A semi-infinite solid is one that extends to infinity except for one identifiable surface.
So a sudden change of conditions at the surface will produce one-dimensional transient conduction.
So there are three different cases of a semi-infinite solid, you have one with constant surface
temperature, case one, case two is constant surface heat flux, and case three is surface
convection.
Note that each one makes use of the error function.
This was obtained by using a similarity variable, where the heat equation is transferred from
a partial differential equation to an ordinary differential equation, and then solved.
We're going to look at an example of a semi-infinite solid, we're going to determine which of the
three cases applies, and then solve for the requested variable.
We're going to start with a thick wall, and by the way when you hear that term or read
that term that usually implies a semi-infinite medium.
This wall has a density of 600 kg/m^3, it has a heat capacity of 2400 J/kg*K, and a
thermal conductivity of 0.2 W/m*K.
By the way, when you see these three properties being given in a transient conduction problem
you should be thinking thermal diffusivity, or alpha.
So this wall has an initial temperature of 40 degrees C, and it's suddenly exposed to
hot gasses at 750 degrees C, and these gasses have a convective heat transfer coefficient
of 25 W/m^2*K, and the question is what is the time at which the surface is equal to
325 degrees C. Now one of the things you have to determine is which direction are we going,
what is our x?
And our x, this top surface is that identifiable surface, so our x would be in this direction
and we call it the depth into the solid.
So here, if we take a look, our x is going to equal 0.
So let's go back to these different cases, and hopefully you can see, especially since
we're looking for the surface temperature after a certain time.
that it's not constant surface temperature, that this is case three, which is surface
convection.
We're looking for the time, and this looks particularly nasty as t is involved in every
single term, so one of the things I do want to point out we talk about the error function,
note that this says erfc, and that's what's known as the complementary error function,
which is just equal to 1-erf.
Let's see if there's another way to solve it rather than solving it this way, and by
the way it can be solved putting it into some kind of program, or something, but let's see
if there is an easier way.
And the answer is yes, what we have for this particular case is a graph that will allow
us to solve for time.
Note that they've shifted the axis, so we had the x-variable going down, they have it
going to the side, either way we can still use this graph.
So if you notice it has this dimensionless temperature versus this x divided 2 times
the square root of the thermal diffusivity times time, and as I said earlier, as you
can see, x is equal to 0, so what we're doing is looking a long this axis.
So if we can solve for that dimensionless temperature we can get this h times the square
root of alpha t divided by k.
We have our h, we have our k, as I mentioned before we can calculate our thermal diffusivity,
and from that we can find the time.
So the first thing we're going to do is calculate that dimensionless temperature.
Now here you need to be a little bit careful because what we've seen before is usually
t minus t(infinity), t(i) minus t(infinity), when you're working with a semi-infinite media
our dimensionless temperature is written this way, and that's going to give us a value of
0.4.
So let's go back to our chart and find where 0.4 is, how handy is that, and what we find
then in this line, that our h times the square root of alpha times t over k is equal to 0.5,
and so now we can use that information to solve for our time, so again the first thing
we're going to want to do is calculate that thermal diffusivity, and so this is equal
to our k divided by our density times the heat capacity, equaling 1.39*10^-7 m^2/s.
So now we have our alpha, let's do our calculation.
So when we solve for our time we end up with 115.2 seconds, and by the way when I put the
time that I got back into this temperature distribution I didn't get the exact same answer,
but I got one very close, and part of that is from being careful about reading the chart,
as well as rounding for my thermal diffusivity.