# Methodological Considerations of Using Thermoelectrics with Fin Heat Sinks for Cooling Applications
## 2 Materials and Methods  
### Thermal Resistance Model  
Figure 3a shows a cross-section of the FHS-TE assembly. The TE module is sandwiched between a conventional FHS and a heat source. An aluminum block is added to the assembly to distribute the heat evenly from the heating element. The associated thermal resistance network is shown in Figure 3b. All interface temperatures are expected to be uniform since the heat flow is one-dimensional. Power (\(P_{te}\)) supplied to the TE module creates a low temperature at the cold side (\(T_{c}\)) of the TE and a high temperature (\(T_{h}\)) at the hot side. Assuming perfect insulation as shown, heat absorbed at the cold side is equal to the heat supplied (\(P_{EH}\)) from the heat source. Heat from the cold surface is dissipated at the hot surface.  
The heat transfer rates at the cold (\(q_{c}\)) and hot sides (\(q_{h}\)) of the TE module are given by:  
\[q_{c}=\alpha_{te}I_{te}T_{c}-\frac{1}{2}I_{te}^{2}R_{te}-K_{te}\Delta T_{te} \tag{1}\]  
and  
\[q_{h}=\alpha_{te}I_{te}T_{h}+\frac{1}{2}I_{te}^{2}R_{te}-K_{te}\Delta T_{te} \tag{2}\]  
where \(\alpha_{te}\), \(I_{te}\), \(R_{te}\), and \(K_{te}\) are the Seebeck coefficient, current, internal resistance, and thermal conductivity of TE, respectively. The temperature difference across the TE module is:  
\[\Delta T_{te}=T_{h}-T_{c} \tag{3}\]  
Figure 3: Thermal resistance model of the FHS-TE assembly.  
where \(T_{h}\) and \(T_{c}\) define the hot side temperature and cold side temperature of TE, respectively. An energy balance (\(P_{te}\)) gives the power to be supplied to the TE module:  
\[P_{te}=q_{h}-q_{c}=\alpha_{te}I_{te}\Delta T_{te}+I_{te}^{2}R_{te} \tag{4}\]  
The applied TE voltage (\(V_{te}\)) is:  
\[V_{te}=\alpha_{te}\Delta T_{te}+I_{te}R_{te} \tag{5}\]  
The cooling coefficient of performance of the TE is:  
\[\text{COP}_{c}=q_{c}/P_{te} \tag{6}\]  
In Figure 3b, the 1-D thermal resistance of the FHS is shown as \(R_{f1D}\) and the thermal resistance of the aluminum block as \(R_{al}\). Contact thermal resistances \(R_{cr}\) between the aluminum block and TE module, as well as between TE and FHS, are assumed to be negligible. From the thermal resistance network the thermal resistance of the FHS is:  
\[R_{f1D}=(T_{h}-T_{a})/q_{h} \tag{7}\]  
where \(T_{a}\) is the ambient temperature. The internal electrical resistance of the TE depends upon the mean operating temperature of the TE. This can be represented by:  
\[R_{te}=a+bT_{mte} \tag{8}\]  
where \(a\) and \(b\) are constants in internal electrical resistance of TE. The mean TE temperature is calculated by:  
\[T_{mte}=(T_{h}+T_{c})/2 \tag{9}\]  
From the above we obtain:  
\[T_{h}=\frac{R_{f1D}\big{(}-bI_{te}^{2}T_{c}-2 aI_{te}^{2}-4K_{te}T_{c}\big{)}-4 T_{a}}{R_{f1D}\big{(}bI_{te}^{2}+4\alpha_{te}I_{te}-4K_{te}\big{)}-4} \tag{10}\]  
Figure 4 shows the flow chart to simulate the TE hot side temperature (\(T_{h}\)) given the cold side temperature (\(T_{c}\)). The procedure involves an iteration process. Ambient (\(T_{a}\)) and cold side temperatures, and applied current (\(I_{te}\)) to the TE module are first specified together with the characteristics of the TE module, such as the Seebeck coefficient (\(\alpha_{te}\)), thermal conductance (\(K_{te}\)), internal electrical resistance (\(R_{te}\)), and the resistance constants \(a\) and \(b\). From the manufacturer's datasheet, \(\alpha_{te}\) = 0.053 V/K, \(K_{te}\) = 0.66 W/K, \(R_{te}\) = 0.016, \(a\) = \(-\)0.36, and \(b\) = 0.006. First, a guessed value of \(T_{h}\) is input into the program. Next, the simulated value of \(T_{h}\)' is calculated from Equation (10). The difference between the predicted and guessed values is then calculated. If the difference is more than 1%, the process is repeated until the difference is less than 1%. The thermal resistance \(R_{f1D}\) under NC and FC is then evaluated using the new value of \(T_{h}\)'.