(a) Explanation Express the mass of refrigerant. m = ν v 1 (I) Here, volume is ν and initial specific volume is v 1 . Express the final quality. x 2 = v 2 − v f , 2 v g , 2 − v f , 2 (II) Here, final specific volume is v 2 and specific volume at saturated liquid and vapor is v f and v g respectively. Express the final specific internal energy. u 2 = u f , 2 + x 2 u f g , 2 (III) Here, specific internal energy at saturated liquid and evaporation is u f and u f g respectively. Express the final specific entropy s 2 = s f , 2 + x 2 s f g , 2 (IV) Here, specific entropy at saturated liquid and evaporation is s f and s f g respectively. Express the initial specific internal energy. u 1 = u f + x 1 u f g (V) Here, initial quality is x 1 . Express initial specific entropy. s 1 = s f + x 1 s f g (VI) Express initial specific volume. v 1 = v f + x 1 ( v g − v f ) (VII) Express the amount of heat transfer between the source and the refrigerant using the energy balance for the closed tank. E in − E out = Δ E system Q in = Δ U = m ( u 2 − u 1 ) (VIII) Here, the total energy entering the system is E in , the total energy leaving the system is E out , the change in the total energy of the system is Δ E system , amount of heat transfer is Q in , change in internal energy is Δ U , and initial and final specific internal energy is u 1 and u 2 respectively. Conclusion: Refer Table A-12E, “saturated refrigerant-134a-pressure table” and write the properties corresponding to initial pressure ( P 1 ) of 30 psia and initial quality ( x 1 ) of 0.55 . u f = 16.929 Btu / lbm u f g = 79.799 Btu / lbm s f = 0.03792 Btu / lbm ⋅ R s f g = 0.18595 Btu / lbm ⋅ R v f = 0.01209 ft 3 / lbm v g = 1.5506 ft 3 / lbm Since the volumes are constant, therefore v 1 = v 2 Here, final specific volume is v 2 . Refer Table A-12E, “saturated refrigerant-134a-pressure table” and write the properties corresponding to final pressure ( P 2 ) of 50 psia . u f , 2 = 24.824 Btu / lbm u f g , 2 = 75.228 Btu / lbm s f , 2 = 0.05412 Btu / lbm ⋅ R s f g , 2 = 0.16780 Btu / lbm ⋅ R v f , 2 = 0.01252 ft 3 / lbm v g , 2 = 0.79361 ft 3 / lbm Substitute 0.01209 ft 3 / lbm for v f , 0.55 for x 1 and 1.5506 ft 3 / lbm for v g in Equation (VII). v 1 = 0.01209 ft 3 / lbm + 0.55 ( 1.5506 − 0.01209 ) ft 3 / lbm = 0 (b) To determine The exergy destroyed during the process.

8th Edition

ISBN: 9780073398174

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

Publisher: McGraw-Hill Education

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Chapter 8.8, Problem 43P

(a)

To determine

The amount of heat transfer between the source and the refrigerant.

(b)

To determine

The exergy destroyed during the process.

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Chapter 8.8, Problem 42P

Chapter 8.8, Problem 44P

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Thermodynamics: An Engineering Approach

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