Question

Consider airflow over a plate surface maintained at a temperature of \(220^{\circ} \mathrm{C}\). The temperature profile of the airflow is given as $$ T(y)=T_{\infty}-\left(T_{\infty}-T_{s}\right) \exp \left(-\frac{V}{\alpha_{\text {faid }}} y\right) $$ The airflow at 1 atm has a free stream velocity and temperature of $0.08 \mathrm{~m} / \mathrm{s}\( and \)20^{\circ} \mathrm{C}$, respectively. Determine the heat flux on the plate surface and the convection heat transfer coefficient of the airflow.

Short answer

In this exercise, we analyzed the profile of airflow temperature over a heated plate to find the heat flux on the plate surface and the convection heat transfer coefficient of the airflow. We first calculated the temperature gradient at the surface and then used it to determine the heat flux. Finally, we found the convection heat transfer coefficient using the heat flux and the temperature difference between the surface and the free stream. The exact numerical result for both heat flux and the convection heat transfer coefficient depends on the value of \(\alpha_{\text{faid}}\), which needs to be specified or determined through experimental data.

Step-by-step solution

Identify the problem parameters

We are given the temperature profile of the airflow: $$ T(y)=T_{\infty}-\left(T_{\infty}-T_{s}\right) \exp \left(-\frac{V}{\alpha_{\text {faid }}} y\right) $$ Also, the given problem parameters are as follows: - Plate surface temperature, \(T_s = 220^{\circ}\)C - Free stream airflow temperature, \(T_{\infty} = 20^{\circ}\)C - Free stream velocity, \(V = 0.08\) m/s - \(y\) is the distance measured perpendicular to the plate surface We need to find the heat flux and the convection heat transfer coefficient of the airflow.

Calculate the temperature gradient at the surface

To find the heat flux on the surface of the plate, we need to evaluate the temperature gradient at the surface. We can do this by finding the derivative of the temperature profile with respect to the distance y and then evaluating it at y = 0: $$\frac{dT}{dy} = -\frac{d}{dy}\left[ T_{\infty}-\left(T_{\infty}-T_{s}\right) \exp \left(-\frac{V}{\alpha_{\text {faid }}} y\right) \right]$$ Applying the Chain rule, we get: $$\frac{dT}{dy} = \frac{V}{\alpha_{\text {faid }}}\left(T_s - T_{\infty}\right)\exp{\left(-\frac{V}{\alpha_{\text {faid }}} y\right)}$$ Now we will evaluate this expression at y = 0: $$\left.\frac{dT}{dy}\right|_{y=0} = \frac{V}{\alpha_{\text {faid }}}(T_s - T_{\infty})$$

Calculate the heat flux on the surface

According to Fourier's law of heat conduction, the heat flux \((q_s)\) at the surface is given by: $$q_s = -k \left.\frac{dT}{dy}\right|_{y=0}$$ where \(k\) is the thermal conductivity of the air at the given temperature. The values of thermal conductivity for air can be obtained from the literature or engineering handbooks. For this problem, we will assume a thermal conductivity: $$k = 0.0262 \,\text{W} / (\text{m}\cdot \text{K})$$ Using the temperature gradient obtained in Step 2: $$q_s = -0.0262 \times \frac{V}{\alpha_{\text {faid }}}(T_s - T_{\infty})$$ We have the given values for \(V\), \(T_s\), and \(T_{\infty}\), substituting them and solving for \(q_s\): $$q_s = -0.0262 \times \frac{0.08}{\alpha_{\text {faid }}}(220 - 20)$$ $$q_s = -0.43784\frac{(200)}{\alpha_{\text {faid }}}$$

Calculate the convection heat transfer coefficient

The heat transfer coefficient (\(h\)) can be related to the heat flux and the temperature difference between the surface and the free stream by: $$q_s = h (T_s - T_{\infty})$$ Rearranging the expression to calculate h: $$h = \frac{q_s}{(T_s - T_{\infty})}$$ Now substituting the value of \(q_s\) obtained in Step 3: $$h = \frac{-0.43784(200) / \alpha_{\text {faid }}}{(220 - 20)}$$ $$h = -0.43784 \frac{200}{\alpha_{\text {faid }}(200)}$$ So, the convection heat transfer coefficient of the airflow is given by: $$h = -\frac{0.43784}{\alpha_{\text {faid }}}\, \text{W}/(\text{m}^2 \cdot \text{K})$$ The value of \(\alpha_{\text{faid}}\) needs to be specified or determined through experimental data to obtain a numerical result for the convection heat transfer coefficient. With this value, we will be able to find the exact numerical solution for both heat flux and the convection heat transfer coefficient.