This screencast shows how to determine the temperature in something that can be modeled
as a semi-infinite solid, and a solid is considered semi-infinite if it extends to infinity in
all but one direction. In other words, there is what we call a single identifiable surface,
and if this surface should change because of a change in outside conditions, it will
undergo transient one-dimension conduction. So we're going to look at an example of this.
We're going to model a slab of steel, so we have our steel slab, it has a density of 7800
kilograms per meter cubed, its heat capacity is 480 joules per kilogram K, its thermal
conductivity is 50 watts per meter K. This slab is initially at 300 degrees C, so we
have water that is flowing over it at a temperature of 25 degrees C provides a constant surface
temperature. So our surface temperature is 25 degrees C. This is constant throughout
the cooling, and we want to know, after 500 seconds of cooling, what is the temperature
at 50 millimeters from the surface? So if you notice, we are looking at a problem that
looks at not only time, but also length. So the governing equation for a temperature as
a function of both x and t in a transient conductive procedure in a semi-infinite medium
is that T minus our T surface divided by T initial minus our T surface, and this is going
to equal the error function of x, which is our depth, divided by 2 times alpha, thermal
diffusivity, times t raised to the 1/2. So when deriving the exact solution for these
kinds of problems, the technique that is used is the error function, and there is tabulated
data such that if you have some w, you can then look up the corresponding error function.
So in our case, our w is x divided by 2 times alpha t to the 1/2, and when we calculate
that, we will have some error function. So that's the first thing that we should do.
Our x is 50 times 10 to the minus 3 meters, and we divide this by 2, however, we don't
have alpha, but alpha is just k divided by rho times C, or thermal conductivity divided
by the density times the heat capacity, times our time which is 500 seconds, and this is
to the 1/2. And so when we calculate that number, we come up with 0.3. And when we look
at the tables of error functions, we find that the error function of 0.3 is equal to
0.32863. Now all we have to do is put that back into our original equation. So this T
of x and t equals 0.32863 times our Ti minus T surface, 300 minus 25 degrees C, plus our
temperature at the surface, 25 degrees C, and that's going to equal 115.4 degrees C.
What if instead you wanted to find out how deep you would go at the same time, our t
= 500 seconds, but we want to know what the x is for a temperature equaling 90 degrees
C. So now the way we set it up is T, 90 degrees, minus T of the surface, 5 degrees C, divided
b the initial temperature minus T at the surface, and this equals 0.24, which is now going to
equal that error function of x times 2 alpha times t to the 1/2, and what we're trying
to solve for here is that x. So what we do is, we don't have our w, but we do have our
error function of w, which is 0.24, so when we look up our error function of w, of 0.24,
we can find a corresponding w which is 0.22, and that's going to equal x divided by 2 times
alpha t to the 1/2. So using the same alpha and t that we had before, our x is going to
equal 0.22 times 2 times 1.3 times 10 to the minus 5 meters squared per second times our
time, 500 seconds, and this is to the 1/2. So our x equals 0.35 meters, or 35 millimeters,
which makes sense because if it's a temperature at 90 degrees C, it shouldn't be as far down
as where we would find a temperature of 115 degrees C.