>> Maybe instead of coupling here, maybe there's
a flange, a bolted flange with bolts going
through there. Same thing. What force must
the bolts have, the bolts in place? That's
a big question. What-- if this is a big steam
pipe on a power plant, nuclear power plant,
what is the force in the bolts to hold this
thing in place? [Inaudible] very important,
very important. Once you find the force divided
by the bolt area that gives the bolt stress
and now you pick a bolt out with the right
stress, factor of safety. All these things
are coupled together, you know, when you're
building things. OK. So angular, we'll come
back to that now, 536, 539, we'll do 550 next.
But I want to finish up the last part of Chapter
6 first. OK. Rewind and we had a talk about
in Chapter 6, we divide the continuity equation,
we divide the differential form of a continuity
equation. OK. There's-- The only other equation
in Chapter 6 which goes through derivation
is going to lead us to-- it's really linear
momentum, linear momentum equation in differential
form. We start off with a differential system
of mass delta M. So, a differential system
of mass delta M. Now, here's the flow field
and we take a little differential system of
mass delta M and we watch it as it moves through
the flow field. Now, that was fruitful, applied
Newton's law to it. The summation of the force,
it was differential forces. Differential forces
on that little element of mass. Here is a
time rate of change, look at the big D, OK?
The big D, OK, material derivative of velocity
times delta M. What's this momentum? Momentum,
mass times velocity. The forces on the object,
the time rate of change of momentum that the
object has. OK. It can be considered to be
[inaudible] because if you replace-- pull
mass outside the material group, pull the
mass outside, why can you do that? Well, don't
forget, a system has a constant mass. So mass
is a constant, pull the mass outside the derivative.
Then we have delta M times capital DvDt. Now
we've defined capital DvDt previously as acceleration.
It looks like Newton's Law, F equal ma. That's
why I set up here Newton's Law. F equal ma.
Where did it come from? A change in momentum
caused by the forces acting on the different
mass delta M. These forces on the left hand
side can be, what we call body forces, that
means it's distributed throughout the body
uniformly. Typically, the weight, the weight
is a body force. And these forces on the surface
of that little differential mass delta M.
Surface forces can be pressured, and what's
pressure of normal stress. And if there's
friction, what are those guys called? Shearing
stresses. So the forces on the surface can
be either normal stress like pressure or forces
like shearing stresses that act on the sides
of this little object right here. So, those
guys there can be put together in the equation
and we know what the capital DMVT is we had
that DVD-- capital DvDt, we've have that previously.
And so if we do that, so this gives rho gx
partial with respect to x, sigma xx, shear
stress partial tau yx with respect to y, shear
stress tau zx with respect to z plus rho.
We had this before. This guy here is just
DvDt. That was in chapter, I think 4. What's
on the left hand side? The forces. What kind
of forces? And [inaudible] that gravity and
mass, rho is mass, that's gravity. That's
the weight. That's not the weight. This guy
right here, sigma is a compressive stress,
that's the pressure. This guy right here,
a shearing stress, on the top and bottom.
This guy right here, a shearing stress, on
the front and back sides, all those subscripts
mean something. You know that if you took
MA 218, strict materials, you know what those
subscripts mean. The first one means the shearing
stress acts on a surface perpendicular to
the y direction and in acts in a direction
in the plot in the x direction. So one subscript
tells you what surface it acts on, the other
one tells you what direction it acts in. That's
the double subscript. The normal stress sigma,
x in a plane perpendicular to the x direction,
that's the x direction, my hand is the plane.
Second subscript, x in the x direction. Yeah,
there it is. First subscript, x in a plane
normal to the x direction, my hand. Second
subscript, it acts in a direction, x direction,
that way. That tells us how the stresses act
on a surface. So and then, I'm back to remind
them all up, there is one for y and there's
one for z. Three equations. One, the x direction.
One, the y direction. One the z direction.
They are equation 6.15 in textbook. OK. These
are called general differential-- -- equations
of motion. There's three. One x direction,
one y, one z. OK. Let's see. Anyone saw that
right now? I don't think so. OK. Now, we're
going to make some assumptions. Number one
assumption. If the flow is inviscid, don't
forget the word inviscid means frictionless--
-- 
or nonviscous. They are all the same thing
in our nomenclature. If there's no friction,
all these shear stress guys here go, everything
with the shear stress goes. So we're left
within-- -- where you replace the sigma xx
which is a normal stress and what's pressure?
