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[{"Name":"Lectures And Exercises","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Unit Conversion","Duration":"7m 43s","ChapterTopicVideoID":11932,"CourseChapterTopicPlaylistID":84745,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/11932.jpeg","UploadDate":"2018-03-29T15:36:28.3570000","DurationForVideoObject":"PT7M43S","Description":null,"MetaTitle":"Unit Conversion: Video + Workbook | Proprep","MetaDescription":"Fluids And Pressure - Lectures And Exercises. Watch the video made by an expert in the field. Download the workbook and maximize your learning.","Canonical":"https://www.proprep.uk/general-modules/all/physics-1-mechanics-waves-and-thermodynamics/fluids-and-pressure/lectures-and-exercises/vid12362","VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.145","Text":"Hello. In this lesson,"},{"Start":"00:02.145 ","End":"00:05.265","Text":"we\u0027re going to be learning about unit conversion."},{"Start":"00:05.265 ","End":"00:06.990","Text":"Our question here is,"},{"Start":"00:06.990 ","End":"00:12.105","Text":"what is water\u0027s density in grams per cubic inch?"},{"Start":"00:12.105 ","End":"00:16.755","Text":"We want the density of water in terms of gram per cubic inch,"},{"Start":"00:16.755 ","End":"00:21.810","Text":"instead of using our standard MKS units."},{"Start":"00:21.810 ","End":"00:27.720","Text":"First of all, water\u0027s density in MKS is"},{"Start":"00:27.720 ","End":"00:35.980","Text":"equal to 1,000 kg/m^3."},{"Start":"00:36.790 ","End":"00:43.025","Text":"Now, everything that you see inside of these square brackets is just the units."},{"Start":"00:43.025 ","End":"00:47.935","Text":"We denote units often through square brackets."},{"Start":"00:47.935 ","End":"00:51.065","Text":"In order to get to where I want,"},{"Start":"00:51.065 ","End":"00:54.920","Text":"what I want to do is I want to multiply this value"},{"Start":"00:54.920 ","End":"00:59.260","Text":"over here by 1 as many times as I want,"},{"Start":"00:59.260 ","End":"01:02.285","Text":"and soon you\u0027ll see exactly what I mean by this."},{"Start":"01:02.285 ","End":"01:04.420","Text":"Such that at the end,"},{"Start":"01:04.420 ","End":"01:06.865","Text":"because I\u0027ve multiplied by 1,"},{"Start":"01:06.865 ","End":"01:09.650","Text":"so I haven\u0027t changed anything,"},{"Start":"01:09.650 ","End":"01:12.860","Text":"but my units will be different,"},{"Start":"01:12.860 ","End":"01:18.270","Text":"but this will be exactly the same."},{"Start":"01:18.270 ","End":"01:22.227","Text":"First of all, let\u0027s multiply by 1."},{"Start":"01:22.227 ","End":"01:28.510","Text":"What I want to get rid of is this kilogram sign over here."},{"Start":"01:28.510 ","End":"01:36.140","Text":"What I\u0027m going to do is I\u0027m going to multiply this by 1,000 grams."},{"Start":"01:36.140 ","End":"01:43.580","Text":"As we know, 1,000g=1kg."},{"Start":"01:43.580 ","End":"01:48.870","Text":"I\u0027ll divide this by 1 kilogram,"},{"Start":"01:50.330 ","End":"01:56.585","Text":"so 1,000 grams is a kilogram divided by a kilogram is 1."},{"Start":"01:56.585 ","End":"02:02.100","Text":"What I\u0027ve done here is I haven\u0027t actually changed anything, have I?"},{"Start":"02:04.070 ","End":"02:09.530","Text":"Now, what we can see is that not only have we multiply it by 1,"},{"Start":"02:09.530 ","End":"02:14.415","Text":"and therefore not changed the original value that we had,"},{"Start":"02:14.415 ","End":"02:17.525","Text":"but we can see that here in the numerator,"},{"Start":"02:17.525 ","End":"02:20.820","Text":"we have 1 kilogram up here,"},{"Start":"02:20.820 ","End":"02:23.085","Text":"1,000 times 1 kilogram,"},{"Start":"02:23.085 ","End":"02:25.700","Text":"and here in the denominator,"},{"Start":"02:25.700 ","End":"02:27.865","Text":"I have 1 kilogram."},{"Start":"02:27.865 ","End":"02:33.360","Text":"What I can do is I can cancel both of these out."},{"Start":"02:33.580 ","End":"02:38.330","Text":"Now, I\u0027m left with the exact same value."},{"Start":"02:38.330 ","End":"02:43.130","Text":"1,000 kilograms per meters cubed is the density of water,"},{"Start":"02:43.130 ","End":"02:46.270","Text":"which is the same as 1,000 times 1,000,"},{"Start":"02:46.270 ","End":"02:49.950","Text":"which is a million grams per meter cubed."},{"Start":"02:49.950 ","End":"02:51.990","Text":"That\u0027s the exact same thing,"},{"Start":"02:51.990 ","End":"02:56.680","Text":"because 1,000 grams is 1 kilogram."},{"Start":"02:57.470 ","End":"03:01.625","Text":"1,000 kilograms per meter cubed is the exact same thing"},{"Start":"03:01.625 ","End":"03:06.630","Text":"as 1,000,000 grams per meter cubed."},{"Start":"03:07.130 ","End":"03:09.740","Text":"Now, the next thing is,"},{"Start":"03:09.740 ","End":"03:12.100","Text":"we want it in cubic inches."},{"Start":"03:12.100 ","End":"03:16.415","Text":"We have to get rid of this meter cube business."},{"Start":"03:16.415 ","End":"03:19.400","Text":"Again, I\u0027m going to multiply by 1,"},{"Start":"03:19.400 ","End":"03:25.105","Text":"where this time the subject of the units is going to be my meters cubed."},{"Start":"03:25.105 ","End":"03:28.750","Text":"If I write here 1,"},{"Start":"03:28.750 ","End":"03:33.520","Text":"then I can have here meters cubed,"},{"Start":"03:33.520 ","End":"03:36.875","Text":"and now because I want to get this in inches,"},{"Start":"03:36.875 ","End":"03:40.895","Text":"let\u0027s see what 1 meter cubed is in inches."},{"Start":"03:40.895 ","End":"03:43.190","Text":"You can find this information online,"},{"Start":"03:43.190 ","End":"03:47.060","Text":"but 1 meter cubed converted to inches cubed is equal"},{"Start":"03:47.060 ","End":"03:55.120","Text":"to 39.3 inches cubed."},{"Start":"03:55.280 ","End":"03:59.565","Text":"This is again, just multiplying by 1."},{"Start":"03:59.565 ","End":"04:01.610","Text":"The same value, what we have in"},{"Start":"04:01.610 ","End":"04:05.269","Text":"the numerator is the exact same thing as we have in the denominator."},{"Start":"04:05.269 ","End":"04:09.020","Text":"1 meter cubed is 39.3 inches cubed,"},{"Start":"04:09.020 ","End":"04:13.770","Text":"just like 1,000 grams is equal to 1 kilogram."},{"Start":"04:13.770 ","End":"04:15.540","Text":"Again, we\u0027ve multiplied by 1."},{"Start":"04:15.540 ","End":"04:18.530","Text":"Once again, we haven\u0027t changed at all what we"},{"Start":"04:18.530 ","End":"04:21.560","Text":"have here from our original value that we had,"},{"Start":"04:21.560 ","End":"04:24.050","Text":"it just looks slightly different because we\u0027re"},{"Start":"04:24.050 ","End":"04:27.590","Text":"multiplying by things that look a bit different but in actual fact,"},{"Start":"04:27.590 ","End":"04:30.125","Text":"if you look at each fraction,"},{"Start":"04:30.125 ","End":"04:31.630","Text":"these fractions are just 1."},{"Start":"04:31.630 ","End":"04:35.510","Text":"As we know, anything multiplied by 1 and multiplied by 1 again,"},{"Start":"04:35.510 ","End":"04:38.760","Text":"doesn\u0027t change. It\u0027s the same number."},{"Start":"04:40.130 ","End":"04:43.810","Text":"Once again, we\u0027ve multiplied by 1 and again,"},{"Start":"04:43.810 ","End":"04:49.805","Text":"what we can see here is that here we have 1 meter cubed in the numerator,"},{"Start":"04:49.805 ","End":"04:52.040","Text":"and here in the denominator,"},{"Start":"04:52.040 ","End":"04:54.550","Text":"we have 1 meter cubed as well."},{"Start":"04:54.550 ","End":"05:01.400","Text":"Just like before, we can cross out 1 meter cubed"},{"Start":"05:01.400 ","End":"05:05.180","Text":"by 1 meter cubed because we\u0027re multiplying by"},{"Start":"05:05.180 ","End":"05:08.960","Text":"something and then dividing by it simultaneously,"},{"Start":"05:08.960 ","End":"05:11.935","Text":"so we can just cross those 2 out."},{"Start":"05:11.935 ","End":"05:15.855","Text":"Now, we can get the final answer."},{"Start":"05:15.855 ","End":"05:21.105","Text":"We\u0027re going to have, 1,000,000 times 1."},{"Start":"05:21.105 ","End":"05:28.110","Text":"That\u0027s just 1,000,000 in the numerator,"},{"Start":"05:28.110 ","End":"05:34.020","Text":"and then the units in the numerator are grams."},{"Start":"05:34.020 ","End":"05:36.560","Text":"Then in the denominator,"},{"Start":"05:36.560 ","End":"05:42.260","Text":"we have 39.3 and"},{"Start":"05:42.260 ","End":"05:48.605","Text":"then this is in cubic inches."},{"Start":"05:48.605 ","End":"05:58.220","Text":"Sorry, I made a mistake because 39.3 inches is equal to 1 meter,"},{"Start":"05:58.220 ","End":"06:05.465","Text":"so if I\u0027m trying to find inches cubed to match this volume in meters cubed,"},{"Start":"06:05.465 ","End":"06:08.150","Text":"I have to cube the 39.3."},{"Start":"06:08.150 ","End":"06:10.010","Text":"I made a mistake here,"},{"Start":"06:10.010 ","End":"06:12.065","Text":"and so I\u0027m cubing it here as well."},{"Start":"06:12.065 ","End":"06:15.620","Text":"39.3 inches is 1 meter,"},{"Start":"06:15.620 ","End":"06:17.465","Text":"but I want 1 meter cubed,"},{"Start":"06:17.465 ","End":"06:19.430","Text":"so I have to cube all of this as well,"},{"Start":"06:19.430 ","End":"06:22.435","Text":"so 39.3 is also cubed."},{"Start":"06:22.435 ","End":"06:29.660","Text":"Now, when we plug in to our calculator 1,000,000 divided by 39.3^3,"},{"Start":"06:29.660 ","End":"06:35.150","Text":"we\u0027ll get that water\u0027s density in grams per"},{"Start":"06:35.150 ","End":"06:43.235","Text":"cubic inch is equal to 16.5 to 1 decimal place,"},{"Start":"06:43.235 ","End":"06:50.490","Text":"grams per cubic inch."},{"Start":"06:51.730 ","End":"06:57.065","Text":"This is water\u0027s density in grams per cubic inch."},{"Start":"06:57.065 ","End":"07:00.410","Text":"The important thing to take away from this lesson"},{"Start":"07:00.410 ","End":"07:04.715","Text":"is to know that when we\u0027re converting between different units,"},{"Start":"07:04.715 ","End":"07:09.200","Text":"we\u0027re always trying to multiply our original value"},{"Start":"07:09.200 ","End":"07:14.600","Text":"by 1s so that we don\u0027t change what we actually have,"},{"Start":"07:14.600 ","End":"07:18.230","Text":"but all we\u0027re doing is manipulating the units to,"},{"Start":"07:18.230 ","End":"07:21.685","Text":"in the end, end up with what we want."},{"Start":"07:21.685 ","End":"07:28.025","Text":"That is what is important and also equally important not to make the mistake that I made."},{"Start":"07:28.025 ","End":"07:32.360","Text":"If you need something in cubic inches or cubic centimeters,"},{"Start":"07:32.360 ","End":"07:37.010","Text":"don\u0027t forget to cube or square if you need square inches,"},{"Start":"07:37.010 ","End":"07:39.860","Text":"to square this value."},{"Start":"07:39.860 ","End":"07:43.080","Text":"Okay, that\u0027s the end of this lesson."}],"ID":12362},{"Watched":false,"Name":"Buoyancy and Lift Force","Duration":"15m 14s","ChapterTopicVideoID":11933,"CourseChapterTopicPlaylistID":84745,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.250","Text":"Hello. In this lesson we\u0027re going to be learning about buoyancy and lift force."},{"Start":"00:05.250 ","End":"00:09.930","Text":"This is the equation for buoyancy and now let\u0027s see how to derive it."},{"Start":"00:09.930 ","End":"00:15.435","Text":"Imagine we take a box of a certain mass and we place it in the water."},{"Start":"00:15.435 ","End":"00:18.120","Text":"As we know in a lot of cases,"},{"Start":"00:18.120 ","End":"00:22.890","Text":"some of the box will submerge into the water."},{"Start":"00:22.890 ","End":"00:28.965","Text":"However, the entire box won\u0027t always sink to the bottom."},{"Start":"00:28.965 ","End":"00:36.190","Text":"Sometimes it will be in this states where semi floating and semi sinking."},{"Start":"00:36.190 ","End":"00:42.040","Text":"This happens, the sinking happens due to the force of gravity acting on the box,"},{"Start":"00:42.040 ","End":"00:48.725","Text":"mg pointing downwards, but also the water itself or liquids in general,"},{"Start":"00:48.725 ","End":"00:51.890","Text":"will also apply some opposite force."},{"Start":"00:51.890 ","End":"00:53.960","Text":"Force in the upwards direction,"},{"Start":"00:53.960 ","End":"00:58.238","Text":"which is aiming to make the box float."},{"Start":"00:58.238 ","End":"01:00.540","Text":"Will find ourselves in"},{"Start":"01:00.540 ","End":"01:07.040","Text":"some no man\u0027s land where our box is both sinking and floating at the same time."},{"Start":"01:07.040 ","End":"01:11.310","Text":"Imagine if my box weighed a 100 kilograms on Earth,"},{"Start":"01:11.310 ","End":"01:16.110","Text":"I place it on a scale and I measured that once I placed it in water,"},{"Start":"01:16.110 ","End":"01:18.665","Text":"and due to this buoyancy force,"},{"Start":"01:18.665 ","End":"01:20.960","Text":"if I place the scale under the box,"},{"Start":"01:20.960 ","End":"01:26.870","Text":"then I\u0027ll weigh that the box is actually now 90 kilograms."},{"Start":"01:26.870 ","End":"01:29.555","Text":"The box has the exact same mass,"},{"Start":"01:29.555 ","End":"01:34.475","Text":"but due to this buoyancy force pushing the box upwards,"},{"Start":"01:34.475 ","End":"01:37.805","Text":"away the box as less."},{"Start":"01:37.805 ","End":"01:44.765","Text":"Now in the case where our buoyancy is greater than our force due to gravity,"},{"Start":"01:44.765 ","End":"01:47.870","Text":"only then our box will float where"},{"Start":"01:47.870 ","End":"01:52.520","Text":"its bottom edge is right on the surface of the water or the liquid."},{"Start":"01:52.520 ","End":"01:54.800","Text":"But in all other cases,"},{"Start":"01:54.800 ","End":"02:00.420","Text":"our box will be submerged either a little bit or a lot,"},{"Start":"02:00.430 ","End":"02:05.970","Text":"depending on the value of our buoyancy force."},{"Start":"02:06.620 ","End":"02:11.180","Text":"What\u0027s important to know is that buoyancy is acting in"},{"Start":"02:11.180 ","End":"02:16.100","Text":"the opposite direction to weight or the force due to gravity."},{"Start":"02:16.100 ","End":"02:20.285","Text":"Now let\u0027s talk about these values in the equation."},{"Start":"02:20.285 ","End":"02:23.915","Text":"Fb is of course, buoyancy force."},{"Start":"02:23.915 ","End":"02:27.230","Text":"G is, of course,"},{"Start":"02:27.230 ","End":"02:30.380","Text":"our acceleration due to gravity."},{"Start":"02:30.380 ","End":"02:34.715","Text":"G on Earth is 9.8 meters per second squared."},{"Start":"02:34.715 ","End":"02:37.805","Text":"However, we can use this equation also on the moon"},{"Start":"02:37.805 ","End":"02:41.660","Text":"or any other planet where there\u0027ll be different values for g,"},{"Start":"02:41.660 ","End":"02:44.540","Text":"we just substitute in the different value."},{"Start":"02:44.540 ","End":"02:48.190","Text":"Now, let\u0027s talk about Rho."},{"Start":"02:48.190 ","End":"02:51.420","Text":"Rho is speaking about density,"},{"Start":"02:51.420 ","End":"02:53.660","Text":"but what\u0027s important to know is that we\u0027re speaking about"},{"Start":"02:53.660 ","End":"02:56.701","Text":"the density of the liquid specifically."},{"Start":"02:56.701 ","End":"02:57.800","Text":"Look at the diagram."},{"Start":"02:57.800 ","End":"03:00.170","Text":"The Rho is the density of the liquid."},{"Start":"03:00.170 ","End":"03:05.060","Text":"We don\u0027t care what the density of box or object is."},{"Start":"03:05.060 ","End":"03:07.940","Text":"They don\u0027t care if it\u0027s metal or wood or plastic."},{"Start":"03:07.940 ","End":"03:13.415","Text":"We\u0027re speaking specifically here in this equation about the density of the liquid."},{"Start":"03:13.415 ","End":"03:18.560","Text":"Now, there\u0027s this V for volume."},{"Start":"03:18.560 ","End":"03:23.120","Text":"Here we are speaking about the volume to do with the box,"},{"Start":"03:23.120 ","End":"03:25.040","Text":"not to do with the liquid."},{"Start":"03:25.040 ","End":"03:31.460","Text":"However, it\u0027s specifically the volume of the box that is underwater."},{"Start":"03:31.460 ","End":"03:33.410","Text":"If we look in this diagram,"},{"Start":"03:33.410 ","End":"03:38.240","Text":"we see that we have this whole box over here and when we speak about V,"},{"Start":"03:38.240 ","End":"03:43.460","Text":"the volume, it\u0027s only this section of the box which is submerged."},{"Start":"03:43.460 ","End":"03:45.920","Text":"Where we can see the gray lines."},{"Start":"03:45.920 ","End":"03:52.490","Text":"Not including this volume of box that is outside of the water."},{"Start":"03:52.490 ","End":"03:55.890","Text":"That\u0027s very important to note as well."},{"Start":"03:56.450 ","End":"04:03.390","Text":"This is the volume of part of the object which is submerged."},{"Start":"04:03.500 ","End":"04:07.610","Text":"Of course, this value for the volume can be"},{"Start":"04:07.610 ","End":"04:12.865","Text":"the entire volume of the box if the entire box is submerged."},{"Start":"04:12.865 ","End":"04:18.260","Text":"Or if none of the boxes submerged and it\u0027s the bottom of the box is"},{"Start":"04:18.260 ","End":"04:23.310","Text":"floating at the top of the water line of the water level."},{"Start":"04:23.310 ","End":"04:26.005","Text":"Then nothing is submerged,"},{"Start":"04:26.005 ","End":"04:28.280","Text":"we have 0 volume."},{"Start":"04:30.050 ","End":"04:36.420","Text":"Now let\u0027s take an example where I have a box."},{"Start":"04:36.730 ","End":"04:41.900","Text":"The total volume of the box is L,"},{"Start":"04:41.900 ","End":"04:46.280","Text":"that\u0027s the total volume of the box and it\u0027s being submerged"},{"Start":"04:46.280 ","End":"04:50.615","Text":"in some liquid which has a density of Rho."},{"Start":"04:50.615 ","End":"04:54.950","Text":"Now we\u0027re being told that 70 percent of the box is"},{"Start":"04:54.950 ","End":"05:01.590","Text":"submerged and 30 percent is above the water level."},{"Start":"05:01.660 ","End":"05:04.625","Text":"This is a common question,"},{"Start":"05:04.625 ","End":"05:09.245","Text":"which is to find given all these parameters,"},{"Start":"05:09.245 ","End":"05:12.320","Text":"what mass the box has."},{"Start":"05:12.320 ","End":"05:15.440","Text":"Let\u0027s see how we solve this."},{"Start":"05:15.440 ","End":"05:19.160","Text":"The first thing that we\u0027re going to do is we\u0027re going to draw,"},{"Start":"05:19.160 ","End":"05:21.930","Text":"a free force diagram."},{"Start":"05:21.930 ","End":"05:24.945","Text":"Here we have our box,"},{"Start":"05:24.945 ","End":"05:27.820","Text":"and it\u0027s a volume L, I\u0027m just redrawing this."},{"Start":"05:27.820 ","End":"05:34.850","Text":"Then we know that over here is our water level and what\u0027s important to"},{"Start":"05:34.850 ","End":"05:43.205","Text":"note is that 70 percent is underwater,"},{"Start":"05:43.205 ","End":"05:45.870","Text":"70 percent of the box."},{"Start":"05:46.520 ","End":"05:50.190","Text":"Now let\u0027s draw our diagram."},{"Start":"05:50.190 ","End":"05:54.409","Text":"First of all we know that we have in the downwards direction,"},{"Start":"05:54.409 ","End":"05:56.135","Text":"our force due to gravity,"},{"Start":"05:56.135 ","End":"05:59.690","Text":"which is M. I\u0027m going to draw a tilde on top because this is our unknown,"},{"Start":"05:59.690 ","End":"06:01.495","Text":"this is what we\u0027re trying to find,"},{"Start":"06:01.495 ","End":"06:06.675","Text":"multiplied by g. Then in the opposite direction."},{"Start":"06:06.675 ","End":"06:09.049","Text":"Because our box is semi floating,"},{"Start":"06:09.049 ","End":"06:12.115","Text":"we have our buoyancy force."},{"Start":"06:12.115 ","End":"06:16.700","Text":"Let\u0027s write over here Fb our buoyancy force,"},{"Start":"06:16.700 ","End":"06:19.970","Text":"which is equal to what we have over here,"},{"Start":"06:19.970 ","End":"06:22.855","Text":"this equation, so it\u0027s equal to Rho."},{"Start":"06:22.855 ","End":"06:25.115","Text":"What\u0027s Rho? The density of the liquid,"},{"Start":"06:25.115 ","End":"06:28.165","Text":"which were given multiplied by,"},{"Start":"06:28.165 ","End":"06:30.110","Text":"let\u0027s go over here,"},{"Start":"06:30.110 ","End":"06:33.020","Text":"or G, acceleration due to gravity."},{"Start":"06:33.020 ","End":"06:35.165","Text":"Let\u0027s just write g over here."},{"Start":"06:35.165 ","End":"06:38.135","Text":"On Earth, it\u0027s 9.8 and then"},{"Start":"06:38.135 ","End":"06:43.385","Text":"multiplied by the volume of the part of the object which is submerged."},{"Start":"06:43.385 ","End":"06:48.960","Text":"We know that the total volume is L. But we"},{"Start":"06:48.960 ","End":"06:54.735","Text":"know that only 70 percent of this total volume L is submerged."},{"Start":"06:54.735 ","End":"06:58.620","Text":"We\u0027re going to multiply this by 70 percent of L,"},{"Start":"06:58.620 ","End":"07:02.580","Text":"which in decimal form is 0.7"},{"Start":"07:02.580 ","End":"07:08.720","Text":"L or 7 over 10 L. This is our buoyancy force,"},{"Start":"07:08.720 ","End":"07:13.130","Text":"and of course, it\u0027s in the opposite direction to our force due to gravity,"},{"Start":"07:13.130 ","End":"07:17.490","Text":"and this is always correct."},{"Start":"07:17.490 ","End":"07:21.104","Text":"We don\u0027t have any forces in the x-direction."},{"Start":"07:21.104 ","End":"07:24.115","Text":"So what we\u0027re going to do is we\u0027re going to write an equation"},{"Start":"07:24.115 ","End":"07:28.490","Text":"for the sum of all of the forces in the y-direction."},{"Start":"07:29.010 ","End":"07:33.295","Text":"We can choose an arbitrary positive direction,"},{"Start":"07:33.295 ","End":"07:36.895","Text":"let\u0027s say that this is the positive y-direction."},{"Start":"07:36.895 ","End":"07:42.985","Text":"In that case, we have Rho g multiplied by"},{"Start":"07:42.985 ","End":"07:49.610","Text":"0.7L minus m tilde g,"},{"Start":"07:49.610 ","End":"07:51.970","Text":"where m tilde is our unknown."},{"Start":"07:51.970 ","End":"07:56.905","Text":"Because we know that our object is at equilibrium,"},{"Start":"07:56.905 ","End":"07:58.555","Text":"it\u0027s not bobbing up and down,"},{"Start":"07:58.555 ","End":"08:00.100","Text":"sometimes it will be bobbing up and down,"},{"Start":"08:00.100 ","End":"08:02.770","Text":"but in our example it\u0027s just stationary,"},{"Start":"08:02.770 ","End":"08:08.590","Text":"at equilibrium, so that means that the sum of all the forces is equal to 0."},{"Start":"08:08.590 ","End":"08:13.870","Text":"Now we can first of all divide everything by g,"},{"Start":"08:13.870 ","End":"08:16.780","Text":"and then we can rearrange the equation."},{"Start":"08:16.780 ","End":"08:18.340","Text":"We\u0027re dividing everything by g,"},{"Start":"08:18.340 ","End":"08:25.449","Text":"so we\u0027ll be left with 0.7L multiplied by the density of the liquid,"},{"Start":"08:25.449 ","End":"08:29.935","Text":"which is equal to our unknown m tilde."},{"Start":"08:29.935 ","End":"08:32.455","Text":"This is exactly what we\u0027re trying to find,"},{"Start":"08:32.455 ","End":"08:34.300","Text":"and there we have our answer."},{"Start":"08:34.300 ","End":"08:39.400","Text":"Our Rho and our volume L was given to us in the question,"},{"Start":"08:39.400 ","End":"08:41.840","Text":"so this is our mass."},{"Start":"08:43.410 ","End":"08:47.755","Text":"It\u0027s very straightforward to find the mass of an object."},{"Start":"08:47.755 ","End":"08:54.520","Text":"Now we can see that the units of this answer really are kilograms for mass."},{"Start":"08:54.520 ","End":"08:56.995","Text":"We know that Rho is density,"},{"Start":"08:56.995 ","End":"09:00.985","Text":"so that is simply mass divided by volume,"},{"Start":"09:00.985 ","End":"09:03.490","Text":"and L is volume."},{"Start":"09:03.490 ","End":"09:07.885","Text":"Our volume will cancel out in the numerator and denominator,"},{"Start":"09:07.885 ","End":"09:11.840","Text":"and then we\u0027re just left with units for mass."},{"Start":"09:12.090 ","End":"09:15.700","Text":"Now another question that can be asked is,"},{"Start":"09:15.700 ","End":"09:20.500","Text":"how much liquid needs to be poured into the box such that"},{"Start":"09:20.500 ","End":"09:27.715","Text":"the top side of the box is at water level?"},{"Start":"09:27.715 ","End":"09:33.535","Text":"We\u0027re trying to find this volume of liquid inside the box,"},{"Start":"09:33.535 ","End":"09:38.350","Text":"or not necessarily at water level but at liquid level such that"},{"Start":"09:38.350 ","End":"09:43.375","Text":"the top of the box is right at that level with the liquid."},{"Start":"09:43.375 ","End":"09:44.995","Text":"It\u0027s still floating."},{"Start":"09:44.995 ","End":"09:47.664","Text":"It hasn\u0027t sunk to the bottom of the tank."},{"Start":"09:47.664 ","End":"09:54.410","Text":"However, the box is still completely submerged under the liquid."},{"Start":"09:54.990 ","End":"09:59.050","Text":"Of course, we\u0027re dealing with the same density of"},{"Start":"09:59.050 ","End":"10:03.235","Text":"liquid that we worked in our first question, so that\u0027s Rho."},{"Start":"10:03.235 ","End":"10:05.875","Text":"Again, like in our first question,"},{"Start":"10:05.875 ","End":"10:12.620","Text":"I\u0027m going to draw a free-body diagram and see which forces are acting on my box."},{"Start":"10:13.140 ","End":"10:19.540","Text":"Which forces are acting in the downwards direction?"},{"Start":"10:19.540 ","End":"10:23.575","Text":"That\u0027s, of course, our forces due to gravity."},{"Start":"10:23.575 ","End":"10:26.830","Text":"What is that? That\u0027s the weight of the box,"},{"Start":"10:26.830 ","End":"10:30.819","Text":"but also we have to take into account the weight of the liquid"},{"Start":"10:30.819 ","End":"10:35.575","Text":"that we\u0027ve poured into the box because that\u0027s what we\u0027re trying to find."},{"Start":"10:35.575 ","End":"10:40.765","Text":"The weight of the box is our mg,"},{"Start":"10:40.765 ","End":"10:43.675","Text":"where m is what we found in our previous question."},{"Start":"10:43.675 ","End":"10:46.600","Text":"This time I won\u0027t draw a tilde above the m because we\u0027ve"},{"Start":"10:46.600 ","End":"10:50.440","Text":"already found it\u0027s value in the previous question."},{"Start":"10:50.440 ","End":"10:55.045","Text":"Mg, I\u0027m just using the tilde to symbolize what our unknown is,"},{"Start":"10:55.045 ","End":"11:00.175","Text":"plus our force due to gravity due to the liquid inside."},{"Start":"11:00.175 ","End":"11:04.165","Text":"I need to take the mass of the liquid inside"},{"Start":"11:04.165 ","End":"11:09.520","Text":"multiplied by g. What is the mass of the liquid inside?"},{"Start":"11:09.520 ","End":"11:12.655","Text":"What\u0027s another way of calculating mass?"},{"Start":"11:12.655 ","End":"11:19.165","Text":"It\u0027s taking the volume of the liquid multiplied by the density."},{"Start":"11:19.165 ","End":"11:22.540","Text":"If we take the volume,"},{"Start":"11:22.540 ","End":"11:25.180","Text":"so we can also say that mass,"},{"Start":"11:25.180 ","End":"11:28.900","Text":"you know what I\u0027ll do a M because here we\u0027re speaking about mass in general,"},{"Start":"11:28.900 ","End":"11:35.185","Text":"is equal to the density of the liquid multiplied by its volume;"},{"Start":"11:35.185 ","End":"11:40.235","Text":"when we\u0027re dealing with volume and this is correct also for area."},{"Start":"11:40.235 ","End":"11:43.470","Text":"Our density it\u0027s the same liquid that we have over here,"},{"Start":"11:43.470 ","End":"11:45.090","Text":"so that\u0027s our Rho."},{"Start":"11:45.090 ","End":"11:50.010","Text":"Then the volume of the water that we have to add which is what we\u0027re trying to find,"},{"Start":"11:50.010 ","End":"11:53.975","Text":"this is our unknown, how much liquid that\u0027s referring to volume."},{"Start":"11:53.975 ","End":"11:57.190","Text":"I\u0027ll put a tilde on top of this as our unknown."},{"Start":"11:57.190 ","End":"12:03.055","Text":"This is going to be Rho which is known multiplied by this volume which is unknown,"},{"Start":"12:03.055 ","End":"12:04.255","Text":"we\u0027re working it out,"},{"Start":"12:04.255 ","End":"12:10.890","Text":"multiplied by g. Now in the opposite direction,"},{"Start":"12:10.890 ","End":"12:14.555","Text":"we have our buoyancy force."},{"Start":"12:14.555 ","End":"12:19.360","Text":"Here we have our buoyancy force which is equal to Rho,"},{"Start":"12:19.360 ","End":"12:22.855","Text":"so that\u0027s the density of the liquid which we have Rho,"},{"Start":"12:22.855 ","End":"12:31.720","Text":"multiplied by g, multiplied by the volume of the part of the object which is submerged."},{"Start":"12:31.720 ","End":"12:35.800","Text":"We\u0027re being told that our whole box is being submerged,"},{"Start":"12:35.800 ","End":"12:39.010","Text":"and let\u0027s imagine it\u0027s the same box that we saw in our previous question,"},{"Start":"12:39.010 ","End":"12:42.682","Text":"so its total volume of the box is L,"},{"Start":"12:42.682 ","End":"12:44.020","Text":"from our previous question."},{"Start":"12:44.020 ","End":"12:45.730","Text":"The volume of the part submerged,"},{"Start":"12:45.730 ","End":"12:49.190","Text":"we can see that 100 percent of the box is submerged,"},{"Start":"12:49.190 ","End":"12:53.095","Text":"so it\u0027s going to be equal to L. Now,"},{"Start":"12:53.095 ","End":"12:56.320","Text":"again, we\u0027re going to write out our forces."},{"Start":"12:56.320 ","End":"12:58.720","Text":"We have no forces in the x-direction,"},{"Start":"12:58.720 ","End":"13:02.515","Text":"and let\u0027s again say that this is our positive y-direction."