Question

Steady heat conduction occurs through a \(0.3\)-m-thick \(9 \mathrm{~m} \times 3 \mathrm{~m}\) composite wall at a rate of \(1.2 \mathrm{~kW}\). If the inner and outer surface temperatures of the wall are \(15^{\circ} \mathrm{C}\) and \(7^{\circ} \mathrm{C}\), the effective thermal conductivity of the wall is (a) \(0.61 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) (b) \(0.83 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) (c) \(1.7 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) (d) \(2.2 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) (e) \(5.1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\)

Short answer

Answer: The effective thermal conductivity of the composite wall is approximately 1.7 W/m·K.

Step-by-step solution

Write Down Given Information

Heat transfer rate, q = 1.2 kW = 1200 W (Convert to watts) Inner surface temperature, \(T_a = 15^{\circ} \mathrm{C}\) Outer surface temperature, \(T_b = 7^{\circ} \mathrm{C}\) Thickness of the wall, d = 0.3 m Dimensions of the wall, 9m x 3m

Calculate the Surface Area of the Wall

A = 9m * 3m = 27 \(\mathrm{m}^2\)

Rearrange the Steady-State Heat Conduction Formula to Solve for k

\(k = \frac{q \cdot d}{A \cdot (T_a - T_b)}\)

Plug in the Known Values and Calculate k

\(k = \frac{1200 \cdot 0.3}{27 \cdot (15 - 7)} = \frac{360}{216} = 1.66666667\)

Check which option is closest

We can see that the closest option among the choices is \(1.7 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) The effective thermal conductivity of the wall is (c) \(1.7 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

Key concepts

Thermal Conductivity

When studying heat transfer, thermal conductivity is a vital concept that measures a material's ability to conduct heat. It is represented by the symbol k and is typically measured in watts per meter kelvin (W/m·K). A higher value of thermal conductivity indicates that the material is a good conductor of heat, such as metals like copper or aluminum, while a lower value suggests that the material is more of an insulator, such as rubber or wood.

For example, when we say a wall has a thermal conductivity of \(1.7\ \mathrm{W/m}\cdot \mathrm{K}\), it means that for every meter thickness of the wall and for every degree temperature difference across it, \(1.7\ \mathrm{W}\) of heat will flow through each meter squared of its surface area. The thermal conductivity is useful in calculations involving steady-state heat conduction, where temperature differences remain constant over time, allowing us to use formulas to solve for unknown variables.

Heat Transfer Rate

The heat transfer rate, denoted as \(q\), is the amount of heat energy transferred per unit of time. Its SI unit is the watt (W), which is equivalent to one joule per second. In practical terms, if a composite wall conducts \(1.2 \text{kW}\) of heat, this is the same as saying that \(1200\ \text{W}\) of energy is being transferred from the hot side to the cold side each second.

The heat transfer rate plays a crucial role in determining the effectiveness of insulation materials, the efficiency of home heating systems, or the requirements for cooling devices. Understanding how to manage and calculate this rate is integral to designing structures and equipment that regulate temperature effectively. For instance, a wall's insulating properties can be characterized by how much heat it allows to pass through, which is directly related to the heat transfer rate.

Surface Area Calculation

In the context of heat transfer, surface area calculation is important because the larger the area through which heat can travel, the greater the total amount of heat that can be conducted. The surface area (\(A\)) of an object is the measure of how much exposed area the object has. It is usually measured in square meters (\(m^2\)).

For example, to determine the heat transfer through a wall, we must calculate the wall's surface area which is simply the product of its height and length. If a wall measures \(9 m\) by \(3 m\), its surface area is \(27 m^2\). This area plays a key role in the formula to calculate the heat transfer rate or thermal conductivity in steady-state heat conduction scenarios. In summary, knowing how to accurately determine surface area is essential for predicting and managing heat flow in various engineering and architectural applications.