Answer:
[tex]\eta _{max} = 0.2413 = 24.13%[/tex]
[tex]\eta' _{max} = 0.5061 = 50.61%[/tex]
Given:
[tex]T_{1max} = 100^{\circ} = 273 + 100 = 373 K[/tex]
operating temperature of heat engine, [tex]T_{2} = 10^{\circ} = 273 + 10 = 283 K[/tex]
[tex]T_{3max} = 300^{\circ} = 273 + 300 = 573 K[/tex]
Solution:
For a reversible cycle, maximum efficiency, [tex]\eta _{max}[/tex] is given by:
[tex]\eta _{max} = 1 - \frac{T_{2}}{T_{1max}}[/tex]
[tex]\eta _{max} = 1 - \frac{283}{373} = 0.24[/tex]
[tex]\eta _{max} = 0.2413 = 24.13%[/tex]
Now, on re designing collector, maximum temperature, [tex]T_{3max}[/tex] changes to [tex]300^{\circ}[/tex], so, the new maximum efficiency, [tex]\eta' _{max}[/tex] is given by:
[tex]\eta' _{max} = 1 - \frac{T_{2}}{T_{3max}}[/tex]
[tex]\eta _{max} = 1 - \frac{283}{573} = 0.5061[/tex]
[tex]\eta _{max} = 0.5061 = 50.61%[/tex]