},{"Start":"13:02.515 ","End":"13:07.615","Text":"Now we can say that the sum of the forces in our y-direction is equal to,"},{"Start":"13:07.615 ","End":"13:13.120","Text":"in the positive direction we have Rho g multiplied by the volume of"},{"Start":"13:13.120 ","End":"13:15.775","Text":"the part of the object which is submerged"},{"Start":"13:15.775 ","End":"13:19.435","Text":"which here is L. Then in the negative direction,"},{"Start":"13:19.435 ","End":"13:28.075","Text":"we have the mass of our box multiplied by g plus the mass of our liquid,"},{"Start":"13:28.075 ","End":"13:30.130","Text":"so that\u0027s Rho multiplied by"},{"Start":"13:30.130 ","End":"13:32.680","Text":"the volume of the liquid which is what we\u0027re trying to find out,"},{"Start":"13:32.680 ","End":"13:37.750","Text":"multiplied by g. We\u0027re taking this to be at equilibrium,"},{"Start":"13:37.750 ","End":"13:41.425","Text":"meaning it\u0027s not bobbing up and down in this question specifically,"},{"Start":"13:41.425 ","End":"13:44.960","Text":"so the sum of all of the forces is equal to 0."},{"Start":"13:45.420 ","End":"13:49.345","Text":"Now let\u0027s rearrange all of these,"},{"Start":"13:49.345 ","End":"13:54.828","Text":"and we can see that we have this common factor g. We\u0027re going to divide everything by g,"},{"Start":"13:54.828 ","End":"14:04.045","Text":"and we\u0027ll be left with Rho L is equal to m plus Rho v tilde,"},{"Start":"14:04.045 ","End":"14:06.590","Text":"which is what we\u0027re trying to find out."},{"Start":"14:06.600 ","End":"14:10.315","Text":"Now let\u0027s scroll down to give us some more space,"},{"Start":"14:10.315 ","End":"14:13.645","Text":"and let\u0027s do some algebra."},{"Start":"14:13.645 ","End":"14:18.910","Text":"We have Rho L minus m,"},{"Start":"14:18.910 ","End":"14:21.985","Text":"the mass of our box when it\u0027s empty,"},{"Start":"14:21.985 ","End":"14:30.370","Text":"which we saw over here is equal to 0.7L Rho."},{"Start":"14:30.370 ","End":"14:35.860","Text":"This is equal to Rho multiplied by the unknown volume,"},{"Start":"14:35.860 ","End":"14:37.690","Text":"which is what we\u0027re trying to find out."},{"Start":"14:37.690 ","End":"14:41.120","Text":"Now what we can do is we can divide both sides by Rho,"},{"Start":"14:41.120 ","End":"14:43.545","Text":"we can see it\u0027s a common factor,"},{"Start":"14:43.545 ","End":"14:50.125","Text":"and then we get how much liquid needs to be poured into the box."},{"Start":"14:50.125 ","End":"14:54.200","Text":"Our volume, how much liquid needs to be poured into the box,"},{"Start":"14:54.200 ","End":"14:59.570","Text":"is equal to the volume of the box minus 0.7,"},{"Start":"14:59.570 ","End":"15:02.765","Text":"so 70 percent of the volume of the box."},{"Start":"15:02.765 ","End":"15:07.655","Text":"That is simply equal to 0.3L,"},{"Start":"15:07.655 ","End":"15:10.500","Text":"and this is our answer."},{"Start":"15:11.190 ","End":"15:14.780","Text":"That\u0027s the end of this question."}],"ID":12363},{"Watched":false,"Name":"Surface Tension, Adhesion and Cohesion","Duration":"14m 20s","ChapterTopicVideoID":11934,"CourseChapterTopicPlaylistID":84745,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.145","Text":"Hello. In this lesson,"},{"Start":"00:02.145 ","End":"00:04.365","Text":"we\u0027re going to be learning about surface tension,"},{"Start":"00:04.365 ","End":"00:07.170","Text":"adhesion, and cohesion."},{"Start":"00:07.170 ","End":"00:13.455","Text":"Surface tension is responsible for a lot of interesting phenomena that we see."},{"Start":"00:13.455 ","End":"00:18.681","Text":"For instance, water droplets are the shape that they are,"},{"Start":"00:18.681 ","End":"00:23.445","Text":"this spherical shape, due to surface tension."},{"Start":"00:23.445 ","End":"00:25.590","Text":"All the little molecules,"},{"Start":"00:25.590 ","End":"00:28.440","Text":"the water molecules inside this droplet have"},{"Start":"00:28.440 ","End":"00:33.220","Text":"forces that are interacting between one another."},{"Start":"00:33.590 ","End":"00:40.230","Text":"These forces hold this droplet in this spherical shape."},{"Start":"00:40.230 ","End":"00:43.430","Text":"That\u0027s what stops the droplet from spreading out into"},{"Start":"00:43.430 ","End":"00:49.255","Text":"some disk or rectangle or any other shape."},{"Start":"00:49.255 ","End":"00:57.725","Text":"Now, what we can see over here is that we have 2 bugs that are resting on water."},{"Start":"00:57.725 ","End":"01:01.015","Text":"Note that they\u0027re resting and not floating."},{"Start":"01:01.015 ","End":"01:06.125","Text":"They\u0027re literally standing on the water and their legs aren\u0027t getting wet."},{"Start":"01:06.125 ","End":"01:07.835","Text":"Now, how is this happening?"},{"Start":"01:07.835 ","End":"01:10.940","Text":"This is also due to surface tension."},{"Start":"01:10.940 ","End":"01:14.930","Text":"The little molecules in this puddle of water are all"},{"Start":"01:14.930 ","End":"01:18.860","Text":"have forces that are interacting with one another,"},{"Start":"01:18.860 ","End":"01:26.340","Text":"such that the upper layer of this puddle is still made out of water molecules."},{"Start":"01:27.160 ","End":"01:32.195","Text":"However, these water molecules form some film"},{"Start":"01:32.195 ","End":"01:36.356","Text":"or membrane that we can see over here in this picture,"},{"Start":"01:36.356 ","End":"01:38.180","Text":"that the bugs are standing on."},{"Start":"01:38.180 ","End":"01:41.630","Text":"This film and membrane is still made out of water molecules."},{"Start":"01:41.630 ","End":"01:44.750","Text":"However, the legs aren\u0027t getting wet because"},{"Start":"01:44.750 ","End":"01:48.840","Text":"the bugs are literally standing on this film/membrane."},{"Start":"01:50.200 ","End":"01:55.240","Text":"They\u0027re not floating, they\u0027re standing on this."},{"Start":"01:55.240 ","End":"02:01.130","Text":"Surface tension is denoted by the Greek letter Gamma."},{"Start":"02:01.130 ","End":"02:05.795","Text":"The first type of question that can be asked is,"},{"Start":"02:05.795 ","End":"02:09.835","Text":"what is the pressure inside the droplets?"},{"Start":"02:09.835 ","End":"02:15.335","Text":"Let\u0027s imagine that we have this water droplet and it\u0027s located"},{"Start":"02:15.335 ","End":"02:21.870","Text":"inside this pool of water or some kind of liquid."},{"Start":"02:21.970 ","End":"02:28.715","Text":"The pressure inside this liquid in general is denoted by P_1."},{"Start":"02:28.715 ","End":"02:34.685","Text":"What we\u0027re trying to find is what this pressure P_2 is equal to."},{"Start":"02:34.685 ","End":"02:40.220","Text":"What we have is lots of tiny little water molecules inside this droplet,"},{"Start":"02:40.220 ","End":"02:43.040","Text":"which have forces which are pulling all of"},{"Start":"02:43.040 ","End":"02:48.115","Text":"the outer molecules of the water droplet inwards."},{"Start":"02:48.115 ","End":"02:57.810","Text":"It\u0027s these forces which are going to dictate what this pressure P_2 is."},{"Start":"02:57.810 ","End":"03:04.700","Text":"Our pressure inside the droplet, P_2,"},{"Start":"03:04.700 ","End":"03:10.265","Text":"is defined as the pressure outside of the droplet, P_1,"},{"Start":"03:10.265 ","End":"03:17.885","Text":"plus the pressure caused by the forces of the molecules pulling each other in."},{"Start":"03:17.885 ","End":"03:21.609","Text":"This is to do with surface tension."},{"Start":"03:21.609 ","End":"03:24.390","Text":"Plus 2 multiplied by Gamma,"},{"Start":"03:24.390 ","End":"03:27.420","Text":"where Gamma is the surface tension,"},{"Start":"03:27.420 ","End":"03:28.835","Text":"divided by r,"},{"Start":"03:28.835 ","End":"03:36.329","Text":"where r is the radius of the droplet."},{"Start":"03:36.710 ","End":"03:39.440","Text":"This is the same equation,"},{"Start":"03:39.440 ","End":"03:44.450","Text":"whether we\u0027re dealing with a drop of water in a pond of water or oil in oil,"},{"Start":"03:44.450 ","End":"03:48.080","Text":"or a drop of water in air like where we"},{"Start":"03:48.080 ","End":"03:52.540","Text":"saw in that picture where we had drops of water on a spider\u0027s web."},{"Start":"03:52.540 ","End":"03:56.360","Text":"This is the equation that you should write down."},{"Start":"03:56.360 ","End":"03:59.400","Text":"Now, another question that they could ask is where will"},{"Start":"03:59.400 ","End":"04:02.480","Text":"there be a larger internal pressure,"},{"Start":"04:02.480 ","End":"04:05.220","Text":"a larger value of P_2,"},{"Start":"04:05.220 ","End":"04:09.839","Text":"in the small droplets or in a large droplet?"},{"Start":"04:10.310 ","End":"04:14.340","Text":"Let\u0027s draw our 2 droplets."},{"Start":"04:14.340 ","End":"04:19.810","Text":"This is the large water droplet with the radius r_1."},{"Start":"04:19.810 ","End":"04:26.365","Text":"This is the smaller water droplet with radius r_2."},{"Start":"04:26.365 ","End":"04:29.740","Text":"I\u0027ll just write it out here, r_1 is greater than r_2."},{"Start":"04:31.310 ","End":"04:37.370","Text":"We know that we\u0027re using this equation in order to find P_2."},{"Start":"04:37.370 ","End":"04:42.530","Text":"Remember where P_2 is the pressure inside the droplet."},{"Start":"04:42.530 ","End":"04:44.795","Text":"That\u0027s equal to P_1,"},{"Start":"04:44.795 ","End":"04:47.090","Text":"which is the pressure outside the droplet."},{"Start":"04:47.090 ","End":"04:53.045","Text":"We can assume that the 2 droplets are either in the air or in water, it doesn\u0027t matter,"},{"Start":"04:53.045 ","End":"04:56.340","Text":"they\u0027ll have the exact same pressure outside,"},{"Start":"04:56.340 ","End":"05:00.020","Text":"plus this value over here,"},{"Start":"05:00.020 ","End":"05:02.880","Text":"2 multiplied by Gamma,"},{"Start":"05:02.880 ","End":"05:05.055","Text":"where Gamma is the surface tension."},{"Start":"05:05.055 ","End":"05:09.415","Text":"Both of the droplets are made out of water or the same liquid,"},{"Start":"05:09.415 ","End":"05:13.025","Text":"so their value for surface tension will be the same."},{"Start":"05:13.025 ","End":"05:16.440","Text":"Then we\u0027re dividing by r,"},{"Start":"05:16.440 ","End":"05:20.610","Text":"the radius of the droplet."},{"Start":"05:20.610 ","End":"05:23.875","Text":"If we have a larger denominator,"},{"Start":"05:23.875 ","End":"05:28.975","Text":"that means that the fraction is a smaller number."},{"Start":"05:28.975 ","End":"05:32.155","Text":"Then we\u0027re adding onto P_1 a smaller number,"},{"Start":"05:32.155 ","End":"05:35.350","Text":"which means that P_2 will be smaller."},{"Start":"05:35.350 ","End":"05:39.220","Text":"However, if our denominator is smaller,"},{"Start":"05:39.220 ","End":"05:42.710","Text":"then this fraction as a whole will be larger."},{"Start":"05:42.710 ","End":"05:45.620","Text":"Then we\u0027re adding a larger number onto our P_1,"},{"Start":"05:45.620 ","End":"05:49.095","Text":"so as a whole, our P_2 will be bigger."},{"Start":"05:49.095 ","End":"05:56.570","Text":"We can see that the pressure inside a smaller droplet is greater."},{"Start":"05:56.570 ","End":"06:03.780","Text":"Another interesting phenomena due to surface tension is capillary action."},{"Start":"06:03.910 ","End":"06:10.310","Text":"Capillary action is the ability of a liquid to flow in narrow tubes"},{"Start":"06:10.310 ","End":"06:16.250","Text":"without and in spite of external forces such as gravity."},{"Start":"06:16.250 ","End":"06:19.860","Text":"We see this a lot of times in nature."},{"Start":"06:19.860 ","End":"06:22.310","Text":"For instance, with the roots of a plant."},{"Start":"06:22.310 ","End":"06:23.900","Text":"A lot of the time,"},{"Start":"06:23.900 ","End":"06:29.400","Text":"the roots of a tree or a flower are very thin, narrow tubes."},{"Start":"06:29.400 ","End":"06:32.870","Text":"Here we have the stem and here we have the ground,"},{"Start":"06:32.870 ","End":"06:35.405","Text":"and then we have lots of tiny,"},{"Start":"06:35.405 ","End":"06:36.950","Text":"little roots coming out,"},{"Start":"06:36.950 ","End":"06:39.080","Text":"which are narrow, little tubes."},{"Start":"06:39.080 ","End":"06:47.484","Text":"Capillary action says that water will go into the tube and without any external forces,"},{"Start":"06:47.484 ","End":"06:50.605","Text":"just due to surface tension,"},{"Start":"06:50.605 ","End":"06:55.505","Text":"the liquid or the water will be able to flow up these natural tubes,"},{"Start":"06:55.505 ","End":"07:01.603","Text":"even if it\u0027s flowing in the opposite direction to gravity."},{"Start":"07:01.603 ","End":"07:08.540","Text":"Now, you can also see that in the wick of a candle,"},{"Start":"07:09.180 ","End":"07:13.225","Text":"but specifically an oil candle."},{"Start":"07:13.225 ","End":"07:21.085","Text":"The oil will travel up the wick of its own accord and this is due to capillary action,"},{"Start":"07:21.085 ","End":"07:25.555","Text":"which is due to surface tension."},{"Start":"07:25.555 ","End":"07:32.935","Text":"You can see this capillary reaction in other day to day life examples."},{"Start":"07:32.935 ","End":"07:38.600","Text":"Now we\u0027re going to speak about cohesion and adhesion."},{"Start":"07:39.060 ","End":"07:41.965","Text":"Let\u0027s first speak about cohesion."},{"Start":"07:41.965 ","End":"07:49.165","Text":"Now, cohesion is the property of a material that says that like molecules stick together."},{"Start":"07:49.165 ","End":"07:54.880","Text":"If we have a water molecule and another water molecule,"},{"Start":"07:54.880 ","End":"07:58.430","Text":"they will stick together because they\u0027re the same."},{"Start":"07:58.590 ","End":"08:02.274","Text":"Now, these attractions are in actual fact,"},{"Start":"08:02.274 ","End":"08:10.930","Text":"electrical attractions between the electrons in the 2 like molecules."},{"Start":"08:10.930 ","End":"08:14.110","Text":"The electrons between the 2 water molecules"},{"Start":"08:14.110 ","End":"08:19.090","Text":"form an electrical attraction between the 2 water molecules,"},{"Start":"08:19.090 ","End":"08:23.120","Text":"which is what causes them to stick together."},{"Start":"08:24.000 ","End":"08:31.490","Text":"Cohesion actually is what brings about surface tension."},{"Start":"08:31.710 ","End":"08:37.780","Text":"It creates that film or membrane that we were speaking about that allows"},{"Start":"08:37.780 ","End":"08:41.740","Text":"either extremely lightweight or low-density objects to"},{"Start":"08:41.740 ","End":"08:46.850","Text":"be placed upon it and for a liquid to behave like a solid."},{"Start":"08:47.220 ","End":"08:51.070","Text":"Of course I\u0027m putting speech marks,"},{"Start":"08:51.070 ","End":"08:55.915","Text":"quotation marks around solid because that isn\u0027t the exact physical term here,"},{"Start":"08:55.915 ","End":"08:58.180","Text":"but just so that you can imagine."},{"Start":"08:58.180 ","End":"09:02.050","Text":"That\u0027s why those bugs could stand on that puddle of"},{"Start":"09:02.050 ","End":"09:06.620","Text":"water and not be submerged in the puddle of water."},{"Start":"09:06.780 ","End":"09:13.450","Text":"Adhesion is when we have different types of molecules which stick together."},{"Start":"09:13.450 ","End":"09:19.375","Text":"For instance, water to some glass beaker."},{"Start":"09:19.375 ","End":"09:21.940","Text":"Cohesion is in fact,"},{"Start":"09:21.940 ","End":"09:27.050","Text":"what is responsible for the surface tension Gamma."},{"Start":"09:27.330 ","End":"09:33.580","Text":"If we look at the example of the spiderweb with water droplets,"},{"Start":"09:33.580 ","End":"09:37.930","Text":"we can see that the cohesion is the forces that hold"},{"Start":"09:37.930 ","End":"09:42.800","Text":"the water droplet itself together in this spherical shape and"},{"Start":"09:42.800 ","End":"09:46.540","Text":"the adhesion is that the water droplets"},{"Start":"09:46.540 ","End":"09:53.150","Text":"stick to the spiderweb and don\u0027t just fall to Earth."},{"Start":"09:54.480 ","End":"10:02.770","Text":"Now, depending on what liquid we\u0027re dealing with and what it\u0027s sticking to,"},{"Start":"10:02.770 ","End":"10:06.070","Text":"sometimes our cohesive forces will be bigger than"},{"Start":"10:06.070 ","End":"10:08.544","Text":"our adhesive forces and sometimes"},{"Start":"10:08.544 ","End":"10:13.310","Text":"our adhesive forces will be bigger than our cohesive forces."},{"Start":"10:13.860 ","End":"10:18.925","Text":"How can we tell in which case we\u0027re looking at?"},{"Start":"10:18.925 ","End":"10:26.845","Text":"Over here, we can see that our liquid has some bump protruding from its top."},{"Start":"10:26.845 ","End":"10:34.250","Text":"In this case, we have a case of cohesion bigger than adhesion."},{"Start":"10:35.340 ","End":"10:37.570","Text":"On the other end,"},{"Start":"10:37.570 ","End":"10:45.220","Text":"we have here that we have some bump which is intruding in on the top of the liquid."},{"Start":"10:45.220 ","End":"10:52.490","Text":"In this case, we have adhesive forces which are greater than cohesive forces."},{"Start":"10:53.460 ","End":"10:59.410","Text":"Why does our liquid look like this in both of these cases?"},{"Start":"10:59.410 ","End":"11:00.925","Text":"It\u0027s pretty obvious."},{"Start":"11:00.925 ","End":"11:07.225","Text":"Let\u0027s look at this case where our cohesive force is greater than our adhesive force."},{"Start":"11:07.225 ","End":"11:11.920","Text":"That means that the forces between the like molecules are"},{"Start":"11:11.920 ","End":"11:16.769","Text":"stronger than the forces between the different type of molecules,"},{"Start":"11:16.769 ","End":"11:22.420","Text":"and that\u0027s why this causes this spherical shape because the like molecules are trying to"},{"Start":"11:22.420 ","End":"11:28.895","Text":"get in their lowest energy state because their cohesive forces are greatest."},{"Start":"11:28.895 ","End":"11:30.795","Text":"In this case over here,"},{"Start":"11:30.795 ","End":"11:36.535","Text":"where the adhesive forces are greater than the cohesive forces,"},{"Start":"11:36.535 ","End":"11:41.860","Text":"there are still cohesive forces between the like molecules, however,"},{"Start":"11:41.860 ","End":"11:48.350","Text":"the adhesive forces are so strong that they\u0027re pulling up the liquid."},{"Start":"11:49.860 ","End":"11:53.950","Text":"Now something interesting that we can do over here is that"},{"Start":"11:53.950 ","End":"11:57.220","Text":"we can measure this angle Theta over here,"},{"Start":"11:57.220 ","End":"12:03.920","Text":"which is called the degree of wetting or wettability."},{"Start":"12:04.110 ","End":"12:08.290","Text":"A measure of the degree of wetting is the contact angle,"},{"Start":"12:08.290 ","End":"12:11.830","Text":"which is in fact this Theta over here and over here,"},{"Start":"12:11.830 ","End":"12:14.210","Text":"this is the contact angle."},{"Start":"12:15.180 ","End":"12:21.134","Text":"It\u0027s the measured angle between the side of the vessel"},{"Start":"12:21.134 ","End":"12:28.660","Text":"and the tangent of our liquid."},{"Start":"12:29.400 ","End":"12:32.920","Text":"In order to see the size of this angle,"},{"Start":"12:32.920 ","End":"12:34.750","Text":"we\u0027re going to look at our vessels,"},{"Start":"12:34.750 ","End":"12:41.740","Text":"so here at some tube and then where we see the gradient of our drops,"},{"Start":"12:41.740 ","End":"12:46.855","Text":"so here we can see that the gradient is in this round protruding way."},{"Start":"12:46.855 ","End":"12:49.720","Text":"At the tangent over here, at the meeting point,"},{"Start":"12:49.720 ","End":"12:52.810","Text":"we\u0027re going to draw a line tangential to"},{"Start":"12:52.810 ","End":"12:56.365","Text":"this point over here and then we\u0027ll measure that angle."},{"Start":"12:56.365 ","End":"13:01.405","Text":"Same over here, here we have our vessel and here\u0027s the meeting point between the liquid,"},{"Start":"13:01.405 ","End":"13:05.410","Text":"the uppermost corner of the liquid molecule and the vessel."},{"Start":"13:05.410 ","End":"13:10.675","Text":"We can see that the gradient of this intrusion,"},{"Start":"13:10.675 ","End":"13:14.305","Text":"it\u0027s an intrusion, so the gradient is going down."},{"Start":"13:14.305 ","End":"13:19.510","Text":"Then we\u0027ll draw a tangent to this dotted line,"},{"Start":"13:19.510 ","End":"13:23.120","Text":"and then we\u0027ll measure this angle over here."},{"Start":"13:24.540 ","End":"13:27.495","Text":"When we have a case of cohesion,"},{"Start":"13:27.495 ","End":"13:31.945","Text":"that means that the cohesive forces are greater than the adhesive forces,"},{"Start":"13:31.945 ","End":"13:36.325","Text":"our Theta, which is our contact angle or our degree of wetting,"},{"Start":"13:36.325 ","End":"13:39.625","Text":"will be greater than 90 degrees."},{"Start":"13:39.625 ","End":"13:40.930","Text":"On the flip side,"},{"Start":"13:40.930 ","End":"13:44.904","Text":"when our adhesive forces are greater than our cohesive forces,"},{"Start":"13:44.904 ","End":"13:46.675","Text":"our angle Theta,"},{"Start":"13:46.675 ","End":"13:52.555","Text":"which is our degree of wetting or our contact angle will be less than 90 degrees."},{"Start":"13:52.555 ","End":"13:56.650","Text":"If our contact angle is equal to 90 degrees,"},{"Start":"13:56.650 ","End":"13:59.815","Text":"then that means we have some flat plane,"},{"Start":"13:59.815 ","End":"14:08.190","Text":"which means that our adhesive forces and our cohesive forces are equal."},{"Start":"14:08.520 ","End":"14:16.190","Text":"Of course, it\u0027s this cohesion and adhesion which is responsible for the capillary action."},{"Start":"14:16.190 ","End":"14:20.040","Text":"Okay, that\u0027s the end of this lesson."}],"ID":12364},{"Watched":false,"Name":"Continuity Equation","Duration":"8m 20s","ChapterTopicVideoID":11935,"CourseChapterTopicPlaylistID":84745,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.965","Text":"Hello. In this lesson,"},{"Start":"00:01.965 ","End":"00:04.965","Text":"we\u0027re going to be learning about the continuity equation."},{"Start":"00:04.965 ","End":"00:09.285","Text":"Before we take a look at the equation and speak about exactly what it means,"},{"Start":"00:09.285 ","End":"00:11.550","Text":"the first thing that we need to know about"},{"Start":"00:11.550 ","End":"00:15.570","Text":"continuity equations is that there is no energy loss."},{"Start":"00:15.570 ","End":"00:20.445","Text":"The second thing that we need to know is that liquid is incompressible."},{"Start":"00:20.445 ","End":"00:27.585","Text":"That means that there\u0027s no volume change if we\u0027re transferring water or another liquid."},{"Start":"00:27.585 ","End":"00:30.285","Text":"Technically this isn\u0027t exactly true."},{"Start":"00:30.285 ","End":"00:32.610","Text":"You can slightly compress a liquid."},{"Start":"00:32.610 ","End":"00:37.530","Text":"However, it can only be very, very slightly compressed."},{"Start":"00:37.530 ","End":"00:42.045","Text":"That means that we can discount the compression because we can barely"},{"Start":"00:42.045 ","End":"00:47.265","Text":"notice it even with very expensive scientific equipment."},{"Start":"00:47.265 ","End":"00:51.195","Text":"Now that we have these 2 things that we have to remember,"},{"Start":"00:51.195 ","End":"00:54.690","Text":"let\u0027s take a look at what the continuity equation says."},{"Start":"00:54.690 ","End":"00:59.730","Text":"It says that if we take some hose with a liquid going through it,"},{"Start":"00:59.730 ","End":"01:07.223","Text":"so long as the cross-sectional surface area of the hose is constant,"},{"Start":"01:07.223 ","End":"01:13.710","Text":"the rate of flow of the liquid through this tube will be constant."},{"Start":"01:13.710 ","End":"01:16.560","Text":"Why is that? Let\u0027s say we have"},{"Start":"01:16.560 ","End":"01:26.460","Text":"some small volume of water or liquid which is passing through this hose at some velocity."},{"Start":"01:26.460 ","End":"01:30.330","Text":"If the next small volume of water is traveling at"},{"Start":"01:30.330 ","End":"01:36.030","Text":"a slower velocity to the first tiny volume of water,"},{"Start":"01:36.030 ","End":"01:43.050","Text":"there will be some vacuum in space between the 2 tiny volumes of water."},{"Start":"01:43.050 ","End":"01:46.395","Text":"That can\u0027t be. If this piece,"},{"Start":"01:46.395 ","End":"01:50.250","Text":"this tiny volume is traveling slower and the tiny"},{"Start":"01:50.250 ","End":"01:54.413","Text":"volume after it is traveling at a faster velocity,"},{"Start":"01:54.413 ","End":"02:00.360","Text":"then at this point when the 2 tiny volumes meet each other,"},{"Start":"02:00.360 ","End":"02:07.215","Text":"there\u0027ll be some pressure that will develop here and that also cannot happen."},{"Start":"02:07.215 ","End":"02:14.160","Text":"That means that as long as the cross-sectional area inside the hose is constant,"},{"Start":"02:14.160 ","End":"02:18.270","Text":"is the same throughout the hose then the velocity or"},{"Start":"02:18.270 ","End":"02:23.820","Text":"the flow rate of the liquid inside the hose is also going to be constant."},{"Start":"02:23.820 ","End":"02:29.370","Text":"What happens if I have my cross-sectional area of the hose and I"},{"Start":"02:29.370 ","End":"02:34.890","Text":"suddenly either increase the cross-sectional area or decrease the cross-sectional area."},{"Start":"02:34.890 ","End":"02:37.380","Text":"What will happen to the rate of flow?"},{"Start":"02:37.380 ","End":"02:39.615","Text":"In our diagram over here,"},{"Start":"02:39.615 ","End":"02:42.870","Text":"because the arrows are pointing and this leftwards direction,"},{"Start":"02:42.870 ","End":"02:45.990","Text":"we can see that we\u0027re decreasing for"},{"Start":"02:45.990 ","End":"02:51.105","Text":"our original cross-sectional area to a smaller cross-sectional area."},{"Start":"02:51.105 ","End":"02:54.945","Text":"If say I have my cross-sectional area,"},{"Start":"02:54.945 ","End":"02:58.800","Text":"then we\u0027ll see that the rate of flow will double."},{"Start":"02:58.800 ","End":"03:01.830","Text":"If I divide this cross-sectional area by 2,"},{"Start":"03:01.830 ","End":"03:09.360","Text":"then the velocity of the flow rate is going to increase. It\u0027s going to double."},{"Start":"03:09.360 ","End":"03:12.315","Text":"We\u0027re going to talk about why this happens."},{"Start":"03:12.315 ","End":"03:16.020","Text":"Here we can see that I have my hose."},{"Start":"03:16.020 ","End":"03:22.275","Text":"First, I have a slightly larger hose with a cross-sectional area of A_1."},{"Start":"03:22.275 ","End":"03:24.855","Text":"When we go into the smallest section of the hose,"},{"Start":"03:24.855 ","End":"03:27.630","Text":"I have a cross-sectional area of A_2."},{"Start":"03:27.630 ","End":"03:33.210","Text":"What are we going to do is we\u0027re going to look at the 2 volumes that are"},{"Start":"03:33.210 ","End":"03:39.315","Text":"working over here of little volumes of this liquid that\u0027s flowing through the hose."},{"Start":"03:39.315 ","End":"03:42.270","Text":"Now, what\u0027s important to note is that at"},{"Start":"03:42.270 ","End":"03:46.785","Text":"any given time through both of the cross-sectional areas,"},{"Start":"03:46.785 ","End":"03:48.165","Text":"A_1 and A_2,"},{"Start":"03:48.165 ","End":"03:55.320","Text":"there\u0027s going to be the exact same volume of water passing at a specific point."},{"Start":"03:55.320 ","End":"04:01.215","Text":"If at a moment in time this volume of liquid passes this point,"},{"Start":"04:01.215 ","End":"04:04.665","Text":"then at the same point in time,"},{"Start":"04:04.665 ","End":"04:07.455","Text":"in the same Delta t,"},{"Start":"04:07.455 ","End":"04:13.020","Text":"this volume of liquid is going to pass through this point of the hose,"},{"Start":"04:13.020 ","End":"04:18.660","Text":"and this volume and this volume must be equal."},{"Start":"04:18.660 ","End":"04:22.950","Text":"Why is that? If this volume is moving leftwards,"},{"Start":"04:22.950 ","End":"04:27.960","Text":"so then the liquid that is located over here must also move leftwards."},{"Start":"04:27.960 ","End":"04:30.570","Text":"Then the liquid that\u0027s located over here must also move"},{"Start":"04:30.570 ","End":"04:34.410","Text":"leftwards and so on and so forth throughout the hose,"},{"Start":"04:34.410 ","End":"04:39.345","Text":"which means that it\u0027s the same volume that is moving leftwards all the time."},{"Start":"04:39.345 ","End":"04:42.465","Text":"In other words, at every second through"},{"Start":"04:42.465 ","End":"04:46.125","Text":"every point of cross-sectional area in this hose,"},{"Start":"04:46.125 ","End":"04:49.260","Text":"be it through A_1 or A_2,"},{"Start":"04:49.260 ","End":"04:54.030","Text":"the same amount of mass is going to be passing."},{"Start":"04:54.030 ","End":"04:58.365","Text":"If we look in some time-frame,"},{"Start":"04:58.365 ","End":"05:00.495","Text":"some amount of time t,"},{"Start":"05:00.495 ","End":"05:04.470","Text":"our liquid inside this hose is going to move"},{"Start":"05:04.470 ","End":"05:11.115","Text":"this distance of L_1 and it\u0027s traveling at this velocity V_1."},{"Start":"05:11.115 ","End":"05:13.783","Text":"What is this distance L_1?"},{"Start":"05:13.783 ","End":"05:16.635","Text":"Velocity multiplied by time."},{"Start":"05:16.635 ","End":"05:19.215","Text":"That means that L_1,"},{"Start":"05:19.215 ","End":"05:20.565","Text":"this distance over here,"},{"Start":"05:20.565 ","End":"05:22.155","Text":"is equal to velocity,"},{"Start":"05:22.155 ","End":"05:27.716","Text":"which in this section of the hose is equal to V_1 multiplied by time,"},{"Start":"05:27.716 ","End":"05:28.908","Text":"so the time taken,"},{"Start":"05:28.908 ","End":"05:33.120","Text":"some t. Then if I multiply all of this by A_1,"},{"Start":"05:33.120 ","End":"05:35.400","Text":"the cross-sectional area over here,"},{"Start":"05:35.400 ","End":"05:40.155","Text":"that\u0027s going to be the volume of this block of liquid."