00:01
So in this problem, we have water flowing in a pipe and the pipe actually converges or tapers to a smaller diameter.
00:11
So we're given the starting diameter of the pipe and i'll label that d1 and the diameter at the end of the pipe or i guess as the pipe tapers i'll call a d2.
00:24
And so these corresponding diameters are given the problem.
00:27
I'm just going to go ahead and convert them to si units.
00:35
We're also given the pressures at these particular points in the pipe.
00:41
I went ahead and wrote these down over here.
00:43
Note that the unit is in pascal's.
00:46
So now let's go ahead and write down some useful equations in fluid mechanics.
00:51
So let's write down bernouille's equation and particularly in which we can compare pipe flow in the portion of the diameter 1.
01:02
The pipe flow within diameter 2.
01:05
So i'm going to say in our bernoulli's equation we have static pressure plus row gh1 plus one half b1 squared is going to be equal to static pressure at 2 plus row gh2 plus 1 half row v2 squared.
01:30
And so we can actually simplify some terms within this problem or within this bernolese equation because if we assess a streamline that is going along this pipe, you'll notice that in the streamline there's no height change.
01:46
It continues along a linear horizontal path.
01:50
And because of this, we can neglect any pressure or pressure difference associated with height.
01:57
So now we have a more simplified version of bernoulli's equation for our problem.
02:02
Another equation that's going to be useful is the continuity equation, and this essentially satisfies continuous flow within a circular cross -section, or in this case our pipe.
02:16
So a1 times v1 is equal to a2 times v2.
02:22
And based on this continuity equation, i just wrote, we can actually develop a relationship between the corresponding velocities.
02:30
So let's go ahead and calculate the area of the thicker section or the part of the pipe with a thicker diameter.
02:40
So a1 is going to be equal to pi over 4 times d1 squared.
02:48
And likewise, a2 will be equal to pi over 4 times d2 squared.
02:57
And so if we want to assess a ratio using this continuity equation, we can read.
03:02
Rewrite and i guess essentially represent a ratio in a relationship of v2 to v1.
03:10
And this will correspond with a little bit of rearrangement in which we get v2 over v1 is equal to a1 over a 2.
03:22
So let's go ahead and plug in what we have for a1, so pi over 4 times d1 squared over pi over 4 d2 squared.
03:38
So right off the bat you can see that the pi over four terms can cancel out...