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A thermodynamic cycle consists of a linked sequence of thermodynamic processes that involve transfer of heat and work into and out of the system, while varying pressure, temperature, and other state variables within the system, and that eventually returns the system to its initial state.^{[1]} In the process of passing through a cycle, the working fluid (system) may convert heat from a warm source into useful work, and dispose of the remaining heat to a cold sink, thereby acting as a heat engine. Conversely, the cycle may be reversed and use work to move heat from a cold source and transfer it to a warm sink thereby acting as a heat pump. At every point in the cycle, the system is in thermodynamic equilibrium, so the cycle is reversible (its entropy change is zero, as entropy is a state function).
During a closed cycle, the system returns to its original thermodynamic state of temperature and pressure. Process quantities (or path quantities), such as heat and work are process dependent. For a cycle for which the system returns to its initial state the first law of thermodynamics applies:
$\Delta E=E_{out}-E_{in}=0$
The above states that there is no change of the energy of the system over the cycle. E_{in} might be the work and heat input during the cycle and E_{out} would be the work and heat output during the cycle. The first law of thermodynamics also dictates that the net heat input is equal to the net work output over a cycle (we account for heat, Q_{in}, as positive and Q_{out} as negative). The repeating nature of the process path allows for continuous operation, making the cycle an important concept in thermodynamics. Thermodynamic cycles are often represented mathematically as quasistatic processes in the modeling of the workings of an actual device.
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Carnot cycle and Carnot engine | Thermodynamics | Physics | Khan Academy
The Internal Combustion Engine - stop motion animations and the PV cycle (Otto cycle)
Transcription
Let's start with this classic
system that I keep referring
to in our thermodynamics
videos.
I have a cylinder.
It's got a little piston on the
top of it, or it's got a
ceiling that's movable.
The gas, and we're thinking of
monoatomic ideal gases in
here, they're exerting pressure
onto this ceiling.
And the reason why the ceiling
isn't moving all the way up,
is because I've placed a bunch
of rocks on the top to offset
the force per area of
the actual gas.
And I start this gas when
it's in equilibrium.
I can define its macrostates.
It has some volume.
It has some pressure that's
being offset by these rocks.
And it has some well-defined
temperature.
Now, what I'm going to do is,
I'm going to place this system
here-- I'm going to place it
on top of a reservoir.
And I talked about what a
reservoir was either in the
last video or a couple
of videos ago.
You can view it as an infinitely
large object, if
you will, of a certain
temperature.
So if I put it next to-- if I
put our system next to this
reservoir-- and let's say I
start removing pebbles from
our system.
We learned a couple of videos
ago that if we did it
adiabatically-- what does
adiabatically mean?
If we removed these pebbles
in isolation, without any
reservoir around, the volume
would increase, the pressure
would go down, and actually
the temperature would
decrease, as well.
We showed that a couple
of videos ago.
So by putting this big reservoir
there that's a lot
larger than our actual canister,
this will keep the
temperature in our
canister at T1.
You can kind of view a reservoir
as-- say I had a cup
of water in a stadium.
And the air conditioner in the
stadium is at 60 degrees.
Well, no matter what I do to
that water, I could put it in
the microwave and warm it up,
but if I put it back in that
stadium, that stadium
is going to keep
that water at 60 degrees.
And you might say, oh, won't
the reservoir's temperature
decrease if it's throwing
off heat?
Well, it would, but it's so much
larger that its impact
isn't noticeable.
For example, if I put a cup of
boiling water into a super
large covered-dome stadium, the
water will get colder to
the ambient temperature
of the stadium.
The stadium will get warmer, but
it will be so marginally
warmer than you won't
even notice it.
So you can kind of view
that as a reservoir.
And theoretically, this
is infinitely large.
So the effect of this is, as we
remove these little rocks,
we're going to keep the
temperature constant.
And remember, if we're keeping
the temperature constant,
we're also keeping the internal
energy constant,
because we're not changing
the kinetic
energy of the particles.
So let me see what happens.
So I keep doing that.
And so I get to a point-- let
me see-- where my volume has
increased-- so let me delete
some of my rocks here.
Delete some of the rocks.
So some of the rocks are gone.
And now my overall volume
is going to be larger.
Let me move this up
a little bit.
And then let me color
this in black.
Oh, whoops.
Let me color this in, just
to give an idea.
So our volume has gotten
a bit larger.
And let me get my
pen correctly.
