00:02
Okay, so what we have here is a closed pipe system, and they want us to find out what p1 is on the left end of this pipe system.
00:13
They've been given us some information, a pretty good bit too.
00:18
This is a bernoulli's problem.
00:21
Bernoulli's problems are notorious for having a lot of variables involved, and we do.
00:27
If you look, the diameter of the left end of this pipe system is 0 .27 meters, and it looks like the the pipe system narrows down as it goes to the right and it drops down to a lower position.
00:42
D2, diameter 2 is 0 .05 meters.
00:45
The height decreases by 7 .88 meters as you go from right to left in the closed pipe system.
00:52
The velocity at the left end of the pipe on the larger end is 1 .21 meters per second.
00:58
They did not give us the velocity on the right side.
01:01
We'll have to find that in order to continue this problem.
01:03
Problem.
01:04
The atmospheric pressure is 1 .01 3 times 10 to the 5th pascal's for this whole problem.
01:11
That's the same thing as one atmosphere, which i have that conversion factor over here to the right, because we'll need it as we need to be able to convert our answer to pascal's, or make sure that it is in pascal's.
01:28
That brings me to the pressure two, that the pressure two on the right end of the pipe system is actually 1 .4 atmospheres as it's given.
01:40
Let's go ahead and convert that into pascal.
01:45
So in order to do that, i know that one atmosphere is 1 .03 times 10 to the 5th pascal.
01:54
And if i do the math, the atmospheres cancel, and i get 1 .4 times 1 .1010.
02:07
What that gives me is 1 .4, let's see, 1 .41820 times 10 to the 5th.
02:23
I'm just going to leave the number like that for now because it's, i don't want to round too much in the middle of my problem.
02:30
I'd rather round at the end just to keep our answer as accurate as possible.
02:34
Gravity, we need to use 9 .8 meters per second squared here for this problem.
02:39
The density of the fluid flowing through this closed pipe system is 1 ,392 kilograms for cubic meter.
02:46
I think that's about it.
02:48
Okay, so what we need to do first, in order to use bernoulli's principle to solve for p1, we need to know how fast the fluid is flowing through the p2 side of the pipe.
02:59
So we need to find out what is, by finding out what is p1, we need to first, we need to first find v2 using continuity of flow principle.
03:15
Continuity of flow, which says that the fluid flowing through one area of a closed pipe system, times its speed through that area is equal to the fluid flowing through another area of the closed pipe system at that speed.
03:36
It's an inverse relationship between area and velocity as you decrease the diameter or the cross -sectional area of the pipe, the velocity through that cross -sectional area increases.
03:49
And as you increase the cross -sectional area, the velocity decreases.
03:53
So we fully expect because we go from a larger end of the pipe to a smaller end of the pipe cross -sectionally, that the velocity should speed up because continuity flow says that however much mass you have flowing through one end of your closed pipe system, that must equal the same amount of mass per unit time flowing through the other, a different end of your closed pipe system with a different cross -section.
04:23
So let's solve this for velocity to b2.
04:27
We want to know what that is.
04:28
That's going to be the ratio of a1 over a2 times v1.
04:35
Okay.
04:36
So we can go ahead and do this.
04:39
If we know that the area that we're using, these are cylinders, these pipes are cylinders.
04:44
So we have a pi r squared kind of scenario here for the area.
04:49
All right.
04:50
So what we're going to do is just kind of carry this over and we'll figure out what v2 is.
04:54
We know that the area is based on the radius, and the radius is going to be half the diameter.
05:01
So half of d1 is 0 .135.
05:04
So we'll put that in right up here, 0 .135 squared, it's r squared, times pi over, and then the diameter at the right side is 0 .05, so that's 0 .025.
05:20
By the way, that's meters.
05:22
I forgot to throw that in.
05:23
There we go.
05:26
Squared and let's see, pi.
05:30
And then all of that is going to be multiplied by the velocity v1.
05:34
1 .21 meters per second is what they gave us for that velocity.
05:39
Okay.
05:41
So if we run the math and things cancel, pies cancel out, the meters will cancel, and we need to just simply square the radii here and do the division, then multiply by 1 .21.
05:56
When we do that, we end up getting a much faster speed through that cross -sectional area.
06:03
The pipe, 35 .3 meters per second.
06:06
It should be faster because, again, we saw the diameter, the cross -sectional area, decreased.
06:15
And because it decreased, in fact, it decreased by a lot.
06:19
0 .27 decreased all the way to 0 .05 in terms of diameter...