>> One last time, Chapter 5. OK. We're covering
three different basic laws of fluid mechanics in Chapter 5. Using the Reynold's transport
theorem, taking the basic laws written for systems and converting those laws using engineering
based on the control volume and control surface approach. All right, we talked about continuity.
We talked about momentum. So today we start energy. We'll kind of derive the correct equations
to use with energy. And then maybe later today, a brochure on Monday will include examples
on energy. The last two homeworks that you have go through the end of Chapter 5 which
is the energy problems. OK. Obviously, a very important equation, we got a little bit of
that back in Chapter 3 in Bernoulli's [inaudible] was something about energy in fluids. But
this is the next step in that. We start off again-- Let's just get a show of hands. I
know some of you had, some haven't. How many of you in here have had ME301? Show of hands.
Let me just do a quick count. Keep up. So I have 2, 4, 6, 7, 8, 9, 10, 11, 12, 13, 14,
15. All right, roughly 17 out 34 people. That's about a half, OK. I understand that. You really
get a good dose of that in ME301. You get a really good derivation. So, for some people,
it's going to be totally new in this class and for some people it will be a review on
some of that. We're not going to go through as much in detail of this derivation as we
do in thermo that we see half the class have been through this thing all the time. So,
we're going to do maybe what you call a more casual derivation of this, not mathematically
rigorous, OK? Now, we start out with the energy equation written in system terms. This is
what you get in physics in ME301, OK? This is the equation written in terms of a system
where the rate of change of the energy in the system is equal to-- we're going to shift
gears temporary for about two class periods. This q, q dot now stands for heat transfer.
No longer by the [inaudible] flow rate. So, it's q [inaudible] flow rate. So you got to
kind of shift gears. W is work. So q is heat transfer, w is work. The dot above them means
with respect to time. W is work. Work is Newton meters divided by time, seconds. Newton meters
per second, what's a Newton meter, a joule? What's a joule per second, a watt? OK. It's
watts. If it's a British computation of the work, pound foot of work, pound feet of work
divided by time, seconds. Foot pounds per second. That's the power. You convert it into
horsepower divided by 550, divided by 550. You did in horsepower if you're in British
computation. If you're in SI which we said joules per second, what's that? Watts. OK.
So, those are power terms. W all by itself stands for work. W dot is the rate at which
work is done and that's power. OK, that-- the left hand side looks something-- that
should be done. It looks something like the left hand side of our previous Reynolds transport
theorem. OK. So, here they are right here. So, we're going to use Reynolds transport
theorem now. So, this right hand side here equals this right hand side here. Since the
left hand sides are the same, system-- e stands for energy, e stands for energy. What's that
energy? Joules, OK, foot pounds, OK. So now we equate the left hand-- the right hand side
of the top equation with the right hand side of the bottom equation. OK, we do that and
we end up with this equation. OK. So we have left hand side. So again, the system equation
says the time rate of change of energy of a system is equal to how much heat transfer
occurs, and net in, net in means in minus out. The net heat transfer in plus the net
work out. OK. Oops, that should be in. Sorry about that. We're going to define that as
positive, some which call it negative, we'll call it positive. OK. Now, remember what we
did when we had-- when we had continuity. This was written originally, Reynolds transport
theorem with a capital v, capital v was here, capital v was here, capital v was here, and
we said let v equal mass. Little v was big v divided by mass. So when we did continuity,
that became 1. Mass divided by mass with 1. When we had momentum that was mass times velocity,
big v. Divide that by mass. That was just the velocity. When we have energy, capital
e is the energy. If we take and divide e by m, we get little e. So that's energy per mass.
So little e is the energy per mass, but we know it is. OK. And what's rho dv? That's
mass. That's the mass. Energy per mass times the mass give me what energy with respect
to time gives me what? The rate of change of energy in the system with respect to time
down here. And over here, it's control volume. OK. Yeah. Let's go ahead and look at what
kind-- In the third one, they talk about work for about a week and describe different kinds
of work. I'll just give you a little flavor to what kind of work we're talking about here.
