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ME-6950 Thermoelectric-I
Summer-II 2015
Final Project Report
Impact of Thomson Effect and Contact Resistance on a
Thermoelectric Cooler
Names:
Pooja Iyer Mani-835408571
Shripad Dhoopagunta-166754583
Rajeev Chippa-750223327
Faculty: Dr. HoSung Lee
Date of Submission: August 19th, 2015
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Table of Contents
Abstract ................................................................................................................................ 4
Nomenclature ....................................................................................................................... 4
Introduction .......................................................................................................................... 6
Problem Definition ................................................................................................................ 6
Results and discussion ........................................................................................................... 8
Comparison between ideal solutions obtained by Mathcad and ANSYS .................................................................. 8
Comparison between Thomson effect obtained by ANSYS and ideal solution obtained by Mathcad ................... 10
Comparison between Contact Resistance obtained by ANSYS and ideal solution obtained by Mathcad .............. 11
Comparison between Contact Resistance and Thomson Effect obtained by ANSYS and ideal solution obtained by
Mathcad .................................................................................................................................................................. 12
Comparison between Contact Resistance obtained by Mathcad and Contact Resistance obtained by ANSYS ..... 14
Role of Figure of Merit (Z) in the variation of Cooling Power (QC) with respect to current and temperature
difference. ............................................................................................................................................................... 15
Conclusion .......................................................................................................................... 16
Future Work ........................................................................................................................ 16
References .......................................................................................................................... 16
Appendix ............................................................................................................................. 17
1). ANSYS Thermoelectric Cooler (TEC) Tutorial With Thomson effect and Contact Resistance ............................ 17
Preparing the ANSYS Workbench ....................................................................................................... 17
Specifying the Materials and Properties ............................................................................................. 17
Creating the Geometry ....................................................................................................................... 20
Setting up the Model .......................................................................................................................... 24
2). Mathcad solution for contact resistance. .......................................................................................................... 31
3). Mathcad solution for ideal equations. ............................................................................................................... 33
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List of Figures
1: Figure 1 (a). Thermoelectric couple .......................................................................................................... 7
2: Figure 1 (b): Dimensions in a thermoelectric couple ................................................................................ 7
3: Figure 2(a). Mathcad Vs ANSYS ideal solutions comparison at 298K ....................................................... 9
4: Figure 2(b). Mathcad Vs ANSYS ideal solutions comparison at 420K ....................................................... 9
5: Figure 3(a). Thomson ANSYS solution Vs Mathcad ideal solutions at 298K ........................................... 10
6: Figure 3(b). Thomson ANSYS solution Vs Mathcad ideal solutions at 420K ........................................... 11
7: Figure 4(a). Contact Resistance solution Vs Mathcad ideal solutions at 298K ....................................... 11
8: Figure 4(b). Contact Resistance solution Vs Mathcad ideal solutions at 420K ....................................... 12
9: Figure 5(a). Contact Resistance and Thomson Effect solution Vs Mathcad ideal solutions at 298K ...... 13
10: Figure 5(b). Contact Resistance and Thomson Effect solution Vs Mathcad ideal solutions at 420K.... 13
11: Figure 6(a). Mathcad Contacts Vs ANSYS Contacts at 298K ................................................................. 14
12: Figure 6(b). Mathcad Contacts Vs ANSYS Contacts at 420K ................................................................. 14
13: Figure 7. ZT Vs T graph showing variation of material properties with temperature .......................... 15
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Abstract
According to literature, there have been experimental works on the Thomson Effect and Contact
Resistance individually to check the performance of a Thermoelectric Cooler. All results so far stated and
obtained have been mostly calculated experimentally. An attempt has been made in this project to
numerically obtain the impact of both the Thomson Effect as well as the Contact Resistance. A comparison
is then made of this impact with the ideal equation solutions. An individual comparison is made and a
simultaneous comparison is also made. According to the results obtained, the cooling power is reduced
when the Thomson Effect and the Contact Resistance is included. It can be determined that Contact
Resistance plays an important role when the leg length is small. The Thomson Effect has its importance
when the average of hot side and cold side temperature is away from the temperature value obtained
when the figure of merit is maximum.
Keywords: Thomson Effect, Contact Resistance, Thermoelectric Cooler, ideal equation, cooling power.
