Question

Heat Transfer Question

Answer 1

The boundary layer scales like \(\large v^{1/2}\) and the thermal layer scales like \(\large \alpha ^{1/2} \approx (\frac{v}{Pr})^{1/2}\) so the thermal layer is much thicker than the velocity boundary layer at small Prandtl Numbers

So we can say \(\large u \approx U_\infty\) and \(\large v \approx 0\) all across the thermal layer. So we have

\[\large u\frac{\partial T}{\partial x} + v\frac{\partial T}{\partial y} = \alpha \frac{\partial ^2 T}{\partial y^2}\] which now can be written as

\[\large U_\infty\frac{\partial T}{\partial x} = \alpha \frac{\partial ^2 T}{\partial y^2}\]

Where the boundary conditions at \(\large T = T_w\) on \(\large y = 0\)
and \(\large T \rightarrow T_\infty\) for \(\large y \rightarrow \infty\)

Answer 2

We note this is the same as \(\large \frac{\partial ^2 T}{\partial x^2} = \frac{1}{\alpha} \frac{\partial T}{\partial t}\) if \(\large t = \frac{x}{U_\infty}\)

From this, we find the solution as

\[\large \frac{T - T_s}{T_i - T_s} = erf(\eta)\]

Answer 3

From that...I am to show that
\[\large \frac{T - T_w}{T_\infty - T_w} = erf(\frac{y}{2\sqrt{\frac{\alpha x}{U_\infty}}})\]

And this is where I'm stuck...any help would be greatly appreciated!

Answer 4

I was trying to see if using the fact that \(\large \eta = y\sqrt{\frac{U_\infty}{v x}}\) would get me anywhere but nothing yet

Answer 5

And just for the sake of having everything...general definition for the error function:

\[\large erf(x) = \frac{2}{\sqrt{x}}\int_{0}^{x}e^{-u^2} du\]

Answer 6

That's a fair amount of information, and very thought provoking. It may take a bit longer than a quick glance to get near it ... but wish us luck ...

Answer 7

I'm about to dive in ...
I'm intrigued by the insight into the pretty grim and dangerous environment alluded to in the post. Liquid sodium won't be exactly friendly, even if it is a good coolant and "bad" neutron moderator.
As I think is part of the thrust of this, "what sort of a liquid is liquid sodium, as opposed to water", and can the liquidity properties be matched with the thermal conduction and thermal capacity properties and the liquid flow properties.
Above is a guess at what I think is behind what's being asked.
If I'm wasting your time let me know ...

Answer 8

@osprey


Dunno about you but I've had a crack at this and I can't even get the PDE sorted

I've tried separating the variables and I come out with these initially :

\(\large T(x,y) = C e^{\dfrac{\alpha \Omega^2 x}{U_{\infty}}} \left\{ \begin{matrix}
A \cosh \Omega y \\
B \sinh \Omega y\\
\end{matrix} \right.
\)

Without even trying to fit the BV's, both those are totally the wrong shape. And the alternative sinusoid eigenfunctions are just a non-starter for same reasons :-((

Clever stuff, n'est-ce pas ?!?!