Explanation Sketch the schematic layout of the two-stage compression refrigeration system as in Figure (1). Sketch the T − s diagram of the two-stage compression refrigeration as in Figure (2). Write the relation between the specific enthalpies at the inlet and exit of throttling. h i = h e (I) Here, specific enthalpy of the refrigerant at the inlet of throttling is h i and specific enthalpy of the refrigerant at the exit of throttling is h e . Write the energy balance equation for the separator. m ˙ 6 ( h 8 − h 5 ) = m ˙ 2 ( h 1 − h 4 ) m ˙ 6 = m ˙ 2 h 1 − h 4 h 8 − h 5 (II) Here, mass flow rate of the refrigerant at the end of high pressure compressor is m ˙ 2 , mass flow rate at the inlet of evaporator is m ˙ 6 , specific enthalpy of refrigerant at the compressor inlet of high pressure cycle is h 1 , specific enthalpy at the exit of expansion valve at high pressure cycle is h 4 , specific enthalpy at the exit of compressor at the low pressure cycle is h 8 , and the specific enthalpy at the inlet of expansion valve at the low pressure cycle is h 5 . Write the expression to calculate the rate of cooling produced ( Q ˙ L ) by the two-stage compression refrigeration. Q ˙ L = m ˙ 6 ( h 7 − h 6 ) (III) Here, specific enthalpy at the inlet of low pressure compressor is h 7 and the specific enthalpy at the inlet evaporator is h 6 . Write the expression to calculate the total power input to the compressor ( W ˙ in ) of the system. W ˙ in = m ˙ 6 ( h 8 − h 7 ) + m ˙ 2 ( h 2 − h 1 ) (IV) Here, specific enthalpy at the exit of high pressure compressor is h 2 . Conclusion: From the Table A-11 of “Saturated refrigerant R-134a: Temperature”, obtain the properties of refrigerant at high pressure compressor inlet temperature ( T 1 ) of 8.9 ° C as follows: h 1 = h g = 255.60 kJ / kg s 1 = s g = 0.92711 kJ / kg ⋅ K From the Table A-13 of “Superheated refrigerant R-134a”, obtain the specific enthalpy of high pressure compression exit at pressure ( P 2 ) of 1400 kPa and specific entropy of 0.92711 kJ / kg ⋅ K as 281.56 kJ / kg . From the Table A-12 of “Saturated refrigerant R-134a: Pressure”, obtain the properties of refrigerant at high pressure cycle throttle inlet pressure of 1400 kPa as follows: h 3 = h f = 127.25 kJ / kg Substitute 127.25 kJ / kg for h 3 in Equation (I). h 4 = 127.25 kJ / kg From the Table A-12 of “Saturated refrigerant R-134a: Pressure”, obtain the properties of refrigerant at low pressure cycle throttle inlet temperature of 8.9 ° C as follows: P 5 = P sat = 400 kPa h 5 = h f = 63.91 kJ / kg Here, pressure of refrigerant at the separator is P 5 . Substitute 63.91 kJ / kg for h 5 in Equation (I). h 6 = 63.91 kJ / kg From the Table A-11 of “Saturated refrigerant R-134a: Temperature”, obtain the properties of refrigerant at low pressure compressor inlet temperature ( T 7 ) of − 32 ° C as follows: h 7 = h g = 230

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Chapter 11.10, Problem 116RP

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The rate of cooling produced by the two stage compression system.

The total power input required to the compressor.

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Chapter 11.10, Problem 114RP

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Chapter 11 Solutions

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 rate of cooling produced by the two stage compression system. The total power input required to the compressor.

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The rate of cooling produced by the two stage compression system. The total power input required to the compressor.

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The rate of cooling produced by the two stage compression system.

The total power input required to the compressor.