1 00:00:11,320 --> 00:00:14,240 STUDENT: Have you ever wondered how rivers flow, 2 00:00:14,240 --> 00:00:16,660 or how fluid travels inside pipes? 3 00:00:16,660 --> 00:00:20,620 Watch this video and you will learn how. 4 00:00:20,620 --> 00:00:24,030 If you apply Newton's second law to fluid motion, 5 00:00:24,030 --> 00:00:27,610 and at the same time, consider the viscosity of the fluid, 6 00:00:27,610 --> 00:00:30,925 you will get the equations you need to describe fluid flow. 7 00:00:30,925 --> 00:00:33,610 The equations that you get this way 8 00:00:33,610 --> 00:00:36,410 are called the Navier-Stokes equations. 9 00:00:36,410 --> 00:00:40,310 This is the general form of the Navier-Stokes equations. 10 00:00:40,310 --> 00:00:43,810 Solving this equation in this form is not easy. 11 00:00:43,810 --> 00:00:46,720 Interestingly, mathematicians have not yet 12 00:00:46,720 --> 00:00:50,050 proven that solutions exist to the Navier-Stokes equations 13 00:00:50,050 --> 00:00:51,610 in three dimensions. 14 00:00:51,610 --> 00:00:54,710 To make our life easy, let's consider motion 15 00:00:54,710 --> 00:00:57,150 for fluid in one dimension. 16 00:00:57,150 --> 00:01:00,670 The Navier-Stokes equations then become this. 17 00:01:00,670 --> 00:01:03,850 As you can see, I've broken down the previous problem 18 00:01:03,850 --> 00:01:08,290 in three dimensions into just one dimension, the x dimension. 19 00:01:08,290 --> 00:01:12,520 The main body force that is Fx acting on a fluid 20 00:01:12,520 --> 00:01:14,896 will be the force due to gravity. 21 00:01:14,896 --> 00:01:17,710 If the fluid is flowing at an angle, theta, 22 00:01:17,710 --> 00:01:20,170 to the horizontal, then the component 23 00:01:20,170 --> 00:01:23,110 of gravity in the direction of fluid motion 24 00:01:23,110 --> 00:01:26,890 is going to be g sine theta. 25 00:01:26,890 --> 00:01:28,930 We now have an equation that looks much 26 00:01:28,930 --> 00:01:30,820 easier than a general form. 27 00:01:30,820 --> 00:01:33,760 As you can see, this is just a restatement 28 00:01:33,760 --> 00:01:38,680 of Newton's second law, with the viscosity term included. 29 00:01:38,680 --> 00:01:40,990 On the left side of this equation, 30 00:01:40,990 --> 00:01:46,590 you get the net force, acting on a unit volume of the fluid. 31 00:01:46,590 --> 00:01:49,610 The three terms on the right side of the equation, 32 00:01:49,610 --> 00:01:51,790 which are governed by the viscosity, the pressure 33 00:01:51,790 --> 00:01:56,020 gradient, and the force due to gravity together, 34 00:01:56,020 --> 00:01:58,960 give the net force acting on a unit volume. 35 00:01:58,960 --> 00:02:01,150 As you can see from this term here, 36 00:02:01,150 --> 00:02:04,150 this is a second order differential equation. 37 00:02:04,150 --> 00:02:09,610 To solve this, we will have to impose two boundary conditions. 38 00:02:09,610 --> 00:02:13,150 For the river problem, we will consider the velocity 39 00:02:13,150 --> 00:02:15,760 at the bottom of the river to be 0. 40 00:02:15,760 --> 00:02:20,140 This is true because the river bed imposes resistance 41 00:02:20,140 --> 00:02:25,030 to the fluid layer just above it, preventing its motion. 42 00:02:25,030 --> 00:02:27,610 The velocity of the river reaches a maximum 43 00:02:27,610 --> 00:02:29,830 towards the top of the river. 44 00:02:29,830 --> 00:02:34,510 So we get the condition that dVx over dy equals 0 at the top. 45 00:02:34,510 --> 00:02:37,660 For the pipe problem, using the same argument, 46 00:02:37,660 --> 00:02:41,080 the velocity at the rim of the pipe is 0. 47 00:02:41,080 --> 00:02:44,650 And at the center of the pipe, dVx over dr 48 00:02:44,650 --> 00:02:47,800 equals 0, where r in this case is 49 00:02:47,800 --> 00:02:51,400 the distance from the center of the pipe to the point we 50 00:02:51,400 --> 00:02:53,470 are looking at. 51 00:02:53,470 --> 00:02:55,450 Now that we have our differential equation 52 00:02:55,450 --> 00:02:58,540 in one dimension and the two boundary conditions needed 53 00:02:58,540 --> 00:03:02,770 to solve it, there is nothing left to do but to solve. 54 00:03:02,770 --> 00:03:04,960 We can do this using Mathematica. 55 00:03:04,960 --> 00:03:09,580 I was able to solve this equation using Mathematica. 