Electrical Analogy For Thermal Conduction

The concept of electrical analogy for thermal conduction bridges two fundamental branches of physics: electricity and heat transfer. Just as electric current flows through a conductor due to a potential difference, heat flows through a material driven by a temperature difference. In this analogy, temperature corresponds to voltage, thermal conductivity to electrical conductivity, and heat flow to electric current. This analogy is not just a theoretical tool but finds real-life applications in understanding and designing efficient thermal systems, such as in the insulation of homes, where materials are chosen based on their ability to resist heat flow, much like choosing a resistor in an electric circuit. This approach helps engineers and scientists analyze complex thermal systems using well-understood electrical principles, making it easier to predict and manage heat transfer in various practical scenarios.

Electrical Analogy for Thermal Conduction

Firstly we discuss Electrical vs Thermal Conductivity
Thermal conductivity and electrical conductivity are two extremely important physical properties of matter. The thermal conductivity of the material defines the rate at which a material can conduct thermal energy. The electrical conductivity of the material describes the amount of electrical current that will result due to a given potential difference. These are both fairly well-characterized properties, with huge applications in fields such as power generation and transmission, electrical engineering, electronics, thermodynamics and heat, and many other fields.

Electrical Conductivity

The resistance of a component depends on various parameters. The length of the conductor, the area of the conductor, and the material of the conductor are to name some. The conductivity of a material can be defined as the conductance of a block having unit dimensions made out of the material. The conductivity of a material is the inverse of the resistivity. Conductivity is usually denoted by the Greek letter σ. The SI unit of conductivity is Siemens per meter.

Thermal Conductivity

Thermal conductivity is the ability of a material to conduct thermal energy. The thermal conductivity is a property of the material. The thermal conductance is a property of the object. The most important law behind thermal conductivity is the heat flow equation. In a mathematical form, this can be written as dH/dt = kA(∆T)/l, where k is the thermal conductivity, A is the cross area, ∆T is the temperature difference between the two ends and l is the length of the object. ∆T/l can be termed a temperature gradient. The thermal conductivity is measured in watts per kelvin per meter.

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Solved Examples Based on Electrical Analogy For Thermal Conduction

Example 1: For a rod of length 'l', area of cross-section 'A, and thermal conductivity K, the thermal resistance R is given as

1) $R=\frac{l}{K A}$
2) $R=\frac{l^2}{K A}$
3) $R=\frac{K}{l A}$
4) $R=\frac{A}{l K}$

Solution:

From the Electrical Analogy for Thermal Conduction

For a rod of length '', area of cross-section 'A, and Thermal conductivity K,

Thermal resistance is defined as $R=\frac{l}{K A}$

Hence, the answer is the option (1).

Example 2: What is the dimension of Heat current (H)?

1) $M L^2 T^{-3}$
2) $M L^2 T^{-2}$
3) $M L^{-3} T^{-3}$
4) $M L^3 T^{-2}$

Solution:

According to Electrical Analogy for Thermal Conduction

The rate of flow of heat is called heat current.

i.e., $H=\frac{d Q}{d t}$

$
\text { So }[H]=\frac{[\text { heat }]}{[T]}=\frac{[\text { work }]}{[T]}=\frac{[F . L]}{[T]}=\frac{\left[M L T^{-2} * L\right]}{[T]}=\left[M L^2 T^{-3}\right]
$

Hence, the answer is the option (1).

Example 3: The temperature $\theta$ at the junction of two insulting sheets, having thermal resistances $R_1$ and $R_2$ as well as top and bottom temperatures $\theta_1$ and $\theta_2$ (as shown in the figure) is given by :

1) $\frac{\theta_1 R_2+\theta_2 R_1}{R_1+R_2}$
2) $\frac{\theta_1 R_2-\theta_2 R_1}{R_1-R_2}$
3) $\frac{\theta_2 R_2-\theta_1 R_1}{R_2-R_1}$
4) $\frac{\theta_1 R_1+\theta_2 R_2}{R_1+R_2}$

Solution:

The heat flow rate will be the same throughout both

$
\begin{aligned}
& \therefore \frac{\theta_1-\theta}{\mathrm{R}_1}=\frac{\theta-\theta_2}{\mathrm{R}_2} \\
& \mathrm{R}_2 \theta_1-\mathrm{R}_2 \theta=\mathrm{R}_1 \theta-\mathrm{R}_1 \theta_2 \\
& \theta=\frac{\mathrm{R}_2 \theta_1+\mathrm{R}_1 \theta_2}{\mathrm{R}_1+\mathrm{R}_2}
\end{aligned}
$

Hence, the answer is the option (1).

