91Ó°ÊÓ

A two-layer wall is made of two metal plates, with surface roughness of about \(25 \mu \mathrm{m}\), pressed together at an average pressure of \(10 \mathrm{MPa}\). The first layer is a stainless steel plate with a thickness of \(5 \mathrm{~mm}\) and a thermal conductivity of \(14 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The second layer is an aluminum plate with a thickness of \(15 \mathrm{~mm}\) and a thermal conductivity of \(237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). On the stainless steel side of the wall, the surface is subjected to a heat flux of \(800 \mathrm{~W} / \mathrm{m}^{2}\). On the aluminum side of the wall, the surface experiences convection heat transfer at an ambient temperature of \(20^{\circ} \mathrm{C}\), where the convection coefficient is \(12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the surface temperature of the stainless steel plate.

Step by step solution

01

Calculate the resistance of each layer

Start by calculating the thermal resistance of each layer. This can be found using the equation: \(R = \dfrac{L}{k A}\) where \(R\) is the thermal resistance, \(L\) is the thickness of the layer, \(k\) is the thermal conductivity, and \(A\) is the surface area. In this case, we will assume the surface area is the same for both layers. The resistance of stainless steel, \(R_1\), and aluminum, \(R_2\), can be calculated as follows: \(R_1 = \dfrac{5 \times 10^{-3}~\mathrm{m}}{14~\mathrm{W/m \cdot K} A}\) \(R_2 = \dfrac{15 \times 10^{-3}~\mathrm{m}}{237~\mathrm{W/m \cdot K} A}\)

02

Calculate total resistance of the system

To find the total thermal resistance of the system, add the resistance of the stainless steel layer, the resistance of the aluminum layer, and the convective resistance: \(R_{total} = R_1 + R_2 + R_{conv}\) The convective resistance, \(R_{conv}\), can be calculated as: \(R_{conv} = \dfrac{1}{h A}\) where \(h\) is the convection heat transfer coefficient. Thus, we get: \(R_{conv} = \dfrac{1}{12~\mathrm{W/m^2 \cdot K} A}\) Now, we can calculate the total resistance: \(R_{total} = \dfrac{5 \times 10^{-3}~\mathrm{m}}{14~\mathrm{W/m \cdot K} A} + \dfrac{15 \times 10^{-3}~\mathrm{m}}{237~\mathrm{W/m \cdot K} A} + \dfrac{1}{12~\mathrm{W/m^2 \cdot K} A}\)

03

Determine the heat flux through the system

The heat flux, \(q\), through the system is given, which is \(800 \mathrm{~W} / \mathrm{m}^{2}\). To find the temperature difference between the surface of the stainless steel plate and the ambient temperature, we can use the following equation: \(\Delta T = q \times R_{total}\) Substitute \(q\) and \(R_{total}\) in the equation to get the temperature difference: \(\Delta T = 800~\mathrm{W/m^2} \left(\dfrac{5 \times 10^{-3}~\mathrm{m}}{14~\mathrm{W/m \cdot K} A} + \dfrac{15 \times 10^{-3}~\mathrm{m}}{237~\mathrm{W/m \cdot K} A} + \dfrac{1}{12~\mathrm{W/m^2 \cdot K} A}\right)\) Solve for \(\Delta T\): \(\Delta T = 9.52~\mathrm{K}\)

04

Determine the surface temperature of the stainless steel plate

Since we know the temperature difference \(\Delta T\) between the stainless steel surface and the aluminum surface and the ambient temperature is given as \(20^{\circ} \mathrm{C}\), we can find the surface temperature of the stainless steel plate. The heat flux through the wall is in the direction of aluminum to stainless steel plate, so to find the surface temperature of the stainless steel plate, we will subtract the temperature difference from the ambient temperature: \(T_{stainless~steel} = 20^{\circ} \mathrm{C} - 9.52~\mathrm{K}\) \(T_{stainless~steel} = 10.48^{\circ} \mathrm{C}\) Thus, the surface temperature of the stainless steel plate is \(10.48^{\circ} \mathrm{C}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Heat Transfer

Heat transfer is the movement of thermal energy from one part of an object to another or between objects. It occurs when there is a temperature difference. In this exercise, heat transfer takes place between the stainless steel and aluminum plates. The heat moves due to the temperature gradient, from the hotter region to the cooler one. Understanding the overall process of heat transfer helps us determine how effectively energy is being transferred and at what rate.

• **Conduction:** The main form of heat transfer in solid objects like metals. It's driven by the thermal energy movement through material particles.
• **Convection:** Involves the movement of heat by the physical movement of fluids. It's crucial when one of the surfaces interacts with a fluid.
• **Radiation:** The energy transfer via electromagnetic waves, noticeable when significant temperature variations occur.

In our example of a two-layer wall, we particularly focus on conduction within the metal plates and convection when the heat crosses into the surrounding air.

Convection Coefficient

The convection coefficient, often denoted as 'h', is a crucial factor in determining heat transfer via convection. It tells us how efficiently heat is moved from a surface to a fluid or vice versa.
A higher convection coefficient implies quicker heat transfer, and its value is influenced by factors like:

• **Fluid velocity**: Faster moving fluids enhance convective heat transfer.
• **Fluid properties**: The density, viscosity, and thermal capacity play a role.
• **Surface conditions**: Surface roughness and geometry can affect how easily a fluid can flow over the surface.

In our problem, the aluminum side of the wall has a convection coefficient of 12 W/m²·K, which suggests the level of heat transfer efficiency at the aluminum surface where it meets the surrounding air.

Thermal Conductivity

Thermal conductivity is a measure of a material's ability to conduct heat. In essence, it tells us how quickly heat can move through a material. Higher thermal conductivity indicates a better capacity of heat transfer through conduction.
This structural property changes with the material type, temperature, and the material's physical state.

• **Metals**, like aluminum and stainless steel, generally have high thermal conductivities. Hence, they are good conductors.
• **Material Specifics**: The exercise shows stainless steel with a thermal conductivity of 14 W/m·K and aluminum having a much higher thermal conductivity of 237 W/m·K.
These differences greatly influence the rate of heat transfer through each material, and thus, the final temperature at the interface and surface.

Heat Flux

Heat flux measures the rate of heat energy transfer per unit surface area. It's given in units such as W/m². In simple terms, it tells us how much thermal energy is flowing through a particular area.
Knowing the heat flux allows engineers and scientists to determine the temperature changes in materials over time.

• **Magnitude and Direction**: It's given as positive or negative based on direction and indicates whether energy is entering or leaving a surface.
• **Application**: In our scenario, a heat flux of 800 W/m² is applied to the stainless steel side, influencing the temperature distribution across the metal plates.
Calculating the resulting temperature informs us on the adequacy of insulation and material suitability for structural applications.