Nomenclature
le Length of element (mm)
lcop Length of copper alloy (mm)
Ae Area of element (mm2)
k Thermal conductivity of element (W/cmK)
kc Thermal conductivity of ceramic (W/cmK)
kcop Thermal conductivity of copper (W/mK)
T1 Cold junction temperature (K)
T2 Hot junction temperature (K)
T1c Temperature between copper alloy and element on the cold side (K)
T2c Temperature between copper alloy and element on the hot side (K)
Q1 Heat absorbed on the cold side (W)
Q2 Heat rejected on the hot side (W)
I Current (A)
Re Resistance of element (Ω)
Z Figure of merit (K-1)
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Greek symbols
α Seebeck coefficient (µV/K)
ρ Resistivity of element (Ω cm)
ρcop Resistivity of copper (Ω m)
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Introduction
The direct conversion of temperature difference into a potential difference is called the thermoelectric
effect. For a long time now, there has been a lot of hesitation in the work of thermoelectricity mainly
because of lack of proper materials and because of very low performance shown by the materials. When
there are dissimilar conductors or semi-conductors and when a temperature difference exists between
them, a voltage difference is also generated. This is called the Seebeck Effect and it was introduced in the
year 1821 by Thomas. J. Seebeck. Another discovery was made by Jean Peltier in the year 1834 that was
the opposite of the Seebeck Effect. It said that when a potential difference was applied across dissimilar
conductors or semi-conductors, a temperature difference was developed. This is called the Peltier Effect.
Again, in the year 1854, Sir William Thomas discovered that when a temperature difference is applied
across conductors that carry current, depending on the direction of the current, heat is either absorbed
or liberated. This is called the Thomson Effect. These three effects put together are called the
thermoelectric effects. These effects play a major role in the performance of a thermoelectric system.
Apart from these effects, there is another effect that is of importance and it is called the Contact
Resistance. When dissimilar current carrying conductors are in contact with each other, the resistance
offered by the contact material is called Contact Resistance. Due to this unexpected resistance, there is
an error in reading the temperature difference and the cooling power. Hence in order to calculate the
error, this effect must be considered along with the thermoelectric effects.
Work was already done on the performance of a thermoelectric cooler using effective material properties
[1] and impact of Thomson Effect on a thermoelectric cooler [2]. Conclusions were drawn that a positive
Thomson Effect improved the performance of a thermoelectric cooler while a negative Thomson Effect
reduced it.
Work was also done on the effect of contact resistance of ceramic plates on the overall performance of a
thermoelectric module [3]. It was concluded that the effect of contact resistance varied as a function of
leg length. It was also concluded that the effect of contact resistance reduced as there was an increase in
the leg length of the element.
In this report, we mainly work on a comparison between the results obtained under the ideal conditions
and the results obtained by including the Thomson Effect and the Contact Resistance.
Problem Definition
While calculating the overall performance of a thermoelectric device, we mostly tend to consider only the
most ideal conditions. When we say ideal conditions, it means that there are no other effects involved.
We have an equation called the ideal equation that helps us calculate the overall performance of a
thermoelectric module without having any external effects considered.
(𝑄1) = 𝛼 ∗ (𝑇𝑎𝑣𝑔) ∗ 𝑇1 ∗ 𝐼 − (1
2) 𝐼2𝑅 −
𝐴∗𝑘
𝐿∗ (𝑇2 − 𝑇1)
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This equation is formed with an assumption that the Thomson Effect and all the other external factors are
negligible. This equation is in agreement with the performance obtained by the commercial modules.
Hence, the name ideal is appropriate for this equation. When there is no Thomson Effect it means that
the Seebeck Coefficient α is independent of Temperature.
When we consider a p-type and n-type thermocouple, the thermocouple is sandwiched between a
ceramic made of Aluminum Nitride (AlN) and Copper Alloy as shown in the figure. The dimensions of each
of these is mentioned in figure 1(b).
1: Figure 1 (a). Thermoelectric couple
2: Figure 1 (b): Dimensions in a thermoelectric couple
Copper
alloy
Ceramic
plates
p-leg n-leg
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The main focus of this project is to calculate the output cooling power while including all the above
mentioned effects and compare each solution with ideal equation values obtained numerically as well as
analytically.
Results and discussion
Comparison between ideal solutions obtained by Mathcad and ANSYS
Using the ideal equations, we obtained results for the cooling power at current (I) 1A, 2A and 3A
respectively with the hot side temperature fixed at 298K as well as at 420K. The equations that we used
for the ideal solution were:
A comparison was made between the cooling power obtained at different values of current and at
different temperature difference values. The outcome was plotted in a Qc Vs ΔT graph.
