in this problem we're looking at a
straight pipe and the water is flowing
from the left to the right the pipe has
constant diameter and in the middle of
the pipe there is a vowel and as found
is partly open to slur the water still
flows but it's creating a pressure drop
inside the pipe what we know about the
flow is that the inlet velocity is given
2 meters per second the valve here is
creating a pressure loss of 3.5 kilo
Pascal's and we have two kilograms per
second of water coming in and we're
looking for three informations one is
the outlet velocity v2 the second one is
the net force that's applied on the
fluid as it transits and the last
question is what is the power dissipated
with with the valve as the flow goes
through now let's have a look at this
I'm going to do it wrong I'm going to
show you how not to do the first
question because it's a classic question
that's pretty much every fluid
dynamicists has done wrong at some point
in their early student lives if they
tell you it never did this perhaps take
everything they say with a grain of salt
so this is how to do it wrong you know
so you can see what I do and try to spot
the mistake and then we'll discuss
afterwards what mistake is and I'll show
you have to do it right of course yeah
so what I'm gonna do is I'm going to
write the Bernoulli equation and I'm
going to write it like so I'm going to
say P 1 over Rho 1 yes plus one-half of
V 1 I'm sorry this should not be a row
here first 1/2 of V 1 squared plus GZ 1
here this is equal to P 2 pressure - if
I didn't tie the density - yes plus
one-half of V 2 squared plus GZ 2 and
I'm gonna look at point one point two
here and I'm gonna see if anything
cancels that anything crosses out and
sure enough the altitude is the same so
I can just remove GZ at 1 and GZ 2 here
and I can look at what I have and what I
don't have
the one I know so that's kind of cool
p1 I don't really have real one I have
I don't know p1 I don't know p2 but I
know P 2 minus P 1 because this is the
pressure loss through the valve so p2
minus p1 I have it's minus 3 kilo
Pascal's Rho 2 is equal to Rho 1 this is
a density of water is known yeah how
cool is this here is V 2 so this is the
velocity I'm looking for let's try to
isolate this guy here and put it as a
function of all the other ones so let's
have a look at I'm gonna take a step by
step because I want to do it too fast
1/2 of V 2 squared here put all the rest
on the other side I got then is equal to
P 1 over Rho 1 minus P 2 here over Rho 2
like so and then I got here plus 1/2 of
V 1 squared so just try to work out if I
can isolate V 2 over here I'm gonna
first do a little bit of cleanup I'm
going to do and then say Rho 1 is equal
to Rho 2 this density here and P 2 minus
P 1 so minus of that basically is a
Delta P and I have for the valve and I'm
going to say so this is Delta P of the
valve and I need to put the miles in
there because the Delta P of the valve
is P 2 minus P 1 I have P 1 minus P 2
yes and I'm going to divide by the
density I'm looking for space here so I
mean push the equal sign a little bit
like so 1/2 so 1 over Rho of the Delta P
valve here and then I'm left with plus
1/2 of the 1 squared and sure enough I
can clean this up now so let's have now
just V 2 here let me just put the
multiply everything on this side by 2 so
I got here minus 2 over Rho of the Delta
P valve multiply the 2 over here so I
get plus V 1 square yes and since I have
a squared here I gotta put here a square
root so I'm gonna put everything here to
the power 0.5 and now let's put numbers in.
 
Put here minus 2 over density of
water 1000 yeah 10 to the power 3
kilograms per meter cube. The Delta P of
the valve is given here, it is minus 3.5
kiloPascals, and watch out for the unit,
times 10 to the power of 3 pascals,
put it in the SI units, plus V1 squared
is given here as 2, so
that's 2 to the power 2. So when I
put all of this to the power 0.5,
type this into the calculator, which
I did for you before, this works out
to be just 11, 2 times 3.5, 7, 7 + 4 that's
11, 11 to the power 0.5 turns out to be
three point three one seven. What was i
calculating? Immediately put units. It was
V2 so these are meters per second. Again, watch out, this
is wrong yeah? I don't want you to
believe this is correct. But let's pretend
this is correct for a moment, and I just say
here "oh yeah velocity increase, why did
the velocity increase? because of course
the pressure decreases, pressure
decreases, so velocity’s got to increase"
no? no? Okay so where where is the mistake
in this, in this page here? Let me show you
first why it's wrong, and what's the
problem with this result, what should
immediately ring a bell when you write
this result here, not a little bell
but like annoying beep-beep-beep alarm
with a red light flashing in your head.
