1 00:00:16,100 --> 00:00:20,880 Welcome to this 7 th lecture in this NPTEL course on fluid mechanics for chemical engineering 2 00:00:20,880 --> 00:00:26,290 undergraduate students. In the last lecture, we discuss the fundamental equation of fluid 3 00:00:26,290 --> 00:00:26,910 statics. 4 00:00:26,910 --> 00:00:33,910 We were discussing fluids under static conditions, and we derive a fundamental equation for fluids 5 00:00:37,430 --> 00:00:43,980 that is, that are present and under the influence of a gravitational field. So we started by 6 00:00:43,980 --> 00:00:50,980 taking a volume element, volume element of a fluid under the influence of gravity, and 7 00:00:51,110 --> 00:00:58,110 in the limit when, so we took a co-ordinate system x, y, and z. In the limit when, this 8 00:00:58,360 --> 00:01:05,360 volume element dimensions of the volume element shrinks, we derive a fundamental equation 9 00:01:06,990 --> 00:01:13,990 for fluid statics is equal to 0 minus the gradient of pressure plus density of the fluid 10 00:01:15,520 --> 00:01:22,520 times gravity is equal to 0. This is a fundamental equation that is of use in describing several 11 00:01:23,460 --> 00:01:27,159 features of fluid statics, fluids under static conditions. 12 00:01:27,159 --> 00:01:34,159 It is customer in to point the gravity vector along the negative z direction as I shown 13 00:01:34,280 --> 00:01:41,280 here. So there are three unit vectors i, j, and k, in the along the x, y, and z direction. 14 00:01:43,170 --> 00:01:50,170 So the acceleration due to gravity vector is given by minus g times k, where k is a 15 00:01:50,579 --> 00:01:56,429 unit vector in the positive z direction, but g is pointing the negative direction. So, 16 00:01:56,429 --> 00:02:03,429 minus g happens, because of that. So, when we substitute this, when we referred this 17 00:02:04,149 --> 00:02:09,340 equation to this co-ordinate system, for this particular co-ordinate system, this equation 18 00:02:09,340 --> 00:02:12,819 is very general, because it has no reference to any co-ordinate system. 19 00:02:12,819 --> 00:02:19,819 This is general, when applied to the co-ordinate system shown here, we get minus partial p 20 00:02:32,220 --> 00:02:39,220 partial x. Now, the vector g can be written as g x times i plus g y times j plus g z times 21 00:02:45,170 --> 00:02:51,340 k. It can be dissolved into the three cartesian directions, and in this co-ordinate system 22 00:02:51,340 --> 00:02:58,340 g x is 0, and g y is 0, and g z is minus g. So, we can proceed further by saying that 23 00:03:01,120 --> 00:03:08,120 d p by minus d p dx is 0, because g x is 0 and minus d p dy is 0, because g y is 0, and 24 00:03:08,670 --> 00:03:14,840 minus d p dz minus rho g is 0, in the z direction. 25 00:03:14,840 --> 00:03:21,840 This implies that p is independent of x and y and it varies only in the z direction. And 26 00:03:27,590 --> 00:03:33,300 this implies, since p is independent of x and y, the partial derivative becomes a normal 27 00:03:33,300 --> 00:03:40,300 derivative minus d p dz is rho g or d p dz is minus rho g. We can integrate this, if 28 00:03:50,230 --> 00:03:57,230 rho is a constant and g is normally a constant under terrestrial conditions, the acceleration 29 00:03:59,570 --> 00:04:06,570 due to gravity on the surface of this a constant. Then d p dz is minus rho g can be integrated 30 00:04:09,120 --> 00:04:16,120 as follows, between any two points, minus rho g dz. Since rho and g are constants, we 31 00:04:22,590 --> 00:04:29,590 can pull them out. So, integral d p between any two points p naught and p is minus rho 32 00:04:30,539 --> 00:04:37,539 g integral z is z naught to any z dz. 33 00:04:37,710 --> 00:04:44,710 So, p minus p naught is nothing but minus rho g times z minus z naught or p minus p 34 00:04:48,910 --> 00:04:55,910 naught is rho g times z naught minus z, after sawing the minus sign. 35 00:04:56,860 --> 00:05:03,860 Now this can also be written as p at any z minus p naught is rho g times z naught minus 36 00:05:07,920 --> 00:05:14,920 z. It is customary in fluid mechanics, so you have the z co-ordinate going up like this. 37 00:05:18,160 --> 00:05:25,160 Suppose you have a fluid, suppose you have a water body which is exposed to atmosphere 38 00:05:25,460 --> 00:05:32,460 air. Where the pressure is p atmosphere, the pressure of air in the atmosphere is due to 39 00:05:36,530 --> 00:05:43,530 the weight of the air that is present above a given elevation. So, at c level the pressure 40 00:05:43,630 --> 00:05:47,590 of air is conventionally called the atmospheric pressure. That is precisely because of the 41 00:05:47,590 --> 00:05:52,130 weight of the air that is present above the level, so this is known. 42 00:05:52,130 --> 00:05:59,130 So, if you call this location as z equals z naught which is the free surface, where 43 00:05:59,330 --> 00:06:06,330 p is p naught is p atmosphere. Then p at any location z is p naught which is p atmosphere 44 00:06:10,430 --> 00:06:17,430 plus rho g times z naught minus z. Z is any location and z naught is this location, so 45 00:06:20,080 --> 00:06:25,850 z naught minus z is this depth from the free surface. So this is conventionally denoted 46 00:06:25,850 --> 00:06:32,850 by the letter h, so p at any location in the liquid is p atmosphere plus rho g h. This 47 00:06:37,790 --> 00:06:42,430 is something that you may be familiar with from your earlier classes in physics. Where 48 00:06:42,430 --> 00:06:48,990 the pressure in a column of liquid is an increase with vertical distance in a linear manner 49 00:06:48,990 --> 00:06:53,810 and that is precisely because of the fact that the pressure. Suppose, you take a column 50 00:06:53,810 --> 00:07:00,770 of liquid and this is the atmospheric pressure, so pressure always acts normally to a surface. 