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3: Figure 2(a). Mathcad Vs ANSYS ideal solutions comparison at 298K
The analytical solution and the ANSYS solution exactly matched with one another and there is accurate
agreement of both the solutions at every value at 298K.
4: Figure 2(b). Mathcad Vs ANSYS ideal solutions comparison at 420K
The analytical solution and the ANSYS solution exactly matched with one another and there is accurate
agreement of both the solutions at every value at 420K.
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Co
olin
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ow
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W)
Temperature DefferenceΔT
Mathcad ideal Vs Ansys ideal at 298 K
Mathcad ideal I=1
Mathcad ideal I=2
Mathcad ideal I=3
Ansys ideal I=1
Ansys ideal I=2
Ansys ideal I=3
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Temperature difference ΔT
Mathcad ideal Vs Ansys ideal at 420K
Mathcad ideal I=1
Mathcad ideal I=2
Mathcad ideal I=3
Ansys ideal I=1
Ansys ideal I=2
Ansys ideal I=3
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Comparison between Thomson effect obtained by ANSYS and ideal solution obtained by
Mathcad
The Thomson Effect was added to the ideal thermocouple in ANSYS. In order to do this, we changed the
general material properties in the ANSYS Workbench file. The Seebeck coefficient (α) was varied with
respect to temperature while keeping the electrical resistivity (ρ) and the thermal conductivity (k)
constant at 300K [2]. The results thus obtained were plotted in comparison with each other in Qc Vs ΔT
graph.
5: Figure 3(a). Thomson ANSYS solution Vs Mathcad ideal solutions at 298K
At 298K, at low temperature difference, the solutions are almost the same. But as the temperature
difference increases to 50K, the difference between the values is increasing. At the highest current and at
the highest temperature difference, the change in value is 0.8 W for 127 couples. When the exact values
are compared to each other, the values obtained when Thomson Effect is included are lesser when
compared to values obtained at ideal conditions.
At 420K the deviation of the Thomson Effect from the ideal solution is higher when compared to the
deviation at 298K. The discrepancy reduces with increase in temperature difference. At a constant
temperature difference, there is a higher change in the values of cooling power with increase in current.
At every stage, the value obtained due to the Thomson Effect is less when compared to the Mathcad Ideal
solution.
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Co
olin
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ow
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W)
Temperature difference ΔT
Thomson Vs ideal equation at 298K
Ideal I=1 Ideal I=2 Ideal I=3 Thomson I=1 Thomson I=2 Thomson I=3
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6: Figure 3(b). Thomson ANSYS solution Vs Mathcad ideal solutions at 420K
Comparison between Contact Resistance obtained by ANSYS and ideal solution obtained by
Mathcad
In order to include the contact resistance, the surfaces of contact are selected and a contact resistance
value of 10-6 Ωcm2 is given. The results then obtained are plotted in a Qc Vs ΔT graph against the values
obtained by ideal equations obtained by Mathcad.
7: Figure 4(a). Contact Resistance solution Vs Mathcad ideal solutions at 298K
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olin
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ow
er (
W)
Temperature difference ΔT
Thomson Vs ideal at 420 K
Ideal I=1 Ideal I=2 Ideal I=3
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Temperature differenceΔT
Ideal Vs Contatcs at 298K
Ideal I=1 Ideal I=2 Ideal I=3 Contacts I=1 Contacts I=2 Contacts I=3
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At this temperature, there is very little deviation of the values with respect to the current and the
temperature difference. The deviation is increasing with increase in the current and is almost the same
with increase in the temperature difference at constant current. The values obtained when the contact
resistance is included is less when compared to the ideal solution.
8: Figure 4(b). Contact Resistance solution Vs Mathcad ideal solutions at 420K
As the temperature of the hot side is increased, we can see at that the agreement between the values
reduces with an increase in the current. At the same current, the difference is constant as the temperature
difference increases. The cooling effect is reduced when contact resistance is included at a high hot side
temperature.
Comparison between Contact Resistance and Thomson Effect obtained by ANSYS and ideal
solution obtained by Mathcad
Now both the Thomson Effect and the Contact Resistance are both simultaneously included and results
are obtained at varying current and varying temperature difference. These values are then plotted in a Qc
Vs ΔT graph.
