Explanation Write the expression of specific internal energy ( u ) to obtain the quality ( x ) . u = u f + x u f g (I) Here, specific internal energy at saturated liquid is u f and specific internal energy at mixed state is u f g . Write the expression to obtain the specific volume ( v ) . v = v f + x v f g v = v f + x ( v g − v f ) (II) Here, specific volume at saturated liquid is v f and specific volume at saturated vapor is v g . Conclusion: Refer Table A-6, “Superheated water”, obtain the pressure and specific volume as 300 kPa and 0.7921 m 3 / kg corresponding to temperature at 250 ° C and specific internal energy at 2728.9 kJ / kg . Refer Table A-6, “Superheated water”, obtain the pressure and specific volume as 300 kPa and 0.7921 m 3 / kg corresponding to temperature at 250 ° C and specific internal energy at 2728.9 kJ / kg . Refer Table A-5, “Saturated water: Pressure table”, obtain the following properties corresponding to a pressure at 300 kPa. T = 133.52 ° C u f = 561.11 kJ / kg u f g = 1982.1 kJ / kg v f = 0.001073 m 3 / kg v f g = 0.60582 m 3 / kg Substitute 1560 kJ / kg for u , 561.11 kJ / kg for u f and 1982

8th Edition

ISBN: 9780100257054

Author: CENGEL

Publisher: YUZU

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Properties of Pure Substances

A substance with fixed chemical composition throughout is called a pure substance like water, air, and nitrogen. A pure substance does not have to be a single element or compound. A mixture of two or more phases of a pure substance is also a pure substan…

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Chapter 3.8, Problem 127RP

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EBK THERMODYNAMICS: AN ENGINEERING APPR

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