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What is the thermal efficiency of this regeneration cycle in terms of enthalpies and fractions of total flow?

User Twal
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Answer:


\eta =((h_3-h_4)-(h_2-h_1))/((h_3-h_5))

Step-by-step explanation:

generally regeneration of cycle is used in the case of gas turbine. due to regeneration efficiency of turbine is increased but there is no effect on the on the net work out put of turbine.Actually in regeneration net heta input is decreases that is why total efficiency increase.

Now from T-S diagram


W_(net)=W_(out)-W_(in)


W_(net)=(h_3-h_4)-(h_2-h_1)


Q_(in)=h_3-h_5

Due to generation
(h_5-h_2) amount of energy has been saved.


Q_(generation)=Q_(saved)

So efficiency of cycle
\eta =(W_(net))/(Q_(in))


\eta =((h_3-h_4)-(h_2-h_1))/((h_3-h_5))

Effectiveness of re-generator


\varepsilon =((h_5-h_2))/((h_4-h_2))

So the efficiency of regenerative cycle


\eta =((h_3-h_4)-(h_2-h_1))/((h_3-h_5))

What is the thermal efficiency of this regeneration cycle in terms of enthalpies and-example-1