Let's have a look at the problem, at the
inlet here. You have two meters per
second of water through a certain area
and with a certain density. So you have a
certain mass flow, well actually you have two
kilograms per second of water coming in.
When they exit, the velocity has increased but
the area is still the same, the density
is still the same, so you have more mass
flow exiting your pipe then coming in!
Where is the mass flow coming from? Who
is creating all these kilograms per
second more
than we had at the inlet? And the
answer is of course /nobody/. If we have a
pipe that has constant cross-section you
may have as big a drop of pressure as
you can in the middle and still are
not going to increase the mass flow at
the outlet. The correct answer, and then
we'll look at the *why* it's wrong to use
the Bernoulli equation in this case, the
correct answer is so: the mass flow here
it is Rho1 v1 a1 and it's also equal to
Rho2 V2 a2. And this is equal to 2
kilograms per second and so it doesn't
take long to see that if I have
A1 is equal to A2, and then I have Rho 2 is equal to
Rho 1, then I have V2 is equal to V1,
yes, it's also equal to 2 meters per
second, period. Okay. You cannot create or
destroy mass by just adding a valve
inside the pipe. So now let me let me
show you what's wrong with this equation
let me show you what is wrong with this.
None of the math in here is wrong. This
is all correct, as far as I can see. What’s
incorrect is when in my mind, somewhere
in the back of my mind, there was a
little voice that said it "use the
Bernoulli equation" because this is not,
this is not correct.
The Bernoulli equation does not apply when
you have losses inside the pipe. And so
you may be thinking "well it's it's easy
for you to say, Olivier but there are half a
billion equations in these lecture notes,
and on top of this some of them are
wrong and are gonna push me off the
tracks, so how do I know which is right
and which is wrong at any given time?
Because this looks pretty tempting to me."
And I see your point.
Now let's see what equations are
available for you. There are basically
three equations and they have different
forms, different shapes, they’re the basic
three equations in fluid dynamics, and these
are written here at the start of the
exercise. You have mass conservation, you
have momentum balance, balance of
momentum, which is about
force, and you have balance of energy
which is about power.
So you have three
things going on. One tells you about mass
flow. One tells you about net force. And
one tells you about power. These are,
these three equations here in this form,
the conditions for them are written
below and they are for a "fixed control
volume with a steady flow" okay
Fixed control volume means your pipe
here is not inflating it's not getting
bigger it's not getting smaller,
inflating or deflating. And steady flow
means if you take a picture of this and
you take a picture every second
those pictures are always the same. It
does not change with time. These
two conditions are pretty reasonable.
they are not arbitrary and as long as
these conditions apply then you're safe
using these equations. The Bernoulli
equation has many more restrictions, I'm
not going to go into detail about this
because it is much easier to use just
here the balance for energy equation. If
I had used this equation here, then I
would have had here to think about the
difference between i1 and i2. And since
there are losses inside the pipe, I would
have refrained from crossing them out.
Certainly these gz1 and gz2 would have gone
but now i can see that there are
exchanges between *three* terms that can
happen inside the pipe: i, p and V. I'm not
adding any power as input or extracting
any power as output from the pipe, this
is just a valve no moving part in here,
and not adding or removing heat, so
there's two terms of zero, and so the
fluid is left to exchange between V, p an d i. V is constrained by the fact that
we know the flow is flowing. Two
kilograms per second are coming in, they
have to get out of the pipe, and
since the the area here is not changing,
V has got to be the same,
I know this for sure. And so I can see
now that the Delta P, the changing
pressure created by the valve, will all
create a delta i. A change of internal
energy of the water. In other words I
destroy energy as pressure with my valve
in the middle where this piece of
friction here and I put this energy back
into the flow as internal energy. It's
going to increase its temperature, by a
very small amount because the heat
capacity of the fluid is very high,
but this, all the energy that’s extracted
by the valve, is put back into the water.
