This screencast is going to show you how to solve heat transfer problems for laminar,
fully-developed flow in circular tubes. First, what do we mean by laminar? Well, laminar
refers to the type of flow, and in this case it's a very orderly, highly organized flow.
It's determined by a Reynolds number, which is the ratio of the inertial forces to the
viscous forces. For internal flow, we can calculate the Reynolds number by 4 times the
mass flow rate divided by pi times the diameter of the tube times the viscosity. If the viscous
forces are larger, there are less disturbances in the flow, leading to a more ordered flow.
Depending on the type of flow and geometry, the transitional Reynolds number can change.
For internal flow in a circular tube, the transitional Reynolds number is 2300. So if
you have a flow that has a Reynolds number of less than 2300, you can characterize it
as laminar. How about fully-developed flow? Well, that's when the temperature profile
does not change along the length of the pipe. In order to calculate this length, how much
of the pipe is not fully developed, this is 0.05 times the diameter of the pipe, times
the Reynolds number, times the Prandtl number, and the Prandtl number is another one of these
dimensionless numbers that characterizes heat transfer. What we're looking for is our convective
heat transfer coefficient, and we get that from the Nusselt number, which is equal to
our h times the diameter, divided by the thermal conductivity of the fluid. For this particular
type of flow, we have two different Nusselt number correlations. If we have constant heat
flux along the tube, then the Nusselt number is simply 4.36. If we have constant surface
temperature along the tube, then our Nusselt number is 3.66. Let's take a look at an example.
We'll have a tube that has a length of 25 meters and a diameter of 15 millimeters. We're
going to assume a constant surface temperature of 100 degrees C, and we're going to have
a fluid that has a mass flow rate of 0.25 kilograms per second. It has a Prandtl number
of 1, a thermal conductivity of 0.138 watts per meter Kelvin, it's heat capacity is 1100
joules per kilogram Kelvin, the temperature of the fluid coming in is 40 degrees C, and
because viscosity changes with temperature, we're going to say that the viscosity coming
in is 20 times 10 to the minus 2 kilograms per meter second, and the viscosity coming
out is going to be equal to 1 times 10 to the minus 2 kilograms per meter second. Why
is it important that we consider a change in viscosity? Remember that it's that Reynolds
number that determines whether we have laminar or turbulent flow. The first thing we want
to figure out is, is this flow laminar? So going in, our Reynolds number is going to
equal.... so it's clearly laminar coming in. How about coming out? So the Reynolds number
is going to be exactly the same, only this viscosity is going to be less, and because
it's less, it's going to increase the Reynolds number, so the question is whether it increases
it greater than 2300. So when we do this calculation, we find, fortunately, that the Reynolds number
is 2123, so we know that the flow is laminar throughout the tube. The next question is,
is the flow fully developed? So we try to find out where along the length of the tube
will the flow become fully developed. Using the Reynolds number that we found for the
inlet, this x is 0.8 meters. Remember that the length of our pipe is 25 meters, so we
can assume for all intents and purposes that is is laminar, fully developed flow. Again,
constant surface temperature, so that means that our Nusselt number is going to be 3.66.
And so that will allow us to find our heat transfer coefficient, which is the Nusselt
number times the thermal conductivity of the fluid divided by the diameter, and we end
up with a convective heat transfer coefficient of 33.7 watts per meter squared K. We can
use this heat transfer coefficient in any number of equations, in particular if we want
to find the temperature of the fluid coming out, we would need to use this heat transfer
coefficient. How would we find that? Well, we would use the fact that our q can be written
as m C sub p delta T, Tmo minus Tmi, that's also going to equal h times A times our delta
T log mean. Our delta T log mean is equal to our delta T out, which is the difference
between the surface temperature and the temperature coming out, minus our delta T in, all divided
by the natural log of our delta T out divided by delta T in. We can rearrange this to solve
for our temperature out, which is going to equal our surface temperature, minus our surface
temperature minus Tm in, times the exponential of minus pi times the diameter, times the
length of the pipe, times that h that we just calculated, divided by our mass flow rate
times our heat capacity. And when we put the numbers into this equation, we find that the
mean temperature coming out of the tube is 48 degrees Celsius.