Pressure is a compressive normal stress. What's
compressive mean? It's a negative sign on
sigma. A compressive stress is a negative
sign, so the stress on the x phase is compressive,
the minus means that's equal to pressure.
So in a fluid, like that, that's what we have.
If it's inviscid now, just the inviscid, OK.
There is three of these guys, of course, there's
one x direction, there's one y direction,
one z direction. See equations, these guys
are 6.51. OK. So these are called Euler's
equations of motion. These are nonlinear PDEs.
And they're not easy to solve. They're, you
know, it takes a little bit of math. There's
some simple geometries and some simple approaches.
So, yeah. By the way, when you solve this,
guys, what you're doing is you're solving
for, look over here what you've got, pressure
may not be known. Little u, little v, little
w. So for little u, little v, little w and
pressure. How many equations of the Euler's
equations of motion? One in the x, one in
the y, one in the z, that's three. How many
unknowns? Four, I need one more. Guess what
we did in Chapter 6, first-- first part of
the chapter, continuity, you got it. Continuity
and the three Euler's, four equations, four
unknowns, it could be solved. It could be
solved. It might be typical, but it could
be solved. OK. Oh, by the way, as a function
of-- Look at the variables, time x, y, z.
So you can solve for the pressure as a function
of time x, y, z. You can solve for the x [inaudible]
of the velocity as a function of time x, y
and z and so on and so forth. OK. You can
also reduce this down. They can be-- or let's
just say, it can be reduced to the Bernoulli
equation. It can be reduced to the Bernoulli
equation. It's very, very similar to Bernoulli
equation in the one dimension. In Bernoulli's,
is there a z term? Yeah, g tells you how g
is a function of how up or down you are or
you could-- There's a pressure term, that's
in Bernoulli's. If it's only a function of
x, we're the y and z. If a steady state, where
the t function. This guy here, or is this
guy here, partial u squared with respect to
x. OK, 1/2. OK. So that's 2u, 2u divided by
2 DmDx, those guys are the same. Does that
look like Bernoulli's? Of course, it does.
V squared divided what? Guess what's divided
by 2. Yeah, you can reduce it down to Bernoulli's.
Some people that teach the course derive Bernoulli's
this way. We derived Bernoulli's back in week
three just to kind of get a flavor for energy
in the fluid mechanics. But you can derive
Bernoulli's-- and back in Chapter 3, we derived
Bernoulli's along a streamline and the whole
fluid element along the streamline. Remember,
we took a little fluid element along a streamline
and we derived Bernoulli's equation from that.
Some people don't cover that in Chapter 3
and they do the same thing in Chapter 6. They
stop here and now they derive Bernoulli's
and textbook [inaudible] again. He says, you
want to see us derive Bernoulli's, read that
thing in the textbook. And they derive Bernoulli's
from this and they end up with the same Bernoulli
we have back in Chapter 3. OK. Now, one more
step. Now, this is frictionless. Now consider
friction. OK. It can be shown-- -- that--
Those are all in equation 6.125. They're sometimes
called the constitutive equations. They relate
the stress terms to the velocity field and
the pressure and the viscosity. And the textbook
says, if you want to see how-- where they
came from, here's three references. Go find
and read them. In other words, it's pretty
tough stuff. It's not something you want to
put in undergraduate textbook. OK. So you
have to say, OK, I believe it. There they
are. That's what they are. So we take those
guys there and where do you see these guys
up here, for sigma xx? You put this one. Tau
xy, you put this one. Tau yx, this one, and
so on and so forth. You solve stuff in. That
is complicated. OK. When you're done, you
need equations like this. Et cetera and yes,
the y direction and z direction. Three of
them, x direction, y direction, z direction.
See equations 6.127. These guys are even worse.