One thing, work could be a rotating shaft. How do you drive a pump with a motor? How
are they connected by a shaft? What's the shaft do? It rotates. It powers the pump to
pump for instant water. So one way work can cross the boundaries or the control service,
is by a rotating shaft crossing the boundaries. A turbine or a pump, those are examples of
that. There's other kinds of work-- pressure work. It requires work to push mass into the
control volume through the control service. It requires a work. The pressure work pushes
the fluid let's say into the pump that's going to pump. There might be friction work. We
have friction. Let's say on a conveyer belt with the fluid, there's friction between the
belt and the fluid. So friction work, those are all examples of what this term here might
be. OK. Now, what kind of energy can there be? We'll consider several. We'll start off
with what's called internal energy. We'll add p over rho, energy and the pressure plus
v squared over 2 and gz. Kinetic energy, potential energy, energy due to pressure, and this guy
u, little u is called internal energy. Pretty much it's everything you can't see. If you
heat a cup of coffee up by 30 degrees Fahrenheit, it doesn't start to go red, [inaudible], OK?
But do you think it has more energy? Oh yeah. Why? Because the molecules are moving faster
in there. Oh yeah. It's invisible, it's microscopic. I can't see it. Now, if I throw a baseball,
oh, you better believe, I'll see this guy right here, kinetic energy. I drop a brick
from the building, you better believe, I'll see this guy here, potential energy, but I
don't see the internal energy. That's why it's called that, it happens internally. OK.
So, that's what this guy in this equation is. We're going to sum that in now where we
see little e. So, using this gives the following ddt. And now we sub in little e. And now,
we do-- it says up here. That's e and we have rho and we have our v.n, dA. And then that
has to be equal to-- way out here-- Q net in plus W dot net in. >> Professor? >> Yeah. >> Is there a g net under the v squared over
it? >> Let's see, under the-- Let's check. Let's
check and see. What's u into v squared? What's u into v? What's u into d? What's u into z?
No, they're the same. That's why you do it. I'll just check it that way. OK. Now, in Chapter
5 we pretty much tell-- yeah, I'm sorry. >> Is there another g? >> Yeah. Oh, there's another G here? Oh yeah,
I got to two Gs. Thank you very much. Bernoulli came back in my mind. Chapter 5 says in her
book pretty much the first introductory through its course, we normally are going to consider
this guy to be 0. So normally we consider steady state except sometimes the continuity
we consider increase in mass by water going in and coming out. But generally in Chapter
5, we look at steady state. So we're going to assume make matters simple. I'll start
here. So, assume steady state. We're also going to assume incompressible. We're also
going to assume inviscid are frictionless. So we're not going to worry about this frictional
work right now. You're going to worry about the work done by the pressure forces and then
the work done by rotating shafts. OK, so we have all that in here. And then, so doing
that, we get-- now, this is guy is going to go out, OK? And we're also going to assume
one more, or right now we're going to assume no network in. We'll come back later to put
that in there, but right now assume it's going to 0. So we end up with this, m dot-- now,
the m dot comes from the rho av. This is m dot right here. Rho av, m dot, OK. So we pull
that outside m dot. Rearrange the terms. Rearrange it one more time. We just separate the terms
with some on the left hand side equation, some on the right hand side of the equation.
So I'm going to write here. We have q net in, q dot net in divided by m dot. So, that's
how we got that. >> Professor. >> Hmm? >> Quick question. >> Yeah. >> If I could use the [inaudible]? >> I'm sorry, which one? This one? >> Over there. >> That one? Oh, let's see that. Is there
a gz in here? I've got my Us, my Ps, my Vs. I got my [inaudible]. The elevation can't
change, thank you so much. So, we got plus gz out minus z in and that whole rho is equal
to q dot net. OK. Thanks. I think you'll work again here. OK, we're getting there, we're
getting there. Now, we say-- you know what, that thing looks a little bit like the end
of Chapter 3, something called Bernoulli equation. It looks like it almost. There're some differences.
I'll tell you where it is here. The p2 over rho plus v2 squared divided by 2 plus gz2
equal p1 divided by rho plus v1 squared divided by 2 plus gz1. Now, in Bernoulli's, that was
it, stop, that's Bernoulli right there. When we derive the energy equation from the system
through the Reynolds transport there, we get this equation. This term suddenly appears,
OK? Well, this is then similar to Bernoulli, OK? So, I'm going to put that over here, I
think. And remember, Bernoulli was pretty far steady, incompressible, and inviscid.
So, neglecting viscosity, neglecting friction due to viscosity. They can call the inviscid
frictionless. Of course, we derived this thing already for incompressible because we said
rho is a constant, rho is a constant. So we've already said incompressible. We've already
said steady, that term went to 0. Oh, what's left over here? Oh, the third one, frictionless.
