Fluid Pressure and Laws

Hello everyone, welcome to Gatorcademy Plus. I am Mr. Gopal Sharma and I am teaching you fluid mechanics. In the last video, we had studied about pressure and we covered its units and ended that video. Now we will carry on with the same topic. So, we have seen in the last video what is pressure. Now, let's see how to measure pressure and what kind of pressure we know in the subject of mechanics. So, we will deal with three types of pressure. First is absolute pressure. Second is gauge pressure. And third is... vacuum pressure ok so first let's see what is absolute pressure see, we live on earth in the earth's atmosphere various types of gases are present how is this? we had studied this at school level let me recall you that gases are present in any particular object that is planet After collecting the gas, a particular velocity is to be attained to get out of the gas. That is escape velocity. If the velocity of the gases is less than the escape velocity, then they will never be able to leave the planet. But in space, there are no gases present. Okay? So when there are no gases present, then what can be the pressure there? The pressure can definitely be zero. Now how can it be that the pressure is zero? Let's understand that. If I go to the microscopic discussion, the kinetic theory of gases, then how does the pressure come basically? Whatever the gas or fluid is, The particles of the fluid are free to move. The tendency of the fluid is less and the tendency of the gases is very high. During motion, when these particles collide with the surface, their momentum changes. According to Newton's second law of motion, the impact of force is equal to rate of change of momentum. means when their momentum will change, then this particular area at every point where they will collide, they will exert a particular force okay, so the intensity of force per unit area will be the pressure so where does the pressure come from? when the number of molecules collide on a fluid or a solid surface then due to that collision, we get the pressure here that is according to the kinetic theory of gases okay So, what happens if we are at a place where no gases, no fluid medium is present? In such a situation, when we do not have any molecule, no medium that impacts on the surface, then definitely the pressure will be zero forever. So, when we take pressure as absolute zero, i.e. pressure of... the Universe or we can say outer space. The pressure of outer space is absolute zero pressure. Absolute zero pressure. Note this at outer space. where there is no medium. So, there is no pressure in the outer space. So, if I consider that absolute zero pressure as a reference and calculate the value of pressure above it, then that pressure will be called absolute pressure. So, what is absolute pressure? Absolute pressure is the value of pressure measured above the absolute zero pressure. Okay. So, let's make a small diagram. So, first of all, I got absolute pressure. Absolute pressure that is zero. Okay, now let's see what is Gauge pressure. As I told you that our atmosphere is made up of number of gases. So, if I consider our atmospheric pressure as reference, I consider the pressure here as reference and in respect of that I measure a particular gas or liquid pressure and tell its value. So, that is called Gauge Pressure. So, Gauge Pressure is the pressure measured above the Atmospheric Pressure. Okay, now how will Atmospheric Pressure be measured? First of all, Atmospheric Pressure is measured by Barometer. And the reference to measure Atmospheric Pressure is the absolute zero pressure. So, in absolute scale, the value of atmospheric pressure which we have seen is 101.3 kilopascal, that is P atmospheric. So, the value of atmospheric pressure in absolute scale is 101.3 kilopascal. If I take this pressure as reference line and measure the pressure of any liquid above it, so that is known as Gauge pressure. So, if I take this as a reference, if I measure the pressure from here, then this will be P gauge. Okay. If I measure the pressure from here, then with respect to absolute pressure, then we will call it absolute pressure. Okay. Of point A. So, this is P gauge of A. This is P absolute of A. Okay. So, I think you must have understood gauge pressure. Now, let's see vacuum pressure. What is vacuum pressure? When a liquid's pressure is less than atmospheric pressure, then we call it vacuum pressure. So, this is the line of atmospheric pressure. This line is basically made to discuss these three types of pressures so that the picture we will do in numerical analysis can be clear. Apart from this discussion, there is no use of this formula. So, let's see this vacuum pressure. So, if there is less pressure than atmospheric pressure of any particular point, then we will call that point's pressure vacuum pressure. We will measure vacuum pressure by the value below atmospheric pressure. So, let us say P vacuum. So, P vacuum will be below atmospheric. So, how will I write this? This will be equal to P vacuum at B will be equal to to Atmospheric pressure minus. This is your atmospheric pressure and this is point B's absolute pressure. P absolute at B. If I minus both, then I will get this vacuum pressure. So this is the vacuum pressure. So let's see Pascal's law. Pascal's law says that pressure at a point in a fluid at rest is same in all directions. That means at a fluid at rest, at a particular point, the pressure at all directions is equal. Suppose this is a point, let us say this is point A in a fluid at rest. So, three directions that we know, that is X direction, Y direction, Z direction. So, in all three directions, the value of pressure will be equal at this point. Similarly, the value of pressure at point B will be equal in X direction, Y direction and Z direction. So, Pascal's law says that pressure at a point in a fluid at rest is same in all directions. So, the proof of Pascal's law comes in university exams. Sometimes it is also asked in SSC's Mains. So let's see the proof of Pascal's Law. To prove Pascal's Law, first of all we consider an element of wedge shape. The thickness of element is 1 and we can say it is unity. There is an element of unit thickness. This is x direction and this is y direction. So, in x direction, the width of this element is dx. The height of this element in y direction is dy. The thickness is its unity. The hypotenuse or slant surface of this element is theta. Now, The pressure in the y direction, let us say that is P y and the pressure in the x direction on this element that is P x. Now, on this surface, the pressure is P y and on this surface, the pressure is P x. So, in dono pressure ke karan in dono surface pe pressure force aega. So, pressure force hoga force is equal to pressure into area. Area jo is surface ka hai that is dx into the thickness 1. So, pressure into area becomes force in y direction that is pressure into area dx into 1. Similarly, pressure force in x direction fx will be become equal to Px into area that is dy into 1. Now, I believe that the slant surface has a pressure of P in its normal direction. The length of this slant is ds. So, the force in the normal direction of this slant will be F. So, it will become P into Boom. ds into 1. This slant area will be ds into the thickness 1. So, this is the force normal to the slant surface. Now, this element, the element of which, let us say the density of fluid is equal to rho and the gravitational acceleration at this point is g. So, the density of this fluid The weight will act vertically downwards So the weight of element weight of element will be equal to rho that is mass density into gravitational acceleration into volume Now how will the volume come out? So see this is the shape of the triangle So the area of the triangle half base into height so area mil gaya and area into thickness that is equal to one area into thickness karke aapko mil jayega volumes so volume is equal to mila area or we can say cross section area cross section area into thickness ye formula aap kahin bhi use kar sakte ho bahuti useful formula hai volume nikalne ka that is cross section area into thickness or We can use length in some places. So this is the weight. So the weight will be applied downward. So that is rho g dx into dy divided by 2. This is the weight applied on the lower side. Now, the normal force applied on the slant surface, I can resolve it about x direction and y direction. So let's see, this slant surface, the normal force applied on it, this angle is theta and this angle is also theta. This 90 minus theta, so this becomes theta. Okay, so this is inclined from the x direction to theta degree, so here its component will be P ds cos theta and its vertical component will be P ds sin theta. Okay, so this is the free body diagram of our element. The free body diagram is complete. Now we have to prove that the pressure is the same in all directions. Okay, so So, first of all we see that the fluid is on rest. So, when the fluid is on rest, in this rest condition, the more force is applied on the second particle of the fluid, the more force is applied on the second particle of the fluid. That means, the summation of forces will be zero in all directions. And if an unbalancing force will come due to weight of the fluid, then that unbalancing force which is the container, will be transmitted in it. Then the container will be applied against the force, a reaction force will generate and at the end of the story all the forces will