(a) Explanation The figure below shows the T − s diagram of the Carnot refrigeration cycle. Figure-(1) The process 1-2 is an isentropic compression process, 2-3 is an isothermal heat rejection process, 3-4 is an isentropic expansion process and the process 4-1 is an isothermal heat addition process. Write the expression for rate of heat transfer to the refrigerant passing through the evaporator. Q ˙ = m ˙ ( h 1 − h 4 ) (I) Here, the mass flow rate of the refrigerant is m ˙ , the specific enthalpy at state 1 is h 1 and the specific enthalpy at state 4 is h 4 . Since the process 1-2 is an isentropic process, so s 1 = s 2 . Write the expression for specific entropy at state 1. s 1 = s f + x 1 ( s g − s f ) (II) Here, the specific entropy at saturated vapor is s g , the specific entropy at saturated liquid is s f and quality of refrigerant at state 1 is x 1 . Write the expression for specific enthalpy at state 1. h 1 = h f + x 1 ( h g − h f ) (III) Here, the specific enthalpy at saturated vapor is h g , the specific enthalpy at saturated liquid is s f and quality of refrigerant at state 1 is x 1 . The process 3-4 is an isentropic process hence s 3 = s 4 . Write the expression for specific entropy at state 4. s 4 = s f + x 4 ( s g − s f ) (V) Here, the quality of refrigerant at state 4 is x 4 . Write the expression for specific enthalpy at state 4. h 4 = h f + x 4 ( h g − h f ) (VI) Conclusion: Refer to table “Properties of Saturated Refrigerant 22 (Liquid–Vapor): Temperature Table” to obtain specific enthalpy at saturated liquid as 89.29 kJ / kg , specific enthalpy at saturated vapor as 260.11 kJ / kg and specific entropy at saturated vapor as 0.8790 kJ / kg ⋅ K with respect to 36 ° C . Refer to table “Properties of Saturated Refrigerant 22 (Liquid–Vapor): Temperature Table” to obtain specific enthalpy at saturated liquid as 11 kJ / kg and specific enthalpy at saturated vapor as 237.78 kJ / kg with respect to − 30 ° C . Refer to table “Properties of Saturated Refrigerant 22 (Liquid–Vapor): Temperature Table” to obtain specific entropy at saturated liquid as 0.3265 kJ / kg ⋅ K with respect to 36 ° C . Refer to table “Properties of Saturated Refrigerant 22 (Liquid–Vapor): Temperature Table” to obtain specific entropy at saturated liquid as 0.0460 kJ / kg ⋅ K and specific entropy at saturated vapor as 0.9787 kJ / kg ⋅ K with respect to − 30 ° C . Substitute 0.8790 kJ / kg ⋅ K for s 1 , 0.0460 kJ / kg ⋅ K for s f and 0.9787 kJ / kg ⋅ K for s g in Equation (II). 0.8790 kJ / kg ⋅ K = 0.0460 kJ / kg ⋅ K + x 1 ( 0.9787 kJ / kg ⋅ K − 0.0460 kJ / kg ⋅ K ) x 1 ( 0.9327 kJ / kg ⋅ K ) = 0.8790 kJ / kg ⋅ K − 0.0460 kJ / kg ⋅ K x 1 = 0.833 kJ / kg ⋅ K 0.9327 kJ / kg ⋅ K x 1 = 0.893 Substitute 0.893 for x 1 , 11 kJ / kg for h f and 237 (b) To determine The net power input to the cycle. (c) To determine The coefficient of performance. (d) To determine The refrigeration capacity, in tons.

Fundamentals of Engineering Thermodynamics, Binder Ready Version

8th Edition

ISBN: 9781118820445

Author: Michael J. Moran, Howard N. Shapiro, Daisie D. Boettner, Margaret B. Bailey

Publisher: WILEY

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Chapter 10.7, Problem 2P

(a)

To determine

The rate of heat transfer to the refrigerant passing through the evaporator.

(b)

To determine

The net power input to the cycle.

(c)

To determine

The coefficient of performance.

(d)

To determine

The refrigeration capacity, in tons.

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