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olin
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Temperature difference ΔT
Contact resistance Vs ideal equation at 420 K
Contacts I=1
Contacts I=2
Contacts I=3
Ideal Equations I=1
Ideal Equations I=2
Ideal Equations I=3
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9: Figure 5(a). Contact Resistance and Thomson Effect solution Vs Mathcad ideal solutions at 298K
At 298K, the values are almost equal at lower current but as the current increases the deviation in the
values increases with the increase in the temperature difference. The values obtained by the ideal
equations again is higher when compared to the values obtained when both these effects are included.
10: Figure 5(b). Contact Resistance and Thomson Effect solution Vs Mathcad ideal solutions at 420K
Now when the hot side temperature is at 420K, the deviation in the values increases. The deviation is low
at a fixed temperature difference and increases with increase in the current. The discrepancy is high at a
low temperature difference and reduces as the temperature difference increases at constant current.
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Thomson and contacts Vs Ideal at 298K
Ideal I=1 Ideal I=2 Ideal I=3
Thomson and Contacts I=1 Thomson and Contacts I=2 Thomson and Contacts I=3
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Thomson and Contact Vs Ideal Equations at 420K
Thomson and Contacts I=1
Thomson and Contacts I=2
Thomson and Contacts I=3
Ideal Equations I=1
Ideal Equations I=2
Ideal Equations I=3
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Comparison between Contact Resistance obtained by Mathcad and Contact Resistance
obtained by ANSYS
An analytical solution was obtained to calculate the effect of Contact Resistance on cooling power and a
numerical solution was also obtained using ANSYS at 298K and 420K. The results were then plotted in a
graph that was plotted between Qc and temperature difference ΔT.
11: Figure 6(a). Mathcad Contacts Vs ANSYS Contacts at 298K
The values obtained analytically and numerically are in good agreement at any current and any
temperature difference. So either Mathcad or ANSYS can be used for the comparisons.
12: Figure 6(b). Mathcad Contacts Vs ANSYS Contacts at 420K
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Temperature difference ΔT
Mathcad Contacts Vs Ansys Contacts 298K
Mathcad contacts I=1
Mathcad contacts I=2
Mathcad contacts I=3
Ansys Contacts I=1
Ansys Contacts I=2
Ansys Contacts I=3
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Temperature Difference ΔT
Mathcad Contacts Vs Ansys Contacts 420K
Mathcad Contacts I=1
Mathcad Contacts I=2
Mathcad Contacts I=3
Ansys Contacts I=1
Ansys Contacts I=2
Ansys Contacts I=3
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At a high hot side temperature, the analytical and the numeric solution are more in agreement with each
other than at lower temperature difference. The difference in the cooling power obtained at different
values of current at different temperature differences is very less the maximum difference being 0.74W
for one complete module containing 127 thermocouples. This value can almost be neglected and can be
said that the solutions are in good agreement with each other.
Role of Figure of Merit (Z) in the variation of Cooling Power (QC) with respect to current and
temperature difference. We have the material properties of a commercial module at different temperatures. We consider the
properties at temperature 273K, we vary the temperature for every 25K. Their respective material
properties at that respective temperatures are given. When a graph is plotted between the product of
Figure of Merit (Z) and its Temperature (T) and Temperature (T), Z*T value is highest at a temperature of
300K but, when we take the constant material properties, the values considered are at 300K (Room
Temperature).When we calculate the cooling power by fixing the hot side temperature while varying the
cold side temperature, any value of Qc obtained is always less than the value at 300K. This is because the
average of the hot side temperature and the cold side temperature when multiplied with its respective
value of Z, is always less than the maximum value of Z*T. This is depicted in figure 7. Therefore, when the
Thomson effect is included, though we vary the Seebeck effect and keep the resistivity and the
conductivity constant at 300K, the values obtained are less when compared to when all three of them are
kept constant at 300K i.e. under ideal conditions. If the Z*T value considered at ideal equations is NOT
maximum, then, the result obtained when the Thomson Effect is included is higher than the solutions
obtained at ideal conditions.
13: Figure 7. ZT Vs T graph showing variation of material properties with temperature
0.E+00
1.E-01
2.E-01
3.E-01
4.E-01
5.E-01
6.E-01
7.E-01
8.E-01
9.E-01
0 100 200 300 400 500
Figu
re o
f M
erit
(ZT
)
Temperature (K)
ZT Vs T
ZT Vs T
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Conclusion
The ideal predictions (Mathcad) is in perfect agreement with ANSYS ideal solutions.
