>> All right. We're going on to the last part of this chapter,
it's a very important theorem that we need, we use it in multiple disciplines if you'd
have any 301 trigonamics, you've seen it there. It's called a Reynolds Transport Theorem a
little background first. We're going to be talking about system and
control volume. The first thing is the control volume, like
it says here, is a fixed region in space. So it's a particular region in space. I mean you could analyze your hot water heater
at home in your garage maybe or in your house, as a control volume. You can analyze a boiler as a control volume. All these pieces of equipment are typically
analyzed by a control volume appropriate. So rounding the control volume is a surface,
a boundary. That surface is called a control surface and
these two guys are abbreviated CV and CS, control volume and control surface. So we're going to look at now mass crossing
a boundary, which is going to be maybe control surface. So we'll start off and we'll, here's some
flow passage. So flow passage bounded on both sides. And we'll identify by a dash line. The control volume. The control surface is a boundary. We can call this for instance .1 and this
is .2. And maybe flow enters at 1 and flow will come
out at 2. And we can also draw that, look like this
take that control volume out now, control it like this. Or it might look something like this. So this is our control volume and we could
have a velocity coming out here. This would be v2, this would be v1. I'm going to hop to the side calculation over
here just, I'm going to use that calculation with derivation over here. So let's look at this real quick so you can
get an idea for doing that. Let's say there's a pipe here. And we have flow coming in and flow going
out and we'll just label that v. The pipe is cut at an angle, the angle it makes with
this is theta. We're going to call this area right here a
and we're going to call this inclined area a theta. A is the cross section area of the pipe. So get you a piece of PVC pipe, cut it at
a 45 degree angle and let water run through it. The water's going to come out just like this,
but the area's at a diagonal like that. Thie flow rate for this q would be equal to
v times a. Cross section area, velocity. We mentioned this back away, we had q equals
v times a, m dot equals row times q equals row va so we've talked about that earlier
in class. So the volumetric flow by q equal the average
velocity times the area. Okay let's rewrite this thing a little different. Write it in terms of a theta. Get rid of the a and make it a theta. So a theta times the cosine of theta, same
thing. Same thing if you do a little right triangle
there. This guy right here looks something like this,
magnitude v, magnitude a times cosine theta. That looks like the dot product. Okay so, if we want to deal with flow at q,
it's the velocity vector dot with the area vector. Just want to make sure you know. This area vector always points outward from
the control volume. Like we mentioned, always points outward from
control volume. Okay we'll use that over that there [inaudible]. So, if we want to find q over here it's equal
to v dot a. Don't forget the area vector always points
outward from the control volume. At 2 the area vector points outward from the
control volume. At 1 the area vector points outward from the
control volume. Okay that's how you set it up that way. Oh, that's not, pardon me that's q. If you want m dot, put on the board over there,
m dot equals row times q. Okay. I guess I better put down, let's see here. Those vectors point in the same direction. The angle between them is zero degrees they're
collinear. Okay the angle between them is zero. Cosine is zero, is plus 1. So this sine here, cosine theta equal cosine
0 equal plus 1. Over here, cosine theta equal 2 vectors pointing
opposite, the angle 180. Minus 1. For a dot product has a sine, if the flow
comes in, it's negative, if the flow goes out it's positive. The area vector always points outward from
the control volume. If you want to to label this again we'll be
real precise now, control surface. All right so m dot is that. And then q net equal sum over the control
surface of v dot a and m dot net equal sum over the control surface row v dot a. Now don't forget v dot a is positive if stuff
goes out, v dot a is negative if stuff comes in. So this is the net out flow, m dot net out. What does net mean? Subtract something. The word net means subtract something. The net balance in your checkbook, you got
to subtract something. What came in minus what goes out. In this case is what goes out minus what comes
in. So q net out then, in that case, would mean
the flow rate leaving minus the flow rate entering. The flow rate leaving minus the flow rate
entering. Okay so that's just some background that we'll
need for that. Okay so let's go on. If an extensive property. Leaves the control volume, the dot means with
respect of time. The rate of change of b with respect of time,
b dot. Oh b can be different variables. One of the most common ones is energy. We're going to do that in about two days,
energy. So maybe capital b means energy. Little b, is big b divided by mass. Capital b is extensive. Little b is intensive. So let's just for the sake of giving it some
kind of name. Let's say capital b is energy in si joules. Little b is then joules per kilogram. Okay so that's what we mean by that. And if you want to be real official and if
it's not a uniform flow across the inlet and outlet. If it's not uniform flow. Okay. That's uniform flow. That's not uniform flow. Non-uniform flow. You can use a summation. If the velocity profile varying across the
