Answer
65.4 %
Work Step by Step
The maximum efficiency of this cycle can have is-
$n_{th,carrot} = 1-\frac{T_{L}}{T_{H}}$
$=1-\frac{(80+460)R}{(1100+460)R}$
$=0.654$
=65.4%
Thermodynamics: An Engineering Approach 8th Edition
by Cengel, Yunus; Boles, Michael
Published by
McGraw-Hill Education
ISBN 10:
0-07339-817-9
ISBN 13:
978-0-07339-817-4
Chapter 9 - Gas Power Cycles - Problems - Page 538: 9-11E
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