Explanation The following diagram shows all the processes of the water. Figure-(1) Write the expression for the work done at constant pressure. W 12 = p ( V 2 − V 1 ) . (I) Here, the work done of the system for process 1-2 is W 12 , the volume at point 1 is V 1 , the volume at 2 point is V 2 and the pressure during the process is p . Write the expression for the internal energy of the system at point 1. u 1 = u f + x ( u g − u f ) . (II) Here, the internal energy of the system at point 1 is u 1 , the quality of the process is x , the internal energy at saturated liquid is u f and the internal energy at the saturated vapor is u g . Write the expression for the volume of the process 1-2 at point 2. V 2 = π D 2 4 L 2 (III) Here the volume of the process at point 2 is V 2 , the total length of the cylinder is L 2 , and the diameter of the piston is D . Write the expression for the volume of the process 1-2 at point 1. V 1 = π D 2 4 L 1 (IV) Here the volume of the process at point 1 is V 1 , the length of the cylinder when piston is on the initial state is L 1 , and the diameter of the piston is D . Write the expression for the specific volume of the system at point 1. v 1 = v f + x ( v g − v f ) . (V) Here, the specific volume of the system at point 1 is v 1 , the specific volume at saturated liquid is v f and the specific volume at the saturated vapor is v g . Write the expression for the mass of the water. m = V 1 v 1 . (VI) Here, the mass of the water is m . Write the expression for the specific volume at point 2. v 2 = V 2 m . (VII) Here, the specific volume of the piston cylinder is v 2 . Write the expression for the heat transfer for the overall process. Q 13 = m ( u 3 − u 1 ) + W 13 . (VIII) Here, the heat transfer for the overall process is Q 13 , the mass of the water is m , the internal energy if the process at point 3 is u 3 and the work done for the process is W 13 . Conclusion: Substitute 15 cm 0.15 m for D and 2 cm for L 2 in Equation (IV). V 1 = ( π ( 15 cm ) 2 4 ) ( 2 cm ) = ( π ( ( 15 cm ) ( 1 m 100 cm ) ) 2 4 ) ( 2 cm ) ( 1 m 100 cm ) = ( 0.07068 m 2 4 ) ( 2 cm ) ( 1 m 100 cm ) = 3.53 × 10 − 4 m 3 Substitute 15 cm for D and 8 cm for L 1 in Equation (III). V 2 = ( π ( 15 cm ) 2 4 ) ( 8 cm ) = ( π ( ( 15 cm ) ( 1 m 100 cm ) ) 2 4 ) ( 8 cm ) ( 1 m 100 cm ) = ( 0.07068 m 2 4 ) ( 8 cm ) ( 1 m 100 cm ) = 1.414 × 10 − 3 m 3 Refer to table A-3 “properties of the saturated water (liquid vapor): pressure table” to obtain as 1.0432 × 10 − 3 m 3 / kg for v f and 1.694 × 10 − 3 m 3 / kg for v g . Substitute 0.3 for x in Equation (V). v 1 = 1.0432 × 10 − 3 m 3 / kg + 0.3 ( 1.694 m 3 / kg − 1

Fundamentals of Engineering Thermodynamics, Binder Ready Version

8th Edition

ISBN: 9781118820445

Author: Michael J. Moran, Howard N. Shapiro, Daisie D. Boettner, Margaret B. Bailey

Publisher: WILEY

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Chapter 3.14, Problem 77P

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The heat transfer for overall process of the water.

The work-done for the overall process of water.

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Chapter 3.14, Problem 76P

Chapter 3.14, Problem 78P

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The heat transfer for overall process of the water. The work-done for the overall process of water.