Combination Of Metallic Rods - Practice Questions & MCQ

SERIES COMBINATION OF ROD/SLABS IN HEAT CONDUCTION - 

Series combination: Let n slabs each of cross-sectional area A , lengths $l_1, l_2, l_3 \ldots \ldots l_n$ and conductivities $K_1, K_2, K_3 \ldots \ldots . . K_n$ respectively be connected in series -
- Heat current: In the case of series combination, heat current is the same in all the conductors, So -

$
\begin{aligned}
& \frac{Q}{t}=H_1=H_2=H_3 \ldots \ldots \ldots=H_n \\
& \qquad \frac{K_1 A\left(\theta_1-\theta_2\right)}{l_1}=\frac{K_2 A\left(\theta_2-\theta_3\right)}{l_2}=\frac{K_n A\left(\theta_{n-1}-\theta_n\right)}{l_n}
\end{aligned}
$

So, by law of thermal conductivity -
- Thermal resistance - Net thermal resistance is equal to the sum of thermal resistance of all the slabs/rods. So, -

Equivalent thermal resistance: $R=R_1+R_2+\ldots . . R_n$
- Thermal conductivity - From the above equation of equivalent thermal resistance, equivalent thermal conductivity can be calculated as-

$
\begin{aligned}
& \text { From } R_S=R_1+R_2+R_3+\ldots \\
& \quad \frac{l_1+l_2+\ldots l_n}{K_{e q} A_{e q}}=\frac{l_1}{K_1 A}+\frac{l_2}{K_2 A}+\ldots+\frac{l_n}{K_n A} \\
& \Rightarrow K_{\text {equivalent }}=\frac{l_1+l_2+\ldots \ldots l_n}{\frac{l_1}{K_1}+\frac{l_2}{K_2}+\ldots \ldots+\frac{l_n}{K_n}}
\end{aligned}
$
 

$
A_{e q} L_{e q}=A_1 L_1+A_2 L_2+A_3 L_3+\ldots+A_n L_n
$

For series combination: $L_{e q}=L_1+L_2+L_3+\ldots+L_n$ (for each slab having constant area of cross section)
For parallel combination: $A_{e q}=A_1+A_2+A_3+\ldots+A_n$ (for each slab having same length of slab)
- The temperature of the interface of composite bar: For the calculation of this, let the two bars be arranged in series as shown in the figure -
ie., $\frac{Q}{t}=\frac{K_1 A\left(\theta_1-\theta\right)}{l_1}=\frac{K_2 A\left(\theta-\theta_2\right)}{l_2}$
By solving, we get $\theta=\frac{\frac{K_1}{l_1} \theta_1+\frac{K_2}{l_2} \theta_2}{\frac{K_1}{l_1}+\frac{K_2}{l_2}}$

PARALLEL COMBINATION OF ROD/SLABS IN HEAT CONDUCTION - 

Parallel combination : Let n slabs each of lengths $l$, cross-sectional area $A_1, A_2, A_3 \ldots \ldots . A_n$ and conductivities $K_1, K_2, K_3 \ldots \ldots . K_n$ respectively be connected in parallel -
- Heat current If each slab will have different thermal conductivity, then Net heat current will be the sum of heat currents through individual slabs. i.e.,

$
H=H_1+H_2+H_3+\ldots H_n
$

So, by law of thermal conductivity -

$
\begin{aligned}
& \frac{K\left(A_1+A_2+\ldots,+A_n\right)\left(\theta_1-\theta_2\right)}{l} \\
& =\frac{K_1 A_1\left(\theta_1-\theta_2\right)}{l}+\frac{K_2 A_2\left(\theta_1-\theta_2\right)}{l}+\ldots+\frac{K_1 A_n\left(\theta_1-\theta_2\right)}{l}
\end{aligned}
$

- Equivalent Thermal resistance - Net thermal resistance in parallel combination -

$
\frac{1}{R_s}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+\ldots \cdot \frac{1}{R_n}
$

- Thermal conductivity - From the above equation of equivalent thermal resistance, equivalent thermal conductivity can be calculated as-

$
\frac{K\left(A_1+A_2+\ldots,+A_n\right)\left(\theta_1-\theta_2\right)}{l}
$
  $\Rightarrow K_{\text {equivalent }}=\frac{K_1 A_1+K_2 A_2+K_3 A_3+\ldots \ldots K_n A_n}{A_1+A_2+A_3+\ldots \ldots A_n}$                        

• Temperature of interface of composite bar : Temperature gradient Same across each slab.