be balanced again and summation of forces will be zero. So, as fluid is at rest, summation of F will be equal to 0 So, we have two directions here, one is x direction and the other is y direction. So, first of all, we will see that summation of forces in x direction will be equal to 0. So, let's see the forces in x direction. We will take the forces which will be in the positive x direction as positive. We will take the direction which will be along negative x as negative. So, first of all, let's see the x direction. The first force in x direction is fx, that is px dy into 1. So, you'll get PX. dy second force is pds cos theta x direction this is along the negative x direction so this will be minus pds cos theta there is no other force in x direction so this will equal to zero okay now let's analyze this diagram a little bit so from here we will know if this is dy this is dx and this is ds and this angle is given to us as theta we have considered this angle as theta I'm gonna wear it. So, in this particular right angle triangle, sin theta, this is its base, this is its perpendicular and this is its hypotenuse. So, sin theta is your perpendicular upon hypotenuse. So, here we will get dx is equal to ds sin theta. Similarly, in this right angle triangle, we will put cos theta. So, cos theta will be your base upon hypotenuse that is dy upon ds. So, from here dy will come ds cos theta okay now we will move on so here ds cos theta is visible ds cos theta equal to what dy so here ds cos theta by removing it i can write dy so px dy minus pdy is equal to 0 px dy is equal to pdy dy dy cancel here proof is done px is equal to p let us say this is equation number 1 now Summation of forces in y direction is equal to 0. So, let us see the forces in y direction. There are forces in y direction. Those which are in upward direction, that is along the positive y axis, we will take them as positive. Those which are in downward direction, that is along negative y axis, we will take them as negative. So, along positive y, we have P y into dx. This is positive. Negative is coming first of all, its weight, that is rho. Minus rho g dx dy divided by 2. Then second force is that is minus p ds sin theta is equal to 0. There is no other force. There are only these many forces in y direction. So, Let's look at this equation. Here, we have dx into dy. dx is a very small elemental length and dy is also a very small elemental length. Since they are very small quantities, their product will be very small. And when we divide it by 2, we will get a much smaller term. Since it is a very small term, that's why we neglect this particular term, that is, the weight of this element. So, this equation will convert to be P y dx minus p ds sin theta is equal to 0. Now, we have seen that ds sin theta is equal to dx. So, from here, py dx, I can write ds sin theta as dx. So, p dx will be equal to 0. From here, I will solve it. So, py dx minus p, sorry, equal to p dx. dx dx is cancelled. From here, py is equal to p. So, this is equation number 2. So, from equation number 1 and equation number 2, I can write Px equal to Py equal to P. So, this is equation number 3. Similar if I write Porsche, similar if I write methodology, and y and z then I will get the same type of expressions so I can write for 3 dimensional system I have derived for 2D system for 3 dimensional system also we will get the same type of result that will give px equal to py equal to pz so here our Pascal law is proved which says that pressure in all direction at a point is same. Okay? Let's go. So, we saw that Pascal Law says that pressure at a point in all directions is same. Sometimes, in some questions, I can ask you that at a particular point, the normal stress is same in all directions or not. So, we have already discussed that at a particular fluid at rest, only one type of stress is acting that is normal stress. And normal stress is nothing but the pressure at that point. So, means at this particular point, if I talk about normal stress, normal stress at this point A, so normal stress is equal to pressure at A. Means, the normal stress at A will be equal to pressure at A. So, if we say that Px is equal to Py is equal to Pz, At A, if Px, Py and Pz are equal, then in the same way, Sigma x, i.e. normal stress in x direction will be equal to normal stress in y direction will be equal to normal stress in z direction. So, you can also say that normal stress at a point in a fluid is same in all directions. Okay, so this is the complete information about the Pascal Law. Now, we will see another statement of the Pascal Law. So, we saw that on a fluid at rest, in all directions, the normal stress or pressure is equal. Now, let us see the second statement of the Pascal Law. That is also known as Law of Transmissibility of pressure. Law of transmissibility of pressure. What does this law say? Let's understand. This law basically talks about external pressure, which we exert on the surface of a fluid. So, this law says that the external pressure applied on the surface of a fluid throughout the mass of fluid is equally distributed. This is what it means. Suppose there is a fluid at rest and I am applying pressure on it through a piston or plunger. And the value of this external pressure is P. If I am applying P external pressure on it, The pressure value of this point will be same at every point. Suppose the pressure of this point is Pa and the pressure of this point is Pb. The pressure of both points is Pa and Pb. So, how will we find the pressure of both points? Pa and Pb. The normal pressure of Pb is P of fluid at A. plus the external pressure P which means when we apply external pressure in a fluid then the pressure spreads evenly everywhere so in this law we are basically talking about the external pressure similarly, the pressure at point B will be pressure at B due to the weight of fluid P fluid at B plus the external pressure so from here we can see that at both the points the pressure due to the weight of the fluid itself even though it is changing with depth, both the pressures will be different on Pb and Pa but the external pressure that we are applying is P on point A and P on point B now this principle is basically used in hydraulic jacks this concept is used in hydraulic jacks let's see how it is used in hydraulic jacks, if I go to its simple fundamental So in that case, On one side, the cylinder area is A1 and on the other side, the cylinder area is A2. We install a plunger or a piston on A1 and A2 is called RAM or platform. Now, we use this RAM or platform as lift or jack to lift cars or trucks. Now how does it work? we can see from this law that the external pressure applied let us say the external pressure is P that P is in this particular fluid if the single fluid is filled from here to here then the external pressure will be equal at every point so let us say this pressure is F1 by A1 the pressure on this particular section is F1 by A1 here the area is A2 So here, The pressure will be equal according to the Pascal's law. So I can write F1 divided by A1 will be equal to F2 divided by A2. So from here I can see that if I apply F1 force at this point, then at the second point I am getting F2. F2 is equal to F1 by A1 into A2. Now, if you observe this diagram, you will see that the area is less than A1. The area is less than A2. So, when we go to this expression, we will see that F1 by A1 is multiplied by A2. And because A1 is smaller than A2 or A2 is bigger than A1, so therefore A2 by A1 will be greater than 1. A2 by A1 ratio will be will be greater than 1. So, definitely F2 will be equal to 18. 2 by A1 into F1 or we can say F2 is greater than F1. Means the second section where the cross section area or the area of the ram is big, at that place we will get more force. So in this way we get the amplification of the force. Means if we want mechanical advantage, mechanical advantage, If we want mechanical advantage, that is, we need more force than the force we are applying, then in that case we use this form of Pascal's law in hydraulic jacks. And you see some machines like JCB machines, the diggers machines that work on hydraulic, the same fundamental is applied in them. The brakes etc. or the arms of it, they are hydraulically operated. In a small pipe, whose area is less, we apply pressure in that and that pressure is transferred to a place where the area is more so the force that is there is amplified so we can apply less force and lift more weight so this is the application of Pascal law and this was the second statement which is also refers to the Pascal law now we will see hydrostatics law this law is useful to tell what pressure will be there on a particular depth of fluid at rest Previously, we saw Pascal's law, in which we said that there are two points PA and PB, point A is above and point B is below. So, the pressure at point A will be equal to pressure of fluid at A plus the pressure acting on the surface, that is the external pressure. External pressure plus pressure due to the weight of fluid will be equal to this. So, where do we get the value of pressure due to the weight of fluid? So, this is the law. Hydrostatic law is the law which tells us the value of pressure at a particular depth. So, let's see its expression. Let's derive the hydrostatic law. So, first of all, let's consider a container which is filled with a certain quantity of fluid. This is the free surface of fluid. Now, I'll consider the directions. I will take Cartesian coordinates. This is