},{"Start":"05:40.155 ","End":"05:43.005","Text":"Similarly, this distance L_2,"},{"Start":"05:43.005 ","End":"05:47.370","Text":"so it\u0027s also equal to velocity multiplied by time,"},{"Start":"05:47.370 ","End":"05:51.960","Text":"where here our velocity is equal to V_2 in this section"},{"Start":"05:51.960 ","End":"05:56.364","Text":"of the hose in the same time-frame t. Of course,"},{"Start":"05:56.364 ","End":"05:58.590","Text":"if I multiply this by A_2,"},{"Start":"05:58.590 ","End":"06:03.615","Text":"then I\u0027ll get this volume of this liquid."},{"Start":"06:03.615 ","End":"06:07.649","Text":"Because we said that liquid is incompressible,"},{"Start":"06:07.649 ","End":"06:09.870","Text":"we know that therefore,"},{"Start":"06:09.870 ","End":"06:13.320","Text":"if we\u0027re taking the same time-frame t,"},{"Start":"06:13.320 ","End":"06:16.245","Text":"then that means that this volume over here,"},{"Start":"06:16.245 ","End":"06:18.970","Text":"relating to L_1 and A_1,"},{"Start":"06:18.970 ","End":"06:21.345","Text":"is going to be the same volume over here,"},{"Start":"06:21.345 ","End":"06:23.625","Text":"relating to L_2 and A_2,"},{"Start":"06:23.625 ","End":"06:27.405","Text":"like we explained at the beginning of the video."},{"Start":"06:27.405 ","End":"06:35.580","Text":"This volume is going to be equal to A_1 multiplied by L_1."},{"Start":"06:35.580 ","End":"06:37.320","Text":"This as we said,"},{"Start":"06:37.320 ","End":"06:39.270","Text":"because the liquid is incompressible,"},{"Start":"06:39.270 ","End":"06:42.165","Text":"is going to be equal to this volume over here."},{"Start":"06:42.165 ","End":"06:48.900","Text":"This volume over here is equal to A_2 multiplied by L_2."},{"Start":"06:48.900 ","End":"06:59.295","Text":"We are going to divide both sides by A_1 and by L_2."},{"Start":"06:59.295 ","End":"07:09.075","Text":"We\u0027ll get that L_1 divided by L_2 is equal to A_2 divided by A_1."},{"Start":"07:09.075 ","End":"07:13.770","Text":"We\u0027ve gotten all of our A\u0027s on 1 side and all of our L\u0027s on the other side."},{"Start":"07:13.770 ","End":"07:23.955","Text":"We know that our L_1 is equal to V_1 multiplied by t. This is divided by L_2,"},{"Start":"07:23.955 ","End":"07:31.185","Text":"which is equal to V_2 divided by t. We can see that these ts can just cancel out."},{"Start":"07:31.185 ","End":"07:38.000","Text":"Then we can see that we get the equation that V_1 divided by V_2,"},{"Start":"07:38.000 ","End":"07:43.860","Text":"so the velocity V_1 divided by the velocity V_2, these are velocity,"},{"Start":"07:43.860 ","End":"07:50.280","Text":"is not volumes, is equal to A_2 divided by A_1."},{"Start":"07:50.280 ","End":"07:58.425","Text":"We can see that if I enlarge the cross-sectional area by 10,"},{"Start":"07:58.425 ","End":"08:03.060","Text":"then my velocity is going to decrease by a factor of 10."},{"Start":"08:03.060 ","End":"08:07.430","Text":"If I reduce my cross-sectional area by 10,"},{"Start":"08:07.430 ","End":"08:11.645","Text":"then my velocity is going to increase by a factor of 10."},{"Start":"08:11.645 ","End":"08:18.070","Text":"This is the continuity equation and this is what is important to remember."},{"Start":"08:18.070 ","End":"08:21.100","Text":"That\u0027s the end of this lesson."}],"ID":12365},{"Watched":false,"Name":"Exercise 1","Duration":"4m 39s","ChapterTopicVideoID":11936,"CourseChapterTopicPlaylistID":84745,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.025","Text":"Hello. In this lesson,"},{"Start":"00:02.025 ","End":"00:03.990","Text":"we\u0027re going to be answering this question."},{"Start":"00:03.990 ","End":"00:07.305","Text":"What is the velocity of water flowing out of a reservoir?"},{"Start":"00:07.305 ","End":"00:09.075","Text":"From the previous lesson,"},{"Start":"00:09.075 ","End":"00:16.830","Text":"we saw that the energy at every single point in a reservoir is equal."},{"Start":"00:16.830 ","End":"00:20.309","Text":"It\u0027s divided into different terms,"},{"Start":"00:20.309 ","End":"00:22.590","Text":"so potential energy, kinetic energy,"},{"Start":"00:22.590 ","End":"00:26.445","Text":"or elastic energy, where depending on where the molecule is,"},{"Start":"00:26.445 ","End":"00:29.775","Text":"it will have more of 1 of the energies and less of the other."},{"Start":"00:29.775 ","End":"00:35.505","Text":"However, the total energy of every molecule in this reservoir is going to be the same."},{"Start":"00:35.505 ","End":"00:37.905","Text":"In order to answer this question,"},{"Start":"00:37.905 ","End":"00:41.580","Text":"we have to assume that the reservoir is very,"},{"Start":"00:41.580 ","End":"00:47.975","Text":"very big, and that this aperture over here is very, very small."},{"Start":"00:47.975 ","End":"00:51.605","Text":"Then in that way we can assume that"},{"Start":"00:51.605 ","End":"00:57.613","Text":"the kinetic energy section of the water molecules over here is equal to 0,"},{"Start":"00:57.613 ","End":"01:00.725","Text":"and that will make this calculation much easier."},{"Start":"01:00.725 ","End":"01:03.245","Text":"That\u0027s our assumption for this question."},{"Start":"01:03.245 ","End":"01:11.150","Text":"Let\u0027s say that the depth of this reservoir is h. Let\u0027s get going."},{"Start":"01:11.150 ","End":"01:14.450","Text":"We\u0027re going to take our energy at point number 1,"},{"Start":"01:14.450 ","End":"01:16.850","Text":"which is this point right at the top,"},{"Start":"01:16.850 ","End":"01:19.240","Text":"at the surface of the water."},{"Start":"01:19.240 ","End":"01:21.945","Text":"Let\u0027s write this over here."},{"Start":"01:21.945 ","End":"01:24.285","Text":"The energy at point number 1."},{"Start":"01:24.285 ","End":"01:27.590","Text":"We\u0027re going to have our potential energy, which as we know,"},{"Start":"01:27.590 ","End":"01:33.170","Text":"is equal to Rho g multiplied by the height of the reservoir."},{"Start":"01:33.170 ","End":"01:36.080","Text":"Then we\u0027re going to add in our kinetic energy."},{"Start":"01:36.080 ","End":"01:40.160","Text":"Because of our assumption that we made that the tank is very,"},{"Start":"01:40.160 ","End":"01:43.010","Text":"very large relative to this aperture,"},{"Start":"01:43.010 ","End":"01:48.500","Text":"we can say that our water level is pretty much constant or stationary,"},{"Start":"01:48.500 ","End":"01:51.395","Text":"which means that our water isn\u0027t flowing."},{"Start":"01:51.395 ","End":"01:53.660","Text":"Which means that it has 0 velocity,"},{"Start":"01:53.660 ","End":"01:56.360","Text":"so it has no kinetic energy at point number 1."},{"Start":"01:56.360 ","End":"01:59.900","Text":"Then we have to add in our elastic energy."},{"Start":"01:59.900 ","End":"02:03.110","Text":"Elastic energy was our pressure."},{"Start":"02:03.110 ","End":"02:06.980","Text":"The pressure of the water at the top over here is going to"},{"Start":"02:06.980 ","End":"02:11.450","Text":"be just simply atmospheric pressure P_atm,"},{"Start":"02:11.450 ","End":"02:16.760","Text":"because it\u0027s just experiencing the weight of the atmosphere on top over here,"},{"Start":"02:16.760 ","End":"02:18.395","Text":"it\u0027s the atmospheric pressure."},{"Start":"02:18.395 ","End":"02:19.985","Text":"This, as we know,"},{"Start":"02:19.985 ","End":"02:23.545","Text":"is going to be equal to the energy at point number 2."},{"Start":"02:23.545 ","End":"02:28.015","Text":"Let\u0027s take a look at what is the energy at point number 2, and obviously,"},{"Start":"02:28.015 ","End":"02:33.870","Text":"point number 2 we mean over here at the exit. Let\u0027s go again."},{"Start":"02:33.870 ","End":"02:35.465","Text":"Potential energy."},{"Start":"02:35.465 ","End":"02:38.780","Text":"Because we can see that the aperture is at height 0,"},{"Start":"02:38.780 ","End":"02:44.465","Text":"so we\u0027re multiplying Rho g by h which is equal to 0,"},{"Start":"02:44.465 ","End":"02:46.505","Text":"so there\u0027s no potential energy."},{"Start":"02:46.505 ","End":"02:48.660","Text":"Then we have kinetic energy."},{"Start":"02:48.660 ","End":"02:52.050","Text":"We know that the water is flowing over here."},{"Start":"02:52.050 ","End":"02:54.445","Text":"We\u0027re going to have kinetic energy,"},{"Start":"02:54.445 ","End":"03:01.408","Text":"which as we saw is equal to 1/2 Rho v^2."},{"Start":"03:01.408 ","End":"03:04.760","Text":"Where this v is exactly what we\u0027re trying to find in the question,"},{"Start":"03:04.760 ","End":"03:06.415","Text":"the velocity of the water."},{"Start":"03:06.415 ","End":"03:11.795","Text":"Then we\u0027re going to add our pressure, our elastic energy."},{"Start":"03:11.795 ","End":"03:17.390","Text":"Because we\u0027re located over here right outside of the aperture of this hose,"},{"Start":"03:17.390 ","End":"03:20.687","Text":"we\u0027re not located anymore inside the tank,"},{"Start":"03:20.687 ","End":"03:22.625","Text":"that means that again,"},{"Start":"03:22.625 ","End":"03:24.920","Text":"the water is simply outside."},{"Start":"03:24.920 ","End":"03:29.670","Text":"That means that it\u0027s subject to atmospheric pressure again."},{"Start":"03:30.530 ","End":"03:37.070","Text":"If we were taking the energy at some point inside the tank or the bottom of the tank,"},{"Start":"03:37.070 ","End":"03:40.505","Text":"then of course, we would have to take into account the depth of the tank."},{"Start":"03:40.505 ","End":"03:43.940","Text":"However, it\u0027s located over here right outside,"},{"Start":"03:43.940 ","End":"03:48.660","Text":"so it\u0027s just subject to atmospheric pressures."},{"Start":"03:48.660 ","End":"03:52.955","Text":"We see that these 2 equations are equal to one another."},{"Start":"03:52.955 ","End":"03:59.585","Text":"Let\u0027s rewrite that Rho at gh plus our atmospheric pressure,"},{"Start":"03:59.585 ","End":"04:08.970","Text":"is equal to 1/2 of Rho v^2 plus our atmospheric pressure."},{"Start":"04:08.970 ","End":"04:13.093","Text":"We can subtract our atmospheric pressure from both sides,"},{"Start":"04:13.093 ","End":"04:17.995","Text":"and then we can multiply both sides by 2,"},{"Start":"04:17.995 ","End":"04:20.160","Text":"divide both sides by Rho,"},{"Start":"04:20.160 ","End":"04:23.450","Text":"and then we\u0027ll square both sides to get"},{"Start":"04:23.450 ","End":"04:26.930","Text":"our v. We\u0027ll get that the velocity of the water flowing out"},{"Start":"04:26.930 ","End":"04:34.185","Text":"of the reservoir is equal to the square root of 2gh."},{"Start":"04:34.185 ","End":"04:36.263","Text":"That\u0027s the velocity of the water,"},{"Start":"04:36.263 ","End":"04:39.309","Text":"and this is the end of our question."}],"ID":12366},{"Watched":false,"Name":"Energy Conservation","Duration":"10m 34s","ChapterTopicVideoID":11937,"CourseChapterTopicPlaylistID":84745,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.115","Text":"Hello. In this lesson,"},{"Start":"00:02.115 ","End":"00:04.995","Text":"we\u0027re going to be learning about energy conservation"},{"Start":"00:04.995 ","End":"00:09.840","Text":"relating to hydrostatics and hydrodynamics."},{"Start":"00:09.840 ","End":"00:14.745","Text":"In order to understand how energy conservation works with a liquid,"},{"Start":"00:14.745 ","End":"00:19.435","Text":"first, we have to understand energy conservation with a spring."},{"Start":"00:19.435 ","End":"00:24.740","Text":"Up until now we\u0027ve seen how springs work when they\u0027re massless."},{"Start":"00:24.740 ","End":"00:29.975","Text":"However, now we\u0027re going to learn how springs work when they are not massless."},{"Start":"00:29.975 ","End":"00:35.330","Text":"Then afterwards we\u0027ll see how water acts like a spring with mass."},{"Start":"00:35.330 ","End":"00:38.195","Text":"If this is our spring with mass,"},{"Start":"00:38.195 ","End":"00:41.570","Text":"we\u0027ll see that each rung or spiral in"},{"Start":"00:41.570 ","End":"00:45.800","Text":"the spring has different spacing when we pull this spring up."},{"Start":"00:45.800 ","End":"00:51.500","Text":"The springs towards the bottom or the rungs in the spring towards the bottom will be"},{"Start":"00:51.500 ","End":"00:57.490","Text":"closer together and the rungs towards the top will be more spaced out from one another."},{"Start":"00:57.490 ","End":"00:59.880","Text":"Why is this interesting?"},{"Start":"00:59.880 ","End":"01:02.150","Text":"We\u0027ll see that even though the rungs of"},{"Start":"01:02.150 ","End":"01:05.885","Text":"the spring have different spacing between one another,"},{"Start":"01:05.885 ","End":"01:13.045","Text":"we\u0027ll see that the energy of each rung is going to be the same regardless of the spacing."},{"Start":"01:13.045 ","End":"01:16.520","Text":"I\u0027m not going to go into the full mathematical proof."},{"Start":"01:16.520 ","End":"01:20.570","Text":"However, intuitively, we can see that this rung, which is high up,"},{"Start":"01:20.570 ","End":"01:23.675","Text":"will have barely any elastic energy,"},{"Start":"01:23.675 ","End":"01:25.125","Text":"if any, however,"},{"Start":"01:25.125 ","End":"01:27.900","Text":"it will have energy due to its height."},{"Start":"01:27.900 ","End":"01:30.350","Text":"Potential energy. Whereas the lower rung"},{"Start":"01:30.350 ","End":"01:33.455","Text":"won\u0027t have any potential energy because it\u0027s at the bottom,"},{"Start":"01:33.455 ","End":"01:37.525","Text":"however, it will have a lot of elastic potential energy."},{"Start":"01:37.525 ","End":"01:39.675","Text":"Then this rung in the middle,"},{"Start":"01:39.675 ","End":"01:46.455","Text":"will have some elastic energy and also some energy due to the potential energy."},{"Start":"01:46.455 ","End":"01:50.060","Text":"It doesn\u0027t matter at which rung of the spring I look,"},{"Start":"01:50.060 ","End":"01:53.330","Text":"it\u0027s going to have the same amount of energy,"},{"Start":"01:53.330 ","End":"01:56.410","Text":"just in a slightly different form."},{"Start":"01:56.410 ","End":"01:58.378","Text":"As we go up the rungs,"},{"Start":"01:58.378 ","End":"02:01.580","Text":"we have more potential energy and as we go down the rungs,"},{"Start":"02:01.580 ","End":"02:04.135","Text":"we have more elastic energy."},{"Start":"02:04.135 ","End":"02:10.895","Text":"In that case, we can write that the total amount of energy in a spring is equal to."},{"Start":"02:10.895 ","End":"02:14.660","Text":"First we have our potential energy, which is mgh,"},{"Start":"02:14.660 ","End":"02:16.655","Text":"and then we have our kinetic energy,"},{"Start":"02:16.655 ","End":"02:20.270","Text":"which is equal to 0.5mv^2."},{"Start":"02:20.270 ","End":"02:22.580","Text":"Then we have, because we\u0027re dealing with a spring,"},{"Start":"02:22.580 ","End":"02:27.320","Text":"our spring energy, which is equal to 0.5 multiplied by the spring constant,"},{"Start":"02:27.320 ","End":"02:30.470","Text":"which is k, multiplied by delta x,"},{"Start":"02:30.470 ","End":"02:35.965","Text":"which is the amount the spring has been stretched squared."},{"Start":"02:35.965 ","End":"02:41.135","Text":"With a spring, every single rung has potential energy,"},{"Start":"02:41.135 ","End":"02:43.985","Text":"kinetic energy, and spring energy."},{"Start":"02:43.985 ","End":"02:48.940","Text":"This is analogous to what is happening with liquids."},{"Start":"02:48.940 ","End":"02:51.680","Text":"What\u0027s happening in this liquid is that"},{"Start":"02:51.680 ","End":"02:56.615","Text":"every single particle in the liquid has the exact same amount of energy."},{"Start":"02:56.615 ","End":"03:02.455","Text":"However, the types of energy are in different quantities depending"},{"Start":"03:02.455 ","End":"03:09.645","Text":"on what is happening to that water molecule or any liquid molecule."},{"Start":"03:09.645 ","End":"03:14.810","Text":"Now what we\u0027re going to do is we\u0027re going to convert from this energy that we see"},{"Start":"03:14.810 ","End":"03:19.990","Text":"in the spring into the energy that we see in the liquid."},{"Start":"03:19.990 ","End":"03:22.280","Text":"The energy in a liquid,"},{"Start":"03:22.280 ","End":"03:26.935","Text":"we split up to different sections within the liquid."},{"Start":"03:26.935 ","End":"03:33.050","Text":"If in a spring, we talk about the energy at some rung in the spring or some coil,"},{"Start":"03:33.050 ","End":"03:34.685","Text":"so in the liquid,"},{"Start":"03:34.685 ","End":"03:39.925","Text":"we speak about the energy of some small unit of volume."},{"Start":"03:39.925 ","End":"03:43.655","Text":"Instead of speaking about the entire energy,"},{"Start":"03:43.655 ","End":"03:46.805","Text":"we\u0027re speaking about the energy of some unit volume."},{"Start":"03:46.805 ","End":"03:53.490","Text":"In that case, we\u0027re going to divide our energy by the total volume."},{"Start":"03:53.490 ","End":"03:55.905","Text":"This E divided by V,"},{"Start":"03:55.905 ","End":"03:58.845","Text":"we\u0027re going to be calling Epsilon."},{"Start":"03:58.845 ","End":"04:02.885","Text":"This epsilon is called the energy density."},{"Start":"04:02.885 ","End":"04:07.220","Text":"As we can see, the smaller the volume that we take,"},{"Start":"04:07.220 ","End":"04:11.615","Text":"the more accurate energy density we\u0027re going to get."},{"Start":"04:11.615 ","End":"04:17.165","Text":"We\u0027re going to be using infinitesimally small energy densities,"},{"Start":"04:17.165 ","End":"04:20.570","Text":"which correspond to a tiny v,"},{"Start":"04:20.570 ","End":"04:22.055","Text":"a tiny volume,"},{"Start":"04:22.055 ","End":"04:23.900","Text":"which corresponds to looking at"},{"Start":"04:23.900 ","End":"04:30.240","Text":"the energy density or the energy of a molecule of the liquid."},{"Start":"04:30.860 ","End":"04:37.190","Text":"Now we\u0027re going to convert this equation into this energy density equation."},{"Start":"04:37.190 ","End":"04:39.710","Text":"As we know over here,"},{"Start":"04:39.710 ","End":"04:45.825","Text":"we know that density is equal to mass divided by volume."},{"Start":"04:45.825 ","End":"04:50.270","Text":"Here we have our energy equation divided by volume."},{"Start":"04:50.270 ","End":"04:54.090","Text":"Here we have m. Let\u0027s actually write this out."},{"Start":"04:54.090 ","End":"05:04.445","Text":"We have mgh divided by volume plus half mv^2 divided by volume,"},{"Start":"05:04.445 ","End":"05:13.490","Text":"plus this spring energy,"},{"Start":"05:13.490 ","End":"05:20.375","Text":"or elastic energy divided by this v. Let\u0027s have here"},{"Start":"05:20.375 ","End":"05:29.880","Text":"elastic energy divided by v. Now let\u0027s see what we get."},{"Start":"05:29.880 ","End":"05:32.400","Text":"This is more relevant here,"},{"Start":"05:32.400 ","End":"05:38.020","Text":"so we know that density is equal to mass divided by volume."},{"Start":"05:38.020 ","End":"05:41.000","Text":"In that case we have mass divided by volume."},{"Start":"05:41.000 ","End":"05:50.895","Text":"That\u0027s density, multiplied by gh plus 1/2 mass divided by volume, that\u0027s density."},{"Start":"05:50.895 ","End":"05:56.255","Text":"Multiplied by velocity squared plus our spring energy,"},{"Start":"05:56.255 ","End":"05:58.550","Text":"which is elastic energy divided by volume,"},{"Start":"05:58.550 ","End":"06:03.200","Text":"which is in actual fact, just pressure."},{"Start":"06:03.200 ","End":"06:12.035","Text":"Now, what we can see here is that this is basically analogous to our potential energy."},{"Start":"06:12.035 ","End":"06:15.810","Text":"I\u0027ll write Ep over here."},{"Start":"06:15.820 ","End":"06:22.070","Text":"This over here is basically just our kinetic energy,"},{"Start":"06:22.070 ","End":"06:23.825","Text":"so Ek over here."},{"Start":"06:23.825 ","End":"06:28.330","Text":"This pressure came from our elastic energy."},{"Start":"06:28.330 ","End":"06:32.435","Text":"This is the energy or the energy density of"},{"Start":"06:32.435 ","End":"06:37.490","Text":"every single particle of water in any tank or reservoir,"},{"Start":"06:37.490 ","End":"06:40.040","Text":"whether the water is sitting in"},{"Start":"06:40.040 ","End":"06:44.165","Text":"the reservoir or flowing out of the reservoir or into the reservoir."},{"Start":"06:44.165 ","End":"06:49.775","Text":"We\u0027ll see later on in this chapter that sometimes the energy is different."},{"Start":"06:49.775 ","End":"06:54.215","Text":"However, right now, what we need to know is for most cases,"},{"Start":"06:54.215 ","End":"06:58.100","Text":"the energy of every single particle of liquid in a reservoir,"},{"Start":"06:58.100 ","End":"07:00.050","Text":"whether it\u0027s flowing or not,"},{"Start":"07:00.050 ","End":"07:07.715","Text":"is going to be the same divided by these 3 energy components."},{"Start":"07:07.715 ","End":"07:12.215","Text":"Now, they can ask us different questions to do with this."},{"Start":"07:12.215 ","End":"07:15.800","Text":"One example is asking us what is the pressure"},{"Start":"07:15.800 ","End":"07:20.560","Text":"at point number 1 and what is the pressure at point number 2?"},{"Start":"07:20.560 ","End":"07:25.095","Text":"First of all, the pressure at point number 1,"},{"Start":"07:25.095 ","End":"07:28.610","Text":"we\u0027re talking about the upper layer of the water."},{"Start":"07:28.610 ","End":"07:30.230","Text":"The pressure there is going to be"},{"Start":"07:30.230 ","End":"07:34.595","Text":"the same pressure that we have in the room that we\u0027re sitting in."},{"Start":"07:34.595 ","End":"07:38.785","Text":"That\u0027s a pressure of 1 atmosphere."},{"Start":"07:38.785 ","End":"07:42.800","Text":"That\u0027s just the pressure of our atmosphere on Earth."},{"Start":"07:42.800 ","End":"07:46.805","Text":"Now, when we\u0027re dealing with the pressure at point number 2,"},{"Start":"07:46.805 ","End":"07:51.995","Text":"we know that we\u0027re going to have the pressure from outside, just from the air."},{"Start":"07:51.995 ","End":"07:55.475","Text":"Then as we go deeper and deeper into this tank,"},{"Start":"07:55.475 ","End":"08:01.280","Text":"there\u0027ll be more water pressing down and causing a pressure at our point over here."},{"Start":"08:01.280 ","End":"08:04.610","Text":"Now we have to figure out what the pressure is over here."},{"Start":"08:04.610 ","End":"08:09.260","Text":"What we\u0027re going to do is we\u0027re going to use our energy equation over here."},{"Start":"08:09.260 ","End":"08:14.750","Text":"We know that our energy at point number 1 is equal to our energy at point number 2."},{"Start":"08:14.750 ","End":"08:18.960","Text":"Let\u0027s write out our energy at point number 1."},{"Start":"08:18.960 ","End":"08:21.845","Text":"We have our potential energy,"},{"Start":"08:21.845 ","End":"08:28.445","Text":"which is equal to Rho g and h. It\u0027s at a height of h, so it has that."},{"Start":"08:28.445 ","End":"08:30.950","Text":"Then we can write down our kinetic energy,"},{"Start":"08:30.950 ","End":"08:35.450","Text":"which because our water isn\u0027t flowing or our liquid isn\u0027t flowing,"},{"Start":"08:35.450 ","End":"08:37.520","Text":"so we don\u0027t have kinetic energy."},{"Start":"08:37.520 ","End":"08:42.290","Text":"Then we have to write our elastic energy which is simply our pressure."},{"Start":"08:42.290 ","End":"08:48.760","Text":"We already saw that our pressure at point number 1 is equal to 1 atmosphere."},{"Start":"08:48.760 ","End":"08:53.015","Text":"Let\u0027s just call it Patm."},{"Start":"08:53.015 ","End":"08:55.640","Text":"Then we\u0027re adding Patm,"},{"Start":"08:55.640 ","End":"08:59.030","Text":"which is our elastic energy for point number 1."},{"Start":"08:59.030 ","End":"09:05.930","Text":"Then all of this is equal to our energy at point number 2."},{"Start":"09:05.930 ","End":"09:09.140","Text":"Now let\u0027s take a look at the energy at point number 2."},{"Start":"09:09.140 ","End":"09:11.690","Text":"First we have our potential energy."},{"Start":"09:11.690 ","End":"09:16.135","Text":"Now point number 2 is located at a height of 0."},{"Start":"09:16.135 ","End":"09:18.080","Text":"We don\u0027t have potential energy,"},{"Start":"09:18.080 ","End":"09:19.460","Text":"so we will write that in."},{"Start":"09:19.460 ","End":"09:22.340","Text":"Then, we\u0027ll write in our kinetic energy."},{"Start":"09:22.340 ","End":"09:25.250","Text":"Now again, our liquid isn\u0027t flowing,"},{"Start":"09:25.250 ","End":"09:27.500","Text":"which means that we don\u0027t have any velocity."},{"Start":"09:27.500 ","End":"09:30.200","Text":"Again, our kinetic energy is equal to 0."},{"Start":"09:30.200 ","End":"09:36.140","Text":"Then we have to add in our potential energy for point number 2,"},{"Start":"09:36.140 ","End":"09:39.485","Text":"which we\u0027re trying to find is equal to P2."},{"Start":"09:39.485 ","End":"09:42.335","Text":"But now we have this equation over here."},{"Start":"09:42.335 ","End":"09:46.025","Text":"We know that the energy at 1 is equal to the energy at 2."},{"Start":"09:46.025 ","End":"09:51.275","Text":"We can just write this out and say that P2 is simply equal to"},{"Start":"09:51.275 ","End":"09:58.205","Text":"Rho gh plus our atmospheric pressure."},{"Start":"09:58.205 ","End":"10:01.145","Text":"This is the pressure at point number 2."},{"Start":"10:01.145 ","End":"10:03.110","Text":"Now this also makes sense."},{"Start":"10:03.110 ","End":"10:04.580","Text":"As we said before,"},{"Start":"10:04.580 ","End":"10:07.100","Text":"the pressure point number 2 is going to be"},{"Start":"10:07.100 ","End":"10:13.615","Text":"the atmospheric pressure plus the pressure due to the weight of the water above it."},{"Start":"10:13.615 ","End":"10:19.010","Text":"We have our atmospheric pressure plus this element over here,"},{"Start":"10:19.010 ","End":"10:21.980","Text":"which is due to the weight of the water above."},{"Start":"10:21.980 ","End":"10:25.145","Text":"We\u0027re taking the density and the height,"},{"Start":"10:25.145 ","End":"10:27.905","Text":"the depth that we have above us, which is h,"},{"Start":"10:27.905 ","End":"10:31.220","Text":"multiplying it by g. That makes sense."},{"Start":"10:31.220 ","End":"10:34.710","Text":"Okay, that\u0027s the end of our lesson."}],"ID":12367},{"Watched":false,"Name":"Exercise 2","Duration":"11m 52s","ChapterTopicVideoID":11938,"CourseChapterTopicPlaylistID":84745,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.920","Text":"Hello. In this lesson,"},{"Start":"00:01.920 ","End":"00:07.200","Text":"we\u0027re going to be answering what are the pressures at points 2 and 3."},{"Start":"00:07.200 ","End":"00:09.000","Text":"Here we have 3 points,"},{"Start":"00:09.000 ","End":"00:14.535","Text":"point number 1 is at the surface of the water at a height of h_3 plus h_4,"},{"Start":"00:14.535 ","End":"00:18.240","Text":"point Number 2 is in this tube over here,"},{"Start":"00:18.240 ","End":"00:21.120","Text":"and point number 3 is still in tube,"},{"Start":"00:21.120 ","End":"00:23.950","Text":"but right at the bottom."},{"Start":"00:23.950 ","End":"00:27.900","Text":"The way we\u0027re going to solve this is similar to the way"},{"Start":"00:27.900 ","End":"00:31.035","Text":"that we\u0027ve solved our previous questions,"},{"Start":"00:31.035 ","End":"00:37.670","Text":"which means that we know that the energy at each one of these points has to be equal."},{"Start":"00:37.670 ","End":"00:39.590","Text":"Just so we remember,"},{"Start":"00:39.590 ","End":"00:45.710","Text":"our energy at each point is equal to our potential energy"},{"Start":"00:45.710 ","End":"00:52.970","Text":"plus our kinetic energy plus our elastic energy at that point."},{"Start":"00:52.970 ","End":"00:58.795","Text":"Let\u0027s first write our energy at point number 1."},{"Start":"00:58.795 ","End":"01:00.815","Text":"Now before we begin,"},{"Start":"01:00.815 ","End":"01:05.990","Text":"we\u0027re going to remember that we\u0027re assuming that this reservoir is very"},{"Start":"01:05.990 ","End":"01:11.435","Text":"large and that this aperture here is small relative to the reservoir."},{"Start":"01:11.435 ","End":"01:17.060","Text":"Then we can imagine that the top layer of the water is stationary,"},{"Start":"01:17.060 ","End":"01:20.230","Text":"which means that it doesn\u0027t have any kinetic energy."},{"Start":"01:20.230 ","End":"01:23.640","Text":"First, what\u0027s our potential energy?"},{"Start":"01:23.640 ","End":"01:28.895","Text":"Our potential energy is our density multiplied by g,"},{"Start":"01:28.895 ","End":"01:31.145","Text":"multiplied by our height."},{"Start":"01:31.145 ","End":"01:36.940","Text":"Our height here is equal to h_3 plus h_4."},{"Start":"01:36.940 ","End":"01:43.630","Text":"Then our kinetic energy we already said at the top layer is going to be equal to 0."},{"Start":"01:43.630 ","End":"01:46.220","Text":"Then we\u0027re adding our elastic energy,"},{"Start":"01:46.220 ","End":"01:48.170","Text":"which is simply our pressure."},{"Start":"01:48.170 ","End":"01:56.020","Text":"Over here, the top layer is subject to our atmospheric pressure."},{"Start":"01:56.020 ","End":"02:00.930","Text":"Now let\u0027s look at our energy at point number 2."},{"Start":"02:01.910 ","End":"02:06.530","Text":"First of all, let\u0027s take our potential energy."},{"Start":"02:06.530 ","End":"02:11.660","Text":"Our potential energy is equal to Rho g multiplied by our height,"},{"Start":"02:11.660 ","End":"02:14.620","Text":"which here is just h_4."},{"Start":"02:14.620 ","End":"02:19.125","Text":"Then we\u0027re going to add in our kinetic energy."},{"Start":"02:19.125 ","End":"02:21.905","Text":"Here we do have kinetic energy,"},{"Start":"02:21.905 ","End":"02:23.210","Text":"now why is this?"