Combination Of Metallic Rods MCQ - Practice Questions with Answers

SERIES COMBINATION OF ROD/SLABS IN HEAT CONDUCTION - 

Series combination: Let n slabs each of cross-sectional area $\mathrm{A}_{\text {, lengths }} l_1, l_2, l_3 \ldots \ldots l_n$ and conductivities $K_1, K_2, K_3 \ldots \ldots . K_{n \text { respectively be connected in series - }}$

• - Heat current: In case of series combination, heat current is the same in all the conductors, So -

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

  - Thermal resistance - Net thermal resistance is equal to the sum of the thermal resistance of all the slab 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

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

• The temperature of an 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 $\theta=\frac{\frac{K_1}{1_1} \theta_1+\frac{K_2}{1_2} \theta_2}{\frac{K_1}{l_1}+\frac{K_2}{l_2}}$

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 has different thermal conductivity, then the 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 the 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

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

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