When the Thomson Effect is included, there is a lot of deviation when the average
temperature is not the temperature of maximum Z*T, i.e. the results obtained by ideal
equations would merge into the result obtained when Thomson Effect is included when
the Z*T value is maximum (300K)
When Contact Resistance is added, there is always a decrease in the resulting cooling
power (Qc). This effect is more prominent at small leg lengths. As the leg length increases,
this effect is almost negligible.
When the leg length reduces, the electrical resistance of the leg is of a very low order.
Hence, the effect of contact resistance increases because it is almost equal to the
resistance of the leg. Therefore, in order to increase the performance of the
thermocouple, the contact material must be more suitable. The resistivity of this material
plays a more important role than the material of the leg.
When both the Contact Resistance and the Thomson Effect are included, the cooling
power obtained is less than the ideal solution. This is not just the addition of both these
effects. It is a complicated differential equation that is tedious to solve analytically.
At times, when the Thomson Effect is positive, i.e. when its solution is higher than the
ideal solution, and when the effect of Contact Resistance is lower than the ideal solution,
the overall effect would almost be equal to the ideal solution.
Future Work
The tedious math work can be taken to a new platform so that an analytical solution is more easily
available when both the Thomson Effect and Contact Resistance are involved. This would help us
have a comparison between the numerical solution and the analytical solution.
If the contact material is improvised to improve its resistivity, this work can be redone to get a
better performance at lesser leg length.
References
1. Lee, HoSung, Attar, Alaa, Weera, Sean, Performance Prediction of Commercial Thermoelectric Cooler Modules using the Effective Material Properties, Journal of Electronic Materials, Vol.44, No.6, 2157-2165 (2015).
2. Lee, HoSung, The Thomson Effect and the Ideal Equation on Thermoelectric Coolers, Energy, 56, 61-69 (2013).
3. V. Semenyuk, Thermoelectric Micro Modules for Spot Cooling of High Density Heat Sources, 20th International Conference on Thermoelectrics (2001)
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4. ANSYS Thermoelectric Generator (TEG) Tutorial, http://homepages.wmich.edu/~leehs/ME695/ME695%20SUM%20II.html
Appendix
1). ANSYS Thermoelectric Cooler (TEC) Tutorial With Thomson effect and Contact Resistance
Preparing the ANSYS Workbench
1) Go Start Menu All Programs Simulation ANSYS 12.1 Workbench
2) In the toolbox menu on the left portion of the window, double click Thermal-Electric. A
project will now appear in the project schematic window of Workbench.
3) Right-click Thermal-Electric at the top of the Project Schematic pane and select Rename as
TEG Ex 3.1 Tutorial.
4) Save the project as Thermoelectric-Generator-Workbench.
Specifying the Materials and Properties
1) Double-click on Engineering Data to open the material data. You will see Structural Steel as the
default material in the Outline of Schematic A2: We are going to enter three materials (Copper
Alloy, p-type semiconductor, and n-type semiconductor) in the Engineering Data.
2) In the Data Source of Outline Filter, click on General Materials. In the Outline of General
Materials pane, right-click on Copper Alloy and select Add to Engineering Data or click on th
icon. A symbol will appear once the material has been added.
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3) In the Outline Filter, click on Engineering Data, you will see that Copper Alloy in the Outline
Schematic A2 is newly added material to the Structural Steel.
4) Now we want to add two more materials (p-type and n-type semiconductors). Double-click on
Structural Steel and rename it as p-type. In the Toolbox pane, double-click Isotropic Thermal
Conductivity to include this property to the p-type.
5) In the Properties of Outline Row 5: p-type: The following values are entered as:
Isotropic Thermal Conductivity: 1.51 W m-1 K-1
Isotropic Resistivity: 0.00101 Ω cm
Isotropic Seebeck Coefficient: 202.17e-6 V K-1
For Thomson Effect Seebeck Coefficient varies with respect to Temperature. All these values can
be given in the tabular form in the Table of Properties Window. Continue the same with n-type.
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6) Create n-type by duplicating the p-type. Right-click on p-type in the Outline of Schematic A2
and select Duplicate. A duplicate of the p-type material will appear below named p-type 2.
Rename this material to n-type. The value of the Isotropic Seebeck Coefficient is now changed
to the negative as–202.17e-6.
In the Properties of Outline Row 5: n-type: make sure the final values to be as:
Isotropic Thermal Conductivity: 1.51 W m-1 K-1
Isotropic Resistivity: 0.00101 Ω cm
Isotropic Seebeck Coefficient: –202.17e-6 V K-1
Input all the temperature dependent Seebeck Values for n-type. This is explained in the 5th
point.
