Consider conduction into the ground. The initial temperature of
the ground is *T _{o}* and the air temperature
(suddenly) drops from

position and time?

Check what the value of the Biot number is ($\displaystyle{Bi = \frac{hL}{k}}$). What is L?

Since the Biot number is **not** very small,
we cannot use "lumped". How about using the charts?!

In order to make our problem dimensionless, first we can choose $\displaystyle{\theta}$ to be

$\displaystyle{\theta =
\frac{T-T_\infty}{T_o-T_\infty}}$

That is the easy part...

How do we make the length or time dimensionless? (since there is no obvious length or time scale!)

Last time, there was no obvious time scale, yet we made the problem dimensionless with a combination of the length scale and the properties of the material.

Clearly that won't work this time, but something similar DOES work...

What we do, based on the units, is define a new dimensionless
variable which is a *combination* of *x* and
*t*

(i.e., use our *variables* as the length and time scales!)
so we get

$\displaystyle{\eta = \frac{x}{\sqrt{4\alpha t}}}$

where we stuck the 4 in for convenience...

Hey this gives us $\displaystyle{T=F(\eta)}$. This is a one dimensional problem!

Thinking about the original problem, we had an initial condition
to satisfy: that T= T_{o}. We also had two boundary
conditions: that the surface temperature T_{s}=$T_\infty$
(because our Bi is effectively large) and that sufficiently far
from the surface (i.e., at $\displaystyle{x\to \infty}$) the
temperature has not changed yet so it is still T_{o}.

A little thought tells us that our new variable will have a hard
time satisfying **all** of these conditions unless we have a
third order ODE...

However, it turns out that the relevant differential equation
(after plugging in our new variable) is:

$\displaystyle{-2\eta\frac{d\theta}{d\eta} =
\frac{d^2\theta}{d\eta^2}}$

The good news is that even though we can only satisfy two
conditions our boundary (initial) conditions become

$\displaystyle{\theta = 1}$ at $\displaystyle{\eta \to
\infty}$; $\displaystyle{\theta = 0}$ at $\displaystyle{\eta =
0}$; and $\displaystyle{\theta = 1}$ at $\displaystyle{\eta
\to \infty}$

Two of our boundary (initial) conditions, that used to be
independent, are now the same! This is in fact critical for our
answer to be possible (since we now have a second order ODE, we
only *need* two conditions, so if we *had* three we
couldn't guarantee to satisfy all of them!).

This happens (and the problem's solution works) because this is what is called a self-similar or similarity problem (see here for a brief explanation of self-similarity.

In order to solve this equation, it is easiest to make the
simple substitution that $\displaystyle{p =
\frac{d\theta}{d\eta}}$, so that we get

$\displaystyle{-2\eta p = \frac{dp}{d\eta}}$

which we can easily rearrange and solve

$\displaystyle{\int -2\eta d\eta = \int \frac{dp}{p}}$

gives

$\displaystyle{-\eta^2 = \ln{[p]}+A}$

or

$\displaystyle{p = \frac{d\theta}{d\eta} =
Be^{-\eta^2}}$

which we can integrate *again* to yield

$\displaystyle{\theta = \int_0^\eta Be^{-u^2} du + C}$

Using our BC's, $\displaystyle{\theta = 0}$ at
$\displaystyle{\eta = 0}$ yields *C*=0 (since the integral
from 0 to 0 is 0).

Our other BC $\displaystyle{\theta = 1}$ at
$\displaystyle{\eta \to \infty}$ gives

$\displaystyle{1 = \int_0^\infty Be^{-u^2} du}$

where we can look up the fact that

$\displaystyle{\int_0^\infty Be^{-u^2} du =
\frac{\sqrt{\pi}}{2}}$

so that

$\displaystyle{B = \frac{2}{\sqrt{\pi}}}$,

which yields a final answer of

$\displaystyle{\theta = \frac{2}{\sqrt{\pi}} \int_0^\eta
e^{-u^2} du}$

The integral in this expression cannot be solved analytically,
but it shows up in so many problems that it has been given a name
and values are easily found in tables. It is called the error
function (*erf*)

$\displaystyle{erf[\eta] =\frac{2}{\sqrt{\pi}} \int_0^\eta
e^{-u^2} du}$

So our final answer, given in its most compact form is

$\displaystyle{\theta = erf[\eta]}$

or

$\displaystyle{\frac{T-T_\infty}{T_o-T_\infty} =
erf[\frac{x}{\sqrt{4\alpha t}}]}$

This is an interesting result because, to quote Eddie Murphy in
*The Golden Child*, "There's a ground Monty!". In other
words, the problem really **doesn't** go to infinity, but as
long as the thermal energy does not **know** that the ground is
not infinite, this solution works. Taking this a step further, we
note that $\displaystyle{\sqrt{\alpha t}}$ has units of length.
This quantity can be thought of as the ** penetration
depth** or the distance over which thermal effects have
penetrated during the time, t. So as long as our system size, L is
much larger than the penetration depth, $\displaystyle{L_p =
\sqrt{\alpha t}}$, we can use this solution.

Use the semi-infinite approximation to solve both transient heat and mass problems for "short" times