You know what, that term there must be somehow tied into friction. And if we have friction
in something, like a turbine or a pump, what happens? Friction eats up the available energy.
That's why pump or turbine is not 100% efficient. The friction, what's it do? Well, it heats
the pump case in, it heats the water inside the pump, is that useful energy? No, I don't
want that. The bearings on your-- wheel of your car, the engine drives that, heat up
with friction, what does that do? That takes power away from your engine to drive the car.
Do you want that? No. So what would we call this thing here again? We call that the loss
of energy. That's the loss of energy due to friction. OK. So this guy here is called a
loss of energy. So comparing to Bernoulli is this u out minus u in minus q net in is
equal to 0, OK, compared to Bernoulli. If there is friction-- it's the difference now.
If there is friction, that is not 0. U out minus u in minus q net in is greater than
0. And that's tied into the second law of thermo. When we derive this thing we said
this is called the first law of thermo or the energy equation. We get to this point
here, we say, "You know what, this term always has to be greater than 0." Loss is due to
friction always use up energy, use up available energy. The loss always has to be a positive
term. This thing here always has to be greater than 0. A loss to the w, friction here in
your car gas doesn't make the car go faster. Never, never, never, never. Why not? Well,
I'll tell you, right here. Second law of thermo. That's why we know greater than 0. OK. So,
we can go ahead and say that this is the energy that leaves, this is the energy that comes
in, and this is the energy that's lost due to frictional effects. So we rewrite that
guy over there. I'll put down again at the bottom here. So we have p out over rho plus
v out squared over 2 plus gz out. Now, everything with 1 are in. This is in. We'll change in
1 and 2 in just a minute. We get tired of running in and out of these things. OK, and
then we have losses. So, this thing right here is called the losses. Right there. That's
the losses. And then once we do that, we say, "You know what, let's bring back in the work
term now." OK, so we bring back in the work term. I'll just give you a [inaudible] here.
This pressure came in because of the work done by pressure forces on the boundary of
the other system. So where did that work come from? We said there's three kinds of work,
shaft work, friction work, and pressure work. Where is the pressure work in here? Right
there and right there. Where's the shaft work? Right there. Where's the friction work? There
would be no frictional work on the boundaries, on the boundaries. Friction inside, yeah,
on the boundaries, no. >> Should it be plus losses or minus losses? >> This is minus losses. And I'll tell you
why I have a plus there. I'd rather write this thing the other way around. You know
how it goes. What comes in this [inaudible] in a second. So what comes in-- Now, I'll
be back in a minute. But anyway, when I write it down, that thing is always a plus sign.
I'll show you why in a minute. OK. So let's go back and include shaft work. So now we
have a p out, rho, v out squared over 2, gz out and now we have our in terms. And that's
equal to w shaft net in minus the losses. That's an equal sign there. Let's get rid
of all of these. This guy up here is equal, this guy is plus right here. There we go.
Well, that's what we end up with. Sometimes we call it mechanical energy equation. Sometimes
we call the extended Bernoulli equation Because it includes now losses and work that's why
it's called extended Bernoulli's. And now, once we've got that basic equation, now we
start to look at different forms of it. OK. So, let's take a look at the first form. By
the way, just so you know, where you see a little w, there's a big w, there's a big q,
and there's a little q. They're a different of course. They're different. You see what
you do to get little q? You take q dot divided by m dot. To get little w, make sure you,
you know, use your lower case w and capital w. Little w is equal to w dot net in divided
by m dot. You divide it by m dot to get the little, lower case w and the lower case q,
OK? Now we divide that equation there by g. This is this. Divide that guy by G. Hs. We
call that w shaft net in. And this divided by g. Hl is losses divided by g. Well, you
got to watch the units. You got to really watch the units, OK? Everything in this equation
here, that guy is in meters. That means every term in there better be in meters. That means
hs and hl are in meters. You finish that. This guy up here is power, Newton meters per
second or joules per second. That's power. Newton meters per second divided by kilograms
per second. Newton meters per kilogram. What's Newton meters? What's that thing there over
there? Work per mass, work per mass. What's this guy here? Little w, OK, little w here,
same units. Newton meters per kilogram divided by g meters per second squared. OK. Take that
little w, little q, they're the same thing. Little w and little q, same thing, OK, divided
by g, meters per second squared, what do I get? Cancel, cancel, Newton seconds squared
per kilogram, the same. What is f? That's one of our engineers, OK? That's why we can
think of big box, OK? You say, "What is a Newton?" Let me go back to basics. What is
a Newton? A Newton is a kilogram meter per second squared. Multiplied by what? Second
squared. Divided by what? Kilogram. And what do you get? Guess what you get, of course,
but nobody's going to tell you that. You're expected to know that. That's why we drill
you with units all the time. Units, units, units. I'm telling you right now, this is
the easy one. What is a kilogram? A thousand grams. What's a kilometer? A thousands meters.
