To determine The compressor power. The refrigeration capacity. The coefficient of performance. Explanation The below figure shows the T − s diagram of the process. Figure-(1) Write the expression for the isentropic efficiency of the compressor. η c = h 2 s − h 1 h 2 − h 1 (I) Here, the enthalpy at point 2s is h 2 s , the enthalpy at point 1 is h 1 and the enthalpy at point is 2 is h 2 . Write the expression for mass flow rate of air. m ˙ a i r = ( A V ) a i r p R T o u t (II) Here, the volumetric flow rate is ( A V ) a i r , the pressure is p , the gas constant is R . Write the expression for energy balance in a evaporator. m g ( h 1 − h 4 ) = m ˙ a i r ( T i n − T o u t ) (III) Write the expression for the compressor power. W ˙ c = m ˙ g ( h 2 − h 1 ) (IV) Write the expression for refrigeration capacity. Q ˙ i n = m ˙ g ( h 1 − h 4 ) (V) Write the expression for coefficient of performance. β = h 1 − h 4 h 2 − h 1 (VI) Conclusion: Refer table A-9 “Properties of superheated refrigerant 22 vapor” to obtain the specific enthalpy and the specific entropy at pressure 4 bar and temperature 10 ° C as h 1 = 259.18 kJ / kg and s 1 = 0.9795 kJ / kgK Since the process 1-2s is constant entropy process, so s 1 = s 2 s Refer table A-9 “Properties of superheated refrigerant 22 vapor” to obtain the specific enthalpy pressure 4 bar and entropy 0.9795 kJ / kgK as h 2 s = 280.68 kJ / kg . Substitute 0.7 for η c , 259.18 kJ / kg for h 1 , 280.68 kJ / kg for h 2 s in Equation (I). 0.7 = 280.68 kJ / kg − 259.18 kJ / kg h 2 − 259.18 kJ / kg 0.7 h 2 − 181.426 kJ / kg = 21.5 kJ / kg 0.7 h 2 = ( 181.426 kJ / kg ) + ( 21.5 kJ / kg ) 0.7 h 2 = 202.926 kJ / kg h 2 = 202.926 kJ / kg 0.7 h 2 = 289.8942 kJ / kg Refer table A-8 “Properties of saturated refrigerant 22 (liquid-vapor): Pressure table” to obtain the enthalpy at pressure 9 bar as h 3 = 68.59 kJ / kg . Since the process 3-4 is throttling process, so h 4 = h 3 Substitute 19 m 3 / min for ( A V ) a i r , 1 bar for p , 0.287 kJ / kgK for R and 288 K for T o u t in Equation (II). m ˙ a i r = ( 19 m 3 / min ) ( 1 bar ) ( 0.287 kJ / kgK ) ( 288 K ) = ( 19 m 3 / min ) ( 1 bar ) ( 100 kN / m 2 1 bar ) 82.656 kJ / kg = 1900 kJ / min 82.656 kJ / kg = 22.986 kg / min Substitute 259.18 kJ / kg for h 1 , 68.59 kJ / kg for h 4 , 22

Fundamentals of Engineering Thermodynamics

8th Edition

ISBN: 9781118832301

Author: SHAPIRO

Publisher: JOHN WILEY+SONS,INC.-CONSIGNMENT

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Chapter 10.7, Problem 24P

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The compressor power.

The refrigeration capacity.

The coefficient of performance.

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Chapter 10.7, Problem 23P

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Fundamentals of Engineering Thermodynamics

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