Back on the P-V diagram.
In the segment from state 2 to state 3 we dump heat from the gas to the cold reservoir.
This cools the gas, and reduces the pressure, which allows us to compress the gas using
less work that we obtained from the expansion segment.
Thatís the key to the engineís operation.
We have to transfer that same amount of heat from the hot reservoir to the gas in the heating
segment from state 4 to state 1.
This heat is wasted in the sense that we donít get any work out it.
This motivates us to look for a more efficient way to implement the heating and cooling of
the gas.
The wasted heat is the gas heat capacity, three-halves ëNí ëkí times the temperature
change delta-T. This gets dumped into the cold reservoir.
We can never get this back, because heat only flows from hot to cold, and the cold reservoir
is the coldest object the gas comes into contact with.
The reheating of the gas requires the same amount of heat to be transferred from the
hot reservoir.
And, this will cost us, typically in burned fuel.
We can cut this waste in half by cooling the gas in two steps, each with temperature cha
nge delta-T equal to one-half the total temperature change T-hot minus T-cold.
For the first step we connect the gas to a reservoir at temperature T-medium, midway
between T-hot and T-cold.
The temperature change is half what it was before, so the transferred heat, Q-e, is also
half what it was before.
Then we connect the gas to the cold reservoir to reduce its temperature by delta-T again.
This transfers another Q-e of heat.
So, now only half the total heat extracted from the gas is wasted.
The other half is stored in the T-medium reservoir.
To warm the gas from T-cold, we reconnect it to the medium-temperature reservoir and
extract the heat we stored during the cooling process.
Then we connect the hot reservoir and transfer the second half of the heat.
We have to make up for this by burning fuel.
But, weíve recycled half of the heat required for heating the gas and so cut our fuel cost
for this process in half.
An obvious extension is to add more intermediate heat reservoirs.
Suppose we transfer heat in four steps, each with temperature change delta-T equal to one-fourth
T-hot minus T-cold.
In the first step we attached the gas to a medium-hot reservoir, and transfer one-fourth
of the total heat to it.
Then we transfer this same amount of heat to the medium reservoir.
Then the same amount to a medium-cold reservoir.
And, finally, the same amount to the cold reservoir.
Weíve now only wasted one-fourth of the total heat transferred.
For heating we can extract three-fourths of the required heat from what we stored in the
intermediate reservoirs.
And then generate only the last one-fourth.
A device that implements intermediate heat storage is called a regenerator.
And, itís essential to making a Stirling-cycle engine as efficient as possible.
Generalizing what weíve done, we could heat and cool the gas in ëMí steps, each with
one-over-M of the total temperature change and hence one-over-M of the total heat transfer.
Only one of these heat transfers would cost us, the rest would be recycled.
A practical regenerator implementation might use a tube through which gas is transported
between hot and cold chambers.
The intermediate heat reservoirs are placed one after another along the tube, gradually
heating or cooling the gas, depending on the direction of flow.
In principle, using an arbitrarily large number of intermediate reservoirs we can reduce the
amount of generated heat needed for the heating/cooling process to an arbitrarily small amount.
If so, we end up getting the heating and cooling parts of the cycle for free, so to speak.
Thereís another way to get free heating and cooling of the gas.
We can make use of adiabatic expansion and compression.
In an adiabatic process the gas is thermally insulated.
There is no transfer of heat.
Suppose the piston moves from position x-1 to x2.
From the ideal gas law, the pressure equals ëNí ëkí ëTí over ëVí.
We multiply by the piston cross-sectional area to get the force, which replaces ëVí
with ëxí.
And the internal energy of a monatomic gas is three-halves ëNí ëkí ëTí.
The change in internal energy, dU, equals three-halves ëNí ëkí times the change
in temperature, dT.
By the first law this is the negative of the work done by the gas, minus ëFí ëdxí.
Substituting for ëFí, this becomes minus ëNí ëkí ëTí ëdxí over ëxí.
Consider the second and fourth expressions.
Canceling the common ëNí and ëkí factors, dividing by ëTí, and multiplying by two-thirds,
we get ëdTí over ëTí equals minus two-thirds ëdxí over ëxí.
The corresponding relation between temperature and piston position is T-2 equals T-1 times
x-1 over x-2 to the two-thirds power.
When the gas temperature changes from T-1 to T-2 the internal energy change is three-halves
ëNí ëkí times the temperature change.
