So let's consider a plane wall that has a length, and remember the length is in this
direction, it has a length of 100 millimeters, and it's convectively heated by a fluid that's
at T infinity at 750 degrees C, and it's heated on both sides, so it has the same temperature
and the same heat transfer coefficient.
So the initial temperature of the wall is 200 degrees C, and what's desired is that
the minimum temperature of the wall is no less than 500 degrees C, and so the plane
wall is heated for 10 minutes, which is 600 seconds, and for that time, or at the end
of the time, is this condition met?
And if so, what is the minimum time it would take, if not, how long would it take?
And so we're going to use the following properties.
The thermal conductivity of the wall is 25 watts per meter Kelvin, the density is equal
to 7500 kilograms per meter cubed, the heat capacity is 600 joules per kilogram Kelvin,
and finally, the convective heat transfer coefficient, and if you remember I said it's
the same on both sides of the wall, is equal to 150 watts per meter squared Kelvin.
So the first question is, if we need the minimum temperature to be no less than 500 degrees
C, where is this minimum temperature going to be?
And since the wall is being heated, the last place to be heated is the midplane, or where
x=0, and if you recall, if you have this kind of symmetric heating or cooling, we say that
the wall actually is 2L, and so when we're looking at the length in any one of our calculations,
it's actually 0.05 meters, or 50 millimeters.
So the first thing that we need to do is we need to calculate that Biot number.
The reason we do is if it is less than 0.1, then we can use lumped capacitance to calculate
what we need.
So the Biot number, which is h times L divided by k, and so when we calculate this, we find
that we have a Biot number of 0.3, and that means that we can't use lumped capacitance,
so we're going to need to use the exact solution in the form of an infinite series.
However, if our Fourier number is greater than 0.2, then we can use only the first term
of that infinite series, so we call that the one-term approximation.
Let's calculate that Fourier number.
So the Fourier number is what's known as dimensionless time, so it's equal to the thermal diffusivity
times the time divided by L squared.
And if you notice, we're not given the thermal diffusivity, so we're going to have to calculate
it.
So this alpha is equal to the thermal conductivity divided by the density times the heat capacity.
So our alpha is equal to 5.56 x 10^-6 meters squared per second, so our Fourier number
is equal to 1.333.
So this is clearly greater than 0.2, and we can use the one-term approximation.
In addition, since we're looking for the midplane temperature, in other words where x=0, or
x*=0, then the equation simplifies to the following.
So we're looking for that T0, so we're going to need to find both this constant and this
Eigenvalue, and so where do we find it?
We use a table that has the C1 and the eigenvalue as a function of Biot.
So we look at this table and find what corresponds to our Biot number equaling 0.3.
And so when we put our Fourier number and our C1 and our eigenvalue into the above equation,
what we end up with is that the temperature at the midplane is equal to 350 degrees C.
That is not the 500 that we need, so that means that we have to solve for the time that
it takes to get to 500.
Let's take a look at the same equation, but written in terms of time.
So where do we find the time?
Well, we find it from the Fourier number.
So we're going to have to solve for the Fourier number, then from that, find the time.
So when we write it out like this, we can solve for the Fourier number, which is equal
to 3.06, and again that's greater than 0.2, so we can still use this first-term or one-term
approximation, and from that, we get that the time is equal to 1376 seconds, or almost
23 minutes, which is quite a long time.
So how can we shorten this time?
Really, what we're going to have to work with is that heat transfer coefficient, or perhaps
changing the fluid or even having a higher bulk temperature.