(a) Explanation Write the expression for energy balance. m 2 u 2 − m 1 u 1 = h 1 ( m 2 − m 1 ) (I) Here, the initial mass is m 1 , initial internal energy is u 1 , initial enthalpy is h 1 , final mass is m 2 and final internal energy is u 2 . Conclusion: The tank is initially evacuated so the initial mass is equal to zero. Substitute 0 kg for m 1 in Equation (I). m 2 u 2 − ( 0 kg ) u 1 = h 1 ( m 2 − 0 kg ) m 2 u 2 = m 2 h 1 u 2 = h 1 (II) Refer to Table A-4 “Properties of superheated water vapor” to obtain enthalpy ( h 1 ) as 3445.3 kJ/kg . Substitute 3445.3 kJ/kg for h 1 in Equation (II). u 2 = 3445.3 kJ/kg Refer to Table A-4 “Properties of superheated water vapor”, at internal energy ( u 2 ) as 3445.3 kJ/kg , obtain internal energy ( u 21 ) as 3362.2 kJ/kg corresponding to temperature ( T 21 ) as 640 ° C , internal energy ( u 22 ) as 3470 (b) To determine The final mass in the tank. (c) To determine The amount of entropy production.

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 114P

(a)

To determine

The final temperature in the tank.

(b)

To determine

The final mass in the tank.

(c)

To determine

The amount of entropy production.

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Chapter 6.13, Problem 113P

Chapter 6.13, Problem 115P

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Chapter 6 Solutions

Fundamentals of Engineering Thermodynamics

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