What's a centimeter? About 10 millimeters. What's a liter? A thousand cubic centimeters.
Everything in SI structure of 1 times 10 to some power, everything. Easy to memorize?
Of course it is. I thought about it before. And how about [inaudible]? Barrel, 55 gallons.
How many quarts in a galloon? How many pints in a quart? You go on and on. How many feet
in a mile? How many square feet in acre? Oh my gosh, nothing's structured to 1. Nothing
structured to 1. What do you want to work in? Of course, you want to work in SI. Bad
news, not in this country. Some places, oh yeah, some places. It's getting there, but
really, really slowly. A gallon of milk and a liter of soda, right? We're getting there
but it's slow. So anyway, what I'm telling you is when you're doing homework or exam,
you better pay attention to units because they will kill you. They will kill you. And
you better know how to do stuff if I say "What in the world is a Newton second squared per
kilogram?" "I must have made mistake somewhere." Oh no, you didn't. It's perfectly right. Yeah,
it's important. Don't play it. That's the easy one. That's the easy one. You know, try
to go to g sub z in English Engineering, use pounds, mass and pounds force. Oh wow. Now
let's [inaudible]. What do I divide by 32.2? What do I multiply by 32.2? You better know
that or it can get embarrassing, I guarantee you that. It can get downright embarrassing.
OK, back to here again. Here we are. Everything is meters now. Isn't that nice? So what's
Bernoulli? And that's what a nice form of Bernoulli's. OK. So, we end up with-- let's
put down the w dot. If we got pump-- now, this guy up here, mean shaft, OK? That's this
guy right here, OK? Let's do a pump first or a pump. Because pretty much in our class,
in these energy equations, we probably will have a pump or turbine or at least the possibility
of that in a problem. So, for a pump, w dot pump is equal to gamma qhp. For our turbine,
we have w dot t-- t for turbine, p for pump-- gamma qht. OK, try it again. I'll do the SI.
Gamma. Anytime you see gamma is Newtons. If you see rho, it's kilograms. Newtons per cubic
meter, cubic meters per second. Ht is-- all these hs are in meters. Oh, guess what we've
got here. Newton meters per second. Guess what that is, joule per second. Guess what
that is, a watt. One Newton meter equal 1 joule, 1 joule per second equal 1 watt. Oh,
1, 1, 1 everywhere, a wonderful system. Now, [inaudible] English Engineering, probably
better convert it to horsepower because this will [inaudible] horsepower. Most pumps are
specified in that, your power, maybe horsepower. OK. Anyway, these guys now are all going to
be in watts. OK. I'll erase that right now. OK. And then, the efficiency of the pump,
we'll talk about pump efficiency. Pump efficiency, the power is the gamma qhp divided by the
power into the pump. It always takes more power in and comes out. This is the power
that comes out, but-- the power into the pump, yeah, right. So, if you're driving a pump,
the pump starts to get warm. The water starts to get warm. There are losses in their friction.
So, turbine different. Well, the turbine-- we have w dot turbine. That's the out divided
by the gamma qh turbine. That's what you put in. This is what comes out. It's always less
than 1. This is what comes out. That's what you put in. It's always less than 1. The efficiency
is always less than 1. OK. And now, I'm going to write the three-- there's different equations
you can write into the energy equation. So, remember Bernoulli's, we have three Bernoulli's
back in Chapter 3. I put on the board, three forms of the same equation, the Bernoulli
equation. Here are three forms of the energy equation. OK. The first form. This is on the
board awhile ago. We're going to use one now. Yeah, this is what I want to do. We're now
going to use subscript 1 and 2, not in and out. Here's the second form. And finally
the third form. OK. So the first equation, these terms should be Newton meters per second.
That's joules per second, that's watts. Everything is in watts. Second one, we divide it by mass.
