The objective of a carnot engine, or a carnot heat engine is to convert heat into work continuously.
And this is a schematic representation of behavior.
We take heat at high temperatures, so Qh is energy in the form of heat being transferred
into this carnot cycle.
Some the the heat goes to work and the rest is rejected at low heat to a low temperature
reservoir.
For this ideal case, and of course the carnot cycle is ideal in itself, but we are looking
at the case where the low temperature of the carnot cycle is the same as the low temperature
reservoir and the high temperature reservoir is the same temperature in the carnot cycle.
Of course we need a temperature difference for heat transfer and we're looking at the
extreme so as the temperature difference get's very small the heat transfer of course would
take a very long time.
But we can still do the calculations treating this as one approach.
And what we want to do is then apply the first law and then the second law to this process.
And what we're going to do apply it to the cycle.
The first law we can apply is Qh plus Qc plus W equals 0.
That's just applying it around the carnot cycle where we have to be careful of sign
notation.
Qh is now energy being added, that's positive, Qcis energy being taken away, that's negative,
and W is also energy being taken away, so that's negative.
The second law says delta S for this carnot engine is zero. and the reason is we have,
for example, some gas going through this cycle, and it starts here, moves through the cycle,
later time it's back where it started.
State function, entropy cannot change and it keeps going around and around many times.
So delta S for the cycle is zero, we now calculate delta S, heat over temperature.
The high temperature, 600 degrees for the cycle and then the heat over low temperature,
that must add to zero.
So what we're going to look at is the numbers or the case where we're going to assume that
we have 500 joules.
Make it easier to see how this works.
500 joules is being transferred in, we have the two temperatures.
And so first law says 500 and that's positive because we are adding energy plus Qc plus
W that's zero.
The second law says 500 divided by, so that's joules and this is kelvin.
Has to be absolute temperature.
Doesn't make sense unless it's absolute temperature.
Qc we don't know, the low temperature is 300 kelvin that's zero.
So here's the two entropy changes that means we can solve for Qc is going to be minus 5,
6, 300.
So it's minus 250 joules.
So this is -250 joules.
We can go back to the first law, 500 minus 250 plus W equals zero.
W is also -250.
So for this particular cycle, and these choices of temperature, we're saying is that we can
put 500 joules in and we get half of it out as work and the rest we reject to the low
temperature.
Of course the first point is that we can't convert all the heat into work we have to
reject heat at the low temperature.
We can talk about a thermodynamic efficiency as the work over Qh.
So one way of looking at this, this is what we desire, this is what we could sell.
This is what we have to pay for to create the process.
We like this number to be as large as possible.
In our case it's 250 divided by 500 or .5.
Ok so this is our thermodynamic efficiency for a carnot cycle, for this particular carnot
cycle.
And in general we look at this definition we can talk about the efficiency, work over
Qh.
Well work is Qc plus Qh.
This will be a positive number over Qh.
so this is 1 plus Qc over Qh.
We can show that this is equal to 1- Tc over Th so efficiency for a carnot cycle is just
related to the temperatures.
Absolute temperature must be used in this calculation.
so let's look at why we can make this substitution.
So second law we said Qh over Th plus Qc over Tc is equal to zero.
So this means Qh over Th is -Qc over Tc.
Or Qc over Qh is Tc over Th with a minus sign.
This then is the substitution that we made here to get to efficiency just in terms of
the temperatures.
So note that this is an ideal carnot cycle.
Doesn't mean we can actually run this in real life.
It's indication is that it's the best.
So this is the best efficiency between these two temperatures.
Right we need a high temperature energy source.
We need a low temperature to reject the heat.
We are not going to do as well as this but this gives us a benchmark of how well we can
do.
So let's go back and look at our cycle and let's make a change to see what happens if
we were to look at conditions that might be more realistic.
So the first thing let's assume this is 700, and this is 250 kelvin.
Right so call this Th prime to distinguish it from the Th of the cycle, Tc of the cycle
and then Tc'.
And now I want to look at the energy being removed from this is a high temperature reservoir,
stays at this temperature.
So I'm going to look at Qh'.
And now the only thing different is now I look at my reservoir as my system and I'm
removing energy from it so that's minus.
I'm going to go down here and look at Qc'.
Now that's plus.
And we've already calculated these numbers, the cycle here has not changes so all these
numbers are the same.
This is positive when I look at the low temperature reservoir as my system.
I can look at the carnot cycle as my system, this is of course then is negative.
If I now analyze this, where now we've introduced irreversibility.
Irreversible of course because there is significant temperature difference.
Delta S for this engine Hs not changed.
It's still zero.
It's still running around in a cycle.
The first law has not changed.
But delta S for the reservoir has changed.
For the high temperature reservoir Qh' over Th' and then for the low temperature reservoir
again these values.
So Qh minus 500 joules, high temperature reservoir is 700 K. Qc is positive and the temperature
for the low temperature reservoir is 250 K.
So this is minus 0.71 plus 1.0 J/K.
So delta S for the reservoirs is 0.29 J/K.
This is positive.
Delta S for the engine is zero because it's running around the cycle.
Delta S for the universe is 0.29 J/k.
It's greater than zero because it's irreversible.
The idea is that we could not put 700 K and 250 K into this equation here for the efficiency.
And so now we have a higher temperature in order to make the process work.
But we didn't' get a higher efficiency we are getting the efficiency of the high temperature
of the carnot cycle.