Question

Heat Transfer Question...Math background

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

@sooobored if you have time and want to have some fun haha

Answer 5

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 6

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 7

;-;

Answer 8

pop this in physics and we'll have a go

looks semantic

@osprey

Answer 9

so you're modeling it as a heat penetration in a semi infinite medium?

Answer 10

Correct! Since it states the solution will be exactly like that of transient heat flow in a semi-infinite medium