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Question 3.14: A thermoelectric unit is made of bismuth telluride p-n pairs...

A thermoelectric unit is made of bismuth telluride p-n pairs. Each pair is made of p- and n-type elements having a square cross section with each side a = 1 mm and each element having a length L = 1.5 mm. The elements are placed between two ceramic plates as shown in Figure Each side of the ceramic plates is a_{c} = 30 mm which allows for 14 elements along each side. This results in N = 7 × 14 pairs over the cross section a_{c} ×a_{c}.
(a) Determine the maximum cooling power Q_{c,max} for T_{h} − T_{c} = 0.
(b) Determine the maximum cooling power Q_{c,max} for T_{h} − T_{c} = 50^{\circ }C.
(c) Determine the corresponding current J_{e}(Q_{c,max}).
The cold plate is at T_{c} = 290 K.

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The maximum cooling power is given by Q_{c,max}=-\frac{\alpha ^{2}_{S}T_{c}^{2}}{2R_{e,h-c}}+R^{-1}_{k,h-c}(T_{h}-T_{c}) and for N pairs we have

\mid Q_{c} \mid  _{max} =N\left[ \frac{\alpha ^{2}_{S}T_{c}^{2}}{2R_{e,h-c}}+R^{-1}_{k,h-c}(T_{h}-T_{c}) \right] 

We need to determine \alpha _{S}, R_{e,h-c}, and R_{k,h-c}. From (\dot{S}_{e,p})_{c}=-\alpha _{S}J_{e}T_{c} ,        \alpha _{S}=\alpha _{S,p}-\alpha _{S,n} for \alpha _{S} we have

\alpha _{S}=\alpha _{s,p}-\alpha _{S,n}

For R_{e,h-c}, from \frac{1}{R_{k,h-c}}=\frac{1}{(R_{k,h-c})_{p}}+\frac{1}{(R_{k,h-c})_{n}}=\left(\frac{A_{k}k}{L} \right)_{p} +\left(\frac{A_{k}k}{L} \right)_{n},        R_{e,h-c}=\left(\frac{\rho _{e}L}{A_{k}} \right)_{p}+\left(\frac{\rho _{e}L}{A_{k}} \right) _{n} we have, based on the variables defined above

R_{e,h-c}=\left(\frac{\rho _{e}L}{A} \right)_{p}+\left(\frac{\rho _{e}L}{A} \right) _{n} =\frac{\rho _{e,p}L}{a^{2}}+\frac{\rho _{e,n}L}{a^{2}}

For R_{k,h-c}, from \frac{1}{R_{k,h-c}}=\frac{1}{(R_{k,h-c})_{p}}+\frac{1}{(R_{k,h-c})_{n}}=\left(\frac{A_{k}k}{L} \right)_{p} +\left(\frac{A_{k}k}{L} \right)_{n},        R_{e,h-c}=\left(\frac{\rho _{e}L}{A_{k}} \right)_{p}+\left(\frac{\rho _{e}L}{A_{k}} \right) _{n} we have

R_{k,h-c}^{2}=\left(k\frac{A}{L} \right)_{p}+\left(k\frac{A}{L} \right) _{n} =\frac{k_{p}a^{2}}{L} +\frac{k_{n}a^{2}}{L}

Now using the numerical values, we have

\alpha _{s}=2.3\times 10^{-4}(V/^{\circ }C)+2.1\times 10^{-4}(V/K)=4.4\times 10^{-4}V/^{\circ }C     or   V/K      Table

   R_{e,h-c}=2\times \frac{1.0\times 10^{-5}(ohm-m)\times 1.5\times 10^{-3}(m)}{10^{-3}(m)\times 10^{-3}(m)}=0.030 ohm          Table

R_{k,h-c}^{-1}=\frac{1.7(W/m-K)\times 10^{-3}(m)\times 10^{-3}(m)}{1.5\times 10^{-3}(m)}+\frac{1.45(W/m-K)\times 10^{-3}(m)\times 10^{-3}(m)}{1.5\times 10^{-3}(m)}=2.100\times 10^{-3 }W/K       Table

Substituting these numerical values in the above expressions for   \mid Q_{c,max}\mid  , we have the following.
(a) The maximum heat flow rate, for   T_{h} − T_{c} = 0 K  , is

\mid Q_{c} \mid  _{max}=98\left[\frac{(4.4\times 10^{-4})^{2}(V/K)^{2}\times (290)^{2}(k)^{2}}{2\times 0.030 (ohm)}-2.100\times 10^{-3}(W/K)\times (0)(k) \right]=26.60 W          for          T_{h} − T_{c}

(b) The maximum heat flow rate, for       T_{h} − T_{c}=50  K , is

\mid Q_{c}\mid  _{max} =98\left[\frac{(4.4\times 10^{-4})^{2}(V/K)^{2}\times (290)^{2}(k)^{2}}{2\times 0.030 (ohm)}-2.100\times 10^{-3}(W/K)\times (50)(k) \right]=16.30 W

(c) The current corresponding to the maximum cooling power is given by ,i.e.,

J_{e}(Q_{c,max})=\frac{\alpha _{S}T_{c}}{R_{e,h-c}}

Using the numerical values, we have

J_{e}(Q_{c,max})=\frac{4.4\times 10^{-4}(V/K)\times (290)(K)}{0.03{ohm}}=4.253 A  

 

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