Question

Q9. You are analysing the performance of thermal engine, where the heat is supplied to the system at 800 °C. Heat is removed from the engine’s cycle at 300 °C. Improvements to the combustion system in the engine could push the temperature up to 920 °C. Other proposed modifications to the cooling part of the circuit could see the temperature at which heat is removed lowered to 220 °C. Analyse the performance using an idealised Carnot Cycle.

A. Calculate the current Carnot efficiency of this engine

B. Calculate how the improvement to the combustion system would change the Carnot efficiency

C. Calculate how the improvement to the cooling circuit alone would change the Carnot efficiency

D. What could be done to make the Engine fully reversible? Fully explain your answer.

Answer 1

part-(A)

Current carnot efficiency = 1 - ( TL / TH )

where TL is temperature in Kelvin at which heat is removed, TH is temperature in Kelvin at which Heat is supplied to system.

Current carnot efficiency = 1 - [ ( 220 + 273 ) / ( 920 + 273) ] = 0.587

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part-(B)

Carnot efficiency of previous unimproved system is given by ,

= 1 - ( TL / TH ) = 1 - [ ( 300 + 273 ) / ( 800 + 273) ] = 0.466

Due to improvement of combustion system , temperature TH at which heta supplied to system becomes 920oC.

Hence improved efficiency = 1 - ( TL / TH ) = 1 - [ ( 300 + 273 ) / ( 920 + 273) ] = 0.520

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part-(C)

Improved Efficiency due to change in cooling system alone

= 1 - ( TL / TH ) = 1 - [ ( 220 + 273 ) / ( 800 + 273) ] = 0.541

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part-(D)

If the process undergone by the working fluid during the cycle is to be reversible,

the heat transfer must take place with no temperature difference, i.e. it should be isothermal.