Calculating Heat Flow Ratio in Walls with Different Thermal Conductivities

This problem involves understanding how heat flows through materials based on their properties and the temperature difference across them. We are given information about two walls and asked to find the ratio of heat flow through them.

The rate of heat flow (\(H\)) through a wall by conduction is described by Fourier's Law:

\[ H = \frac{kA\Delta T}{L} \]

Where:

• \(k\) is the thermal conductivity of the material.
• \(A\) is the cross-sectional area of the wall.
• \(\Delta T\) is the temperature difference (drop) across the wall.
• \(L\) is the thickness of the wall.

We are given the following information for the two walls (let's call them Wall 1 and Wall 2):

• Thickness: Same for both walls, so \(L_1 = L_2\).
• Cross-sectional area: Same for both walls, so \(A_1 = A_2\).
• Ratio of thermal conductivities: \(k_1 : k_2 = 1 : 2\), which means \(\frac{k_1}{k_2} = \frac{1}{2}\).
• Ratio of temperature drop: \(\Delta T_1 : \Delta T_2 = 2 : 3\), which means \(\frac{\Delta T_1}{\Delta T_2} = \frac{2}{3}\).

We need to find the ratio of heat flow, \(H_1 : H_2\).

Using the formula for heat flow for both walls:

For Wall 1: \( H_1 = \frac{k_1 A_1 \Delta T_1}{L_1} \)

For Wall 2: \( H_2 = \frac{k_2 A_2 \Delta T_2}{L_2} \)

Now, let's find the ratio of \(H_1\) to \(H_2\):

\[ \frac{H_1}{H_2} = \frac{\frac{k_1 A_1 \Delta T_1}{L_1}}{\frac{k_2 A_2 \Delta T_2}{L_2}} \]

This can be rewritten as:

\[ \frac{H_1}{H_2} = \frac{k_1}{k_2} \times \frac{A_1}{A_2} \times \frac{\Delta T_1}{\Delta T_2} \times \frac{L_2}{L_1} \]

Now we substitute the given ratios and equalities:

\[ \frac{H_1}{H_2} = \left(\frac{k_1}{k_2}\right) \times \left(\frac{A}{A}\right) \times \left(\frac{\Delta T_1}{\Delta T_2}\right) \times \left(\frac{L}{L}\right) \]

\[ \frac{H_1}{H_2} = \left(\frac{1}{2}\right) \times (1) \times \left(\frac{2}{3}\right) \times (1) \]

\[ \frac{H_1}{H_2} = \frac{1}{2} \times \frac{2}{3} \]

\[ \frac{H_1}{H_2} = \frac{1 \times 2}{2 \times 3} = \frac{2}{6} = \frac{1}{3} \]

So, the ratio of heat flow through the two walls is 1 : 3.

Using the ratio relation for heat flow:

\[ \frac{H_1}{H_2} = \frac{k_1}{k_2} \times \frac{\Delta T_1}{\Delta T_2} \] (since \(A\) and \(L\) are the same)

\[ \frac{H_1}{H_2} = \left(\frac{1}{2}\right) \times \left(\frac{2}{3}\right) = \frac{1}{3} \]

The ratio of heat flow \(H_1 : H_2\) is 1 : 3.