let us say x direction, this is y direction and this is z direction. Okay, I have taken z direction vertically downward in this derivation. At the end of derivation, I will tell you what consequences can be there if you take it upward. Okay, so from here, I will consider an element at h depth whose cross section area is dA. This particular The pressure on the position is Pz. Now the dimension of the element along the z axis is dz. Now its equal and opposite pressure will be applied from below. That will be Pz plus dz. Now there is a particular volume of this element. The volume of element dv will be equal to da into dz. The mass density of this fluid is rho and the gravitational acceleration at this point is g. So, we will discuss the free body diagram of the element in detail. So, because the pressure at the upper surface is pz and the area of that region is dA, then the pressure force applied on this particular surface will be pressure into area dA. Now, the pressure applied below is Pz plus dz. Now, let us see how the value of Pz plus dz will be. In mathematics, there is a series called Taylor series. Taylor series, in Taylor series if you expand fx plus dx then it comes to fx plus del by del x of fx into dx then del square by del x square fx dx and so on. the Taylor series runs. So, the higher orders of this series we eliminate it, it is of no use as it is very small. So, we consider this term as starting. We can write the pressure change in the form of Pz plus dz. So, we can write Pz plus dz. From this Taylor series expansion, we will get Pz plus del by del z Pz into dz. If you want to understand this in words, So how will we understand? See, point number, I call this A and this B. The pressure on point number A was Pz, how much distance did I move? Dz. Okay? So suppose I write this, that if there is a length of del z, right? The pressure on del z length is Pz. If the pressure on del z is Pz, then how much pressure will there be on moving a distance of 1 unit? so that will be equal to del p z by del z sorry del z if the pressure changes in the depth if that is del p z so in one unit distance the pressure that will change in the pressure will be del p z by del z okay now the previous pressure on the surface is p z so let us write Now, how much distance I have moved? DZ How much pressure increases when moving 1 unit distance? Del PZ by Del Z So, how much pressure increases when moving DZ distance? So, that will be PZ How much pressure increases? Del PZ by Del Z into DZ So, this will be pressure at point number B PZ plus DZ This is the value of your pressure Okay So, Here we can write pressure force as Pz plus del Pz by del z into dz into area that is da. Now its weight will also come here. Why? Because we have considered its density and it has a finite volume that is equal to area into dz. So here its weight will come down, small weight that is dw. is equal to density into gravitational acceleration into volume that is dv so this co-lick satya row G D a into D Z so that is the weight of the element that is acting downward so sorry forces amara vertical direction may okay so above look summation of forces in z direction. summation of forces in z direction now because this is on fluid rest that's why all the particles of fluid will exert equal and opposite force on each other there will be no disturbing force so therefore the summation of force in z direction will be equal to 0 ok so now see the reference axis is taken in downward direction ok so the forces in downward direction we will take them positive So, in downward direction, first of all, the force is Pz into Da. This is positive. Second, its weight. Weight is also applied in downward direction. So, plus weight is its rho G Da into DZ. Now, the force coming from below is in negative direction. The force is Pz plus del Pz by del Z into DZ into Da. This will be equal to... Solve it. So, it will be PZ DA plus Rho G DA DZ minus PZ into DA plus del PZ by del Z into DZ into DA is equal to 0. From here, PZ into DA minus PZ into ADA is cancelled. Now minus del PZ by del Z DZ into DA will be equal to minus rho G DA DZ. Here DA DZ, DA ZDAT is cancelled. Minus, minus, plus. So, del PZ by del Z comes out to be rho into G. This has rho into G as your gamma. So this is the Expression for hydrostatic law. So, this law tells that the rate, that is change of pressure with respect to... depth or z direction is equal to the weight density of fluid okay so this expression that is rate of change in pressure or we can say change in pressure change in pressure along depth that is nothing but the z direction is equal to the weight density of the fluid okay so here the derivation we did in this we took the direction of z that was downward if in this derivation we take z upward So the upward forces will be positive, that is this term, Pz plus del Pz by del z into dz into da. This term will be positive and this term will be negative. And to solve this we will get del Pz by del z is equal to minus gamma. When will we get this? When z axis is taken. vertically upward and this first expression we saw when we will get it when we will consider it vertically downward both the expressions are correct but in IES once there was a question that what is the expression for hydrostatic law you were given four different options in which there is confusion between two main options that is del P by del z is equal to minus gamma or del P by del z is equal to gamma. If you want this, then dP by dZ is equal to gamma and dP by dZ is equal to minus gamma. You can write this. The reason is that the pressure here is changing only in the z direction. So, pressure is only function of z. Pressure is not function of x and y direction. So, that's why the pressure is because it is only depending on one variable so we can represent this partial differential in an exact differential ok so there is no difference between the two because pressure changes along z axis now how did I say that pressure is only function of z direction pressure does not depend on x and y direction so similar there is a simple concept as we saw earlier in Pascal law that the pressure which will come at a particular point let us say the pressure Pz is coming here The remaining two normal directions, that is, x direction and y direction, i.e. Px, pressure acting in x direction and pressure acting in y direction, all these three will be equal as per the Pascal's law. So, if you take out Pz, then Px and Py will be the same. And, at the same time, pressure will not change in x and y direction. Because, if two points are at the same level, because if two points are at same level here two points are point A and point B their depth is also h so here also Pz will be same as Pz at A will be equal to Pz at B so if pressure is equal to A and B Px at A will be equal to Px at B then Px at A sorry Py at A will be equal to Py at B So, the plane whose depth is same as the surface, throughout the plane, the pressure is equal to every point. So, in simple terms, we can say that the pressure is equal to all the points present on any plane. Okay, let's see, this question which was asked, now which option is correct in this? DPI dP by dz is equal to minus gamma or dP by dz is equal to plus gamma so in the derivation we have taken z direction downward and if we take upward direction then minus gamma will come in right hand side now we know that the standard is that x direction is like this and y direction is like this and in its perpendicular upward direction z direction is there so this is the standard case so when someone will ask you which expression is correct, then definitely this expression will be correct because z direction technically upwards we have considered it downwards for our convention so till question may mention now ho case that is in downward direction vertically downward top the key a wall expression go home hydrostatic no expression Kahengi otherwise the home numerical solve Karangi to hum a wall expression use Karangi koki a wall and expression of ojada convenient with a as per this this relation that is dp by d z is equal to minus gamma okay so that was all about hydrostatic law up the hydrostatic loss How pressure is taken out at a particular point? Okay? So, this is fluid at rest. Let's assume this is the free surface. This is the free surface. Let's assume the free surface is the dot. I have taken the datum of z here, which is 0. From here, there is a point at the edge distance where I need pressure, that is point A. The density of the fluid is rho and the gravitational acceleration is g. So, I can write the hydrostatic law here. dp by dz is equal to gamma. So, dp is equal to gamma into dz. So, I have to get the pressure up to h def, so I integrate it. So, that is pressure on free surface, that is atmospheric pressure to pressure at the point A, Pa. So, I integrate dp with atmospheric pressure, Pa will be equal to integration from, the height here is 0 and here is h, gamma, dz. I will get P A minus P atmospheric is equal to gamma into H minus 0. Okay. So, P A that is pressure at point A will be equal to atmospheric pressure plus gamma into H. Ya, isko aise liht sakte hain P atmospheric plus rho G H. This will be the pressure at point A. Okay. So, ab maine If I observe this expression, then I have taken atmospheric pressure at point A and pressure due to the depth. That is also considered. So this is actually the absolute pressure at point A. Or at any depth, this is the absolute pressure. If we talk about gauge pressure, what