In this problem, we're looking at a pipe
which has a contraction in it it, has a
large diameter at the inlet and a small
diameter at the outlet. What we know
about the pipe is the inlet area A1, 2
meter squared, the velocity inlet is v1,
3 meters per second, and at the outlet we
have a smaller diameter which results in
an area 2 of 1 meter squared. And we're
looking for three things in this problem.
One is the mass flow that's going
through the pipe. Two is the outlet
velocity, and  Three is the change in pressure
across the pipe between here and there.
So let's have a look at how to solve
this problem. And I'm going to put this
to the side so we can keep an eye on the thing. And let's work through the
problem. The first question is answered
relatively easily. The mass flow is the
density times the area times the
velocity, and we can apply this anywhere
inside the pipe, but we choose to apply
it where we know the most information,
which is at the inlet, 1,  let's put
a 1 over here and the 2 over there so we
have a clear idea about what we have. A1
and V1 belong to one so we have A1 and
V1, so let's take Rho1 A1 and V1
like so. And it's very easy, water has a
density of a thousand kilograms per
meter cubed, so 10 to the power 3.
the area 1 is given here as 2, and the
velocity 1 is given here as 3, yes. So
this gives us 6 times 10 to the power 3.
6 times 10 to the power 3. And these are
kilograms
every second, kilograms per second like
so, yes? I could also express this in
tonnes, there's a lot of, it's a lot of
mass per second, and this is 6 tons per
second. So about the weight of 6 cars
every second passing through the inlet
at this point. Now, this water is
coming in, six times per second are
coming in at the inlet here, and are
travelling through the pipe, and by the
time they exit they're still here, no mass
is created, no mass is destroyed, and so
since the area 2 is smaller,
the flow has got to exit quicker. It has
got increased velocity and this is what
we’re after. What is the velocity at the
outlet? Well the mass flow has not
changed between the inlet and the outlet. mdot here through the system steady
flow
is Rho 1 a1 v1 but it's also equal
to Rho 2 V 2 a 2 or rho 2 a2 v2, as you
prefer.
So those two things are equal. We know
that quantity here, and at the outlet
here; we know A2, we know the density has
not changed, it's still water, so there’s
a thousand kilograms per meter cube,
and so using this we can find V2. Now we
can just put the
number for m dot which we had here,
and try to get V2 by dividing m dot by
Rho2 and A2, but a nicer, more elegant
way is to just re-express V2 as a
function of V1.
so I write V2 is equal to V1 like so, V2 is equal to V1, and what do I have to
put on the side, I put A1 divided by A2, and I have Rho1
divided by Rho2. Then of course the two densities
are the same, it is still water, water flows
in incompressible way. So we still
have your v1 a1 / a2 like so and then I
can just, say, put numbers now, V1 is 3
meters per second so this is 3 times, A1
is 2 and A2 is 1 here, so I have 6,
and immediately I write units, and I have
to remember what I was calculating was a
velocity, it's gotta be 6 meters per
second. V2 like so. Now pretty much every
time you calculate the velocity using
the mass conservation equation, you have
to ask yourself: is this true or not? Is
this realistic or not? Because if you
look at the velocity of the water, it
changed quite a lot. You come in at 3
meters per second and you exit
at six meters per second. The velocity
has doubled. The water has got more
kinetic energy, and immediately comes the
question: Who pays? Who pays for this
increase in kinetic energy, yeah?
So every time we calculate the velocity,
six meters per second, you have to ask
yourself is this true, is this realistic,
and who, where does this energy come from?
Who is paying for this increase in
kinetic energy? And this, we answer with
the question "what is the change in
pressure" here, and let me show you how to
calculate the change in pressure. Well we
start with the energy balance equation.
We could start with a Bernoulli equation,
I don't like the Bernoulli equation too
much, I'll show you why in a different
example, so we'll start with the energy
balance equation. And the energy balance
equation is written like so. It says that
the net power as heat provided to the
pipe plus the net power as work provided
to the pipe, so this is equal to the
change of energy of the flow, and this
change of energy we can write it like so:
take the mass flow here, and this is now
all the properties at the inlet. And
there are different forms of energy that
the flow can have. One is the internal
energy how much heat it has inside,
inside itself, it's gonna be i1. In
thermodynamics we call it u, but in fluid
mechanics we prefer i because u is used for
velocity. So i1 plus then the pressure
p1 divided by the density Rho1, yes. And
I take very much care to always
differentiate the way I write pressure
and density because I don't want to goof
up and mix them up, yes. Plus then
one-half of the velocity square, this
would be the kinetic energy, and this is
the thing that's going to increase now,
yes?
And then this g z one, the
altitude potential energy. And
this is what's coming in. And what’s
coming in usually is written with a
negative term like so and then what's
leaving is here. And it's the same mass
flow here multiplied by the same terms
but with index 2. This is i2 plus P 2 over Rho 2 plus 1/2
of V 2 squared plus g z 2 like so. All
right.
