(a) Explanation Write the expression for dryness fraction. x = s − s f s g − s f (I) Here, entropy is s , entropy of saturated liquid is s f , entropy of saturated vapor is s g and the dryness fraction is x . Write the expression for enthalpy. h = h f + x ( h g − h f ) (II) Here, enthalpy is h , enthalpy of saturated liquid is h f and enthalpy of the superheated liquid is h g . Conclusion: Refer to the table A-3, “Properties of Saturated Water (Liquid–Vapor): Pressure Table” to obtain the value of s f as 0.8320 kJ / kg ⋅ K and s g as 7.9085 kJ / kg ⋅ K at 0.20 bar pressure. Substitute 0.8320 kJ / kg ⋅ K for s f , 7.9085 kJ / kg ⋅ K for s g and 4.3703 kJ / kg ⋅ K for s in Equation (I). x = 4.3703 kJ / kg ⋅ K − 0.8320 kJ / kg ⋅ K 7.9085 kJ / kg ⋅ K − 0.8320 kJ / kg ⋅ K = 3 (b) To determine The entropy. The T − v diagram The T − s diagram. (c) To determine The specific entropy. The T − v diagram. The T − s diagram. (c) To determine The specific internal energy. The T − v diagram. The T − s diagram.

8th Edition

ISBN: 2818440116926

Author: MORAN

Publisher: WILEY CONS

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Chapter 6.13, Problem 3P

(a)

To determine

The enthalpy.

The T−v diagram

The T−s diagram.

(b)

To determine

The entropy.

The T−v diagram

The T−s diagram.

(c)

To determine

The specific entropy.

The T−v diagram.

The T−s diagram.

(c)

To determine

The specific internal energy.

The T−v diagram.

The T−s diagram.

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Chapter 6.13, Problem 2P

Chapter 6.13, Problem 4P

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