will be the gauge pressure at point A? So we know that the pressure at free surface is equal to atmospheric pressure. Now gauge scale, in gauge scale atmospheric pressure is considered as datum. P gauge atmospheric pressure at gauge is taken as zero. Atmospheric pressure is considered as zero, it is considered as a reference point for measurement. So if I write the gauge pressure of point A, then it will be Pa equal to rho G H. So this will be the gauge pressure. So this is the absolute pressure, you can call this as total pressure at point A and this is the gauge pressure at point A. So, we saw that on some flute, on some particular H-depth, Total pressure or absolute pressure is P is equal to atmospheric pressure that is the pressure acting above the surface. Atmospheric pressure is acting above the free surface. So that is the pressure of atmosphere above the free surface plus rho gh. This is the depth of the point. So this is also known as absolute pressure or absolute or total pressure then similarly if I remove this if I take the atmosphere as reference point then it will be rho gh that is the gauge pressure okay now if you observe this expression let's talk about gauge pressure generally the numerical problems that will come which is solved through hydrostatic law majorly it is for pressure measurement we will discuss manometer questions in future so for pressure measurement we use gauges gauges measure pressure and indicate the value that is also called gauge pressure in that case we don't need this value majorly we use this value we deal with the value of pressure which is RhoGH so this formula of pressure is used in the future in the form of manometry so you should remember this formula so now if you observe this expression you will see that there is no extra term other than H so we can simply say that pressure changes along the depth only now a new concept comes out from here which is called Hydrostatic Paradox Hydrostatic Paradox Okay, now what happens is that generally we think that if a container is big then on a particular depth suppose if the depth is X then the pressure will be high on that X depth and if the vessel is narrow or thin then the pressure will be low on the same depth but this does not happen How? This law P is equal to rho gs that is gauge pressure is equal to rho gs there is no area of any term in this so suppose I take 2-3 vessels one is frustum, one is cylinder whose diameter is d1 and second one is cylinder whose diameter is d2 and d1 is greater than d2 this is frustum its bottom surface area is a1 its bottom surface area is a2 and the area of its bottom surface is A3. Now, if I fill the fluid in all three vessels to the same height, this one is of H height and this one is of H height, I have poured the fluid of H height in all three vessels. Now, all three of these fluids are water, that is, the density for all three vessels is rho. For all three vessels, the density is same. That means, for all three vessels whose size is different, whose volume is different, I have poured the same liquid and the same height. So, from the hydrostatic law, we can understand what the pressure will be at this particular point. So, look at the hydrostatic law, rho gh. There is no term for area in this. That means, at this point, at the base, that is pressure at base, for this also, it will be rho gh. For this also, pressure at base will be rho g h and for this also, pressure is equal to rho g h. The pressure at base on all three will be same. That means, the pressure at a particular vessel at a particular point does not depend on the shape of the vessel. It does not depend on its cross section. No matter how wide the cylinder is or how much its diameter is. no matter how much the diameter is, if we are talking about pressure at the same depth then in both cases the value of pressure will be same so pressure does not depend on the cross section area but when I will talk about the force that will come on the base due to pressure so force is exerted by the liquid at the base will be different because the area here is A1, A2, A3 and the pressure force F that is pressure into area. Pressure is same for all but area is different for all. So, in the first case F1 will not be equal to F2 and F2 will not be equal to F3. The values of all three will be different because the pressure is same but area is different. So, this is known as hydrostatic paradox theory. Now let's see when a fluid starts moving from rest. There are two conditions that we are discussing now, that is fluid statics and the fluid dynamics. So how does a fluid move at rest? For example, when we talk about solid body and Newton's law of motion, the first law of motion states that the body will remain in its state of motion unless there is an external force, okay? until we don't apply any external force on it the body will remain in its state of motion if it is stopped, it will remain stopped and if it is moving, it will keep moving similarly, if I initially want to move a fluid at rest so, first of all, I need the pressure difference a fluid at rest moves only when it gets the pressure difference so, if at two points, the fluid that there will be a difference in pressure and the fluid will start flowing there so this creates a confusion that this point was A and this point was B, here the pressure is PA and here the pressure is PB. Here the height is H1 and here the height is H2. So, definitely we can see that H2 is bigger than H1. So, definitely PB will be bigger than PA. Right? So, why is there no fluid flow here? There is no fluid flow here because this fluid is kept in a container. The equal and opposite pressure force applied on the container is also applied on it. So that's why all the fluid particles are applying mutual forces on each other. That is equal and opposite in direction. And all the forces due to pressure are neutralizing each other. And the unbalanced force that is coming, whatever is the difference in the magnitude of the force, and the unbalance that is being created, unbalancing force go This wall is carrying it and then the unbalancing force is going towards the wall and this wall is neutralizing the unbalancing force by giving a reaction. So basically summation of forces is zero here. That's why this fluid is not flowing. But what happens is, if I make a hole in this container, if I make a hole in this container, then the external pressure is atmospheric pressure and the pressure here, let us say, pressure at point A, Pa, Pa is greater than P-atmospheric. So definitely we have seen that generally if there is a hole in a vessel, then the water starts flowing out. So from here one thing is understood that the fluid at rest, it flows from high pressure to low pressure. Now let's talk about fluid in motion which is already moving. Like in its initial condition, the fluid was at rest. If the fluid in motion, in dynamic condition so in that case the flow that is due to energy difference which Bernoulli and Euler had told us we will discuss in fluid dynamics but if there is a fluid on rest then it comes under motion condition when we get pressure difference ok now till when this flow will be there so see we have already studied these things at school level let's discuss it once more suppose this is a vessel in which the liquid up to the edge height is filled Now I will put a pipe in this and connect it with another vessel. Observation is that its cross section area is A1 and its cross section area is A2. And A1 is very big compared to A2. So, in the last theory that we had read, that was hydrostatic paradox, we saw that the pressure does not depend on the cross section area. So, what will happen? Initially, I say that when I connected this pipe here, there was no fluid in it. Slowly, the liquid that is filled in it will start flowing in it. Now, as I just told you that flow occurs when there is a difference in pressure. So, when will the flow occur here? It will flow until there is a difference in pressure. And at the point where these two points meet, Pressure will be equal in vessels at this particular base and at this particular point and then flow will stop. So, how long will the flow be? Until the height of these two is measured from the base, it does not become equal. The height to which water or any other fluid is filled in it, until the height to which the fluid does not fill in it, the flow will be. How will it happen? Suppose its height is let us say H1, and final height is h1 so here also the pressure will be rho gh1 and on this side also it will be rho gh1 so basically there will be an equilibrium condition in which pressure will be equal in both vessels and there will be no full plausible because flow can be possible when there is pressure difference so this will be the condition for equilibrium when two vessels are connected by a pipe so today we have discussed Pascal's law, hydrostatic's law and We have covered all the important aspects of fluid statics today. In the last videos, we have started the fluid statics. If you are watching our video for the first time, then don't forget to subscribe to our channel and watch our previous videos. If you liked today's lecture, and if you have any doubts on today's topic, then you can put your doubts in the comment section below. To subscribe to our channel, press the subscribe button below and the bell icon. Don't forget to press the bell icon so that you get our latest updates from time to time. Thank you from me and my Gate Academy Plus team.