51 00:07:00,770 --> 00:07:07,139 So if you look at the pressure on this side, this pressure will have to be greater than 52 00:07:07,139 --> 00:07:10,730 atmospheric pressure. Because of the fact that under static conditions, the fluid will 53 00:07:10,730 --> 00:07:15,919 have to balance the weight of this liquid column, which is precisely given by rho g 54 00:07:15,919 --> 00:07:22,919 h times the area of the element. So that is the precise physical meaning of this equation. 55 00:07:24,830 --> 00:07:31,830 And this is valid only for incompressible fluids, where rho is constant. Now, let me 56 00:07:39,020 --> 00:07:46,020 just spend a couple of minutes commenting on the nature of atmospheric pressure. P atmosphere, 57 00:07:46,580 --> 00:07:53,580 the atmospheric pressure is precisely the pressure of the air that is present in the 58 00:08:00,449 --> 00:08:07,449 atmosphere. And so, if you consider c level that is ground level, then if you take a cylindrical 59 00:08:08,720 --> 00:08:15,720 column of air, the weight of this air is precisely the pressure that you will feel at the ground 60 00:08:17,630 --> 00:08:24,199 level. And at pure vacuum, that is when there is no air, when you go far away from the ground 61 00:08:24,199 --> 00:08:30,080 level, far into the atmosphere there is the density of air will decrease significantly 62 00:08:30,080 --> 00:08:35,449 the pressure will also decrease. So at pure vacuum, the pressure of air is 0. 63 00:08:35,449 --> 00:08:39,399 So there is no air molecules, there is no pressure and the pressure at the ground level 64 00:08:39,399 --> 00:08:45,870 is called the atmospheric pressure. And this atmospheric pressure is roughly 10 to the 65 00:08:45,870 --> 00:08:52,870 5 pascals, is 10 to 5 newton per meter square in S I units. The other thing we discussed 66 00:08:53,490 --> 00:09:00,490 was the role of compressibility. So, if air is treated compressible, to be an ideal gas, 67 00:09:08,730 --> 00:09:15,730 so we said that p is rho the specific gas constants time T, then you had minus d p dz 68 00:09:19,949 --> 00:09:26,949 is d p dz is minus rho g. Instead of rho, so this implies rho is p by R g T, so I can 69 00:09:31,269 --> 00:09:38,269 eliminate d p by dz is minus p by R g T times g, acceleration due to gravity. 70 00:09:42,119 --> 00:09:49,119 So I can integrate this in the following way d p by p integral, this is minus g by R g 71 00:09:51,300 --> 00:09:58,300 T integral dz. So, logarithm of p is nothing but minus g by R g T z plus some constant. 72 00:10:05,189 --> 00:10:12,189 Which can be simplified to write as p is p naught times exponential of minus g by R g 73 00:10:18,309 --> 00:10:24,749 T and T is assumed to be constant in this analysis. The air is a constant temperature, 74 00:10:24,749 --> 00:10:31,749 so let us call that constant T naught times z. This p naught is the value, pressure at 75 00:10:32,410 --> 00:10:39,079 z equals 0. That is the condition, boundary condition we use to fix this constant. This 76 00:10:39,079 --> 00:10:46,079 constant is fixed by saying that the pressure at a z equal to 0 is p naught. 77 00:10:46,179 --> 00:10:53,179 So p is p naught times exponential of minus g by R g T naught z. This is an equation that 78 00:10:56,579 --> 00:11:03,579 is valid, if air is treated compressible. But we also said that, if the value of this 79 00:11:04,189 --> 00:11:11,189 exponent g z by R g T naught is small compared to 1. This is a dimensional less a group, 80 00:11:13,490 --> 00:11:20,490 so if this number is small compare to 1, you can Taylor expand and write p is p naught 81 00:11:23,170 --> 00:11:29,559 times 1 minus. So Taylor expand, e to the minus x is approximately 1 minus x, if x is 82 00:11:29,559 --> 00:11:36,559 small. So 1 minus g minus z by R g T naught, so p is p naught minus p naught by R g T naught 83 00:11:44,679 --> 00:11:51,679 g z. From ideal gas law, this is nothing but rho naught, so p is p naught minus rho naught 84 00:11:52,009 --> 00:11:57,819 g z. So, this is similar to the incompressible equation that we derived by treating density 85 00:11:57,819 --> 00:11:58,689 to be constant. 86 00:11:58,689 --> 00:12:04,209 So, this linear variation is a simplification of this exponential variation of pressure 87 00:12:04,209 --> 00:12:11,209 and is valid if g by R g T naught is g z by R g T naught is less than 0.1 or if z is less 88 00:12:16,389 --> 00:12:21,980 than 800 meter. We can treat the pressure variation even in air which is in general 89 00:12:21,980 --> 00:12:28,980 a compressible system to be a linear variation. So, the other thing the next thing we will 90 00:12:32,360 --> 00:12:39,360 do is to apply the fundamental equation of hydrostatics to what is called manometry. 91 00:12:45,779 --> 00:12:52,779 Manometry is that branch of fluid mechanics, which deals with measurement of pressures. 