Okay so if I had used this equation then
it would have allowed me to stop and see
that there's no possible way to
calculate the velocity with this energy
equation in this case and the mass flow
equation is enough to solve the flow. So
in case you're lost,
ask yourself what you want to calculate
and remember that these three equations
are almost always good. The pipe
is not expanding or contracting yes,
your pipe is not accumulating mass, and it's a steady flow? these three
equations are good for you. All right. Okay,
so we're done with the first question
the other two questions are going to
be much easier and much simpler to solve,
you're going to see. So the second thing
we're after is net force applying on the
pipe. Well for this I can draw my pipe
like so, and the valve in the middle, and
I'm going to draw a control volume which
is basically the area I'm interested in,
the volume of interest. Let me draw
it like this. This is a control
volume, I call it CV, control volume here.
I look at my vectors coming in and I have
V one here as a vector, and I have V 2
here exiting, and the length of V 2 and
the length of V 1 are the same. And the
directions are the same. So that I have
in fact V2 here is equal to V1 as a
vector, like so, and so it doesn't take
long when I look back on my momentum
balance equation which is here, to
write it like so: the net force
which is equal to the mass flow
multiplied by V2 minus V1, yeah the
mass flow may not be 0 but V2 minus V1
is 0 or more precisely vector 0,
that's going to be here, 0, vector 0, F
net.
Just to be clear, this is a vector and so
just for you to see and to use, remember
that arrows, equations with arrows, have
many dimensions. I could also write it
like this: I could write it as the three
components F net X, F net Y, and F net
Z,  this is a vector and
this vector is equal to 0 0 0.
You don't have to do this, I just want to
make clear these this is actually three
equations.  The last question
we are asked is what is the energy
dissipated as friction? What is the power
lost here through the
presence of the valve, as the water
passes through the valve, how much power?
You can ask yourself, if you want, you can
formulate the question in a slightly
different way. You could say you have
your pipe here with your valve, and then you
plug this pipe here back into the inlet,
then you put a little pump, should work
like th  is, yeah, and the pump here pushes
the water through the valve, what is the
power that you need to put into this
pump to compensate for the friction
inside the pipe? It's just a conceptual
idea, yeah.  And this is told, power is told
using a energy balance equation. So I go
back to the script over here and I just
copy the energy balance equation now.
Let's just again take time to formulate
each of the terms, and not in a hurry, I'm
not trying to save 20 seconds, so let's
break them out. I have power as heat, net
given to the flow, plus power as work
net given or taken away from the flow, is
equal to the mass flow here and I'm
gonna condense this into a Delta
equation, so I'm gonna say delta i
plus Delta P
over Rho (Rho remains constant in this
equation) plus one-half of Delta of V
squared —I'm sorry V squared like this—
plus g Delta z like so. And so I
can see that the net power given as heat,
net power given as work, causes a change
in different forms of energy in the flow.
And in this case we have a simple
valve here, there is no power given from
the outside or taken to the outside,
either as work or as heat, so both of
those terms are zero here, and I'm just
left with the addition of those terms:
Delta Z, the change in altitude that's
zero over here,
Delta V squared, zero here, and I can see
now here that I have zero is equal to
mass flow here multiplied by Delta i
plus Delta P over Rho. And this is what I
meant earlier when I said the entire
change in pressure is going to a change in
internal energy Delta i.  The power
that goes to increase the temperature of
the flow, the internal energy of the flow,
here is M Delta i. Power
dissipation is equal to m dot delta i
here. And I can see through this here it
is equal to m dot times minus Delta P
over Rho like so. Let's put numbers
into this, and we'll say P dissipation
is m dot minus Delta P over rho
m dot is in this case 2 kg/s, the Delta P
is minus 3.5, so minus Delta P will be
plus 3.5 times ten to the
power 3 (because it's kilopascals)
and then divide this by the density
which is 1000, or ten to the power of 3,
and so I can see that I have two
times 3.5, and this is 7 watts.
Seven watts of power that's not a lot of power
So again when you're stuck not knowing
what equation you need to look for, remember
what those three basic equations say.
Read the conditions. If they apply you're
always safe. Better take a big equation
here and cross out the terms one by one,
than take shortcuts and apply equations
that will run you into the wall. Good luck!
