Question

A \(0.2-\mathrm{m}\)-thick plane wall is constructed of concrete. At steady state, the energy transfer rate by conduction through a \(1-\mathrm{m}^{2}\) area of the wall is \(0.15 \mathrm{~kW}\). If the temperature distribution is linear through the wall, what is the temperature difference across the wall, in \(\mathrm{K}\) ?

Short answer

The temperature difference across the wall is approximately 17.65 K.

Step-by-step solution

Understand the heat conduction equation

The heat conduction through a plane wall can be described by Fourier's law: \[Q = \frac{kA(T_1 - T_2)}{L}\]where, - \(Q\) is the energy transfer rate (0.15 kW),- \(k\) is the thermal conductivity of the concrete,- \(A\) is the area of the wall (1 m²),- \(T_1\) and \(T_2\) are the temperatures on the two sides of the wall,- \(L\) is the thickness of the wall (0.2 m).

Rearrange the equation to solve for the temperature difference

We need to solve for the temperature difference \(\Delta T = T_1 - T_2\). Rearrange Fourier's law to solve for \(\Delta T\):\[\Delta T = \frac{Q \cdot L}{k \cdot A}\]

Substitute known values into the equation

Substitute the known values into the rearranged equation. Note that we need the value for the thermal conductivity \(k\). Let's assume an average value for thermal conductivity of concrete: \(k = 1.7 \mathrm{W}/(\mathrm{m}\cdot\mathrm{K})\). Now plug in the numbers:\[\Delta T = \frac{0.15~\text{kW} \cdot 0.2~\text{m}}{1.7 \frac{\text{W}}{\text{m}\cdot\text{K}} \cdot 1~\text{m}^2}\]Convert 0.15 kW to watts (since 1 kW = 1000 W):\[\Delta T = \frac{150~\text{W} \cdot 0.2~\text{m}}{1.7 \frac{\text{W}}{\text{m}\cdot\text{K}} \cdot 1~\text{m}^2} = \frac{30}{1.7}~\text{K}\]

Calculate the temperature difference

Finally, perform the calculation:\[\Delta T = \frac{30}{1.7} \approx 17.65~\text{K}\]

Key concepts

Fourier's Law of Heat Conduction

In heat transfer, Fourier's Law of Heat Conduction is fundamental. It provides a mathematical model to describe how heat energy transfers through a material. The equation is given by: \[Q = \frac{kA(T_1 - T_2)}{L} \] where:- Q is the rate of heat transfer (in Watts or Kilowatts). - k is the thermal conductivity of the material (W/m·K). - A is the cross-sectional area perpendicular to the direction of heat flow (in m²). - T₁ and T₂ are the temperatures on both sides of the wall (in K). - L is the thickness of the material (in m). This law highlights that heat transfer ( Q ) is directly proportional to the temperature difference ( ΔT=T₁−T₂ ), cross-sectional area ( A ), and thermal conductivity ( k ), but inversely proportional to the thickness ( L ) of the material.Understanding this relationship allows us to control the heat flow in different engineering applications, from building insulation to electronic cooling systems.

Thermal Conductivity

Thermal conductivity ( k ) is a property that measures a material's ability to conduct heat. Substances with high thermal conductivity, like metals, conduct heat very efficiently. Conversely, materials with low thermal conductivity, like wood or fiberglass, are good thermal insulators. In the given exercise, the concrete wall's thermal conductivity, k , is assumed to be 1.7 W/(m·K) . This value signifies that for each meter thickness of concrete, a temperature difference of 1 Kelvin results in a heat transfer rate of 1.7 Watts per square meter. When designing systems for heating or cooling, selecting materials with appropriate thermal conductivities is critical to achieving desired heat transfer rates.

Temperature Difference Calculation

The key goal of the exercise is to calculate the temperature difference ( ΔT ) across the wall. Based on Fourier's Law, we rearrange the equation to solve for ΔT :\[ ΔT = \frac{Q·L}{k·A} \]In this specific problem: - Q = 0.15 kW = 150 W (as 1 kW = 1000 W) - L = 0.2 m - k = 1.7 W/(m·K) - A = 1 m²Plugging these values into the equation provides: \[ ΔT = \frac{150 W·0.2 m}{1.7 W/(m·K)·1 m²} = \frac{30}{1.7} ≈ 17.65 K \]This result tells us that approximately 17.65 Kelvin is the temperature difference needed to drive the given heat transfer rate through the concrete wall with the specified parameters.

Steady State Heat Transfer

In thermodynamics, steady-state heat transfer occurs when the temperature field in the conducting material does not change over time. In other words, the system has reached an equilibrium, and the rate of heat entering any part of the system equals the rate of heat exiting it.In the provided exercise, the temperature distribution is said to be linear, indicating steady-state conditions. Under steady-state conditions, heat transfer calculations become more straightforward and predictable. This is because the temperature gradient ( T₁−T₂ ) remains consistent over time, allowing for precision in designing and assessing thermal systems.Recognizing whether a system is in steady-state or transient (where temperatures change over time) is crucial for selecting the correct analytical approach to solve heat transfer problems.