A) 10.2%
B) 8.6%
C) 6.9%
D) 5.7%
Correct Answer: D
Solution :
\[{{y}_{th}}\frac{{{W}_{t}}-{{W}_{c}}}{{{Q}_{s}}}=\frac{120-60}{200}=\frac{60}{200}=0.3\,\,or\,\,30%\]Heat in exhaust gases \[={{Q}_{s}}-{{W}_{t}}=200-120=80\,kJ\] Heat recovered in regenerator \[=0.4\times 30=32\,kJ\]. Heat supplied \[=200-32-168\,\,kJ\] New efficiency \[=\frac{200-60}{168}=35.7%\] Increase in efficiency \[=-35.7-30.0=5.7\]You need to login to perform this action.
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