# Peltier heat engine

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```					     University of Surrey, Department of Physics

2nd Year Solid State/Thermal Physics Laboratory

Heat Engines and Heat Pumps — 3 weeks

1. Introduction

A heat engine is the name given to a device which converts heat into work (or vice
versa) by a cyclic process. The device takes in heat from a so-called reservoir at a high
temperature, converts part of it into work — it will eventually end up as some other form
of energy — and outputs the rest of the heat into a second reservoir at a lower
temperature.
An important example of a heat engine is a power station. The fuel (be it coal, oil, gas
or nuclear) maintains a hot reservoir at some temperature. The turbines that generate
electricity are the ―engine‖ part and waste heat is rejected into a ―cold‖ reservoir
maintained at this temperature by some coolant. The coolant is kept at as low a
temperature as possible by circulating it in the massive cooling towers, which are the
most prominent feature of the power station.
A heat pump is simply a heat engine run in reverse. Instead of generating work as heat
passes from a hot to a cold reservoir, we use work to make heat flow from the cold to the
hot reservoir (i.e., against the thermal gradient). Heat pumps are perhaps even more
common than heat engines, the typical examples being refrigerators and air conditioners,
where heat is removed from the cold inside and ―dumped‖ into the already warm exterior.
You can find out more about heat engines in general and a special type of device called
the Carnot Engine, by reading Appendix 3 of these notes.
In this experiment you will determine the efficiency of a simple heat engine as a function
of its hot junction temperature, and compare this with the efficiency of a Carnot engine
working between the same two temperatures. The measured efficiency will then be adjusted
to compensate for heat energy losses in the apparatus.

2. Aims of the Practical

The practical consists of two main experiments: (i) finding the variation in efficiency of
the apparatus with temperature and (ii) making compensations for heat losses in the
apparatus. The particular skills you will start to acquire by performing these experiments
include:
 Familiarity with a Seebeck/Peltier thermoelectric converter;
 Use of thermistors for accurate steady state temperature measurement;
 Adjustment of a heat engine to obtain thermal equilibrium;
 Graphing skills, plotting and interpretation.

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3. Theory

3.1 Principles of a Heat Engine
The Pasco thermal efficiency apparatus consists of a simple heat engine in the form of a
semiconductor Seebeck/Peltier thermoelectric converter, in which electrical energy and
heat energy can be directly interconverted. (Appendix 1 explains in more detail the theory
of the Seebeck/Peltier effect.)

Q2                         Q1
Cold                                                    Hot
Engine
Reservoir                                               Reservoir
Psink                     Pin
T2                                            T1

W, Pout
Figure 1: Principle of a heat engine

Heat energy Q1 is extracted from a hot reservoir, and then part of this energy W is used to
perform work, while the remainder Q2 is exhausted to a cold reservoir. Throughout this lab
script, the sign convention is that energy flowing in the direction of the arrows is positive.
Since in the experiment, we will be dealing all the time with power transferred at
equilibrium, it will be more useful to use the alternative notation with P’s.
In Figure 1:
T1        is the temperature of the hot reservoir;
T2        is the temperature of the cold reservoir;
Pin       is the rate of extraction of heat energy from the hot reservoir, i.e., the
power input to the system. Note Pin = dQ1/dt.
Psink     is the rate of exhaustion of heat energy to the cold reservoir dQ2/dt.
Pout      is the rate of working, i.e., the useful power output of the system dW/dt.

Conservation of energy tells us that

Pin  Pout  Psink .                             [1]

In our case, because we are using a Seebeck-Peltier heat engine, the work done by the
engine is in the form of an electrical power output. We could make this electrical power
do a number of jobs (e.g., run a motor or light a bulb), but, paradoxically, the easiest way

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to measure it is to turn it back into heat! We pass the electrical current created by the heat
engine through a load resistor, R. The power output by the heat engine is then given by

Pout  Vout 2 / R .                                         [2]

(The reason for using  and not = will become clear later.) By knowing R and measuring
Vout, we can calculate the power generated.

3.2 Reservoirs
We have previously said that heat flows into the system from a ―reservoir‖ at
temperature T1 and is ejected to a cold reservoir at constant temperature T2. Implicit is the
assumption that taking heat out of or putting heat into a reservoir does not alter its
temperature. This can be achieved in two ways: (i) have an infinite heat reservoir (not
very practical!), or (ii) maintain the reservoir temperature using another heat source/sink.
In the Pasco heat engine, both the hot and cold reservoirs consist of small aluminium
blocks plus a system which maintains these blocks at a constant temperature.

