To determine The heat transfer for all the process of the system. Explanation Write the expression for the heat transfer for the process 1-2. Q 12 = m ( u 2 − u 1 ) + W 12 . (I) Here, the heat transfer for the process 1-2 is Q 12 , the mass of the substance is m , the internal energy if the process at point 2 is u 2 , the heat transfer of the process 1-2 at point 1 is u 1 , the work done for the process is W 12 . Write the expression for the specific volume of the process 1-2. v 1 = V 1 m (II) Here the specific volume of the process at the 1 is v 1 and the volume of the process at point 1 is V 1 . Write the expression for the specific volume of the process 1-2. v 2 = V 2 m (III) Here the specific volume of the process at point 2 is v 2 and the volume of the process at the point 2 is V 2 . Write the expression for the specific volume of the process 2-3. v 3 = V 3 m (IV) Here the specific volume of the process at point 3 is v 3 and the volume of the process at point 3 is V 3 . Write the expression for the heat transfer for the process 2-3. Q 23 = m ( u 3 − u 2 ) + W 23 . (V) Here, the heat transfer for the process 2-3 is Q 23 , the mass of the substance is m , the internal energy if the process at point 3 is u 3 , the heat transfer of the process 2-3 at point 2 is u 2 , the work done for the process is W 23 . Write the expression for the work done for the process 1-2. W 12 = p 1 ( v 2 − v 1 ) . (VI) Here, the pressure at point 1 is p 1 . Conclusion: Substitute 71.7 ft 3 for V 1 and 3 lb for m in Equation (II). v 1 = 71.7 ft 3 3 lb = 23.9 ft 3 / lb Thus, the state 1 is the superheat vapor region. Refer to table A-4E “properties of the superheated water vapor” to obtain the pressure and internal energy at specific volume 23.9 ft 3 / lb and temperature 350 ° F as 20 lbf / in 2 for p 1 and 1126.9 Btu / lb for u 1 . Substitute 85.38 ft 3 for V 2 and 3 lb for m in Equation (III). v 2 = 85.38 ft 3 3 lb = 28.46 ft 3 / lb Thus, the state 2 is the superheat vapor region. Refer to table A-4E “properties of the superheated water vapor” to obtain pressure and internal energy at specific volume 28.46 ft 3 / lb and temperature 350 °F as 20 lbf / in 2 p 2 and 1181.5 Btu / lb for u 2 . Since the pressure is constant for the process 1-2, so p 1 = p 2 Substitute 85.38 ft 3 for v 2 , 71.7 ft 3 for v 1 and 20 lbf in 2 for p 1 in Equation (VI)

8th Edition

ISBN: 9781119391630

Author: MORAN

Publisher: WILEY

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Chapter 3.14, Problem 73P

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The heat transfer for all the process of the system.

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Chapter 3.14, Problem 72P

Chapter 3.14, Problem 74P

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