Question

A 0.5-m-long thin vertical copper plate is subjected to a uniform heat flux of \(1000 \mathrm{~W} / \mathrm{m}^{2}\) on one side, while the other side is exposed to air at \(5^{\circ} \mathrm{C}\). Determine the plate midpoint temperature for \((a)\) a highly polished surface and \((b)\) a black oxidized surface. Hint: The plate midpoint temperature \(\left(T_{L / 2}\right)\) has to be found iteratively. Begin the calculations by using a film temperature of \(30^{\circ} \mathrm{C}\).

Short answer

Answer: The final midpoint temperatures, after following the iterative method for both the polished and oxidized surfaces, are the values of \(T_{L/2,p}\) and \(T_{L/2,o}\), which can be determined after the iteration process converges to a negligible difference between the current and previous calculated temperatures (e.g., less than 0.001).

Step-by-step solution

Calculate heat transfer through convection

First, we will find the heat transfer rate, \(q\), through convection for the given heat flux, \(q''\). Since the heat flux is given in watts per square meter, we need to consider the surface area of the copper plate, \(A\). The plate's length is given as 0.5 m, and it is a thin vertical plate, so its height would be negligible. Hence, the area of the copper plate can be taken as 0.5 m²: $$q = q'' \cdot A$$ $$q = 1000 \frac{\text{W}}{\text{m}^2} \cdot 0.5 \text{m}^2 = 500 \text{W}$$ So, the heat transfer rate through convection is 500 W.

Calculate temperature difference between plate and air

For both polished and oxidized copper surfaces, we will now find the temperature difference between the plate and the surrounding air using the given air temperature of 5°C. This value will be used in the iteration process to find the midpoint plate temperature: Polished surface: $$\Delta T_{p} = T_{L/2,p} - T_{\text{air}}$$ Oxidized surface: $$\Delta T_{o} = T_{L/2,o} - T_{\text{air}}$$

Find the midpoint temperature using iteration

We need to find the midpoint temperature iteratively for both polished and oxidized surfaces using a starting film temperature of 30°C. This requires calculating the convection heat transfer coefficient, \(h\), using empirical correlations for natural convection with the specified film temperature. For a vertical plate, an empirical correlation for the Nusselt number (\(Nu\)) can be used: $$Nu = C \times Ra_L^n$$ Where \(C\) and \(n\) are constants depending on the surface type (polished or oxidized), and \(Ra_L\) is the Rayleigh number. The Rayleigh number can be calculated as: $$Ra_L = \frac{g \beta (\Delta T) L^3}{\nu \alpha}$$ Where \(g\) is the acceleration due to gravity, \(\beta\) is the thermal expansion coefficient, \(\Delta T\) is the temperature difference between the plate and the air, \(L\) is the length of the plate, \(\nu\) is the kinematic viscosity, and \(\alpha\) is the thermal diffusivity. Most of these properties depend on the film temperature and can be found using a thermodynamic properties table. Now, we can use the Nusselt number to find the convection heat transfer coefficient \(h\): $$h = \frac{k}{L} Nu$$ Where \(k\) is the thermal conductivity of the fluid (air). This leads to the following equation for convection heat transfer: $$q = hA(\Delta T)$$ We will now use this equation and iterate using the initial film temperature of 30°C to find the midpoint temperature \(T_{L/2,p}\) and \(T_{L/2,o}\) for both the polished and oxidized surfaces. This process will be repeated until the difference between the current and previous calculated temperatures is negligible (say, less than 0.001). After reaching the desired convergence, we can obtain the midpoint temperatures for both the polished and oxidized surfaces.

Key concepts

Convection

Convection is a key heat transfer mechanism that occurs when a fluid, like air, moves over a surface, transferring energy from one place to another.
It can be natural or forced, depending on whether the fluid movement is due to natural differences in temperature (natural) or driven by external forces like fans (forced).
In the given exercise, convection plays a crucial role as we calculate how heat is transferred from a vertical copper plate to the surrounding air.

• The heat transfer rate per unit area, also known as the heat flux, is known. In this case, it's 1000 W/m².
• We then use that information to calculate the total heat transfer rate using the plate's surface area.

This helps us understand how convection affects the temperature difference needed for future calculations.

Empirical Correlations

Empirical correlations are mathematical relationships derived from experimental data.
They help estimate complex physical phenomena, like heat transfer, under specific conditions.
For convection, one important empirical correlation involves the Nusselt number (\(Nu\)), which is related to the heat transfer coefficient:- The Nusselt number is calculated using \(Nu = C \times Ra_L^n\), where the constants \(C\) and \(n\) vary based on the surface type and conditions.- The Rayleigh number (\(Ra_L\)) is a dimensionless number representing the ratio of buoyancy to viscous force and depends on factors such as the temperature difference (\(\Delta T\)), fluid properties, and plate dimensions.These correlations offer a reliable way to predict convection heat transfer, which is crucial in determining the precise midpoint temperature of the plate.

Thermodynamic Properties

Thermodynamic properties of a fluid, such as air in this problem, are essential in calculations related to heat transfer.
These include density, specific heat, thermal conductivity, viscosity, and the thermal expansion coefficient.
To calculate the Rayleigh number and subsequently the Nusselt number, we rely heavily on these properties:

• Thermal conductivity (\(k\)): A measure of a material's ability to conduct heat.
• Thermal expansion coefficient (\(\beta\)): Describes how the volume of a material changes with temperature.
• Other properties like viscosity (\(u\)) and thermal diffusivity (\(\alpha\)).

These values depend on the film temperature, which is why we often refer to tables or charts to find them.
Understanding these properties allows us to make accurate calculations related to the heat transfer from the plate.

Iteration Process

The iteration process is a method used when the solution to a problem is approached through successive approximations.
It's an essential technique when direct calculation is complicated by the interdependency of variables.
In this exercise, we begin with an initial guess for the film temperature, such as 30°C, and use it to calculate certain parameters like the convection heat transfer coefficient (\(h\)):

• We start with an initial assumption of the midpoint temperature.
• With this, the Rayleigh and Nusselt numbers are updated, leading to a new heat transfer coefficient.
• New temperature estimates are continuously computed until the changes between successive iterations are minimal (e.g., less than 0.001°C).

This method allows us to home in on the true midpoint temperature of the plate by refining our calculations iteratively.