00:01
We have a pipe.
00:02
And we have location one and location two.
00:07
And we're trying to figure out the change and pressure between the two locations.
00:12
We're told the density of the fluid flowing through this pipe is 1 ,330 kilogram per meter cubed.
00:20
And at location one on the pipe, the velocity, is 9 .97 meters per second.
00:32
And our diameter of the pipe at that point is 12 .9 centimeters, which if we divide this by 100 to make meters, it would be 0 .129 meters.
00:45
At location 2, the diameter is now 15 .1 centimeters, again, divide by 100.
00:54
You get 0 .151 meters.
00:59
But we don't know the velocity at this point.
01:01
We will have to find that.
01:03
So this is everything we need to solve.
01:06
The first thing we need to do is figure out how fast it's moving in the second part of that pipe.
01:11
We're going to use the volume flow rate equation where the area times the velocity in one point, let's make this look a little smaller so it looks more like velocity, is equal to the area and velocity at the other point.
01:25
The volume flow rates remains constant throughout.
01:28
And so area of the pipe is going to be pi r squared on both sides.
01:39
And you can see that if you divide both sides by pie, that will cancel out.
01:43
We don't want diameter.
01:45
We want radius.
01:46
And so we're going to divide these both by two.
01:49
So one point, so it's the radius of one.
01:52
We'll put that right here.
01:54
It's 0 .129 divided by 2.
02:00
And when we do this, we get 0 .0645.
02:06
We want the radius at the other part of the pipe, which is the .151 divided by 2 because radius is half a diameter.
02:17
And we get 0 .0755.
02:22
And these are both in meters.
02:24
So radius 1 is 0 .0645.
02:29
This is squared times our velocity, which is 9 .97, is equal to our r2, which is 0 .0 .0 .0 .0 .5.
02:41
0755 squared times velocity at section 2, which is what we're looking for.
02:45
So we're going to have that first radius squared times the 9 .97 will give us 0 .0415...