flow of area then you have to be more official and use the interval definition. Because then v is going to depend on a, da. Okay so two possible forms of it. One for uniform flow, one for non-uniform
flow. We'll pretty much deal with uniform flow for
a while. So it's going to be uniform flow that we're
looking at. All right let's draw our picture one more
time. I'll change it a little bit. The flow passes right. I don't know if you want a subtle line coming
out. What's inside the solid line? Solid line is system at time t equals 0. That's a system, is the control volume. Control volume, a fixed region of space. A system, a collection of matter, a bunch
of molecules, of fixed identity. So if I had. Fixed system. Like this where the flow comes in here and
the flow goes out here somewhere. If I want to do a system, I'll say okay, I'm
going to take. These particles and I'm going to spray paint
them maybe. I don't know green. So here is, there's a system at time t equals
0. This water in the pipeline. Wait 15 seconds, it's up there now, wait 30
seconds, it's down here now. Wait a minute and a half, it's up here now. Wait a minute and 3/4's, comes out here. There it is. Identify every particle I'm going to make. Particle a, 1, b, 2, c, 10. Everyone's tagged, named. Fixed identity. We don't do that generally in engineering. If I'm analyzing my hot water heater in my
garage, I'm not going to watch the water molecule come from the city water supply, go into the
water heater, get heated and end up going out the faucet in the kitchen, no. I'm going to draw a system boundary around
that hot water heater and I'm going to analyze what comes in cold water and what goes out
hot water. Now over here I'm going to draw a control
arm, I might say I'm going to look at this pipe fitting, it's called an elbow a 90 degree
elbow. I'm going to analyze that 90 degree elbow. That now becomes a control arm. Okay, back to here again. That's like this. I'm going to follow those particles. But. Reynolds Transport Theorem is a missing link
type equation. Most of the laws in science, physics and so
on are written for a system. Okay. In engineering though we do a lot of control
volume studies. So this theorem allows us to take basic laws
written for a system and convert it to one more useful to us the engineers which are
in control volume form. Okay so what I'm going to do now, is I'm going
to draw and relate the control volume to the system. All right, I'm going to call that same volume
at time 0, my control volume. Okay so here's my control volume. Okay that is also going to be my control volume. This is the control volume for anytime, remember
we said control volume fixed in space, not moving in space, fixed, that's the control
volume. They choose to make the coincident at times
0. Okay, wait 30 seconds, does the control volume
move? No, I'll say it again fixed in space. Does a system move? Yeah I just showed you. Now the system moves to a new location, down
that way, cause the velocity is here. Okay now, where's it now? Okay that was a solid line, now it's going
to be a dash line. Dash line, system at time t, I think I'll
call that, just dt, yeah helta t. Okay now way back over here. All right, next step for
the system. Db dt in the system. The amount of b, hows the amount of b change
with respect to time in the system? Don't forget if you get confused just say
to yourself energy. Okay, energy. If you watch a cubic inch of water come into
your cold water pipe in hot water heater and you watch it go through the hot water heater. Do you think it's going to change energy? Oh yeah. Why? Cause you're adding heat from the natural
gas burner. Yeah as it moves through the hot water heater,
it's going to change its energy. Well b is like energy, b can be various properties. Energy, momentum, mass. So this is the time rate of change, the amount
of b in the system. Start off, basic definition, what is a derivative
limit as delta t approaches 0. The amount of b in there at time t plus delta
t minus the amount of b in there at time t equals 0. T equal t, we won't call that 0. We won't use a 0 just call it time t. Divided by, what was the time change, delta
t. Equals. Okay the amount of b in there at time t plus
delta t. I'm going to call this region 1, this region
2, this region 3. At time t equals 0, there's the system, black
line. The amount of b in region 1 plus the amount
of b in region 2. Let's put that here. Let's do it this way. I'll put this down first and explain it. I'm going to keep this in the right order. Okay the amount of b in region t, was the
amount b1 plus b2. T plus dt, system at t plus dt. Dash lines 2 plus 3, 2 plus 3. Okay. Keep going rearrange. Db, dt system. You can see where the terms came from. Nothing new was added, it was just moving
the terms around. This term is this term, this term is that
term, this term is that term, this term is that term. Yeah just rearranging the two. Okay let's look at this term first. The amount of b in region 2, lets see what
is region 2? Region 2, that's right here, see in the middle
here. In region 2 at time t plus dt time t. Okay. All right so this guy right here in the [inaudible]
db cv dt. This guy right here, what went out minus what
came in. What went out minus what came in, the amount
of b that went out minus the amount of b that came in. Okay, so this term right here. Okay. So, I just erased it unfortunately, maybe
it's still up here, yeah okay here right here. B row v dot a is what came out minus what
came in. So that's what that term is. Let's then combine these two into one equation. We'll put that here. Okay db system dt, d dt. Okay now we did this guy right here. This is the amount of b in the control volume,