So our volume has gotten larger
by roughly this amount.
We have the same number
of particles.
They're going to bump into
the ceiling a little less
frequently, so my pressure
would have gone down.
But because I kept this
reservoir here, because this
reservoir was here the whole
time during this process, the
temperature stayed at T1.
And that was only because
of this reservoir.
And I want to make that clear.
And also, just as review, this
is a quasi-static process,
because I'm doing
it very slowly.
The system is in equilibrium
the whole time.
So let's draw what we have so
far on our famous PV diagram.
So this is the P-axis.
That's the V-axis.
You label them.
This is P.
This is V.
Let me call this-- I'm going
to do it in a good color.
This is state A of the system.
This is state B of the system.
So state A starts at some
pressure and volume-- I'll do
it like that.
That's state A.
And it moves to state B.
And notice, I kept the
temperature constant.
And what did we learn in,
I think it was one
or two videos ago?
Well, we're at a constant
temperature, so we're going to
move along an isotherm,
which is just
a rectangular hyperbola.
Because when your temperature
is constant, your pressure
times your volume is going to
equal a constant number.
And I went over that before.
So we're going to move over--
our path is going to look
something like this, and I'll
move here, to state B.
I'll move over here
to state B.
And the whole time, this was at
a constant temperature T1.
Now, we've done a bunch
of videos now.
We said, OK, how much work
was done on this system?
Well, the work done on
the system is the
area under this curve.
So some positive work was-- not
done on the system, sorry.
How much work was done
by the system?
We're moving in this
direction.
I should put the direction
there.
We're moving from
left to right.
The amount of work done
by the system is
pressure times volume.
We've seen that multiple
times.
So you take this area of the
curve, and you have the work
done by the system
from A to B.
Right?
So let's call that
work from A to B.
Now, that's fair and everything,
but what I want to
think more about, is
how much heat was
transferred by my reservoir?
Remember, we said, if this
reservoir wasn't there, the
temperature of my canister
would have gone down as I
expanded its volume, and as
the pressure went down.
So how much heat came into it?
Well, let's go back to our basic
internal energy formula.
Change in internal energy is
equal to heat applied to the
system minus the work
done by the system
Now, what is the change
in internal
energy in this scenario?
Well, it was at a constant
temperature
the whole time, right?
And since we're dealing with a
very simple ideal gas, all of
our internal energy is due
to kinetic energy, which
temperature is a measure of.
So, temperature didn't change.
Our average kinetic energy
didn't change, which means our
kinetic energy didn't change.
So our internal energy did not
change while we moved from
left to right along
this isotherm.
So we could say our internal
energy is zero.
And that is equal to the heat
added to the system minus the
work done by the system.
Right?
So if you just-- we put the work
done by the system on the
other side, and then switch the
sides, you get heat added
to the system is equal to the
work done by the system.
And that makes sense.
The system was doing some work
this entire time, so it was
giving energy to-- well,
you know, it was giving
essentially maybe
some potential
energy to these rocks.
So it was giving energy away.
It was giving energy outside
of the system.
So how did it maintain
its internal energy?
Well, someone had to give
it some energy.
And it was given that energy
by this reservoir.
So let's say, and the convention
for doing this is
to say, that it was given--
let me write this down.
It was given some energy Q1.
We just say, we just put this
downward arrow to say that
some energy went into
the system here.
Fair enough.
Now let's take this state B and
remove the reservoir, and
completely isolate ourselves.
So there's no way that heat can
be transferred to and from
our system.
And let's keep removing
some rocks.
So if we keep removing some
rocks, where do we get to?
Let me go down here.
So let's say we remove a
bunch of more rocks.
So let me erase even more
rocks than we had in B.
Maybe I only have
one rock left.
And obviously, the overall
volume would have increased.
So let me make our piston go up
like that, and I can make
our piston is maybe
a lot higher now.
And let me just fill in the rest
of our, just so that we
don't have some empty
space there.
So if I fill that in right
there-- OK Let
me fill that in.
And then I just use the blue--
I should be talking about
thermodynamics, not drawing.
But you get the idea.
And then I have some
more-- you know, I
shouldn't add particles.
But my volume has increased
a good bit.
My pressure will have gone down,
they're going to bump
into the walls less.
And because I removed the
reservoir, what's going to
happen to the temperature?
My temperature is going
to go down.
This was an adiabatic process.
So an adiabatic just means
we did it in isolation.
There was no exchange of heat
from one system to another.
