To determine The final pressure in bar for the expansion process. The work in kJ for the expansion process. Explanation The figure below shows the T − s diagram of the given reversible process. Figure-(1) In Figure-(1), the process 1 − 2 is an isothermal process. Write the heat transfer for the process 1 − 2 . Q = T ( S 2 − S 1 ) (I) Here, the constant temperature is T , the entropy at state 1 is S 1 and the entropy at state 1 is S 2 . Write the change in entropy for the process. S 2 − S 1 = m ( s 2 − s 1 ) 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 (I). Q = m T ( s 2 − s 1 ) s 2 = s 1 + Q T m (II) Write the expression for dryness fraction at state 2 . x 2 = s 2 − ( s f ) 2 ( s g ) 2 − ( s f ) 2 (III) Here, the specific entropy at saturated vapor at state 2 is ( s f ) 2 and the specific entropy at saturated liquid at 2 is ( s g ) 2 . Write the expression for work. W = Q + Δ U (IV) Here, the heat transfer is Q and the change in internal energy is Δ U . Write the change in internal energy. Δ U = m ( u 1 − u 2 ) Here, the specific internal energy at 1 is u 1 and the specific internal energy at 2 is u 2 . Substitute m ( u 1 − u 2 ) for Δ U in Equation (IV). W = Q + m ( u 1 − u 2 ) (V) Write the specific internal energy at 2 . u 2 = ( u f ) 2 + x 2 [ ( u g ) 2 − ( u f ) 2 ] (VI) Here, the specific internal energy of saturated liquid at 2 is ( u g ) 2 and the specific internal energy of saturated vapor at 2 is ( u f ) 2 . Conclusion: Refer to Table A-2 “Properties of Saturated Water (Liquid–Vapor): Temperature Table” to obtain specific entropy at saturated vapor as 1.9427 kJ / kg ⋅ K , specific internal energy at saturated vapor as 674.86 kJ / kg with respect to 10 bar pressure and 160 ° C temperature. Refer to Table A-2 “Properties of Saturated Water (Liquid–Vapor): Temperature Table” to obtain specific entropy at saturated liquid as with respect to 10 bar pressure and 160 ° C temperature. Substitute 1.9427 kJ / kg ⋅ K for s 1 , 2700 kJ for Q , 2 kg for m and 160 ° C for T in Equation (II). s 2 = 1.9427 kJ / kg ⋅ K + 2700 kJ ( 160 ° C ) ( 2 kg ) = 1.9427 kJ / kg ⋅ K + 1350 kJ / kg ( 160 ° C + 273 ) K = 1

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Chapter 6.13, Problem 22P

To determine

The final pressure in bar for the expansion process.

The work in kJ for the expansion process.

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Chapter 6.13, Problem 21P

Chapter 6.13, Problem 23P

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