Earlier we looked at a few correlations to find the Nusselt number for radial systems
in crossflow, ultimately finding h and our convective heat transfer.
In this series of screencasts we're going to look at the three different correlations
for the same problem, and to determine how much of a difference there is between the
three answers.
So the first one we're going to look at is what's known as the Hilpert correlation, as
shown.
Here we determine our Nusselt number by using two constants, the C and the m that we look
up using the Reynolds number, and then calculate it, multiplying those by the Prandtl number
to the one third.
Let's consider a thin wire, and it has a diameter that's going to be equal to 0.4 millimeters,
and heat from the wire is dissipated to air flowing around it by convection.
So the velocity of the air is going to be equal to 75 meters per second, the temperature
of the air is 25 degrees C, the heat transfer from the wire per unit length is equal 25
Watts per meter, and what we want to know is what is the surface temperature of the
wire for those conditions.
So our governing equation, and how we're going to find that surface temperature, is that
that q' is equal to our convective heat transfer coefficient times 2 pi r, times T(surface)
minus T(infinity).
The hard part of this problem is going to be calculating that h(bar).
So we'll start with our correlation for the Nusselt number, and in order to do that we're
going to have to calculate the Reynolds number.
So the Reynolds number is equal to the velocity times the diameter divided by the kinematic
viscosity of the air surrounding it.
So it looks like we're going to have to look up some properties, this kinematic viscosity
as well as the Prandtl number and the thermal conductivity of the air.
So for this correlation we look up our properties, which is at the film temperature, which is
the average of the two temperatures in Kelvin.
However, we don't have the surface temperature, so let's make a guess at it, and I am going
to guess that this surface temperature is equal to about 70 degrees C, and so what that
means is that we look up our properties at 320 Kelvin, and if we do that what we find
is.
So let's put those numbers into our Reynolds number, find our constants, C and m, and then
calculate the Nusselt number, and that gives us a Reynolds number of 2632, which leads
us to the constants, such that C is going to equal to 0.683, and our m is going to equal
0.466.
So when we go back, and let's look at that correlation, when we put those C's and our
Reynolds number, as well as our Prandtl number in let's see what we get for that Nusselt
number.
So our Nusselt number, and remember that the bar over the Nusselt number means that it's
an average Nusselt number, and our h also is an average heat transfer coefficient.
So out Nusselt number is equal to 2384, and from that we can find our average convective
heat transfer coefficient.
So we get an h of 1668.8 Watts per meter squared Kelvin.
So now let's put that into out equation for the surface temperature.
So our surface temperature is equal to our q' divided by h times pi times D added to
our bulk temperature, and that's equal to 36.9 degrees C. So you can see that's pretty
far off from our original guess of a surface temperature at 70 degrees C, so what we're
going to do is we're going to reevaluate it such that the surface temperature is about
40 degrees C, and therefore our film temperature is going to be 304 K.
So let's see what the properties look like at that temperature, and from that we get
a Reynolds number that's slightly smaller, 1887, however our constants C and m will stay
the same because they're in the same range, and it gives us a Nusselt number of 20.4,
which then gives us an h(bar) of 1351 Watts per meters squared K, and our surface temperature
is equal to 39.7.
So using the properties at 36.9 degrees C gives us the surface temperature very close
to where we guessed.
In the next screencast we're going to look at the Zukauskas correlation, which is very
similar to this one, but slightly more comprehensive.