In this problem, we're going to consider a spherical container.
It has an inside diameter of 1 meter, it has a stainless steel wall here, and so the wall
has a thermal conductivity of 17 watts per meter K. It's exposed to air, so there's a
T infinity, at 25 degrees C with a convective heat transfer coefficient equaling 6 watts
per meter squared K. And what we want to do is find the thickness of this wall such that
the inside temperature, so our T inside, is equal to 50 degrees C, if the heat loss is
equal to 500 watts.
So we're going to start with the governing equation for 1 dimensional steady-state conduction
with no heat generation, and that's that q_r is equal to the change for our delta T divided
by the sum of the resistances between those two temperatures.
And in this case, we're looking at the inside wall and the outside bulk temperature, so
our delta T is going to be 25 degrees C, or we can also call it 25 K.
So we have two thermal resistances, we have a conductive resistance, and for a sphere,
that's equal to 1 over 4 pi k, where k is the thermal conductivity, times 1 over r1
minus 1 over r2.
Our convective heat transfer resistance is going to be equal to 1 over our h, our convective
heat transfer coefficient, times 4 pi, and now here the outside r2 squared.
Our q_r is going to be equal to that delta T, 25 Kelvin, divided by the sum of these
resistances, so first we're going to have the conductive one, and what we add to that
is our convective resistance, and this has to equal 500 watts.
So let's simplify this some.
So you're going to need to put this in a solver to find r2.
You should get two possibilities, but remember, the number cannot be negative and it has to
be greater than 0.5, and the reason for that, we get an r2 that's equal to 0.517 meters.
this is the outside radius, but we need to subtract from that that inside radius in order
to get the thickness of the wall.
In this case, the thickness of the wall is going to be 17 millimeters.
Let's take a look when we solve for the two resistances separately.
So when we put in the numbers for our conductive resistance, we end up with 3.08 x 10^-4 kelvin
per watt.
When we solve for our convective resistance, we come up with 0.049 kelvin per watt.
Clearly the convective resistance dominates in the scenario.
In another screencast, I'll show you the effect of insulation on the temperature of the inside
wall.