Thermal Conductivity and Heat Flow in Composite Slabs

To analyze the heat flow and temperature distribution across the composite block in the steady state, we can use the concept of thermal resistance, which is given by:
$R = \frac{L}{K A}$
where $L$ is the length of the slab along the direction of heat flow, $K$ is the thermal conductivity, and $A$ is the cross-sectional area perpendicular to the heat flow. Let $w$ be the width of all the slabs (perpendicular to the plane of the diagram).

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1. Geometrical and Thermal Parameters of Each Slab

From the given figure:

• Slab A:

  • Length $L_A = 1L - 0 = L$
  • Height $h_A = 4L - 0 = 4L \implies A_A = 4L \cdot w$
  • Conductivity $K_A = 2K$
  • Thermal Resistance:
    $R_A = \frac{L}{(2K)(4L w)} = \frac{1}{8 Kw}$

• Slab B:

  • Length $L_B = 5L - 1L = 4L$
  • Height $h_B = 1L - 0 = 1L \implies A_B = 1L \cdot w$
  • Conductivity $K_B = 3K$
  • Thermal Resistance:
    $R_B = \frac{4L}{(3K)(1L w)} = \frac{4}{3 Kw}$

• Slab C:

  • Length $L_C = 5L - 1L = 4L$
  • Height $h_C = 3L - 1L = 2L \implies A_C = 2L \cdot w$
  • Conductivity $K_C = 4K$
  • Thermal Resistance:
    $R_C = \frac{4L}{(4K)(2L w)} = \frac{1}{2 Kw}$

• Slab D:

  • Length $L_D = 5L - 1L = 4L$
  • Height $h_D = 4L - 3L = 1L \implies A_D = 1L \cdot w$
  • Conductivity $K_D = 5K$
  • Thermal Resistance:
    $R_D = \frac{4L}{(5K)(1L w)} = \frac{4}{5 Kw}$

• Slab E:

  • Length $L_E = 6L - 5L = L$
  • Height $h_E = 4L - 0 = 4L \implies A_E = 4L \cdot w$
  • Conductivity $K_E = 6K$
  • Thermal Resistance:
    $R_E = \frac{L}{(6K)(4L w)} = \frac{1}{24 Kw}$

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2. Evaluation of Options

Analysis of Option (A):

Slab A, the combination of slabs (B, C, D), and slab E are connected in series along the heat flow path.
In a series arrangement, the total rate of heat flow through each section must be equal to the total heat current $H$:
$H_A = H_{BCD} = H_E = H$
Hence, the heat flow through slab A and slab E is the same ($H_A = H_E$).

• Option (A) is CORRECT.

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Analysis of Option (C):

Let $H$ be the total heat flow rate through the block.

• Temperature difference across slab A:
  $\Delta T_A = H \cdot R_A = \frac{H}{8 Kw}$

• Equivalent resistance $R_{BCD}$ of the parallel combination of slabs B, C, and D:
  $\frac{1}{R_{BCD}} = \frac{1}{R_B} + \frac{1}{R_C} + \frac{1}{R_D} = \frac{3 Kw}{4} + 2 Kw + \frac{5 Kw}{4} = 4 Kw$
  $R_{BCD} = \frac{1}{4 Kw}$
  Thus, the temperature difference across the parallel section (BCD) is:
  $\Delta T_{BCD} = H \cdot R_{BCD} = \frac{H}{4 Kw}$

• Temperature difference across slab E:
  $\Delta T_E = H \cdot R_E = \frac{H}{24 Kw}$

Comparing the temperature differences:
$\Delta T_E = \frac{1}{3} \Delta T_A = \frac{1}{6} \Delta T_{BCD}$
Therefore, the temperature difference across slab E is the smallest.

• Option (C) is CORRECT.

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Analysis of Option (D):

Since slabs B, C, and D are connected in parallel, they experience the same temperature difference $\Delta T_{BCD}$. The heat flow through each individual slab is:
$H_B = \frac{\Delta T_{BCD}}{R_B} = \Delta T_{BCD} \cdot \left(\frac{3 Kw}{4}\right)$
$H_C = \frac{\Delta T_{BCD}}{R_C} = \Delta T_{BCD} \cdot \left(2 Kw\right) = \Delta T_{BCD} \cdot \left(\frac{8 Kw}{4}\right)$
$H_D = \frac{\Delta T_{BCD}}{R_D} = \Delta T_{BCD} \cdot \left(\frac{5 Kw}{4}\right)$

Adding the heat flows through slab B and slab D:
$H_B + H_D = \Delta T_{BCD} \cdot \left( \frac{3 Kw}{4} + \frac{5 Kw}{4} \right) = \Delta T_{BCD} \cdot (2 Kw) = H_C$
Hence, heat flow through C = heat flow through B + heat flow through D.

• Option (D) is CORRECT.

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Conclusion

The correct options are A, C, and D.