Explanation The below figure shows the Carnot cycle on T − s diagram. Figure-(1) In Figure-(1), the adiabatic compression of refrigerant is 2 − 3 , the heat removal process from the condenser is 2 − 3 , the adiabatic expansion process is 4 − 1 and hear absorption process is 1 − 2 . Write the expression of entropy for the process 2 − 3 . s 2 = s 3 (I) Write the expression for conversion of low temperature in carnot cycle from ° C to K . T C = ( T C ° C+273 ) K (I) Here, the low temperature in ° C is T C ° C . Write the expression for conversion of high temperature in carnot cycle from ° C to K . T H = ( T H ° C+273 ) K (II) Here, the high temperature in ° C is T H ° C . Write the expression for coefficient of performance of reversible refrigeration cycle in terms of temperature. β max = T C T H − T C (III) Conclusion: Refer to table A-10, “Properties of saturated refrigerant 134 a (liquid-vapor): temperature table.” to obtain the entropy at the temperature of 0 ° C as s 2 = 0.9190 kJ / kg ⋅ k Substitute 0.9190 kJ / kg ⋅ k for s 2 in Equation (I). s 3 = 0.9190 kJ / kg ⋅ k Refer to table A-10, “Properties of superheated refrigerant 134 a vapor.” to obtain the temperature at the pressure of p 3 = 16 bar by using interpolation method of two variable. 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 (II) Here, the variables denote by x and y are entropy and temperature. Show the entropy at 60 ° C and 70 ° C as in Table (1). Temperature, °C Specific entropy, kJ / kg ⋅ k 60 ° C 0.9069 kJ / kg ⋅ k ? 0.9190 kJ / kg ⋅ k 70 ° C 0

Fundamentals of Engineering Thermodynamics

8th Edition

ISBN: 9781118832318

Author: MORAN

Publisher: WILEY

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Chapter 6.13, Problem 31P

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The coefficient of performance of the cycle.

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Chapter 6.13, Problem 30P

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Fundamentals of Engineering Thermodynamics

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