7) Similarly add the third material Ceramic.
Isotropic Thermal Conductivity: 180 W m-1 K-1
Isotropic Resistivity: 10e14 Ω cm
8) Click on the icon in the menu bar to return to the Project.
9) Save the project.
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Creating the Geometry
We are creating the following model, all the dimensions are mentioned here.
Dimensions are to the left hand side of the fig. The model is extruded to 0.41mm in the perpendicular
direction.
Example for creating the model. (The dimensions Mentioned from this section are just for an
Example)
1) In the project A under the Project Schematic, double-click on Geometry to launch the Design
Modeler.
2) Select Millimeter as the desired length unit and click OK.
3) In the Tree Outline pane, right-click on the XY Plane and select Look at. Add a new sketch by
clicking on the icon in the menu bar.
4) Sketch1 will appear below the XY Plane. Click on the Sketch1 and select the Sketching tab at
the bottom of the Tree Outline pane.
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5) Click on the Settings tab and Grid and check the boxes of both Show in 2D and Snap options as
6) Click on the Draw tab and select Rectangle by 3 points.
7) Once the Rectangle by 3 points is selected, click on the origin (indicated by )of the
Graphics pane to add the first point of the rectangle. Place the second point of the rectangle at
(X=36,Y=0). Place the third and final point of the rectangle at (36, 5). This is the sketch for the
base of the p-type element.
Note: The coordinate position is displayed in millimeter at the bottom right of the Graphics. 8) With the Rectangle by 3 points still selected place another first point at (24, 5). Place the next
point at (36, 5). Place the final point at (36, 15). This is the sketch for the p-element.
9) Place another first point at (24,15), the second point at (24, 20) and the final point at (72, 20).
This is the sketch for the top plate.
10) Place another first point at (72, 15), the second point at (60, 15) and the final point at (60, 5).
This is the sketch for the n-element.
11) Place another first point at (60, 5), the second point at (96, 5) and the final point at (96, 0).
This is the sketch for the base of the n-element. Upon completion all rectangles will form the
shape below.
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12) In the Tree Outline, click on the Modeling tab and select Sketch1 and extrude it by clicking the
icon in the menu bar or by going to Create Extrude on the menu bar.
13) Extrude1 will appear in the Tree Outline. Click on it. In the Details View pane, change the
Operation from Add Material to Add Frozen. Change the Depth to 10 mm.
Note: The ‘Add Frozen’ option keeps the various elements from merging. The ‘Add Material’ option will perform merging to one element.
14) To see an isometric view, click on the ball of the global coordiante axes.
15) Click on the icon in the menu bar to generate the extrusion or right-click on Extrude1 and
click Generate.
Note: The symbol will change to a symbol and, indicating that the command is successful.
16) In the Tree Outline, there will now be 5 Parts and 5 Bodies as a result of the extrusion.
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17) Under the 7 Parts, 7 Bodies in the Tree Outline, select each body and rename it according to
the figure shown below. Rename each part by right-clicking on it in the Tree Outline and
selecting Rename. The currently selected part will become highlighted in the Graphics pane.
18) Ensure that each part is a Solid (by default) in the Details View pane under the Fluid/Solid
option. Select 7 bodies in the Tree Outline by clicking on them while holding down the Ctrl key.
Right-click and select Form a new part. Right-click and rename the part to TEC.
Finally the following bodies are created and have to be made into a single part.
19) Close the Design Modeler. Save the project from the Workbench.
p - base
top
n - base
p - leg
n - leg
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Setting up the Model
Input all the parameters mentioned in the fig. You can see all the inputs in the graphics window in the
fig.
And finally Calculate the heat reaction at the cold junction.
Creating contacts
In the model tab, expand connections, contacts. You will find all the contacts in the system.
1. Click on any one of the contacts and look for the details of the contact.
2. Search for the contacts which connect the p-type and the copper and also the n-type and
the copper.
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3. You will find 4 such contacts. Select them individually and scroll down the details of the
contacts window.
4. Under Advanced tab set the Electrical conductance to manual and enter the value 1e+10
S/m^2.
Example for setting up the model.
This just an example the output of the program would be similar with different input and output
parameters.
In the Workbench, double-click on Model to launch the solver. This may take several seconds up to a
minute.
1) In the Outline pane, expand Geometry TEG.
Specify the material for each body by clicking on it and changing the Assignment under the
Material section in the Details of “ ” pane.
Click on n-base and change the Assignment in the Details of “n-base” to Copper Alloy.