Now we will carry on with the same topic. So, we have seen in the last video what is pressure. Now, let's see how to measure pressure and what kind of pressure we know in the subject of mechanics. So, we will deal with three types of pressure. First is absolute pressure.

Second is gauge pressure. And third is... vacuum pressure ok so first let's see what is absolute pressure see, we live on earth in the earth's atmosphere various types of gases are present how is this?

we had studied this at school level let me recall you that gases are present in any particular object that is planet After collecting the gas, a particular velocity is to be attained to get out of the gas. That is escape velocity. If the velocity of the gases is less than the escape velocity, then they will never be able to leave the planet. But in space, there are no gases present.

Okay? So when there are no gases present, then what can be the pressure there? The pressure can definitely be zero.

Now how can it be that the pressure is zero? Let's understand that. If I go to the microscopic discussion, the kinetic theory of gases, then how does the pressure come basically? Whatever the gas or fluid is, The particles of the fluid are free to move.

The tendency of the fluid is less and the tendency of the gases is very high. During motion, when these particles collide with the surface, their momentum changes. According to Newton's second law of motion, the impact of force is equal to rate of change of momentum. means when their momentum will change, then this particular area at every point where they will collide, they will exert a particular force okay, so the intensity of force per unit area will be the pressure so where does the pressure come from?

when the number of molecules collide on a fluid or a solid surface then due to that collision, we get the pressure here that is according to the kinetic theory of gases okay So, what happens if we are at a place where no gases, no fluid medium is present? In such a situation, when we do not have any molecule, no medium that impacts on the surface, then definitely the pressure will be zero forever. So, when we take pressure as absolute zero, i.e. pressure of... the Universe or we can say outer space. The pressure of outer space is absolute zero pressure.

Absolute zero pressure. Note this at outer space. where there is no medium.

So, there is no pressure in the outer space. So, if I consider that absolute zero pressure as a reference and calculate the value of pressure above it, then that pressure will be called absolute pressure. So, what is absolute pressure?

Absolute pressure is the value of pressure measured above the absolute zero pressure. Okay. So, let's make a small diagram. So, first of all, I got absolute pressure.

Absolute pressure that is zero. Okay, now let's see what is Gauge pressure. As I told you that our atmosphere is made up of number of gases.

So, if I consider our atmospheric pressure as reference, I consider the pressure here as reference and in respect of that I measure a particular gas or liquid pressure and tell its value. So, that is called Gauge Pressure. So, Gauge Pressure is the pressure measured above the Atmospheric Pressure. Okay, now how will Atmospheric Pressure be measured?

First of all, Atmospheric Pressure is measured by Barometer. And the reference to measure Atmospheric Pressure is the absolute zero pressure. So, in absolute scale, the value of atmospheric pressure which we have seen is 101.3 kilopascal, that is P atmospheric. So, the value of atmospheric pressure in absolute scale is 101.3 kilopascal.

If I take this pressure as reference line and measure the pressure of any liquid above it, so that is known as Gauge pressure. So, if I take this as a reference, if I measure the pressure from here, then this will be P gauge. Okay.

If I measure the pressure from here, then with respect to absolute pressure, then we will call it absolute pressure. Okay. Of point A.

So, this is P gauge of A. This is P absolute of A. Okay.

So, I think you must have understood gauge pressure. Now, let's see vacuum pressure. What is vacuum pressure? When a liquid's pressure is less than atmospheric pressure, then we call it vacuum pressure.

So, this is the line of atmospheric pressure. This line is basically made to discuss these three types of pressures so that the picture we will do in numerical analysis can be clear. Apart from this discussion, there is no use of this formula. So, let's see this vacuum pressure. So, if there is less pressure than atmospheric pressure of any particular point, then we will call that point's pressure vacuum pressure.

We will measure vacuum pressure by the value below atmospheric pressure. So, let us say P vacuum. So, P vacuum will be below atmospheric.

So, how will I write this? This will be equal to P vacuum at B will be equal to to Atmospheric pressure minus. This is your atmospheric pressure and this is point B's absolute pressure.

P absolute at B. If I minus both, then I will get this vacuum pressure. So this is the vacuum pressure. So let's see Pascal's law. Pascal's law says that pressure at a point in a fluid at rest is same in all directions.

That means at a fluid at rest, at a particular point, the pressure at all directions is equal. Suppose this is a point, let us say this is point A in a fluid at rest. So, three directions that we know, that is X direction, Y direction, Z direction.

So, in all three directions, the value of pressure will be equal at this point. Similarly, the value of pressure at point B will be equal in X direction, Y direction and Z direction. So, Pascal's law says that pressure at a point in a fluid at rest is same in all directions. So, the proof of Pascal's law comes in university exams.

Sometimes it is also asked in SSC's Mains. So let's see the proof of Pascal's Law. To prove Pascal's Law, first of all we consider an element of wedge shape. The thickness of element is 1 and we can say it is unity.

There is an element of unit thickness. This is x direction and this is y direction. So, in x direction, the width of this element is dx. The height of this element in y direction is dy.

The thickness is its unity. The hypotenuse or slant surface of this element is theta. Now, The pressure in the y direction, let us say that is P y and the pressure in the x direction on this element that is P x.

Now, on this surface, the pressure is P y and on this surface, the pressure is P x. So, in dono pressure ke karan in dono surface pe pressure force aega. So, pressure force hoga force is equal to pressure into area.

Area jo is surface ka hai that is dx into the thickness 1. So, pressure into area becomes force in y direction that is pressure into area dx into 1. Similarly, pressure force in x direction fx will be become equal to Px into area that is dy into 1. Now, I believe that the slant surface has a pressure of P in its normal direction. The length of this slant is ds. So, the force in the normal direction of this slant will be F. So, it will become P into Boom.

ds into 1. This slant area will be ds into the thickness 1. So, this is the force normal to the slant surface. Now, this element, the element of which, let us say the density of fluid is equal to rho and the gravitational acceleration at this point is g. So, the density of this fluid The weight will act vertically downwards So the weight of element weight of element will be equal to rho that is mass density into gravitational acceleration into volume Now how will the volume come out? So see this is the shape of the triangle So the area of the triangle half base into height so area mil gaya and area into thickness that is equal to one area into thickness karke aapko mil jayega volumes so volume is equal to mila area or we can say cross section area cross section area into thickness ye formula aap kahin bhi use kar sakte ho bahuti useful formula hai volume nikalne ka that is cross section area into thickness or We can use length in some places.

So this is the weight. So the weight will be applied downward. So that is rho g dx into dy divided by 2. This is the weight applied on the lower side. Now, the normal force applied on the slant surface, I can resolve it about x direction and y direction.

So let's see, this slant surface, the normal force applied on it, this angle is theta and this angle is also theta. This 90 minus theta, so this becomes theta. Okay, so this is inclined from the x direction to theta degree, so here its component will be P ds cos theta and its vertical component will be P ds sin theta. Okay, so this is the free body diagram of our element. The free body diagram is complete.

Now we have to prove that the pressure is the same in all directions. Okay, so So, first of all we see that the fluid is on rest. So, when the fluid is on rest, in this rest condition, the more force is applied on the second particle of the fluid, the more force is applied on the second particle of the fluid. That means, the summation of forces will be zero in all directions. And if an unbalancing force will come due to weight of the fluid, then that unbalancing force which is the container, will be transmitted in it.

