Thermodynamics of thermocouple:
consider a thermocouple made of two conductors (A) and (B) at the junction (a) and (b) as shown in figure.
Thermocouple is an electrical device that can be used as a thermometer. It is comprised of two metals, A, B, that form a simple circuit with two junctions, J, between the metals. efficient of conductors a and b and π+dπ be the Peltier coefficient at the junction a and b respectively.
let I be the current flowing in the circuit when a and b are maintained at temperature T + d t and T respectively. then due to Peltier effact alone , it is observed at and a and rejected at and be the source of heat and b at as a sink of heat . Also due to Thomson effact heat is rejected along the conductor which act as the sink and heat is gained along b which act as a source.
heat absorbed at heat junction at Peltier effect:
\[Q_{1}=( \pi+d \pi)I\]
Heat absorbed at cold junction at peltier effect
\[Q_{2}= \pi I \]
Net heat absorbed or evolved
\[Q_{p} = Q_{1} - Q_{2} \Rightarrow ( \pi + d \pi ) I - \pi I \\\ Q_{p}= d \pi I \rightarrow (1).. \]
Similarly
Heat absorbed at hot and Cold junction of Thomson effect
\[Q_{T}= \sigma_{b} Idr- \sigma_{a}Idt \rightarrow (\sigma_{b}- \sigma_{a})Idt\Rightarrow (2)...\]
Net change in Peltier effect + Thomson effect
\[Q=d \pi I+ (\sigma_{b}- \sigma_{a})Idt\Rightarrow (3)...\]
Net electron energy dissipated/second:
\[Q=dEI\Rightarrow (4)... \\\ Compare \ eq \ (3) \ and \ (4):- \\\ dEI= d \pi I+( \sigma_{b} - \sigma_{a})Idt \\\ dE=d \pi +( \sigma_{b}-\sigma_{a})dt\Rightarrow \frac{d \pi}{dt}+ (\sigma_{b}- \sigma_{a})\rightarrow (5)...\]
According to reversible engine
\[\frac{ Q_{1}}{T{1}}=\frac{ Q{2}}{ T2{}} i.e \ \frac{heat \ absorved}{temprature \ of \ source}=\frac{heat \ evolved}{temprature \ of \ sink } \\\ \rightarrow \frac{ \pi I}{T}+\frac{I \sigma _{a}dt}{T}=\frac{ ( \pi +d \pi )I}{T+dt} +\frac{I \sigma _{b}dt}{T} \\\ \rightarrow \frac{ \sigma _{a}dt}{T}-\frac{ \sigma _{b}dt}{T} =\frac{ ( \pi +d \pi)}{T+dt}-\frac{ \sigma_{a}dt}{T} \\\ I( \sigma_{a}- \sigma_{b}) \frac{dt}{T}=( \frac{ \pi +d \pi }{T+dt}- \frac{ \pi}{T})I \\\ \rightarrow ( \sigma_{a} - \sigma_{b}) \frac{dt}{T}= \frac{ \pi +d \pi T- \pi T- \pi dt}{T} \\\ LetT+dt=t \ and \ T(T+dt)= t^{2} \\\ ( \sigma_{a} -\sigma_{b}) \frac{dt}{t}=\frac{d \pi}{T}-\frac{ \pi dt}{ T^{2}} \\\ (\sigma_{a}-\sigma_{b})=\frac{d \pi}{dt}-\frac { \pi}{T}\rightarrow -\frac{ d \pi}{dt}= ( \sigma_{b} -\sigma_{a})- \frac{ \pi}{T} \rightarrow (6) \\\ from \ eq \ 5 \\\ \frac{dE}{dt} = d \pi +( \sigma_{b}- \sigma_{a})dt\Rightarrow -\frac{ d \pi}{dt}= ( \sigma_{b} -\sigma_{a})- \frac{ dE}{ dt} \rightarrow (7)\]
On comparing eq (6) and (7) we get:
\[\frac {( \sigma_{b}- \sigma_{a})}{}- \frac{dE}{dt}=( \sigma_{b}- \sigma_{a})- \frac{ \pi}{t} \\\ \frac{dE}{dt}=\frac{ \pi}{T} \\\ \frac{dE}{dt}T= \pi\rightarrow (8) \\\ differenciate \ w.r.t(t) \\\ \frac{d \pi}{dt}=\frac{ T d^{2}E}{d t^{2}}+\frac{dE}{dt} \\\ \frac{d \pi}{dt}-\frac{dE}{dt}=\frac{ T d^{2}E}{d t^{2}} \\\ \frac{ T d^{2}E}{d t^{2}} = ( \sigma_{a}-\sigma_{b}) \ \ \rightarrow [( \sigma_{a}- \sigma_{b})= \frac {d \pi}{dt}- \frac{dE}{dt} \ according \ to \ (7) \ eq.]\]
Also click here to see working of carnot engine
