So we're going to consider a plane composite wall composed of three materials with different
thermal conductivites. We're given the thicknesses of the three sections, as well as contact
resistances that exist at the interface between A and B, as well as the interface between
B and C. The left face of the wall is insulated. The right side is exposed to convective conditions
such that, given our h and our T infinity, which is 20 degrees C. Thermal energy is generated
within material A at a rate of 5000 watts per meter cubed. So we are asked to determine
the maximum temperature within this composite wall. So let's take a look here at our picture.
It gives us all the information, and there we go. And if you recall, the equation for
one-dimensional, steady-state temperature is given to us by this equation. And so, if
we solve for T, we find that T(x) equals minus q dot/2k * x squared plus C1x plus C2.
Again, this temperature distribution is always for this type of situation. The only reason it would
change is because C1 and C2 will change depending on the boundary conditions. So the first thing
we do as we look at the boundary between the heat generation, which is occurring in A,
and the beginning of our conductive part of the heat transfer, which starts at B. So we
can find the flux at that point has to equal the heat generation times the length of that
wall. So that equals 5000 watts per meter cubed times 20 millimeters, which is La, which
equals 100 watts per meter squared. Now we can start our thermal circuit where our flux,
100 watts per meter squared, stays constant. Remember that you cannot include generation
within our circuit. So let's start with Ta, which occurs at x=La, and the first resistance
it encounters is our thermal contact resistance. Then we have our next temperature that we
can call Ta2, and this is the temperature that also occurs at x=La because of the contact
resistance between (the interface between) A and B. Now we have conduction through B,
so our resistance is Lb over Kb, and now we have our next temperature, which is at x = La+Lb.
Again, we have another contact resistance between B and C, and we have our temperature
after that contact resistance, which is also at x = La+Lb. Now we have conduction again
through wall C, so that's Lc over Kc, and when we're finished going through C, we're
at the surface, so that's our surface temperature. And finally, we have our convective resistance
and our final temperature, T infinity. So if you can recall from a thermal circuit,
this q'' stays constant throughout. So you can find any temperature as long as you add
up the resistances between the two temperatures, so that's what we're going to do here. First
we'll find R'' total, and that involves adding the Rtc, the Lb over Kb, the next Rtc, the
Lc over Kc, and 1 over h. So if you put those all together, you come out with 0.26 meters
squared K per watt. So our q'' is Ta minus T infinity over our R'' total, and what we're
really looking for is Ta. We have q'', we have our T infinity, which is 20 degrees,
and we have our R'' total. So that allows us to find Ta, which equals 46 degrees C.
So why are we interested in Ta? Well, we need to find the temperature at T0, and we're going
to need Ta, as you'll see, to be able to solve it. By the way, we could have solved for any
temperature along this thermal circuit. So let's take a look at what we do. So we're
looking for T at x = 0, so we go back to our temperature distribution, which is minus q/2k
times x squared plus C1x plus C2. So we know the temperature at x = La, so T at La, which
equals 46 degrees C, equals minus q/2k times L squared, plus C1 times L, plus C2. So that's
our first boundary condition, that's a constant surface temperature of 46 degrees C. Now let's
look at T=0, where we know because its insulated that dT/dx also equals zero, which equals
minus q/2k *x plus C1. Since x equals zero, C1 has to equal zero. Therefore, our T at
0 is going to equal minus q over 2k, x squared,
plus C1x plus C2. Since x equals zero, our T at zero is just going to equal C2. So how
to we find C2? We go back up to here, remembering that that's zero, and we find that C2 equals
q dot over 2Ka times La squared, because remember we looked at this temperature at La, plus
Ta, which equals T at zero. Therefore, T of zero equals 46
degrees C plus our 5000 watts per meter cubed, times .02 meters squared, divided by 2 times
0.24 watts per meter K, and that equals 50.2 degrees C.