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Derive the expression for Carnot engine efficiency. |
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Answer» Solution :Efficiency of a Carnotenginr: Efficiency is DEFINED as the ratio of WORK done by the working substance in one cycle to the amount of heat extrcted from the source. `eta=("work done" )/("heat extracted")=(W)/(Q_(H))` From the FIRST law of thermodynamics, `W=Q_(h)-Q_(L)` `eta=(Q_(H)-Q_(L))/(Q_(H))=(Q_(L))/(Q_(H))` Applying isothermal conditions, we GET `Q_(H)=muRT_(H) In((V_(2))/(V_(1)))` `Q_(L)=muRT_(L) In((V_(3))/(V_(4)))` `therefore (T_(L))/(T_(H))=(Q_(L) In(V_(3))/(V_(4)))/(Q_(H)In(V_(2))/(V_(1)))` By applying adiabatic conditions we get , `T_(H)V_(2)^(y-1) = T_(L) V_(3)^(y-1)` `T_(H)V_(1)^(y-1) = T_(L) V_(4)^(y-1)` By dividing the above two equations we get, `(V_(2)/(V_(1)))^(y-1)=(V_(3)/(V_(4)))^(y-1)` Subsitituting EQUATION (5) in (4), we get `(Q_(L))/(Q_(H)) =(T_(L))/(T_(H))` `therefore` The effiency `mu=1-(T_(L))/(T_(H))` |
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