# Electrical Analogy for Thermal Conduction

## Electrical Analogy for Thermal Conduction

Thermal Conductivity is defined as the quantity of heat (Q) transmitted through a unit thickness (l) in a direction normal to a surface of unit area due to a unit temperature gradient (ΔT) under steady state condition and when the heat transfer is dependent only on the temperature gradient.

From Ohm’s Law, if current (I) flows through a resistance at potential difference (V₁ – V₂), then (V₁ – V₂) = IR (Or) $$I=\frac{\left( {{V}_{1}}-{{V}_{2}} \right)}{R}$$.

Electrical Resistance $$\left( R \right)=\frac{\rho l}{A}=\frac{l}{\sigma A}$$.

Where,

R = Resistance of conductor,

l = Length of conductor,

σ = Conductivity of the material.

$$Current\left( I \right)=\frac{\Delta Q}{\Delta t}=\frac{\left( {{V}_{1}}-{{V}_{2}} \right)}{\left( \frac{l}{\sigma A} \right)}$$.

$$Current\left( I \right)=\frac{\Delta Q}{\Delta t}=\frac{\sigma A\left( {{V}_{1}}-{{V}_{2}} \right)}{l}$$ … (1)

If a metal rod of length (l), cross – section area (A) and thermal conductivity (K) connects two bodies at temperature θ₁ and θ₂ (θ₁ > θ₂). The rate of heat flow or heat current:

$$H=\frac{\Delta Q}{\Delta t}=\frac{\left( {{\theta }_{1}}-{{\theta }_{2}} \right)}{{{R}_{T}}}$$.

Thermal resistance of the rod:

$${{R}_{T}}=\frac{l}{KA}$$.

∴ $$H=\frac{\Delta Q}{\Delta t}=\frac{\left( {{\theta }_{1}}-{{\theta }_{2}} \right)}{\left( \frac{l}{KA} \right)}=\frac{KA\left( {{\theta }_{1}}-{{\theta }_{2}} \right)}{l}$$ … (2)

Equation (2) is mathematically equivalent to Equation (1), with temperature. Hence, results derived from Ohm’s law are also valid for thermal conduction.