Qu.      How can we keep the blocks at a constant temperature? You can see that there are
tubes for connecting a chilled water supply up to the cold reservoir. Why can we
not simply connect a hot water supply to the hot reservoir?

Ans.     The cold reservoir is cooled using water from a dewar containing a mixture of ice
and water. You should know that as long as it is properly stirred, a mixture of ice
and water at thermal equilibrium will be at 0C.† Any heat ―dumped‖ into the
reservoir by the heat engine will be absorbed as the latent heat required to melt a
part of the ice. No matter how much heat‡ is output by the engine, the reservoir
will remain at 0C until all the ice has melted.
Now consider what happens with the hot reservoir. If you had a dewar containing
hot water, there would be two main problems. Suppose there is a mass m of hot
water at temperature T1. If the heat engine takes out energy at a rate Pin, then

†
To be strictly accurate, we need to define a few conditions a bit better, for example, concerning the
pressure of the water vapour with which the ice-water mixture is in equilibrium, but this doesn’t make more
than a hundredth or so of a degree’s difference.
‡
Again, subject to a few caveats, such as requiring good mixing to take place in the cooling water and
requiring that the rate of heat output is slow enough so that there is time to transfer it to the ice before local
heating occurs at the aluminium block.

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dT1    P
  in ,                                     [3]
dt    mc

where c is the specific heat capacity of the water. The reservoir cools down. If m is
large (our ―infinite reservoir‖ scenario), then the temperature change will be small,
but for a small dewar of hot water, the ―thermal mass‖ is not large enough. The
second factor to take into consideration is that heat losses will occur in the pipes
between the dewar and the heat engine.
Instead, we adopt method (ii) above. If a power Pin is removed from the hot
reservoir, we put back in Pin. The easiest way to do this is to heat the aluminium
block with an electrical heater. The power input to the heater is simply VinIin.

3.3 Efficiency and the Carnot Engine
The efficiency,, of a heat engine is defined as the rate of work performed divided by
the rate of heat extraction from the hot source:

Pout
        .                               [4]
Pin

The Frenchman Sadi Carnot showed that the efficiency of a heat engine depends on the
temperatures of the hot and cold reservoirs. He proposed a type of heat engine (the Carnot
engine) that would have the maximum possible efficiency. He showed that this maximum
theoretical efficiency is related to the Kelvin temperatures of the reservoirs by

T1  T2     T
 Carnot             1 2 .                          [5]
T1       T1

(See Appendix 3 or the 2TP lecture notes for the derivation of this relationship.)
While there are other limitations on thermal efficiency, such as heat energy losses,
friction, etc., the Carnot efficiency provides an upper limit, and this efficiency can clearly
only approach unity if the hot reservoir is exceedingly hot or if the cold sink temperature
is close to absolute zero. Both these conditions are difficult to achieve. In a power station,
we make T1 as hot as possible by burning the fuel in a boiler, while T2 is as close to room
temperature as the cooling towers will let the water coolant get.

3.4 Compensation for Heat Losses
In the first part of the experiment, no allowance is made in determining the efficiency for
energy losses in the input and output powers of the heat engine. In the second part, the
values of Pin and of Pout will be adjusted to allow for heat energy losses and for wasted
work. When this is done the measured efficiency of the engine should be close to that of a
Carnot engine operating between the same two reservoirs.

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Input Power
In calculating the input power Pin to the engine no allowance was made for the energy
lost by conduction, convection and radiation. In practice, these losses mean that, in order
to maintain the temperature of the hot block at T1, we have to put in more energy than that
removed by the heat engine. Thus:

Vin I in  Pin  Ploss ,                       [6]

where Ploss is the extra power required to compensate for the losses. We can work out
what these losses are by running the engine with the load across the output disconnected
(―open‖ mode). If a complete circuit does not exist at the output terminal, then no current
flows and the engine does no work. This means that no energy is extracted from the hot
reservoir and Pin = 0. By adjusting Vin and Iin until the same constant hot junction
temperature is attained as in the ―engine‖ mode the true power input to the device may be
found. Mathematically,

Ploss  (Vin I in ) open mode
.         [7]
Pin, true  (Vin I in ) engine mode  Ploss

(We are assuming that the losses are the same whether the device acts in engine or open
mode.)
Output power
The total output power developed is not in fact given by

Vout 2
Pout  I out 2 R                               [8]
R
but by
2
V 
Pout, true  I out 2 ( R  r )   out  ( R  r ) ,        [9]
 R 

where Iout = Vout/R and r is the internal resistance of the thermoelectric converter. It is
thus necessary to measure this internal resistance, and since it is likely to be temperature-
dependent in the case of a semiconductor, this measurement must be made under
operating conditions.
If we assume that the emf E developed in the converter is constant, then we can write

E  I out ( R  r ) .                       [10]

To measure the internal resistance, we take measurements of Vout at two different values
R1 and R2 of the load resistance.