So let me just-- this arrow
continues down here.
I'll say adiabatic.
Now, since I'm moving from
one temperature to
another, this is at T2.
So I will have moved to
another isotherm.
This is the isotherm for T1.
If I keep my temperature
constant, I
move along this hyperbola.
And I would have kept moving
along this hyperbola.
But now that we didn't keep our
temperature constant, we
now move like this.
We move to another isotherm.
So let's say I have another
isotherm at T2.
It looks something like this.
So let me draw like that.
So let's say I have another-- it
should actually curve up a
little bit.
So let's say, everything at
temperature T2, depending on
its pressure and volume, is
someplace along this curve
that asymptotes up like
that, and then goes to
the right like that.
Now, I would have moved down
to this isotherm, and my
pressure would have kept going
down, and my volume would have
kept going down.
So this move, from B to state
C, will look like this.
Let me do to it in
another color.
Let me do it in the orange
color of this arrow.
So it will look like this.
And now we're at state C.
Now, this was adiabatic.
Which means, there is
no exchange of heat.
So I don't have to figure out
how much heat got transferred
into the system.
Now, there's something
interesting here.
We still did do some work.
We can take the area
under this curve.
And we're going to leave it to
a future video to think about
where that work energy-- well,
the main thing is, is what was
reduced by that work energy.
And, well, if you think
[UNINTELLIGIBLE]
to leave it to future video.
Our internal energy was
reduced, right?
Because our temperature
went down.
So our internal energy
went down.
We'll talk more about that
in the future video.
So now that we're at state C,
and we're at temperature T2.
Let's put back another
sink here.
But this sink, what it's going
to have is a reservoir.
So let me put two things
right here.
So I'm going to add--
let me erase some of
these blocks in black.
So now I'm going to
add blocks back.
I'm going to add little
pebbles back into it.
But I'm going to do it as
an isothermic process.
I'm going to do it with
a reservoir here.
But this reservoir here, it's
not going to be the same
reservoir that I put up there.
I swapped that one out.
I got rid of any reservoir
when I went from B to C.
And now I'm going to swap
in a new reservoir.
Actually, let me make it blue.
Because it's going to be--
Because here's what's
happening.
I'm now adding pebbles in.
I'm compressing the gas.
If this was an adiabatic
process, the gas would
want to heat up.
So what I'm doing is, I need to
put a reservoir to keep it
at T2, to keep it along
this isotherm.
So this is T2.
Remember, this reservoir is
kind of a cold reservoir.
It keeps the temperature down.
As opposed to here.
This was a hot reservoir.
It kept the temperature up.
So you can imagine.
The heat generated in the
system, or internal energy
being generated in the
system-- well, no, I
shouldn't say that.
The temperature of the system
will want to go up, but it's
being released, because it's
able to transfer that heat
into our new reservoir.
And that amount of heat is Q2.
So I move along this.
This is right here.
I'm moving along another
isotherm, I'm moving along
this isotherm.
Until I get to state D.
We're almost there.
This is state D.
So state D will be someplace
here, along this isotherm
right here.
Maybe this is state D.
And once again, you can make
the argument that we moved
along an isotherm Our
temperature did not change
from C to D.
We know that our internal energy
went down from B to C,
because we did some work.
But from C to D, our temperature
stayed the same.
It was at temperature-- let me
write it down-- T2, right?
Because we had this
reservoir here.
It stayed the same.
If your temperature stays the
same, then your internal
energy stays the same.
At least for the system we're
dealing with, because it's a
very simple gas.
It's actually the system you'll
deal with most of the
time, in an intro thermodynamics
course.
So.
Given our internal energy didn't
change, we can apply
the same argument that the heat
added to the system is
equal to the work done
by the system.
Right?
Same math as we did up here.
Now, in this case, the work
wasn't done by the system.
The work was done
to the system.
We compressed this piston.
The force times distance
went the other way.
So given that work was done to
the system, the heat added to
the system was negative,
right?
We're just applying
the same thing.
If our internal energy is 0, the
heat added to the system
is equal to the work
done by the system.
The work done by the
system is negative.
Work was done to it.
So the heat added to the system
would be negative.
Or another way to think
about it is that the
system gave away heat.
We put that with Q2.
And where did it
give that heat?
It gave it to this reservoir
that we put here, this kind of
cold reservoir.
You could almost view
it as a-- well, it's
accepting the heat.
OK.
We're almost there.