We divide it by the m dot. Everything now is in Newton meters per kilogram which means
work per kilogram of fluid flowing. Or if you try to think about it, divide both numerator
and denominator by time. Newton meters per second divided by kilograms per second. This
then becomes power per mass flow rate or you use this one where now everything is in meters
or feet. This is called the head loss in feet, the turbine, the pump. If there's no pump,
hp goes to 0. If there's no turbine, ht goes to 0. If there's no losses, hl goes to 0.
Yeah. >> Will those m dots always be the same value? >> In Chapter 5, what we're doing, yeah. It
can get-- It doesn't have to be, but normally what comes in goes out. Take a pump. Of course,
what goes in goes out. A turbine, what goes in goes out, of course. Sometimes, no. You
can have mixing chambers. Two things go in, they mix together, one thing goes out. Uh-oh,
changing now. How do you change it? What do we do? We went around the control surface,
didn't we? The control surface, there might be other terms in there, yeah. OK. So now
the bottom line is [inaudible], the bottom line is you're probably going to choose one
of these three to work most of the problems on energy in Chapter 5, OK? And what-- And
we'll work different forms, we'll probably mostly use the last one. Just like in Bernoulli,
we mostly use the last one. It was kind of convenient and simple. Everything is in meters
or feet. That makes life a little bit easier. When you can do it, then normally you do it
that way. OK. So, I'm going to do one problem today and then we're going to do-- next class
meeting, we're going to do-- that day going through a lot of those problems in energy,
and maybe even start in Chapter 6. And by the end of next Monday's lecture, that will
be the end of what you're responsible for, for the next exam which is a week from next
Monday. I give you a week to study after the exam after we finish and say stop here. OK,
that's our plan. Anyway, let's try to see your homework. You've got 102 and 105. I'm
going work 103. You have 102 and 105 homework. I'm going to work 103. All right, there's
a picture in the textbook on this one. We don't need this anymore. All right, here's
an inclined pipe over here and here. There's a manometer here. It's a piece of metric tube.
And it goes up. There's a piece of metric tube here. It goes out, it goes up. It's a
clear tube. This is a review of Chapter 2. We use symmetric tubes to measure static pressure.
If we can't, sometimes you have to do it if you can. We've got several of those of with
the fluids hydraulics with, several of them fluids hydraulics with. OK. The left hand
one-- let me see I got this thing right. And by the way, there's again, ton of work examples
in your textbooks. So, take advantage of the textbook and look at those examples especially
when you're studying for the midterm because it really help, it quite varies. OK. So going
from that left hand point to the right hand point, we have1.5 meters. So from here to
here is 1.5 meters. Now, let's see where our manometer levels are. What's the [inaudible]
the fluid. It doesn't say what the fluid is on that page. So then we got-- there it is--
1 meter and 3 meters. So, the 3 meters is up high. We got 1, also 3 meters, 1 meter
and 3 meter. One meter here, 3 meters here, atmosphere. This is 1 meter. This is 3 meters.
OK. Incompressible liquid. We don't know what it is, water boil, whatever. Flow is along
the pipe. Determine-- Oh, let's see. I don't know if you need-- 0.75 meter diameter. Part
B, [inaudible] part B. It's OK. Determine the direction of flow. So my question is,
is it going down the pipe or is it going up the pipe? That's what we have to find, the
direction of flow. OK. Let's see which one [inaudible], but I'd say, normally with this
guy, so I'll start with him. That's fine. Now, you have to assume-- can't assume point
1 somewhere. So, my assumption is the flow goes from point 1 to 2. And here's point 1
and point 2. So, I'm going to see if that's right, am I right? OK. So, p1 over gamma plus
v1 squared over 2g plus z1. No pump, hp is 0. No turbine, ht is 0. P2 over gamma plus
v2 squared over 2g plus z2 plus losses. There were losses, yeah. A matter of fact he says,
what's the head loss over a 6-meter pipe? So the distance from 1 to 2 is 6 feet. That's
6 meters apart. Yeah. So there's losses, impulsive losses. Now, you know, we're really doing
this for our control line. We derive this thing for our control line. So, if you want--
if you want, you could draw the dash lines around your control volume and all the control
surface. But it's not nearly as critical here as it is on momentum. There, you got to draw
the control volume and control surface. Here, it's like a modified Bernoulli equation. So
as we write it from point 1 to point 2, that's why it's called modified Bernoulli. But if
you want to be official and do it by control volume or a control surface which you probably
would for a pump returning, then it's still [inaudible] there's a point 1 and point 2.