92 00:13:01,220 --> 00:13:08,220 And the specific device we are going to use is called a U-tube manometer. So what is the 93 00:13:14,439 --> 00:13:21,439 construction? Well, it consists of U shaped tube and one end is exposed to atmosphere. 94 00:13:32,139 --> 00:13:39,029 The other end is joined to a region, whose pressure we want to know. So here, they let 95 00:13:39,029 --> 00:13:46,029 us call this point as A the pressure here is not known, so pressure unknown. The manometer 96 00:13:50,040 --> 00:13:57,040 is filled with working liquid is called the manometric liquid. Now this is open to atmosphere. 97 00:14:01,579 --> 00:14:07,379 Now the idea is to relate this pressure, unknown pressure to the atmospheric pressure which 98 00:14:07,379 --> 00:14:14,379 is known. So, how do we go about doing this? Let us now draw some label some heights here, 99 00:14:15,199 --> 00:14:22,199 let us call this height as h 3, let us call this height as h 4, let us call this height 100 00:14:36,149 --> 00:14:43,149 as h 2 and between point A and this interfaces h 1. So we want to now apply the principle 101 00:14:51,860 --> 00:14:58,649 of the result from fundamental equation of hydrostatics, that p is p atmosphere plus 102 00:14:58,649 --> 00:15:05,649 rho g h. Now, if you go from point here to here, the pressure will increase, this is 103 00:15:08,649 --> 00:15:13,170 atmospheric pressure, the pressure will increase because of the weight of air, but that is 104 00:15:13,170 --> 00:15:20,170 negligible, so we will not worry about this. From this point, now we are going to between 105 00:15:20,179 --> 00:15:23,779 these two points, between these two levels the pressure at this point and the pressure 106 00:15:23,779 --> 00:15:28,670 at this point is the same. Because the pressure in this manometric liquid is a function only 107 00:15:28,670 --> 00:15:33,540 of the elevation and the elevations are the same to pressure at this point, which is called, 108 00:15:33,540 --> 00:15:40,540 let us call it B and the pressure at this point D must be the same. So p B must be equal 109 00:15:40,540 --> 00:15:47,540 to p D, because the pressure in the manometric liquid is the function only of the elevation. 110 00:15:47,709 --> 00:15:54,239 And since these two points are the same elevation, p B must be equal to p D. If p B is, so how 111 00:15:54,239 --> 00:16:01,239 do we get p B in terms of p A. Well, p B is nothing but p A plus rho 1. 112 00:16:01,869 --> 00:16:08,059 Let us call this liquid rho 1, equate the density rho 1, acceleration due to gravity 113 00:16:08,059 --> 00:16:15,059 times, this column height h 1, rho 1, g h 1. This is the pressure at point B. Now p 114 00:16:19,209 --> 00:16:24,879 D, the pressure at this point D is given by the pressure at this point which is approximately 115 00:16:24,879 --> 00:16:31,879 atmospheric pressure plus rho manometric liquid. Let us call it rho m, so density of this liquid 116 00:16:31,910 --> 00:16:38,910 the manometric liquid is rho m g. This total height is h 4, this is h 4 and so this total 117 00:16:42,359 --> 00:16:48,749 height is h 4, this is h 2, so that this weight of this, the height of this column is h 4 118 00:16:48,749 --> 00:16:55,749 minus h 2. So, we can write p A minus p atmosphere. The 119 00:16:56,749 --> 00:17:01,819 difference in the value of the pressure at the point A minus atmospheric pressure is 120 00:17:01,819 --> 00:17:08,819 nothing but rho m g h 4 minus h 2 minus rho 1 g h 1. So, by measuring these two heights, 121 00:17:16,579 --> 00:17:23,139 by measuring the height difference h 4 minus h 2, h 4 minus h 2 is basically this height. 122 00:17:23,139 --> 00:17:30,139 I am measuring this height that is this and typically the density of the manometric liquid 123 00:17:30,899 --> 00:17:37,899 is very large compared to density of the working fluid. So this is usually negligible, so we can get the difference 124 00:17:47,080 --> 00:17:53,169 in the value of pressure at point A from the atmospheric pressure to be the density of 125 00:17:53,169 --> 00:17:57,700 the manometric liquid times, the acceleration due to gravity times, and the height difference 126 00:17:57,700 --> 00:18:03,649 between the two lengths of the manometer. So this is a fundamental equation of manometry. 127 00:18:03,649 --> 00:18:10,649 And by just simply measuring the height, we measure this to obtain the unknown pressure. 128 00:18:12,740 --> 00:18:19,740 Now, this difference is called the gage pressure, as I mentioned in the last lecture. The difference 129 00:18:19,870 --> 00:18:24,389 between the values of a pressure at a point from the atmospheric pressure is called the 130 00:18:24,389 --> 00:18:29,269 gage pressure, and because that is what is measured by a pressure measuring devices, 131 00:18:29,269 --> 00:18:32,820 such as the manometers. 132 00:18:32,820 --> 00:18:39,179 Now how do, how does one measure atmospheric pressure itself? In order to that, we have, 133 00:18:39,179 --> 00:18:46,179 what is called a barometer. Usually the manometric liquid is mercury rho m is the typically mercury, 134 00:18:51,639 --> 00:18:58,639 which is 13.6 times 10 to the 3 kg per meter cube, so very large density liquid. Now, a 135 00:19:00,399 --> 00:19:07,399 barometer is used to measure atmospheric pressure. So, this is the device that is used to obtain 136 00:19:16,590 --> 00:19:22,409 what to estimate the atmospheric pressure. How is it done? Well, the construction of 137 00:19:22,409 --> 00:19:29,409 the barometer is very simple. You take a trough of mercury and then in this trough, we invert 138 00:19:34,289 --> 00:19:41,289 a tube which already has mercury in it. Now the mercury in this tube will raise. 