},{"Start":"02:23.210 ","End":"02:25.955","Text":"Because we know that our water is flowing,"},{"Start":"02:25.955 ","End":"02:28.715","Text":"although at a very small rate,"},{"Start":"02:28.715 ","End":"02:32.140","Text":"it is flowing through this aperture over here,"},{"Start":"02:32.140 ","End":"02:35.930","Text":"which means that if water is exiting over here,"},{"Start":"02:35.930 ","End":"02:39.980","Text":"that means that water over here is also flowing in this direction,"},{"Start":"02:39.980 ","End":"02:45.830","Text":"which means that water over here is also flowing in this direction and that means that"},{"Start":"02:45.830 ","End":"02:52.680","Text":"the water over here at point number 2 is also flowing in this direction."},{"Start":"02:52.680 ","End":"02:56.600","Text":"That means that we have kinetic energy at this point."},{"Start":"02:56.600 ","End":"03:03.140","Text":"Our kinetic energy over here is half Rho V^2."},{"Start":"03:03.140 ","End":"03:05.165","Text":"Now I\u0027m writing V,"},{"Start":"03:05.165 ","End":"03:09.730","Text":"and this is the same velocity at point number 2 and at point number 3."},{"Start":"03:09.730 ","End":"03:15.320","Text":"This is what we learned in our first lesson to do with this and we saw that if we"},{"Start":"03:15.320 ","End":"03:21.230","Text":"have some volume of water moving in a specific velocity over here,"},{"Start":"03:21.230 ","End":"03:31.105","Text":"unless our tube is the cross-sectional area of the tube,"},{"Start":"03:31.105 ","End":"03:36.655","Text":"unless it is either enlarged or reduced in size,"},{"Start":"03:36.655 ","End":"03:41.515","Text":"the velocity over here will be the same velocity over here,"},{"Start":"03:41.515 ","End":"03:46.375","Text":"so long as the cross-sectional area of the tube remains constant,"},{"Start":"03:46.375 ","End":"03:48.895","Text":"which is the case over here."},{"Start":"03:48.895 ","End":"03:53.245","Text":"It\u0027s the same velocity over here and over here."},{"Start":"03:53.245 ","End":"03:57.055","Text":"Now we\u0027re going to add in our elastic energy,"},{"Start":"03:57.055 ","End":"04:01.900","Text":"which is simply the pressure at this point over here."},{"Start":"04:01.900 ","End":"04:06.410","Text":"This is P_2 and this is an unknown."},{"Start":"04:06.410 ","End":"04:15.289","Text":"Of course P_2 is going to be the pressure at this point due to the depth,"},{"Start":"04:15.289 ","End":"04:20.250","Text":"and due to the atmospheric pressure."},{"Start":"04:20.250 ","End":"04:25.465","Text":"Now let\u0027s take a look at our energy at point number 3."},{"Start":"04:25.465 ","End":"04:28.095","Text":"Let\u0027s look at the potential energy."},{"Start":"04:28.095 ","End":"04:32.210","Text":"Point number 3 is located at h=0."},{"Start":"04:32.210 ","End":"04:35.780","Text":"If we plug in 0 into Rho gh,"},{"Start":"04:35.780 ","End":"04:39.560","Text":"we\u0027ll get 0 so there\u0027s no potential energy here."},{"Start":"04:39.560 ","End":"04:42.785","Text":"Now let\u0027s add on our kinetic energy."},{"Start":"04:42.785 ","End":"04:47.640","Text":"That is equal to half Rho V^2."},{"Start":"04:47.640 ","End":"04:51.320","Text":"Of course, this is the same velocity is at point Number 2,"},{"Start":"04:51.320 ","End":"04:54.515","Text":"plus our elastic energy,"},{"Start":"04:54.515 ","End":"04:59.585","Text":"which is simply our pressure at point 3 over here."},{"Start":"04:59.585 ","End":"05:03.950","Text":"Now of course, the pressure includes the pressure of the water"},{"Start":"05:03.950 ","End":"05:09.540","Text":"being pushed down here and our atmospheric pressure."},{"Start":"05:10.160 ","End":"05:17.580","Text":"Now we can equate the energy at point 1 to the energy at point 2,"},{"Start":"05:17.580 ","End":"05:20.670","Text":"and the energy at point 1 to the energy at point 3."},{"Start":"05:20.670 ","End":"05:26.855","Text":"Then we\u0027ll rearrange our equations in order to find our unknowns, P_2 and P_3."},{"Start":"05:26.855 ","End":"05:28.460","Text":"Now before we do that,"},{"Start":"05:28.460 ","End":"05:32.570","Text":"we can remember from previous lessons that our velocity"},{"Start":"05:32.570 ","End":"05:38.075","Text":"is simply equal to the square root of 2g"},{"Start":"05:38.075 ","End":"05:45.410","Text":"multiplied by the total depth of our reservoir or"},{"Start":"05:45.410 ","End":"05:53.350","Text":"the total height that it is at which here is at a height of h_3 plus h_4."},{"Start":"05:53.350 ","End":"05:56.620","Text":"Whenever you have water flowing out of a tank,"},{"Start":"05:56.620 ","End":"06:01.380","Text":"in order to find the velocity at which it is exiting,"},{"Start":"06:01.490 ","End":"06:05.840","Text":"we just take the total height from the top of"},{"Start":"06:05.840 ","End":"06:09.760","Text":"the water level until the point where the water is exiting,"},{"Start":"06:09.760 ","End":"06:12.065","Text":"this entire height over here,"},{"Start":"06:12.065 ","End":"06:15.410","Text":"and you multiply it by 2g and take"},{"Start":"06:15.410 ","End":"06:19.630","Text":"the square root and that is the velocity of the water exiting over here."},{"Start":"06:19.630 ","End":"06:22.350","Text":"This is actually good to know."},{"Start":"06:22.350 ","End":"06:28.715","Text":"Now if we say that our energy at point 1 is equal to our energy at point 2,"},{"Start":"06:28.715 ","End":"06:31.655","Text":"then let\u0027s write out this equation."},{"Start":"06:31.655 ","End":"06:40.120","Text":"We\u0027ll get that Rho g h_3 plus h_4,"},{"Start":"06:40.640 ","End":"06:49.800","Text":"here plus our atmospheric pressure is going to be equal to,"},{"Start":"06:49.800 ","End":"06:53.145","Text":"let\u0027s move slightly to the side, here,"},{"Start":"06:53.145 ","End":"07:01.830","Text":"Rho gh_4 plus half Rho V^2"},{"Start":"07:01.830 ","End":"07:06.520","Text":"plus our pressure at point 2."},{"Start":"07:07.020 ","End":"07:10.810","Text":"Now we already said that our pressure at point 2"},{"Start":"07:10.810 ","End":"07:17.080","Text":"includes the pressure due to the depth plus our atmospheric pressure."},{"Start":"07:17.080 ","End":"07:22.700","Text":"That means that from both sides we can cross off this atmospheric pressure."},{"Start":"07:22.700 ","End":"07:25.800","Text":"I\u0027ll cross it off over here and here I\u0027ll write"},{"Start":"07:25.800 ","End":"07:30.295","Text":"P_2 Tilda so that we know that this is the pressure at this point,"},{"Start":"07:30.295 ","End":"07:34.760","Text":"excluding the addition of the atmospheric pressure,"},{"Start":"07:34.760 ","End":"07:36.630","Text":"because we already took it away."},{"Start":"07:36.630 ","End":"07:38.955","Text":"That\u0027s what that Tilda means."},{"Start":"07:38.955 ","End":"07:44.030","Text":"Now what we can do is we can rearrange this equation."},{"Start":"07:45.030 ","End":"07:48.785","Text":"First of all, we can see that we have Rho g,"},{"Start":"07:48.785 ","End":"07:51.010","Text":"and if we open up these brackets,"},{"Start":"07:51.010 ","End":"07:56.500","Text":"we get Rho g multiplied by h_4 and here we also have Rho g multiplied by h_4."},{"Start":"07:56.500 ","End":"07:59.695","Text":"We\u0027ll subtract that from both sides."},{"Start":"07:59.695 ","End":"08:01.330","Text":"Then what we\u0027ll have here is"},{"Start":"08:01.330 ","End":"08:09.390","Text":"Rho g h_3 = 1/2 Rho V^2"},{"Start":"08:09.390 ","End":"08:12.215","Text":"plus the pressure at point 2,"},{"Start":"08:12.215 ","End":"08:16.325","Text":"which is excluding our atmospheric pressure."},{"Start":"08:16.325 ","End":"08:21.325","Text":"Now all we have to do is we have to isolate our P_2."},{"Start":"08:21.325 ","End":"08:30.015","Text":"That\u0027s simply going to be equal to Rho gh_3 minus half of Rho,"},{"Start":"08:30.015 ","End":"08:34.000","Text":"and then a V^2 is simply this."},{"Start":"08:36.260 ","End":"08:42.680","Text":"This is our pressure at point 2 excluding our atmospheric pressure."},{"Start":"08:42.680 ","End":"08:45.990","Text":"Now let\u0027s see what the pressure is at point 3."},{"Start":"08:45.990 ","End":"08:52.065","Text":"Now the energy at point 1 is equal to the energy at point 3."},{"Start":"08:52.065 ","End":"08:59.075","Text":"Therefore, Rho g h_3 plus h_4."},{"Start":"08:59.075 ","End":"09:03.035","Text":"I\u0027m already going to rewrite this without our atmospheric pressure."},{"Start":"09:03.035 ","End":"09:07.080","Text":"Over here, I\u0027ll add a Tilda to my P_3,"},{"Start":"09:07.540 ","End":"09:12.285","Text":"is equal to, and then that means that it\u0027s the pressure at point 3,"},{"Start":"09:12.285 ","End":"09:15.365","Text":"excluding the atmospheric pressure,"},{"Start":"09:15.365 ","End":"09:22.100","Text":"is equal to 1/2 Rho V^2 where this is"},{"Start":"09:22.100 ","End":"09:30.485","Text":"our V plus our P_3 without our atmospheric pressure added on."},{"Start":"09:30.485 ","End":"09:37.960","Text":"Now I can write that my pressure at point 3 without that atmospheric pressure is equal to"},{"Start":"09:37.960 ","End":"09:44.869","Text":"Rho g h_3 plus h_4"},{"Start":"09:44.869 ","End":"09:52.865","Text":"minus half Rho V^2."},{"Start":"09:52.865 ","End":"09:56.060","Text":"Here we have our pressures at point 2 and 3,"},{"Start":"09:56.060 ","End":"10:02.220","Text":"and we can see that our pressure at point 2 is relating to the height."},{"Start":"10:02.220 ","End":"10:07.900","Text":"Then we\u0027re minusing from both our pressure at point 2 and our pressure at point 3,"},{"Start":"10:07.900 ","End":"10:11.710","Text":"this same value over here,"},{"Start":"10:11.710 ","End":"10:15.545","Text":"half Rho V^2 from both of these,"},{"Start":"10:15.545 ","End":"10:21.040","Text":"which makes sense because the water or the liquid traveling through this tube"},{"Start":"10:21.040 ","End":"10:23.560","Text":"has the same kinetic energy"},{"Start":"10:23.560 ","End":"10:27.050","Text":"because it\u0027s traveling at the same velocity throughout the tube."},{"Start":"10:27.050 ","End":"10:31.110","Text":"Then our pressure at point number 3, also,"},{"Start":"10:31.110 ","End":"10:37.160","Text":"we\u0027re taking away this kinetic energy due to there being an opening over here."},{"Start":"10:37.160 ","End":"10:41.420","Text":"We\u0027re taking into account the pressure due to the amount of"},{"Start":"10:41.420 ","End":"10:46.655","Text":"water that is resting on this point over here."},{"Start":"10:46.655 ","End":"10:53.520","Text":"Here we have all of the liquid that is taking up this height, h_4 plus h_3."},{"Start":"10:53.630 ","End":"10:58.820","Text":"All the depth of water that is sitting on top of"},{"Start":"10:58.820 ","End":"11:04.375","Text":"this point over here is causing pressure so that\u0027s Rho g h_3."},{"Start":"11:04.375 ","End":"11:09.110","Text":"Over here, all the water that\u0027s sitting on top of point 3 is causing a pressure,"},{"Start":"11:09.110 ","End":"11:11.750","Text":"so that\u0027s going to be Rho g,"},{"Start":"11:11.750 ","End":"11:15.805","Text":"and then taking into account our h_3 plus our h_4."},{"Start":"11:15.805 ","End":"11:20.780","Text":"Now we can see that if I put a cap over here to close this aperture,"},{"Start":"11:20.780 ","End":"11:28.385","Text":"we would see that the pressure would be larger than this over here."},{"Start":"11:28.385 ","End":"11:31.565","Text":"Here we\u0027re subtracting some of the pressure"},{"Start":"11:31.565 ","End":"11:36.394","Text":"because some of the liquid or some of the water can exit through this hole,"},{"Start":"11:36.394 ","End":"11:38.360","Text":"releasing some of the pressure."},{"Start":"11:38.360 ","End":"11:40.460","Text":"If this hole was closed,"},{"Start":"11:40.460 ","End":"11:42.350","Text":"if I put a cap over here,"},{"Start":"11:42.350 ","End":"11:45.320","Text":"then there would be no releasing of"},{"Start":"11:45.320 ","End":"11:49.430","Text":"pressure or over here and so the pressure would be slightly larger."},{"Start":"11:49.430 ","End":"11:53.400","Text":"Okay, that\u0027s the end of this lesson."}],"ID":12368},{"Watched":false,"Name":"Volumetric Flow Rate","Duration":"2m 3s","ChapterTopicVideoID":11939,"CourseChapterTopicPlaylistID":84745,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.875","Text":"Hello. In this lesson,"},{"Start":"00:01.875 ","End":"00:04.950","Text":"we\u0027re going to be learning about volumetric flow rate."},{"Start":"00:04.950 ","End":"00:09.210","Text":"Volumetric flow rate just means how much water or"},{"Start":"00:09.210 ","End":"00:14.355","Text":"liquid is exiting some aperture in a given time."},{"Start":"00:14.355 ","End":"00:18.960","Text":"The flow rate is denoted by the letter"},{"Start":"00:18.960 ","End":"00:24.450","Text":"Q and volumetric is denoted by the subscript v for volume."},{"Start":"00:24.450 ","End":"00:29.010","Text":"This is equal to the velocity that the water"},{"Start":"00:29.010 ","End":"00:36.655","Text":"has multiplied by the cross-sectional area of the hose or of the aperture."},{"Start":"00:36.655 ","End":"00:43.740","Text":"Now, usually the hose or the aperture is some cylindrical hose,"},{"Start":"00:43.740 ","End":"00:45.890","Text":"so the aperture will be a circle."},{"Start":"00:45.890 ","End":"00:47.960","Text":"A lot of the time,"},{"Start":"00:47.960 ","End":"00:50.405","Text":"the volumetric flow rate is given by"},{"Start":"00:50.405 ","End":"00:54.905","Text":"the velocity of the water multiplied by the area of a circle,"},{"Start":"00:54.905 ","End":"00:58.200","Text":"which is Pi r^2."},{"Start":"00:58.370 ","End":"01:07.150","Text":"This Pi r^2 is the cross-sectional area of a cylindrical hose."},{"Start":"01:07.150 ","End":"01:10.520","Text":"Some questions that can be asked to do with"},{"Start":"01:10.520 ","End":"01:14.600","Text":"volumetric flow rate is if we have some tank,"},{"Start":"01:14.600 ","End":"01:17.615","Text":"how long will it take for the tank to empty,"},{"Start":"01:17.615 ","End":"01:20.430","Text":"and questions along those lines."},{"Start":"01:20.430 ","End":"01:23.070","Text":"Let\u0027s say that we have this tank,"},{"Start":"01:23.070 ","End":"01:25.679","Text":"and let\u0027s say that we have 2 apertures,"},{"Start":"01:25.679 ","End":"01:28.655","Text":"and imagine that they\u0027re both at the same height,"},{"Start":"01:28.655 ","End":"01:32.255","Text":"which means that the pressure on both of them is going to be the same,"},{"Start":"01:32.255 ","End":"01:35.630","Text":"that means that the water traveling through this aperture"},{"Start":"01:35.630 ","End":"01:39.715","Text":"and through this aperture will have the same volume."},{"Start":"01:39.715 ","End":"01:44.930","Text":"However, the cross-sectional areas of each aperture are different."},{"Start":"01:44.930 ","End":"01:47.780","Text":"We can see that the cross-sectional area of this aperture is"},{"Start":"01:47.780 ","End":"01:50.660","Text":"much larger than the cross-sectional area of this aperture,"},{"Start":"01:50.660 ","End":"01:55.580","Text":"which means that the volumetric flow rate through the larger aperture will be"},{"Start":"01:55.580 ","End":"02:00.320","Text":"much larger than the volumetric flow rate through the smaller aperture."},{"Start":"02:00.320 ","End":"02:03.970","Text":"Okay, that\u0027s the end of this lesson."}],"ID":12369},{"Watched":false,"Name":"Carburetor and Heart","Duration":"7m 36s","ChapterTopicVideoID":11940,"CourseChapterTopicPlaylistID":84745,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.935","Text":"Hello. In this lesson,"},{"Start":"00:01.935 ","End":"00:08.105","Text":"we\u0027re going to be looking at volumetric flow rate and pressure in real-life examples."},{"Start":"00:08.105 ","End":"00:12.480","Text":"These 2 diagrams are relating to something called a carburetor."},{"Start":"00:12.480 ","End":"00:19.620","Text":"A carburetor mixes air and fuel inside engines,"},{"Start":"00:19.620 ","End":"00:22.230","Text":"such as car engines."},{"Start":"00:22.230 ","End":"00:25.320","Text":"This leads to the fuel combusting."},{"Start":"00:25.320 ","End":"00:29.100","Text":"Let\u0027s take a look at how a carburetor works."},{"Start":"00:29.100 ","End":"00:32.550","Text":"As we can see, we\u0027re looking at this tube over here,"},{"Start":"00:32.550 ","End":"00:37.510","Text":"which is this tube over here in this diagram."},{"Start":"00:37.510 ","End":"00:42.185","Text":"What we can see is that air enters the tube in this direction."},{"Start":"00:42.185 ","End":"00:45.920","Text":"It enters a section of the tube which is slightly wider,"},{"Start":"00:45.920 ","End":"00:49.960","Text":"and then it goes through a narrower part of the tube."},{"Start":"00:49.960 ","End":"00:56.150","Text":"As we know, we have energy conservation throughout,"},{"Start":"00:56.150 ","End":"00:58.070","Text":"which means that the energy of"},{"Start":"00:58.070 ","End":"01:02.810","Text":"all the particles in the wider part of the tube has to be equal"},{"Start":"01:02.810 ","End":"01:09.725","Text":"to the energy of the particles in the narrow part of the tube or every particle here,"},{"Start":"01:09.725 ","End":"01:14.840","Text":"has to have the same energy as any particle here."},{"Start":"01:15.700 ","End":"01:21.815","Text":"We know that the energy is equal to"},{"Start":"01:21.815 ","End":"01:30.155","Text":"our potential energy plus our kinetic energy plus our elastic energy."},{"Start":"01:30.155 ","End":"01:35.510","Text":"Well, we know that our elastic energy is simply the pressure."},{"Start":"01:35.510 ","End":"01:41.540","Text":"We learned a few lessons ago when we were talking about the continuity equation,"},{"Start":"01:41.540 ","End":"01:44.950","Text":"that if we have some fluid,"},{"Start":"01:44.950 ","End":"01:47.990","Text":"gas, air is also a fluid,"},{"Start":"01:47.990 ","End":"01:53.585","Text":"not just liquids, which is flowing through a tube,"},{"Start":"01:53.585 ","End":"01:59.780","Text":"if we go from a larger cross sectional area to a smaller cross-sectional area"},{"Start":"01:59.780 ","End":"02:02.480","Text":"then the velocity of the fluid through"},{"Start":"02:02.480 ","End":"02:06.820","Text":"the smaller cross-sectional area is going to increase."},{"Start":"02:06.820 ","End":"02:11.990","Text":"We can see that the kinetic energy therefore is going to"},{"Start":"02:11.990 ","End":"02:16.789","Text":"be higher in the thinner section of the cylinder."},{"Start":"02:16.789 ","End":"02:21.500","Text":"However, we know that the energy here is the same as the energy here."},{"Start":"02:21.500 ","End":"02:25.805","Text":"In order for that to balance out with a higher kinetic energy here,"},{"Start":"02:25.805 ","End":"02:30.775","Text":"that means that here we\u0027re also going to have a lower pressure."},{"Start":"02:30.775 ","End":"02:34.760","Text":"The lower pressure that we have over here as a result of"},{"Start":"02:34.760 ","End":"02:40.685","Text":"the higher kinetic energy and our need to have the exact same energy total,"},{"Start":"02:40.685 ","End":"02:42.515","Text":"over here and over here,"},{"Start":"02:42.515 ","End":"02:44.285","Text":"so, we have a lower pressure."},{"Start":"02:44.285 ","End":"02:49.550","Text":"This tube over here is attached to the fuel tank."},{"Start":"02:49.550 ","End":"02:58.910","Text":"Fuel is in this tube and this field is pumped into this space due to this lower pressure."},{"Start":"02:58.910 ","End":"03:02.300","Text":"The lower pressure over here in the thinner section or"},{"Start":"03:02.300 ","End":"03:05.840","Text":"the narrowest section of the cylinder,"},{"Start":"03:05.840 ","End":"03:14.690","Text":"causes the petrol or the fuel to get suctioned up into this vacuum."},{"Start":"03:14.690 ","End":"03:18.530","Text":"Then this field gets mixed in with the air,"},{"Start":"03:18.530 ","End":"03:21.320","Text":"which is exactly the carburetor\u0027s job,"},{"Start":"03:21.320 ","End":"03:25.655","Text":"and then it can go into the combustion engine."},{"Start":"03:25.655 ","End":"03:28.760","Text":"Of course, the potential energy throughout"},{"Start":"03:28.760 ","End":"03:32.030","Text":"the cylinder is not changing because as we can see,"},{"Start":"03:32.030 ","End":"03:34.100","Text":"the cylinder is horizontal."},{"Start":"03:34.100 ","End":"03:37.745","Text":"That means it\u0027s at the same height at every single point,"},{"Start":"03:37.745 ","End":"03:41.345","Text":"which means that the potential energy is constant throughout."},{"Start":"03:41.345 ","End":"03:42.755","Text":"It has to change."},{"Start":"03:42.755 ","End":"03:49.650","Text":"From here to here is the balance between the kinetic energy and the pressure."},{"Start":"03:49.730 ","End":"03:53.870","Text":"Let\u0027s look at what this is."},{"Start":"03:53.870 ","End":"03:56.820","Text":"This thing over here,"},{"Start":"03:56.820 ","End":"03:59.415","Text":"this device, is called the throttle."},{"Start":"03:59.415 ","End":"04:05.095","Text":"The throttle controls the flow of fuel to an engine."},{"Start":"04:05.095 ","End":"04:10.710","Text":"When you push down the gas pedal or you hit the gas,"},{"Start":"04:10.710 ","End":"04:18.305","Text":"the throttle turns and allows a much larger volumetric flow rate."},{"Start":"04:18.305 ","End":"04:21.320","Text":"Right now we can see that it\u0027s almost closed."},{"Start":"04:21.320 ","End":"04:24.710","Text":"That means that the volumetric flow rate is going to"},{"Start":"04:24.710 ","End":"04:29.080","Text":"be relatively small as we can see from here."},{"Start":"04:29.900 ","End":"04:34.670","Text":"What happens is that our volumetric flow rate is very"},{"Start":"04:34.670 ","End":"04:39.110","Text":"small because it is equal to the velocity of"},{"Start":"04:39.110 ","End":"04:41.390","Text":"the fluid multiplied by"},{"Start":"04:41.390 ","End":"04:48.050","Text":"the cross-sectional area or the area of the opening over here and also over here."},{"Start":"04:48.050 ","End":"04:51.500","Text":"When we hit the gas or when we push the gas pedal,"},{"Start":"04:51.500 ","End":"04:54.740","Text":"this section of the throttle moves downwards,"},{"Start":"04:54.740 ","End":"04:57.335","Text":"this section of the throttle moves upwards,"},{"Start":"04:57.335 ","End":"05:02.045","Text":"and then we can see that although the fluid is traveling with the same velocity,"},{"Start":"05:02.045 ","End":"05:05.225","Text":"the cross-sectional area is now much bigger."},{"Start":"05:05.225 ","End":"05:09.385","Text":"We have a much larger volumetric flow rate."},{"Start":"05:09.385 ","End":"05:11.810","Text":"It doesn\u0027t alter the velocity of the air,"},{"Start":"05:11.810 ","End":"05:18.035","Text":"but it simply alters the cross-sectional area that the air can flow through."},{"Start":"05:18.035 ","End":"05:20.210","Text":"That\u0027s a carburetor."},{"Start":"05:20.210 ","End":"05:24.430","Text":"Let\u0027s see what happens with our heart."},{"Start":"05:24.430 ","End":"05:29.210","Text":"Here we have our heart and we can see that we have veins and arteries"},{"Start":"05:29.210 ","End":"05:33.785","Text":"coming in and out of our heart that start off when we\u0027re close to the heart,"},{"Start":"05:33.785 ","End":"05:40.350","Text":"quite wide, so a cross-sectional area that could be similar to a finger let\u0027s say."},{"Start":"05:40.350 ","End":"05:42.130","Text":"As we move away from the heart,"},{"Start":"05:42.130 ","End":"05:45.035","Text":"we eventually get to capillaries at the end,"},{"Start":"05:45.035 ","End":"05:47.615","Text":"which are extremely narrow."},{"Start":"05:47.615 ","End":"05:51.230","Text":"We can see that that is very similar to the carburetor how we start off"},{"Start":"05:51.230 ","End":"05:55.460","Text":"with a wider section and end up with this smaller,"},{"Start":"05:55.460 ","End":"05:57.260","Text":"narrower section over here,"},{"Start":"05:57.260 ","End":"06:02.085","Text":"where here is analogous to our capillaries."},{"Start":"06:02.085 ","End":"06:07.610","Text":"What we can see is that the pressure differences between the pressure over here close to"},{"Start":"06:07.610 ","End":"06:10.535","Text":"the height and the pressure over here in the capillaries"},{"Start":"06:10.535 ","End":"06:14.710","Text":"far away from the heart can be very, very large."},{"Start":"06:14.710 ","End":"06:18.350","Text":"Aside from just knowing that little fact,"},{"Start":"06:18.350 ","End":"06:22.745","Text":"when people say that you shouldn\u0027t be eating food items with a lot of cholesterol,"},{"Start":"06:22.745 ","End":"06:25.100","Text":"this is the exact reason because"},{"Start":"06:25.100 ","End":"06:28.765","Text":"the correct cholesterol affects the volumetric flow rate."},{"Start":"06:28.765 ","End":"06:31.895","Text":"What happens is that somewhere in the vein or the artery,"},{"Start":"06:31.895 ","End":"06:35.360","Text":"the cholesterol will build up along the walls."},{"Start":"06:35.360 ","End":"06:41.015","Text":"What we\u0027ll get is a vein or artery that looks something like this with a wider section."},{"Start":"06:41.015 ","End":"06:43.700","Text":"Then here the cholesterol will build up along"},{"Start":"06:43.700 ","End":"06:47.690","Text":"the walls and so we\u0027ll get a narrower section in the vein,"},{"Start":"06:47.690 ","End":"06:49.775","Text":"which is what we don\u0027t want."},{"Start":"06:49.775 ","End":"06:52.730","Text":"Then that will create a difference in"},{"Start":"06:52.730 ","End":"06:59.270","Text":"pressure and some vacuum in the middle of the vein or the artery,"},{"Start":"06:59.270 ","End":"07:01.910","Text":"which is something which isn\u0027t very healthy."},{"Start":"07:01.910 ","End":"07:05.450","Text":"That\u0027s why there\u0027s a very strong relationship"},{"Start":"07:05.450 ","End":"07:10.820","Text":"between cholesterol buildup in the veins and arteries and heart problems."},{"Start":"07:10.820 ","End":"07:13.550","Text":"Because as soon as there is this cholesterol buildup,"},{"Start":"07:13.550 ","End":"07:16.470","Text":"the heart has to work much"},{"Start":"07:16.470 ","End":"07:20.810","Text":"harder to get the same amount of blood pumped through the system,"},{"Start":"07:20.810 ","End":"07:22.550","Text":"pumped through this area."},{"Start":"07:22.550 ","End":"07:25.960","Text":"That\u0027s what leads to heart problems."},{"Start":"07:25.960 ","End":"07:31.520","Text":"These were 2 examples from everyday life that deal with volumetric flow rate."},{"Start":"07:31.520 ","End":"07:34.550","Text":"They\u0027re both very important and very useful."},{"Start":"07:34.550 ","End":"07:37.590","Text":"That\u0027s the end of this lesson."}],"ID":12370},{"Watched":false,"Name":"Exercise 3","Duration":"18m 35s","ChapterTopicVideoID":11941,"CourseChapterTopicPlaylistID":84745,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.840","Text":"Hello. Below, we have a reservoir"},{"Start":"00:03.840 ","End":"00:07.560","Text":"of water pipe and another pipe for measuring the pressure."},{"Start":"00:07.560 ","End":"00:13.155","Text":"We\u0027re being asked to find the pressures at the different points of the system."},{"Start":"00:13.155 ","End":"00:17.325","Text":"Soon we\u0027ll speak about this pipe for measuring the pressure,"},{"Start":"00:17.325 ","End":"00:18.405","Text":"but then in the meantime,"},{"Start":"00:18.405 ","End":"00:22.440","Text":"let\u0027s just look at the reservoir and this water pipe."},{"Start":"00:22.440 ","End":"00:29.955","Text":"Here we\u0027re assuming that the reservoir is very large in respect to this water pipe."},{"Start":"00:29.955 ","End":"00:33.065","Text":"Then we can say that the water over here,"},{"Start":"00:33.065 ","End":"00:35.645","Text":"the water level is stationary."},{"Start":"00:35.645 ","End":"00:39.290","Text":"That\u0027s the first thing that we have to consider."},{"Start":"00:39.290 ","End":"00:41.435","Text":"Then aside from that,"},{"Start":"00:41.435 ","End":"00:47.120","Text":"we know that the water at the top over here is under"},{"Start":"00:47.120 ","End":"00:51.050","Text":"atmospheric pressure and the water that"},{"Start":"00:51.050 ","End":"00:57.350","Text":"exits this water pipe over here is also under atmospheric pressure."},{"Start":"00:57.350 ","End":"01:01.160","Text":"Now another thing that we have to know is"},{"Start":"01:01.160 ","End":"01:04.805","Text":"that we\u0027re not losing energy to the environment."},{"Start":"01:04.805 ","End":"01:12.180","Text":"Which means that every single water molecule in the system has the exact same energy."},{"Start":"01:13.040 ","End":"01:17.210","Text":"If we know that the energy at every single point is the same,"},{"Start":"01:17.210 ","End":"01:20.120","Text":"then that means that the energy at point number 1"},{"Start":"01:20.120 ","End":"01:22.970","Text":"over here on the surface of the water and"},{"Start":"01:22.970 ","End":"01:25.760","Text":"the energy at point number 4 over here where the water"},{"Start":"01:25.760 ","End":"01:29.775","Text":"is flowing out of the pipe is going to be the same."},{"Start":"01:29.775 ","End":"01:35.145","Text":"We can say that Epsilon 1 is equal to Epsilon 4."},{"Start":"01:35.145 ","End":"01:39.550","Text":"Now let\u0027s see what Epsilon 1 is equal to."},{"Start":"01:42.380 ","End":"01:48.815","Text":"We already said that the reservoir is very big relative to the aperture of the pipe,"},{"Start":"01:48.815 ","End":"01:50.720","Text":"which means that its velocity is 0,"},{"Start":"01:50.720 ","End":"01:52.730","Text":"so it has no kinetic energy."},{"Start":"01:52.730 ","End":"01:54.650","Text":"However, it does have height,"},{"Start":"01:54.650 ","End":"01:56.375","Text":"so it has potential energy."},{"Start":"01:56.375 ","End":"01:59.765","Text":"That\u0027s going to be equal to Rho g and its height,"},{"Start":"01:59.765 ","End":"02:03.500","Text":"which as we can see here, is h_5."},{"Start":"02:03.500 ","End":"02:06.020","Text":"Then we\u0027re going to have our elastic energy,"},{"Start":"02:06.020 ","End":"02:07.625","Text":"which comes from the pressure,"},{"Start":"02:07.625 ","End":"02:09.950","Text":"which we said over here at the top,"},{"Start":"02:09.950 ","End":"02:14.110","Text":"is simply atmospheric pressure."