(Brief Silence)
- [Speaker] Welcome. This lesson is
on thermal boundary layers
and thermal boundary layers occur when there's
a heated surface and then there's flow past that surface.
And in particular we're gonna look at an example
of laminar flow over a flat plate that's heated
and there's a velocity V infinity far away from the plate
and also a temperature T infinity far away from the plate.
And then the plate is heated at temperature TS.
And we form both a momentum boundary layer
and a thermal boundary layer.
In the momentum boundary layer.
The velocity goes from zero due
to Nusselt slip on the surface
of the plate up to the infinity outside the boundary layer.
And there's both visus effects
and inertia important in the boundary layer.
And then in the thermal boundary layer
the temperature goes from T at ts at the surface down
to T infinity outside the boundary layer.
And both conduction and convection are important
in the thermal boundary layer, which has thickness Dell T.
Now an important parameter is the Reynolds number
and the Reynolds number has to be large
for there to be a boundary layer.
And it's, as you probably know, it's proportional
to V infinity times a characteristic length
which is the length
of the plate and divided by the kinematic viscosity
which is the sheer viscosity divided by the density.
Another important parameter is the PR number
which is the kinematic viscosity divided
by the thermal diffusivity.
Thermal diffusivity is the thermal connectivity
of the fluid divided by the density
of the fluid and the specific heat.
And the pre number is a little bit over one
for simple liquids like water and a little bit under one.
For gases, it's much larger than one.
If you have a very viscous fluid like an oil
or a polymer melt, it's very rarely much less than one.
The only exception would be a liquid metal
like liquid mercury.
Now another parameter of importance is the Nusselt number
and the Nusselt number is a dimensionless heat transfer
coefficient made dimensionless with the thermal connectivity
in the denominator and the link scale of the plate
and the numerator.
And that's our goal is to find the Nusselt number
because then we can find the area average heat transfer
coefficient H with a bar over it.
So we'll start by looking at the heat flux
from the surface Q double prime, which has units
will say watts for meter squared.
According to Fourier's law, it's the thermal conductivity
of the fluid and the fluid in particular because the heat
transfers through the fluid times the temperature gradient
and the temperature gradient.
The component we're looking at is in the Y direction
because that's normal to the plate and we're evaluating it
on the plate, which is at y equals zero.
Okay. And this can also be set equal to a heat
transfer coefficient times the temperature difference
between the plate and the ambient fluid.
And that heat transfer coefficient would be to be found
- And called Newton's law of cooling.
We'll start with a scaling estimate.
And so we're gonna estimate that temperature difference
or temperature derivative as the temperature difference
at the top of the boundary layer
minus the temperature on the plates divided
by the thickness of the boundary layer.
And that's an order
of magnitude estimate for that derivative.
And then we can plug that
into Fourier's law and we get an estimate
for the heat flux and it's inversely proportional
to the thickness of the boundary layer.
That also allows us to get an expression
for the heat transfer coefficient H
which would be proportional to the thermal boundary layer
the thermal connectivity divided
by the boundary layer thickness.
And you get that by plugging this result into here.
And what you see is is that since the
Since the boundary layer varies
with distance and generally increases
the heat transfer coefficient will decrease
with distance from the edge of the plate.
And therefore the heat transfer coefficient is
a local heat transfer coefficient.
And the Nusselt el number, which is based
on the area average heat transfer coefficient
we can see scales as the length divided
by the thickness of the boundary layer.
And this is often much greater
than one because the boundary layer is quite thin.
And that would mean that the convection is very important
and we have high heat transfer rates.
Now, in order to evaluate the Nusselt number
we need to know the thermal boundary layer thickness.
And to do so, we look at the thermal energy equation
and we'll assume constant properties and steady state
and no generation of heat in the fluid
and therefore we just have convection equals conduction.
We'll do a scaling of the convection terms
and we'll look at the first term of a scaling
for the velocity in the X direction is V infinity.
The temperature difference goes from Ts on the right hand
side after the fluid has been heated up to T infinity
on the left hand side where it comes in.
And that takes place over distance L in the X direction.
The convection in the Y direction is actually
a similar magnitude tell Y is smaller, but VY is
also smaller.
And then finally for conduction, the order of magnitude
for conduction would be the thermal diffusivity
times the temperature difference again, and then divided
by the link scale squared.
And here the link scales much smaller for conduction
across the boundary layer than along the boundary layer.
So the second term dominates and that link scale is delta T
and it's squared because we have a second derivative.
Now we can equate conduction and convection
in the thermal boundary layer.
And that gives rise to this expression here
which tells us the boundary layer thickness grows
with length to the one half power, so it grows
along the plate.
It also grows with increasing thermal diffusivity alpha
stronger heat conduction means the boundary layer will be
thicker and it's inversely proportional to the square root
of the velocity.
Stronger velocity means stronger convection
that would convect the heat away.
And this scaling
- Holds for panel numbers of order unity.
So simple liquids and gases.
There's a slightly different scaling
if the panel number is large
because then the boundary layer is thinner.
The thermal boundary layer is thin compared
to the momentum boundary layer
and the velocity in that thermal boundary layer is lower.
And there's a different scaling
I'll say a little bit more about that in a minute.
And then finally our dimensionless heat transfer rate
which is the Nusselt number scales as the length
of the plate divided by the thermo boundary layer thickness.
We can use this result
in that equation and that gives us the following expression
This quantity here
it looks a little bit like a Reynolds number
but it actually is called the P lane number.
And of course it's raised to the one half power
and it shows that heat transfer increases with velocity
increases with the length
of the plate and then decreases with thermal diffusivity.
That's actually the muscle number.
The actual heat transfer rate will increase
with thermal connectivity as you would see when
you calculate the heat transfer coefficient.
Now this, the ling number is equal
to the rentals number times the PR number.
And so this scaling gives us rentals number times PR number
the one half, and this is an approximation
or order a magnitude estimate.
If we do an exact analysis using a similarity solution
you'd get a solution of about oh 0.65 times Reynolds
to the one half panel than one half.
An empirical correlation is given here
which also is oh 0.65 rentals to the half.
But in this case
NEL numbers multiplied to the one third power.
And that's because the experiments were done
for NEL numbers near one and much greater than one.
And where we would have that different scaling
and the one third power gives the different scaling.
And then finally, once you have the Nusselt number
you can calculate the area average heat transfer coefficient
from the equation above.
It's just the Nusselt number times the thermal connectivity
divided by the length and then we're done.