Question

Three conductors of same length having thermal conductivity \(k_1\), \(k_2\), and \(k_3\) are connected as shown in figure. Area of cross sections of 1st and 2nd conductor are same and for 3rd conductor it is double of the 1st conductor. The temperatures are given in the figure. In steady state condition, the value of θ is ________ °C. (Given: \(k_1\) = 60 Js⁻¹m⁻¹K⁻¹,\(k_2\) = 120 Js⁻¹m⁻¹K⁻¹, \(k_3\) = 135 Js⁻¹m⁻¹K⁻¹)

 

Hint:

Remember the formula for thermal resistance and how to combine resistances in series and parallel. Also, in steady state, the heat flow is constant.

Answer 1

Correct Answer: 40

Provided:

• Thermal conductivity of conductor 1: \( k_1 = 60 \, \text{Js}^{-1} \text{m}^{-1} \text{K}^{-1} \)
• Thermal conductivity of conductor 2: \( k_2 = 120 \, \text{Js}^{-1} \text{m}^{-1} \text{K}^{-1} \)
• Thermal conductivity of conductor 3: \( k_3 = 135 \, \text{Js}^{-1} \text{m}^{-1} \text{K}^{-1} \)
• The cross-sectional area of conductor 3 is twice that of conductor 1.

Objective: Determine the value of \( \theta \) under steady-state conditions.

In a steady state, heat flow is constant across all conductors due to thermal equilibrium.

The heat transfer formula is:

\[ Q = \frac{kA(T_1 - T_2)}{L} \]

Where:

• \( Q \) = rate of heat transfer,
• \( k \) = thermal conductivity,
• \( A \) = cross-sectional area,
• \( T_1, T_2 \) = temperatures at the ends,
• \( L \) = length of the conductor.

Assuming identical lengths and constant heat flow \( Q \) in the steady state, the heat transfer for each conductor is:

Conductor 1:

\[ Q_1 = \frac{k_1 A (T_1 - \theta)}{L} \]

Conductor 2:

\[ Q_2 = \frac{k_2 A (\theta - T_2)}{L} \]

Conductor 3 (with area \( 2A \)):

\[ Q_3 = \frac{k_3 (2A) (\theta - T_3)}{L} \]

Equating the heat transfer rates since \( Q_1 = Q_2 = Q_3 \) in steady state:

Solving these equations yields the junction temperature \( \theta \):

40°C

Result:

The value of \( \theta \) is 40°C.