This screencast is going to go through an example of heat conduction in a cooling fin,
specifically a rectangular fin. When we think of a real-world example where this takes place,
the easiest thing that came to mind is that in the microprocessing business, if you open
up the back of a computer, you'll see that mainly in the microchips, they'll have some
kind of strange looking array of metal fins, and these fins serve a purpose of reducing
heat from the microprocessor, which can get very hot, and if it overloads, it can fry
out. So along with the fan and convection, these fins create more surface area for that
cooling to occur. And so, we can break this problem down and try to come up with something
that's kind of more realistic than the microprocessing business, here's some values that we could
pull from that. So I wanted to try and pull some real-world values to make this a somewhat
realistic problem. Looking at air temperature in the room, we can assume it's about 25 degrees
Celsius, and I looked up the value on some of the - as hot as these microprocessors can
get - and I've seen it upwards of 100 degrees Celsius. So these fins are mainly made of
aluminum, we could say that thermal conductivity of the fin would be about 160 watts per meter
Kelvin. We can look up the thermal conductivity of air, which is about 0.025 watts per meter
degree Kelvin. The heat transfer coefficient in this situation, we'll just come up with
80 watts per meter squared Kelvin. That may or may not be too realistic, but hopefully
in the ballpark. The length of the fin, about 5 centimeters, the width of the fin, about
the same, and then the thickness itself is a whole different order of magnitude, about
maybe 2 millimeters, maybe a little larger, but that seems pretty good. So the problem
in this case that we're posing is, how much heat can be removed from the microprocessor
using these kinds of fins with these properties? So first thing we want to do is start with
a diagram that we will fully label. Once we have our schematic, we're going to choose
an origin, and that origin here is going to be in the center of our fin, so that our x
direction will be down the face of the wall, our y direction will be along the other face
of the wall parallel to the edge of the fin, and our z direction will be down the fin.
So our wall temperature, Tw, will be at where x=0, y=0,z=0, but also for all of y and all
of x, so really we're just looking at z=0 in that case. Our air temperature we said
was 25 degrees, so then we need to set our - in this case our thickness - and we're going
to say it's 2b, and the reason we're going to say it's 2b is since we chose x in the
center of the fin, b will be our thickness in both directions from that origin, so that's
2b. We also need to label the width of our fin and the length of our fin, and we have
values for those, but we'll just label them W and L for now so we can set up our equations.
So we have our labeled schematic at this point, next thing to do is to think about what the
question is asking. We want to calculate the heat loss from the faces of the fin, and to
simplify this problem, we should probably list our assumptions. So our first assumption,
to make it easier to calculate, is that although we know temperature is going to be a function
of position, x, y, and z, it's predominantly going to be changing in the z direction, so
we can kind of disregard any changes in the x and y directions, just focus on z. The second
assumption that we're going to make is where the heat is coming out of the plate. We know
that it will leave on all of the edges, but because the dimensions of the thickness are
so much smaller than the other dimensions, we can say that the edges of the fin, the
2BW and 2BL are much smaller in area than the face of the fin, the width times the length,
so we're going to say that those are negligible in comparison. And then the third and final
assumption that we're going to make is, we know that the cooling is predominantly from
the surface to air, so we're going to go ahead and use Newton's Law of Cooling and say that,
although the heat coefficient may be a function of position, we're just going to say that
the heat flux in the z direction is going to be equal to a constant heat transfer coefficient,
h, times the difference in temperature, T, a function of z, minus T of air, so this will
help us figure out our total heat loss later when we multiply that flux by an area. So
now that we have our schematic fully labeled, we can move on to doing our balances. So we
take a small fraction that we want to do our balance over in the z direction, some length
delta z, and we want to look at the heat going in and out of that area, and so the first
thing we're going to do is the heat flux, qz, and multiply it by the area, 2 BW, and
the 2 is there because it's on both sides, and we're going to subtract out the same heat
loss at the z plus delta z position, and this would typically be zero, but we have to account
for the heat loss that is on the face of the plate to the air, and it's going to be related
to Newton's Law of Cooling, so that's just h times the area, which is 2W and it would
be L, but since we're doing the delta z, we need to account for the delta z area. and
then that's just times the temperature difference T minus T air, and then this is all going
to be equal to zero. So at this point now we have our energy balance, we're going to
just divide all sides by the 2BW delta z term, and as we take the limit as delta z approaches
zero, that's going to leave us with an equation that looks like this following. So we have
a first order differential of the heat flux in the z direction with respect to z. We've
got to insert our constitutive equation, Fourier's Law, which relates the heat flux to a thermal
conductivity and a temperature gradient, so when we plug this into the differential equation,
we're going to get a second order differential equation that looks like the following. We
have a second order differential equation, and to solve this, we would need two boundary
conditions. So the first boundary condition that we're going to use is just going to be
at z=0, so right in this area that I'm highlighting, that's going to be at a temperature equivalent
to the wall temperature, which is Tw, so that's our first boundary condition. Our second boundary
condition I'm going to label with highlighting on the end of the fin, so for z = L, the temperature
flux at that point is now zero. So at this point we can go ahead and integrate the second
order differential equation once, and in doing that we'll get dT/dz equals the following.
