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Ruby_338
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Homework Statement
Carnot theorem states that no engine working between two temperatures T1 of source and T2 of sink can have a greater efficiency than that of the Carnot engine.
Second law of thermodynamics:it is impossible for a self acting machine to transfer heat from a body at a higher temperature to a body at a lower temperature.
Consider a reversible engine R and an irreversible engine I. The engines are coupled such that as I runs forwards, it drives R backwards. I absorbs heat Q1from source, performs work W and and rejects heat Q2= Q1 - W to the sink.
Efficiency of I, μI = W/Q1
The engine R absorbs heat Q'2 from the sink, work W is done on it and it rejects heat Q'1to the source
Efficiency of engine R, μR= W/Q'1
Suppose, I more efficient than R. Then
μI > μR or W/Q1> W/ Q'1 so
Q'1>Q1
The source loses heat Q1 to I and gains heat Q'1 from R
Net heat gained by source per cycle is. Q'1- Q1 which is positive since Q'1 greater than Q1
The sink gains heat Q1- W from I and loses Q'2 to R.
Net heat lost by sink per cycle = Q'2 - (Q1-W). = Q'2- {Q1 -(Q'1-Q'2)}
= Q'1- Q1
The compound engine IR is a self acting machine which transfers heat from the sink(at lower temperature) to the source (at higher temperature) without any external agency, which is against the second law of thermodynamics.
My question is: if we put R in I's place and
, assume R more efficient than I, then using the same procedure as above, won't we find that heat would still flow from sink to source? Does this not also violate the second law of thermodynamics? Is the proof valid?
Homework Equations
Efficiency of an engine, μ= W/Q1= (Q1-Q2)/Q1
Where W is work done by engine, Q1 is heat drawn from source by engine and Q2 is heat rejected to sink by engine.
The Attempt at a Solution
Putting R in I's place,
Consider a reversible engine R and an irreversible engine I. The engines are coupled such that as R runs forwards, it drives I backwards. R absorbs heat Q1from source, performs work W and and rejects heat Q2= Q1 - W to the sink.
Efficiency of R, μI = W/Q1
The engine I absorbs heat Q'2 from the sink, work W is done on it and it rejects heat Q'1to the source
Efficiency of engine I, μI= W/Q'1
Suppose, R more efficient than I. Then
μI < μR or W/Q1> W/ Q'1 so
Q'1>Q1
The source loses heat Q1 to R and gains heat Q'1 from I
Net heat gained by source per cycle is. Q'1- Q1 which is positive since Q'1 greater than Q1
The sink gains heat Q1- W from R and loses Q'2 to I
Net heat lost by sink per cycle = Q'2 - (Q1-W). = Q'2- {Q1 -(Q'1-Q'2)}
= Q'1- Q1
The compound engine IR is a self acting machine which transfers heat from the sink(at lower temperature) to the source (at higher temperature) without any external agency, which is against the second law of thermodynamics.