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 \(30^{\circ} \mathrm{C}\).

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.

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: $$\Delta T_{o} = T_{L/2,o} - T_{\text{air}}$$

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.