},{"Start":"02:14.450 ","End":"02:19.035","Text":"That\u0027s Epsilon 1 and this is equal to Epsilon 4."},{"Start":"02:19.035 ","End":"02:22.500","Text":"Now let\u0027s see what Epsilon 4 is equal to."},{"Start":"02:22.500 ","End":"02:26.935","Text":"That\u0027s over here, the water coming out of the aperture."},{"Start":"02:26.935 ","End":"02:31.730","Text":"First of all we know that we have kinetic energy over here,"},{"Start":"02:31.730 ","End":"02:33.830","Text":"because the water is exiting the aperture,"},{"Start":"02:33.830 ","End":"02:35.795","Text":"so it has some velocity."},{"Start":"02:35.795 ","End":"02:40.970","Text":"That\u0027s going to be 1/2 Rho velocity squared."},{"Start":"02:40.970 ","End":"02:48.605","Text":"Then we know that it has no potential energy because its height here is equal to 0."},{"Start":"02:48.605 ","End":"02:52.835","Text":"But we know it has elastic energy which is due to the pressure."},{"Start":"02:52.835 ","End":"02:57.450","Text":"Here again, the water is subject to atmospheric pressure."},{"Start":"02:59.630 ","End":"03:04.820","Text":"Now if we minus our atmospheric pressure from"},{"Start":"03:04.820 ","End":"03:10.335","Text":"both sides and then we can isolate out our velocity."},{"Start":"03:10.335 ","End":"03:17.685","Text":"We\u0027ll get that our velocity is equal to the square root of 2g and the height,"},{"Start":"03:17.685 ","End":"03:19.680","Text":"which here is h_5."},{"Start":"03:19.680 ","End":"03:23.810","Text":"Now, this we got both by doing the algebra over here."},{"Start":"03:23.810 ","End":"03:26.180","Text":"But we could have also said this automatically."},{"Start":"03:26.180 ","End":"03:30.650","Text":"We\u0027ve seen in previous questions that the velocity of the water exiting"},{"Start":"03:30.650 ","End":"03:35.885","Text":"an aperture is always equal to the square root of 2 multiplied by g,"},{"Start":"03:35.885 ","End":"03:40.915","Text":"multiplied by the height that the water is falling from originally."},{"Start":"03:40.915 ","End":"03:46.000","Text":"The top, the uppermost height is at this height h_5."},{"Start":"03:46.000 ","End":"03:49.410","Text":"That\u0027s why we have an h_5 over here."},{"Start":"03:49.410 ","End":"03:56.340","Text":"Now let\u0027s see what our energy is at Point 2 over here, E_2."},{"Start":"03:57.230 ","End":"04:02.090","Text":"First of all, we\u0027re using the idea of energy conservation."},{"Start":"04:02.090 ","End":"04:05.435","Text":"We know that the energy at Point 2 over here at"},{"Start":"04:05.435 ","End":"04:10.165","Text":"this intersection is also going to be the same as the energy at Point 1."},{"Start":"04:10.165 ","End":"04:14.490","Text":"We can again say that Epsilon 1 is equal to Epsilon 2."},{"Start":"04:14.490 ","End":"04:17.390","Text":"Now let\u0027s see what they\u0027re equal to."},{"Start":"04:17.390 ","End":"04:21.245","Text":"Epsilon 1 is equal to,"},{"Start":"04:21.245 ","End":"04:23.190","Text":"as we already calculated,"},{"Start":"04:23.190 ","End":"04:29.010","Text":"Rho gh_ 5 plus the atmospheric pressure."},{"Start":"04:29.010 ","End":"04:32.210","Text":"This is equal to the energy at Point 2."},{"Start":"04:32.210 ","End":"04:34.970","Text":"Now let\u0027s see what the energy at Point 2 is."},{"Start":"04:34.970 ","End":"04:42.095","Text":"First of all, we also have kinetic energy over here."},{"Start":"04:42.095 ","End":"04:48.303","Text":"If we know that the velocity of the water flowing out of here is equal to this,"},{"Start":"04:48.303 ","End":"04:53.180","Text":"we know that therefore the velocity here has to be exactly the same."},{"Start":"04:53.180 ","End":"04:56.060","Text":"We saw this at 1 of the beginning lessons in"},{"Start":"04:56.060 ","End":"04:59.585","Text":"this chapter because the velocity has to be equal."},{"Start":"04:59.585 ","End":"05:02.900","Text":"Because if it\u0027s slower than this velocity,"},{"Start":"05:02.900 ","End":"05:04.820","Text":"then there\u0027ll be some vacuum here,"},{"Start":"05:04.820 ","End":"05:08.108","Text":"which we know isn\u0027t happening and if it\u0027s faster,"},{"Start":"05:08.108 ","End":"05:10.790","Text":"then it\u0027s going to cause some added pressure,"},{"Start":"05:10.790 ","End":"05:13.280","Text":"which we also know isn\u0027t the case."},{"Start":"05:13.280 ","End":"05:17.090","Text":"We know that the velocity here is the exact same as the velocity here."},{"Start":"05:17.090 ","End":"05:21.575","Text":"We can write this as equal to 1/2 Rho V^2,"},{"Start":"05:21.575 ","End":"05:26.365","Text":"where we\u0027ll just substitute in this for V. Then"},{"Start":"05:26.365 ","End":"05:31.790","Text":"we can also see that there\u0027s potential energy because there\u0027s some height over here."},{"Start":"05:31.790 ","End":"05:39.930","Text":"That is going to be equal to Rho g and then the height over here, so that\u0027s h_6."},{"Start":"05:40.040 ","End":"05:42.620","Text":"We have our elastic energy,"},{"Start":"05:42.620 ","End":"05:45.680","Text":"of course, which is simply the pressure over here."},{"Start":"05:45.680 ","End":"05:48.590","Text":"What is the pressure over here?"},{"Start":"05:48.590 ","End":"05:50.270","Text":"This is what we don\u0027t know."},{"Start":"05:50.270 ","End":"05:54.030","Text":"Add in plus P_2."},{"Start":"05:54.030 ","End":"05:58.590","Text":"Now we want to solve to find out what P_2 is equal to."},{"Start":"05:58.880 ","End":"06:03.665","Text":"Now, all we have to do is isolate out our P_2."},{"Start":"06:03.665 ","End":"06:05.930","Text":"This is very easy."},{"Start":"06:05.930 ","End":"06:08.854","Text":"Now, this is going to be Rho g,"},{"Start":"06:08.854 ","End":"06:18.890","Text":"and then we have h_5 minus h_6 and then we have plus our pressure from the atmosphere,"},{"Start":"06:18.890 ","End":"06:26.640","Text":"which is always acting minus 1/2 Rho V^2,"},{"Start":"06:26.640 ","End":"06:32.415","Text":"where this is our V. Now let\u0027s go over this answer."},{"Start":"06:32.415 ","End":"06:36.470","Text":"First of all, the pressure at Point 2 is always going to"},{"Start":"06:36.470 ","End":"06:40.400","Text":"include this component of pressure from the atmosphere."},{"Start":"06:40.400 ","End":"06:42.725","Text":"We\u0027re always subject to"},{"Start":"06:42.725 ","End":"06:47.590","Text":"at least atmospheric pressure unless we\u0027re located in some vacuum."},{"Start":"06:47.590 ","End":"06:50.480","Text":"This is in most questions is going to be effective."},{"Start":"06:50.480 ","End":"06:58.025","Text":"Then over here, we\u0027re subject to the pressure due to the height that we\u0027re located at."},{"Start":"06:58.025 ","End":"07:01.520","Text":"We can see that the difference in height"},{"Start":"07:01.520 ","End":"07:06.120","Text":"between our 2 levels of water is really h_5 minus h_6."},{"Start":"07:06.120 ","End":"07:09.095","Text":"It\u0027s this height over here."},{"Start":"07:09.095 ","End":"07:12.305","Text":"We have the pressure from the amount of"},{"Start":"07:12.305 ","End":"07:16.040","Text":"water at a certain height which is resting on this point,"},{"Start":"07:16.040 ","End":"07:17.675","Text":"which is exactly that,"},{"Start":"07:17.675 ","End":"07:21.290","Text":"then that would simply be the pressure at"},{"Start":"07:21.290 ","End":"07:25.370","Text":"point number 2 if there was a cap over here at the aperture."},{"Start":"07:25.370 ","End":"07:28.010","Text":"However, there isn\u0027t a cap at the aperture."},{"Start":"07:28.010 ","End":"07:31.175","Text":"We know that water is released from here."},{"Start":"07:31.175 ","End":"07:33.365","Text":"Water is flowing out through here."},{"Start":"07:33.365 ","End":"07:35.090","Text":"If the water is released from here,"},{"Start":"07:35.090 ","End":"07:39.880","Text":"that means that there\u0027s some pressure being released over here as well."},{"Start":"07:39.880 ","End":"07:42.350","Text":"That\u0027s why we have a minus over here because we\u0027re"},{"Start":"07:42.350 ","End":"07:46.115","Text":"reducing the pressure because some water can flow through."},{"Start":"07:46.115 ","End":"07:49.010","Text":"This is what we\u0027re taking out."},{"Start":"07:49.010 ","End":"07:54.060","Text":"It\u0027s the kinetic energy of the water that\u0027s going past here."},{"Start":"07:55.190 ","End":"07:59.110","Text":"Now we found the pressure at point number 2 over here."},{"Start":"07:59.110 ","End":"08:05.270","Text":"Now what we want to do is we want to find this height h_3."},{"Start":"08:05.360 ","End":"08:09.010","Text":"Let\u0027s just imagine we can switch these numbers."},{"Start":"08:09.010 ","End":"08:10.390","Text":"It\u0027s a bit confusing."},{"Start":"08:10.390 ","End":"08:15.830","Text":"We found the pressure over here at Point 2 and here we have a pressure P_1."},{"Start":"08:16.370 ","End":"08:23.400","Text":"Now this section of the diagram is actually a section of a much larger diagram,"},{"Start":"08:23.400 ","End":"08:24.780","Text":"and that\u0027s why these numbers,"},{"Start":"08:24.780 ","End":"08:29.685","Text":"we have 1,2,6,8 over here."},{"Start":"08:29.685 ","End":"08:31.800","Text":"You don\u0027t have to know it now,"},{"Start":"08:31.800 ","End":"08:36.120","Text":"but later we\u0027re going to understand the larger version of this diagram."},{"Start":"08:36.120 ","End":"08:40.330","Text":"But right now we\u0027re looking just one small section."},{"Start":"08:41.090 ","End":"08:45.615","Text":"Now what we\u0027re going to do is we\u0027re looking at this little intersection here,"},{"Start":"08:45.615 ","End":"08:48.075","Text":"and here we have a zoom in."},{"Start":"08:48.075 ","End":"08:52.800","Text":"This section corresponds to this and this bit going up,"},{"Start":"08:52.800 ","End":"08:57.960","Text":"which is the pipe for measuring pressure is this section over here."},{"Start":"08:57.960 ","End":"09:02.325","Text":"Now there\u0027s something about this intersection that we need to know."},{"Start":"09:02.325 ","End":"09:09.135","Text":"What happens at this intersection is that there\u0027s no energy conservation over here."},{"Start":"09:09.135 ","End":"09:13.500","Text":"Everything we learned about energy conservation throughout the reservoir and"},{"Start":"09:13.500 ","End":"09:19.335","Text":"shield at this intersection is not correct anymore."},{"Start":"09:19.335 ","End":"09:22.950","Text":"What happens is that we already saw that there\u0027s"},{"Start":"09:22.950 ","End":"09:26.565","Text":"some velocity to the water flowing down this tube."},{"Start":"09:26.565 ","End":"09:29.700","Text":"Can we? It\u0027s equal to this over here."},{"Start":"09:29.700 ","End":"09:34.260","Text":"However, the velocity in the pipe for measuring the pressure,"},{"Start":"09:34.260 ","End":"09:36.915","Text":"over here, is equal to 0."},{"Start":"09:36.915 ","End":"09:39.405","Text":"The velocity over here is equal to 0,"},{"Start":"09:39.405 ","End":"09:42.840","Text":"and we know that the energy at this point,"},{"Start":"09:42.840 ","End":"09:46.830","Text":"or any point in this tube or in the reservoir is not"},{"Start":"09:46.830 ","End":"09:51.360","Text":"equal to the energy in this pipe for measuring pressure."},{"Start":"09:51.360 ","End":"09:53.910","Text":"Those are two important things to note."},{"Start":"09:53.910 ","End":"09:55.725","Text":"The energy here,"},{"Start":"09:55.725 ","End":"09:58.905","Text":"and here is not conserved and is not the same,"},{"Start":"09:58.905 ","End":"10:02.925","Text":"and the velocity here is equal to some value,"},{"Start":"10:02.925 ","End":"10:07.570","Text":"and the velocity in this pipe over here is equal to 0."},{"Start":"10:07.940 ","End":"10:11.775","Text":"What do these two points have in common?"},{"Start":"10:11.775 ","End":"10:14.655","Text":"One, there at the same height,"},{"Start":"10:14.655 ","End":"10:16.560","Text":"they are located at the same height."},{"Start":"10:16.560 ","End":"10:19.005","Text":"This tiny difference between"},{"Start":"10:19.005 ","End":"10:23.535","Text":"this line over here and here is really doesn\u0027t make a difference."},{"Start":"10:23.535 ","End":"10:26.745","Text":"We can say that the two heights are the same,"},{"Start":"10:26.745 ","End":"10:29.790","Text":"and another thing that these two points have in"},{"Start":"10:29.790 ","End":"10:33.135","Text":"common is that the pressures are the same."},{"Start":"10:33.135 ","End":"10:40.170","Text":"P_1=p_2, and the height"},{"Start":"10:40.170 ","End":"10:47.160","Text":"of 0.2 is equal to the height of 0.1."},{"Start":"10:47.160 ","End":"10:50.430","Text":"That is what these two points have in common."},{"Start":"10:50.430 ","End":"10:53.745","Text":"They have different energies and different velocities,"},{"Start":"10:53.745 ","End":"10:57.220","Text":"but the same pressure and the same height."},{"Start":"10:57.920 ","End":"11:04.230","Text":"As we\u0027ve said, the point of this hose or a pipe is to measure the pressure."},{"Start":"11:04.230 ","End":"11:08.730","Text":"As we know, the pressure in this pipe is"},{"Start":"11:08.730 ","End":"11:13.665","Text":"going to be the same as the pressure in this pipe over here."},{"Start":"11:13.665 ","End":"11:15.840","Text":"That\u0027s what we just said."},{"Start":"11:15.840 ","End":"11:23.160","Text":"What we know is that if the pressure p_2 that we just worked out goes higher."},{"Start":"11:23.160 ","End":"11:24.990","Text":"If we have a higher pressure over here,"},{"Start":"11:24.990 ","End":"11:30.930","Text":"that means that the pressure of p_1 will similarly also increase."},{"Start":"11:30.930 ","End":"11:35.145","Text":"Because we\u0027re dealing with thin pipes and we\u0027re losing energy,"},{"Start":"11:35.145 ","End":"11:38.835","Text":"that means that if there is a pressure increase at 0.2,"},{"Start":"11:38.835 ","End":"11:43.095","Text":"there\u0027ll be a pressure increase in the pipe for measuring pressure,"},{"Start":"11:43.095 ","End":"11:49.395","Text":"and that means that the height of the water is going to increase as well."},{"Start":"11:49.395 ","End":"11:51.780","Text":"That means that this height over here,"},{"Start":"11:51.780 ","End":"11:58.450","Text":"h_3 with a tilde on top is going to increase with an increase of pressure."},{"Start":"11:59.390 ","End":"12:05.925","Text":"We can see that our height 3 is dynamic and is dependent on the pressure."},{"Start":"12:05.925 ","End":"12:08.925","Text":"What we want to do is we want to measure this height."},{"Start":"12:08.925 ","End":"12:12.089","Text":"Here we have 0.5,"},{"Start":"12:12.089 ","End":"12:14.985","Text":"which is the point at the top,"},{"Start":"12:14.985 ","End":"12:18.660","Text":"and then here we have 0.8 over here,"},{"Start":"12:18.660 ","End":"12:24.130","Text":"which is the point lower down in the pipe for measuring pressure."},{"Start":"12:25.820 ","End":"12:29.940","Text":"Because they\u0027re connected to the same pipe,"},{"Start":"12:29.940 ","End":"12:36.450","Text":"we know that the energy at 0.5 will be equal to the energy at 0.8."},{"Start":"12:36.450 ","End":"12:40.470","Text":"Here within this pipe for measuring the pressure,"},{"Start":"12:40.470 ","End":"12:43.275","Text":"we have energy conservation,"},{"Start":"12:43.275 ","End":"12:48.150","Text":"and within the reservoir and this pipe that the water is flowing through,"},{"Start":"12:48.150 ","End":"12:50.025","Text":"we also have energy conservation."},{"Start":"12:50.025 ","End":"12:52.485","Text":"But because they\u0027re both different pipes,"},{"Start":"12:52.485 ","End":"12:56.085","Text":"the energy isn\u0027t conserved between the two."},{"Start":"12:56.085 ","End":"13:01.800","Text":"We have energy conservation between 0.5 and 0.8."},{"Start":"13:01.800 ","End":"13:05.505","Text":"We can write that Epsilon 5 is equal to Epsilon 8."},{"Start":"13:05.505 ","End":"13:09.000","Text":"Now let\u0027s see what the energy at 0.5 is,"},{"Start":"13:09.000 ","End":"13:12.540","Text":"which is at this top point over here."},{"Start":"13:12.540 ","End":"13:17.125","Text":"First of all, we have 0 kinetic energy."},{"Start":"13:17.125 ","End":"13:19.520","Text":"Why do we have 0 kinetic energy?"},{"Start":"13:19.520 ","End":"13:23.975","Text":"We just said that in this pipe over here we have velocity,"},{"Start":"13:23.975 ","End":"13:26.465","Text":"and in this pipe for measuring the pressure,"},{"Start":"13:26.465 ","End":"13:27.965","Text":"we have a 0 velocity,"},{"Start":"13:27.965 ","End":"13:29.980","Text":"which means 0 kinetic energy."},{"Start":"13:29.980 ","End":"13:31.845","Text":"What do we have?"},{"Start":"13:31.845 ","End":"13:35.080","Text":"First of all, it has height."},{"Start":"13:35.360 ","End":"13:37.695","Text":"Let\u0027s take a look."},{"Start":"13:37.695 ","End":"13:39.825","Text":"It\u0027s going to have potential energy."},{"Start":"13:39.825 ","End":"13:44.715","Text":"That\u0027s going to be equal to Rho g multiplied by the height than it\u0027s at."},{"Start":"13:44.715 ","End":"13:48.195","Text":"At what height does it? It\u0027s at a height of h6."},{"Start":"13:48.195 ","End":"13:56.430","Text":"That\u0027s h6, plus this height h3."},{"Start":"13:56.430 ","End":"14:00.420","Text":"The height h3 and might be at the exact same height as h6,"},{"Start":"14:00.420 ","End":"14:03.105","Text":"in which case h3 is equal to 0,"},{"Start":"14:03.105 ","End":"14:04.560","Text":"and then we\u0027re located here."},{"Start":"14:04.560 ","End":"14:07.725","Text":"Or because we said it\u0027s something that\u0027s more dynamic,"},{"Start":"14:07.725 ","End":"14:10.950","Text":"it could be higher up like we can see in the diagram,"},{"Start":"14:10.950 ","End":"14:13.095","Text":"and then we just added on."},{"Start":"14:13.095 ","End":"14:16.950","Text":"Plus h3, which is the unknown that we\u0027re trying to find."},{"Start":"14:16.950 ","End":"14:18.705","Text":"This is our high potential energy,"},{"Start":"14:18.705 ","End":"14:22.590","Text":"and then we know that we have to add in our elastic energy,"},{"Start":"14:22.590 ","End":"14:30.330","Text":"which is simply due to the pressure that this 0.5 over here at the top is subject to."},{"Start":"14:30.330 ","End":"14:33.900","Text":"We know that 0.5 is at the top of this water level,"},{"Start":"14:33.900 ","End":"14:38.700","Text":"which means that it\u0027s only subject to the atmospheric pressure."},{"Start":"14:38.700 ","End":"14:41.415","Text":"Just like this point 1,"},{"Start":"14:41.415 ","End":"14:45.615","Text":"at the water level of the reservoir is subject only to atmospheric pressure."},{"Start":"14:45.615 ","End":"14:48.885","Text":"This is also the water level just at a different area,"},{"Start":"14:48.885 ","End":"14:53.025","Text":"but it\u0027s also just subject to the atmosphere."},{"Start":"14:53.025 ","End":"14:58.845","Text":"Then this is equal to the energy at 0.8 over here,"},{"Start":"14:58.845 ","End":"15:03.630","Text":"which is at the bottom of the pipe for measuring the pressure,"},{"Start":"15:03.630 ","End":"15:07.185","Text":"it\u0027s at this intersection over here."},{"Start":"15:07.185 ","End":"15:10.710","Text":"That\u0027s 0.8. We know that it\u0027s equal to the energy over here."},{"Start":"15:10.710 ","End":"15:13.530","Text":"Let\u0027s see what the energy is over here."},{"Start":"15:13.530 ","End":"15:18.915","Text":"First of all, we know that we\u0027re going to have potential energy due to the height."},{"Start":"15:18.915 ","End":"15:22.455","Text":"That\u0027s equal to Rho g multiplied by the height,"},{"Start":"15:22.455 ","End":"15:25.125","Text":"which here is just h_6."},{"Start":"15:25.125 ","End":"15:27.105","Text":"It\u0027s not changing."},{"Start":"15:27.105 ","End":"15:30.450","Text":"It\u0027s at this intersection over here, h_6."},{"Start":"15:30.450 ","End":"15:35.865","Text":"Then we know that again at this bottom point over here,"},{"Start":"15:35.865 ","End":"15:38.355","Text":"this is 0.8 over here."},{"Start":"15:38.355 ","End":"15:43.140","Text":"We know that there\u0027s 0 velocity over here"},{"Start":"15:43.140 ","End":"15:48.510","Text":"as well because we\u0027re still located inside the pipe for measuring pressure."},{"Start":"15:48.510 ","End":"15:50.805","Text":"That\u0027s what we\u0027re looking at over here,"},{"Start":"15:50.805 ","End":"15:55.245","Text":"and we know that the velocity inside the pipe for measuring pressure is equal to 0."},{"Start":"15:55.245 ","End":"15:57.645","Text":"That means 0 kinetic energy,"},{"Start":"15:57.645 ","End":"16:02.025","Text":"and then we have to add on our pressure."},{"Start":"16:02.025 ","End":"16:04.500","Text":"We know that this is the p_1,"},{"Start":"16:04.500 ","End":"16:06.690","Text":"which we said is just equal to p_2."},{"Start":"16:06.690 ","End":"16:11.250","Text":"Because we said that the pressure in this pipe inside,"},{"Start":"16:11.250 ","End":"16:13.200","Text":"not on the surface of the pipe over here,"},{"Start":"16:13.200 ","End":"16:15.930","Text":"but rather inside where the liquid is,"},{"Start":"16:15.930 ","End":"16:20.565","Text":"is equal to the pressure over here for the water pipe."},{"Start":"16:20.565 ","End":"16:23.325","Text":"That\u0027s going to be equal to p_2,"},{"Start":"16:23.325 ","End":"16:27.250","Text":"which is what we calculated in the previous section."},{"Start":"16:28.280 ","End":"16:33.105","Text":"Our unknown over here is our height here, h_3."},{"Start":"16:33.105 ","End":"16:34.845","Text":"We want to isolate that out."},{"Start":"16:34.845 ","End":"16:39.770","Text":"First of all, we can see that on both sides of the equation we have Rho g(h_6)."},{"Start":"16:39.770 ","End":"16:43.980","Text":"We can subtract it from both sides of the equation,"},{"Start":"16:44.150 ","End":"16:53.885","Text":"and now we have Rho g h_3 plus our atmospheric pressure,"},{"Start":"16:53.885 ","End":"16:58.105","Text":"which is equal to our p_2."},{"Start":"16:58.105 ","End":"17:00.000","Text":"Now what\u0027s our p_2?"},{"Start":"17:00.000 ","End":"17:01.200","Text":"We saw it over here,"},{"Start":"17:01.200 ","End":"17:02.490","Text":"let\u0027s substitute that in."},{"Start":"17:02.490 ","End":"17:08.840","Text":"It\u0027s equal to Rho g( h5-h_6)"},{"Start":"17:08.840 ","End":"17:17.450","Text":"plus our atmospheric pressure minus 1/2 Rho velocity squared,"},{"Start":"17:17.450 ","End":"17:19.070","Text":"where this is our velocity."},{"Start":"17:19.070 ","End":"17:24.190","Text":"In both sides of the equation we have our atmospheric pressure so we can cross them off,"},{"Start":"17:24.190 ","End":"17:27.685","Text":"and now let\u0027s scroll down a bit to give us some more space."},{"Start":"17:27.685 ","End":"17:32.150","Text":"Now all we have to do is we have to isolate our h_3,"},{"Start":"17:32.150 ","End":"17:34.040","Text":"because this is the unknown that we\u0027re trying to find."},{"Start":"17:34.040 ","End":"17:37.190","Text":"What we\u0027re going to do is we\u0027re just going to divide both sides by"},{"Start":"17:37.190 ","End":"17:41.690","Text":"Rho g. Then Rho g multiplied"},{"Start":"17:41.690 ","End":"17:49.590","Text":"by here in brackets divided by rho g is simply going to be equal to h_5 minus h_6,"},{"Start":"17:49.590 ","End":"17:53.340","Text":"and then we have minus 1/2 Rho V^2,"},{"Start":"17:53.340 ","End":"17:58.605","Text":"divided by Rho g. We\u0027ll just be left with -1/2."},{"Start":"17:58.605 ","End":"18:00.270","Text":"The Rhos will cancel out,"},{"Start":"18:00.270 ","End":"18:03.770","Text":"and then we\u0027ll have v^2 divided by"},{"Start":"18:03.770 ","End":"18:12.875","Text":"g. This will be the height of the water level in the tube for measuring pressure,"},{"Start":"18:12.875 ","End":"18:16.970","Text":"and that\u0027s how we can also use this detail in"},{"Start":"18:16.970 ","End":"18:21.335","Text":"order to know the pressure in the water pipe."},{"Start":"18:21.335 ","End":"18:24.350","Text":"If we know the height of this water,"},{"Start":"18:24.350 ","End":"18:27.250","Text":"then we can know the pressure in this tube over here,"},{"Start":"18:27.250 ","End":"18:32.415","Text":"and we can know the pressure in the water pipe over here."},{"Start":"18:32.415 ","End":"18:35.590","Text":"That\u0027s the end of this question."}],"ID":12371},{"Watched":false,"Name":"Exercise 4","Duration":"18m 10s","ChapterTopicVideoID":11942,"CourseChapterTopicPlaylistID":84745,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.830","Text":"Hello. In this question,"},{"Start":"00:01.830 ","End":"00:05.520","Text":"we\u0027re being told that water is flowing out of a reservoir."},{"Start":"00:05.520 ","End":"00:07.620","Text":"We\u0027re being told that at point 3,"},{"Start":"00:07.620 ","End":"00:13.710","Text":"the cross-sectional area is 1/3 the cross-sectional area of the rest of the tube."},{"Start":"00:13.710 ","End":"00:18.530","Text":"We\u0027re being asked to find the pressure at each of the points 1-5,"},{"Start":"00:18.530 ","End":"00:19.680","Text":"so 1, 2,"},{"Start":"00:19.680 ","End":"00:22.605","Text":"3, 4, and 5,"},{"Start":"00:22.605 ","End":"00:27.270","Text":"and to find the height of the water above point 3."},{"Start":"00:27.270 ","End":"00:36.120","Text":"As we know, we\u0027re going to have energy conservation in this question and we know that"},{"Start":"00:36.120 ","End":"00:39.120","Text":"we\u0027re going to assume that the reservoir is"},{"Start":"00:39.120 ","End":"00:44.955","Text":"large relative to the cross-sectional area of the tube."},{"Start":"00:44.955 ","End":"00:51.480","Text":"That way we can assume that the surface of the water is stationary over here."},{"Start":"00:51.480 ","End":"00:54.540","Text":"First of all, at point 1,"},{"Start":"00:54.540 ","End":"00:56.520","Text":"that\u0027s at the surface of the water,"},{"Start":"00:56.520 ","End":"01:03.490","Text":"we know what the pressure is going to be over there and that is atmospheric pressure."},{"Start":"01:04.040 ","End":"01:08.655","Text":"That we know and for the same reason,"},{"Start":"01:08.655 ","End":"01:13.095","Text":"the pressure at point number 5 is also going to be atmospheric pressure"},{"Start":"01:13.095 ","End":"01:18.975","Text":"because the water is out of the tube and open to the atmosphere."},{"Start":"01:18.975 ","End":"01:23.730","Text":"We can say this over here."},{"Start":"01:23.730 ","End":"01:26.250","Text":"Now we\u0027re trying to find the pressure in 2,"},{"Start":"01:26.250 ","End":"01:28.540","Text":"3, and 4."},{"Start":"01:28.820 ","End":"01:33.975","Text":"Now, we can look at the pressures at points 2, 3, and 4."},{"Start":"01:33.975 ","End":"01:37.245","Text":"Now we know that 2 and 4,"},{"Start":"01:37.245 ","End":"01:42.780","Text":"both of these points will have the exact same values for"},{"Start":"01:42.780 ","End":"01:45.600","Text":"pressure because we can see that they"},{"Start":"01:45.600 ","End":"01:48.540","Text":"are the same height so they have the same pressure,"},{"Start":"01:48.540 ","End":"01:51.090","Text":"the same height and the water flowing through"},{"Start":"01:51.090 ","End":"01:54.660","Text":"these 2 points is also going to have the same velocity."},{"Start":"01:54.660 ","End":"01:59.430","Text":"Then, we have to calculate the pressure at point number 3."},{"Start":"01:59.430 ","End":"02:05.550","Text":"That we\u0027re going to have to do and then through that we\u0027re going to find this height h_3."},{"Start":"02:05.550 ","End":"02:08.400","Text":"Just like in the previous question,"},{"Start":"02:08.400 ","End":"02:13.350","Text":"this pipe over here is for measuring the pressure and"},{"Start":"02:13.350 ","End":"02:18.960","Text":"the water level over here is open to the outside air."},{"Start":"02:18.960 ","End":"02:23.040","Text":"I\u0027ll just extend this tubing so that it\u0027s clear that"},{"Start":"02:23.040 ","End":"02:28.810","Text":"the tube here doesn\u0027t have a cap on it."},{"Start":"02:29.840 ","End":"02:36.945","Text":"Now we know that we\u0027re going to use energy conservation in order to solve this question."},{"Start":"02:36.945 ","End":"02:42.180","Text":"Before we do that, we know that the velocity of the water which is"},{"Start":"02:42.180 ","End":"02:48.540","Text":"exiting here at point number 5 but it\u0027s going to be the same velocity at 4 and 2,"},{"Start":"02:48.540 ","End":"02:52.875","Text":"is simply equal to the velocity of free flow."},{"Start":"02:52.875 ","End":"02:56.490","Text":"We\u0027ve already seen in questions that it\u0027s just equal to"},{"Start":"02:56.490 ","End":"03:00.675","Text":"the square root of 2 multiplied by g,"},{"Start":"03:00.675 ","End":"03:03.705","Text":"multiplied by the total height of the system."},{"Start":"03:03.705 ","End":"03:09.600","Text":"Here the total height of the system is h_1 plus h_2,"},{"Start":"03:09.600 ","End":"03:13.030","Text":"it\u0027s from the bottom all the way to the top."},{"Start":"03:18.590 ","End":"03:22.530","Text":"This velocity is correct for points 2,"},{"Start":"03:22.530 ","End":"03:24.660","Text":"4, and 5."},{"Start":"03:24.660 ","End":"03:29.865","Text":"However, this isn\u0027t the velocity at point 1 or at point 3."},{"Start":"03:29.865 ","End":"03:33.015","Text":"The velocity at point 1 is equal to 0"},{"Start":"03:33.015 ","End":"03:36.450","Text":"because we know that right at the beginning of the lesson we"},{"Start":"03:36.450 ","End":"03:39.300","Text":"said that we have this reservoir which is very"},{"Start":"03:39.300 ","End":"03:43.200","Text":"large in comparison to the cross-sectional area of the tubing."},{"Start":"03:43.200 ","End":"03:48.840","Text":"That means that we can consider this water level to just be stationary, constant,"},{"Start":"03:48.840 ","End":"03:52.035","Text":"which means that there\u0027s no velocity here and"},{"Start":"03:52.035 ","End":"03:56.055","Text":"we know that this isn\u0027t the velocity at point number 3."