Film Pool Boiling Video Lecture | Heat Transfer - Mechanical Engineering

In an earlier screencast, we looked at the
regime of pool boiling known as nucleate.
In this screencast, we're going to look at
the regime known as film pool boiling. Here's
where a vapor film covers the surface of the
liquid, and radiation becomes significant
so we're going to need to include the heat
transfer coefficient for radiation in any
calculations for our heat transfer. We're
going to use correlations that are similar
to those of film condensation. For example,
the Nusselt number, which allows us to find
the convective heat transfer coefficient,
can be determined using the following correlation.
The C, right here, is a constant and its equal
to 0.62 if you have a cylinder, and 0.67 if
you're looking at a sphere. The corrected
latent heat accounts for the sensible heat
that keeps the temperatures in the vapor film
about T sat, where it's calculated using the
heat of vaporization. So the vapor properties
that are contained in there, we look at the
film temperature, which is just the average
of the surface temperature and the saturated
temperature, and we look up the liquid density
at the saturated temperature. Eventually,
our goal is going to be to use Newton's Law
of Cooling to find our variable of interest.
Certainly finding the Nusselt number and solving
for the convective heat transfer coefficient
is tedious, but not necessarily difficult.
Which makes this situation more challenging
is that, as our surface temperature gets greater
than 300, radiation needs to be included,
so we need a radiative heat transfer coefficient.
And we're going to use the expression that
h radiation is equal to epsilon, which is
the emissivity, the Stefan-Boltzmann Constant,
Ts to the fourth, minus T sat to the fourth,
so that's the surface temperature and the
saturation temperature, where these temperatures
have to be in Kelvin, divided by this difference
between the surface temperature and the saturation
temperature. So the best way to show how to
calculate film pool boiling is by using an
example. So let's say we have a sphere. It
has a surface temperature of 400 degrees C,
and this sphere is immersed in a water bath
that is 100 degrees C, and we want to know,
what is the heat transfer rate from the sphere.
Here's our plan. First, we're going to look
up the properties. Then we're going to calculate,
because we're going to need it for our Nusselt
Number, our corrected latent heat. We're going
to calculate out Nusselt Number, which will
give us our convective heat transfer coefficient,
then we're going to calculate our radiation
heat transfer coefficient. These, by the way,
are both average heat transfer coefficients.
Then we're going to find our heat transfer
coefficient that uses both radiation and convection,
and finally, we're going to find our q using
Newton's Law of Cooling. So let's start by
looking up the properties. One thing to be
really careful of is units, because this is
given to us in kilojoules per kilogram, and
everything else is in joules per kilogram,
so this is actually going to equal 2257 times
10^3 joules per kilogram, and this is going
to equal 2733 times 10^3 joules per kilogram.
So when we put that and everything else in
our expression for the Nusselt Number, we
come out with a value of 1.071. So now let's
use that number to calculate our convective
heat transfer coefficient. Our heat transfer
coefficient for convection is equal to our
Nusselt Number, times the thermal conductivity
of the vapor, divided by the diameter of our
sphere. And so we're left with a convective
heat transfer coefficient of 36.3 watts per
meter squared K. Now we have to calculate
our radiation heat transfer coefficient. That's
equal to 14.05 watts per meter squared K.
Now, if h convection is greater than that
of radiation, we can find our heat transfer
coefficient from the expression h convection
plus 3/4 h radiation, which is equal to 46.8
watts per meter squared K. And now finally,
we can find our q, which is equal to that
h, 46.8 watts per meter squared K, times our
area, which is 4 pi r^2, times our change
in temperature, and that's equal to 4.41 watts.