In this screencast, we're going to look at a composite wall, which is a wall that is
made up of more than one material, and in this case, we'll call this material A and
this material B. Whenever you have a composite wall, there's usually an interface between
the materials, and this causes a resistance that creates a temperature drop.This is known
as a thermal contact resistance, and it's caused by surface roughness and a combination
of gaps and contact spots.
So here we're going to look at this composite wall and determine what that contact resistance
is.
And we will look at convective heat transfer.
So we have a bulk fluid outside the wall, a wall made of two separate materials with
a gap between them, and finally a bulk fluid on the other side of the wall.
And we're going to use the equation that the flux going through a planar wall is going
to equal the change in temperature between those two bulk fluids divided by the total
resistances between the two temperatures.
So let's say we're given a flux, and this is heat transfer per unit area, that's 20
watts per meter squared, and we're given the following properties.
In order to find our contact resistance, we have to find the total resistances and then
use a thermal circuit to determine what that contact resistance is.
So our total resistance is going to be equal to our change in temperature divided by our
flux, which is 250 degrees C minus 50 degrees C, divided by our 20 watts per meter squared,
and that's going to equal 10 meters squared kelvin, because remember when we're talking
about a temperature difference, 1 kelvin is equal to 1 degree Celsius, divided by watts.
So that's our total resistance.
Now let's divide it up into its different parts.
So let's draw a thermal circuit to see exactly where these resistances are.
So we start with our outside temperature, and the first resistance that the flux goes
through is a convective resistance.
Then the second one is this conductive resistance, then we have another resistance between the
two materials, which is this contact resistance which we are looking for.
Our next resistance is our second conductive resistance, then we have our final convective
resistance, and our T infinity 2.
Notice that every one of these nodes indicates a temperature.
So as we go through this composite wall, the temperature is going to drop accordingly as
you go through a resistance.
Our 10.0 meters squared K per watt is going to equal the sum of these resistances.
And we have the information so that we can solve for every single one of these resistances
except for this final thermal contact resistance.
And if we do the math, we find that this thermal contact resistance is 2.1 meters squared kelvin
per watt.
Now let's take a look and see what would happen to the flux if we did not include that contact
resistance, but had the same temperatures.
So we're going to use the same equation that q is equal to our change in temperature, 250
degrees C minus 50 degrees C, over our total resistance, but here, our total resistance
is going to to equal our 10 minus our 2.1, or 7.9, meters squared kelvin per watt.
And when we calculate our flux, we find that it's equal to 25.3 watts per meter squared.
So not having, or not including, that contact resistance, gives us a higher flux, because
we don't take into account the temperature drop that's caused by it.