},{"Start":"03:56.055 ","End":"04:01.650","Text":"This we know due to the continuity equation that we learned,"},{"Start":"04:01.650 ","End":"04:05.670","Text":"where we saw that the velocity is going to be constant so long"},{"Start":"04:05.670 ","End":"04:09.420","Text":"as the cross-sectional area of the tube remains constant."},{"Start":"04:09.420 ","End":"04:11.460","Text":"But if the cross-sectional area changes,"},{"Start":"04:11.460 ","End":"04:14.280","Text":"if it increases or in our case decreases,"},{"Start":"04:14.280 ","End":"04:19.930","Text":"then the velocity will decrease or increase."},{"Start":"04:21.670 ","End":"04:25.865","Text":"Now let\u0027s find out what our pressure is at point number 2."},{"Start":"04:25.865 ","End":"04:29.120","Text":"We\u0027re going to use the fact that we know that the energy at"},{"Start":"04:29.120 ","End":"04:32.915","Text":"point number 1 is equal to the energy at point number 2."},{"Start":"04:32.915 ","End":"04:38.190","Text":"Let\u0027s write that out and now let\u0027s do the calculation."},{"Start":"04:38.190 ","End":"04:40.275","Text":"The energy at point number 1,"},{"Start":"04:40.275 ","End":"04:45.150","Text":"we know that the water level here is static which means that there\u0027s no velocity,"},{"Start":"04:45.150 ","End":"04:47.770","Text":"which means that there\u0027s no kinetic energy."},{"Start":"04:47.770 ","End":"04:50.105","Text":"We can see, however,"},{"Start":"04:50.105 ","End":"04:54.365","Text":"that our reservoir is at a height so we have potential energy."},{"Start":"04:54.365 ","End":"04:59.315","Text":"That is going to be equal to Rho g multiplied by the height,"},{"Start":"04:59.315 ","End":"05:06.225","Text":"where the height is equal to h_1 plus h_2, that\u0027s the total height."},{"Start":"05:06.225 ","End":"05:09.030","Text":"Then we have elastic energy, of course,"},{"Start":"05:09.030 ","End":"05:11.640","Text":"which comes from the pressure which we know that"},{"Start":"05:11.640 ","End":"05:15.135","Text":"the pressure at this point is atmospheric pressure."},{"Start":"05:15.135 ","End":"05:17.265","Text":"We can write that down."},{"Start":"05:17.265 ","End":"05:21.310","Text":"Then this is equal to the energy at point number 2."},{"Start":"05:21.890 ","End":"05:27.315","Text":"First of all, we know that the water inside the tube has a velocity."},{"Start":"05:27.315 ","End":"05:29.730","Text":"Which means that it has kinetic energy."},{"Start":"05:29.730 ","End":"05:34.905","Text":"Our kinetic energy is going to be equal to 1/2 Rho v^2,"},{"Start":"05:34.905 ","End":"05:39.765","Text":"where this is the velocity so we can just substitute that in."},{"Start":"05:39.765 ","End":"05:42.660","Text":"Then we also have height over here,"},{"Start":"05:42.660 ","End":"05:44.445","Text":"so there\u0027s potential energy."},{"Start":"05:44.445 ","End":"05:49.110","Text":"The height over here is equal to h_2,"},{"Start":"05:49.110 ","End":"05:51.705","Text":"so this is Rho g h_2."},{"Start":"05:51.705 ","End":"05:54.060","Text":"Then, we of course,"},{"Start":"05:54.060 ","End":"05:58.170","Text":"also know that we have elastic energy which is simply the pressure"},{"Start":"05:58.170 ","End":"06:02.550","Text":"at point number 2 which is what we\u0027re trying to find out."},{"Start":"06:02.550 ","End":"06:05.430","Text":"This is what we need to isolate out."},{"Start":"06:05.430 ","End":"06:09.660","Text":"First of all, we can see that on both sides of the equation,"},{"Start":"06:09.660 ","End":"06:12.120","Text":"we have Rho g h_2,"},{"Start":"06:12.120 ","End":"06:15.870","Text":"so we can subtract them from both sides of the equation."},{"Start":"06:15.870 ","End":"06:18.750","Text":"Now, all we have to do is isolate out"},{"Start":"06:18.750 ","End":"06:22.935","Text":"our pressure at point number 2 which is unknown that we\u0027re trying to find."},{"Start":"06:22.935 ","End":"06:31.125","Text":"We\u0027ll see that our pressure point number 2 is equal to Rho g h_1 plus"},{"Start":"06:31.125 ","End":"06:39.835","Text":"our atmospheric pressure minus our kinetic energy,"},{"Start":"06:39.835 ","End":"06:42.735","Text":"where the velocity is this velocity over here,"},{"Start":"06:42.735 ","End":"06:45.090","Text":"and this of course makes sense."},{"Start":"06:45.090 ","End":"06:49.650","Text":"We have the pressure here due to the amount"},{"Start":"06:49.650 ","End":"06:54.720","Text":"of water sitting above this point, the potential;"},{"Start":"06:54.720 ","End":"07:01.050","Text":"that\u0027s Rho g multiplied by this height above this point over here which is h_1,"},{"Start":"07:01.050 ","End":"07:05.160","Text":"plus we\u0027re always subject to the atmospheric pressure unless we\u0027re"},{"Start":"07:05.160 ","End":"07:10.920","Text":"located inside of some vacuum which for most of the questions that isn\u0027t the case,"},{"Start":"07:10.920 ","End":"07:12.990","Text":"so that\u0027s our atmospheric pressure."},{"Start":"07:12.990 ","End":"07:15.465","Text":"Then, if we had a cap over here,"},{"Start":"07:15.465 ","End":"07:18.435","Text":"that would simply be the pressure at point number 2."},{"Start":"07:18.435 ","End":"07:22.770","Text":"However, there\u0027s no cap so there\u0027s some releasing of water or releasing of"},{"Start":"07:22.770 ","End":"07:28.030","Text":"pressure and that comes from this over here, the kinetic energy."},{"Start":"07:28.850 ","End":"07:35.760","Text":"Great. Now, we\u0027re going to find the pressure at point number 3 over here."},{"Start":"07:35.760 ","End":"07:38.550","Text":"We know that the pressure at point number 3 is going to be"},{"Start":"07:38.550 ","End":"07:42.480","Text":"different to the pressure at point number 2 because we learned"},{"Start":"07:42.480 ","End":"07:45.270","Text":"about the continuity equation and we saw that"},{"Start":"07:45.270 ","End":"07:49.290","Text":"the pressure changes if the cross-sectional area of the tubing changes."},{"Start":"07:49.290 ","End":"07:53.640","Text":"We\u0027re going to use the idea of energy conservation and we\u0027re going to say"},{"Start":"07:53.640 ","End":"07:58.870","Text":"that the energy at point number 2 is equal to the energy at point number 3."},{"Start":"08:03.230 ","End":"08:08.115","Text":"Now let\u0027s write out the energy at point number 2."},{"Start":"08:08.115 ","End":"08:10.710","Text":"We of course did that over here."},{"Start":"08:10.710 ","End":"08:20.130","Text":"It\u0027s equal to 1/2 Rho v^2 plus Rho g h_2,"},{"Start":"08:20.130 ","End":"08:23.370","Text":"plus our pressure at point number 2,"},{"Start":"08:23.370 ","End":"08:26.265","Text":"so let\u0027s just substitute this in over here."},{"Start":"08:26.265 ","End":"08:27.930","Text":"I\u0027m just going to write it here in short but"},{"Start":"08:27.930 ","End":"08:30.180","Text":"it\u0027s this over here, we\u0027ve already found it."},{"Start":"08:30.180 ","End":"08:34.110","Text":"This is equal to our energy at point number 3."},{"Start":"08:34.110 ","End":"08:37.410","Text":"What energy do we have a point number 3?"},{"Start":"08:37.410 ","End":"08:40.800","Text":"First of all, we have our potential energy because we can see"},{"Start":"08:40.800 ","End":"08:44.940","Text":"that point number 3 is located at a height."},{"Start":"08:44.940 ","End":"08:52.110","Text":"That\u0027s going to be Rho g multiplied by the height over here, so that\u0027s h_2."},{"Start":"08:52.110 ","End":"08:57.030","Text":"Then we\u0027re going to add in our kinetic energy."},{"Start":"08:57.030 ","End":"08:59.910","Text":"We know that if there\u0027s kinetic energy at point"},{"Start":"08:59.910 ","End":"09:02.805","Text":"2 then there\u0027s also kinetic energy at point 3."},{"Start":"09:02.805 ","End":"09:04.800","Text":"Now from the continuity equation,"},{"Start":"09:04.800 ","End":"09:09.360","Text":"we learned that if the cross-sectional area is reduced by"},{"Start":"09:09.360 ","End":"09:14.790","Text":"a certain factor then the velocity is increased by that factor."},{"Start":"09:14.790 ","End":"09:17.040","Text":"Here we\u0027re being told that at point number 3,"},{"Start":"09:17.040 ","End":"09:22.200","Text":"the cross-sectional area is 1/3 the area of the rest of the tube."},{"Start":"09:22.200 ","End":"09:25.410","Text":"That means that the velocity at this point is going to be"},{"Start":"09:25.410 ","End":"09:29.460","Text":"3 times the velocity of the water in the rest of the tube."},{"Start":"09:29.460 ","End":"09:34.223","Text":"That means that our kinetic energy is 1/2 Rho,"},{"Start":"09:34.223 ","End":"09:37.245","Text":"and then instead of v^2,"},{"Start":"09:37.245 ","End":"09:42.840","Text":"we\u0027re going to be multiplying it by (3v)^2 because we\u0027re"},{"Start":"09:42.840 ","End":"09:49.920","Text":"multiplying the velocity factor by 3 because the cross-sectional area is 1/3."},{"Start":"09:49.920 ","End":"09:56.415","Text":"Of course, we also have pressure at point number 3 which is our unknown,"},{"Start":"09:56.415 ","End":"09:58.365","Text":"this is what we\u0027re trying to find out."},{"Start":"09:58.365 ","End":"10:03.700","Text":"Let\u0027s add in our elastic energy p_3."},{"Start":"10:04.960 ","End":"10:08.405","Text":"Now we can see that at both sides of the equation,"},{"Start":"10:08.405 ","End":"10:12.575","Text":"we have the same potential energy Rho g h_2."},{"Start":"10:12.575 ","End":"10:17.870","Text":"We can subtract Rho g h_2 from both sides of the equation."},{"Start":"10:17.870 ","End":"10:20.665","Text":"Then, we are left with"},{"Start":"10:20.665 ","End":"10:29.685","Text":"1/2 Rho v^2 plus our p_2 which we can substitute in later,"},{"Start":"10:29.685 ","End":"10:33.960","Text":"is equal to 1/2 Rho,"},{"Start":"10:33.960 ","End":"10:42.310","Text":"3^2 is 9 multiplied by v^2 plus our p_3."},{"Start":"10:43.910 ","End":"10:48.345","Text":"Now all that\u0027s left to do is to isolate out our p_3."},{"Start":"10:48.345 ","End":"10:50.340","Text":"Which is the pressure at point number 3,"},{"Start":"10:50.340 ","End":"10:52.125","Text":"which is what we\u0027re trying to find out."},{"Start":"10:52.125 ","End":"10:59.970","Text":"This is simply going to be equal to half Rho v^2 minus 4.5 Rho v^2."},{"Start":"10:59.970 ","End":"11:02.623","Text":"The 9 divided by 2 is 4.5,"},{"Start":"11:02.623 ","End":"11:10.620","Text":"so that\u0027s going to be equal to negative 4 Rho v^2."},{"Start":"11:10.620 ","End":"11:15.975","Text":"Then, we have our pressure at 0.2. I didn\u0027t."},{"Start":"11:15.975 ","End":"11:20.604","Text":"Now, if we want to substitute in what our p_2 is equal to,"},{"Start":"11:20.604 ","End":"11:23.260","Text":"to get the full answer."},{"Start":"11:23.330 ","End":"11:30.180","Text":"The final answer will be Rho gh_1 plus our atmospheric pressure."},{"Start":"11:30.180 ","End":"11:33.495","Text":"Then we had negative 1/2 Rho v^2,"},{"Start":"11:33.495 ","End":"11:35.715","Text":"negative 4 Rho v^2."},{"Start":"11:35.715 ","End":"11:41.025","Text":"In total that\u0027s negative 9 over 2 Rho v^2."},{"Start":"11:41.025 ","End":"11:44.295","Text":"This is the pressure at point number 3."},{"Start":"11:44.295 ","End":"11:48.795","Text":"Now, we want to find the pressure at point number 4."},{"Start":"11:48.795 ","End":"11:50.970","Text":"Now we already said that the pressure at point number"},{"Start":"11:50.970 ","End":"11:54.075","Text":"2 is equal to the pressure at point number 4."},{"Start":"11:54.075 ","End":"11:59.865","Text":"Because we have the same velocity of water flowing through these 2 points."},{"Start":"11:59.865 ","End":"12:05.715","Text":"They are located at the exact same heights with the same cross-sectional area."},{"Start":"12:05.715 ","End":"12:07.665","Text":"That means that our pressure,"},{"Start":"12:07.665 ","End":"12:11.530","Text":"0.2 is equal to our pressure at 0.4."},{"Start":"12:12.410 ","End":"12:17.145","Text":"We were meant to find the pressures of each of the points 1-5."},{"Start":"12:17.145 ","End":"12:21.013","Text":"You have the pressure at 0.1,"},{"Start":"12:21.013 ","End":"12:24.945","Text":"0.2, 0.3, 0.4, and 0.5."},{"Start":"12:24.945 ","End":"12:29.670","Text":"Now what we want to do is we want to find the height of the water above 0.3."},{"Start":"12:29.670 ","End":"12:33.610","Text":"I tried to find this height h_3 over here."},{"Start":"12:33.980 ","End":"12:37.380","Text":"What we\u0027re going to do is we\u0027re going to do a little"},{"Start":"12:37.380 ","End":"12:40.935","Text":"zoom-in into this intersection over here."},{"Start":"12:40.935 ","End":"12:47.370","Text":"We saw in the previous video that when we\u0027re dealing with 2 interconnecting hoses,"},{"Start":"12:47.370 ","End":"12:50.115","Text":"k_2 separate hoses like this."},{"Start":"12:50.115 ","End":"12:54.150","Text":"Then we know that there is no energy conservation between"},{"Start":"12:54.150 ","End":"12:59.140","Text":"the energy in this hose and the energy in this hose over here."},{"Start":"13:00.290 ","End":"13:06.371","Text":"Aside from the fact that there\u0027s no energy conservation between these 2 hoses,"},{"Start":"13:06.371 ","End":"13:12.260","Text":"that means a particle here will have a different energy to a water particle here."},{"Start":"13:12.260 ","End":"13:13.910","Text":"We do know, however,"},{"Start":"13:13.910 ","End":"13:18.290","Text":"that the pressure of this 0.3 in this hose that we"},{"Start":"13:18.290 ","End":"13:22.530","Text":"worked out just now is going to be the same as the pressure at"},{"Start":"13:22.530 ","End":"13:26.940","Text":"this 0.3 with the star because"},{"Start":"13:26.940 ","End":"13:31.800","Text":"this 0.3 with a star is right at the connection between the 2 hoses."},{"Start":"13:31.800 ","End":"13:35.280","Text":"Because we\u0027re assuming that this hose and also this hose,"},{"Start":"13:35.280 ","End":"13:40.725","Text":"but that these hoses are small enough or are narrow enough,"},{"Start":"13:40.725 ","End":"13:43.830","Text":"then this slight change in height,"},{"Start":"13:43.830 ","End":"13:46.230","Text":"we don\u0027t have to consider."},{"Start":"13:46.230 ","End":"13:50.760","Text":"We can say that our 0.3 and 0.3 star have"},{"Start":"13:50.760 ","End":"13:55.845","Text":"the exact same height and we know that they have the same pressure."},{"Start":"13:55.845 ","End":"14:00.119","Text":"What we also know is that they don\u0027t share the same energy"},{"Start":"14:00.119 ","End":"14:05.820","Text":"and that if water is flowing through point number 3 with a certain velocity,"},{"Start":"14:05.820 ","End":"14:10.815","Text":"the velocity of the water in this hose which is perpendicular."},{"Start":"14:10.815 ","End":"14:13.485","Text":"This hose, which is for measuring the pressure,"},{"Start":"14:13.485 ","End":"14:18.285","Text":"we know that the velocity of the water in this hose is equal to 0."},{"Start":"14:18.285 ","End":"14:22.600","Text":"This is what we saw in the previous lesson as well."},{"Start":"14:24.050 ","End":"14:29.535","Text":"Let\u0027s just write this out so we know that the pressure at 0.3"},{"Start":"14:29.535 ","End":"14:35.340","Text":"over here is equal to the pressure at 0.3 star over here."},{"Start":"14:35.340 ","End":"14:38.730","Text":"Now we\u0027re going to define this extra point number 6,"},{"Start":"14:38.730 ","End":"14:42.930","Text":"which is exactly at the surface of the water in this tube."},{"Start":"14:42.930 ","End":"14:47.370","Text":"It\u0027s right over here, this is 0.6."},{"Start":"14:47.370 ","End":"14:49.080","Text":"Now what we\u0027re going to do,"},{"Start":"14:49.080 ","End":"14:53.970","Text":"we know that within this tube there is energy conservation so that means that"},{"Start":"14:53.970 ","End":"15:00.420","Text":"the energy at 0.6 is the same as the energy at 0.3 star."},{"Start":"15:00.420 ","End":"15:04.020","Text":"What we\u0027re going to do is we\u0027re going to use our energy, conservation."},{"Start":"15:04.020 ","End":"15:11.355","Text":"Energy at 3 star is equal to the energy at 0.6."},{"Start":"15:11.355 ","End":"15:15.520","Text":"Now, we\u0027re going to write out our equations."},{"Start":"15:15.770 ","End":"15:19.800","Text":"What is the energy at 0.3 star?"},{"Start":"15:19.800 ","End":"15:22.380","Text":"Let\u0027s write out the equation."},{"Start":"15:22.380 ","End":"15:28.080","Text":"First of all, we know that the velocity of the water over here is equal to 0."},{"Start":"15:28.080 ","End":"15:30.450","Text":"That\u0027s something that we know."},{"Start":"15:30.450 ","End":"15:35.445","Text":"That means that there\u0027s no kinetic energy over here."},{"Start":"15:35.445 ","End":"15:39.420","Text":"Then we have this slight height, h_2."},{"Start":"15:39.420 ","End":"15:42.015","Text":"We\u0027re going to have potential energy."},{"Start":"15:42.015 ","End":"15:45.480","Text":"That\u0027s going to be Rho g multiplied by the height,"},{"Start":"15:45.480 ","End":"15:48.300","Text":"which is equal to H_2."},{"Start":"15:48.300 ","End":"15:51.885","Text":"Then, we\u0027re going to have our elastic energy,"},{"Start":"15:51.885 ","End":"15:53.415","Text":"which comes from the pressure."},{"Start":"15:53.415 ","End":"16:00.640","Text":"We know that the pressure at p_3 star is equal to the pressure at p_3."},{"Start":"16:01.190 ","End":"16:04.140","Text":"We can just substitute that in."},{"Start":"16:04.140 ","End":"16:07.965","Text":"I\u0027m just going to write p_3 over here because it\u0027s quite a long-expression."},{"Start":"16:07.965 ","End":"16:10.590","Text":"But, this is this even just sub it in."},{"Start":"16:10.590 ","End":"16:13.845","Text":"Now let\u0027s look at the energy at 6,"},{"Start":"16:13.845 ","End":"16:17.595","Text":"so that\u0027s at the top of this tube over here."},{"Start":"16:17.595 ","End":"16:22.740","Text":"First of all, we know that again, there\u0027s no velocity."},{"Start":"16:22.740 ","End":"16:25.260","Text":"The water has no velocity within the tubes,"},{"Start":"16:25.260 ","End":"16:27.270","Text":"so there\u0027s no kinetic energy."},{"Start":"16:27.270 ","End":"16:29.790","Text":"There is going to be potential energy, however,"},{"Start":"16:29.790 ","End":"16:32.385","Text":"because we see that there\u0027s this height over here."},{"Start":"16:32.385 ","End":"16:37.650","Text":"The potential energy is equal to Rho g multiplied by the total height,"},{"Start":"16:37.650 ","End":"16:41.685","Text":"which is h_2 plus this height over here,"},{"Start":"16:41.685 ","End":"16:46.380","Text":"h_3 so h_2 plus h_3."},{"Start":"16:46.380 ","End":"16:51.525","Text":"Then, we also know that there is elastic energy from the pressure."},{"Start":"16:51.525 ","End":"16:56.100","Text":"We know that point number 6 is on the surface of the water in this tube."},{"Start":"16:56.100 ","End":"16:59.910","Text":"Which means that it\u0027s subject to atmospheric pressure."},{"Start":"16:59.910 ","End":"17:05.970","Text":"Now, we want to find out what this height h_3 is over here."},{"Start":"17:05.970 ","End":"17:09.179","Text":"First of all, we can see that on both sides of the equation,"},{"Start":"17:09.179 ","End":"17:11.535","Text":"we have Rho g h_2."},{"Start":"17:11.535 ","End":"17:15.824","Text":"We can cancel them out from both sides of the equation."},{"Start":"17:15.824 ","End":"17:19.905","Text":"Then all we want to do is we know what our p_3 is,"},{"Start":"17:19.905 ","End":"17:21.420","Text":"it\u0027s this over here."},{"Start":"17:21.420 ","End":"17:24.930","Text":"All we\u0027re trying to do is isolate out this h_3."},{"Start":"17:24.930 ","End":"17:32.805","Text":"We\u0027ll get that p_3 minus our atmospheric pressure."},{"Start":"17:32.805 ","End":"17:36.105","Text":"Then we\u0027re going to divide both sides by Rho g."},{"Start":"17:36.105 ","End":"17:40.350","Text":"Then we\u0027ll have isolated out our height h_3."},{"Start":"17:40.350 ","End":"17:43.950","Text":"Then if you want to simplify it a bit more."},{"Start":"17:43.950 ","End":"17:46.620","Text":"We can substitute in our p_3."},{"Start":"17:46.620 ","End":"17:48.510","Text":"That\u0027s our pressure at point number 3,"},{"Start":"17:48.510 ","End":"17:50.070","Text":"which we have over here."},{"Start":"17:50.070 ","End":"17:53.760","Text":"Then when we minus our atmospheric pressure and"},{"Start":"17:53.760 ","End":"17:57.720","Text":"everything and divide by Rho g will get that our height,"},{"Start":"17:57.720 ","End":"18:03.555","Text":"h_3 over here is simply equal to h_1 minus"},{"Start":"18:03.555 ","End":"18:11.500","Text":"9 v^2 divided by g. That\u0027s the end of our question."}],"ID":12372},{"Watched":false,"Name":"Exercise 5","Duration":"13m 1s","ChapterTopicVideoID":11943,"CourseChapterTopicPlaylistID":84745,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.295","Text":"Hello. In this lesson,"},{"Start":"00:02.295 ","End":"00:06.705","Text":"we\u0027re going to learn about energy loss due to viscosity."},{"Start":"00:06.705 ","End":"00:10.605","Text":"Up until now, we\u0027ve been speaking about energy conservation,"},{"Start":"00:10.605 ","End":"00:14.670","Text":"where every single point within a fluid when we\u0027re"},{"Start":"00:14.670 ","End":"00:19.095","Text":"dealing with energy conservation is going to have the same energy."},{"Start":"00:19.095 ","End":"00:22.800","Text":"The energy is going to be divided differently between the elastic,"},{"Start":"00:22.800 ","End":"00:27.045","Text":"kinetic, and potential energy depending on the point that we\u0027re looking at."},{"Start":"00:27.045 ","End":"00:32.535","Text":"However, the total energy of each point is going to be the same."},{"Start":"00:32.535 ","End":"00:38.675","Text":"Of course, that is a theoretical, an idealized scenario."},{"Start":"00:38.675 ","End":"00:41.570","Text":"However, we know that in reality,"},{"Start":"00:41.570 ","End":"00:44.870","Text":"there is going to be energy loss in everything."},{"Start":"00:44.870 ","End":"00:49.010","Text":"Not only is there going to be energy loss due to the friction of"},{"Start":"00:49.010 ","End":"00:53.680","Text":"the fluid on the side of the reservoir or the tube,"},{"Start":"00:53.680 ","End":"01:02.310","Text":"but there\u0027s also going to be energy loss due to the friction of the fluid on itself."},{"Start":"01:03.860 ","End":"01:12.005","Text":"This friction of a fluid between itself is called viscosity,"},{"Start":"01:12.005 ","End":"01:17.410","Text":"and viscosity is symbolized by the Greek letter Eta."},{"Start":"01:17.690 ","End":"01:24.830","Text":"Viscosity is how much energy is lost from the fluid when it is"},{"Start":"01:24.830 ","End":"01:33.370","Text":"moving due to the forces of friction between the different molecules inside the fluid."},{"Start":"01:33.370 ","End":"01:40.655","Text":"We can also think of viscosity as how difficult it is for a fluid to change its shape."},{"Start":"01:40.655 ","End":"01:44.090","Text":"For instance, if we look at honey and water."},{"Start":"01:44.090 ","End":"01:46.505","Text":"If we have a drop of honey,"},{"Start":"01:46.505 ","End":"01:50.510","Text":"it takes a relatively long time for the drop to flatten out."},{"Start":"01:50.510 ","End":"01:53.630","Text":"Whereas water, the droplet will almost"},{"Start":"01:53.630 ","End":"02:00.735","Text":"instantly take up some flatter shape or change its shape very easily."},{"Start":"02:00.735 ","End":"02:05.540","Text":"Of course, the honey has a higher viscosity than the water."},{"Start":"02:05.540 ","End":"02:09.065","Text":"It\u0027s simply how thick a fluid is,"},{"Start":"02:09.065 ","End":"02:12.990","Text":"so a thicker fluid will have a higher viscosity."},{"Start":"02:13.340 ","End":"02:21.315","Text":"This changing of shape due to a higher level of viscosity takes a lot of energy."},{"Start":"02:21.315 ","End":"02:24.880","Text":"A higher viscosity takes more energy,"},{"Start":"02:24.880 ","End":"02:27.885","Text":"and that\u0027s how we get energy loss."},{"Start":"02:27.885 ","End":"02:31.684","Text":"Now of course, energy loss can be to do with losing"},{"Start":"02:31.684 ","End":"02:36.445","Text":"either kinetic energy or potential energy or our elastic energy."},{"Start":"02:36.445 ","End":"02:38.525","Text":"When we\u0027re dealing with viscosity,"},{"Start":"02:38.525 ","End":"02:41.420","Text":"we\u0027re going to be speaking specifically about"},{"Start":"02:41.420 ","End":"02:46.280","Text":"our elastic energy decreasing due to this viscosity."},{"Start":"02:46.280 ","End":"02:51.000","Text":"Because our elastic energy we know is to do with pressure."},{"Start":"02:51.140 ","End":"02:56.690","Text":"That means that a higher viscosity means a lower pressure."},{"Start":"02:56.690 ","End":"02:58.910","Text":"That means a lower elastic energy,"},{"Start":"02:58.910 ","End":"03:03.180","Text":"and that means a lower total energy overall."},{"Start":"03:03.860 ","End":"03:08.150","Text":"This question, we\u0027re going to be answering the question with"},{"Start":"03:08.150 ","End":"03:12.160","Text":"energy loss or pressure loss due to viscosity."},{"Start":"03:12.160 ","End":"03:17.780","Text":"Now we\u0027re taking the viscosity of the fluid into account during our calculations."},{"Start":"03:17.780 ","End":"03:23.360","Text":"A fluid of known viscosity is flowing through the following system."},{"Start":"03:23.360 ","End":"03:28.055","Text":"Find the flow rate through the tube given the following information,"},{"Start":"03:28.055 ","End":"03:31.180","Text":"so we have a known viscosity of the fluid,"},{"Start":"03:31.180 ","End":"03:34.940","Text":"we know the length and the radius of the tube,"},{"Start":"03:34.940 ","End":"03:38.405","Text":"and we know the height of the reservoir."},{"Start":"03:38.405 ","End":"03:41.590","Text":"Let\u0027s see how we answer this type of question."},{"Start":"03:41.590 ","End":"03:44.030","Text":"If we didn\u0027t have any viscosity,"},{"Start":"03:44.030 ","End":"03:47.810","Text":"we would have a classic case of energy conservation."},{"Start":"03:47.810 ","End":"03:50.365","Text":"That would mean that the energy here at"},{"Start":"03:50.365 ","End":"03:54.740","Text":"Epsilon 2 is equal to the energy over here, Epsilon 1."},{"Start":"03:54.740 ","End":"03:59.360","Text":"Then, all we would have to do is to use that equation in order to find"},{"Start":"03:59.360 ","End":"04:05.065","Text":"the velocity of the water or the fluid flowing out through this tube."},{"Start":"04:05.065 ","End":"04:10.475","Text":"Which as we saw in previous lessons in the case of energy conservation,"},{"Start":"04:10.475 ","End":"04:12.810","Text":"the fluid coming out over here,"},{"Start":"04:13.240 ","End":"04:18.875","Text":"its velocity is always going to be equal to the velocity of free fall."},{"Start":"04:18.875 ","End":"04:20.675","Text":"However, in this case,"},{"Start":"04:20.675 ","End":"04:24.700","Text":"we know that there\u0027s energy loss because our fluid has some viscosity,"},{"Start":"04:24.700 ","End":"04:29.010","Text":"so we have to answer this question in a different way."},{"Start":"04:29.540 ","End":"04:32.350","Text":"The first thing that we have to do,"},{"Start":"04:32.350 ","End":"04:39.080","Text":"is we have to see how much energy is lost due to our fluid flowing through this tube."},{"Start":"04:39.080 ","End":"04:41.464","Text":"If you\u0027ll remember in the previous questions,"},{"Start":"04:41.464 ","End":"04:44.750","Text":"the length of the tube was of no importance,"},{"Start":"04:44.750 ","End":"04:48.410","Text":"and the only thing that was important was this height over here."},{"Start":"04:48.410 ","End":"04:52.880","Text":"But here the length is important because the longer this tube is,"},{"Start":"04:52.880 ","End":"04:58.660","Text":"the more energy is lost due to this friction and the change in pressure."},{"Start":"04:58.660 ","End":"05:03.540","Text":"Now we\u0027re going to see which equation to use in a case like this."},{"Start":"05:03.580 ","End":"05:07.310","Text":"This is the equation that you have to write down."},{"Start":"05:07.310 ","End":"05:09.245","Text":"Let\u0027s go over what this means."},{"Start":"05:09.245 ","End":"05:12.350","Text":"We have Delta p, which is the change in pressure,"},{"Start":"05:12.350 ","End":"05:16.195","Text":"and as we know that means the change in elastic energy."},{"Start":"05:16.195 ","End":"05:18.530","Text":"That means the energy loss,"},{"Start":"05:18.530 ","End":"05:22.190","Text":"which is equal to this constant over here, 8."},{"Start":"05:22.190 ","End":"05:26.630","Text":"In some textbooks, you\u0027ll see a different coefficient over here,"},{"Start":"05:26.630 ","End":"05:30.115","Text":"but we\u0027re going to be using 8."},{"Start":"05:30.115 ","End":"05:32.525","Text":"It just depends on the units that you\u0027re using,"},{"Start":"05:32.525 ","End":"05:35.450","Text":"so 8 multiplied by Eta,"},{"Start":"05:35.450 ","End":"05:38.960","Text":"which is our viscosity of the fluid,"},{"Start":"05:38.960 ","End":"05:43.955","Text":"multiplied by L, the length of the tube."},{"Start":"05:43.955 ","End":"05:50.300","Text":"Now notice this has nothing to do with the height or the horizontal length."},{"Start":"05:50.300 ","End":"05:54.840","Text":"This is dealing with just this entire length of the tube."},{"Start":"05:54.840 ","End":"06:02.695","Text":"This length here multiplied by the velocity of the liquid or fluid,"},{"Start":"06:02.695 ","End":"06:07.675","Text":"and then divide it by r^2 which is simply the radius,"},{"Start":"06:07.675 ","End":"06:11.275","Text":"r is the radius of the tube squared."},{"Start":"06:11.275 ","End":"06:15.895","Text":"We can see that our change in pressure is going to be greater"},{"Start":"06:15.895 ","End":"06:21.340","Text":"for a greater viscosity or a greater length of tube,"},{"Start":"06:21.340 ","End":"06:25.525","Text":"or a greater velocity of the fluid flowing through."