JANET: In the previous screencast we discussed
what a thermal boundary layer is
and how we can use its properties
to determine a convective heat transfer coefficient.
So here we'll look at a thermal boundary
layer that forms on a flat plate and determine
not only that convective heat transfer coefficient
but also the temperature gradient
through the plate itself and at the surface.
So let's start by considering a flat plate,
it has a thickness and conduction that thickness
that it goes through we call L of zero point four meters.
And it has a thermal conductivity of 60 watts
per meter kelvin and we have a fluid here
that comes in, it's at a T infinity of 30 degrees C,
and it has a thermal conductivity
of zero point six watts per meter kelvin.
And then as soon as it hits the plate,
a boundary layer is going to form.
And in this case, we'll get a thermal boundary layer
because the temperature of the surface,
which we'll call, T-S one, the surface that
the fluid flows over, is equal to 50 degrees C.
In addition, we have convection through that plate,
because it has a length of zero point four meters,
and the temperature at the bottom of that plate
is equal to 150 degrees C.
So what we want to find, in this problem,
is net convective heat transfer coefficient.
We want to find the temperature gradient within the plate.
And we wanna find the temperature gradient at the surface.
And in this case, our Y is going in this direction,
therefore, this is going to be
the temperature gradient at Y equals zero.
So the first thing we'll do is use the energy balance
around the surface of the plate
to find our convective heat transfer coefficient.
And what that says is the energy that leaves that surface,
which is convection, and we'll use Newton's Law of Cooling,
to find the convection, is equal
to the conduction that enters the surface.
So since we have everything except for that H,
let's go ahead and solve for it.
And what we end up with is a convective heat transfer
coefficient of 750 watts per meter squared K.
The next thing that we want to find
is the temperature gradient through the plate.
And because the mode of heat transfer through the plate
is conduction, we can write that our temperature gradient
D-T-D-Y is equal to minus T two minus T one
divided by that thickness of the plate which is L.
And so this temperature gradient is equal
to minus 250 degrees C per meter.
If we look at the surface of the plate, however,
we have to use a different expression.
Here, our D-T-D-Y as we showed in that previous
screencast, is equal to minus our heat transfer coefficient
divided by the thermal conductivity of the fluid
'cause if you recall the conduction there at the surface
is through the fluid and this is going to be multiplied
by T-S one minus T infinity.
And if we put in the numbers from above,
that's equal to minus two point five
times ten to the fourth degrees Celsius per meter.
And so, see what we mean about boundary layers being
characterized by sometimes extreme temperature gradients.
So remember that here, at the surface,
that's where those gradients are going to be the greatest.