OK. P1 over gamma minus p2 over gamma plus v1 squared over 2g minus v2 squared over 2g
plus z1 minus z2 equal losses. First of all, v1 equal v2. OK, so why? Let's see, momentum
equation. No, of course not. Momentum is for forces. Continuity? Oh, of course it is. Steady
state? Yeah. Incompressible? Yeah. Continuity? M dot 1 equal m dot 2. What's m dot? Rho 1q1
equal rho 2q2, but the rhos are equal. Cancel out, q1 equal q2. What's q? A1v1 equal a2v2.
A1 though is equal to a2. Conclusion, v1 equal v2, got it. Z1 minus z2. Z1 is down here minus
z2. OK. So, that's 1.5. Let's [inaudible]. OK. Let's see-- Oops. There, it's right there.
So, my z1 minus z2 minus 1.5. OK. P1 over gamma, p1 here over gamma. Now remember, how
we get p1 from Chapter 2. We did this in Chapter 2. OK. P1 is equal to gamma of the fluid times
the height from where you are to the free surface. Three-- OK, so that guy is 3-- so
take 3 gamma divided by gamma and you get 3 minus p2. P2 is gamma times 1, 1 times gamma.
One times gamma divided by gamma, 1 minus 1. Three minus 2.5. OK. The losses have to
be positive. It can never be negative. If you would go the other way, go and try it.
Start up here and go down. Call this point up here 1. Call this point down here 2. Use
the same equation and you find out that this number is negative. So impossible, losses
can't be negative. Friction never does create more energy. Friction dissipates energy. That's
why [inaudible] thermo, it has to be held on. Yeah. >> Was p1 given to you or was that derived? >> No. It was from the picture. >> OK. >> P1 is gamma of the fluid in the pipe times
the distance from the pipe centerline to the top of the fluid out there. How far is that?
One. OK. Now, just so we know over here that's why I like to write this equation this way.
The one on the left side and two on the right, because this is the energy that comes in with
the fluid, the energy that comes in with the fluid. This is the energy added to the fluid
from the pump. The pump adds energy to fluid. This is the energy that the fluid has. This
is where the-- this is where the energy goes. It goes out, the location too, it can drive
a turbine to create power, and it has to overcome losses. So, this is the available energy first
three terms. It's used up, it's used up. Some of it at least, some of it maybe of turbine
and the rest of it losses by friction. So it's easy to think that way. The losses are
unavailable energy. You run the car it gets hot. It powers the car. You know, can you
not use that heat from the engine getting warm and the fluid getting warm to drive your
car? >> No. >> No. Gosh, no. It's lost. It's lost. Don't
try and say, "Well, yeah. No, I could put that back in [inaudible] desk." No, we can't.
It's lost to what? The atmosphere and the air around the engine in the hood of the car,
it's lost. You can't recover it. So that's why this is called the available energy. This
is the loss of energy. L stands for loss. OK. So, we'll come back and we'll work problems
next time on that. But, I want to go over a couple other problems that I've had in office
hour and also after class. We didn't get it just by the windows over there. All right,
let's do problem 5. OK. Water flows through-- out through a thin closely space [inaudible]
shown with a speed around the entire circumference of the outlet. Determine the mass flow rate
through the end of the pipe. Look like this. Fluid comes in here and it comes out here.
I'll look down and up above now. Looking down and up above. We have the picture given to
you in the textbook showing this. And the angle here that goes out is 60 degrees. That's
the velocity. It gives that 10 feet per second. Go back to continuity equation in Chapter
5 like [inaudible] board. OK. Rho av, rho av is m dot. So, if I want to know how much
mass flow rate comes out of this thing, I do that. Here's my picture. My control volume
is inside here. Where is my area vector? Take the dot product. V times A times 60. Is--
Continuity says steady state m dot in equal m dot out. You're supposed to solve for m
dot in. That's what you do, OK? Oh by the way, remember now, the area vector always
points outward from the control surface, the control volume, it does. What is the area?