139 00:19:42,380 --> 00:19:49,380 So this is mercury, will this raise to in this tube will rise to a particular height, 140 00:19:49,769 --> 00:19:56,769 this is air atmospheric pressure, p atmosphere. Now in this part, it is largely vacuum, it 141 00:19:58,919 --> 00:20:04,730 has some molecules of mercury vapor. But it is since a vapor pressure of mercury is very 142 00:20:04,730 --> 00:20:10,799 very small, this is essentially a vacuum, there is no pressure here. So the pressure 143 00:20:10,799 --> 00:20:17,799 at this point is atmospheric pressure, from the fact that this liquid is exposed to air. 144 00:20:19,039 --> 00:20:23,860 Now, the pressure at this point must be the same as the pressure at this point, because 145 00:20:23,860 --> 00:20:28,850 this is connected by the same liquid is connecting these two points. 146 00:20:28,850 --> 00:20:33,429 And since they are at the same elevation, there cannot be any pressure difference. So 147 00:20:33,429 --> 00:20:40,429 the pressure at this point is also atmospheric. So the pressure at this point, which is atmospheric 148 00:20:41,500 --> 00:20:48,500 pressure is the pressure at point A, which is let us call it p A which is 0, because 149 00:20:49,039 --> 00:20:54,950 it is a vacuum plus rho mercury g times h. Where h is the column height of the column 150 00:20:54,950 --> 00:21:01,950 of mercury that is present in the tube. So, p atmosphere which is what we want to calculate 151 00:21:04,909 --> 00:21:11,909 want this. Can be obtained by this measuring what is the height of mercury. So, typically 152 00:21:13,440 --> 00:21:20,440 this height is 760 mm at normal conditions of temperature and altitudes or 76 centimeters. 153 00:21:25,179 --> 00:21:32,179 So, the atmospheric pressure is nothing but density of mercury, this is 13.6 times 10 154 00:21:35,679 --> 00:21:42,679 to the 3 times 9.8 meter per second square times 0.76 meters, this is height. When we 155 00:21:54,049 --> 00:22:01,049 do all these, we get atmospheric pressure to be 1.03 times 10 to the 5 newton per meter 156 00:22:03,179 --> 00:22:10,179 square or 1.03 times 10 to the 5 pascals. So the barometer is the simple construct that 157 00:22:11,130 --> 00:22:18,090 is used to calculate the atmospheric pressure. So, the atmospheric pressure in several text 158 00:22:18,090 --> 00:22:25,090 books or even hand books is denoted in terms of S I units as 1.013 times 10 to the 5 pascals 159 00:22:27,139 --> 00:22:34,139 or it is written as 76 mm Hg. Because that is the height of the mercury column that raises 160 00:22:35,230 --> 00:22:42,230 to counter balance atmospheric pressure. So, it is sometimes refer in terms of mm h g and 161 00:22:42,690 --> 00:22:49,690 this is also trivially called as one atmosphere. Because it is a normal pressure that is encountered 162 00:22:50,009 --> 00:22:54,320 in atmosphere, it is of the order of 10 to the 5 pascals. So, one atmosphere is essentially 163 00:22:54,320 --> 00:23:01,320 1.013 times 10 to the 5 pascals is also 760 mm of Hg, so mercury column. So, all these 164 00:23:01,980 --> 00:23:08,980 are used interchangeably while reporting the values of pressure. Now, that we have done 165 00:23:11,110 --> 00:23:18,110 all this, now we are going to worry about the next topic, which is forces under static 166 00:23:25,259 --> 00:23:32,259 as forces on solid surfaces or forces on sub merged, surfaces under static conditions. 167 00:23:46,100 --> 00:23:52,100 So the issue that we are going to understand is the following. Suppose, you have a liquid 168 00:23:52,100 --> 00:23:59,100 surface, where it is exposed to atmosphere, p atmosphere, and within the liquid, there 169 00:23:59,269 --> 00:24:05,830 is a surface, a planer surface, so we will look at planer surface for simplicity, so 170 00:24:05,830 --> 00:24:12,830 we look at a planer surface. So you have a plane and this plane extends in the third 171 00:24:16,370 --> 00:24:23,370 direction. So let me put co-ordinate system, this is the z co-ordinate, this is the y co-ordinate, 172 00:24:24,379 --> 00:24:29,250 and the x co-ordinate runs in the direction perpendicular to the board. 173 00:24:29,250 --> 00:24:36,250 So, if you look at the x y, this surface may look like this. Now, since and in the y z 174 00:24:47,519 --> 00:24:52,230 plane, this will look like a line, because this is a planer surface, this is like a plate 175 00:24:52,230 --> 00:24:59,230 with some arbitrary shape. Now, we want to know and this is let us say liquid like water. 176 00:25:02,309 --> 00:25:08,289 We want to know and gravity is acting in this direction and this is the liquid surface, 177 00:25:08,289 --> 00:25:13,720 where the pressure is atmospheric. We want to know, what is the force that is exerted 178 00:25:13,720 --> 00:25:19,110 by the fluid on one side of this solid surface, on this planer surface. 179 00:25:19,110 --> 00:25:26,110 The question that we are asking, the question that we want to answer is what is the force? 