So we'll use our second boundary condition that for dT/dz = 0, so that we can solve for
C1, our constant in this case, and if we do so, that's going to end up equaling the following.
So we can plug C1 back into the equation, and we can integrate a second time so that
we get our equation relating temperature to z. When we do that, that's what it looks like,
and we can use our first boundary condition for z = 0 to cancel out the firs two big terms,
and that'll just mean that C2, our other constant, is just going to be equal to the wall temperature,
Tw. So now we have our temperature equation in the z direction. I'm going to put that
on a new page, and I kind of want to show you something when we look at this equation.
You can see that there's alot of different variables that have similar dimensions, and
so to make this equation simpler to use, and also simpler to solve, especially since we
have temperature on both sides, we're going to try to group some of those variables together
and use dimensionless quantities, and you'll see kind of later on in this video why that
becomes important, and how it will actually help us solve this problem. So let's take
temperature and make it dimensionless. We subtract out Ta from both sides and group
that together, all our temperature values now look somewhat similar. What we're going
to do now is divide the entire equation by Tw minus Ta, and what this is going to do
is make all our temperatures dimensionless, since we'll have a temperature over a temperature.
So on the left hand side, that's what we're going to get, and then I'll go ahead and fill
that in for the right hand side. So now we have all our temperatures divided by temperatures,
so we can go ahead and give this value a new character, and we're going to call this theta.
So theta is going to equal T minus Ta, over Tw minus Ta, and so we can plug this back
into the equation and you get the simplified version, which hopefully looks a little more
manageable at this point. We're going to go ahead on move on to our length terms, the
z and the L, and if we bunch these together such that it is z over L, or some fraction
down that fin, call that zeta, we'll plug that into our equation and we'll get something
that may or may not look as simplified, since we had to rearrange the terms a little bit.
But when we do so, we get the following equation. Now, if I pull the L squared out, we'd have
hL squared over kb on that left side. So we're going to go ahead and create that third dimensionless
quantity, and you can check all the units on those terms, and you'll see that it is
dimensionless. We're going to say that that is equal to N squared, and you'll see why
we do that later. So when we do that, we get a final dimensionless equation which will
relate temperature to z as the following. What you'll notice is that theta is on the
left side and the right side, and even dividing through that is still going to give us a fairly
complicated equation to solve. So, what may be easier in this case, and why we would want
to start with dimensionless quantities at the beginning, is if we go back to our second
order differential, d squared T over dz squared equals h/Bk times T minus Ta, if we would
go ahead and make this equation dimensionless, plug in these dimensionless variables into
that such that we get a d squared theta over d zeta squared, what would that look like?
