Question
Download Solution PDFA heat engine is producing 20 kW of work, but it can produce 40 kW if it will operate on a reversible cycle. what can be the second law efficiency of the engine?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
The second law efficiency of an engine is given by:
\(\eta_{II}=\frac{W_{actual}}{W_{maximum}}\)
where, Wactual = Actual work produced by the engine, Wmaximum = Maximum work an engine can produce
A reversible engine produced the maximum work.
Explanation:
Given:
Wactual = 20 kW, Wmaximum = 40 kW
\(\eta_{II}=\frac{20}{40}\)
\(\eta_{II}=\frac{1}{2}\)
\(\eta_{II}=0.5\)
Hence the second law efficiency of the heat engine is 0.5
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