of the saturated vapor is s g . The specific entropy at the end of isentropic compression ( s 2 ) is calculated as s 2 = s 1 = 0.22192 Btu / lbm ⋅ R Refer to Table A-13E, “Superheated R-134a”, and obtain the values of R-134a at pressure of 120 psia and specific entropy of 0.22192 Btu / lbm ⋅ R by using interpolation method. Write the formula of interpolation method of two variables. y 2 = ( x 2 − x 1 ) ( y 3 − y 1 ) ( x 3 − x 1 ) + y 1 (VI) Here, the variables denoted by x and y are specific entropy and specific enthalpy. Show temperature and enthalpy values from the Table A-13E. Entropy ( s 2 ) , in Btu / lbm ⋅ R Enthalpy ( h 2 ) , in Btu / lbm 0.21928 115.18 0.22192 ? 0.22364 117.59 Substitute 0.21928 for x 1 , 0.22192 for x 2 , 0.22364 for x 3 , 115.18 for y 1 , and 117.59 for y 3 in Equation (VI) to calculate the specific enthalpy at the isentropic exit of compressor ( h 2 )

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Chapter 11.10, Problem 45P

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The electric power saved from using a heat pump in place of resistance heater.

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Chapter 11.10, Problem 44P

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Thermodynamics: An Engineering Approach

Ch. 11.10 - Why is the reversed Carnot cycle executed within...Ch. 11.10 - Why do we study the reversed Carnot cycle even...Ch. 11.10 - 11–3 A steady-flow Carnot refrigeration cycle uses...Ch. 11.10 - Does the ideal vapor-compression refrigeration...Ch. 11.10 - Why is the throttling valve not replaced by an...Ch. 11.10 - It is proposed to use water instead of...Ch. 11.10 - In a refrigeration system, would you recommend...Ch. 11.10 - Does the area enclosed by the cycle on a T-s...Ch. 11.10 - Consider two vapor-compression refrigeration...Ch. 11.10 - The COP of vapor-compression refrigeration cycles...

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The power input required to the heat pump. The electric power saved from using a heat pump in place of resistance heater.

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The power input required to the heat pump.

The electric power saved from using a heat pump in place of resistance heater.