These guys now are non-linear second order
partial differential equation, second order,
partial differential equations. Oh yeah, that's
for the advanced group, OK? Yeah, those guys,
because they include viscosity. Euler's do
not include viscosity. The ones that work
all the time are the general differential
equations of motion. If you neglect viscosity,
you get these guys. If you include viscosity,
you get these guys. These guys are called
the Navier-Stokes equations. So now we 
have three equations on the board, the general
equations of motion of a fluid in a fluid
field. The equations which are inviscid which
you're-- end up what we call Euler's equation
of motions. And if you include viscosity,
you end up with what's called the Navier-Stokes
equations. Oh yeah. Oh yeah. You know now,
if you finished, typically, civil engineers
and [inaudible] take two quarters of fluids,
OK. But if this is going to be your area of
interest and you want to get a master's degree,
let's say, here at Cal Poly, you would first
of all, in a master's program, take Advanced
Fluids, EGR 535, Advanced Fluids, four unit
class, one quarter. Then if you want to go
further, you would take potential flow four
unit class in the graduate program. Potential
flow neglects viscosity, OK. And then if you
want to look at these guys, you take a course
called boundary layer theory, which talks
about the boundary layers on objects, between
an object in a fluid like an air flow or a
turbine blade. You take a course which focuses
on these guys. So, there's at least three
graduate courses worth 4 units each, 12 units
of graduate work, OK, in Advanced Fluids.
And what did we-- done this thing here, probably
about 12 to 15 minutes. We just, you know,
when we scratch like that. Nothing, nothing.
It's advanced stuff, it's heavy stuff. OK.
We got our own problems, OK. So anyway, just
so you know. That's what you get into if you're
interested in fluid mechanics, thermo sciences,
you get equations like that. They're called
field equations. They apply at every point
in a field. They're not a controlled volume
approach, there is no controlled surfaces.
You solve for the velocities and the pressure
as a function of time x, y and z analytically
if you can, which you many times can. Numerically,
if you can, many times you can. OK. That's
where numerical methods comes in, courses
like that. OK. So, I just want to give a flavor
of that so you know what those words mean.
If somebody throws out the word Navier-Stokes,
you say, who's that guy? No, no, it's not
a guy's name, OK? They're advanced and the
key thing is to know when they apply. Friction,
considering friction, Euler's, frictionless,
these guys, that's why they call it general.
You start here and you derive these two guys
from that one. And the textbook takes about
eight pages with that stuff, eight pages.
We don't have time in a quarter course. OK.
Now let's go back, so if we did that, let's
go back to our problems we were working and
you've got four problems for homework in Chapter
6, all right? That homework will now be due
with the Chapter 5 homework on Monday. So,
add the Chapter 6 homework. So the Chapter
5 homework that was due anyway on Monday.
OK. We're going through the problems I assigned
for homework in Chapter 5 talking about the
energy equation. We discussed two already,
test homework of that, it's over there if
you missed it. We talked about 36 and 39,
so we'll quickly look at 50 if I think a test--
something is important, then we just look
at it real quick, 6.50. Down the-- probably
not, but I'll do additional. Oh yeah, we will--
6, that's quite difficult. OK. I'll read it.
And now to attach to a virtual plane, discharges
water to the atmosphere, all these were major
case. Discharge is we know Q, we know the
pressure at the plant, so discharge and we
know the Q. We know the pressure at the plant.
We know Q. The pressure here, we know, zero.
He said atmosphere. Those were all his. Determine
the vertical component of the anchoring force
required to hold the nozzle in place. The
nozzle has a weight of 200 Newtons and the
volume of water in the nozzle is 0.012, is
the anchoring force of horizontal. OK. So
again, we always show our points through all
your control line, OK, control surface. Call
this 1, call this 2. OK now but, you know,
we got to be careful now because there is
a weight acting down of the nozzle and there's
a weight acting down of the water. OK. We
got to include those two guys in there. And
now you might say, well, I'm going to assume
that the force here is a positive to make
it easy in my equation. Here is my Fy. So,
and my x and y, like this, left hand side
of momentum, plus Fy minus weight of nozzle
minus weight of the water plus the pressure
force up here in the y direction. Zero, zero,
equal to change in momentum, I'm not going
to move to it again but, you know, rho B times
B dot A at one, rho B times B dot A at two.
You know the Q, you know, big times A because
it's given. If he gives you the areas A1 and
A2, done deal. Plug them in, crank it up,
you got it. And as I told you, when you're
studying for exam, don't stop there. What
if I ask you to find a force in the x direction?
