2.5.2. Theoretical Analysis

In order to confirm the validity of the numerically predicted values of all performance characteristics of the thermoelectric module, the same performance characteristics were evaluated with a theoretical approach. The voltage at maximum power indicates optimum voltage and is calculated from Equation (6) [1,31,32]. Here, α is the Seebeck coefficient, which is measured as the amount of thermoelectric voltage induced in any material due to temperature difference across that material. Each thermoelectric material has its own unique value of Seebeck coefficient. The Seebeck coefficient is also recognized as thermopower, thermoelectric power, and thermoelectric sensitivity:

$$V\_{\text{theoretical},opt} = \alpha \left( T\_H - T\_{\text{C}} \right) \tag{6}$$

Current depends on voltage, external load resistance, and internal resistance—Equation (7) [10]. *R* = *R<sup>L</sup>* is the condition to obtain optimal current wherein power is maximum [4,31]:

$$I\_{\text{theoretical},opt} = \frac{V\_{\text{theoretical,opt}}}{R + R\_L} = \frac{\alpha \left(T\_H - T\_{\odot}\right)}{2R} \tag{7}$$

By manipulating the equations of *I* = α (*TH*−*TC*) *R*+*R<sup>L</sup>* and *P* = *I* 2 (*R* + *RL*), the power equation was obtained. To obtain maximum power, constraint of <sup>∂</sup>*<sup>P</sup>* ∂*R<sup>L</sup>* = 0 was applied on the power equation, which gave the maximum power condition of *R* = *RL*. When *R* = *R<sup>L</sup>* is used in the power equation, Equation (8) is derived, which calculates the theoretical maximum power of the thermoelectric module [4,33]:

$$P\_{\text{theoretical,max}} = \frac{\alpha^2}{4R}(T\_H - T\_C)^2\tag{8}$$

The theoretical maximum efficiency was calculated using Equation (9) under the same condition of *R* = *R<sup>L</sup>* in the theoretical efficiency equation of η*theoretical* = ∆*T*·*R<sup>L</sup> TH*(*R*+*RL*)− ∆*T*·*R* <sup>2</sup> + (*R*+*RL* ) 2 *ZR* [4].

$$
\eta\_{\text{theoretical},\text{max}} = \Delta T \frac{\sqrt{1 + Z\overline{T}} - 1}{T\_H \sqrt{1 + Z\overline{T}} + T\_{\text{C}}} \tag{9}
$$

Resistance for square prism and cylindrical legs was calculated with Equation (10) [1].

$$R = \frac{(\rho\_p + \rho\_n)L}{A} \tag{10}$$

Resistance of the trapezoidal legs cannot be calculated using Equation (10) because of the non-uniform cross-sectional area, which varies along the height of the thermoelectric legs. In the trapezoidal legs, an unevenly distributed area along the centerline of leg in height was converted to an evenly distributed area along the same centerline of the leg such that volume of the leg was same in both cases. Using the uniform area approach, the resistance for the trapezoidal legs was calculated with Equation (11) [11].

$$R = \frac{\sigma\_p + \sigma\_n}{2\sigma\_p \sigma\_n \frac{A\_0}{L} \left(\frac{R\_A - 1}{R\_A + 1}\right)} \text{ } \ln(R\_A) \tag{11}$$

where *R<sup>A</sup>* is the ratio of the top side area of the trapezoidal leg (*AT*) to the bottom side area of the trapezoidal leg (*AB*) from Equation (12). *A*<sup>0</sup> is the equivalent uniform cross-section area of the trapezoidal leg:

$$R\_A = \frac{A\_T}{A\_B} \tag{12}$$

In order to evaluate the maximum power and maximum efficiency of the thermoelectric module with the two-stage arrangement, it is necessary to derive the intermediate temperature between the two stages. This intermediate temperature can be obtained using voltage and temperature difference correlation equation of *V* = α (*T<sup>H</sup>* − *Ti*) or α (*T<sup>i</sup>* − *TC*). In the present study, the voltages applied to both the stages were different; hence, the equations were modified to *VTop* = α (*T<sup>H</sup>* − *Ti*1) and *VBottom* = α (*Ti*<sup>2</sup> − *TC*). *Ti*<sup>1</sup> is the cold side temperature for the first stage, whereas *Ti*<sup>2</sup> is the hot side temperature of the second stage. The difference of *Ti*<sup>1</sup> − *Ti*<sup>2</sup> shows the temperature loss due to the thermal resistance of the intermediate plate. The voltage boundary conditions applied to the high potential sides of both the stages were inserted in the corresponding voltage equations, obtaining *Ti*<sup>1</sup> and *Ti*<sup>2</sup> temperatures. These temperatures are used in Equations (8) and (9) in order to calculate the maximum power and maximum efficiency of the two-stage arrangement of the thermoelectric module. The segmented thermoelectric module consists of a combination of two materials in a single thermoelectric leg. Hence, the concept of equivalent Seebeck coefficient, equivalent resistivity, equivalent thermal conductivity, and equivalent Figure of Merit was used. Using this equivalent parameter concept, maximum power and maximum efficiency of the segmented thermoelectric module were calculated theoretically from the following equations [22,34]:

$$P\_{\text{theoretical},\max} = \frac{\alpha\_{\text{equivalent}}^2}{R\_{\text{equivalent}}} (T\_H - T\_\text{C})^2 \tag{13}$$

$$
\alpha\_{equivalent} = \frac{\alpha\_2 k\_1 + \alpha\_1 k\_2}{k\_1 + k\_2} \tag{14}
$$

$$\eta\_{\text{theoretical},\text{max}} = \Delta T \frac{\sqrt{1 + Z\_{\text{equivalent}}\overline{T}} - 1}{T\_H \sqrt{1 + Z\_{\text{equivalent}}\overline{T}} + T\_\mathbb{C}} \tag{15}$$

$$Z\_{equivalent} = \frac{\alpha\_{equivalent}^2}{(\rho\_1 + \rho\_2) \* \left(\frac{k\_1 k\_2}{k\_1 + k\_2}\right)}\tag{16}$$

Thermal stress induced in the thermoelectric modules due to various temperature conditions, as well as due to mismatch in thermal properties of different materials, were calculated in the form of von Mises stress. This is the equivalent of the second deviatory stresses and computed using the following equation [1,3]:

$$\sigma\_{\upsilon} = \sqrt{\frac{\left(\sigma\_{11} - \sigma\_{22}\right)^2 + \left(\sigma\_{22} - \sigma\_{33}\right)^2 + \left(\sigma\_{33} - \sigma\_{11}\right)^2 + 6\left(\sigma\_{12}^2 + \sigma\_{23}^2 + \sigma\_{31}^2\right)}{2}} \tag{17}$$

Normal stress in the longitudinal direction can be calculated using the following equation. In this equation, normal strain ε*<sup>n</sup>* = 0 and curvature K = 0 for the present study [3] are:

$$
\sigma\_{\rm xx} = \frac{E}{1 - v^2} \left( \varepsilon\_{\rm II} + Z\_{\rm c} K \right) - \frac{E \alpha\_{\rm T}}{1 - v} (T\_H - T\_{\rm C}) = -\frac{E \alpha\_{\rm T}}{1 - v} (T\_H - T\_{\rm C}) \tag{18}
$$

Thermal stress in the direction of temperature gradient can be calculated using the following equations, which are based on temperature difference and thermal properties.

$$
\sigma\_T = \alpha\_T \mathbb{E}(T\_H - T\_\mathbb{C}) \tag{19}
$$

$$
\sigma\_T = \alpha\_T E \Delta T \tag{20}
$$

Thermal strain is the ratio of thermal stress to the Young modulus, which is represented by Equation (21).

$$
\varepsilon = \frac{\sigma\_T}{E} \tag{21}
$$