56 00:03:09,580 --> 00:03:14,360 In the pipe, the velocity at a distance, r, from the center 57 00:03:14,360 --> 00:03:16,680 is given by this formula. 58 00:03:16,680 --> 00:03:20,830 And in the river, the velocity the height, y, in the riverbed 59 00:03:20,830 --> 00:03:23,270 is given by this formula. 60 00:03:23,270 --> 00:03:26,410 If you want to know the average velocity inside the pipe 61 00:03:26,410 --> 00:03:28,660 or inside the river, all you have to do 62 00:03:28,660 --> 00:03:33,260 is add up each individual velocity at all possible radii, 63 00:03:33,260 --> 00:03:35,480 or heights, you can think of. 64 00:03:35,480 --> 00:03:39,700 And then divide that sum by the area that you added it over. 65 00:03:39,700 --> 00:03:42,930 It's just integrating and dividing. 66 00:03:42,930 --> 00:03:46,300 Mathematica gave me this result. You 67 00:03:46,300 --> 00:03:49,380 can see that the average velocity of the fluid in a pipe 68 00:03:49,380 --> 00:03:52,510 and in a river increases if you increase 69 00:03:52,510 --> 00:03:54,910 the density of the fluid, or if you increase 70 00:03:54,910 --> 00:03:57,670 the inclination of the pipe or the slope of the river 71 00:03:57,670 --> 00:03:59,470 to the horizontal. 72 00:03:59,470 --> 00:04:03,210 Also, if you increase the negative pressure gradient, 73 00:04:03,210 --> 00:04:06,280 the velocity increases. 74 00:04:06,280 --> 00:04:08,200 A lower viscosity of the fluid will 75 00:04:08,200 --> 00:04:10,630 increases the average velocity. 76 00:04:10,630 --> 00:04:12,760 This is all intuitive, but wouldn't it 77 00:04:12,760 --> 00:04:14,770 be cool if we could actually model 78 00:04:14,770 --> 00:04:17,910 how the velocity profile looks like inside a pipe 79 00:04:17,910 --> 00:04:19,390 and inside a river? 80 00:04:19,390 --> 00:04:21,399 I'll do this using Mathematica. 81 00:04:21,399 --> 00:04:23,440 I will start with the river problem. 82 00:04:23,440 --> 00:04:26,220 First, I will create a grid of coordinates. 83 00:04:26,220 --> 00:04:29,800 For each coordinate on this grid, I will assign a vector. 84 00:04:29,800 --> 00:04:33,220 This vector will correspond to the velocity at that point. 85 00:04:33,220 --> 00:04:35,620 We know how this velocity looks like because we 86 00:04:35,620 --> 00:04:38,350 solved the differential equation for velocity. 87 00:04:38,350 --> 00:04:40,750 Once we have vectors for each of these points, 88 00:04:40,750 --> 00:04:44,480 you'll get a picture that looks like this. 89 00:04:44,480 --> 00:04:48,190 As you can see, at the bottom of the river, where y equals 0, 90 00:04:48,190 --> 00:04:49,870 you have 0 velocity. 91 00:04:49,870 --> 00:04:54,860 As you increase y, the velocity increases. 92 00:04:54,860 --> 00:04:58,900 Let's look at just one slice of the previous image. 93 00:04:58,900 --> 00:05:02,380 We'll see what happens when you change parameters. 94 00:05:02,380 --> 00:05:07,480 When you change mu, the average velocity decreases. 95 00:05:07,480 --> 00:05:10,540 Think of these dark blue dots as fixed. 96 00:05:10,540 --> 00:05:12,450 These are the reference frame. 97 00:05:12,450 --> 00:05:16,000 With higher mu, you have lower velocities. 98 00:05:16,000 --> 00:05:18,040 You can do the same for other parameters. 99 00:05:22,760 --> 00:05:25,970 I used a similar method to model the velocity 100 00:05:25,970 --> 00:05:28,280 profile inside a pipe. 101 00:05:28,280 --> 00:05:30,980 I first created a circular grid, and then 102 00:05:30,980 --> 00:05:34,490 I assigned each point on the grid a vector. 103 00:05:34,490 --> 00:05:38,320 As you can see, on the rim of the pipe, you have 0 velocity. 104 00:05:38,320 --> 00:05:40,580 And towards the center, this dark blue line 105 00:05:40,580 --> 00:05:42,655 here, you have the highest velocity. 106 00:05:42,655 --> 00:05:45,390 [MUSIC PLAYING] 107 00:05:45,390 --> 00:05:47,760 I hope you now understand how fluids 108 00:05:47,760 --> 00:05:50,280 flow in pipes and rivers, and that you 109 00:05:50,280 --> 00:05:53,190 will be able to apply Navier-Stokes equations 110 00:05:53,190 --> 00:05:54,902 to solve similar problems. 111 00:05:54,902 --> 00:05:55,860 Thank you for watching. 112 00:05:55,860 --> 00:06:00,110 [MUSIC PLAYING]