180 00:25:34,940 --> 00:25:41,940 F R that is exerted by the fluid. Now, we also want to know the point at which the point 181 00:25:55,070 --> 00:26:02,070 x prime, y prime, at which the resultant force acts. So we want to calculate these two things. 182 00:26:09,529 --> 00:26:15,159 Now, why is this thing important? Well, this thing is important in several applications, 183 00:26:15,159 --> 00:26:22,159 where suppose you are interested in construction of a dam. So, a dam is something that stores 184 00:26:22,990 --> 00:26:29,990 or obstructs water, this is water. And this surface has to be constructed in such a manner, 185 00:26:31,350 --> 00:26:37,379 that it withstands the force due to the water. And the reason why hydrostatics is different 186 00:26:37,379 --> 00:26:43,000 is, because the pressure varies with depth. So, the pressure at this point is merely atmospheric, 187 00:26:43,000 --> 00:26:47,100 but as you go down the pressure will increase linearly. 188 00:26:47,100 --> 00:26:51,460 So the force will not be a, the force cannot be obtained by simply multiplying the pressure 189 00:26:51,460 --> 00:26:56,659 by the area. It as to be obtained by integrating the pressure with respect to the vertical 190 00:26:56,659 --> 00:27:02,990 co-ordinate. So this what we want to do. So in order to do this, the way we are going 191 00:27:02,990 --> 00:27:09,990 to proceed is by taking a tiny strip in the, you take a tiny area element of length dx 192 00:27:17,889 --> 00:27:19,259 dy. 193 00:27:19,259 --> 00:27:25,720 So it will appear like a strip here, this basically a tiny area element. And this tiny 194 00:27:25,720 --> 00:27:32,429 area element is at a distance vertical distance h. And on this area element; the pressure 195 00:27:32,429 --> 00:27:38,429 force will be acting purely normally, because the fluid is static. So the differential force 196 00:27:38,429 --> 00:27:42,840 acting on this area element is purely normal. And what is this differential force? This 197 00:27:42,840 --> 00:27:49,840 is the pressure at this vertical location h from the free surface times the area dx 198 00:27:50,840 --> 00:27:57,840 dy. This is the essential idea of doing the whole thing. So, dx dy is dA this all we will 199 00:28:03,370 --> 00:28:08,100 do, in order to calculate the effective force. 200 00:28:08,100 --> 00:28:14,590 So the resultant force is nothing but you take the differential force p dA acting on 201 00:28:14,590 --> 00:28:20,710 a tiny slice and integrate over the entire area that will give you the resultant force. 202 00:28:20,710 --> 00:28:27,710 Now, p is nothing but p naught plus rho g h. So, F R is nothing but integral A p naught 203 00:28:31,679 --> 00:28:38,679 plus rho g h times dA. Now, h so this is y, this is h, this is theta, so h is basically 204 00:28:51,419 --> 00:28:58,419 like this angle, let me this rules slightly differently. So, this angle is theta so this 205 00:29:04,730 --> 00:29:11,730 is the surface, this angle theta, this is h, this is y, h is nothing but y sin theta. 206 00:29:17,419 --> 00:29:24,419 So, we want the value at point h. So instead of h, we will do y, because the co-ordinate 207 00:29:26,460 --> 00:29:33,460 is along the surface that is y. So, we will say F R is nothing but integral a p 0 plus 208 00:29:36,019 --> 00:29:43,019 rho g y sin theta times dA. This is how one calculates, the force the resultant force 209 00:29:47,580 --> 00:29:53,919 on a surface that is submerged in a fluid. Well traditionally, the way this force is 210 00:29:53,919 --> 00:29:58,499 done, calculated is you could calculate either by just carrying out this integral or you 211 00:29:58,499 --> 00:30:05,499 define the centroid of a plane. A centroid of a plane, the plane is in the x y this entered 212 00:30:12,480 --> 00:30:19,480 of a surface which is on x y plane is one over area integral area dA x y. 213 00:30:21,809 --> 00:30:28,210 This is the centroid of the co-ordinates of the centroid of a surface. So, integral y 214 00:30:28,210 --> 00:30:35,210 dA is nothing but y c times A. So we have from here, F R is p 0 A upon integration, 215 00:30:38,399 --> 00:30:45,340 p 0 is a constant; it’s usually an atmospheric pressure plus rho g sin theta. Since, y dA 216 00:30:45,340 --> 00:30:52,340 integral is y c, so we will write this as y c A or F R is nothing but p 0 plus rho g. 217 00:30:56,740 --> 00:31:03,740 This is sometime refer to us h c rho g h c A, this is the pressure at the centroid of 218 00:31:09,100 --> 00:31:15,139 the area of the surface. 219 00:31:15,139 --> 00:31:22,139 So F R, the resultant force is the pressure at the centroid times the area. So this is 220 00:31:23,950 --> 00:31:30,950 the simple result for flow R for the forces that have been exerted on a planar surface 221 00:31:31,039 --> 00:31:38,039 that is submerged into inside a liquid like water under static conditions. Now two comments, 222 00:31:39,820 --> 00:31:46,820 firstly, this force acts only on one side. So we are looking at, recall that the geometry 223 00:31:47,570 --> 00:31:53,659 is like this, this is theta, this is the free surface, this is liquid. This is the force 224 00:31:53,659 --> 00:31:59,259 only acting on one side, the other side is also comprise of the same liquid, a same amount 225 00:31:59,259 --> 00:32:05,279 of force will act on this side also. But if the other side is open to some other, it may 226 00:32:05,279 --> 00:32:09,879 be open to atmosphere, then this is the force that is because of the liquid that is present 227 00:32:09,879 --> 00:32:15,649 on one side of the surface. So it depends on the problem and context has to what the 228 00:32:15,649 --> 00:32:21,110 other side is. If it is open to atmosphere, then this will be the force due to the liquid 229 00:32:21,110 --> 00:32:25,240 that is present, and the pressure variation in the liquid under static conditions. What 230 00:32:25,240 --> 00:32:32,240 is the point of action of the force? 