Today’s Lecture
Carnot Cycle
Carnot Refrigerator
Reversibility - Entropy
What is the net work and efficiency of a
Carnot cycle?
Carnot Cycle
Wtotal = WAB + WCD
A
B
D
C
Since ThVBγ−1=TcVCγ-1
and ThVAγ-1=TcVDγ-1
Qc Tc
=
Qh Th
We have VB / VA = VC / VD and
Efficiency
W Qh − Qc Th − Tc
=
=
e=
Qh
Qh
Th
Carnot Cycle
Efficiency
W Qh − Qc Th − Tc
e=
=
=
Qh
Qh
Th
Irreversible engines are necessarily less efficient,
but so are many reversible engines!
In a Carnot cycle all of the heat exchange takes place
only between the highest and lowest temperatures,
hence the Carnot cycle is the most efficient cycle.
Carnot Cycle
Efficiency
W Qh − Qc Th − Tc
=
=
e=
Qh
Qh
Th
• For highest efficiency we would want to run our engine
between a very hot and a very cold reservoirs.
• Large temperature difference, Th-Tc, and low temperature of
the cold reservoir, Tc, are very helpful.
• Efficiency can in principle reach 100% for Tc = 0, but we
normally do not have such reservoirs available…
Example: Carnot Cycle
During a Carnot cycle the isothermal
expansion takes place at 227oC while taking in
1500J from the hot reservoir. The isothermal
compression takes place at 27oC.
(a) What is the energy expelled to the
cold reservoir?
(b) What is the net work done by the
gas during each cycle?
(c) What is the efficiency of this cycle?
Adiabatic, Isothermal Cyclic Process Versus Carnot
We have already examined this cyclic
process ABCA. This time we will examine
it with some care in the reverse order,
ACBA. Assume that the initial conditions,
VA, PA, and TA are known as well as the
compressed volume VB=VC.
The temperature at B is found from the adiabatic expansion to A.
The work done and heat absorbed by the gas during the adiabatic expansion from
B to A are:
The work done and heat absorbed by the gas during the isothermal compression
from A to C are:
Adiabatic, Isothermal Cyclic Process Versus Carnot
We have already examined this cyclic
process ABCA. This time we will examine
it with some care in the reverse order,
ACBA. Assume that the initial conditions,
VA, PA, and TA are known as well as the
compressed volume VB=VC.
The work done and heat absorbed by the gas during the isochoric process from C
to B are:
The net work done and net heat absorbed by the gas during one cycle are:
They are equal. Why??
Adiabatic, Isothermal Cyclic Process Versus Carnot
Finally what is the efficiency of this
cycle versus a Carnot cycle?
The efficiency of
this cycle is:
The efficiency of a
Carnot cycle is:
For a volume ratio of 4:1 and
γ = 1.4 these efficiencies are:
We see that the Carnot cycle represents ~ 70% improvement!
Stirling Engine
A Stirling engine is based on a fixed amount of gas in a sealed engine
with a hot and cold reservoir. There are two cylinders one is heated by
an external heat source while the other is cooled (radiator etc.) The
Stirling engine comes as close to following the Carnot cycle as is
practically possible.
http://www.keveney.com/Vstirling.html
Since it can run on any source of heat, it holds promise for alternative
fuel engines (burning garbage), solar heating (no exhaust), geothermal,
etc.
Demonstration of a Stirling Engine.
Stirling Engine
In a Stirling cycle there are two isothermal
processes and two isochoric processes. (a) Find
the net work done by n moles of an inert gas during
this cycle. (b) Find the efficiency of this cycle. (c)
Compare this efficiency to that of a Carnot cycle.
(a) The only work is performed during the
isothermal processes.
(b) To find the efficiency we need to determine the heat input.
Stirling Engine
In a Stirling cycle there are two isothermal
processes and two isochoric processes. (a) Find
the net work done by n moles of an inert gas during
this cycle. (b) Find the efficiency of this cycle. (c)
Compare this efficiency to that of a Carnot cycle.
(b) The efficiency is Wnet / Qin:
(c) The Carnot efficiency is:
Carnot Refrigerator
Isothermal
contraction
adiabatic
expansion
adiabatic contraction
Isothermal
expansion
The key part is the isothermal process DC – the only time the system
is in contact with the cold reservoir. The gas is made to expand taking
heat Qc from the cold reservoir.
