Question 10 / 33:  Combustion of a fuel at 1200°K at a rate of 3 kW produces steam at 550°K.The steam then produces 2 kW of work and rejects some heat to 310°K. What is the second-law efficiency of the process?
A  66%
B  74%
C  39%
D  89%
E  100%
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Explanation:

The thermal or rational efficiency of the process is simply 2 kW/3kW = 66%. The second law efficiency takes into account the maximum efficiency possible for any system working between the two temperature limits; it is the quotient of the thermal efficiency and the Carnot efficiency. The Carnot efficiency in this case is 1 - 310/1200 = 0.74. Hence, the second-law efficiency for the process at hand is a remarkable 66/74 = 89%.

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Thermal-Fluid Systems ME303

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Attribution:  Dr. Steve Gibbs. Thermal-Fluid Systems. The Saylor Academy 2014, http://www.saylor.org/courses/me303/
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