Now, let's say we remove this
reservoir from under our
system again, so it's completely
isolated from
everything else, at least
in terms of heat.
And what we do is, we start
adding-- so state D, we still
had a few less pebbles.
But we start adding more
pebbles again.
We start adding more pebbles
to get it to state A.
So let me change my
pebble color.
So we start adding more
pebbles again to
get it to state A.
So that's this process
right here.
Let me do a different color.
Let's say this is green.
So as we add pebbles, that's
this movement right here.
We're moving from one isotherm
up to another isotherm at a
higher temperature.
And remember, this whole
time we went
this clockwise direction.
So a couple of interesting
things are going on here.
Because we're assuming an ideal
scenario, nothing was
lost of friction.
This piston just moves
up and down.
No heat loss due to that.
What we can say is that we've
achieved-- we are back at our
original internal energy.
In fact, this is one of the
properties of a state
variable, is that if we're at
the same point on the PV
diagram, the same exact
point, we have
the same state variable.
So now we have the same
pressure, volume, temperature,
and internal energy as
what we started with.
So we've done here is
completed a cycle.
And this particular cycle, it's
an important one, it's
called the Carnot cycle.
It's named after a French
engineer who was trying to
just optimize engines
in the early 1800s.
So Carnot cycle.
And we're going to study this
a lot in the next few videos
to really make sure we
understand entropy correctly.
Because in a lot of chemistry
classes, they'll
throw entropy at you.
Oh, it's measure of disorder.
But you really don't know what
they're talking about, or how
can you quantify it, or
measure it anyway.
And we really need to deal with
the Carnot cycle in order
to understand where the first
concepts of entropy really
came from, and then relate
it to kind of more
modern notions of it.
Now, a system that completes
a Carnot cycle is called a
Carnot engine.
So our little piston here that's
moving up and down, we
can consider this
a Carnot engine.
You might say, oh, Sal,
this doesn't seem
like a great engine.
I have to move pebbles
and all of that.
And you're right.
You wouldn't actually implement
an engine this way.
But it's a useful engine, or
it's a useful theoretical
construct, in order for
understanding how heat is
transferred in an engine.
I mean, if you think about
what's happening here, is this
first heat sink transferred some
heat to the system, and
then the system transferred a
smaller amount of heat back to
the other reservoir.
Right?
So this system was transferring
heat from one
reservoir to another
reservoir.
From a hotter reservoir
to a colder reservoir.
And in the process, it was
also doing some work.
And what was the work
that it did?
Well, it's the area under this
curve, or the area inside of
this cycle.
So this is the work done
by our Carnot engine.
And the way you think about it
is, when you're going in the
rightward direction with
increasing volume, it's the
area under the curve is the
work done by the system.
And then when you move in the
leftward direction with
decreasing volume, you subtract
out the work done to
the system, and then you're
left with just
the area in the curve.
So we can write this Carnot
engine like this.
It's taking, it's starting-- so
you have a reservoir at T1.
And then you have your
engine, right here.
And then it's connected--
so this takes Q1
in from this reservoir.
It does some work, all right?
The work is represented by the
amount of-- the work right
here is the area inside
of our cycle.
And then it transfers Q2, or
essentially the remainder from
Q1, into our cold reservoir.
So T2.
So it transfers Q2 there.
So the work we did is really
the difference
between Q1 and Q2, right?
You say, hey.
If I have more heat coming in
than I'm letting out, where
did the rest of that heat go?
It went to work.
Literally.
So Q1 minus Q2 is equal to the
amount of work we did.
And actually, this is a good
time to emphasize again that
heat and work are not
a state variable.
A state variable has to be the
exact same value when we
complete a cycle.
Now, we see here that we
completed a cycle, and we had
a net amount of work done, or a
net amount of heat added to
the system.
So we could just keep going
around the cycle, and keep
having heat added
to the system.
So there is no inherent heat
state variable right here.
You can't say what the
value of heat is at
this point in time.
All you could say is what amount
of heat was added or
taken away from the system, or
you can only say the amount of
work that was done to, or
done by, the system.
Anyway, I want to leave
you there right now.
We're going to study
this a lot more.
But the real important thing is,
and if you never want to
get confused in a thermodynamics
class, I
encourage you to even
go off on your
own, and do this yourself.
Kind of-- you can almost take
a pencil and paper, and redo
this video that I just did.
Because it's essential that
you understand the Carnot
engine, understand this
adiabatic process, understand
what isotherms are.