Well, I'll make one. There's the area, the fluid comes out. What is this area? Circumference
times w. So that's the area, 2 pi r times w dimension, OK? So, just so you know how
to properly do it efficiently the correct way, use the product. OK. See that one. Oh,
and by the way, the second problem just so you know
that's this container with water coming in and there's water in here and there's gasoline
up here. Gasoline, the water comes in, it pushes the gasoline out. I think they give
you water in. They want to know how much-- yeah, 100 liter per second, something like
that coming in. They want to know how much gasoline goes out. Well, you know, if I push
a galloon of water in here, guess what's going to happen, a galloon of gasoline goes out
here. That's the equation you need. That's the only equation you need. M dot gasoline
equal m dot water, and the water they gave you q. Rho times q, OK? Rho of water times
q, OK. And that's the change of the weight of gasoline in the tank. That's the weight
of change of the gasoline-- this is gasoline by the way, fuel water. It's kind of-- you
know, fuel water, fuel gasoline, OK. OK. So that's the kind of an easier way out of that
problem just to realize that if you push a galloon of water in the bottom, a galloon
of gasoline is coming out the top. All right, now, let's take the last few minutes and you
get the problem for, well, for homework. And I had two or three people coming on office
hour want to know. >> Professor? >> Yes. >> So, my question on my pipe flow is that
pipe conservation for the continuity for the steady state, m dot gasoline and m dot water
that since the different densities you'll have a larger-- >> Yeah. Yeah, yeah, yeah. Yeah, yeah, yeah.
This is a Q-- pardon me. Thanks. Q water. Now, once you get q water-- once you get q
gasoline [inaudible]. OK. Now, I've got q gas. Now, I put q gas down here multiply it
by rho gas, now I get that gas now. >> OK. All right. >> Thanks. OK, work 12, it's a thinking problem
[inaudible]. I did one in class, go simple. I said, "I'll do a simple one in class and
let the hard one-- you try it." All right, I'm going to do an intermediate one now. Hard
in class but simple in my homework, very similar though. Here we go. One more time. Just--
Again, this is problem 4.12 modified. All right, smoke plume lasts for five hours. Smoke
comes out of a smokestack for five hours. I'm going to put several perhaps on the board.
So, [inaudible] phase. All right, there are three different segments. First, for the time,
t for three hours. The wind speed is in vector terms 10i minus 5j.
Plot where the plume is, show the streakline of the plume at three hours. OK. The origin.
And by the way, the plume in three hours is emitted at 0, 0. Here's what's emitted at,
I don't know, minus something and plus something. That's OK. All right, here's 0, 0. 5, 10,
15, 20, 25, 30, OK, that's fine. 10, 20, 30. Let's see here, 5, 10, 15, 20, 5, 10, 15,
20. All right, the first bit of smoke that comes out of a smokestack when it comes out,
it has that velocity. It goes to the right and it goes down. OK, got it. After three
hours, where is that first smoke that came out, the first smoke emitted after three hours?
Velocity in the x direction, 10 miles per hour. This one, 5 miles per hour. After three
hours, 3 times 10, 30 miles, OK? Yeah, yeah, 30 miles. OK. [Inaudible], that's OK. No,
that's way too big. Sorry about that. I'll be out of the window if I do that. Make this
20, 40, 60, 10, 30, 50. Oh we got-- OK, that's fine. All right, after three hours, 10 times
3, 30, 30 miles it went. Which way? East, 30 miles here. But also has southeast records
of plume. Plume is moving southeast. It went down 5 miles per hour times 3. 15, 15 down.
30 to the right, 15 down. There is the first bit of smoke that left a chimney at time equals
0 after three hours. Where is the smoke that left the chimney at three hours? Right here,
connect the dots. There's the streakline of the plume after three hours. OK. I'll erase
this. For time three to five hours. Now, the wind speed changes. Now, it's a 15 miles an
hour i, that's x direction, plus 10 miles an hour in the j direction. Show the screen--
show the streakline at t equal five hours. All right, 5, 10, 15, 20. OK. Let's see. I
just want to go up to maybe 8, 10, 20, 30, 40, 50, 60, 70, there goes 80. First of all,
where did this line of smoke go? Well, at time equal three hours, it was here. Then
the wind change for how long? From three hours to five hours. What's the difference in time?
Two hours. So, the smoke particle right here. In two hours, two hours times 15 miles an
hour, it went 30 miles to east, 30 miles to east. He's sitting on 30 right now. He went
to 30. He's out there now at 60. How about the y direction? He went north 10 miles an
hour times how many hours? Two hours, two hours because 3 to 5, two hours. Twenty, which
way? North, 20 north. Where did he start? Fifteen. Where is he now? Up here. Where is
that? That's 5. And so right up here. Just so you know where it is. OK. So over here.