231 00:32:32,860 --> 00:32:37,119 What is the point of action? Well, in order to find the point of action, we simply take 232 00:32:37,119 --> 00:32:44,119 the moment. Let us call that point of action as y prime. So y prime, the moment of the 233 00:32:44,119 --> 00:32:49,240 force about the point of action must be equal to the distributed moment y times p times 234 00:32:49,240 --> 00:32:56,240 dA over the entire area. So, y times F R is nothing but y is rho g sin theta. Now, if 235 00:33:02,909 --> 00:33:09,059 on one side you have liquid and the other side you have atmospheric air, then you need 236 00:33:09,059 --> 00:33:15,749 not be worry about the atmospheric pressure. So, p is simply written as the gage pressure, 237 00:33:15,749 --> 00:33:19,860 because the atmospheric, the contribution due to atmospheric pressure on this side and 238 00:33:19,860 --> 00:33:24,419 this side will cancel. So we can neglect by atmospheric pressure and write only the gage 239 00:33:24,419 --> 00:33:31,419 pressure. So this is rho g h c times A, so p is p g is rho g times h which is rho g y 240 00:33:40,700 --> 00:33:47,700 sin theta. So, integral rho g sin theta A y square dA. 241 00:33:50,720 --> 00:33:57,720 But F R is nothing but p c times the pressure at this centroid times A. So this is rho g 242 00:33:59,549 --> 00:34:06,549 sin theta times integral y squared. So F R is nothing but p c times A which is rho g 243 00:34:16,970 --> 00:34:23,970 y c sin theta integral A y square dA. So p c is nothing but rho g y c sin theta, if the 244 00:34:28,460 --> 00:34:35,460 other side is surrounded by air. So this implies y prime is nothing but 1 by A y c integral, 245 00:34:41,630 --> 00:34:48,630 so there is this A here is coming, their denominator integral y square times dA. This is the line 246 00:34:48,700 --> 00:34:55,700 of action of the force that is going to act on us submerged planer surface. So this is 247 00:34:59,070 --> 00:35:06,070 of use in several applications where you are interested in at the forces that have been 248 00:35:06,360 --> 00:35:13,070 exerted by the fluid under static conditions on solid surfaces. And this is primarily of 249 00:35:13,070 --> 00:35:20,070 interest in storage of water in dams and so on. We can also generate, generalize these 250 00:35:20,200 --> 00:35:25,310 two curved surfaces, but I will not go through this for want of time. 251 00:35:25,310 --> 00:35:32,310 So I will go to the next topic, which is buoyancy, which is also related to forces exerted on 252 00:35:33,830 --> 00:35:40,830 curved surfaces. Suppose, you have an object that is a solid object, that is completely 253 00:35:42,770 --> 00:35:49,770 immersed in a liquid. So you have a free surface that is atmosphere, you have a liquid like 254 00:35:50,540 --> 00:35:57,540 water, so you have a solid object is completely immersed. Now, let us say you are coating 255 00:36:00,990 --> 00:36:07,990 a co-ordinate z like this and gravity is acting like this. Now, this density has this liquid 256 00:36:08,210 --> 00:36:13,640 has density rho. Now, because of the fact that is this liquid has a density and acceleration 257 00:36:13,640 --> 00:36:18,730 due to gravity is acting downwards the pressure here is p atmosphere, the pressure here is 258 00:36:18,730 --> 00:36:25,730 more. So, the pressure exerted on this side on this submerged solid surface is more than 259 00:36:28,240 --> 00:36:32,930 the pressure that will be exerted on this side, because the pressure is less. So, this 260 00:36:32,930 --> 00:36:37,450 net force will act up wards is called the buoyancy force. 261 00:36:37,450 --> 00:36:44,450 So, how do we estimate or derive an expression for a buoyancy force? It is not very difficult. 262 00:36:44,930 --> 00:36:51,930 Simply we have to take a thin cylindrical volume element, let us call this height as 263 00:36:54,860 --> 00:37:01,860 h. Now, p at h at the bottom is basically p naught, the pressure at the top plus this 264 00:37:05,380 --> 00:37:12,000 is p naught let us say plus rho g h. This is the fundamental equation of hydrostatics. 265 00:37:12,000 --> 00:37:19,000 So, the net vertical force on this volumes cylindrical volume is nothing but p naught plus rho g, let’s call this 266 00:37:27,700 --> 00:37:34,700 h equals h 2 and lets call the top surface h equals h 1 so p naught plus rho g h 2 times 267 00:37:36,430 --> 00:37:43,430 dA minus p naught plus rho g h 1 times dA, so h 2 minus h 1 let us call it h so its rho 268 00:37:49,750 --> 00:37:56,750 g were h dA. This is h dA is the differential volume of the cylindrical volume element. 