the change of it with respect of time. This is the amount of b in the control volume. So this is b in the control volume, integrate
over the control volume. Let's see what the integration is. I'll just pretend it's si. What is b? Okay let's just say b is joules again to make
something up, energy in joules. So b is joules per kilogram. What is row? Kilograms per cubic meter. What's the volume? Cubic meters. When you're done, what do you have? Joules. That's the amount of b in the control volume. That's a function of time. That's what the interval represents. If it's steady state of course, goodbye nothing
but the time, that term goes out if we're steady state. Okay we're almost there. This one, time rate of change of b in the
system. This one time rate of change. Of b in the control volume. This one, net outflow rate of b. Okay, this guy here, if it's not uniform flow,
okay remember not uniform flow or if not uniform flow. The we have the interval of b row v dot n
da. This guy unit vector in the n direction. Okay I can't botch it in, there's to much
up there. That final equation right there, in symbolic
terms and word terms that is the Transport Theorem, Reynolds Transport Theorem. What is it relate? Well it relates on the left hand side, so
there's the board of course. Changes in the system, that's the basic laws
of science or generally expressed it. We engineers want to express in control volume
terms. So the right hand side is the control volume
terms. Control surface there, control volume here,
give a steady state that term goes out. Okay so that's what you use for mass, momentum,
and energy in chapter 5. Will that capital b be mass first of all,
then momentum and finally energy? And develop three equations from that Reynolds
Transport Theorem. Okay so you have an example for homework. Let me see which one you got assigned. Yeah you have 68 and 69. I'm going to work problem 70. It's somewhat similar to 69. So let me, I don't need this. I need that over there maybe, we'll see. Okay see what problem 470. Let me draw a picture first. Here's the picture given in the textbook. There's two plates pulled in opposite directions. Let's see here, the speed of the plates are
1 foot per second. So this plate will be pulled this way, velocity
equal 1 foot per second. And the other one is being pulled to the left
at 1 foot per second. Okay, the oil is between the two plates. I'll just put it there. Okay, velocity is 10y times I. Y is measured from the middle. And the distances are 0.1 feet, [inaudible]
talking about 2 feet. That's, let's just make sure that's what it
is. Uh 470. No, 1/10th. Yeah 1/10th, 1/10th. Okay now let's see if I can find. Okay. Fixed control volume a, b, c, d. 2/10ths, 2/10ths, okay fixed control volume. Okay see a, b, c, d. Good. That is showing this coincides, cv coincides
with system at t equals 0. So, here's the system at times 0. Okay system at time. And the control volume, plenty of time. Control volume is not moving, fixed control
volume. Okay. Okay let's see that's still given heat reading. Indicate the system at time t equal .2 seconds. So system at time .2 seconds. It's going to be a solid blue line. The velocity, okay let's draw the velocity. At the center y equals 0 velocity equals 0. At the top y equal 1/10th, v equal 1. I already put it up there for you. At the bottom y equal minus 1, points left
yeah v is 1. Between the two, v is the function of y only,
linear. Linear. So it looks something like this. There's the velocity field for the oil. After 2/10ths of a second, the oil molecule
in the middle between the plates move. No, because the velocity is 0. Did the oil particle at the very top move? Yes. Which way? To the right. How much? It's going at 1 foot per second. For how long? 2/10ths of a second. 1 times 2/10ths, it moved over 2/10ths of
a foot. By the way I didn't put this on here but it
started out at 2/10ths and 2/10ths here. This was given. Okay so that top molecule is right there times
0. 2/10ths of a second, it moved over 2/10ths
times 1, 2 times second time 1 foot per second, 2/10ths of a foot. We're going to start out at 2/10ths of a foot. Where did it end up? Another 2/10ths out. This particle moved out here, to there. Where did the middle part move? It didn't move. How about the one down here? It went the other direction. This guy right here, where the big blue point
is. He went to the left. How much? 1 foot per second time how long, 2/10ths of
a second. He went 2/10ths that way. He's back here. Now the guy up here, at point b on the plate
or molecule. Which way is here going? To the right. At what rate? 1 foot per second. For how long? 2/10ths of a second. He moved 2/10ths of a foot that way. He's up here. The guy in the middle, did he move? No. It's linear of course. The guy at the bottom at point a, the oil
molecule, on in that plate. Did he move? Yes he did. What's his velocity? 1 foot per second. Which way? To the left. For how long? 2/10ths of a second. 2/10ths times 1. Okay he's up here now. 2/10ths, he's right here. Now the blue on this chapter here. Better have multicolored pens and catching
techniques because you're going to need it. This is not, don't make these things postage
stamp size, you'll be crazy when you get done, make them big. So that's the answer. Where is they system after 2/10ths of a second? In the blue boundaries. It got skewed like this. For the last part, the last part says, identify
the amount that was leaving and the amount that came in. Okay this is 1. This is 1, this is 2, oh let's make that 3