Because if you understand that,
then a lot of what we're
about to do in the next few
videos with regard to entropy
will be a little bit more
intuitive, and not too confusing.
Two primary classes of thermodynamic cycles are power cycles and heat pump cycles. Power cycles are cycles which convert some heat input into a mechanical work output, while heat pump cycles transfer heat from low to high temperatures by using mechanical work as the input. Cycles composed entirely of quasistatic processes can operate as power or heat pump cycles by controlling the process direction. On a pressure–volume (PV) diagram or temperature–entropy diagram, the clockwise and counterclockwise directions indicate power and heat pump cycles, respectively.
Relationship to work
The net work equals the area inside because it is (a) the Riemann sum of work done on the substance due to expansion, minus (b) the work done to re-compress.
Because the net variation in state properties during a thermodynamic cycle is zero, it forms a closed loop on a PV diagram. A PV diagram's Y axis shows pressure (P) and X axis shows volume (V). The area enclosed by the loop is the work (W) done by the process:
${\text{(1)}}\qquad W=\oint P\ dV$
This work is equal to the balance of heat (Q) transferred into the system:
${\text{(2)}}\qquad W=Q=Q_{in}-Q_{out}$
Equation (2) makes a cyclic process similar to an isothermal process: even though the internal energy changes during the course of the cyclic process, when the cyclic process finishes the system's energy is the same as the energy it had when the process began.
If the cyclic process moves clockwise around the loop, then W will be positive, and it represents a heat engine. If it moves counterclockwise, then W will be negative, and it represents a heat pump.
Each point in the cycle
Description of each point in the thermodynamic cycles.
2→3: Isochoric cooling: Constant volume(v), Decrease in pressure (P), Decrease in entropy (S), Decrease in temperature (T)
3→4: Isentropic compression: Constant entropy (s), Increase in pressure (P), Decrease in volume (v), Increase in temperature (T)
4→1: Isochoric heating: Constant volume (v), Increase in pressure (P), Increase in entropy (S), Increase in temperature (T)
A list of thermodynamic processes
Adiabatic : No energy transfer as heat (Q) during that part of the cycle would amount to δQ=0. Energy transfer is considered as work done by the system only.
Isothermal : The process is at a constant temperature during that part of the cycle (T=constant, δT=0). Energy transfer is considered as heat removed from or work done by the system.
Isobaric : Pressure in that part of the cycle will remain constant. (P=constant, δP=0). Energy transfer is considered as heat removed from or work done by the system.
Isochoric : The process is constant volume (V=constant, δV=0). Energy transfer is considered as heat removed from or work done by the system.
Isentropic : The process is one of constant entropy (S=constant, δS=0). Energy transfer is considered as heat removed from the system only; no physical work is done by/to the system.
Isenthalpic: process that proceeds without any change in enthalpy or specific enthalpy
Polytropic: process that obeys the relation: $pv^{\,n}=C$
Reversible:process where entropy production is zero $dS-{\frac {dQ}{T}}=0$
Thermodynamic power cycles are the basis for the operation of heat engines, which supply most of the world's electric power and run the vast majority of motor vehicles. Power cycles can be organized into two categories: real cycles and ideal cycles. Cycles encountered in real world devices (real cycles) are difficult to analyze because of the presence of complicating effects (friction), and the absence of sufficient time for the establishment of equilibrium conditions. For the purpose of analysis and design, idealized models (ideal cycles) are created; these ideal models allow engineers to study the effects of major parameters that dominate the cycle without having to spend significant time working out intricate details present in the real cycle model.
The clockwise thermodynamic cycle indicated by the arrows shows that the cycle represents a heat engine. The cycle consists of four states (the point shown by crosses) and four thermodynamic processes (lines).
For example :--the pressure-volume mechanical work output from the ideal Stirling cycle (net work out), consisting of 4 thermodynamic processes, is^{[citation needed]}^{[dubious – discuss]}:
Thermodynamic heat pump cycles are the models for household heat pumps and refrigerators. There is no difference between the two except the purpose of the refrigerator is to cool a very small space while the household heat pump is intended to warm a house. Both work by moving heat from a cold space to a warm space. The most common refrigeration cycle is the vapor compression cycle, which models systems using refrigerants that change phase. The absorption refrigeration cycle is an alternative that absorbs the refrigerant in a liquid solution rather than evaporating it. Gas refrigeration cycles include the reversed Brayton cycle and the Hampson–Linde cycle. Multiple compression and expansion cycles allow gas refrigeration systems to liquify gases.