269 00:38:03,800 --> 00:38:10,800 So, the net vertical force on this infinitesimal cylindrical volume element is rho g times 270 00:38:12,740 --> 00:38:15,530 the differential volume. 271 00:38:15,530 --> 00:38:22,220 So the net vertical force on a cylindrical volume element that is present in a solid 272 00:38:22,220 --> 00:38:29,220 that is sub merged in a fluid the differential force is called d F z is rho g times dv. This 273 00:38:31,660 --> 00:38:38,660 what we just derive, where dv is the differential volume of this volume element. And this force 274 00:38:39,580 --> 00:38:45,900 is precisely, because of the fact that the pressure here and the pressure here are different. 275 00:38:45,900 --> 00:38:50,990 And because of the fact that fluid is under a gravitational feel and the pressure varies 276 00:38:50,990 --> 00:38:56,720 due to hydrostatic equation, a hydrostatic force balance. Now to get the force on the 277 00:38:56,720 --> 00:39:01,300 entire object, we simply have to integrate this differential force over the entire object. 278 00:39:01,300 --> 00:39:08,300 Which is nothing but integrating over the entire volume rho g dv. Since, rho g is constants, 279 00:39:09,430 --> 00:39:16,430 so you get rho g integral of dv which is nothing but rho g times the volume of the object. 280 00:39:16,990 --> 00:39:23,990 But let us try to understand this slightly differently, this is rho times v times g, 281 00:39:24,060 --> 00:39:31,060 here rho is density of the liquid in which this object is present. So, this is the mass 282 00:39:32,980 --> 00:39:39,980 of the liquid that is displaced by the solid. 283 00:39:48,190 --> 00:39:55,190 So the net vertical force on a solid substrate solid object that is completely submerged 284 00:39:56,070 --> 00:40:02,700 under in a liquid which is present under gravitational field, this is called the buoyancy force, 285 00:40:02,700 --> 00:40:09,700 and this is nothing but the weight of the fluid that is displaced by the solid object. 286 00:40:24,550 --> 00:40:31,550 This of course, the famous Archimedes principle, so this is called the Archimedes principle. 287 00:40:36,130 --> 00:40:42,950 This is again a consequence of a basic equation force balance in hydrostatics. And we merely 288 00:40:42,950 --> 00:40:49,950 have to apply this to the context of object that is immersed. So this is for a fully, 289 00:40:53,090 --> 00:40:56,730 what we derived is for a fully submerged object. 290 00:40:56,730 --> 00:41:03,730 Suppose you have a floating object, suppose you have an object that is partially submerged, 291 00:41:05,900 --> 00:41:12,900 this is floating. So you have this is atmosphere, air at atmospheric pressure, this is liquid. 292 00:41:14,170 --> 00:41:21,170 Suppose, you have partially submerged objects, so only this part is submerged. So this portion 293 00:41:26,770 --> 00:41:33,770 has displaced, this submerged object has displaced a volume of fluid. And the buoyancy force 294 00:41:35,890 --> 00:41:41,690 will be because of the fact that of due to that weight of the displaced fluid, so this 295 00:41:41,690 --> 00:41:45,620 is gravity. Now this solid object is under stable equilibrium 296 00:41:45,620 --> 00:41:51,130 that is its not sinking, it is not moving down. That means that, the net downward force 297 00:41:51,130 --> 00:41:58,130 on the solid object, this is mass of the solid object, times acceleration due to gravity 298 00:41:59,880 --> 00:42:05,910 must be equal to the buoyancy force, which is acting upward. This is the downward force, 299 00:42:05,910 --> 00:42:12,420 this is the upward force, which is buoyancy, this is the density of the liquid, times acceleration 300 00:42:12,420 --> 00:42:19,420 due to gravity, times the displaced volume. Because only the displaced volume will contribute 301 00:42:19,620 --> 00:42:26,330 to the net upward force. So, this is the condition for floating. That the net downward force 302 00:42:26,330 --> 00:42:32,300 must be equal to the net upward force, which is the buoyancy force which is nothing but 303 00:42:32,300 --> 00:42:36,060 rho times the density of the liquid times acceleration due to gravity times at displaced 304 00:42:36,060 --> 00:42:37,040 volume. 305 00:42:37,040 --> 00:42:44,040 Now, let us consider a simple example of an ice berg. Ice icebergs are found in oceans, 306 00:42:45,690 --> 00:42:52,690 these are huge chunks of ice that are present in sea, in ocean. So this is the water level, 307 00:42:57,110 --> 00:43:04,110 and this whole thing is submerged under water, this whole thing is submerged under water. 308 00:43:07,220 --> 00:43:13,910 So, let us call this submerged volume V under water V u w. Let the mass of the ice berg 309 00:43:13,910 --> 00:43:20,910 be M, this is the ice berg mass and the total volume of ice berg is V total, this is total 310 00:43:22,060 --> 00:43:29,060 volume of ice berg. Now density of ice is smaller than rho ice is smaller than density 311 00:43:31,820 --> 00:43:38,820 of water. Density of water is 1 gram per cc and density of ice is 0.92 grams per cc, centimeter 312 00:43:41,360 --> 00:43:47,810 cube. So, if this ice berg is under stable equilibrium, then the mass of the ice berg 313 00:43:47,810 --> 00:43:51,940 times is the acceleration due to gravity. The weight of gravity by is nothing but the 314 00:43:51,940 --> 00:43:57,040 buoyancy force; this is the weight of the displaced fluid which is nothing but the volume 315 00:43:57,040 --> 00:44:03,250 that is submerged under water of the ice berg times density times acceleration due to gravity. 