Example 4: Two metallic blocks $\mathrm{M}_1$ and $\mathrm{M}_2$ of the same area of the cross-section are connected to each other(as shown in the figure). If the thermal conductivity of $\mathrm{M}_2$ is K then the thermal conductivity of $\mathrm{M}_1$ will be :
[Assume steady state heat conduction]

1) 10 K
2) 8 K
3) 12.5 K
4) 2 K

Solution:

Since the two blocks are in series, the heat current through both of them will be the same as the steady state
$
\begin{aligned}
& \mathrm{H}_1=\mathrm{H}_2 \\
& \frac{\Delta \mathrm{T}_1}{\mathrm{R}_1}=\frac{\Delta \mathrm{T}_2}{\mathrm{R}_2} \\
& \frac{(100-80)}{\left(\frac{\mathrm{l}_1}{(\mathrm{k}) \mathrm{A}_1}\right)}=\frac{(80-0)}{\left(\frac{\mathrm{l}_2}{\mathrm{k}_1 \mathrm{~A}_2}\right)}
\end{aligned}
$
Here, $\mathrm{A}_1=\mathrm{A}_2=$ Area of cross-section (given) $\mathrm{k}_1 \rightarrow$ Thermal conductivity of block $\mathrm{M}_1$ $\mathrm{l}_1=16 \mathrm{~cm}, \mathrm{l}_2=8 \mathrm{~cm}$

$
\begin{aligned}
& \frac{20}{\left(\frac{16}{k_1}\right)}=\frac{80}{\frac{8}{k_0}} \\
& \mathrm{k}_1=8 \mathrm{k}
\end{aligned}
$

Hence, the answer is the option (2).

Example 5: As per the given figure, two plates $A$ and $B$ of thermal conductivity K and 2 K are joined together to form a compound plate. The thickness of the plates are 4.0 cm and 2.5 cm respectively and the area of the cross-section is $120 \mathrm{~cm}^2$ for each plate. The equivalent thermal conductivity of the compound plate is $\left(1+\frac{5}{\alpha}\right) \mathrm{K}$, then the value of $\alpha$ will be_____________.

1) 21

2) 56

3) 47

4) 6

Solution:

$\begin{aligned} & \text { Thermal resistance }=\mathrm{R}=\frac{1}{\mathrm{KA}} \\ & \mathrm{R}_{\mathrm{eff}}=\left(\mathrm{R}_1+\mathrm{R}_2\right) \\ & \frac{\left(\mathrm{L}_1+\mathrm{L}_2\right)}{\mathrm{Keff} \times \mathrm{A}}=\frac{\mathrm{L}_1}{\mathrm{~K}_1 \mathrm{~A}}+\frac{\mathrm{L}_2}{\mathrm{~K}_2 \mathrm{~A}} \\ & \frac{6.5}{\mathrm{Keff}}=\frac{4}{\mathrm{k}}+\frac{2.5}{2 \mathrm{~K}} \\ & \mathrm{Keff}=\frac{13 k}{10.5}=\left[1+\frac{5}{21}\right] \mathrm{k} \\ & \alpha=21\end{aligned}$

Hence, the answer is the option (1).

Summary

The electrical conduction analogy is a flow of heat through a material analogous to the flow of electric current through a conductor. The temperature gradient is similar to the voltage difference; the continuous flow of heat refers to the electric current; and thermal resistance becomes equivalent to electrical resistance. The electricity flows from the region of higher voltage to that of lower voltage. Similarly, heat flows from a higher to a lower temperature. This analogy helps in understanding and calculating the heat transfer more easily.