Question

In an experiment, the local heat transfer over a flat plate was correlated in the form of the local Nusselt number as expressed by the following correlation $$ \mathrm{Nu}_{x}=0.035 \mathrm{Re}_{x}^{0.8} \mathrm{Pr}^{1 / 3} $$ Determine the ratio of the average convection heat transfer coefficient \((h)\) over the entire plate length to the local convection heat transfer coefficient \(\left(h_{x}\right)\) at \(x=L\).

Short answer

Based on the correlation for the local Nusselt number, we found the expression for the ratio of the average convection heat transfer coefficient over the entire plate length to the local convection heat transfer coefficient at \(x=L\), which is: $$ \frac{h_{\mathrm{avg}}}{h_x(L)} = \frac{0.01944 L^{0.8}}{0.035}. $$

Step-by-step solution

Express the local convection heat transfer coefficient

We are given the local Nusselt number, \(\mathrm{Nu}_{x}=0.035 \mathrm{Re}_{x}^{0.8} \mathrm{Pr}^{1/3}\). The Nusselt number is defined as the ratio of convective to conductive heat transfer, which also relates the local heat transfer coefficient to the thermal conductivity: $$ \mathrm{Nu}_{x} = \frac{h_x \cdot x}{k}, $$ where \(h_x\) is the local convection heat transfer coefficient, \(x\) is the distance from the leading edge, and \(k\) is the thermal conductivity. From this, we can find the local convection heat transfer coefficient \(h_x\): $$ h_x = \frac{\mathrm{Nu}_{x} \cdot k}{x}. $$

Find an expression for the average Nusselt number

To find the average heat transfer coefficient, we will need to find the average Nusselt number. The average Nusselt number can be obtained by integrating the local Nusselt number over the entire plate length and dividing by the length: $$ \mathrm{Nu}_{\mathrm{avg}} = \frac{1}{L} \int_0^L \mathrm{Nu}_{x} \, dx. $$ Now, substitute the given correlation for the local Nusselt number: $$ \mathrm{Nu}_{\mathrm{avg}} = \frac{1}{L} \int_0^L \left(0.035 \mathrm{Re}_{x}^{0.8} \mathrm{Pr}^{1/3}\right) \, dx. $$

Solve the integral

In the previous step, we expressed the average Nusselt number in terms of an integral. To find the average heat transfer coefficient, we need to solve this integral with respect to \(x\). The variables that depend on \(x\) are contained in the Reynolds number. Let's rewrite the integral with \(\mathrm{Re}_x\) replaced by the expression \(\frac{u_\infty x}{\nu}\), where \(u_\infty\) is the fluid velocity and \(\nu\) is the kinematic viscosity. $$ \mathrm{Nu}_{\mathrm{avg}} = \frac{1}{L} \int_0^L \left(0.035 \left(\frac{u_\infty x}{\nu}\right)^{0.8} \mathrm{Pr}^{1/3}\right) \, dx. $$ The variables \(u_\infty\), \(\nu\), and \(\mathrm{Pr}\) are constants with respect to \(x\). Therefore, we can rewrite the integral as: $$ \mathrm{Nu}_{\mathrm{avg}} = \frac{0.035 u_\infty^{0.8} \nu^{-0.8} \mathrm{Pr}^{1/3}}{L} \int_0^L x^{0.8} \, dx. $$ Now we integrate with respect to \(x\): $$ \mathrm{Nu}_{\mathrm{avg}} = \frac{0.035 u_\infty^{0.8} \nu^{-0.8} \mathrm{Pr}^{1/3}}{L} \left[\frac{x^{1.8}}{1.8}\right]_0^L. $$ Substitute the limits of integration: $$ \mathrm{Nu}_{\mathrm{avg}} = 0.01944 u_\infty^{0.8} \nu^{-0.8} \mathrm{Pr}^{1/3} L^{0.8}. $$

Find the average convection heat transfer coefficient

Now that we have the average Nusselt number, we can find the average convection heat transfer coefficient \(h_{\mathrm{avg}}\) using the definition: $$ h_{\mathrm{avg}} = \frac{\mathrm{Nu}_{\mathrm{avg}} \cdot k}{L}. $$ Substitute the results from Step 3: $$ h_{\mathrm{avg}} = \frac{0.01944 u_\infty^{0.8} \nu^{-0.8} \mathrm{Pr}^{1/3} L^{0.8} \cdot k}{L}. $$

Find the ratio of \(h_{\mathrm{avg}}\) to \(h_x\) at \(x=L\)

We are asked to find the ratio of the average convection heat transfer coefficient \(h_{\mathrm{avg}}\) over the entire plate length to the local convection heat transfer coefficient \(h_x\) at \(x=L\). To do this, we will divide \(h_{\mathrm{avg}}\) by \(h_x\) and evaluate at \(x=L\): $$ \frac{h_{\mathrm{avg}}}{h_x(L)} = \frac{0.01944 u_\infty^{0.8} \nu^{-0.8} \mathrm{Pr}^{1/3} L^{0.8} \cdot k \div L}{0.035 u_\infty^{0.8} \nu^{-0.8} \mathrm{Pr}^{1/3} k}. $$ After cancelling common terms, we can simplify the expression: $$ \frac{h_{\mathrm{avg}}}{h_x(L)} = \frac{0.01944 L^{0.8}}{0.035}. $$ This is the final expression for the required ratio of the average convection heat transfer coefficient over the entire plate length to the local convection heat transfer coefficient at \(x=L\).