And this is converted into work done on the environment.
For compression we simply reverse the direction of the force and move the piston from x-2
to x-1.
The gas returns to its original temperature T-1, and the work done on the environment
during expansion is now done on the gas.
An expansion-compression cycle returns everything to its original state.
We saw that adiabatic expansion and compression are reversible.
Replacing the heating and cooling segments of the Stirling cycle with adiabatic processes,
we obtain the Carnot cycle.
We start with the gas attached to a hot reservoir and the piston in position x-1.
The gas undergoes isothermal expansion to piston position x-2.
This is identical to the Sterling cycle, and produces the same work on the environment.
An equal amount of energy flows as heat from the hot reservoir to the gas.
Now, instead of attaching the cold reservoir, we insulate the gas.
And expand the gas adiabatically to position x-3, causing the temperature to drop from
T-hot to T-cold.
This produces additional work three-halves ëNí ëkí T-hot minus T-cold.
Now we attach the cold reservoir.
Since the gas has already cooled to this temperature, there is no heat transfer.
The gas is isothermally compressed from piston position x-3 to x-4.
This requires work ëNí ëkí T-cold log x-3 over x-4 to be done on the system.
An equal amount of heat energy is transferred to the cold reservoir.
The gas is insulated again.
And adiabatically compressed back to the initial piston position x-1 with the gas temperature
increasing back to T-hot.
This increases the internal energy by three-halves ëNí ëkí T-hot minus T-cold, which equals
the work done on the system.
On a P-V diagram the first segment of the cycle is identical to the Stirling cycle.
Isothermal expansion at T-hot takes the system from state 1 to state 2.
Then adiabatic expansion takes the system from state 2 to state 3 and reduces the temperature
from T-hot to T-cold.
Isothermal compression at T-cold takes the system from state 3 to state 4.
Finally, adiabatic compression heats the gas from T-cold to T-hot, and takes the system
from state 4 back to state 1.
Isothermal expansion converts heat Q-s to work W-e equal to ëNí ëkí T-hot log x-2
over x-1, just as in the first segment of the Stirling cycle.
Adiabatic expansion converts internal energy to work W-e, equal to three-halves ëNí ëkí
T-hot minus T-cold as the gas cools to T-cold.
Isothermal compression converts work W-s to heat Q-e equal to ëNí ëkí T-cold log x-3
over x-4.
And, adiabatic compression converts work W-s, equal to three-halves ëNí ëkí T-hot minus
T-cold, into internal energy as the gas warms back up to T-hot.
This is equal to the work we got out of the system during adiabatic expansion.
So, we get this reheating for free.
From the relation between volume and temperature for the adibatic segments, we find that x-3
over x-4 equals x-2 over x-1.
So the heat-flow and work for the isothermal compression segment are also identical to
the Stirling cycle.
As for the Stirling cycle, the area enclosed by these four curves equals the net work produced
by the cycle.
If the temperatures and the piston positions x-1 and x-2 are the same, then the Stirling
and Carnot cycles generate the same net work.
Letís summarize our findings for the Carnot cycle.
The net work generated is ëNí ëkí T-hot minus T-cold times log x-2 over x-1.
To get this our hot reservoir has to provide heat energy ëNí ëkí T-hot log x-2 over
x-1 during isothermal expansion.
The cycle efficiency is the ratio of the output work to the input heat energies.
These expressions have factors of ëNí ëkí log x-2 over x-1 in common.
So the ratio depends only on the temperature factors, which we can write as one minus T-cold
over T-hot.
The efficiency is less than one (or 100 percent) by the ratio of the cold to hot temperatures.
For the Stirling cycle.
The net work generated is the same as for the Carnot cycle.
But the heat supplied by the hot reservoir is greater, because in addition to the isothermal
expansion term, we also have a term for the heating of the gas.
It follows that the efficiency is less than that of the Carnot cycle.
Now, as we saw, in principle the heating term can be made arbitrarily small by using a regenerator.
So, weíve found two heat-engine cycles having different efficiencies.
And weíve seen how the efficiency of one can be increased.
This raises the question.
Is there a cycle with greater efficiency than the Carnot cycle?
Or, can we find a trick, like the regenerator, to improve the Carnot cycle?
The answer to this question leads us to the Second Law of Thermodynamics, which is the
topic of the next video in this series.