Explanation The system of water is a closed system. The figure below shows the T − s diagram of the given reversible process. Figure-(1) The process 1 − 2 is a constant pressure process. The figure below shows the P − v diagram of the given reversible process. Figure-(2) Write the specific volume at state 2 . v 2 = v f + x 2 ( v g − v f ) (I) Here, the specific volume at saturated vapor is v f , the specific volume at saturated liquid is s g and the dryness fraction at state 2 is x 2 . Write the work transfer for process 1 − 2 . W = ∫ 1 2 p ⋅ d V = p ∫ 1 2 d V = p ( V 2 − V 1 ) (II) Write the expression for volume change. V 2 − V 1 = m ( v 2 − v 1 ) Here, the specific volume at 1 is v 1 . Substitute m ( v 2 − v 1 ) for V 2 − V 1 in Equation (II). W = p ⋅ m ( v 2 − v 1 ) (III) Write the heat transfer for process 1 − 2 . Q = ∫ 1 2 T ⋅ d S = T ( S 2 − S 1 ) (IV) Here, the entropy at 2 is S 2 , the entropy at 1 is S 1 and the constant temperature is T . Write the change in entropy for the process. S 2 − S 1 = m ( s 2 − s 1 ) (V) Here, the mass of water is m , the specific entropy at 1 is s 1 and the specific entropy at 2 is s 2 . Substitute m ( s 2 − s 1 ) for S 2 − S 1 in Equation (II). Q = T m ( s 2 − s 1 ) (VI) Write the specific entropy at final state. s 2 = s f + x 2 ( s g − s f ) (VII) Here, the specific entropy at saturated vapor is s f and the specific entropy at saturated liquid is s g . Conclusion: Convert the pressure of 1 atm from atm into lbf / in 2 . p = 1 atm = ( 1 atm ) ( 14.696 lbf / in 2 1 atm ) = 14.696 lbf / in 2 Refer to Table A-3E “Properties of Saturated Water (Liquid–Vapor): Pressure Table” to obtain specific volume at saturated vapor as 0.01672 ft 3 / lb , specific volume at saturated liquid as 26.8 ft 3 / lb , specific entropy at saturated liquid as 1.7567 Btu / lb ⋅ ° R and specific entropy at saturated vapor as 0.3121 Btu / lb ⋅ ° R with respect to 14.696 lbf / in 2 pressure. Substitute 0.01672 ft 3 / lb for v f , 0.9 for x 2 and 26.8 ft 3 / lb for v g in Equation (I). v 2 = 0.01672 ft 3 / lb + 0.9 ( 26.8 ft 3 / lb − 0.01672 ft 3 / lb ) = 0.01672 ft 3 / lb + 0.9 ( 26.78 ft 3 / lb ) = 0

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It is generally related to thermal engineering which tells about the use, creation, conversion and transfer of heat energy between the systems. It is also considered that to transfer heat from one source to another, transfer of masses is also required. T…

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Chapter 6.13, Problem 23P

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The heat transfer in the expansion process in Btu . The work transfer in the expansion process in Btu .

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The heat transfer in the expansion process in Btu.

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