So it's a long equation it's longer than
the Bernoulli equation, but it's a very
safe equation because everything is in
there.
This works every time you have a steady
flow and every time you have fixed
control volume, so as long as your pipe
is not inflating or deflating, yeah? And
as long as you take many pictures of the
flow and they remain the same this
equation will hold. The Bernoulli
equation has many more constraints added
to it, and they're hard to remember, and
they're hard to work with, so I
always prefer spending 20 seconds more
writing the whole equation, and then
crossing out terms that cancel out,
rather than being too quick and then
risking making big mistakes by applying
an equation that doesn’t apply. So in here,
what is applying, and what is not
applying? Well what is the heat transfer
added or subtracted to the flow Q net
here? This is zero. It's not heating up,
it's not cooling down, the same thing
goes for the power, there is no here,
there is no input of a propeller or an
impeller, there is no extraction
through a turbine or any moving part
inside, it's just a pipe, so the fluid is
just left to itself to exchange energy
that it possesses already, and so these
terms are over here. Let's start
with the easy part,
the altitude does not change, this is a
horizontal pipe so this goes to 0, both
of those cancel out. An then I’m left with
velocity, pressure, and internal energy.
Internal energy when the flow is perfect
and when there is no friction and the
pipe has very smooth corners,
this internal energy will not change.
This mean the fluid will not convert
friction work as increase in internal
energy, or increase in temperature.
So as long as the flow is perfectly
smooth, which is the case in this example,
the i will not change. And I'm left with
figuring out now, we can see
where the increase in kinetic energy
comes from! V2 is greater than V1, why?
because there's a decrease in p2
compared to p1. And this is what we want
to calculate. So let's take this equation
now, and I will cross out all the
annoying terms, and let's look at the
comparison between P and V. So this
becomes now 0 is equal to — I'm going to
divide both sides now, the 0 on the left
side and the right side, by the mass flow
here. So I just have now here 0 is P 1
over Rho 1 plus 1/2 of v1 squared,
yes, and this is in the parenthesis that
has a minus in front yes,
plus on the other side, here this
part here, P 2 over Rho 2 plus one-half
of v2 squared like so. And now let's try
to get p2 minus p1, which is what we
want to have. Let's try to isolate them
on one side of the equation. And so I'm
gonna have p2 divided by Rho 2
yes, minus P 1 divided by Rho 1, so this
is fine.
And now I've got on the other side here
I've got this plus becomes a minus so
I've got here 1/2 of V 1 squared yes
minus one-half of V 2 squared like so.
Rho2 and Rhro1 in this case are the
same because this water, a case where it
would not be the same would be for
example air when there is a lot of heat
transfer, or if I added heat or a very
inefficient turbine, then for sure the
density of the water would
change, density of the fluid would change,
but in this case, it’s water which is
incompressible, so Rho is this thing. So I
can just say now P 2 minus P 1 and I
take the rho here and I put them on the
other side and it's just called Rho now.
it's gonna be one half of Rho of v1
squared minus v2 squared. V1 is high, V2
is even higher, so V1 squared minus v2
squared, it's going to be a negative
number.
And so P 2 minus P 1 is gonna be
negative number, which means P 2 here is
going to be smaller than P 1. So we
expect a negative number coming up
here. So this I call the
Delta P, let's call it Delta P here yeah,
Delta P is equal to — I mean let me write
it aligned so I do not have
terms of an equation that are all over
the place— let's have here Delta P here
and put numbers now. 1/2, Rho is 10 to
the power of 3
yes 1,000 kilograms per meter cube
multiplied by V1 squared, V1 is 3
meters per second, 3 squared minus V2
here squared, which is 6 squared like this. And
then I can have this in my calculator
which I did for you before, and it shows
up as minus one point three five times
10 to the power of 4, and immediately I
write the units, and the units of
pressure are Pascals or Newtons per
meter squared, yes? so Newton per meter
squared or Pascals. And in engineering
we like always to write Pascals, kilopascals,
megaPascals, and so on and so
forth, so I'm going to convert this here
to kiloPascals, so it's going to be
minus thirteen point five kiloPascals.
And it's the Delta P, this is P2
minus P1. So we see here that when the
flow accelerates, and there is nobody to
pay for this from the external
environment, and there is no dissipation
due to friction losses, then, the increase
in kinetic energy is paid for by a
decrease in pressure. Who pays for the
kinetic energy increase? Pressure. It would be the reverse if we had
an expansion in the pipe, we would expect
the pressure to increase — Again, providing
there is no input from the external
environment, and there is no dissipation
due to friction.
So this is how you calculate mass flow,
velocity, and changing pressure, in a very
simple case.
