Explanation Given information: For tank A , Initial temperature, ( T A ) 1 = 280 F Initial pressure, ( P A ) 1 = 1 a t m = 14.696 p s i Volume, V A = 7 f t 3 For tank B , Initial pressure, ( P B ) 1 = 40 p s i a Volume, V B = 11 f t 3 Final pressure, p 2 = 40 p s i a Assumptions: For tank A , According to the super heated water tables At ( p A ) 1 = 14.696 p s i ( T A ) 1 = 280 F Specific volume, ( v A ) 1 = 29.687 f t 3 / l b m Specific internal energy, ( u A ) 1 = 1102.1 B t u / l b m For tank B , According to saturated water tables At ( p B ) 1 = 40 p s i a Specific volume, ( v B ) 1 = 10.501 f t 3 / l b m Specific internal energy, ( u B ) 1 = 1092.27 B t u / l b m Calculations: For tank A , Initial mass is calculated as follows: ( m A ) 1 = V A ( v A ) 1 ( m A ) 1 = 7 29.687 ( m A ) 1 = 0.236 l b m For tank B , Initial mass is calculated as follows, ( m B ) 1 = V B ( v B ) 1 ( m B ) 1 = 11 10.501 ( m B ) 1 = 1.04752 l b m So that by using continuity eq, m 2 = ( m A ) 1 + ( m B ) 1 m 2 = 0.263 + 1.04752 m 2 = 1.28352 l b m Total volume is calculated as follows: V 2 = ( V A ) 1 + ( V B ) 1 V 2 = 7 + 11 V 2 = 18 f t 3 Final specific volume is calculated as follows: v 2 = V 2 m 2 v 2 = 18 1

10th Edition

ISBN: 9781119494966

Author: Borgnakke, C. (claus), Sonntag, Richard Edwin, Author.

Publisher: Wiley,

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To determine: The property values of final heat transfer and final state.

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