P21_029 - W = p V , we have Q abc = 13 W/ 2 = 14 . 8 kJ....

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29. (a) The net work done is the rectangular “area” enclosed in the pV diagram: W =( V V 0 )( p p 0 )=(2 V 0 V 0 )(2 p 0 p 0 )= V 0 p 0 . Inserting the values stated in the problem, we obtain W =2 . 27 kJ. (b) We compute the energy added as heat during the “heat-intake” portions of the cycle using Eq. 20-39, Eq. 20-43, and Eq. 20-46: Q abc = nC V ( T b T a )+ nC p ( T c T b ) = n µ 3 2 R T a µ T b T a 1 + n µ 5 2 R T a µ T c T a T b T a = nRT a µ 3 2 µ T b T a 1 + 5 2 µ T c T a T b T a = p 0 V 0 µ 3 2 (2 1) + 5 2 (4 2) = 13 2 p 0 V 0 where, to obtain the last line, the gas law in ratio form has been used (see Sample Problem 20-1). Therefore, since
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Unformatted text preview: W = p V , we have Q abc = 13 W/ 2 = 14 . 8 kJ. (c) The eciency is given by Eq. 21-9: = W | Q H | = 2 13 = 0 . 154 = 15 . 4% . (d) A Carnot engine operating between T c and T a has eciency equal to = 1 T a T c = 1 1 4 = 0 . 750 = 75 . 0% where the gas law in ratio form has been used. This is greater than our result in part (c), as expected from the second law of thermodynamics....
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