Fluid Mechanics: Reynolds Transport Theorem, Conservation of Mass, Kinematics Examples (9 of 34)
>>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
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