make it sensible. We'll make that 2. This is 4. 3 left, 2 out. So 1 and 3 out. All these fluid molecules moved out of the
control volume. What came into the control volume? These guys along the edge they came in, these
guys that were on the edge they came in. 2 and 4 came in. So some fluid came in some fluid went out
carrying the property capital b. So again what did you start out with? The control volume and the system were coincident. Then what happened? You wait 2/10ths of a second. Draw the new system about these. I did the solid blue line. Third part, identify the amount of material
that came in and went out of the control volume. 1 came out of the control volume, 3 came out
of the control volume, 2 came in to the control volume, 4 came in to the control volume. Your homework is just about like that but
a different geometry. Identify the system after a certain time,
show the stuff that came in, the stuff that left, sketch it. And again just trying to explain to you about
the Reynolds Transport Theorem, what it means, what it means. Okay, now our last step today. Our last step is, we start with number 1,
we're going to let b be a certain quantity. I told you there was 3, it would be mass,
momentum and energy. We're going to do all 3. Today we'll take a easy one, mass. Better need it for your hot water heater. If you got a gallon of cold water coming in
from the city, guess what's going out to your dishwasher in the kitchen. A gallon. Why? It's a steady state. What comes in equal what goes out. So yeah you know some things are pretty straightforward. So, let's look at our mass then. We'll call this continuity equation or conservation
mass. Let capital b equal the mass of system. Then little b equal mass of the system divided
by mass. So of course little b just becomes 1. Kilograms per kilogram, just 1. And Transport Theorem. Says the following. Don't forget now capital b is mass. D mass, dt of the system equal d dt this is
the control volume. Don't forget little b is 1, there was a little
b in there, the little b was 1 plus interval over the control surface row v dot da. Okay, remember on the board we had, still
on there, yeah okay good. Our board, a system is a collection of matter
of fixed identity. I showed you the pipe. If there is 1 pound of water at 1 minute and
you tag them with some kind of spray paint or something. After another 2 minutes, did their mass change? No because we're following the same particles. So the mass is not changing for what? A system. So this term here is 0. That's the system approach, it set conservation
of mass. If we don't lose particles or gain particles,
we have the same particles, then their mass didn't change of the system. Okay, so now this becomes d dt, interval control
volume row dv. Don't forget d slash is volume, v no slash
is velocity. Some people aren't to good about that and
they don't make a little v very good, with a big v. I saw an example. There's a difference. Little v is the velocity in the, time form
of velocity, it's in the y direction. What's big v? It's a velocity, it might be the average velocity
we don't know, it's a velocity. But this guy is a specific velocity and the
v slash is a volume. Okay anyway back to here. I'm going to assume uniform flow again to
make things simple as we go through. So our flow in whatever it is, in pipes whatever
it is. Is assumed to be uniform. It doesn't change with the cross section area. So get rid of these intervals and make it
summation over the control surface of row v dot a. Okay now if the flow is 1 dimensional and
steady state. Okay. Okay then if that's the case, the steady state
this guy 0, so this term here is 0, okay. Of row v dot a. All right don't forget those two terms at
0. I'll put it down here. So if it's steady and uniform flow 1 dimensional,
then we got this guy right here. Well don't forget about there, I had it before
it's in your notes. This guy here is a mass flow rate. Don't forget if it leaves it's positive, if
it comes in it's negative. So this is m dot in equal m dot out. If you want to do it officially and it looks
like this. Outflow is positive so row out v out a out
minus, if it comes in it's negative sign, row in v in a in equals 0. M dot out minus m dot in equals 0. Which gives that guy right there. Okay. Next step if it's incompressible. Also besides b, 1d instead. Row in equal row out for steady flow. Okay so here we go cancel out the rows, cancel
out the rows. So v out a out equal v in a in. Our q out equal q in. Okay. Okay brought some energy in. Okay those equations carry with them some
assumptions and they better be true or don't use them. Number 1, steady flow, said it up here. Number 2, uniform flow, 1d flow. If those 2 are the assumptions you can make
then m dot in equal m dot out. Of what? The control volume. If the 2 is incompressible or can be treated
as incompressible, the density of this thing in and out. Then you can say the volumetric flow rate
in equal the volumetric flow rate out. That's what I said if one gallon of water
goes in to your hot water heater cold water. Guess what happens? One gallon of hot water comes out at the other
end. Why? Because then your hot water heater, the flow
is steady 1d and the fluid is incompressible. If those aren't true you have to go back to
the basics like this. If it's not steady you got to use the interval. If it's not uniform, pardon me, use the interval. If it's not steady keep him in there. Okay. Okay so that is the conservation of mass. Okay I just a good stopping point, I don't
want to do an example because I won't get finished with it in the time we have. So we'll stop for today and pick it up then
on Monday.
