(a) Explanation Figure below shows the schematic of problem with water as system. Figure-(1) Write the expression for specific volume of a substance in Liquid-vapour mixture. v = v f + x ( v g − v f ) (I) Here, the specific volume of liquid content is v f , the specific volume of vapour content is v g , the fraction of vapour content in mixture is x . Write the expression for specific internal energy of a substance in liquid-vapour mixture. u = u f + x ( u g − u f ) (II) Here, the specific internal energy of liquid content is u f , the specific internal energy of vapour content is u g , the fraction of vapour content in mixture is x . Write the expression for energy balance in a system. Δ U + Δ KE + Δ PE = Q − W (III) Here, the change in internal energy is Δ U , the change in kinetic energy is Δ KE , the change in potential energy is Δ PE , heat supplied to the system is Q and the work done by system is W . Assume that the change in kinetic and potential energy of the system is negligible. Write the expression for change in internal energy. Δ U = m ( u 2 − u 1 ) Here, mass is m , specific internal energy of state 1 is u 1 and the specific internal energy of state 2 is u 2 . Substitute 0 for Δ K E , 0 for Δ P E and − m ( u 2 − u 1 ) for Δ U in Equation (III). m ( u 2 − u 1 ) = Q − W m ( u 2 − u 1 ) + W = Q (IV) Write the expression for work done for a constant pressure process in a closed system. W = ∫ V 1 V 2 p d V = m p ( v 2 − v 1 ) (V) Here, the pressure is p , the specific volume at state 1 is v 1 , the specific volume at state 2 is v 2 , the mass of system is m and the work done is W . Conclusion: State 1 is saturated liquid and hence, specific volume v 1 will be liquid specific volume at pressure equals to 20 lbf / in 2 and specific internal energy u 1 will be liquid specific internal energy at pressure equals to 20 lbf / in 2 . Refer to Table-3E “Properties of Saturated Water (Liquid–Vapor): Pressure Table” to obtain values of liquid specific volume as 0.01683 ft 3 / lb and vapour specific volume as 20.09 ft 3 / lb at pressure equals to 20 lbf / in 2 . Substitute 0.01683 ft 3 / lb for v f , 20.09 ft 3 / lb for v g and 0.8 for x in Equation (I) v 2 = ( 0.01683 ft 3 / lb ) + 0.8 ( 20.09 ft 3 / lb − 0.01683 ft 3 / lb ) = ( 0.01683 ft 3 / lb ) + ( 16.0585 ft 3 / lb ) = 16 (b) To determine The change of entropy in the enlarged system.

Fundamentals of Engineering Thermodynamics

8th Edition

ISBN: 9781118412930

Author: Michael J. Moran, Howard N. Shapiro, Daisie D. Boettner, Margaret B. Bailey

Publisher: WILEY

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Chapter 6.13, Problem 52P

(a)

To determine

Work transfer in Btu for water as system.

Heat transfer in Btu for water as system.

(b)

To determine

The change of entropy in the enlarged system.

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Chapter 6.13, Problem 51P

Chapter 6.13, Problem 53P

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Work transfer in Btu for water as system. Heat transfer in Btu for water as system.