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I out1 ( R1  r )  I out 2 ( R2  r )
.          [11]
I R I R                 Vout1  Vout 2
 r  out1 1 out 2 2 
I out 2  I out1 Vout 2 / R2  Vout1 / R1

The compensated efficiency is simply the ratio of the compensated input power to the
compensated output power, i.e.

Pout, true                  (Vout / R) 2 ( R  r )
 comp                                                                      .   [12]
Pin, true        (Vin I in ) engine mode  (Vin I in ) open mode

3.5 The Heat Pump

Q2                           Q1
Cold                                                                Hot
Engine
Reservoir                                                           Reservoir

Pextract                     Pdump
T2                                                       T1

W, Pused

Figure 2: Principle of a heat pump

Figure 2 shows a schematic view of a heat pump. Thermodynamically, the situation is
very similar to Fig. 1: we still have our quantities Q1, Q2 and W, but now the flows of
energy are going in the opposite direction. To take account of the different roles of the
energy flow, we re-label the corresponding power terms as:
Pextract   the rate of extraction of heat energy from the cold reservoir, i.e., the
―refrigeration power‖ of the system. Note Pextract = dQ2/dt.
Pdump      is the rate at which unwanted heat is dumped to the hot reservoir,
dQ1/dt.
Pused      is the rate at which we use power to pump heat out of the cold reservoir,
dW/dt.
Similarly, we re-label the current going through the Seebeck-Peltier device and the
voltage across it as Vused and Iused to reflect the fact that we are now using this electrical
power to do the pumping, rather than getting power out of the engine. Equation [1] is
valid, as before, but now we interpret power as being positive when it flows in the
direction of the arrows in Fig. 2. Instead of defining an efficiency, as for a heat engine,
one defines a coefficient of performance 

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Pextract 1
              1.                                      [13]
Pused   
Like the efficiency, the coefficient of performance is a measure of what is achieved
divided by the energy required to achieve it. (You can check the relation of to by
replacing the P terms in Eqs. [13] and [4] with the corresponding Q’s and W’s. As with
the Carnot efficiency, we can derive a maximum value for the coefficient of performance:

1                 T2
 max               1            .                          [14]
 Carnot          T1  T2

Qu.      To find the coefficient of performance experimentally, we need to measure how
much energy is being removed from the cold reservoir. How can we do this?

Ans.     We make an indirect measurement. Conservation of energy tells us that if we
know Pdump and Pused, then we know Pextract, too. We can measure Pused simply by
the voltage across the Peltier device and the current which we pass through it. For
Pdump, we need to be a little more subtle. Recall that when the engine was being
run in open mode, the power being supplied by the hot junction at equilibrium was
equal to the power Ploss lost by conduction and radiation. So, if we can adjust the
heat pump in such a way that the hot junction is at the same temperature as we had
for the engine in open mode, then at thermal equilibrium, we know that the hot
reservoir is being supplied with a power Ploss, i.e., Pdump = Ploss.

Hence,

Pextract Pdump  Pused Ploss  Pused
                                                                   [15]
Pused       Pused         Pused
In the same way that one calculates a compensated heat engine efficiency, which takes
into account the internal resistance of the Seebeck-device, so it is possible to calculate a
compensated coefficient of performance. The power we use in running the machine is

Pused  Vused I used                                   [16]
However, because some of this power is dissipated in the device’s internal resistance, the
power the Peltier device actually ―sees‖ is

Pused, true  Vused I used  I used r
2
[17]

Pextract     Pdump  Pused, true Ploss  (Vused I used  I used r )
2
  comp                                                                         [18]
Pused, true       Pused, true         (Vused I used  I used r )
2

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4. Experimental Procedure

4.1 Experiment 1: Variation of the efficiency of the heat engine with temperature
1.      Connect the power supply and four multi-meters to the heat engine as shown in
Figure 3.
2.      Place cold water and ice into the dewar and stir the mixture to obtain a mixture at
0°C. Check that this temperature has been reached using a thermometer. Throughout
the experiment, keep checking to ensure that your ice has not all melted. If it has, you
will get the wrong results!
3.      Measure the initial temperature of the ―cold‖ aluminium block using the resistance
meter connected to the thermistor. Make sure the toggle switch is in the correct
position.
4.      Connect up the cooling tubes to the Pasco apparatus and the water bath. Start the
pump running. You may need to ―prime‖ the pump by initially filling the tubes with
water.
5.      Measure the temperature of the ―cold‖ block for several minutes. How quickly does it
come to equilibrium?



