Question

At steady state the temperature profile in a laminated system appears as in figure. What is the thermal conductivity of material 2 if steady state flux is 12.6$\times$10$^3$ W/m$^2$ and conductivity of material 1 is 52 W/mK ?

• A. 25.4 W/mK
• B. 106.315 W/mK
• C. 38.3 W/mK
• D. 102.3 W/mK

Hint:

Treat composite wall heat transfer problems exactly like electric circuits in series:
\[ \text{Current } (I) \leftrightarrow \text{Heat flux } (q) \]
\[ \text{Voltage } (V) \leftrightarrow \text{Temperature difference } (\Delta T) \]
\[ \text{Resistance } (R) \leftrightarrow \text{Thermal resistance } (x/k) \]
This analogy helps structure calculations and prevent errors.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The question asks us to calculate the thermal conductivity (\( k_2 \)) of the second layer in a two-layer laminated composite wall under steady-state heat conduction.
We are given the heat flux (\( q \)), the thickness of both layers (\( x_1 = 0.5 \text{ m} \) and \( x_2 = 0.3 \text{ m} \)), the thermal conductivity of the first layer (\( k_1 = 52 \text{ W/mK} \)), and the boundary temperatures (\( T_1 = 530 \text{ K} \) and \( T_3 = 310 \text{ K} \)).

Step 2: Key Formula or Approach:
For steady-state, one-dimensional heat conduction through a composite wall, the heat flux is constant through all layers.
Using the thermal resistance concept, the heat flux is:
\[ q = \frac{T_1 - T_3}{R_{\text{th, total}}} \]
where the total thermal resistance per unit area is the sum of individual resistances in series:
\[ R_{\text{th, total}} = R_{\text{th, 1}} + R_{\text{th, 2}} = \frac{x_1}{k_1} + \frac{x_2}{k_2} \]

Step 3: Detailed Explanation:

• Identify the given values:
Heat flux, \( q = 12.6 \times 10^3 \text{ W/m}^2 = 12600 \text{ W/m}^2 \)
Thickness of Material 1, \( x_1 = 0.5 \text{ m} \)
Thermal conductivity of Material 1, \( k_1 = 52 \text{ W/mK} \)
Thickness of Material 2, \( x_2 = 0.3 \text{ m} \)
Overall temperature difference, \( \Delta T = T_1 - T_3 = 530 - 310 = 220 \text{ K} \)

• Express the heat flux equation:
\[ q = \frac{T_1 - T_3}{\frac{x_1}{k_1} + \frac{x_2}{k_2}} \]

• Substitute the known values:
\[ 12600 = \frac{220}{\frac{0.5}{52} + \frac{0.3}{k_2}} \]

• Calculate the resistance of the first layer:
\[ R_{\text{th, 1}} = \frac{0.5}{52} \approx 0.009615 \text{ m}^2\text{K/W} \]

• Solve for the total resistance:
\[ R_{\text{th, total}} = \frac{220}{12600} \approx 0.017460 \text{ m}^2\text{K/W} \]

• Determine the resistance of the second layer:
\[ R_{\text{th, 2}} = R_{\text{th, total}} - R_{\text{th, 1}} \]
\[ \frac{0.3}{k_2} = 0.017460 - 0.009615 = 0.007845 \text{ m}^2\text{K/W} \]

• Solve for \( k_2 \):
\[ k_2 = \frac{0.3}{0.007845} \approx 38.24 \text{ W/mK} \]

• This matches option (C) which is 38.3 W/mK.


Step 4: Final Answer:
The thermal conductivity of material 2 is approximately 38.3 W/mK.