(a) To determine The temperature of water. The internal energy of water. Explanation Write the expression for quality of the two-phase liquid vapor mixture. x = υ − υ f υ g − υ f (I) Here, the given volume is υ , specific volume at saturated liquid state is υ f and volume at saturated vapor is υ f Write the expression for internal energy. u = u f + x ( u g − u f ) (II) Here, the given internal energy is u , quality of the mixture is x , internal energy at saturated liquid sate is u f and internal energy at saturated vapor state is u g . Conclusion: Refer to Table A-3E “Properties of saturated water-pressure table”, to obtain 0.01683 ft 3 / lb for υ f , 20.09 ft 3 / lb for υ g , 196.19 Btu / lb for u f , 227.96 ° F for saturation temperature and 1082 Btu / lb for u g corresponding to the pressure 20 lbf / in 2 . Thus, the temperature of water is 227.96 ° F . Substitute 0.01683 ft 3 / lb for υ f , 20.09 ft 3 / lb for υ g and 16 ft 3 / lb for υ in Equation (I) (b) To determine The specific volume of water. The specific enthalpy of water. (c) To determine The pressure of water. The internal energy of water. (d) To determine The specific pressure of water. The specific enthalpy of water. (e) To determine The specific volume of water. The specific internal energy of water.

Fundamentals of Engineering Thermodynamics

8th Edition

ISBN: 9781118832301

Author: SHAPIRO

Publisher: JOHN WILEY+SONS,INC.-CONSIGNMENT

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Chapter 3.14, Problem 44P

(a)

To determine

The temperature of water.

The internal energy of water.

(b)

To determine

The specific volume of water.

The specific enthalpy of water.

(c)

To determine

The pressure of water.

The internal energy of water.

(d)

To determine

The specific pressure of water.

The specific enthalpy of water.

(e)

To determine

The specific volume of water.

The specific internal energy of water.

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Chapter 3.14, Problem 43P

Chapter 3.14, Problem 45P

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