316 00:44:03,250 --> 00:44:09,480 What is the total mass of the ice berg? it is nothing but rho ice times V total volume 317 00:44:09,480 --> 00:44:16,480 times g is V under water times rho liquid, which is let us say water here of course, 318 00:44:17,200 --> 00:44:24,200 times g, g cancels. So V under water by V total, this the fraction of volume that is 319 00:44:25,710 --> 00:44:32,710 under water is nothing but rho ice divided b y rho water. This is nothing but 0.92 divided 320 00:44:34,500 --> 00:44:41,400 by 1 is 0.92. So, what this simple example is telling you is that. When our ice berg 321 00:44:41,400 --> 00:44:48,400 is floating, in a floating ice berg, 92 percent of the solid mass is under water. So this 322 00:44:59,060 --> 00:45:06,060 is straight forward consequences the buoyancy principle, the 92 percent of the ice in an 323 00:45:06,530 --> 00:45:10,730 ice berg is completely under water. 324 00:45:10,730 --> 00:45:17,730 We can also derive simple criteria for floating, when does an object float? When can object 325 00:45:19,170 --> 00:45:26,170 float? Let us consider a simple geometry here, let us consider a very simple geometry to 326 00:45:35,040 --> 00:45:37,860 get this result. 327 00:45:37,860 --> 00:45:44,860 Which is, take a cylindrical object and this is water, let us say this cylindrical object 328 00:45:47,570 --> 00:45:54,570 is floating, this is water, and this is air. So floating happens, when the weight downward 329 00:45:57,110 --> 00:46:01,730 force that is weight acting due to acceleration due to gravity on the solid object is equal 330 00:46:01,730 --> 00:46:07,950 to the buoyancy force. which is nothing but so let us call this height that is submerged 331 00:46:07,950 --> 00:46:13,600 as h, so the volume of cylinder that is submerged is 8 times h, times the density of the fluid 332 00:46:13,600 --> 00:46:20,600 is rho f times g. Now m is nothing but suppose let us call this whole height as l A l times 333 00:46:24,940 --> 00:46:31,940 rho solid times g is nothing but A h times rho fluid times g. So, if you cancel A and 334 00:46:32,330 --> 00:46:39,330 g so floatingly can happen. By definition, floating means h is less than l otherwise 335 00:46:43,680 --> 00:46:46,000 the object will completely immerse under water. 336 00:46:46,000 --> 00:46:51,920 So floating means h is less than l, so if h is less than l then this equation tells 337 00:46:51,920 --> 00:46:58,920 you that rho s l is rho f h, this implies that rho s is less than rho l. This is a necessary 338 00:47:02,990 --> 00:47:09,990 condition for floating. So, this completes the basic concepts that are that can be obtained by 339 00:47:24,150 --> 00:47:31,150 simple considerations of a fluid under static conditions. So just to recapitulate, we first 340 00:47:32,040 --> 00:47:38,000 derived the governing equation for fluid under static conditions, which was simply minus 341 00:47:38,000 --> 00:47:45,000 del p plus rho g is 0. And using this, in this lecture we derive the fundamental equation 342 00:47:45,740 --> 00:47:50,790 for manometers, the principle of manometry. And then we introduce a notion of atmospheric 343 00:47:50,790 --> 00:47:55,910 pressure and barometers. Then we proceeded to derive the forces that 344 00:47:55,910 --> 00:48:01,940 are experienced by a planar surface that is submerged inside a liquid. And we found that 345 00:48:01,940 --> 00:48:07,360 it can be very easily obtained by integrating the pressure on a small area element. And 346 00:48:07,360 --> 00:48:11,940 by integrating this over the entire area we can get the force and we can also find the 347 00:48:11,940 --> 00:48:17,850 line of action of this resultant force on a solid surface. Next, we proceeded to discuss 348 00:48:17,850 --> 00:48:24,710 the notion of buoyancy on a completely submerged solid surface. And we can derive the Archimedes 349 00:48:24,710 --> 00:48:31,710 principles from the basic equation of hydrostatics. By realizing that, the net force down downward 350 00:48:33,280 --> 00:48:37,820 on a on the solid surface is greater than the net force on the upper surface of the 351 00:48:37,820 --> 00:48:43,100 solid. So, this results to the buoyancy and we derived the Archimedes principle from the 352 00:48:43,100 --> 00:48:48,640 basic equation of fluid statics. And we also saw, when can an object float and what are 353 00:48:48,640 --> 00:48:50,980 the necessary conditions under which an object floats. 354 00:48:50,980 --> 00:48:57,630 So, this completes the discussion on fluid statics, which is one of the simplest topics 355 00:48:57,630 --> 00:49:02,990 in fluid mechanics, because the subject of fluid mechanics deals with fluid flow, but 356 00:49:02,990 --> 00:49:07,540 fluid statics is an integral part, because even under static conditions, the forces that 357 00:49:07,540 --> 00:49:13,540 are being exerted are not simple, because of the fact that the pressure in a fluid varies 358 00:49:13,540 --> 00:49:19,660 with vertical distance. So, we will stop here, and we will continue with a new topic in the 359 00:49:19,660 --> 00:49:26,660 next lecture. So, we will see in the next lecture.