Explanation Write the initial mass of the steam. m 1 = V v 1 (I) Here, the initial specific volume of the steam is v 1 and the volume of the tank is V . Write the initial specific volume of the steam. v 2 = V m 2 (II) Here, the final mass of the steam is m 2 Write the dryness fraction of steam at final state. x 2 = v 2 − v f v g − v f (III) Here, the specific volume at saturated vapor is v f and the specific volume at saturated liquid state is v g . Write the specific internal energy at final state. u 2 = u f + x 2 ( u g − u f ) (IV) Here, the specific internal energy at saturated vapor is v f and the specific internal energy at saturated liquid state is u g . Write the expression for mass rate balance. d m d t = m ˙ t d m = m ˙ t ⋅ d t (V) Here, the change in mass of the system from initial state to final state is d m and the rate of flow of mass is m ˙ . Write the change in mass of the system from initial state to final state. d m = m 2 − m 1 Substitute m 2 − m 1 for d m in Equation (V). m 2 − m 1 = ∫ m ˙ ⋅ d t = m i − m e (VI) There is no mass at the exit hence, m e = 0 . Substitute 0 for m e in Equation (VI). m 2 − m 1 = m i − 0 m i = m 2 − m 1 (VII) Write the energy balance equation for the tank. Q + m i h 1 = m 2 u 2 − m 1 u 1 Q = m 2 u 2 − m 1 u 1 − m i h 1 (VIII) Here, the heat transfer from the tank is Q , the specific internal energy at final state is u 2 , the specific internal energy at initial state is u 1 and the specific enthalpy at initial state is h 1 . Conclusion: Refer to Table A-15 “Properties of Superheated Ammonia Vapor” to obtain specific volume as 0.9382 m 3 / kg , specific internal energy as 1372 kJ / kg , specific enthalpy as 1512.8 kJ / kg with respect to pressure 1.50 bar and temperature 20 ° C . Refer to Table A-15 “Properties of Superheated Ammonia Vapor” to obtain specific volume as 0.12378 m 3 / kg , specific internal energy as 1404.54 kJ / kg , specific enthalpy as 1553.07 kJ / kg with respect to pressure 12 bar and temperature 60 ° C . Substitute 0.5 m 3 for V and 0.9382 m 3 / kg for v 1 in Equation (I). m 1 = 0.5 m 3 0.9382 m 3 / kg = 0.5329 kg Substitute 0.5 m 3 for V and 143.36 kg for m 2 in Equation (II). v 2 = 0.5 m 3 143.36 kg = 3.48 × 10 − 3 m 3 / kg Refer to Table “Properties of Saturated Ammonia (Liquid–Vapor): Temperature” to obtain specific volume at saturated vapor as 1

Fundamentals of Engineering Thermodynamics

8th Edition

ISBN: 9781118832318

Author: MORAN

Publisher: WILEY

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Chapter 4.12, Problem 115P

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Chapter 4.12, Problem 114P

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Fundamentals of Engineering Thermodynamics

Ch. 4.12 - Prob. 1E Ch. 4.12 - 2. When a drip coffeemaker on-off switch is turned...Ch. 4.12 - Prob. 3E Ch. 4.12 - Prob. 4E Ch. 4.12 - Prob. 5E Ch. 4.12 - Prob. 6E Ch. 4.12 - Prob. 7E Ch. 4.12 - Prob. 8E Ch. 4.12 - Prob. 9E Ch. 4.12 - 10. How does the operator of a pumper-tanker fire...

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The heat transfer between the tank contents and the surroundings in kJ.

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