Click on n-leg and change the Assignment in the Details of “n-leg” to n-type.
Click on top and change the Assignment in the Details of “top” to Copper Alloy.
Click on p-leg and change the Assignment in the Details of “p-leg” to p-type.
Click on p-base and change the Assignment in the Details of “p-base” to Copper Alloy.
Click on Cer and change the Assignment in the Details of “Cer” to Ceramic.
2) In the Outline pane right-click on Mesh and select Update. This may take several seconds.
26
3) In the Outline pane, right-click on Stead-State Thermal-Electric Conduction (A5), select Insert and click on Temperature.
Click on the newly added Temperature under the Steady-State Thermal-Electric
Conduction (A5) to ensure that it is selected. Right-click anywhere in the graphics pane on
the right, select Cursor Mode and click Face. This cursor mode allows you to select
surfaces.
27
Click on the top surface of the top body to select it. Highlighted surfaces are indicated in
green. In the Details of “Temperature” pane, click Apply in the Geometry option. The
applied surface will change the color which confirms the application.
Change the Magnitude in the Details pane to 452 °C (ramped).
Right-click on Temperature in the Outline pane and rename it to Hot Junction. This is the hot
junction boundary condition.
28
4) In the Outline pane, right-click on Stead-State Thermal-Electric Conduction (A5), select Insert and click on Temperature again.
Note: Right-click anywhere in the Graphics pane and select Cursor Mode and click Rotate. Switch the cursor mode back to Face, click on the bottom surface of n-base and p-base while holding down Ctrl.
In the Details of “Temperature” pane, click Apply in the Geometry option. The applied
surface will change the color. Change the Magnitude in the Details pane to 22 °C
(ramped). Right-click on Temperature in the Outline pane and rename it to Cold
Junction. This is the cold junction boundary condition.
5) In the Outline pane, right-click on Stead-State Thermal-Electric Conduction (A5), select
Insert and click on Voltage .In the Graphics pane, change the view back to isometric by
clicking the small ball in the global coordinate axes. Select the outer face of the base of
the n-leg (parallel to the y-z plane).
In the Details of “Voltage” pane, click Apply in the Geometry option. The applied surface
will change the color. Ensure the Magnitude in the Details pane is 0 V (ramped). Right-
click on Voltage in the Outline pane and rename it to Low Potential. This is the lower
electric potential boundary condition. If current value has to be given, the low Voltage
(0V) is the same but the instead of high potential input current with desired value.
6) In the Outline pane, right-click on Stead-State Thermal-Electric Conduction (A5), select
Insert and click on Voltage again.
29
Note: Right-click anywhere in the Graphics pane and select Cursor Mode and click Rotate
and rotate the thermocouple until the other-side outer face of the base of the p-leg
(parallel to the y-z plane) can be seen. And select the outer face.
In the Details of “Voltage” pane, click Apply in the Geometry option. The applied surface
will change the color. Change the Magnitude in the Details pane is 0.08 V (ramped). Right-
click on Voltage in the Outline pane and rename it to High Potential. This is the higher
electric potential boundary condition.
7) In the Outline pane, right-click on Steady-State Thermal-Electric Conduction (A5), select
Insert and click on Convection. In the Graphics pane, select all faces excluding those
which have been assigned boundary conditions previously. Select multiple faces by
holding down the Ctrl key.
In the Details of “Convection” pane, click Apply in the Geometry option. The applied surface
will change the color. The Geometry option should read 21 Faces upon application. Change
the Magnitude in the Details pane is 1e-6 W/m2·°C (ramped) for negligible convection. Right-
click on Convection in the Outline pane and rename it to Insulation.
8) In the Outline pane, click on Steady-State Thermal-Electric Conduction (A5) to view the
boundary conditions selected as shown below.
30
9) In the Outline pane, right-click on Solution (A6), select Insert Thermal Temperature.
Make sure that All Bodies is default under Geometry in the Details of “Temperature”
pane. And leaves others as they are.
10) In the Outline pane, right-click on Solution (A6), select Electric Total Current Density.
Make sure that All Bodies is default under Geometry in the Details of “Total Current
Density” pane. And leaves others as they are. These render the desired results for display.
11) In the Outline pane, right-click on Solution (A6), select Probe Current Reaction. In the
Details of “Current reaction” pane, select Low Potential in the Boundary Condition
option. Right-click on Reaction Probe in the Outline pane and rename it to Current.
