Explanation The figure below shows the T − v diagram of the system. Figure-(1) In Figure-(1), x -axis shows the specific volume, y -axis shows the temperature of the system, the process 1 − 2 shows the constant volume process. Write the expression of heat transfer per unit mass. Q m = ( u 2 − u 1 ) . (I) Here, the heat transfer is Q , the mass of the refrigerant is m and the specific internal energy at point 1 is u 1 and the specific internal energy at point 2 is u 2 . Write the expression of the quality of the water. x = v − v f v g − v f . (II) Here, the quality of the water is x , the specific volume of mixture is v and the specific volume of water in liquid state is v f and the specific volume of water in vapor state is v g . Write the expression for the specific internal energy at state 2 . u = u f + x ( u g − u f ) (III) Here, the specific internal energy at state 2 is u , specific internal energy of water in liquid phase is u f , and specific internal energy of water in vapor phase is u g . Conclusion: Refer to table A-2E “Properties of Saturated Water” to obtain critical volume as 0.05053 ft 3 / lb , critical pressure as 3402 lbf / in 2 and specific internal energy as 872.6 Btu / lb at 705.4 ° F temperature. Refer to table A-2 “Properties of Saturated Water” to obtain specific volume in vapor phase as 1.866 ft 3 / lb , specific volume in liquid phase as 0.01864 ft 3 / lb , specific internal energy in liquid phase as 374.3 Btu / lb and specific internal energy in vapor phase as 1116.6 Btu / lb at 400 ° F temperature. Substitute 4 bar for p , 0.06872 m 3 / kg for v 2 and 0.7672 × 10 − 3 m 3 / kg for v 1 in Equation (II)

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 3.14, Problem 49P

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