Then the container will be applied against the force, a reaction force will generate and at the end of the story all the forces will be balanced again and summation of forces will be zero. So, as fluid is at rest, summation of F will be equal to 0 So, we have two directions here, one is x direction and the other is y direction. So, first of all, we will see that summation of forces in x direction will be equal to 0. So, let's see the forces in x direction.

We will take the forces which will be in the positive x direction as positive. We will take the direction which will be along negative x as negative. So, first of all, let's see the x direction. The first force in x direction is fx, that is px dy into 1. So, you'll get PX. dy second force is pds cos theta x direction this is along the negative x direction so this will be minus pds cos theta there is no other force in x direction so this will equal to zero okay now let's analyze this diagram a little bit so from here we will know if this is dy this is dx and this is ds and this angle is given to us as theta we have considered this angle as theta I'm gonna wear it.

So, in this particular right angle triangle, sin theta, this is its base, this is its perpendicular and this is its hypotenuse. So, sin theta is your perpendicular upon hypotenuse. So, here we will get dx is equal to ds sin theta.

Similarly, in this right angle triangle, we will put cos theta. So, cos theta will be your base upon hypotenuse that is dy upon ds. So, from here dy will come ds cos theta okay now we will move on so here ds cos theta is visible ds cos theta equal to what dy so here ds cos theta by removing it i can write dy so px dy minus pdy is equal to 0 px dy is equal to pdy dy dy cancel here proof is done px is equal to p let us say this is equation number 1 now Summation of forces in y direction is equal to 0. So, let us see the forces in y direction.

There are forces in y direction. Those which are in upward direction, that is along the positive y axis, we will take them as positive. Those which are in downward direction, that is along negative y axis, we will take them as negative. So, along positive y, we have P y into dx.

This is positive. Negative is coming first of all, its weight, that is rho. Minus rho g dx dy divided by 2. Then second force is that is minus p ds sin theta is equal to 0. There is no other force. There are only these many forces in y direction. So, Let's look at this equation.

Here, we have dx into dy. dx is a very small elemental length and dy is also a very small elemental length. Since they are very small quantities, their product will be very small.

And when we divide it by 2, we will get a much smaller term. Since it is a very small term, that's why we neglect this particular term, that is, the weight of this element. So, this equation will convert to be P y dx minus p ds sin theta is equal to 0. Now, we have seen that ds sin theta is equal to dx. So, from here, py dx, I can write ds sin theta as dx.

So, p dx will be equal to 0. From here, I will solve it. So, py dx minus p, sorry, equal to p dx. dx dx is cancelled. From here, py is equal to p.

So, this is equation number 2. So, from equation number 1 and equation number 2, I can write Px equal to Py equal to P. So, this is equation number 3. Similar if I write Porsche, similar if I write methodology, and y and z then I will get the same type of expressions so I can write for 3 dimensional system I have derived for 2D system for 3 dimensional system also we will get the same type of result that will give px equal to py equal to pz so here our Pascal law is proved which says that pressure in all direction at a point is same. Okay? Let's go. So, we saw that Pascal Law says that pressure at a point in all directions is same.

Sometimes, in some questions, I can ask you that at a particular point, the normal stress is same in all directions or not. So, we have already discussed that at a particular fluid at rest, only one type of stress is acting that is normal stress. And normal stress is nothing but the pressure at that point.

So, means at this particular point, if I talk about normal stress, normal stress at this point A, so normal stress is equal to pressure at A. Means, the normal stress at A will be equal to pressure at A. So, if we say that Px is equal to Py is equal to Pz, At A, if Px, Py and Pz are equal, then in the same way, Sigma x, i.e. normal stress in x direction will be equal to normal stress in y direction will be equal to normal stress in z direction.

So, you can also say that normal stress at a point in a fluid is same in all directions. Okay, so this is the complete information about the Pascal Law. Now, we will see another statement of the Pascal Law.

So, we saw that on a fluid at rest, in all directions, the normal stress or pressure is equal. Now, let us see the second statement of the Pascal Law. That is also known as Law of Transmissibility of pressure.

Law of transmissibility of pressure. What does this law say? Let's understand.

This law basically talks about external pressure, which we exert on the surface of a fluid. So, this law says that the external pressure applied on the surface of a fluid throughout the mass of fluid is equally distributed. This is what it means.

Suppose there is a fluid at rest and I am applying pressure on it through a piston or plunger. And the value of this external pressure is P. If I am applying P external pressure on it, The pressure value of this point will be same at every point.

Suppose the pressure of this point is Pa and the pressure of this point is Pb. The pressure of both points is Pa and Pb. So, how will we find the pressure of both points? Pa and Pb. The normal pressure of Pb is P of fluid at A.

plus the external pressure P which means when we apply external pressure in a fluid then the pressure spreads evenly everywhere so in this law we are basically talking about the external pressure similarly, the pressure at point B will be pressure at B due to the weight of fluid P fluid at B plus the external pressure so from here we can see that at both the points the pressure due to the weight of the fluid itself even though it is changing with depth, both the pressures will be different on Pb and Pa but the external pressure that we are applying is P on point A and P on point B now this principle is basically used in hydraulic jacks this concept is used in hydraulic jacks let's see how it is used in hydraulic jacks, if I go to its simple fundamental So in that case, On one side, the cylinder area is A1 and on the other side, the cylinder area is A2. We install a plunger or a piston on A1 and A2 is called RAM or platform. Now, we use this RAM or platform as lift or jack to lift cars or trucks. Now how does it work?

we can see from this law that the external pressure applied let us say the external pressure is P that P is in this particular fluid if the single fluid is filled from here to here then the external pressure will be equal at every point so let us say this pressure is F1 by A1 the pressure on this particular section is F1 by A1 here the area is A2 So here, The pressure will be equal according to the Pascal's law. So I can write F1 divided by A1 will be equal to F2 divided by A2. So from here I can see that if I apply F1 force at this point, then at the second point I am getting F2. F2 is equal to F1 by A1 into A2. Now, if you observe this diagram, you will see that the area is less than A1.

The area is less than A2. So, when we go to this expression, we will see that F1 by A1 is multiplied by A2. And because A1 is smaller than A2 or A2 is bigger than A1, so therefore A2 by A1 will be greater than 1. A2 by A1 ratio will be will be greater than 1. So, definitely F2 will be equal to 18. 2 by A1 into F1 or we can say F2 is greater than F1.

Means the second section where the cross section area or the area of the ram is big, at that place we will get more force. So in this way we get the amplification of the force. Means if we want mechanical advantage, mechanical advantage, If we want mechanical advantage, that is, we need more force than the force we are applying, then in that case we use this form of Pascal's law in hydraulic jacks.

And you see some machines like JCB machines, the diggers machines that work on hydraulic, the same fundamental is applied in them. The brakes etc. or the arms of it, they are hydraulically operated. In a small pipe, whose area is less, we apply pressure in that and that pressure is transferred to a place where the area is more so the force that is there is amplified so we can apply less force and lift more weight so this is the application of Pascal law and this was the second statement which is also refers to the Pascal law now we will see hydrostatics law this law is useful to tell what pressure will be there on a particular depth of fluid at rest Previously, we saw Pascal's law, in which we said that there are two points PA and PB, point A is above and point B is below.

So, the pressure at point A will be equal to pressure of fluid at A plus the pressure acting on the surface, that is the external pressure. External pressure plus pressure due to the weight of fluid will be equal to this. So, where do we get the value of pressure due to the weight of fluid?

So, this is the law. Hydrostatic law is the law which tells us the value of pressure at a particular depth. So, let's see its expression. Let's derive the hydrostatic law. So, first of all, let's consider a container which is filled with a certain quantity of fluid.

