Question

The inner and outer surfaces of a 4-m \(\times 7-\mathrm{m}\) brick wall of thickness \(30 \mathrm{~cm}\) and thermal conductivity \(0.69 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) are maintained at temperatures of \(26^{\circ} \mathrm{C}\) and \(8^{\circ} \mathrm{C}\), respectively. Determine the rate of heat transfer through the wall, in W.

Short answer

Answer: The rate of heat transfer through the wall is approximately 9476 W.

Step-by-step solution

Identify given values and constants

Given values are: - Wall dimensions: 4 m x 7 m - Wall thickness: 30 cm - Thermal conductivity: \(0.69 \frac{\text{W}}{\text{m} \cdot \text{K}}\) - Inner surface temperature \(T_1\): \(26^{\circ}\text{C}\) - Outer surface temperature \(T_2\): \(8^{\circ}\text{C}\)

Use Fourier's law of heat conduction

The rate of heat transfer through a plane wall can be determined through Fourier's law of heat conduction: $$ Q = kA\frac{T_1-T_2}{L} $$ where \(Q\) is the rate of heat transfer, \(k\) is the thermal conductivity, \(A\) is the area of the wall, \(T_1\) and \(T_2\) are the temperatures of the inner and outer surfaces of the wall, and \(L\) is the thickness of the wall.

Calculate wall area

The area of the wall is the product of its height and length: $$ A = 4\;\text{m} \cdot 7\;\text{m} = 28\;\text{m}^2 $$

Convert wall thickness into meters

The wall thickness is given in centimeters, which we need to convert into meters for consistency in our calculations: $$ L = 30\;\text{cm} \cdot \frac{1\;\text{m}}{100\;\text{cm}} = 0.3\;\text{m} $$

Calculate the rate of heat transfer

Using the values from the previous steps, substitute them into the Fourier's law equation: $$ Q = 0.69 \frac{\text{W}}{\text{m} \cdot \text{K}} \cdot 28\;\text{m}^2 \cdot \frac{26^{\circ}\text{C} - 8^{\circ}\text{C}}{0.3\;\text{m}} $$ Calculate the result: $$ Q = 0.69 \frac{\text{W}}{\text{m} \cdot \text{K}} \cdot 28\;\text{m}^2 \cdot \frac{18 \text{K}}{0.3\;\text{m}} = 9476\;\text{W} $$ So, the rate of heat transfer through the wall is approximately \(9476\;\text{W}\).

Key concepts

Fourier's Law

Fourier's Law is a fundamental principle in heat transfer. It describes how heat energy moves through materials. The rule says that the rate of heat transfer is proportional to the negative gradient of temperature and the area through which it flows. This means that heat will naturally flow from warmer areas to cooler ones until equilibrium is reached. This principle helps us understand how efficient a material is at transferring heat. The formula used in Fourier's Law looks like this:

• \( Q = kA \frac{T_1-T_2}{L} \)

Here, \( Q \) is the rate of heat transfer, \( k \) is the thermal conductivity, \( A \) is the area, \( T_1 \) and \( T_2 \) are the temperatures on either side of the wall, and \( L \) is the thickness of the wall. By using this formula, you can calculate how much heat is transferred through a material over a specific period of time.

Thermal Conductivity

Thermal conductivity is a material property that indicates how well a material can conduct heat. It's symbolized by \( k \) in the Fourier's Law formula. The higher the thermal conductivity, the better the material is at transmitting heat. It's measured in watts per meter-kelvin (\( \text{W/m} \cdot \text{K} \)).

• High thermal conductivity: Materials like metals, which quickly transfer heat.
• Low thermal conductivity: Materials such as wood or bricks, which are better at insulating.

In the given exercise, the wall's thermal conductivity is \( 0.69 \text{W/m} \cdot \text{K} \). This relatively low value indicates that the wall provides some level of insulation between the inside and outside temperatures. Understanding thermal conductivity helps in material selection for construction and city planning scenarios, ensuring temperature regulations inside buildings.

Rate of Heat Transfer

The rate of heat transfer refers to the quantity of heat energy that passes through the wall over a certain period. It is denoted by \( Q \) and measured in watts (\( \text{W} \)). In our exercise:

• The inner surface temperature is \( 26\text{°C} \).
• The outer surface temperature is \( 8\text{°C} \).

The difference in temperatures drives the rate at which heat moves through the wall. Using the calculated rate of heat transfer, i.e., \( 9476 \text{W} \), we learn how efficiently or inefficiently heat moves through the wall. This rate also helps us determine how effective our insulation may be in real-world applications, such as walls in buildings, or even insulation materials used in spaceships.

Wall Dimensions

Wall dimensions play a critical role in calculating heat transfer. They determine the surface area \( A \) through which the heat flows. In our problem, the wall measures \( 4 \text{ m} \times 7 \text{ m} \), giving us an area of \( 28 \text{ m}^2 \). Additionally, the wall's thickness \( L \) impacts heat conduction. The thicker a wall is, the more resistance it provides to heat flow. In this example:

• The thickness is \( 30 \text{ cm} \) or \( 0.3 \text{ m} \) after conversion.

Together, these dimensions are pivotal in applying Fourier's Law to find the rate of heat transfer. Adjusting either the area or the thickness changes how much heat energy moves through. It's crucial for settings involving insulation design or thermal management in spaces like homes or workplaces.