},{"Start":"06:25.525 ","End":"06:30.950","Text":"All of these things mean a larger energy loss."},{"Start":"06:31.880 ","End":"06:37.180","Text":"Now, if we go back to our lesson on volumetric flow rate,"},{"Start":"06:37.180 ","End":"06:40.619","Text":"we saw that we got this equation."},{"Start":"06:40.619 ","End":"06:46.955","Text":"We got that our volumetric flow rate is equal to the velocity of the fluid"},{"Start":"06:46.955 ","End":"06:53.760","Text":"multiplied by the cross-sectional area of the tube that it is passing through."},{"Start":"06:53.760 ","End":"06:57.050","Text":"We saw if we have a circular cross-sectional area,"},{"Start":"06:57.050 ","End":"06:59.690","Text":"that this is equal to the velocity of"},{"Start":"06:59.690 ","End":"07:03.515","Text":"the water multiplied by this circular cross-sectional area,"},{"Start":"07:03.515 ","End":"07:08.155","Text":"which is equal to Pi r^2."},{"Start":"07:08.155 ","End":"07:13.985","Text":"In that case, we could get an equation for our change in pressure is"},{"Start":"07:13.985 ","End":"07:18.950","Text":"equal to 8 Eta multiplied by L,"},{"Start":"07:18.950 ","End":"07:23.510","Text":"multiplied by the volumetric flow rate"},{"Start":"07:23.510 ","End":"07:29.475","Text":"divided by Pi r^4."},{"Start":"07:29.475 ","End":"07:32.630","Text":"That\u0027s just by isolating over here,"},{"Start":"07:32.630 ","End":"07:35.715","Text":"my V, or over here."},{"Start":"07:35.715 ","End":"07:39.350","Text":"That means that I get the volumetric flow rate divided by Pi r^2,"},{"Start":"07:39.350 ","End":"07:43.430","Text":"and then I just substituted that into this equation over here."},{"Start":"07:43.430 ","End":"07:46.370","Text":"This is another way that we can write out this equation for"},{"Start":"07:46.370 ","End":"07:51.815","Text":"Delta p. Both of these equations are correct,"},{"Start":"07:51.815 ","End":"07:54.670","Text":"and it\u0027s just depending what they\u0027re asking you,"},{"Start":"07:54.670 ","End":"07:56.690","Text":"and then you\u0027ll see which one to use."},{"Start":"07:56.690 ","End":"08:00.425","Text":"For instance, if they\u0027re asking you to find the volumetric flow rates,"},{"Start":"08:00.425 ","End":"08:05.130","Text":"then obviously you\u0027ll use this equation that has the volumetric flow rate,"},{"Start":"08:05.130 ","End":"08:06.510","Text":"and if they\u0027re asking you,"},{"Start":"08:06.510 ","End":"08:11.480","Text":"such as an odd question about the velocity of the water,"},{"Start":"08:11.480 ","End":"08:16.350","Text":"so then we\u0027ll be using this version of the equation."},{"Start":"08:16.730 ","End":"08:19.545","Text":"Now that we know the equation,"},{"Start":"08:19.545 ","End":"08:22.770","Text":"let\u0027s begin with answering this question."},{"Start":"08:22.770 ","End":"08:27.445","Text":"Here we have the energy at the top and we have the energy at the bottom."},{"Start":"08:27.445 ","End":"08:34.255","Text":"We know that as our fluid flows down to the bottom, we\u0027re losing energy."},{"Start":"08:34.255 ","End":"08:36.775","Text":"That means that the energy at the top,"},{"Start":"08:36.775 ","End":"08:41.860","Text":"which is Epsilon 2 is equal to the energy that we have at the bottom,"},{"Start":"08:41.860 ","End":"08:44.125","Text":"which here we just called it Epsilon 1,"},{"Start":"08:44.125 ","End":"08:51.585","Text":"plus all of the energy that our fluid lost traveling from the top to the bottom."},{"Start":"08:51.585 ","End":"08:54.100","Text":"What does this energy loss we just saw?"},{"Start":"08:54.100 ","End":"08:55.975","Text":"It\u0027s this change in pressure,"},{"Start":"08:55.975 ","End":"09:01.750","Text":"this Delta p. The energy at the top where we have the total energy is"},{"Start":"09:01.750 ","End":"09:04.525","Text":"equal to the energy that we have at the bottom"},{"Start":"09:04.525 ","End":"09:08.720","Text":"plus the energy loss on the way to the bottom."},{"Start":"09:09.480 ","End":"09:12.820","Text":"Now let\u0027s write out this equation."},{"Start":"09:12.820 ","End":"09:17.840","Text":"What is our energy at this point to over here?"},{"Start":"09:17.840 ","End":"09:20.335","Text":"First of all, we know that we have"},{"Start":"09:20.335 ","End":"09:23.830","Text":"our potential energy because we have a height over here."},{"Start":"09:23.830 ","End":"09:27.110","Text":"That\u0027s going to be equal to Rho gh,"},{"Start":"09:27.110 ","End":"09:30.180","Text":"then we have our kinetic energy."},{"Start":"09:30.180 ","End":"09:33.280","Text":"We know that usually when we\u0027re dealing with a reservoir,"},{"Start":"09:33.280 ","End":"09:40.230","Text":"the size of the reservoir is much larger than the size of the tube,"},{"Start":"09:40.230 ","End":"09:42.124","Text":"than the radius of the tube,"},{"Start":"09:42.124 ","End":"09:44.990","Text":"which means that we can consider"},{"Start":"09:44.990 ","End":"09:50.150","Text":"the velocity of the water over here inside the reservoir is equal to 0."},{"Start":"09:50.150 ","End":"09:52.610","Text":"That means that we have no kinetic energy."},{"Start":"09:52.610 ","End":"09:54.560","Text":"Then we have our elastic energy,"},{"Start":"09:54.560 ","End":"09:56.435","Text":"which is of course due to pressure,"},{"Start":"09:56.435 ","End":"09:59.085","Text":"so at 0.2 over here,"},{"Start":"09:59.085 ","End":"10:03.305","Text":"our water or our fluid is subject to atmospheric pressure."},{"Start":"10:03.305 ","End":"10:08.220","Text":"We can write plus P atmospheric."},{"Start":"10:08.720 ","End":"10:11.415","Text":"That\u0027s our energy here."},{"Start":"10:11.415 ","End":"10:15.080","Text":"This is equal to the energy at point Number 1."},{"Start":"10:15.080 ","End":"10:17.420","Text":"That\u0027s the energy over here,"},{"Start":"10:17.420 ","End":"10:20.240","Text":"plus our Delta P. First of all,"},{"Start":"10:20.240 ","End":"10:22.745","Text":"let\u0027s see what the energy over here is equal to,"},{"Start":"10:22.745 ","End":"10:25.430","Text":"just like we\u0027ve done in previous lessons."},{"Start":"10:25.430 ","End":"10:28.835","Text":"First of all, we\u0027re going to have our potential energy,"},{"Start":"10:28.835 ","End":"10:32.690","Text":"now here we can see that point number 1 is at height 0."},{"Start":"10:32.690 ","End":"10:35.480","Text":"That means that we have no potential energy."},{"Start":"10:35.480 ","End":"10:37.780","Text":"Then we have our kinetic energy,"},{"Start":"10:37.780 ","End":"10:43.000","Text":"so we know that the fluid over here is flowing with some velocity."},{"Start":"10:43.000 ","End":"10:44.950","Text":"That means that we have kinetic energy."},{"Start":"10:44.950 ","End":"10:48.915","Text":"That\u0027s half Rho V^2,"},{"Start":"10:48.915 ","End":"10:53.360","Text":"and then it\u0027s also subject to elastic energy,"},{"Start":"10:53.360 ","End":"10:55.070","Text":"which is of course, pressure."},{"Start":"10:55.070 ","End":"10:58.790","Text":"Because our fluid over here is outside of the tube,"},{"Start":"10:58.790 ","End":"11:01.865","Text":"that means that it\u0027s subject to atmospheric pressure,"},{"Start":"11:01.865 ","End":"11:09.500","Text":"just like the fluid at the top over here, plus P atmospheric."},{"Start":"11:09.500 ","End":"11:12.545","Text":"Up until now this is exactly what we\u0027ve been doing."},{"Start":"11:12.545 ","End":"11:15.065","Text":"However, now we also have this term over here,"},{"Start":"11:15.065 ","End":"11:18.940","Text":"due to our energy loss due to viscosity."},{"Start":"11:18.940 ","End":"11:22.115","Text":"Because in the question we\u0027re being asked to find"},{"Start":"11:22.115 ","End":"11:28.790","Text":"the flow rate or the velocity of the water at this point over here,"},{"Start":"11:28.790 ","End":"11:34.280","Text":"so we\u0027re going to use this version of the equation that incorporates the velocity term."},{"Start":"11:34.280 ","End":"11:38.300","Text":"Now we\u0027re going to add plus 8, Eta,"},{"Start":"11:38.300 ","End":"11:43.530","Text":"LV divided by r^2."},{"Start":"11:44.480 ","End":"11:47.820","Text":"Now we can see that first of all,"},{"Start":"11:47.820 ","End":"11:50.945","Text":"we have our atmospheric pressure on both sides of the equation,"},{"Start":"11:50.945 ","End":"11:55.205","Text":"so we can subtract our atmospheric pressure from both sides of the equation."},{"Start":"11:55.205 ","End":"11:58.540","Text":"Then we\u0027re simply left, we can see,"},{"Start":"11:58.540 ","End":"12:04.755","Text":"if we rearrange this equation with a quadratic equation."},{"Start":"12:04.755 ","End":"12:10.275","Text":"If we subtract Rho gh from both sides,"},{"Start":"12:10.275 ","End":"12:18.660","Text":"we\u0027ll have 0 is equal to half Rho V^2 plus"},{"Start":"12:18.660 ","End":"12:30.220","Text":"8 Eta L divided by r^2 multiplied by V minus our Rho gh."},{"Start":"12:30.350 ","End":"12:35.615","Text":"Now we see that we have exactly a quadratic equation."},{"Start":"12:35.615 ","End":"12:39.855","Text":"We have a V^2 term,"},{"Start":"12:39.855 ","End":"12:45.060","Text":"a V term, and something without our variable over here."},{"Start":"12:45.060 ","End":"12:48.740","Text":"All we have to do is solve this using a simple quadratic equation,"},{"Start":"12:48.740 ","End":"12:51.815","Text":"which I\u0027m assuming that you all know how to solve."},{"Start":"12:51.815 ","End":"12:57.810","Text":"That is how you can find the velocity of the liquid over here."},{"Start":"12:58.280 ","End":"13:02.140","Text":"That\u0027s the end of this question."}],"ID":12373},{"Watched":false,"Name":"Exercise 6","Duration":"12m 12s","ChapterTopicVideoID":11944,"CourseChapterTopicPlaylistID":84745,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.145","Text":"Hello. In this question,"},{"Start":"00:02.145 ","End":"00:05.065","Text":"we are being given the following system."},{"Start":"00:05.065 ","End":"00:07.980","Text":"We\u0027re being told that the cross-sectional area at"},{"Start":"00:07.980 ","End":"00:11.970","Text":"0.3 is twice that of the rest of the tube."},{"Start":"00:11.970 ","End":"00:13.860","Text":"Question number 1 is,"},{"Start":"00:13.860 ","End":"00:18.075","Text":"what is the velocity of the flow if there is no energy loss?"},{"Start":"00:18.075 ","End":"00:21.555","Text":"No energy loss just translates to"},{"Start":"00:21.555 ","End":"00:27.465","Text":"a 0 viscosity and we answer it just like we\u0027ve seen in many questions prior."},{"Start":"00:27.465 ","End":"00:29.135","Text":"In question number 2 is,"},{"Start":"00:29.135 ","End":"00:33.790","Text":"what is the velocity of the flow if a coefficient of viscosity is given?"},{"Start":"00:33.790 ","End":"00:37.730","Text":"That means that we\u0027re dealing with viscous flow and that means that there"},{"Start":"00:37.730 ","End":"00:41.929","Text":"is energy loss and so then we answer the question a little bit differently,"},{"Start":"00:41.929 ","End":"00:45.450","Text":"just like we did in the previous question."},{"Start":"00:45.590 ","End":"00:48.445","Text":"Let\u0027s start with question number 1,"},{"Start":"00:48.445 ","End":"00:51.425","Text":"the velocity of the flow if there is no energy loss."},{"Start":"00:51.425 ","End":"00:52.790","Text":"Now, here we\u0027ve seen,"},{"Start":"00:52.790 ","End":"00:55.205","Text":"we don\u0027t even have to do a calculation."},{"Start":"00:55.205 ","End":"01:01.875","Text":"We know that the velocity of the flow is simply the velocity due to free fall."},{"Start":"01:01.875 ","End":"01:07.760","Text":"We already saw that that is going to be equal to the square root of"},{"Start":"01:07.760 ","End":"01:13.910","Text":"2_g multiplied by the total height that the water is falling."},{"Start":"01:13.910 ","End":"01:16.701","Text":"Because we\u0027re going from point number 1 or point number 2,"},{"Start":"01:16.701 ","End":"01:20.240","Text":"they\u0027re at the same height until point number 4,"},{"Start":"01:20.240 ","End":"01:25.460","Text":"we can see that the change in height is simply this h_1 over here,"},{"Start":"01:25.460 ","End":"01:28.700","Text":"so we can write in h_1."},{"Start":"01:28.700 ","End":"01:34.955","Text":"Now I don\u0027t care if my fluid has gone up and then down."},{"Start":"01:34.955 ","End":"01:42.480","Text":"All I care about is the original position that the fluid was at and its final position."},{"Start":"01:42.480 ","End":"01:47.590","Text":"We can see that the total change in height is just this h_1,"},{"Start":"01:47.590 ","End":"01:50.540","Text":"not including the h_2, and that\u0027s it,"},{"Start":"01:50.540 ","End":"01:54.380","Text":"that\u0027s our answer to question number 1, very easy."},{"Start":"01:54.380 ","End":"01:58.010","Text":"Now we\u0027re going to be answering question number 2."},{"Start":"01:58.010 ","End":"02:01.865","Text":"Question number 2 is telling us that we have a coefficient of viscosity,"},{"Start":"02:01.865 ","End":"02:05.690","Text":"which means that there isn\u0027t any energy conservation,"},{"Start":"02:05.690 ","End":"02:08.305","Text":"so there\u0027s energy loss."},{"Start":"02:08.305 ","End":"02:10.640","Text":"What I have to do is,"},{"Start":"02:10.640 ","End":"02:16.700","Text":"I have to say that my energy at point number 1 is equal to"},{"Start":"02:16.700 ","End":"02:20.780","Text":"my energy at point number 4 plus"},{"Start":"02:20.780 ","End":"02:26.365","Text":"all the energy that was lost on its route to point number 4."},{"Start":"02:26.365 ","End":"02:29.120","Text":"The energy or in other words,"},{"Start":"02:29.120 ","End":"02:34.530","Text":"the energy at point number 4 plus the energy lost"},{"Start":"02:34.530 ","End":"02:42.125","Text":"for the fluid traveling up until point number 4 is equal to the original energy."},{"Start":"02:42.125 ","End":"02:46.055","Text":"How did we say we express this lost energy?"},{"Start":"02:46.055 ","End":"02:48.320","Text":"We call that Delta P,"},{"Start":"02:48.320 ","End":"02:52.375","Text":"the change in pressure or the change in elastic energy."},{"Start":"02:52.375 ","End":"02:57.450","Text":"First of all, let\u0027s see what our energy is at point number 1."},{"Start":"02:57.470 ","End":"03:02.888","Text":"First of all, we have potential energy,"},{"Start":"03:02.888 ","End":"03:06.125","Text":"are point number 1 has some height,"},{"Start":"03:06.125 ","End":"03:07.940","Text":"which means it has potential energy,"},{"Start":"03:07.940 ","End":"03:11.360","Text":"so that\u0027s equal to Rho g multiplied by its height,"},{"Start":"03:11.360 ","End":"03:14.560","Text":"which we can see here is h_1."},{"Start":"03:14.560 ","End":"03:21.140","Text":"Then of course, we\u0027re dealing with the water or the fluid at the top,"},{"Start":"03:21.140 ","End":"03:25.070","Text":"which means that it has some elastic energy"},{"Start":"03:25.070 ","End":"03:30.080","Text":"that we\u0027re going to give as the pressure over here,"},{"Start":"03:30.080 ","End":"03:33.545","Text":"which of course, if it\u0027s subject to the atmosphere,"},{"Start":"03:33.545 ","End":"03:36.190","Text":"it\u0027s just our atmospheric pressure."},{"Start":"03:36.190 ","End":"03:39.275","Text":"Then we have to write in our kinetic energy."},{"Start":"03:39.275 ","End":"03:43.250","Text":"Now we\u0027re using the same assumption that the size of the reservoir is"},{"Start":"03:43.250 ","End":"03:48.545","Text":"very large relative to the cross-sectional area of our tube,"},{"Start":"03:48.545 ","End":"03:54.425","Text":"which means that we can take the velocity of the water in the reservoir as equaling to 0,"},{"Start":"03:54.425 ","End":"03:58.790","Text":"which means that we have no kinetic energy."},{"Start":"03:58.790 ","End":"04:03.330","Text":"Now we\u0027re going to move over to the other side of the equation."},{"Start":"04:03.950 ","End":"04:07.520","Text":"What is our energy at point number 4?"},{"Start":"04:07.520 ","End":"04:12.935","Text":"First of all, we can see that the height of point number 4 is 0,"},{"Start":"04:12.935 ","End":"04:15.005","Text":"so we have no potential energy."},{"Start":"04:15.005 ","End":"04:17.000","Text":"Then we have our elastic energy,"},{"Start":"04:17.000 ","End":"04:18.590","Text":"which is of course our pressure."},{"Start":"04:18.590 ","End":"04:22.705","Text":"Now the pressure at point number 4 is simply our atmospheric pressure,"},{"Start":"04:22.705 ","End":"04:26.920","Text":"because we\u0027re not inside the tube anymore."},{"Start":"04:26.920 ","End":"04:34.055","Text":"Then we know that the fluid flowing through point number 4 has some velocity,"},{"Start":"04:34.055 ","End":"04:36.620","Text":"so that means that we have kinetic energy."},{"Start":"04:36.620 ","End":"04:39.540","Text":"That\u0027s equal to 1/2 Rho v^2,"},{"Start":"04:39.540 ","End":"04:44.555","Text":"where of course our v is unknown and then because we\u0027re dealing with a viscous fluid,"},{"Start":"04:44.555 ","End":"04:46.320","Text":"we have this energy loss,"},{"Start":"04:46.320 ","End":"04:48.945","Text":"what is our delta P?"},{"Start":"04:48.945 ","End":"04:52.070","Text":"We\u0027re going to use the version of Delta P that includes"},{"Start":"04:52.070 ","End":"04:55.250","Text":"the velocity because that\u0027s what we\u0027re trying to find."},{"Start":"04:55.250 ","End":"05:00.260","Text":"That\u0027s simply equal to 8 Eta L"},{"Start":"05:00.260 ","End":"05:06.040","Text":"divided by R^2 multiplied by v, our velocity."},{"Start":"05:06.040 ","End":"05:08.585","Text":"Now again, on both sides of the equation,"},{"Start":"05:08.585 ","End":"05:11.420","Text":"we can cross out our atmospheric pressure."},{"Start":"05:11.420 ","End":"05:14.060","Text":"Then if we just rearrange this equation,"},{"Start":"05:14.060 ","End":"05:15.515","Text":"we subtract Rho g,"},{"Start":"05:15.515 ","End":"05:17.045","Text":"h_1 from both sides."},{"Start":"05:17.045 ","End":"05:19.190","Text":"We\u0027ll get just like in the previous question,"},{"Start":"05:19.190 ","End":"05:23.455","Text":"a quadratic equation, which we know how to solve."},{"Start":"05:23.455 ","End":"05:26.275","Text":"Once we\u0027ve solved this quadratic equation,"},{"Start":"05:26.275 ","End":"05:31.265","Text":"we\u0027ll have the velocity of the water for every single point in this tube,"},{"Start":"05:31.265 ","End":"05:33.725","Text":"aside from at point 3,"},{"Start":"05:33.725 ","End":"05:37.220","Text":"where the cross-sectional area is greater."},{"Start":"05:37.220 ","End":"05:41.150","Text":"We know from the continuity equation that if"},{"Start":"05:41.150 ","End":"05:46.715","Text":"the cross-sectional area is double or twice as large as the rest of the tube,"},{"Start":"05:46.715 ","End":"05:54.690","Text":"then the velocity at this point is going to be 1/2 as fast as in the rest of the tube."},{"Start":"05:54.850 ","End":"05:58.475","Text":"If the cross-sectional area is twice as big,"},{"Start":"05:58.475 ","End":"06:02.350","Text":"then the velocity is going to be twice as small."},{"Start":"06:02.350 ","End":"06:06.680","Text":"I can know what the velocity over here at point 3 is,"},{"Start":"06:06.680 ","End":"06:10.955","Text":"however, I can\u0027t know what the pressure over here at point 3 is."},{"Start":"06:10.955 ","End":"06:12.740","Text":"Why can\u0027t I know the pressure?"},{"Start":"06:12.740 ","End":"06:15.565","Text":"Because I know that I\u0027m dealing with the viscous fluid,"},{"Start":"06:15.565 ","End":"06:19.775","Text":"which means that throughout the tube I have energy loss,"},{"Start":"06:19.775 ","End":"06:22.990","Text":"which means that I can\u0027t say that the energy at point"},{"Start":"06:22.990 ","End":"06:26.570","Text":"2 is equal to the energy at point 3 because it isn\u0027t."},{"Start":"06:26.570 ","End":"06:30.835","Text":"Some of the energy has been lost between point 2 and point 3,"},{"Start":"06:30.835 ","End":"06:33.620","Text":"so the pressure isn\u0027t going to be the same."},{"Start":"06:33.620 ","End":"06:37.220","Text":"What I can say is that the velocity of the fluid throughout"},{"Start":"06:37.220 ","End":"06:42.225","Text":"the tube is going to be the same velocity throughout and that is correct,"},{"Start":"06:42.225 ","End":"06:46.145","Text":"whether we have energy loss or we have energy conservation."},{"Start":"06:46.145 ","End":"06:49.340","Text":"The only place that the velocity will change is if"},{"Start":"06:49.340 ","End":"06:52.700","Text":"the cross-sectional area of the tube changes,"},{"Start":"06:52.700 ","End":"06:56.940","Text":"and then we know that that is according to the continuity equation."},{"Start":"06:57.130 ","End":"07:05.045","Text":"That means that we can know the velocity at all the points over here inside this tube,"},{"Start":"07:05.045 ","End":"07:07.100","Text":"including at point 3."},{"Start":"07:07.100 ","End":"07:10.490","Text":"However, another question that can be asked is what is"},{"Start":"07:10.490 ","End":"07:15.900","Text":"the velocity or what is the pressure at point 2 over here."},{"Start":"07:16.070 ","End":"07:19.960","Text":"The first thing that we can see is that point number 2"},{"Start":"07:19.960 ","End":"07:22.910","Text":"is right at the beginning of our tube."},{"Start":"07:22.910 ","End":"07:24.980","Text":"It\u0027s right at the entrance to the tube."},{"Start":"07:24.980 ","End":"07:28.190","Text":"Which means that the energy loss at this point is"},{"Start":"07:28.190 ","End":"07:31.240","Text":"going to be so small that we can discount it."},{"Start":"07:31.240 ","End":"07:34.220","Text":"The energy loss increases as we go through"},{"Start":"07:34.220 ","End":"07:38.165","Text":"the tube but right at the entrance to the tube, it\u0027s virtually 0."},{"Start":"07:38.165 ","End":"07:43.565","Text":"That means that we can use our energy conservation equation,"},{"Start":"07:43.565 ","End":"07:49.320","Text":"which means that the energy at point 1 is equal to the energy at point 2."},{"Start":"07:49.940 ","End":"07:53.105","Text":"What is our energy at point number 1?"},{"Start":"07:53.105 ","End":"07:55.835","Text":"Well, it\u0027s what we just did before,"},{"Start":"07:55.835 ","End":"07:58.490","Text":"so it has height, so potential energy,"},{"Start":"07:58.490 ","End":"08:01.190","Text":"so that\u0027s Rho g multiplied by its height,"},{"Start":"08:01.190 ","End":"08:09.035","Text":"which is h_1 plus it has the elastic energy due to the pressure,"},{"Start":"08:09.035 ","End":"08:13.265","Text":"which of course it\u0027s subject to atmospheric pressure,"},{"Start":"08:13.265 ","End":"08:16.780","Text":"and of course, we\u0027re dealing with point number 1,"},{"Start":"08:16.780 ","End":"08:17.845","Text":"which is in the reservoir,"},{"Start":"08:17.845 ","End":"08:19.645","Text":"which means that it has no velocity,"},{"Start":"08:19.645 ","End":"08:21.040","Text":"so no kinetic energy."},{"Start":"08:21.040 ","End":"08:24.250","Text":"This is equal to the energy at point number 2,"},{"Start":"08:24.250 ","End":"08:26.585","Text":"which is then the entrance to the tube."},{"Start":"08:26.585 ","End":"08:29.355","Text":"Again, it has the exact same height,"},{"Start":"08:29.355 ","End":"08:30.820","Text":"so we have potential energy,"},{"Start":"08:30.820 ","End":"08:33.880","Text":"so that\u0027s going to be Rho gh_1,"},{"Start":"08:33.880 ","End":"08:36.610","Text":"then it\u0027s also subject to a pressure."},{"Start":"08:36.610 ","End":"08:39.790","Text":"Now because we\u0027re right at the entrance to the tube,"},{"Start":"08:39.790 ","End":"08:41.965","Text":"it\u0027s subject to the pressure of the tube,"},{"Start":"08:41.965 ","End":"08:43.345","Text":"which we don\u0027t know what that is,"},{"Start":"08:43.345 ","End":"08:45.740","Text":"so I\u0027ll just add P_2,"},{"Start":"08:45.740 ","End":"08:48.130","Text":"which we have to find, and then of course,"},{"Start":"08:48.130 ","End":"08:49.405","Text":"because we\u0027re in the tube,"},{"Start":"08:49.405 ","End":"08:55.995","Text":"we also have the same velocity of liquid flow through this tube."},{"Start":"08:55.995 ","End":"08:58.170","Text":"That means that we have kinetic energy,"},{"Start":"08:58.170 ","End":"09:03.520","Text":"so that\u0027s going to be equal to 1/2 Rho v^2."},{"Start":"09:04.520 ","End":"09:09.525","Text":"Now we can see that we have Rho gh_1 on both sides,"},{"Start":"09:09.525 ","End":"09:13.235","Text":"so we can subtract this from both sides,"},{"Start":"09:13.235 ","End":"09:19.310","Text":"and then we can rearrange this equation in order to get what our pressure at point 2 is."},{"Start":"09:19.310 ","End":"09:22.775","Text":"P_2 is simply going to be equal to"},{"Start":"09:22.775 ","End":"09:30.120","Text":"our atmospheric pressure minus our kinetic energy, 1/2 Rho v^2."},{"Start":"09:30.120 ","End":"09:32.119","Text":"Now, our velocity,"},{"Start":"09:32.119 ","End":"09:36.560","Text":"because we\u0027re dealing with a viscous fluid going through this tube,"},{"Start":"09:36.560 ","End":"09:41.975","Text":"that\u0027s why we can\u0027t find the pressure here directly because there\u0027s energy loss."},{"Start":"09:41.975 ","End":"09:49.205","Text":"We already found from this quadratic equation what our velocity is and of course,"},{"Start":"09:49.205 ","End":"09:51.695","Text":"our atmospheric pressure is a given,"},{"Start":"09:51.695 ","End":"09:58.565","Text":"so we can just use this in order to calculate our pressure at point 2, like so."},{"Start":"09:58.565 ","End":"10:01.085","Text":"Before we end this lesson,"},{"Start":"10:01.085 ","End":"10:06.110","Text":"let\u0027s ask the question of why is the water leaving the reservoir in the first place?"},{"Start":"10:06.110 ","End":"10:10.205","Text":"Because the water from being over here, this certain heights,"},{"Start":"10:10.205 ","End":"10:13.485","Text":"then has to go up and across,"},{"Start":"10:13.485 ","End":"10:16.965","Text":"only to then finally fall down."},{"Start":"10:16.965 ","End":"10:20.120","Text":"How come the water is going up in the first place?"},{"Start":"10:20.120 ","End":"10:23.135","Text":"How can we get this motion in the first place?"},{"Start":"10:23.135 ","End":"10:28.850","Text":"How this is done is we have some suction over here or a vacuum,"},{"Start":"10:28.850 ","End":"10:32.075","Text":"which sucks this fluid through,"},{"Start":"10:32.075 ","End":"10:35.495","Text":"from the reservoir and then through the tube."},{"Start":"10:35.495 ","End":"10:39.420","Text":"Once there\u0027s fluid flowing out,"},{"Start":"10:39.420 ","End":"10:41.025","Text":"through point number 4,"},{"Start":"10:41.025 ","End":"10:44.300","Text":"even if we remove our vacuum or our suction,"},{"Start":"10:44.300 ","End":"10:47.810","Text":"the fluid is going to continue traveling from"},{"Start":"10:47.810 ","End":"10:51.940","Text":"the reservoir through the tube into this point number 4."},{"Start":"10:51.940 ","End":"10:54.270","Text":"Why is this?"},{"Start":"10:54.270 ","End":"11:00.005","Text":"We know that our fluid is going to travel in a downwards direction."},{"Start":"11:00.005 ","End":"11:03.050","Text":"Fluid always travels downhill and never"},{"Start":"11:03.050 ","End":"11:06.500","Text":"uphill unless it\u0027s subject to some external force."},{"Start":"11:06.500 ","End":"11:10.160","Text":"The vacuum is the initial external force,"},{"Start":"11:10.160 ","End":"11:11.810","Text":"but once we remove it,"},{"Start":"11:11.810 ","End":"11:16.615","Text":"the water carries on going through the tube and falling through point number 4."},{"Start":"11:16.615 ","End":"11:19.010","Text":"Why is this? As we said,"},{"Start":"11:19.010 ","End":"11:23.960","Text":"the fluid always travels downhill and what we can see is that the difference in height"},{"Start":"11:23.960 ","End":"11:29.305","Text":"between points 1 and 2 is still higher than point 4."},{"Start":"11:29.305 ","End":"11:33.480","Text":"Point 4 is located at height 0 and points 1 and 2,"},{"Start":"11:33.480 ","End":"11:35.910","Text":"where the reservoir is,"},{"Start":"11:35.910 ","End":"11:39.545","Text":"is located at height h_1, which means that,"},{"Start":"11:39.545 ","End":"11:44.840","Text":"overall, our water is traveling downhill in a downwards direction,"},{"Start":"11:44.840 ","End":"11:49.190","Text":"and so that is why even after the vacuum has been removed,"},{"Start":"11:49.190 ","End":"11:55.070","Text":"the fluid is going to continue flowing through the tube."},{"Start":"11:55.070 ","End":"12:00.590","Text":"Now, this phenomenon that we\u0027ve been speaking about is called communicating vessels."},{"Start":"12:00.590 ","End":"12:02.300","Text":"I\u0027m not going to go into it now,"},{"Start":"12:02.300 ","End":"12:05.375","Text":"but if you want to read a little bit more about it on the Internet,"},{"Start":"12:05.375 ","End":"12:09.280","Text":"just look up the term communicating vessels."},{"Start":"12:09.280 ","End":"12:12.550","Text":"Okay, that\u0027s the end of this lesson."}],"ID":12374},{"Watched":false,"Name":"Pumps","Duration":"11m 1s","ChapterTopicVideoID":11945,"CourseChapterTopicPlaylistID":84745,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.325","Text":"Hello. In this lesson,"},{"Start":"00:02.325 ","End":"00:06.615","Text":"we\u0027re going to be learning how to answer questions dealing with a pump."},{"Start":"00:06.615 ","End":"00:10.320","Text":"Here, we\u0027re being told that a pump with electrical power N,"},{"Start":"00:10.320 ","End":"00:12.720","Text":"is placed at the bottom of a tube of diameter"},{"Start":"00:12.720 ","End":"00:16.065","Text":"D. We\u0027re being told what the viscosity of the fluid is,"},{"Start":"00:16.065 ","End":"00:19.395","Text":"as well as the heights and lengths of the system."},{"Start":"00:19.395 ","End":"00:23.050","Text":"How do we solve this type of question?"