Isothermal
contraction
adiabatic
expansion
adiabatic contraction
Isothermal
expansion
Inverting the cycle - building a refrigerator
Mind the conventions: Qh is positive when obtained from the Hot Res.;
Qc is positive, when rejected to the Cold Res.; W is positive, when done
by the gas… Now we have: Qh< 0, Qc< 0, W< 0.
W = Qh − Qc
Qc = Qh − W
Qc Tc
=
Qh Th
Refrigerators
Coefficient of performance, COP
Qc
Qc
COP =
=
W Qh − Qc
Inverting the cycle building a refrigerator
For a Carnot cycle
Tc
COP =
Th − Tc
COP goes to 0, when Th goes to infinity;
COP is equal to 1, when Th = 2⋅Tc;
COP becomes infinitely large as Th approaches Tc!
Example: Carnot Refrigerator
A home freezer, working between
-18°C inside and 30°C outside has
Inverting the cycle building a refrigerator
How much electrical energy is required to cool 0.53 kg of water by 1°C?
How much heat will be rejected to the air in the kitchen?
For a Carnot cycle
Tc
COP =
Th − Tc
What is the ratio between the work done by the motor in a heat
pump and heat transferred to the house?
Qh Qc + W
=
= COP + 1
W
W
For 25°C inside the house and 10°C
found 2m under the ground COP = 19
Carnot’s Theorem
No engine operating between temperature baths with temperatures Th and
Tc can have an efficiency greater than the Carnot efficiency, e=(Th-Tc)/Th.
To prove this theorem consider a Carnot
refrigerator
We could design a Carnot refrigerator in
which each process runs in reverse using
the same temperature baths. If we then
had an superefficient engine operating
between the same two temperature baths
with a greater efficiency than the Carnot
efficiency, we could use it in conjunction with
a Carnot refrigerator to extract work from
the hot reservoir without transferring any
heat to the cool reservoir!
Carnot’s Theorem
No engine operating between temperature baths with temperatures Th and
Tc can have an efficiency greater than the Carnot efficiency, e=(Th-Tc)/Th.
The superefficient engine extracts energy Qh + Δ from the high temperature reservoir
and emits energy Qc to the low temperature reservoir. In conjunction a Carnot
refrigerator operates between the same two temperatures. The net effect is to extract
energy, Δ, from a high temperature reservoir and perform an equal amount of work. This
is in violation of the second law of thermodynamics!
Carnot’s Theorem
No engine operating between temperature baths with temperatures Th and
Tc can have an efficiency greater than the Carnot efficiency, e=(Th-Tc)/Th.
Consider the example of a superefficient (70%) Carnot engine
acting as a Carnot refrigerator (when e=(Th-Tc)/Th =.6) reversed
so it acts as a refrigerator and a 60% efficient Carnot engine.
This again violates the 2nd law of thermodynamics. We could
make the same argument for any Carnot engine operating
between two temperature baths. Remember:
Since all of the heat exchange takes place only
between the highest and lowest temperatures,
the Carnot cycle is the most efficient cycle,
Reversibility
Where do we find reversible processes?
In mechanics –
• elastic collisions;
• oscillations with no friction;
http://www.myphysicslab.com/pendulum1.html
• rotation of planets. . .
No mechanical energy is dissipated into heat-internal energy!
You can run the movie backwards and it will still be a plausible
process. For example a Carnot engine and a Carnot refrigerator.
Irreversibility
Where do we find irreversible processes?...
Pretty much everywhere, damn it!..
And we are not getting any younger either!..
You can’t possibly run
that movie back…
Losing, breaking, destroying,
saying stupid things….
Seriously.
Three common scenarios of irreversibility in thermodynamics.
1) Mixing and loosing structural order in general. Two molecularly
mixed fluids never “unmix”.
http://mutuslab.cs.uwindsor.ca/schurko/animations/irreversibility/happy.htm
A broken vase never repairs itself…
2) Conversion of mechanical energy into internal energy (dissipation
into heat).
Ordered motion of an object is converted into disordered motion of
its molecules. Never coming back…
http://mutuslab.cs.uwindsor.ca/schurko/animations/secondlaw/bounce.htm
3) Heat transfer from a hotter to a cooler object – never goes in the
opposite direction.
Irreversibly lost
opportunities...
