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.