12) Repeat the insertion process for Probe Heat Reaction. In the Details of “Heat
Reaction” pane, select Hot Junction in the Boundary Condition option. Right-click on
Reaction Probe in the Outline pane and rename it to Hot Junction Heat Absorbed.
13) In the Outline pane, right-click Solution (A6) and click Solve or click on the
icon. The solver will take up to several minutes to complete.
14) Once the solver has completed its tasks, click on any of the solutions (Temperature, Total
Current Density, etc.) to display the results. For the Total Current Density, you can display
vectors (to indicate the direction of current flow) instead of contours by clicking on the
icon below the menu bar. Zoom in closer by scrolling up on the mouse wheel to
observe the direction of current. The results of the probes (Current and Hot Junction
Heat Absorbed), when selected, are displayed in Tabular Data pane at the bottom right
31
of the window. The computed current and heat absorbed will be 28.942 A and 21.682 W,
which are in good agreement with those results in Example 3.1 in the textbook.
15) Close the Mechanical Solver and Save the project.
2). Mathcad solution for contact resistance.
Given Dimensions
Given Properties
Initial Guess
le 0.2mm lc 0.45mm lcop 0.2mm Ae 0.41mm0.41 mm
2 202.17V
K 2 1.01 10
3 cm k 2 1.51 10
2
W
cmK c 10
6 cm
2
kcop 401W
m K
kc 1.8W
cmK Re
le
Ae
4 c
Ae
Z
2
k
T1 410K T2 420K
T max T21
Z
T21
Z
2
T22
120.333K Imax
Re
T21
Z
2
T22
1
Z
4.587A
Q1 29.38W Q2 3W T1c 290K T2c 300K
I 1A
32
Given
Q1
Ae kc
lc
Ae kcop
lcop
T1 T1c
Q1 I T1cI2
2
le
Ae
4 c
Ae
Ae k
le
T2c T1c
Q2 I T2cI2
2
le
Ae
4 c
Ae
Ae k
le
T2c T1c
Q2
Ae kc
lc
Ae kcop
lcop
T2c T2
T1c I T1 le T2c I T1 le Q1 I T1 le Q2 I T1 le
Find T1c T2c Q1 Q2
T1c I T1 le 409.69K
T1c I T1 le 273K 136.690071K
T2c I T1 le 420.386K
T2c I T1 le 273K 147.385959K
Q1 I T1 le 0.1252982W
Q1 1A 410K le 0.125WQ2 I T1 le 0.1560358W
T1 410K 400K 370K
I 1A 2A 4A
Q1 1A T1 le 0.125
0.096
0.067
0.038
-39.068·10
W
Q1 2A T1 le 0.249
0.216
0.183
0.15
0.117
W
Q1 3A T1 le 0.346
0.309
0.272
0.235
0.198
W
Q1 4A T1 le 0.417
0.376
0.335
0.294
0.253
W
33
3). Mathcad solution for ideal equations.
Given Dimensions
Given Properties
Initial Guess
le 0.2mm lc 0.45mm lcop 0.2mm
2 202.17V
K 2 1.01 10
3 cm k 2 1.51 10
2
W
cmK
kcop 401W
m K
kc 1.8W
cmK Re
le
Ae
4 c
Ae
T1 410K T2 420K
T max T21
Z
T21
Z
2
T22
120.333K
Q1 29.38W Q2 3W T1c 290K T2c 300K
I 1A
Given
Q1
Ae kc
lc
Ae kcop
lcop
T1 T1c
Q1 I T1cI2
2
le
Ae
Ae k
le
T2c T1c
Q2 I T2cI2
2
le
Ae
Ae k
le
T2c T1c
Q2
Ae kc
lc
Ae kcop
lcop
T2c T2
T1c I T1 le T2c I T1 le Q1 I T1 le Q2 I T1 le
Find T1c T2c Q1 Q2
34
T1c I T1 le 409.687K
T1c I T1 le 273K 136.687131K
T2c I T1 le 420.383K
T2c I T1 le 273K 147.383013K
Q1 I T1 le 0.1264868W
Q1 1A 410K le 0.126WQ2 I T1 le 0.15484486W
T1 410K 400K 370K
I 1A 2A 4A
Q1 1A T1 le 0.126
0.097
0.068
0.039
0.01
W
Q1 2A T1 le 0.254
0.221
0.188
0.155
0.122
W
Q1 3A T1 le 0.357
0.32
0.283
0.246
0.209
W
Q1 4A T1 le 0.436
0.395
0.354
0.313
0.272
W