Modeling real systems
Example of a real system modelled by an idealized process: PV and TS diagrams of a Brayton cycle mapped to actual processes of a gas turbine engine
Thermodynamic cycles may be used to model real devices and systems, typically by making a series of assumptions.^{[2]} simplifying assumptions are often necessary to reduce the problem to a more manageable form.^{[2]} For example, as shown in the figure, devices such a gas turbine or jet engine can be modeled as a Brayton cycle. The actual device is made up of a series of stages, each of which is itself modeled as an idealized thermodynamic process. Although each stage which acts on the working fluid is a complex real device, they may be modelled as idealized processes which approximate their real behavior. If energy is added by means other than combustion, then a further assumption is that the exhaust gases would be passed from the exhaust to a heat exchanger that would sink the waste heat to the environment and the working gas would be reused at the inlet stage.
The difference between an idealized cycle and actual performance may be significant.^{[2]} For example, the following images illustrate the differences in work output predicted by an ideal Stirling cycle and the actual performance of a Stirling engine:
Ideal Stirling cycle
Actual performance
Actual and ideal overlaid, showing difference in work output
As the net work output for a cycle is represented by the interior of the cycle, there is a significant difference between the predicted work output of the ideal cycle and the actual work output shown by a real engine. It may also be observed that the real individual processes diverge from their idealized counterparts; e.g., isochoric expansion (process 1-2) occurs with some actual volume change.
Well-known thermodynamic cycles
In practice, simple idealized thermodynamic cycles are usually made out of four thermodynamic processes. Any thermodynamic processes may be used. However, when idealized cycles are modeled, often processes where one state variable is kept constant are used, such as an isothermal process (constant temperature), isobaric process (constant pressure), isochoric process (constant volume), isentropic process (constant entropy), or an isenthalpic process (constant enthalpy). Often adiabatic processes are also used, where no heat is exchanged.
Some example thermodynamic cycles and their constituent processes are as follows:
The Carnot cycle is a cycle composed of the totally reversible processes of isentropic compression and expansion and isothermal heat addition and rejection. The thermal efficiency of a Carnot cycle depends only on the absolute temperatures of the two reservoirs in which heat transfer takes place, and for a power cycle is:
$\eta =1-{\frac {T_{L}}{T_{H}}}$
where ${T_{L}}$ is the lowest cycle temperature and ${T_{H}}$ the highest. For Carnot power cycles the coefficient of performance for a heat pump is:
$\ COP=1+{\frac {T_{L}}{T_{H}-T_{L}}}$
and for a refrigerator the coefficient of performance is:
$\ COP={\frac {T_{L}}{T_{H}-T_{L}}}$
The second law of thermodynamics limits the efficiency and COP for all cyclic devices to levels at or below the Carnot efficiency. The Stirling cycle and Ericsson cycle are two other reversible cycles that use regeneration to obtain isothermal heat transfer.
A Stirling cycle is like an Otto cycle, except that the adiabats are replaced by isotherms. It is also the same as an Ericsson cycle with the isobaric processes substituted for constant volume processes.
TOP and BOTTOM of the loop: a pair of quasi-parallel isothermal processes
LEFT and RIGHT sides of the loop: a pair of parallel isochoric processes
Heat flows into the loop through the top isotherm and the left isochore, and some of this heat flows back out through the bottom isotherm and the right isochore, but most of the heat flow is through the pair of isotherms. This makes sense since all the work done by the cycle is done by the pair of isothermal processes, which are described by Q=W. This suggests that all the net heat comes in through the top isotherm. In fact, all of the heat which comes in through the left isochore comes out through the right isochore: since the top isotherm is all at the same warmer temperature $T_{H}$ and the bottom isotherm is all at the same cooler temperature $T_{C}$, and since change in energy for an isochore is proportional to change in temperature, then all of the heat coming in through the left isochore is cancelled out exactly by the heat going out the right isochore.
State functions and entropy
If Z is a state function then the balance of Z remains unchanged during a cyclic process:
$\oint dZ=0$.
Entropy is a state function and is defined as
$S={Q \over T}$
so that
$\Delta S={\Delta Q \over T}$,
then it appears that for any cyclic process,
$\oint dS=\oint {dQ \over T}=0$
meaning that the net entropy change of the working fluid over a cycle is zero.