Explanation Write the energy balance equation for a closed system. E in − E out = Δ E system − W out = Δ U − W out = m ( u 2 − u 1 ) (I) Here, net energy transfer in to the control volume is E in , net energy transfer exit from the control volume is E out and change in internal energy of system is Δ E system , mass of refrigerant-134a is m , work output from the system is W out and internal energy of refrigerant-134a at inlet and outlet conditions are u 1 and u 2 respectively. Write the formula to calculate actual power input of the expansion process ( w a,out ) . w a,out = η T ( u 1 − u 2 s ) (II) Here, isentropic expansion efficiency is η T and internal energy of refrigerant-134a at state 2s is u 2 s . Write the formula to calculate actual internal energy at the end of the expansion process ( u 2 ) . u 2 = u 1 − w a,out (III) Here, isentropic expansion efficiency is η T . Write the formula to calculate useful work of the expansion process ( w u ) . w u = w a,out − w surr = w a,out − P 0 ( v 2 − v 1 ) (IV) Here, surrounding work is w surr , surrounding pressure is P 0 ,specific volume of refrigerant-134a at inlet and outlet conditions are v 1 and v 2 respectively. Write the formula to difference in exergy between the initial and final states of the expansion process ( Δ ψ ) . Δ ψ = ( ψ 1 − ψ 2 ) w a = ( u 1 − u 2 ) + P 0 ( v 1 − v 2 ) − T 0 ( s 2 − s 1 ) (V) Here, initial and final exergies of refrigerant-134a are ψ 1 and ψ 2 respectively, surrounding temperature is T 0 and specific entropies of refrigerant-134a at the inlet and outlet conditions are s 1 and s 2 respectively. Write the formula to calculate second-law efficiency of the expansion process ( η II ) . η II = w u Δ ψ (VI) Conclusion: Refer Table A-13,“ Superheated refrigerant-134a”, obtain the following properties of the refrigerant-134a at pressure of 1600 kPa and temperature of 80°C . v 1 = 0.014362 m 3 / kg u 1 = 282.11 kJ / kg s 1 = 0.9875 kJ / kg ⋅ K Refer Table A-11 " Saturated refrigerant-134a-Temperature table ", through Table A-13,“ Superheated refrigerant-134a”, obtain the following properties of the refrigerant-134a at pressure of 100 kPa . u 2 s = 223.20 kJ / kg Substitute 0.85 for η T , 282.11 kJ / kg for u 1 and 223.20 kJ / kg for u 2 s in Equation (II). w a,out = 0.85 ( 282.11 kJ / kg − 223.20 kJ / kg ) = 50.57 kJ / kg Substitute 50.57 kJ / kg for w a,out and 282.11 kJ / kg for u 1 in Equation (III). u 2 = 282.11 kJ / kg − 50.57 kJ / kg = 232.04 kJ / kg Refer Table A-13,“ Superheated refrigerant-134a”, obtain the following properties of the refrigerant-134a at pressure of 1600 kPa and internal energy of 232

9th Edition

ISBN: 9781259822674

Author: Yunus A. Cengel Dr., Michael A. Boles

Publisher: McGraw-Hill Education

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Chapter 8.8, Problem 126RP

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The second-law efficiency of the expansion process.

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Chapter 8.8, Problem 125RP

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Chapter 8 Solutions

Thermodynamics: An Engineering Approach

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Chapter 8 Solutions