You know? [Inaudible] you know, that's a y
direction, but in the x direction, what's
the pressure on the top there? Zero. You know,
so let's take [inaudible]. If you got momentum
leading but what's the velocity, you know,
the velocity leaving there is at-- also needed
is rho q times v. That's the easy one there.
All right. Let's go to 5.61, 5.61. Oh yeah,
OK, 5.61 is like this. Stuff comes in, it's
going to go on two locations now. One location
is it goes up and back, turns around your
back and the other location down here. Like
this, out, out, in. Control volume, control
surface inside. Pressure we know, velocity
we know, velocity we know, OK. We call this
one, we know V1. Call this two, we're given
V2. Call this three, we don't know V3. We
weren't given V3. We know P1 and here's a
key thing. Water discharges into the atmosphere,
thank you very much. P3 is zero, P2 is zero.
He told me it discharged atmosphere and I
know P1. I know P1, I know V1, I know V2.
Continuity gives V3. Momentum gives support
in the x direction. OK. Just so you know how
to solve the V3, continuity. V times A, V1A1
equal V2A2 plus V3A3, the only unknown is
V3, solve for it. OK. Let's go on--
>> Professor?
>> Yeah. Oh. Let's see [inaudible] 6.51, OK.
OK, let's do this next one, 65. And then here's
extra plate and let's go down here, here's
weight here and [inaudible]. OK. A horizontal
circular jet of air, air, OK, strikes a stationary
flat place indicating jet velocity is 40 meters
per second, the jet diameter is 30. If the
air velocity magnitude remains constant over
the-- as the air flows over the plate surface
in the direction shown, determine the magnitude
of F, the anchoring force required to hold
in place stationary and then do part B and
then part C. That's air to air now, OK. A
horizontal circular jet of air. Horizontal
tells you, don't worry about the delta Z's,
don't worry about Z1 and Z2 and Z3. These
nicely labeled, 0.1, 0.2 and 0.3. So this
thing is here's my horizontal jet of air and
it hits this thing, see here it is right here.
Here's a horizontal jet of air. The jet of
air hits this, he's saying some goes up, some
goes down. So that's how it looks, OK. It's
something like the one we've worked before
where the jet of water hit the vertical plate,
it's on the ground, the jet of water hits
here and it spreads out in all direction uniformly,
except now the plate is air now, the plate
is like this, it hits here and goes up that
way and down there. He tells you B1, B2 and
B3 I guess. But the key of that problem to
make life easier, to find your coordinate
system this way. And then momentum in the
y direction use the force F. So it changed
your coordinate system to make life easy for
yourself.
[ Inaudible Remark ]
OK. That's the key. That's the key. It's part
C I think, maybe, yeah, maybe. In the x direction,
OK, you're going to find out it's zero. He
told you that there is no foreseen x direction
which means it's frictionless flow. It's frictionless
flow, part C is frictionless flow, OK? OK.
I think that was the last one. Yeah, that
was the last one, momentum. Let's do the energy
ones [inaudible] time. All right, 102, 5.102.
OK. OK. This big diameter going down to a
small diameter, yeah. And we've got the flow
going down, the problem says it flows downward.
Vertical plate now, so you're like the [inaudible]
Z's. Mercury manometer, OK, get a manometer
up here down to here. We have-- This is--
He says oil, that's oil, not water, so oil.
And this guy goes down there and comes up
H. So down here comes up to here. This distance
here is H. And that's gamma of mercury. Yeah.
And it's 0.6 meters from here to here. I'm
going to call this location 1 and I'll call
this location 2. OK. And I'm going to write
energy 1 to 2. I think I'm going to write
P1 over gamma plus V1 squared over 2g plus
Z1. P2 over gamma plus V2 squared over 2g
plus Z2. Let's see what's applying here now.
Determine the volumetric flow rate for our
frictionless flow, OK. So I put the loss is
zero in here. Because it said frictionless
flow, put the last term in the x equation
zero. There is no pump, HP is zero. There
is no turbine, HT is zero. There is no losses,
HL, HL, losses. In what, feet or meters? What's
this in? Feet or meters. What's that in? Feet
or meters. What's that in? Feet or meters.