This is the free surface of fluid. Now, I'll consider the directions. I will take Cartesian coordinates.

This is let us say x direction, this is y direction and this is z direction. Okay, I have taken z direction vertically downward in this derivation. At the end of derivation, I will tell you what consequences can be there if you take it upward.

Okay, so from here, I will consider an element at h depth whose cross section area is dA. This particular The pressure on the position is Pz. Now the dimension of the element along the z axis is dz.

Now its equal and opposite pressure will be applied from below. That will be Pz plus dz. Now there is a particular volume of this element. The volume of element dv will be equal to da into dz. The mass density of this fluid is rho and the gravitational acceleration at this point is g.

So, we will discuss the free body diagram of the element in detail. So, because the pressure at the upper surface is pz and the area of that region is dA, then the pressure force applied on this particular surface will be pressure into area dA. Now, the pressure applied below is Pz plus dz. Now, let us see how the value of Pz plus dz will be. In mathematics, there is a series called Taylor series.

Taylor series, in Taylor series if you expand fx plus dx then it comes to fx plus del by del x of fx into dx then del square by del x square fx dx and so on. the Taylor series runs. So, the higher orders of this series we eliminate it, it is of no use as it is very small. So, we consider this term as starting. We can write the pressure change in the form of Pz plus dz.

So, we can write Pz plus dz. From this Taylor series expansion, we will get Pz plus del by del z Pz into dz. If you want to understand this in words, So how will we understand? See, point number, I call this A and this B.

The pressure on point number A was Pz, how much distance did I move? Dz. Okay? So suppose I write this, that if there is a length of del z, right? The pressure on del z length is Pz.

If the pressure on del z is Pz, then how much pressure will there be on moving a distance of 1 unit? so that will be equal to del p z by del z sorry del z if the pressure changes in the depth if that is del p z so in one unit distance the pressure that will change in the pressure will be del p z by del z okay now the previous pressure on the surface is p z so let us write Now, how much distance I have moved? DZ How much pressure increases when moving 1 unit distance? Del PZ by Del Z So, how much pressure increases when moving DZ distance? So, that will be PZ How much pressure increases?

Del PZ by Del Z into DZ So, this will be pressure at point number B PZ plus DZ This is the value of your pressure Okay So, Here we can write pressure force as Pz plus del Pz by del z into dz into area that is da. Now its weight will also come here. Why? Because we have considered its density and it has a finite volume that is equal to area into dz. So here its weight will come down, small weight that is dw.

is equal to density into gravitational acceleration into volume that is dv so this co-lick satya row G D a into D Z so that is the weight of the element that is acting downward so sorry forces amara vertical direction may okay so above look summation of forces in z direction. summation of forces in z direction now because this is on fluid rest that's why all the particles of fluid will exert equal and opposite force on each other there will be no disturbing force so therefore the summation of force in z direction will be equal to 0 ok so now see the reference axis is taken in downward direction ok so the forces in downward direction we will take them positive So, in downward direction, first of all, the force is Pz into Da. This is positive. Second, its weight.

Weight is also applied in downward direction. So, plus weight is its rho G Da into DZ. Now, the force coming from below is in negative direction. The force is Pz plus del Pz by del Z into DZ into Da. This will be equal to...

Solve it. So, it will be PZ DA plus Rho G DA DZ minus PZ into DA plus del PZ by del Z into DZ into DA is equal to 0. From here, PZ into DA minus PZ into ADA is cancelled. Now minus del PZ by del Z DZ into DA will be equal to minus rho G DA DZ. Here DA DZ, DA ZDAT is cancelled.

Minus, minus, plus. So, del PZ by del Z comes out to be rho into G. This has rho into G as your gamma.

So this is the Expression for hydrostatic law. So, this law tells that the rate, that is change of pressure with respect to... depth or z direction is equal to the weight density of fluid okay so this expression that is rate of change in pressure or we can say change in pressure change in pressure along depth that is nothing but the z direction is equal to the weight density of the fluid okay so here the derivation we did in this we took the direction of z that was downward if in this derivation we take z upward So the upward forces will be positive, that is this term, Pz plus del Pz by del z into dz into da. This term will be positive and this term will be negative.

And to solve this we will get del Pz by del z is equal to minus gamma. When will we get this? When z axis is taken.

vertically upward and this first expression we saw when we will get it when we will consider it vertically downward both the expressions are correct but in IES once there was a question that what is the expression for hydrostatic law you were given four different options in which there is confusion between two main options that is del P by del z is equal to minus gamma or del P by del z is equal to gamma. If you want this, then dP by dZ is equal to gamma and dP by dZ is equal to minus gamma. You can write this. The reason is that the pressure here is changing only in the z direction. So, pressure is only function of z.

Pressure is not function of x and y direction. So, that's why the pressure is because it is only depending on one variable so we can represent this partial differential in an exact differential ok so there is no difference between the two because pressure changes along z axis now how did I say that pressure is only function of z direction pressure does not depend on x and y direction so similar there is a simple concept as we saw earlier in Pascal law that the pressure which will come at a particular point let us say the pressure Pz is coming here The remaining two normal directions, that is, x direction and y direction, i.e. Px, pressure acting in x direction and pressure acting in y direction, all these three will be equal as per the Pascal's law. So, if you take out Pz, then Px and Py will be the same. And, at the same time, pressure will not change in x and y direction.

Because, if two points are at the same level, because if two points are at same level here two points are point A and point B their depth is also h so here also Pz will be same as Pz at A will be equal to Pz at B so if pressure is equal to A and B Px at A will be equal to Px at B then Px at A sorry Py at A will be equal to Py at B So, the plane whose depth is same as the surface, throughout the plane, the pressure is equal to every point. So, in simple terms, we can say that the pressure is equal to all the points present on any plane. Okay, let's see, this question which was asked, now which option is correct in this?

DPI dP by dz is equal to minus gamma or dP by dz is equal to plus gamma so in the derivation we have taken z direction downward and if we take upward direction then minus gamma will come in right hand side now we know that the standard is that x direction is like this and y direction is like this and in its perpendicular upward direction z direction is there so this is the standard case so when someone will ask you which expression is correct, then definitely this expression will be correct because z direction technically upwards we have considered it downwards for our convention so till question may mention now ho case that is in downward direction vertically downward top the key a wall expression go home hydrostatic no expression Kahengi otherwise the home numerical solve Karangi to hum a wall expression use Karangi koki a wall and expression of ojada convenient with a as per this this relation that is dp by d z is equal to minus gamma okay so that was all about hydrostatic law up the hydrostatic loss How pressure is taken out at a particular point? Okay? So, this is fluid at rest. Let's assume this is the free surface.

This is the free surface. Let's assume the free surface is the dot. I have taken the datum of z here, which is 0. From here, there is a point at the edge distance where I need pressure, that is point A. The density of the fluid is rho and the gravitational acceleration is g. So, I can write the hydrostatic law here.

dp by dz is equal to gamma. So, dp is equal to gamma into dz. So, I have to get the pressure up to h def, so I integrate it. So, that is pressure on free surface, that is atmospheric pressure to pressure at the point A, Pa. So, I integrate dp with atmospheric pressure, Pa will be equal to integration from, the height here is 0 and here is h, gamma, dz.

I will get P A minus P atmospheric is equal to gamma into H minus 0. Okay. So, P A that is pressure at point A will be equal to atmospheric pressure plus gamma into H. Ya, isko aise liht sakte hain P atmospheric plus rho G H. This will be the pressure at point A. Okay. So, ab maine If I observe this expression, then I have taken atmospheric pressure at point A and pressure due to the depth.