And if we do that, it may be that it's alot easier to solve. So how do we transform the
left side to the right side? Well, we could do a little bit of chain rule manipulation
of the derivatives, so if we do the first order differential on following differentials,
you should see hopefully that these will cancel themselves out. So d theta over d T, multiply
that by the dT over dz, so that would cancel out our temperatures, and to cancel out z
it would be dz over d zeta. So at this point, if you look at that, we'll get that right
hand side of that second order differential in dimensionless quantities, and that's exactly
what we're shooting for. So now we have to figure out exactly what each of those differentials
are with respect to each other, then multiply that together. So, we start with the first
differential, the d theta over dT, we look at that first equation on the top left, and
it's pretty straightforward. It's just going to be 1 over Tw minus Ta. Second derivative,
the dT over dz, is a little bit more tricky and we have to go back to our original equation,
which I'm posting here. Our third derivative, dz over d zeta, is just going to be equal
to L. And it's L, not 1 over L, because it's the inverse of what you would think initially
by looking at it. So if we multiply these three equations together, then take the derivative
with respect to zeta, we should get what we're looking for. Although this looks like a mess,
we could take our dimensionless quantities on top and substitute that into that equation,
and so when we do that, we're going to get a differential with respect to zeta of our
differential of theta with respect to zeta. If we simply, it's just going to be theta
times our N squared, and this is all going to be multiplied by zeta minus 1. So when
we take that derivative with respect to zeta, it becomes pretty straightforward. The second
order differential of theta with respect to zeta squared is just going to be theta N squared,
so this is a fairly simple second order differential equation that we can solve, but to solve it,
we're going to need our boundary conditions also transferred into dimensionless quantities,
so let's go ahead and start with that. So our first boundary condition where z equals
zero, we have the temperature equaling the wall temperature Tw, so at this point now
instead of z = 0, we're going to say that zeta equals zero, because it's z over L, and
so if T equals Tw, then our theta is going to equal 1. Our second dimensionless quantity
was at z = L, we said dT/dz equals zero, so at this point now we're going to say at zeta
equals 1, our d theta/d zeta is going to equal zero. So if we rearrange our second order
differential into the following form, we look this up in a table or solve using ODE methods,
and what you're going to see is that this second order differential equation in this
form is going to to take on the following solution. Theta is going to equal constant
C1 times a hyperbolic cosine of N times zeta plus some constant C2 times a hyperbolic sine
of N times zeta. So now, at this point, we're going to use our boundary conditions to solve
for C1 and C2. So I'm just going to put up the hyperbolic cosine and hyperbolic sine
just to help remind everybody. It might help you solve this if you're doing this on your
own. So using our boundary condition theta equals 1, this is at zeta equals 0, so we
would plug in zeta equals 0, we're just going to see that C1 is equal to 1, which will simplify
our equation a little bit. For the second boundary condition, we have to do d theta
with respect to zeta, and that was equal to zero, so if we take the derivative of our
theta, it's going to take on the following form, and we're going to plug in zeta equals
1 and then solve for C2. So when we do this, C2 is just going to equal the following. So
when I simplify all this together, we get our solution for theta, so this is our dimensionless
quantity solution, but this doesn't answer our original question of the heat losses from
the fin. So we're going to create a new variable which is just going to be a ratio of the actual
heat loss in our fin compared to that in a theoretical fin, one that's simple to calculate,
say an isothermal fin, where it's just going to be related to the temperature difference
from the wall to the air. Now the reason we do this is that we can basically figure out
what the heat loss is in the isothermal case, multiply this by our effectiveness factor,
and we get an actual rate of loss. So our heat is just going to be related to the heat
flux times the area over which that is occurring, so if we plug this back into our equation,
the way this is going to end up looking is over the entire fin, we're going to have to
integrate it. So it's going to be from 0 to the width of the fin, integrate 0 to the length
of the fin, then it's just related to that Newton's Law of Cooling. I listed that before
in our assumptions, qz which is just equal to h times the difference in temperatures.
So if we plug that back in, it's just going to be T minus Ta, we're going to integrate
this in the y and z directions. So this is our actual rate of loss, and our isothermal
case is going to look very similar. It's just going to be 0 to W, 0 to L, but in this case,
it's just going to be Tw minus Ta, dzdy, and if we plug in what our dimensionless quantities
look like, and our solution from above, we have our effectiveness factor basically equaling
0 to 1 of theta d zeta, and this is just going to be 0 to 1 d zeta. When we solve this, our
solution is going to be the hyperbolic tan of N over N. So now what we can do is go back
to our heat loss problem in the beginning. We can summarize all the information we have,
and remember that our heat is coming from the sides of these fins predominantly, and
we related that heat to the heat flux times area. In an idealized case, our heat is going
to be h times Tw minus Ta, this is where the fin is isothermal, we just mentioned that
effectiveness factor calculation. So we have these values - h, Tw and Ta - and we can plug
this in and solve for q. We also know the area and the area is just going to be equal
to the width of the fin times the length of the fin times 2, since it's happening on both
the top and the bottom, so area is equal to 2 LW. So we plug all this in and we find our
q value, then our actual heat loss is then just going to be multiplied by the effectiveness
factor, which would get rid of the denominator that we set up initially. So when we simplify
this out and plug in some of the values, hopefully this is what you get in terms of numbers.
Then when we simplify this out, you should end up getting about 15000 watts, or otherwise
15 kilowatts. So I can't say for certain that this is a realistic value, it would be interesting
to compare to actual fins using the microprocessor industry, but hopefully this kind of setup
and these estimations for numbers gives you an idea of how to set up this kind of heat-energy
problem and solve for a rectangular fin.