And by the way, where did this guy here go? He started at 0. In two hours he went 30 up--
30 to the right, 20 up, 30 to the right, 20 up, 30, 20. He's right here. This guy went
up here. This guy went up over here. These are molecules of smoke. So here he is. That
line if I had [inaudible] the same scope with my scope is correct. It moved up here. This
line moved up here. But wait, there's still smoke coming out after three hours. It goes
for five hours, OK? We've got five hours smoke [inaudible]. OK. Well, when the smoke came
out of stack at three hours and one second, three hours and one second, which way it go?
Fifteen miles an hour times how long? For two hours. Thirty-- two hours times 10? Twenty.
It went to 30 and 20. Thirty and-- Oh, he's sitting right there now, right next to his
buddy right there. That smoke left after three hours and one second. How about the one who
left after three hours and 30 seconds? He's here. Three hours? Three hours, 30. Three
and a half hours, there. Twenty and half hours, there. Five. The last guy comes out, last
guy comes out, he's right here. OK, over here. OK. Now, it's the same problem, right? So,
60 up to 5, here. OK. He goes at 20 and 30. So here's 30 and there's 20. He's up here.
And this one up is right here. There's the whole smoke in five hours, five hours. Smoke
went out that way for a while, went southeast, then the wind shifted, it went up that way.
And the smoke leaving here in 3 to 5, went straight out that way, northeast. North--
that's north, that's east. OK, after 10. For 5 to 10 hours, v is equal to 5i, the wind
is dying down. It's 5 miles an hour east now, 5 miles an hour east. Show streakline at 10
hours. OK, here we go again. Five, 10, 15, 20. Five, 10, 15, 20. And over here. Let's
see, 10, 20, 30, 40, 50, 60, 70, 80. OK. Take the one at 0. Just make that easy on ourselves.
Take the one at 0. That's the last smoke came out. This was the last smoke particle that
came out of the chimney. That wind, five hours. For the next five hours, what he do? He drifted
east at 5 miles an hour times how long? Five hours. Five times 5, 25 miles. He went to
east 25 miles. He's out here now at 85. So, there he is right there. That's the point.
That was right there. He moved down to here, OK? And then let's see. OK. Up there. Let's
see, the scale of light, I guess I'm out of proportion here. Yeah. I'm out of proportion
here. OK, because-- oh, I'm sorry 25. I'm going to go 25, excuse me. He moved 25 and
gone from 60. He moved 25. He's out here. Let's take this guy up here. Now, he's at
60. He's going to go east 5 miles an hour times five hour, 25, 60, 70, 85. Eighty five.
Did he go anywhere in the y direction? No, he didn't. Where was he? Here at 5. Where
is he here? Five. In the end, [inaudible] down first. Left hand side, right hand side.
Where did this guy go up here? He was at 20 and 30, 20 and 30. He moved due east 5 miles
an hour times five hours, 25 miles from where he was. He was at 30 plus 25, 55. There he
is. Once you got those three guys pinned down, now it's simple. Connect the dots. There it
is, the streakline after 10 hours. Oh, big long explanation, but people want to know
how to do these things, so I thought I spent some time on it. That's how you do them, OK?
Yeah. >> So three separate wind velocity, you have
to make separate graph. >> You don't have to, but life is simpler
if you do. >> OK. >> You go get-- I mean ME214 from statics
and try represent one sketch. I guarantee you will suffer sometimes by doing that. It's
always better to draw diagrams. That's why I did this way. You try do [inaudible] like
let's say this solution [inaudible], oh no you guys-- confusion, mass confusion. Trying
to put these three guys on this over here, my gosh, you're asking for confusion. You
simplify something if you spread out and do a graph for each particular time interval.
Yeah, at the time interval. Uh-huh. >> So you put in the initial just 5 into the
equation and you get-- >> I plug in the-- nope, nope, nope. I plugged
in the differential time. How many hours between these two? >> Five. >> Times what? We got 25 miles. Every point
goes 25 miles to the right. Every point on this line and this line go 25 miles to the
right. I made it a little bit easier than the problem you did with the assignment problem.
But, if you can understand this and do it this way, you'll get that problem fine, I
guarantee you. OK. Good stopping point. So we'll see you on Monday. Homework is due on
Monday.
