Let's consider a constriction that looks like
this.
There is an inviscid fluid flowing from the
left.
It flow through this constriction and it exits
to this widened area off to the right.
If we were to examine the stream lines of
the air through this we would see horizontal
stream lines on the left in parallel and horizontal
stream lines exiting to the right.
In the middle what we would observe is the
fluid flowing on the sides is constricted
in this region.
It flows together through this narrow part
and then expands on the down stream edge of
it.
As you can imagine this constriction causes
the speed of the air to increase as it flows
from left to right essentially taking all
of the air and cramming it into a small surface
area.
What I want you to think about is how the
pressure at this constriction compares to
the upstream pressure and the down stream
pressure at the wider areas.
I want you to think about which pressure might
be greater than the other pressure.
To visualize this I want you to consider an
identical geometry in which we drill two holes
and connected those holes by this U-shaped
pipe.
The bottom of this pipe is filled with water.
At the moment there is no air flowing through
it and the water seeks its own level so the
level on the left side is equal to the level
on the right side.
I want you to think about what would happen
when we start up the flow of air.
We have air flowing through it in these constricting
stream lines there is a relatively fast velocity
here and a relatively slow velocity on either
side of it.
I want you to take a moment and think about
if you were to draw the position of the water
down here after the air has reach steady state.
What would it look like?
Would you expect that the liquid level on
the right side at the constriction is lower
than the liquid level on the left side?
Would you expect that the level on the left
side is lower than that on the right side.
Make you feel differently but what I would
expect when I look at this I think of all
of the fluid flowing through here.
I guess at a glance that the fluid flowing
on the right should be pushing down more on
the water than the fluid on the left.
I would expect that the liquid level on the
right would be lower than the liquid level
on the left side.
What is fascinating to me is that it is actually
the opposite that happens.
There is a higher pressure here than there
is here and the fluid flowing through it actually
sucks up the liquid a little bit.
The high pressure here pushes down on the
water and that causes the liquid level on
the right side to actually be higher than
the liquid level on the left.
To examine this let's consider 2 points.
Point 1 is at the wide entrance and point
2 is at the constriction.
As it turns out if we pick two points on the
same stream line.
Here I am picking point 1 and eventually point
1 would flow into point 2 so we can connect
it with an arrow.
If we pick two points along a stream line
and we will make three major assumptions.
We assume that the fluid is inviscid so we
have something like air and the viscosity
is essentially zero.
We assume that we are at steady state.
Finally we assume that the flow is incompressible
such that the density does not change between
point 1 and point 2.
We can apply an equation known as Bernoullis
equation.
This equation is simply a conservation of
energy where it relates the pressure or flow
energy to the kinetic energy and potential
energy between points 1 and 2.
It says that the sum of these three forms
of energy have to equal the sum of the three
forms of energy at point 2.
If we examine the energy between points 1
and 2 we will find that the potential energy
between those two points is the same because
they are at the same level.
What we find is that the velocity at point
2 is greater because it is flowing through
this constriction.
What happens is the fluid gains kinetic energy
at the expense of pressure or flow energies
as it flows through.
As a consequence of this equation and the
three assumptions that we are making and the
fact that the velocity increases it results
in this decrease in pressure.
Let's demonstrate the use of this equation
by working a quick example problem in which
the pressure is measured at point 1 and point
2.
The pressure at point 1 is 10 kPa greater
than the pressure at point 2.
We want to know what the speed is at point
2 if the cross sectional area at this point
is 1/5 the cross sectional area of point 1.
Let's make some pretty big assumptions and
to apply this equation let's see if they check.
The first assumption that we need to make
is that the fluid is inviscid.
We are dealing with air and the viscosity
of air is relatively low.
We are probably dealing with some pretty big
velocities so the assumption that the fluid
is inviscid is probably pretty good.
The next major assumption that we are making
is that the flow is incompressible.
We have a pressure drop of 10 kPa.
Provided the absolute pressure is big enough
a pressure change of 10 kPa will not be big
enough to change the density or the volume
of air very much as it flows through the channel.
Then the last assumption that we will make
is that the flow is steady which is reasonable
in this case.
Nothing is changing with respect to time.
Since those assumptions check out we can justify
the use of Bernoullis equation and what we
need to find out in this problem is how much
the velocity increases between points 1 and
points 2.
Because the fluid is incompressible the volumetric
flow rate of air at point 1 is equal to the
volumetric flow rate of air at point 2.
We can write that the velocity at point 2
multiplied by its cross sectional area is
equal to the velocity at point 1 times its
cross sectional area.
We can solve for the cross sectional area
A2 is equal to 1/5 of A1.
With these two equations we can deduce that
the velocity at point 1 is 1/5 that of the
the velocity at point 2.
The speed at point 2 is 5 times greater than
the speed at point 1.
If I make this substitution I have that P1
plus 1/2times rho times (1/5*V2)^2 is equal
to P2 plus 1/2 times rho times V2 squared.
When I solve for V2 I get the pressure difference
divided by 1/2 minus 1/50 times rho square
rooted.
If i plug numbers in with units what I come
up with is 10,000 Pa divided by (1/2-1/50)
times rho.
Rho is 1 kg/m^3.
I did not specify the density in the problem
statement but 1 kg/m^3 is reasonable for air.
Then take the square root of the whole thing.
What I should come up with is something in
terms of m/s.
If I examine it I was careful to multiply
10 kPa by 1000.
A Pa is equal to 1 N/m^2 it is also equal
to 1 kg/(m*s^2).
In this square root I have kg/(m*s^2) and
I will invert the denominator so I have m^3/kg.
The kg cancel out and i am left with m^2/s^2.
Taking the square root of that leaves me with
m/s as we would expect.
Making the calculation I am left with to two
significant figures 140 m/s.