That is also considered. So this is actually the absolute pressure at point A. Or at any depth, this is the absolute pressure.

If we talk about gauge pressure, what will be the gauge pressure at point A? So we know that the pressure at free surface is equal to atmospheric pressure. Now gauge scale, in gauge scale atmospheric pressure is considered as datum.

P gauge atmospheric pressure at gauge is taken as zero. Atmospheric pressure is considered as zero, it is considered as a reference point for measurement. So if I write the gauge pressure of point A, then it will be Pa equal to rho G H. So this will be the gauge pressure.

So this is the absolute pressure, you can call this as total pressure at point A and this is the gauge pressure at point A. So, we saw that on some flute, on some particular H-depth, Total pressure or absolute pressure is P is equal to atmospheric pressure that is the pressure acting above the surface. Atmospheric pressure is acting above the free surface. So that is the pressure of atmosphere above the free surface plus rho gh. This is the depth of the point.

So this is also known as absolute pressure or absolute or total pressure then similarly if I remove this if I take the atmosphere as reference point then it will be rho gh that is the gauge pressure okay now if you observe this expression let's talk about gauge pressure generally the numerical problems that will come which is solved through hydrostatic law majorly it is for pressure measurement we will discuss manometer questions in future so for pressure measurement we use gauges gauges measure pressure and indicate the value that is also called gauge pressure in that case we don't need this value majorly we use this value we deal with the value of pressure which is RhoGH so this formula of pressure is used in the future in the form of manometry so you should remember this formula so now if you observe this expression you will see that there is no extra term other than H so we can simply say that pressure changes along the depth only now a new concept comes out from here which is called Hydrostatic Paradox Hydrostatic Paradox Okay, now what happens is that generally we think that if a container is big then on a particular depth suppose if the depth is X then the pressure will be high on that X depth and if the vessel is narrow or thin then the pressure will be low on the same depth but this does not happen How? This law P is equal to rho gs that is gauge pressure is equal to rho gs there is no area of any term in this so suppose I take 2-3 vessels one is frustum, one is cylinder whose diameter is d1 and second one is cylinder whose diameter is d2 and d1 is greater than d2 this is frustum its bottom surface area is a1 its bottom surface area is a2 and the area of its bottom surface is A3. Now, if I fill the fluid in all three vessels to the same height, this one is of H height and this one is of H height, I have poured the fluid of H height in all three vessels. Now, all three of these fluids are water, that is, the density for all three vessels is rho.

For all three vessels, the density is same. That means, for all three vessels whose size is different, whose volume is different, I have poured the same liquid and the same height. So, from the hydrostatic law, we can understand what the pressure will be at this particular point. So, look at the hydrostatic law, rho gh.

There is no term for area in this. That means, at this point, at the base, that is pressure at base, for this also, it will be rho gh. For this also, pressure at base will be rho g h and for this also, pressure is equal to rho g h. The pressure at base on all three will be same.

That means, the pressure at a particular vessel at a particular point does not depend on the shape of the vessel. It does not depend on its cross section. No matter how wide the cylinder is or how much its diameter is. no matter how much the diameter is, if we are talking about pressure at the same depth then in both cases the value of pressure will be same so pressure does not depend on the cross section area but when I will talk about the force that will come on the base due to pressure so force is exerted by the liquid at the base will be different because the area here is A1, A2, A3 and the pressure force F that is pressure into area.

Pressure is same for all but area is different for all. So, in the first case F1 will not be equal to F2 and F2 will not be equal to F3. The values of all three will be different because the pressure is same but area is different.

So, this is known as hydrostatic paradox theory. Now let's see when a fluid starts moving from rest. There are two conditions that we are discussing now, that is fluid statics and the fluid dynamics.

So how does a fluid move at rest? For example, when we talk about solid body and Newton's law of motion, the first law of motion states that the body will remain in its state of motion unless there is an external force, okay? until we don't apply any external force on it the body will remain in its state of motion if it is stopped, it will remain stopped and if it is moving, it will keep moving similarly, if I initially want to move a fluid at rest so, first of all, I need the pressure difference a fluid at rest moves only when it gets the pressure difference so, if at two points, the fluid that there will be a difference in pressure and the fluid will start flowing there so this creates a confusion that this point was A and this point was B, here the pressure is PA and here the pressure is PB.

Here the height is H1 and here the height is H2. So, definitely we can see that H2 is bigger than H1. So, definitely PB will be bigger than PA. Right? So, why is there no fluid flow here?

There is no fluid flow here because this fluid is kept in a container. The equal and opposite pressure force applied on the container is also applied on it. So that's why all the fluid particles are applying mutual forces on each other.

That is equal and opposite in direction. And all the forces due to pressure are neutralizing each other. And the unbalanced force that is coming, whatever is the difference in the magnitude of the force, and the unbalance that is being created, unbalancing force go This wall is carrying it and then the unbalancing force is going towards the wall and this wall is neutralizing the unbalancing force by giving a reaction.

So basically summation of forces is zero here. That's why this fluid is not flowing. But what happens is, if I make a hole in this container, if I make a hole in this container, then the external pressure is atmospheric pressure and the pressure here, let us say, pressure at point A, Pa, Pa is greater than P-atmospheric.

So definitely we have seen that generally if there is a hole in a vessel, then the water starts flowing out. So from here one thing is understood that the fluid at rest, it flows from high pressure to low pressure. Now let's talk about fluid in motion which is already moving.

Like in its initial condition, the fluid was at rest. If the fluid in motion, in dynamic condition so in that case the flow that is due to energy difference which Bernoulli and Euler had told us we will discuss in fluid dynamics but if there is a fluid on rest then it comes under motion condition when we get pressure difference ok now till when this flow will be there so see we have already studied these things at school level let's discuss it once more suppose this is a vessel in which the liquid up to the edge height is filled Now I will put a pipe in this and connect it with another vessel. Observation is that its cross section area is A1 and its cross section area is A2. And A1 is very big compared to A2.

So, in the last theory that we had read, that was hydrostatic paradox, we saw that the pressure does not depend on the cross section area. So, what will happen? Initially, I say that when I connected this pipe here, there was no fluid in it.

Slowly, the liquid that is filled in it will start flowing in it. Now, as I just told you that flow occurs when there is a difference in pressure. So, when will the flow occur here? It will flow until there is a difference in pressure. And at the point where these two points meet, Pressure will be equal in vessels at this particular base and at this particular point and then flow will stop.

So, how long will the flow be? Until the height of these two is measured from the base, it does not become equal. The height to which water or any other fluid is filled in it, until the height to which the fluid does not fill in it, the flow will be. How will it happen?

Suppose its height is let us say H1, and final height is h1 so here also the pressure will be rho gh1 and on this side also it will be rho gh1 so basically there will be an equilibrium condition in which pressure will be equal in both vessels and there will be no full plausible because flow can be possible when there is pressure difference so this will be the condition for equilibrium when two vessels are connected by a pipe so today we have discussed Pascal's law, hydrostatic's law and We have covered all the important aspects of fluid statics today. In the last videos, we have started the fluid statics. If you are watching our video for the first time, then don't forget to subscribe to our channel and watch our previous videos. If you liked today's lecture, and if you have any doubts on today's topic, then you can put your doubts in the comment section below. To subscribe to our channel, press the subscribe button below and the bell icon.

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Transcript for:Fluid Pressure and Laws

Transcript for:
Fluid Pressure and Laws