Everything in there is in feet or meters.
OK. The reason why I assigned this because
I want to start-- I know you got test Monday
but the final exam is two weeks later. OK,
[inaudible] I'm sure. Yeah, you got to go
back and review manometers again. Come back
in week two, might as well do it now. That
way-- Let's study for, for the final. So,
let's just read in there and see, we're supposed
to find Q. Of course, Q1 and Q2, so don't
even call it Q1. I know they're both the same.
Why? Steady flow, incompressible. OK. Can
I find P1 minus P2? Of course, I can, manometer,
Chapter 2. Can I find V1 in terms of V2? Of
course, V1A1 equal V2A2. Got it? Can I find
Z1 minus Z2? Of course, 0.6. What are you
going to solve for? I don't know, V1 or V2,
take your choice. Once you get V1, where do
you most plot by? A1. To get what? Q, end
of problem. But he's not going to be super
kind to you. He's not going to tell you that
you're supposed to call this distance here
x. It's not in the sketch, but he hopes you
realize it, because if you-- I think, he made
a mistake, he didn't put it [inaudible]--
I think he meant to-- I think he meant to
put-- to put this thing up to there. Oh no,
he didn't. No, not way up there, no. Well,
you know, you don't have to do that. If you
just realize it. If you go down in water,
that distance x on the left hand side, going
down is what? Plus gamma x. And go up on this
side, going up and use what? Minus sign, gamma
times x, plus gamma times x of oil minus gamma
times x of oil cancels up, it's gone, x disappears.
But it's nice to know why it disappears. Well,
you know, that's-- The point there was you
shouldn't put an x there when you do the problem
so that I know that you know what you're doing.
That's the point. OK. Now let's take the next
one. 102, we're going to go to 105 I guess.
105, yeah 105. All right 105, we have a siphon
in a tank. And I guess this is water, we'll
see. Yeah, water. OK. We know the diameter,
given. We know all the distances from here.
Let's put it down [inaudible] shown down here.
So here to here, to here. We know the distance
from here to here. Those are on one. Check,
check, check. We know it's water, gamma water.
Got it? And the friction loss is 0.8 V squared
divided by 2. OK. We're supposed to find the
flow rate. OK, which means find the velocity?
What velocity? The velocity in the pipe. OK,
so we start out and say, you know what, I'm
going to call this point 1 and I'm going to
call that point 2. OK, got it. Let's go ahead
and write the energy equation down. Here it
is. Here we go. There is no pump. There is
no turbine. Zero, zero. Pick out the right
equation. I put three on the board for you.
I said sometimes it's easier to use one equation
than it is the other equation. Which equation
am I going to pick now? Here, I'll tell you
which one. What did he give me? He gave me
the losses in terms of what? B squared or
2g? No. That'd be the third equation. B squared--
it's called the head equation. No. He gave
me the loss in terms of V squared over 2.
I try and find in those three equations a
V squared over 2 sitting all by itself. Oh
there, just right there. The second down,
the middle one. So I choose the middle one
here. Why? Because then I'm getting difference
of losses and I better have what's the units
here? What's the units here? What's the units
here? What's the units here? When you can
derive it. This guy better have the same units
that those guys have, OK? Oh, I did again.
Let's say you put him right here. Now we can.
Just so you know, there are three velocities,
V1, V2, and V. They're all different generally.
V1 has a velocity F point 1, on the control
surface. If it's-- if you draw them. V2, the
velocity F point 2, and V, the velocity in
the height. Yeah, got [inaudible].
>> Yeah, so on the midterm problem that you
put us in online for the energy problem, you
gave us the losses had a like-- as a coefficient.
So, what-- when you give us the coefficient,
what is it usually in terms of? And P squared
over 2 and--
>> Coefficients like, let us say P sub 2,
that's the coefficient 2. Do you like mean
coefficient, a subscript? You mean something
in front of something.
>> Yeah, like--
>> Give, give me an example here. What was
it? It was like it-- It gave the-- the lot--
the coefficient of losses due to the typing
in the 2 the 0.4.
>> OK. I don't recall that. I'll look at--
Yeah. Give it some more and look at afterwards.
>> Yeah.
>> Online something.
>> I'll look at as the--
>> OK.