#1 Expanding gas…
On the way from a to b the gas could
be harnessed to do some mechanical
work at expense of its internal energy…
W = −ΔU
Q=0
Instead of that we have
W =0
Maxwell’s
demon
ΔU = 0
#2 Two systems with different temperatures reaching
equilibrium…
There was an opportunity for a spontaneous
process – heat flow from Th to Tc.
It could be used to run a heat engine
between the two reservoirs (hot and cold).
Maxwell’s demon: high speed molecules
go to the right, low speed – to the left.
Maxwell distribution after thermal
equilibrium is established… Order is lost!
There is no way the molecules would spontaneously break
into two groups – with high and low temperatures.
Entropy
Entropy provides a quantitative measure of disorder.
Consider the isothermal expansion of an ideal gas. If we add heat dQ and let
the gas expand just enough to keep the temperature constant, then from the
first law:
The gas becomes more disordered because there is a larger volume and hence
more randomness in the position of the molecules.
We define the infinitesimal entropy change dS during an
infinitesimal process as:
For an isothermal process the change in entropy is ΔS = Q/T. Higher temperature
implies greater randomness of motion. If the substance is initially cold then adding heat
causes a substantial fractional increase in molecular motion (and randomness). But if
the substance is already hot then adding the same quantity heat adds relatively little
molecular motion that what was already present. Hence Q/T characterizes the increase
in randomness (or disorder) when heat flows into a system.
Entropy
Consider the Carnot
Cycle where we found
If we change the definition of Qc so that it is
the heat added vs heat rejected then
Any closed reversible cycle can be made
up of incrementally small isothermal and
adiabatic cycles. This leads to
This means that the integral
representing the change in entropy,
is path independent!
Entropy
If we take a system around a path that is not closed then its entropy does
change. Since the change in ΔS is path independent, entropy is a state
property, like temperature, and is independent of the path between state 1
and state 2.
Consider mixing 1kg of 0oC with 1kg of 100oC water, an irreversible
process. Assume two reversible processes, slowly heating the cold water
to 50oC while slowly cooling the hot water to 50oC.
This example is an irreversible processes. However entropy is a state
property which is independent of how you got to that state. Hence the
results were obtained by reaching the final states via reversible
processes.
Entropy
This definition of entropy is only meaningful for reversible processes. An
irreversible process takes a system out of equilibrium, this means that T
may not be well defined. But entropy is a state variable, it doesn’t care
how you got to a certain state only what the state is!
Consider adiabatic free expansion, an irreversible process, where the
gas is discharged into a vacuum chamber. Neither the temperature nor
internal energy of the gas changes. From the ideal gas law in an
isothermal expansion:
The energy that became unavailable to do work is:
Both of these examples were irreversible processes. Remember entropy is a
state property and as such is independent of how you get to that state. Hence
the results were obtained by reaching the final states via reversible processes.
Entropy
The examples on the previous slide described systems that evolved into systems of
higher disorder. In free expansion and coming to thermal equilibrium, the ability to do
work was lost as the systems evolved into states of higher disorder.
The mixing of colored ink starts from a state of
relative order to a state that is more disordered. A
state of higher entropy. Spontaneous unmixing, a
net decrease in entropy, is never observed!
In general, “When all systems taking part
are included, the change in entropy is
greater than or equal to zero”.
Entropy
In general, “When all systems taking part are
included, the change in entropy is greater than
or equal to zero”.
As an example consider 4 coins that lie on a table
in random orientations.
There is only 1 way that they have either 4 heads
or 4 tails. These are the orientations of highest
order. There are 4 ways that they have either 3
heads and 1 tail or 3 tails and 1 head. These are
the orientations of next highest order. Finally there
are 6 ways in which there are 2 heads and 2 tails.
This is the state of highest disorder!
For any system, the most probable state is the one
with the greatest number of corresponding
microscopic states, which is also the macroscopic
state with the greatest disorder and largest entropy!
Summary of the Second law of Thermodynamics
Cyclic processes:
Second Law of Thermodynamics:
It is impossible to construct a heat engine that extracts
heat from a reservoir and delivers an equal amount of heat.
Irreversible processes decrease the organization of a system. This
is due to statistical reasons.
Efficiency of a reversible
heat engine is
Entropy measures the relative disorder of a system and corresponds to
decreasing energy quality. The entropy difference between two states is
This integral may be done assuming reversible
processes as entropy is a state variable!