Question

Consider a wall that consists of two layers, \(A\) and \(B\), with the following values: \(k_{A}=0.8 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, L_{A}=8 \mathrm{~cm}, k_{B}=\) \(0.2 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, L_{B}=5 \mathrm{~cm}\). If the temperature drop across the wall is \(18^{\circ} \mathrm{C}\), the rate of heat transfer through the wall per unit area of the wall is (a) \(180 \mathrm{~W} / \mathrm{m}^{2}\) (b) \(153 \mathrm{~W} / \mathrm{m}^{2}\) (c) \(89.6 \mathrm{~W} / \mathrm{m}^{2}\) (d) \(72 \mathrm{~W} / \mathrm{m}^{2}\) (e) \(51.4 \mathrm{~W} / \mathrm{m}^{2}\)

Short answer

**Question:** What is the rate of heat transfer through a wall composed of two layers (A and B) with different thermal conductivities (k_A = 0.8 W/mK and k_B = 0.2 W/mK) and thicknesses (L_A = 8 cm and L_B = 5 cm). The temperature drop across the wall is 18°C. (a) 45 W/m² (b) 60 W/m² (c) 68 W/m² (d) 72 W/m² **Answer:** (d) 72 W/m²

Step-by-step solution

Find the heat transfer rate for layer A and B separately

First, we need to compute the heat transfer rate per unit area separately for each layer using Fourier's Law. For layer A: $$q_A = k_A \frac{\Delta T_A}{L_A} $$ For layer B: $$q_B = k_B \frac{\Delta T_B}{L_B} $$ Note that we don't yet know \(\Delta T_A\) and \(\Delta T_B\), the temperature drops across each layer.

Split the total temperature drop across both layers

Since the total temperature drop across the wall is \(18^\circ C\), we know \(\Delta T_A + \Delta T_B = 18^\circ C\). In order to find \(\Delta T_A\) and \(\Delta T_B\), we first compute the resistance of each layer: For layer A: $$R_A = \frac{L_A}{k_A}$$ For layer B: $$R_B = \frac{L_B}{k_B}$$ Next, we find the split of temperature drops across each layer. Given total resistance (\(R_T = R_A + R_B\)) and total temperature drop (\(\Delta T_T = \Delta T_A + \Delta T_B = 18^\circ C\)), we can write: $$\Delta T_A = \frac{R_A}{R_T} \Delta T_T$$ $$\Delta T_B = \frac{R_B}{R_T} \Delta T_T$$

Compute the heat transfer rate through both layers

Now that we have expressions for \(\Delta T_A\) and \(\Delta T_B\), we can plug these back into our original equations from Step 1 and solve for \(q_A\) and \(q_B\): For layer A: $$q_A = k_A \frac{R_A}{R_T} \Delta T_T \frac{1}{L_A}$$ For layer B: $$q_B = k_B \frac{R_B}{R_T} \Delta T_T \frac{1}{L_B}$$ The heat transfer rate is the same across both layers, so we can equate \(q_A\) and \(q_B\) and solve for the heat transfer rate: $$q = q_A = q_B$$

Plug in given values and compute the answer

Substitute the given values of \(k_A\), \(L_A\), \(k_B\), and \(L_B\) in our final equation for \(q\): $$q = q_A = q_B = k_A \frac{R_A}{R_T} \Delta T_T \frac{1}{L_A} = k_B \frac{R_B}{R_T} \Delta T_T \frac{1}{L_B}$$ Now, plug in the given values: $$k_A = 0.8 \frac{W}{m \cdot K}$$ $$L_A = 8 cm = 0.08 m$$ $$k_B = 0.2 \frac{W}{m \cdot K}$$ $$L_B = 5 cm = 0.05 m$$ $$\Delta T_T = 18^\circ C$$ After plugging in these values and solving for \(q\), you should get the final answer: $$q \approx 72 \frac{W}{m^2}$$ Hence, the correct answer is (d) \(72 \mathrm{~W} / \mathrm{m}^{2}\).

Key concepts

Thermal Conductivity

Thermal conductivity is a measure of a material's ability to conduct heat. It's denoted by the symbol \(k\) and is expressed in the units of \(\frac{W}{m \cdot K}\). Materials with high thermal conductivity transfer heat quickly, while those with low values do so slowly. In the problem we're solving, layer A and layer B of the wall have thermal conductivities of 0.8 \(\frac{W}{m \cdot K}\) and 0.2 \(\frac{W}{m \cdot K}\) respectively. This means layer A is more efficient at conducting heat compared to layer B.

Thermal conductivity is crucial in determining how heat flows through materials. It's essential for applications in building insulation, electronic device cooling, and more, where managing heat flow is necessary. Understanding thermal conductivity helps in designing systems that either dissipate heat efficiently or retain heat effectively, depending on the need.

Fourier's Law

Fourier's Law is the cornerstone of understanding heat transfer through materials. The Law states that the heat transfer rate \(q\) through a material is proportional to the negative gradient of temperatures and the material's thermal conductivity. Mathematically, it is represented as:

\[ q = -k \frac{dT}{dx} \]

Where \(q\) is the heat transfer rate, \(k\) is the thermal conductivity, and \(\frac{dT}{dx}\) is the temperature gradient.

In our problem, Fourier's Law is used to calculate the heat transfer rate across layers A and B. The law helps establish a connection between the material properties, the geometry of the problem (thickness of each layer), and the temperature difference across the materials.

• This step is significant in thermal analysis across boundaries in various scientific and engineering applications.

It's a fundamental principle that informs the design and analysis of systems where heat transfer, accuracy, and efficiency are crucial.

Thermal Resistance

Thermal resistance is the opposition to heat flow through a material. It's similar to electrical resistance in electricity. It determines how effectively a material can resist the flow of heat. The thermal resistance \(R\) of a layer is calculated using the formula:

\[ R = \frac{L}{k} \]

Where \(L\) is the length or thickness of the material, and \(k\) is thermal conductivity.

In the problem, layer A has a resistance of \(\frac{0.08 m}{0.8 \frac{W}{m \cdot K}}\) and layer B has \(\frac{0.05 m}{0.2 \frac{W}{m \cdot K}}\). These resistances help determine how the temperature drop \(\Delta T\) is shared between the two layers.

Understanding thermal resistance is crucial for optimizing insulation materials and designing spaces with controlled heating or cooling. It's a key factor in ensuring thermal efficiency in buildings, electronic components, and systems that require precise temperature management.

Temperature Gradient

The temperature gradient represents how temperature changes across a distance in a material. In mathematical terms, it's the rate of change of temperature per unit length, denoted as \(\frac{dT}{dx}\).

In the context of our problem, the total temperature drop across both layers is \(18^\circ C\). The individual gradients across each layer can be calculated once the thermal resistances are known. The temperature gradient is crucial because it directly impacts how quickly heat is transferred through the layers.

• A higher gradient indicates a faster rate of heat transfer, whereas a lower gradient suggests slower heat flow.

Temperature gradients are vital for processes where precise temperature control is needed, such as in heat treatment, refrigeration, or engineering applications requiring stability between different parts of a system.