Answer:
[tex]n_{thermal-efficiency}=0.6[/tex]
Explanation:
Given data
Temperature T1=270K
Temperature T2=650K
Work=4.3 kJ
Heat rejects=8.0 kJ
To find
Thermal efficiency
Solution
The thermal efficiency is given as:
[tex]n_{thermal-efficiency}=\frac{n^{|}_{real-cycle} }{n^{||}_{Carnot-cycle} }[/tex]
For Carnot cycle efficiency
[tex]n^{||}=1-(T1/T2)\\n^{||}=1-(270K/650K)\\n^{||}=0.584\\[/tex]
For real cycle efficiency
[tex]n^{|}=\frac{4.3}{4.3+8.0}\\n^{|}=0.35\\[/tex]
So the thermal efficiency is:
[tex]n_{thermal-efficiency}=\frac{n^{|}_{real-cycle} }{n^{||}_{Carnot-cycle} }\\n_{thermal-efficiency}=(0.35/0.584)\\ n_{thermal-efficiency}=0.6[/tex]
