This screencast will illustrate the use of the Re transport theorem through a water park
example.
So when we are characterizing fluid flow.
In the past we have discussed a control volume approach to simplify a problem and this will
be demonstrated in the following example.
One of my favorite rides at this water park.
Was a lazy river.
It had a series of jets in the middle of the river to propel the fluid and thus move the
lazy river.
Now if we look at one of these areas, drawn in a schematic it would look like the following.
Where we have this block right here.
You are shooting water from the base of the river at some higher velocity.
So what I am interested in looking at is the region 20 foot of the jet.
So lets say that is right here.
We will say that is 20 feet from jet to that point.
We will say that this channel is 15 feet wide, and we will say that the jet shooting the
water is roughly 5 ft.
Right in the middle.
We can also say that the depth of this river is about 5 ft.
Just kind of throwing numbers out there, but it seems some what reasonable.
Now lets say that the water coming out of the jet is at 5 ft/s, and the rest of the
water is at 2 ft/s up to this point, and we will say that 20 feet down river we have uniform
flow and it has reached a point of 3 ft/s.
We will say that the control volume that we are going to look at considers with the system
of interest at a time equals 0.
We want to know what the system of water looks likes 0.5 seconds later, and the volume of
water entering and exiting the control volume after 0.5 seconds.
So we have defined a problem statement.
This point I would actually start with a diagram, and that would be the most useful to solve
this problem.
So here is the problem statement.
Written in words.
So you can stop at this point and try it yourself.
As I mentioned the best place to start is with a diagram, and we will say from the jet
to that down stream location is the control volume of interest, which I have labeled yellow.
Lets go ahead, and put in some relative dimensions that we care about.
We know that this width is 5 ft, which means these two widths are also 5 feet.
So we are curious to find out how this system moves at a time of 0.5 seconds.
So for reference I am going to use a couple of points.
We are going to say this is A, this is B, this is C, that is D, and we have E and F.
If we wanted to calculate the distance that particular part of the system has moved.
We can just take the velocity times time for each of these areas.
So lets do this for section A, B assuming that uniform flow.
We would have 2 ft/s times 0.5 seconds.
So we know that part of the water has moved 1 ft in that amount of time.
So obviously we are going to have to assume that the water doesn't mix even though we
know it would, but we will draw that this has moved 1 ft.
So if we did this for point B, C, at 5 ft/s.
We would have 2.5 ft.
So we can draw that a little bit further down, and the other side is symmetrical so it will
be 1 ft as well.
So if we were interested in that part of the system and what it would look like.
We can see that at 0.5 seconds.
We can see that this system has moved down stream.
We are looking at the region A, B, C, and D. Now for E and F, because it is uniform
on that side.
It is pretty straight forward.
We have 3 ft/s at t equals 0.5 seconds.
We have that it traveled 1.5 ft, and we can highlight this region in light blue.
So now in terms of answering part A. What does the system look like at point 0.5 seconds.
Well in this case it is going to be the light blue region and the yellow region.
The green region indicates what new water has entered the control volume, what was originally
what we have depicted as just the yellow.
The light blue what indicates what came out of the control volume and started flowing
down stream.
So if we are look at the system, the system is what has moved on.
The control volume is still just that region of particular interest that 20 feet down stream.
So to answer part b.
We need to figure out what amount of water has come into the control volume, which would
be indicated in green, and what amount of water has left the control volume, which would
be indicated in light blue.
So lets start going through those calculations.
So at this point we are going to use the Re transport theorem to analysis the mass flow
of both in and out of this control volume.
So we are going to use large B as the mass, which we will indicate as m, and if we do
this then that little b is going to equal 1, and our equation simplifies to the following.
So if we start going though our assumptions that our mass per times is not changing, then
the left side of this equation goes to 0.
If we assume incompressible, and that means that density does not change.
Then that left integral looks like the following.
We have also removed the density in the right integral, because again we can divide though,
and the last one that we are going to make is that our control volume is non-deforming.
So this basically means that we have steady state for the control volume, and this term
drops to 0.
So we are left with a pretty straight forward integral.
That basically says we want to know the total amount of mass that comes into this system,
and that most equal the total amount that leave.
So if we look at what comes into the control volume.
Our original control surface was marked right here.
We have broken that up into 3 regions.
Two of them are identical the bottom and the top.
So we will say that we have 2 times whatever has broken the control surface.
So we will call this region 1, this region 2, and this region 3.
So we are looking for the control surface for 1 and 3.
This is going to be the dot product times the velocity times normal over the area of
interest, and we also need to do this for that second region.
So at this point we can start putting in given velocities, and the areas of cross-sectional
flow, but since we are interested at a very specific times of 0.5 seconds.
We can just take the fluid volumes.
Since we have already calculated the length that the water has traveled into our control
volume.
We know the widths and the depths.
So if we want to calculate that entire volume.
We will say a volume of section 1, plus volume at section 2, plus the volume at section 3,
must equal the volume that has gone into our control volume, and I have plugged in numbers,
and if we do this we calculate that 112.5 ft^3 of water has entered our control volume.
If we do this for the other side.
What is left is a little bit more straight forward, and we will just skip straight to
our volume lets say is 5 our the term out.
If we calculate this we also get 112.5 ft^3.
So in this case we have shown that the volumetric flow rate in is equaling the volumetric flow
rate out, and we have answered the second part of our question by determining what has
entered our control volume and what has left.
Hopefully this gives you an idea of both the system approach and the a control volume approach
if we are looking at flow.