},{"Start":"00:23.630 ","End":"00:26.870","Text":"A pump\u0027s job, given its name,"},{"Start":"00:26.870 ","End":"00:30.605","Text":"is to pump fluid somewhere."},{"Start":"00:30.605 ","End":"00:36.605","Text":"That means to push it through wherever we\u0027re trying to push it through,"},{"Start":"00:36.605 ","End":"00:40.280","Text":"and that means that it is increasing the pressure."},{"Start":"00:40.280 ","End":"00:46.820","Text":"We can see that the pressure at 3 is going to be higher than the pressure at 2."},{"Start":"00:46.820 ","End":"00:49.535","Text":"Now what\u0027s important to note is that the pump"},{"Start":"00:49.535 ","End":"00:52.370","Text":"isn\u0027t changing the height or the potential energy."},{"Start":"00:52.370 ","End":"00:54.200","Text":"We can see that the same height,"},{"Start":"00:54.200 ","End":"00:58.250","Text":"these 2 points and the velocity of"},{"Start":"00:58.250 ","End":"01:03.125","Text":"the fluid flowing through points 2 and points 3 is also the same."},{"Start":"01:03.125 ","End":"01:07.985","Text":"The velocity of the fluid at point 3 is the same as the velocity at point 2."},{"Start":"01:07.985 ","End":"01:11.220","Text":"Again, that means that the kinetic energy at point"},{"Start":"01:11.220 ","End":"01:14.425","Text":"3 is the same as the kinetic energy at point 2."},{"Start":"01:14.425 ","End":"01:17.705","Text":"The only thing that the pump is changing is"},{"Start":"01:17.705 ","End":"01:23.940","Text":"the pressure at point 3 relative to the pressure at point 2."},{"Start":"01:24.740 ","End":"01:28.355","Text":"The pump is just adding pressure,"},{"Start":"01:28.355 ","End":"01:31.550","Text":"which means changing the energy because"},{"Start":"01:31.550 ","End":"01:35.060","Text":"we know that pressure and energy are interchangeable over here."},{"Start":"01:35.060 ","End":"01:40.730","Text":"Let\u0027s look at this. We have this equation which connect"},{"Start":"01:40.730 ","End":"01:47.915","Text":"the electric power to our change in energy or our change in pressure."},{"Start":"01:47.915 ","End":"01:51.190","Text":"Here we have our power,"},{"Start":"01:51.190 ","End":"01:53.945","Text":"which is N. Then here,"},{"Start":"01:53.945 ","End":"01:56.785","Text":"we have our volumetric flow rate."},{"Start":"01:56.785 ","End":"02:02.310","Text":"Of course, our Delta P means our change in pressure."},{"Start":"02:02.310 ","End":"02:05.915","Text":"Where does this equation come from and why does it make sense?"},{"Start":"02:05.915 ","End":"02:09.425","Text":"This Delta P, which is the change in pressure or energy,"},{"Start":"02:09.425 ","End":"02:15.565","Text":"means the pressure change or the energy change per particle."},{"Start":"02:15.565 ","End":"02:21.575","Text":"This Delta P is the addition of energy that every single particle gets."},{"Start":"02:21.575 ","End":"02:24.005","Text":"Now this Q_v is,"},{"Start":"02:24.005 ","End":"02:26.585","Text":"as we know, our volumetric flow rate."},{"Start":"02:26.585 ","End":"02:33.635","Text":"That is talking about the volume of fluid that flows through a specific point."},{"Start":"02:33.635 ","End":"02:39.770","Text":"If I take the amount of water particles that are passing a point at every moment,"},{"Start":"02:39.770 ","End":"02:42.755","Text":"and I multiply it by the addition of energy"},{"Start":"02:42.755 ","End":"02:46.070","Text":"that every single one of these particles receives,"},{"Start":"02:46.070 ","End":"02:54.800","Text":"then I get the total amount of energy that this pump is putting in to the fluid."},{"Start":"02:54.800 ","End":"02:57.185","Text":"That is exactly power."},{"Start":"02:57.185 ","End":"03:00.965","Text":"Power is how much energy I get in a second,"},{"Start":"03:00.965 ","End":"03:05.599","Text":"or how much energy the pump is supplying per second."},{"Start":"03:05.599 ","End":"03:09.944","Text":"Now, of course, this power, this is electrical."},{"Start":"03:09.944 ","End":"03:12.120","Text":"What we\u0027re dealing with here,"},{"Start":"03:12.120 ","End":"03:15.615","Text":"our volumetric flow rate is in liters."},{"Start":"03:15.615 ","End":"03:20.315","Text":"We\u0027re getting this electrical power and it\u0027s been converted into"},{"Start":"03:20.315 ","End":"03:25.335","Text":"this energy per particle over here by the pump."},{"Start":"03:25.335 ","End":"03:29.090","Text":"Let\u0027s see how we answer this type of question."},{"Start":"03:29.090 ","End":"03:34.175","Text":"In this question, we\u0027re being told that there\u0027s viscosity of the fluid."},{"Start":"03:34.175 ","End":"03:38.225","Text":"That means that we have energy loss through the tube."},{"Start":"03:38.225 ","End":"03:40.820","Text":"We know how to deal with this type of question."},{"Start":"03:40.820 ","End":"03:45.050","Text":"We say that the energy at point number 1 is equal to the energy at"},{"Start":"03:45.050 ","End":"03:50.215","Text":"point number 4 plus the energy that was lost on the way there."},{"Start":"03:50.215 ","End":"03:55.630","Text":"That\u0027s great. But now we just have this additional pump over here."},{"Start":"03:55.630 ","End":"03:57.080","Text":"What are we going to do?"},{"Start":"03:57.080 ","End":"03:58.895","Text":"We have to take that into account."},{"Start":"03:58.895 ","End":"04:05.405","Text":"We know that the pump is providing energy to the system. It\u0027s the same thing."},{"Start":"04:05.405 ","End":"04:07.490","Text":"We take the energy at point number 1"},{"Start":"04:07.490 ","End":"04:09.770","Text":"and say that it\u0027s equal to the energy at point number"},{"Start":"04:09.770 ","End":"04:16.430","Text":"4 plus the energy that was lost on the way there."},{"Start":"04:16.430 ","End":"04:23.275","Text":"We take into account the energy that was provided on the way there by the pump."},{"Start":"04:23.275 ","End":"04:29.660","Text":"What we\u0027re going to do is we\u0027re going to write out the energies at each point."},{"Start":"04:29.660 ","End":"04:33.570","Text":"Let\u0027s begin with the energy at point number 1."},{"Start":"04:34.040 ","End":"04:36.970","Text":"Again, using the idea that this is"},{"Start":"04:36.970 ","End":"04:42.665","Text":"a very large reservoir relative to the cross-sectional area of our tube."},{"Start":"04:42.665 ","End":"04:45.640","Text":"That means that the velocity of"},{"Start":"04:45.640 ","End":"04:50.385","Text":"the water particles or the fluid particles at point number 1=0,"},{"Start":"04:50.385 ","End":"04:52.440","Text":"so there\u0027s no kinetic energy."},{"Start":"04:52.440 ","End":"04:56.890","Text":"We can see that our reservoir is located at height 0,"},{"Start":"04:56.890 ","End":"04:59.125","Text":"which means that there\u0027s no potential energy."},{"Start":"04:59.125 ","End":"05:02.035","Text":"We do know, however, that there\u0027s elastic energy"},{"Start":"05:02.035 ","End":"05:05.995","Text":"because the fluid is subject to atmospheric pressures."},{"Start":"05:05.995 ","End":"05:11.290","Text":"The energy at point number 1 is simply the atmospheric pressure."},{"Start":"05:11.290 ","End":"05:16.355","Text":"Now let\u0027s look at the energy at point number 2 over here."},{"Start":"05:16.355 ","End":"05:19.550","Text":"Again, we\u0027re at height 0,"},{"Start":"05:19.550 ","End":"05:21.290","Text":"so there\u0027s no potential energy."},{"Start":"05:21.290 ","End":"05:25.370","Text":"However, because we\u0027re located at the beginning of the tube,"},{"Start":"05:25.370 ","End":"05:29.650","Text":"we know that there\u0027s going to be some kinetic energy."},{"Start":"05:29.650 ","End":"05:33.990","Text":"Let\u0027s add that, that\u0027s going to be equal to 1/2 Rho V^2,"},{"Start":"05:33.990 ","End":"05:36.185","Text":"where of course the velocity at this point,"},{"Start":"05:36.185 ","End":"05:37.625","Text":"we don\u0027t know yet."},{"Start":"05:37.625 ","End":"05:41.780","Text":"Then we\u0027re also going to be subject to elastic energy,"},{"Start":"05:41.780 ","End":"05:43.070","Text":"some kind of pressure."},{"Start":"05:43.070 ","End":"05:47.195","Text":"Now we don\u0027t know what this pressure is because it\u0027s going to be a combination"},{"Start":"05:47.195 ","End":"05:52.240","Text":"of atmospheric pressure and pressure inside the tube."},{"Start":"05:52.240 ","End":"05:54.600","Text":"Let\u0027s write here plus P_2,"},{"Start":"05:54.600 ","End":"05:58.005","Text":"where P_2 is also an unknown."},{"Start":"05:58.005 ","End":"06:03.060","Text":"Now let\u0027s take a look at the energy at point 3."},{"Start":"06:03.060 ","End":"06:07.340","Text":"Before we begin, we are given this length L over here."},{"Start":"06:07.340 ","End":"06:09.605","Text":"Because we know that this is a major length."},{"Start":"06:09.605 ","End":"06:13.895","Text":"However, here, because we\u0027re not given the length between points 2 and 3,"},{"Start":"06:13.895 ","End":"06:19.460","Text":"we can assume that the distance between the two is small and insignificant."},{"Start":"06:19.460 ","End":"06:23.059","Text":"That means that although we\u0027re located inside"},{"Start":"06:23.059 ","End":"06:28.900","Text":"this tube and we\u0027re being told that this fluid has some viscosity."},{"Start":"06:28.900 ","End":"06:35.465","Text":"We learned that viscosity means energy loss because this length is so small,"},{"Start":"06:35.465 ","End":"06:38.645","Text":"we can disregard the energy loss due to"},{"Start":"06:38.645 ","End":"06:43.840","Text":"the frictional forces between the fluid and itself and the tube."},{"Start":"06:43.840 ","End":"06:50.280","Text":"That\u0027s number 1. Now let\u0027s write out the energy."},{"Start":"06:50.280 ","End":"06:53.150","Text":"We can see that we\u0027re located at the same heights and"},{"Start":"06:53.150 ","End":"06:56.560","Text":"everything inside the tube as point 2."},{"Start":"06:56.560 ","End":"06:59.705","Text":"That means that, we can say that the energy at point 3"},{"Start":"06:59.705 ","End":"07:03.260","Text":"is equal to the energy at point 2."},{"Start":"07:03.260 ","End":"07:08.195","Text":"But we can see that point 3 is after the pump."},{"Start":"07:08.195 ","End":"07:16.380","Text":"We know that the pump is providing energy or extra pressure to point number 3."},{"Start":"07:17.450 ","End":"07:21.525","Text":"We\u0027re going to add on that energy."},{"Start":"07:21.525 ","End":"07:27.375","Text":"The energy at point 3 is the energy at point 2 plus the energy gain by the pump."},{"Start":"07:27.375 ","End":"07:33.100","Text":"That\u0027s plus the energy gain Delta P by the pump."},{"Start":"07:34.940 ","End":"07:42.660","Text":"Of course, this Delta P due to the pump is also unknown."},{"Start":"07:43.220 ","End":"07:49.085","Text":"Now let\u0027s look at the energy at point number 4."},{"Start":"07:49.085 ","End":"07:53.150","Text":"As we can see, the energy at point number 4 is going to"},{"Start":"07:53.150 ","End":"07:57.900","Text":"be equal to the energy that was at point number 3"},{"Start":"07:58.480 ","End":"08:04.445","Text":"minus this energy loss that we have over here"},{"Start":"08:04.445 ","End":"08:10.040","Text":"from the fluid traveling from point number 3 to point number 4,"},{"Start":"08:10.040 ","End":"08:12.025","Text":"due to the viscosity."},{"Start":"08:12.025 ","End":"08:16.520","Text":"We have a viscous fluid traveling from point 3 to point 4,"},{"Start":"08:16.520 ","End":"08:21.690","Text":"which means that between these two points, it\u0027s losing energy."},{"Start":"08:23.690 ","End":"08:29.190","Text":"We have the energy at point 4 is equal to our energy at point"},{"Start":"08:29.190 ","End":"08:35.025","Text":"3 minus the energy that was lost in this direction."},{"Start":"08:35.025 ","End":"08:37.245","Text":"We know that that\u0027s the viscous energy."},{"Start":"08:37.245 ","End":"08:39.725","Text":"We saw this equation in the previous lesson."},{"Start":"08:39.725 ","End":"08:41.150","Text":"It\u0027s equal to 8,"},{"Start":"08:41.150 ","End":"08:44.120","Text":"some coefficient multiplied by Eta,"},{"Start":"08:44.120 ","End":"08:46.130","Text":"which is the viscosity of the fluid,"},{"Start":"08:46.130 ","End":"08:53.240","Text":"multiplied by L, which is the length of the tube or the distance traveled."},{"Start":"08:53.240 ","End":"08:57.180","Text":"That\u0027s the distance between point 3 and point 4."},{"Start":"08:57.180 ","End":"09:03.319","Text":"Then all of this is divided by the radius of the tube squared,"},{"Start":"09:03.319 ","End":"09:05.075","Text":"which is given to us in the question,"},{"Start":"09:05.075 ","End":"09:09.460","Text":"multiplied by the velocity of the fluid."},{"Start":"09:09.460 ","End":"09:12.740","Text":"The velocity of the fluid is an unknown."},{"Start":"09:12.740 ","End":"09:17.855","Text":"This is also a change in energy,"},{"Start":"09:17.855 ","End":"09:22.675","Text":"Delta P, but due to the viscosity of the fluid."},{"Start":"09:22.675 ","End":"09:29.405","Text":"This is a different Delta P to the energy provided by the pump."},{"Start":"09:29.405 ","End":"09:34.050","Text":"This is energy lost due to friction."},{"Start":"09:34.180 ","End":"09:39.035","Text":"Our unknowns in this question is the velocity which also appears here,"},{"Start":"09:39.035 ","End":"09:44.790","Text":"and also the added energy provided by the pump."},{"Start":"09:46.100 ","End":"09:51.020","Text":"In this question, it\u0027s just how do we solve questions with pumps."},{"Start":"09:51.020 ","End":"09:56.840","Text":"You see how to incorporate the energy added from the pump into your equations."},{"Start":"09:56.840 ","End":"09:58.340","Text":"This is energy added,"},{"Start":"09:58.340 ","End":"10:02.120","Text":"that\u0027s why we have a plus and this is energy subtracted between these two points,"},{"Start":"10:02.120 ","End":"10:04.190","Text":"so that\u0027s why there\u0027s a minus."},{"Start":"10:04.190 ","End":"10:10.280","Text":"Now any question that is asked which contains some viscous fluid"},{"Start":"10:10.280 ","End":"10:16.700","Text":"or a pump or both in some reservoir tube system."},{"Start":"10:16.700 ","End":"10:20.270","Text":"If we use this equation for power,"},{"Start":"10:20.270 ","End":"10:26.330","Text":"these equations or these type of equations for the energy at each point,"},{"Start":"10:26.330 ","End":"10:30.380","Text":"taking into account the energy added to the system by"},{"Start":"10:30.380 ","End":"10:35.355","Text":"a pump and the energy subtracted from a system due to the viscosity."},{"Start":"10:35.355 ","End":"10:41.030","Text":"If we also include our equation for volumetric flow rate,"},{"Start":"10:41.030 ","End":"10:42.590","Text":"which is Q_v,"},{"Start":"10:42.590 ","End":"10:46.640","Text":"which I\u0027m reminding you is equal to the velocity of the liquid"},{"Start":"10:46.640 ","End":"10:52.190","Text":"multiplied by the cross-sectional area of the tube."},{"Start":"10:52.190 ","End":"10:58.970","Text":"We can calculate any question that may be asked in this subject."},{"Start":"10:58.970 ","End":"11:02.070","Text":"That\u0027s the end of this lesson."}],"ID":12375},{"Watched":false,"Name":"Exercise 7","Duration":"10m 37s","ChapterTopicVideoID":11946,"CourseChapterTopicPlaylistID":84745,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.100","Text":"Hello. In this question,"},{"Start":"00:02.100 ","End":"00:06.480","Text":"we\u0027re being told that there\u0027s a tube of diameter D and length L,"},{"Start":"00:06.480 ","End":"00:09.765","Text":"which leads out from the bottom of a shallow pool."},{"Start":"00:09.765 ","End":"00:14.670","Text":"A man of mass M stands on a surface of area s,"},{"Start":"00:14.670 ","End":"00:18.255","Text":"which fits the pools dimensions perfectly."},{"Start":"00:18.255 ","End":"00:25.275","Text":"The surface with a man standing on top then pushes down on the air underneath it,"},{"Start":"00:25.275 ","End":"00:29.715","Text":"causing an increase in pressure on the pool water."},{"Start":"00:29.715 ","End":"00:32.895","Text":"How long will it take the pool to empty?"},{"Start":"00:32.895 ","End":"00:36.510","Text":"If we are told that the initial volume of the pool is"},{"Start":"00:36.510 ","End":"00:41.290","Text":"K and that the viscosity of the water is Eta."},{"Start":"00:42.800 ","End":"00:45.325","Text":"Before we begin the question,"},{"Start":"00:45.325 ","End":"00:50.575","Text":"we know that the surface of area S perfectly fits the pools dimension,"},{"Start":"00:50.575 ","End":"00:54.845","Text":"which means that no air can escape."},{"Start":"00:54.845 ","End":"00:59.610","Text":"This is completely enclosed,"},{"Start":"00:59.610 ","End":"01:04.360","Text":"and our 2nd assumption is that the pool is very"},{"Start":"01:04.360 ","End":"01:09.800","Text":"large compared to the cross-sectional area of our tube."},{"Start":"01:09.980 ","End":"01:14.905","Text":"That means that we can begin with the energy at Point number 1,"},{"Start":"01:14.905 ","End":"01:18.425","Text":"so that\u0027s at the surface of the water."},{"Start":"01:18.425 ","End":"01:24.985","Text":"First of all, we know that we\u0027re located at height 0 because we\u0027re at a shallow pool,"},{"Start":"01:24.985 ","End":"01:28.610","Text":"and we can see from the diagram it\u0027s approximately at height 0,"},{"Start":"01:28.610 ","End":"01:31.280","Text":"so that means that there\u0027s no potential energy."},{"Start":"01:31.280 ","End":"01:37.565","Text":"Also, because of the size of the pool relative to the cross-sectional area of the tube."},{"Start":"01:37.565 ","End":"01:41.495","Text":"That means that the velocity of the water over here is also 0."},{"Start":"01:41.495 ","End":"01:44.765","Text":"That means that there\u0027s no kinetic energy here."},{"Start":"01:44.765 ","End":"01:48.965","Text":"The next thing that we need to look at is the elastic energy."},{"Start":"01:48.965 ","End":"01:53.330","Text":"We have elastic energy and it comes in the form of pressure,"},{"Start":"01:53.330 ","End":"01:57.895","Text":"so festival we know that there\u0027s air over here."},{"Start":"01:57.895 ","End":"02:02.420","Text":"The first thing we know is that we\u0027re going to have atmospheric pressure."},{"Start":"02:02.420 ","End":"02:04.580","Text":"We already mentioned in,"},{"Start":"02:04.580 ","End":"02:06.200","Text":"and I\u0027ll say it again."},{"Start":"02:06.200 ","End":"02:12.140","Text":"We always have atmospheric pressure unless we\u0027re located in some vacuum."},{"Start":"02:12.140 ","End":"02:19.775","Text":"There\u0027s always atmospheric pressure at least and then in certain cases like we have here,"},{"Start":"02:19.775 ","End":"02:23.570","Text":"there\u0027s an added pressure due to the man standing on the surface,"},{"Start":"02:23.570 ","End":"02:24.935","Text":"which we\u0027re going to add now."},{"Start":"02:24.935 ","End":"02:27.335","Text":"But there\u0027s always at least atmospheric pressure"},{"Start":"02:27.335 ","End":"02:31.075","Text":"unless we\u0027re being told that we\u0027re located in a vacuum."},{"Start":"02:31.075 ","End":"02:34.565","Text":"What exactly is pressure in general?"},{"Start":"02:34.565 ","End":"02:37.100","Text":"It\u0027s force divided by area."},{"Start":"02:37.100 ","End":"02:43.165","Text":"We know that there\u0027s also the pressure due to the man standing on the surface."},{"Start":"02:43.165 ","End":"02:45.170","Text":"We\u0027re going to add in that pressure,"},{"Start":"02:45.170 ","End":"02:47.570","Text":"which we just said is force divided by area."},{"Start":"02:47.570 ","End":"02:53.085","Text":"What is force? We\u0027re using the force due to his weight."},{"Start":"02:53.085 ","End":"02:59.085","Text":"That\u0027s going to be m and g and then divided by area."},{"Start":"02:59.085 ","End":"03:03.465","Text":"We\u0027re being told that he\u0027s standing on the surface of area S,"},{"Start":"03:03.465 ","End":"03:10.010","Text":"so that means that it\u0027s mg divided by S. That\u0027s the energy at 0.1."},{"Start":"03:10.010 ","End":"03:13.015","Text":"Let\u0027s see what the energy is at 0.2."},{"Start":"03:13.015 ","End":"03:19.610","Text":"First of all, we know that the energy at 0.2 is equal to the energy at"},{"Start":"03:19.610 ","End":"03:28.010","Text":"0.1 minus the energy lost in the journey of the water from 0.1-0.2."},{"Start":"03:28.010 ","End":"03:29.795","Text":"Why is there energy loss?"},{"Start":"03:29.795 ","End":"03:33.950","Text":"Because we\u0027re being told that the water has viscosity Eta."},{"Start":"03:33.950 ","End":"03:38.885","Text":"That means minus our energy loss,"},{"Start":"03:38.885 ","End":"03:43.110","Text":"which of course is Delta P is just a change in pressure."},{"Start":"03:45.130 ","End":"03:50.610","Text":"Let\u0027s see what Epsilon 2 is equal to."},{"Start":"03:50.610 ","End":"03:53.230","Text":"Epsilon 2 let\u0027s see over here."},{"Start":"03:53.230 ","End":"03:56.980","Text":"We have height, which means that we have potential energy,"},{"Start":"03:56.980 ","End":"03:59.860","Text":"so that\u0027s equal to Rho gh."},{"Start":"03:59.860 ","End":"04:06.145","Text":"Second of all, we know that our water is flowing through the end of the tube,"},{"Start":"04:06.145 ","End":"04:07.825","Text":"which means that there\u0027s velocity,"},{"Start":"04:07.825 ","End":"04:10.060","Text":"which means that there\u0027s kinetic energy,"},{"Start":"04:10.060 ","End":"04:12.875","Text":"so that\u0027s 1/2Rho v^2,"},{"Start":"04:12.875 ","End":"04:15.240","Text":"where of course v is an unknown."},{"Start":"04:15.240 ","End":"04:26.160","Text":"and the water over here is subject to atmospheric pressures, so that\u0027s P_atm."},{"Start":"04:26.160 ","End":"04:28.890","Text":"What is our energy loss?"},{"Start":"04:28.890 ","End":"04:35.545","Text":"Our Delta P is equal to the equation that we learned a few lessons ago."},{"Start":"04:35.545 ","End":"04:41.600","Text":"That is equal to 8 multiplied by the viscosity of the water,"},{"Start":"04:41.600 ","End":"04:45.140","Text":"multiplied by the length of the tube,"},{"Start":"04:45.140 ","End":"04:49.955","Text":"divided by the radius of the tube^2,"},{"Start":"04:49.955 ","End":"04:54.635","Text":"multiplied by the velocity of the water and again,"},{"Start":"04:54.635 ","End":"04:56.480","Text":"we still don\u0027t know what this velocity is."},{"Start":"04:56.480 ","End":"04:58.210","Text":"This is an unknown."},{"Start":"04:58.210 ","End":"05:04.400","Text":"What we can do is we can plug all of this back into this equation over here."},{"Start":"05:04.400 ","End":"05:06.140","Text":"Let\u0027s write it over here,"},{"Start":"05:06.140 ","End":"05:12.579","Text":"so we have Epsilon 2 which is equal to Rho gh plus"},{"Start":"05:12.579 ","End":"05:20.180","Text":"1/2Rho v^2 plus our atmospheric pressure"},{"Start":"05:20.180 ","End":"05:23.270","Text":"is equal to Epsilon 1,"},{"Start":"05:23.270 ","End":"05:27.275","Text":"which is our atmospheric pressure,"},{"Start":"05:27.275 ","End":"05:29.855","Text":"plus the pressure due to the man,"},{"Start":"05:29.855 ","End":"05:36.500","Text":"so that\u0027s mg divided by S minus our energy loss Delta P,"},{"Start":"05:36.500 ","End":"05:45.990","Text":"so minus 8 Eta L divided by r^2v."},{"Start":"05:47.540 ","End":"05:54.050","Text":"This is just these equations plugged into this equation over here."},{"Start":"05:54.050 ","End":"05:57.590","Text":"The energy at 0.2 is going to be equal to the energy at"},{"Start":"05:57.590 ","End":"06:02.355","Text":"0.1 minus the energy that was lost on the journey,"},{"Start":"06:02.355 ","End":"06:05.620","Text":"so that\u0027s why there\u0027s a minus here."},{"Start":"06:05.960 ","End":"06:09.635","Text":"We can see that on both sides of the equation,"},{"Start":"06:09.635 ","End":"06:13.420","Text":"we have this atmospheric pressure."},{"Start":"06:13.420 ","End":"06:18.470","Text":"We can cross it off from both sides of the equation and then we can see that we"},{"Start":"06:18.470 ","End":"06:24.370","Text":"can simply rearrange this entire equation to have a quadratic equation,"},{"Start":"06:24.370 ","End":"06:28.120","Text":"so our variable is the velocity of the water."},{"Start":"06:28.120 ","End":"06:32.470","Text":"Here we have a coefficient for v^2."},{"Start":"06:32.470 ","End":"06:36.460","Text":"Here we have a coefficient for v and then we"},{"Start":"06:36.460 ","End":"06:42.280","Text":"have our constants which are independent of our variable."},{"Start":"06:42.280 ","End":"06:46.480","Text":"If we subtract everything from 1 of the sides"},{"Start":"06:46.480 ","End":"06:50.065","Text":"that can move the whole equation to 1 side of the equal sign."},{"Start":"06:50.065 ","End":"06:51.894","Text":"We have a quadratic equation,"},{"Start":"06:51.894 ","End":"06:54.430","Text":"which we all know how to solve,"},{"Start":"06:54.430 ","End":"07:00.490","Text":"and then we can have onset for the velocity of the water."},{"Start":"07:01.370 ","End":"07:05.965","Text":"Let\u0027s imagine that we\u0027ve solved the quadratic equation"},{"Start":"07:05.965 ","End":"07:10.950","Text":"and rarely we get a very long expression."},{"Start":"07:10.950 ","End":"07:18.745","Text":"Let\u0027s say that we get that the velocity of the water is equal to x dot,"},{"Start":"07:18.745 ","End":"07:24.705","Text":"so this is the velocity of the water, some value."},{"Start":"07:24.705 ","End":"07:27.465","Text":"Because it\u0027s really some long expression."},{"Start":"07:27.465 ","End":"07:32.510","Text":"The question is, how long will it take the pool to empty if we are told that"},{"Start":"07:32.510 ","End":"07:37.830","Text":"the initial volume of the pool is k and that the viscosity of the water is Eta?"},{"Start":"07:37.830 ","End":"07:41.615","Text":"We\u0027re asking how long will it take for the pool to empty?"},{"Start":"07:41.615 ","End":"07:46.595","Text":"We\u0027re going to use the idea of our volumetric flow rate,"},{"Start":"07:46.595 ","End":"07:49.555","Text":"so that is equal to Q."},{"Start":"07:49.555 ","End":"07:54.220","Text":"That\u0027s our volumetric flow rate and that is equal to the velocity of"},{"Start":"07:54.220 ","End":"07:59.365","Text":"the water multiplied by the cross-sectional area of the tube,"},{"Start":"07:59.365 ","End":"08:01.300","Text":"so in our case,"},{"Start":"08:01.300 ","End":"08:03.575","Text":"our volume or velocity,"},{"Start":"08:03.575 ","End":"08:09.640","Text":"sorry, is x dot and we\u0027re multiplying by the cross-sectional area of the tube."},{"Start":"08:09.640 ","End":"08:12.625","Text":"We know that the tube is of diameter d,"},{"Start":"08:12.625 ","End":"08:16.150","Text":"and we know that diameter is twice the radius."},{"Start":"08:16.150 ","End":"08:20.240","Text":"That means that the area if L tube is some cylinder,"},{"Start":"08:20.240 ","End":"08:22.930","Text":"so the cross-sectional areas that of a circle."},{"Start":"08:22.930 ","End":"08:26.124","Text":"The area of a circle is Pi r^2."},{"Start":"08:26.124 ","End":"08:28.675","Text":"But we\u0027re not given r, we\u0027re given d,"},{"Start":"08:28.675 ","End":"08:30.160","Text":"which is twice r,"},{"Start":"08:30.160 ","End":"08:34.550","Text":"so that\u0027s (Pi d divided by 2)^2."},{"Start":"08:34.790 ","End":"08:40.120","Text":"This is our area, Pi r^2."},{"Start":"08:41.270 ","End":"08:45.195","Text":"We found the volumetric flow rate,"},{"Start":"08:45.195 ","End":"08:50.360","Text":"so scroll down a bit because we\u0027re being asked how long."},{"Start":"08:50.360 ","End":"08:53.900","Text":"We\u0027re trying to find some kind of value for time."},{"Start":"08:53.900 ","End":"08:57.500","Text":"We know that our volumetric flow rate,"},{"Start":"08:57.500 ","End":"09:03.500","Text":"so how much passes in a given time multiplied by a given time,"},{"Start":"09:03.500 ","End":"09:09.590","Text":"is going to be equal to the total volume of water inside the pool,"},{"Start":"09:09.590 ","End":"09:17.140","Text":"which we\u0027re given is equal to K. If we set this to this,"},{"Start":"09:17.140 ","End":"09:21.620","Text":"than what we can do is we can isolate out our time"},{"Start":"09:21.860 ","End":"09:28.660","Text":"and we\u0027ll get that the total volume of the pool divided by our volumetric flow rate,"},{"Start":"09:28.660 ","End":"09:31.550","Text":"which is what we found over here,"},{"Start":"09:32.310 ","End":"09:37.285","Text":"is going to be equal to our onset,"},{"Start":"09:37.285 ","End":"09:40.750","Text":"the time taken for the pool to empty,"},{"Start":"09:41.210 ","End":"09:45.700","Text":"so this is the time taken which is given by K,"},{"Start":"09:45.700 ","End":"09:51.610","Text":"the total volume of the pool divided by the volumetric flow rate."},{"Start":"09:51.610 ","End":"09:56.660","Text":"K was given to us in the question and our volumetric flow rate is simply"},{"Start":"09:56.660 ","End":"10:02.140","Text":"our velocity of the water multiplied by the cross-sectional area of the tube."},{"Start":"10:02.140 ","End":"10:04.790","Text":"The cross-sectional area of the tube we have,"},{"Start":"10:04.790 ","End":"10:06.080","Text":"it\u0027s given to us in the question."},{"Start":"10:06.080 ","End":"10:11.570","Text":"We\u0027re told that the tube is of diameter d and the velocity we had to find."},{"Start":"10:11.570 ","End":"10:19.310","Text":"The velocity we found by working out the energy at 0.1 and the energy at 0.2."},{"Start":"10:19.310 ","End":"10:22.460","Text":"Taking into account the energy loss due to the viscosity of"},{"Start":"10:22.460 ","End":"10:27.920","Text":"the water and then we put that into an equation,"},{"Start":"10:27.920 ","End":"10:31.220","Text":"solved it using the quadratic formula,"},{"Start":"10:31.220 ","End":"10:35.545","Text":"and then we found a value for the velocity."},{"Start":"10:35.545 ","End":"10:38.570","Text":"That\u0027s the end of this lesson."}],"ID":12376}],"Thumbnail":null,"ID":84745}]

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