>> Yeah. Normally, it's given like that. Often
with that with [inaudible]. OK, now we're
supposed to find V, which mean that V. How
about this guy? He's gone. This guy, he's
gone. This guy, he's gone. Z1 minus Z2, I'm
right here, Z2, Z1. This distance minus that
distance. Got it. Got it. Got it. What's V2?
The velocity leaving the pipe. What's V? The
velocity in the pipe. Are they the same? Of
course they have to be. So the equation for
V. Once you get V, multiply it by the area
of the pipe. Pi E squared divided by 4. End
of problem. You've got Q. But make sure you
know what those V's are. Is it V1, is it V2,
and is it V, OK. On Monday, oh hey, I'm telling
you, things like that [inaudible]. What if
I asked you for the pressure here? And the
losses in the pipe up to that point A, the
losses in the pipe up to point A were 0.3
V squared over 3. OK, I'm not going to solve.
I'm just telling you how I think. What if
I asked you for the pressure point B? Oh I'm
going to stop here. You get the point. Don't
stop. Just didn't work the problem you're
assigned. Use your mind to think ahead. What
can the instructor ask me by tweaking the
problem a little bit? Yeah, I'm interested.
Up here, maybe the water's going to vaporize
and stop deciding. That's what I'm worried
about, the vapor pressure of water up there.
And when-- how about the entrance here. Is
the pressure-- Does gamma times that distance?
Oh, no. No. No. Because if there's a velocity
there, what's that velocity? V. OK. So, I
can go from 1 to B or I can go from 1 to A
or from B to A. So, we'll be thinking about
things like that too. All right. Let's go
on, that was 105. Now let's go to 114.
>> Professor [inaudible]. So, the question
was asking what is the last coefficient for
[inaudible] is K equals 0.4.
>> Say it again now, the last one?
>> The last coefficient we're [inaudible]
is K equals 0.4. On this problem in the board
or--
>> No, no, [inaudible].
>> Oh, K equal-- what again now?
>> K equals 0.4.
>> OK. That's all it says?
>> The last coefficient for [inaudible] is
K equal 0.4. Determine--
>> OK?
>> -- our output on the turbine.
>> All right. Typically, that's the coefficient
in front of the V squared or 2g [inaudible].
>> OK.
>> The coefficient in front of the V. I would
be more specific next time I tell you that
because you couldn't tell from that. You couldn't
tell from that. I would put it back, yes,
to make sure there's no air interpretation.
>> [Inaudible] this will be 0.8?
>> Yeah, I would put 0.4 V squared 2g probably.
Because normally it's specified in feet, feet
loss. Yeah. That's a good point. That's for
defining it for us. OK, 114. All right. This
must have a [inaudible]. This one would happen
real quick, 114. Oh, the fire truck, OK. See,
the pressure is there, the diameter is there.
So, we have this fire truck with the pump
here and the water comes in and then it goes
out here. Here is the fire hose. And then
the water goes up here, fire hose and-- The
pumper truck show and delivers-- Q is given.
Sixty feet is given. So, Q-- Q. We have 60
feet. Let's see what we know about the problem.
The pressure. We know the pressure and we
know the diameter up of the hydrant is 10.
The head losses in the pipe are negligibly
small. Determine the power that the pump provides.
So, find W dot comma. OK, let's write down
the energy equation. Leave hp, that's the
pump head. And write the equation in terms
of head. At least I'll do it that way. I prefer
that. If you want to use the one that has
the power in it directly, you can do that
too. I know they look like this though. So,
it is equal to P2 gamma P2 squared over 2g
plus Z2. Let's see, find-- Neglect the losses
and go ahead and find out the power. OK. P1.
Yes, I know what V1, OK, Z,1 I'll call it
zero, Z2 is 60, V2 I don't know, I don't know
that guy there. Determine there. Yeah, just
like the P2 is atmosphere, zero. This point
2. Say you know what point 2 is. And I'll
see what it does there. Yeah, we know Q here.
So, we know Q so we get V1 equal Q over A.
I know that. Q is good. Solve for hp. Then
W dot power of the pump equal gamma Q hp divided
by-- I want this. Let's do it like that. If
you want it in horsepower divided by 550 so
that's in [inaudible] computational. Otherwise
it comes out to be in kilowatt-- kilowatts.