Water
Pump

Cooling water                                               Aluminium
in/out                      Thermistor                     block

A
V     Power
Supply

Seebeck/Peltier                           Heating
device                                element                  0-12 V

0.5   1      2

V

Figure 3: The Pasco thermal efficiency apparatus, set up to run in heat-engine mode

6.      Now turn the power supply on and adjust the voltage to about 11 V. (11 V is just a
suggested value, chosen to make the hot temperature nearly the maximum allowed.
You may use any value less than 12 V, but do not run the apparatus for more than

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5 minutes with the temperature of the hot side exceeding 80°C or you may damage
the apparatus.)
7.   Record the temperature of both the hot and cold blocks as the device settles to
equilibrium. Plot the variations in temperature graphically. Think about a possible
physical law which might govern how the heating/cooling processes occur. When
you are satisfied that the system is at steady state, measure the final temperatures, the
voltage across the heating resistor, the current through the heating resistor and the
7.   Repeat step 6 for four different values of the heater voltage suitably spaced between 0
and 12 V.
8.   Using the appropriate equations from the theory section, calculate the efficiency of
the heat engine for each different heater voltage. Each voltage should correspond to a
different hot junction temperature, T1.
9.   Calculate the efficiency of a Carnot engine at each temperature and plot a suitably
scaled graph of against 1/T1 to compare your two sets of results. Comment on the
relative efficiencies of the two engines and suggest some reasons for any differences.
Express the efficiency of the heat engine as a percentage of Carnot.

4.2 Experiment 2: Measurement of the compensated efficiency
1.   Repeat step 5 of Experiment 1 and leave the apparatus to reach equilibrium. Check
that the readings you obtained previously for Vin, Iin, Vout, T1 and T2 are consistent
with what you obtained previously.
2.   Now quickly remove the connection from the voltmeter to the left-hand side of the
2resistor (as shown in Fig. 2) and replace it in the socket to the left of the 1
resistor. Current is now circulating through a total load resistance of 3. Measure the
new value of Vout at once. Why does this measurement have to be done quickly?
3.   Repeat steps 1 and 2 two more times, on the first occasion re-connecting the lead to
the left of the 0.5 resistor and the second time, connecting it to the leftmost ―open-
circuit‖ socket. You should now have enough data to calculate r several times
independently. Do this and obtain a mean value of r.
4.   With the apparatus in ―open‖ mode (i.e., the last configuration in 3), gradually
decrease the heater voltage until the system comes to equilibrium at the same
temperature as in 1–3. Now read off the new values of Vin and Iin. Multiplying these
values together gives (Vin Iin)open mode, as in Eq. [7].
5.   Using Eq. [12], calculate Comp. How does this new value compare with Carnot?
6.   If you have time, repeat steps 1–5 for a different value of T1.
7.   Make a careful assessment of the errors in all your measurements.

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4.3 Experiment 3: The heat pump
1.   Disconnect the power supply from the heater and put it directly across the Peltier
device, with no load resistor, as shown if Fig. 4 below. Include an ammeter in the
circuit so you can measure the current which is being supplied.
2.   Increase the voltage slowly from zero until the temperature of the hot block is the
same as the value which you measured in Experiment 2 section 4. Record both Vused
and Iused.
3.   You should now be able to calculate the coefficient of performance.
4.   Using Eq. [17], calculate comp and compare it with max.



0.5         1       2

              +   A
Power
Supply

V

Figure 4: The Pasco apparatus configured as a heat pump

5. References
[1] Ohanian, Physics, 2nd Edition, Sec. 21.2
[2] Instruction manual and experiment guide for the PASCO scientific model TD-8564
thermal efficiency apparatus, PASCO scientific, Roseville, CA, USA, 1991
[3] Thermal Physics (2TP) lecture notes
[4] C. B. P. Finn, Thermal Physics

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