# Homework Help: Question about Carnot theorem proof

1. May 4, 2017

### Ruby_338

1. The problem statement, all variables and given/known data

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?
2. Relevant 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.
3. 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.

2. May 4, 2017

### TSny

When an irreversible engine runs backward, the ratio W'/Q'1 is not the same as the ratio W/Q1 for the same engine when it runs forward. (Otherwise, it would be a reversible engine.) In your second analysis you wrote
That's true if both R and I are running in the forward direction. But I is running backward.

3. May 4, 2017

### Ruby_338

The engines I and R in the second case are not the same engines I and R in

4. May 4, 2017

### TSny

OK. But when you say that $R$ is more efficient than $I$ when operating between the same reservoirs, that means that the efficiency $\left(\frac{W}{Q_1}\right)_R$ for $R$ is greater than $\left(\frac{W}{Q_1}\right)_I$ for $I$ when $I$ is running in the forward direction. Efficiency of an engine is defined as $\frac{W}{Q_1}$ when the engine is run in the forward direction (i.e., $Q_1$ is heat input to the engine from the hot reservoir and $W$ is work output by the engine).

If you run $I$ in the reverse direction between the two reservoirs, the ratio $\left(\frac{W'}{Q'_1}\right)_I$ will not be the same as when $I$ is running in the forward direction. (Now, $Q'_1$ is heat output to the hot reservoir and $W'$ is work input to the engine.)

So, although it is true that $\left(\frac{W}{Q_1}\right)_R > \left(\frac{W}{Q_1}\right)_I$ when $I$ is running forward, it does not logically follow that $\left(\frac{W}{Q_1}\right)_R > \left(\frac{W'}{Q'_1}\right)_I$ when $I$ is running backward. But you assumed this latter inequality in your derivation.

5. May 5, 2017

### Ruby_338

Is it possible for efficiency of R to be more than the backwards running efficiency of
I for any engines I and R? Because in that case, net heat would flow from the sink to the source ( I think) which violates the second law of thermodynamics even though Carnot's theorem implies that R should be more efficient than I.
If net heat does not flow from sink to source, how can it be proven?

Last edited: May 5, 2017
6. May 5, 2017

### Ruby_338

K. I got it. Thanks

Last edited: May 5, 2017