OK.
>> Is it OK if we leave it as pounds per second
or which is you prefer [inaudible]?
>> I will say the problem, find the power
and horsepower, what kind of power [inaudible]
to be specific.
>> OK. All right.
>> OK, 5.116. OK. Again, you know, I can tweak
this problem so you-- of losses in the hose
or want to take these [inaudible] too and
stuff like that and all the-- It doesn't change
the problem much, you know. I can say, if
the pump efficiency is 80%, how much power
must be provided to the pump because I'll
tell you, this power I get out of here is
the power that the pump blades provide to
the water. But what's the power required to
rotate the pump blades in a centrifugal pump?
OK. You got to divide that guy by the efficiency.
The equations are in your notes, the efficiency
equations, they're in your notes. So, I can
put efficiency in there. I can put losses
in there. Those are things I can do to the
same problem assigned for homework. OK, 5.116.
Now we got a reservoir with a pump pumping
to another reservoir on top of this reservoir
that's open to the atmosphere. And we take
water running here and it goes through a pump,
then it goes up into a second reservoir which
has its capped on pump. Has air up there at
2 atmospheres. So we know the pressure up
there. So, here is the water and it's capped
on top. This is air, and this is water. OK.
So, we know the flow rise that way. We're
specifically given those [inaudible] what
we know. OK. We know the flow rate, the atmosphere--
We know gpm. We know the Q, so we know Q.
It has 3 horsepower, so. OK, losses. P1, zero,
I'm going to call it point 1 here. I'm going
to call it point 2 up here. P1 zero, V1 zero,
Z1 zero.
>> [Inaudible] the pump.
>> Oh yeah, yeah. Over the pump here [inaudible]
two good things. P2 is not [inaudible] 2 atmospheres,
OK, got it. V2, yeah, that's zero. Z2 got
it. Hp I know what it is. It's right here.
OK. What's the power, somebody, in kilowatts?
>> Three horsepower.
>> OK, three horsepower. Now you got to play
the 550 game. I can just write 50. I know
this value, I've got it. I know gamma of water,
got it. He gave me Q. He gave me Q, got it.
Solve for hp, for hp, there. OK, not hp. I
know, check mark. Now, I solve for the losses.
OK. Now, if I get the loss is positive, it's
OK. If I get a negative, it's impossible.
So, that's what you do. You solve for the
losses. If you get a positive sign, yeah,
it can happen. If you get a negative sign,
it can happen. Losses can never be negative.
Losses are always positive. They take energy
up. Losses don't put energy into a fluid.
They take energy out of the fluid. OK. I'm
not going to do 117 and give you everything
you need. Let's see. Yeah, OK, good. Let's
read 120. Water flows by gravity from one
side of the lake to another, from one lake
to another lake at a steady rate 80. Good,
we know Q. What is the loss of available energy
associated with this flow? And he gives us--
He gives us delta Z. OK. Flow down, losses
equal 50. And this is in feet, so losses are
50 feet of water. P1 zero, P2 zero. There's
no pump. There's no turbine. P1 zero, P2 zero,
V1 zero, V2 zero, Z1 50, Z2 zero. There's
only one thing left in the equation, Z2 equal
losses. OK, easy, you got it, 50. Now comes
in the other side of the coin. Keep reading.
Now, if we want to reverse this flow and pump
water up in the lower part of the higher reservoir,
estimate the pumping power required. So, now,
I put a pump in here and now I change the
flow direction. Now, you better change the
subscripts, if you don't, you're going to
hit really bad. This has to be 1 because that's
where the flow comes from. This has to be
2. That's where the flow ends up. P1 zero,
Z1 zero, V1 zero. Two, P2 zero, V2 zero, Z2
50, Z2 50. Here's the equation. Hp equals
zero, zero, zero equal zero, zero, plus 50
plus 50. Fifty plus 50. OK. Power, you want
power, hp times gamma Q hp, that's the power.
