Consider a laminar ideal gas flow over a flat plate, where the local Nusselt number can be expressed as \(\mathrm{Nu}_{x}=0.332 \mathrm{Re}_{x}^{1 / 2} \operatorname{Pr}^{1 / 3}\). Using the expression for the local Nusselt number, show that it can be rewritten in terms of local convection heat transfer coefficient as \(h_{x}=C[V /(x T)]^{m}\), where \(C\) and \(m\) are constants.

The Nusselt number is a dimensionless number that indicates the relationship between convective heat transfer and conduction heat transfer. It is defined as: $$ \mathrm{Nu}_{x} = \frac{h_x x}{k} $$ where \(h_x\) is the local convective heat transfer coefficient, \(x\) is the distance along the flat plate, and \(k\) is the thermal conductivity of the fluid.

We also have the Reynolds (\(\mathrm{Re}_{x}\)) and Prandtl (\(\mathrm{Pr}\)) numbers in the given expression for the Nusselt number. The Reynolds and Prandtl numbers are defined as follows: $$ \mathrm{Re}_{x} = \frac{Vx}{\nu} $$ $$ \mathrm{Pr} = \frac{\nu}{\alpha} $$ where \(V\) is the velocity of the gas flow, \(\nu\) is the kinematic viscosity, and \(\alpha\) is the thermal diffusivity.

Substitute the given expression for the local Nusselt number, \(\mathrm{Nu}_{x}=0.332 \mathrm{Re}_{x}^{1 / 2} \mathrm{Pr}^{1 / 3}\), into the definition of the Nusselt number: $$ 0.332 \mathrm{Re}_{x}^{1 / 2} \mathrm{Pr}^{1 / 3} = \frac{h_x x}{k} $$

Now, replace the Reynolds and Prandtl numbers with their expressions from Steps 2: $$ 0.332 \left(\frac{Vx}{\nu}\right)^{1 / 2} \left(\frac{\nu}{\alpha}\right)^{1 / 3} = \frac{h_x x}{k} $$

Now, we need to isolate the term \(h_x\) in the equation and rewrite it in the form \(h_{x}=C[V /(x T)]^{m}\), where \(C\) and \(m\) are constants: $$ h_x = \frac{k}{x} \cdot 0.332 \left(\frac{Vx}{\nu}\right)^{1 / 2} \left(\frac{\nu}{\alpha}\right)^{1 / 3} $$ Since we have the product of the inverse of \(x\) and \(x^{1/2}\), we can write \(x^{-1/2}\): $$ h_x = k \cdot 0.332 \left(\frac{V}{\nu}\right)^{1 / 2} \left(\frac{\nu}{\alpha}\right)^{1 / 3} x^{-1 / 2} $$ Now, let's rewrite \(\frac{\nu}{\alpha}\) as \(\frac{\nu}{k / \rho c_p} = \frac{\rho c_p \nu}{k}\), where \(\rho\) is the density and \(c_p\) is the specific heat: $$ h_x = k \cdot 0.332 \left(\frac{V}{\nu}\right)^{1 / 2} \left(\frac{\rho c_p \nu}{k}\right)^{1 / 3} x^{-1 / 2} $$ Now, we can assign the constants: $$ C = k \cdot 0.332 \left(\frac{\rho c_p}{k}\right)^{1 / 3} $$ and $$ m = -\frac{1}{2}+\frac{1}{2}+\frac{1}{3} = \frac{1}{6} $$ Thus, $$ h_{x} = C \left(\frac{V}{x \nu}\right)^{m} $$