OK. So, what [inaudible] my gravity. That's
how these-- How many stations in the [inaudible]
work sometimes. They let the water flow all
the way down to a generator, in a lower elevation
during the day to generate electricity. Then
when it's cheap to run electricity, they used
electricity to pump the water back up late
at night, and [inaudible] pump storage. And
a lot of-- a lot of utilities use that because
electricity at night is cheap because nobody
has the lights on in that. So, they pump the
water back up to the upper way in the [inaudible]
at night. They let it run down the day to
power the lights in Fresno and blah blah blah
and so on and so forth. So it happens, and
it happens all the time. That's why it's an
important [inaudible]. OK. Last one, we're
going to finish on time, that's great. All
right, 5.131. OK. Let's look at this guy now.
All right, 130-- Boy, that's [inaudible].
I got to take my book and sketch [inaudible].
It's a little bit different than we've done
in the past. It looks like this. All right,
so we have water. Water, we think it's water.
Yes, it is. Water flows steadily down. Down,
down. There is a manometer. So, practice again
from Chapter 2. OK, like this, like this,
and this and this, like this. And we've got
the fluid level here and here and we know
this distance from here to here, checked.
We know this is mercury, gamma Hg. We know
this is water, gamma water. The diameter is
the same. The difference in the first part
is it says go back to Chapter 2 and find P1
minus P2. There's 1, there's 2. Part A, find
P1 minus P2, Chapter 2. I'm not going to do
this, we've done enough. OK, part B, the log
between sections 1 and 2. OK, it is over there
again. It's over there. Is there a pump? No.
Is there a turbine? No. Are there losses?
Yes, leave it in. Do I know P1 minus P2? Of
course I do, I found that from part A. 
Conservation of mass. Q1 equal Q2. V1A1 equal
V2A2 and it's the same diameter inclusion
V2 equal V1, OK. Cancel, cancel. Difference
in elevation, it tells us it's 5, 5 feet.
It tells us 30 degrees. Yeah, I got it. The
vertical, the vertical part of that, OK. Five
times the sign of 30, I dropped the delta
Z. Matter of fact, I'm going to call Z2 zero,
my choice, I call it zero. So solve for Z1,
5 sign 30. That's it, solve for losses, part
B. OK, done. Got that. We got that. Now let's
do part C. OK, here's my control volume. Find
the friction velocity in the pipe. OK.
>> Is it part C asked for RX?
>> Yeah.
>> What did it ask for?
>> For the extra--
>> The force in [inaudible].
>> OK. That's right. Thank you. Yeah. OK.
OK, it comes in, [inaudible] dot product dot.
I'm writing it this way. OK. In the x direction,
that's what I want. I want the force to the
friction in the x direction called the axial
direction. OK, top product out, air vector
out, plus vector in minus dot product, area
vector out, velocity vector out positive,
velocity here V1, velocity there V2, OK. This
sign negative, that sign positive. Let's see
here. OK, we got that guy and then we've got
the V2 rho 2 A2V2. Force be the x direction,
of course the friction force. Which way you
want to assume friction acts? Well, you know,
most folks believe friction acts against the
flow. Of course it does, so I'm plugging it
to minus F. My intuition tells me that's the
way it goes. It acts against the flow. It
tends to slow the flow down. Pressure force
here. Don't try and get. I told you, don't
try and do too much on one picture. Three
forces, there they are. Don't try to [inaudible]
your mind. You put too much stuff on one diagram,
you're asking for trouble, I guarantee you.
P1, the force of pressure force-- Your pressure,
P1. P1A1, positive, P2 minus P2A2. The weight,
the actual weight times the sign of 30 acting
in the axial direction. The weight times the
weight of water. The weight of what? Don't
include the pipe, look at my dash lines. Did
my dash lines include the pipe wall? No, they
only included the water. They didn't include
the pipe wall. There is an identical problem
of this. It's a long problem. It's the longest
problem in Chapter 5. It starts on page 216,
it goes through page 217, and ends up on page
218. And on page 218, they draw a nozzle with
just the water shown. So the example to follow
is on page 218, it's example 511. That's what
you need to do to follow to do what I did
to the problem right here. OK, this is an
interesting problem. It's a great review problem.
You know, chapter-- We would go to Chapter
2 for part A, we would go to Chapter 5 on
energy for part B and we go into Chapter 5
for momentum on part C. Oh yeah, it gets you
ready for-- for the exam, yeah, it will. OK,
let's not be blank so see you then.
