Today I want to start on the second topic in fluid mechanics, which is fluid statics. Here, the topics we will deal with first are, what is the pressure variation in a stationary fluid? Then show how this applies to measurement of pressure by manometers, and pressure by barometers. Then we'll look at calculating forces on submerged plane surfaces. And finally, look at issues of buoyancy and stability. So, the topic for today is pressure variation in a stationary, or a static fluid. So the questions we want to address are, let's sup,pose that we have an example of water, sitting in a beaker of water and we look at some point in this fluid right here. How does the pressure at that point vary with direction? Then secondly, how does the pressure in this beaker vary with position in the horizontal direction, this way? And also in the vertical direction, this way. So these are the questions that we're going to address in this segment. The first question, how does pressure vary at a point with direction? To address this question, I'm going to take an element of fluid, which I'll take to be a wedge shape, and here is a blow up of that wedge shaped element, and I'm going to balance the forces which are acting on it. So, the forces which are acting are py, in the y direction. Px in the vertical or z direction. And also, the pressure on the sloping plane here, I will denote by ps. The dimensions of this element are delta z in the z direction, delta x in the x direction, and delta y in the y direction. And the length of the sloping face is delta s. And the angle, or the orientation of the sloping plane, to the horizontal is theta. First, I'll balance the forces in the x direction or the horizontal direction. The forces which we have acting are, on the sloping plane we have the pressure force, pf, and the component of this pressure force in the horizontal, or the x direction, is ps times the area, which is delta-x delta-s, multiplied by sine theta. And this is balanced by the pressure force pushing in the y direction, which is py times its area, delta x, delta z. So, simple bit of geometry. The slope, the angle of the slope, sine theta, is delta z over delta s. So plugging that in and canceling out, we get the very simple answer that ps, the pressure on the sloping surface, is equal to py, the pressure on the vertical surface. In the y direction. Next, we apply equilibrium and balance the forces in the vertical direction. The forces are the same or similar. We have the component of the force on the sloping surface which is now multiplied by cosine theta. But not in the vertical direction, we have a new force which is gravity, which is equal to the weight of the fluid element. And the weight of the fluid element is the specific weight gamma times the volume delta x, delta y, delta z, over 2. And that is balanced by the upward pressure force in the z direction, pz, delta x, delta y. So again, simplifying this equation, we get ps + 1/2 gamma delta z = pz. But now, as I shrink this element down to a point, down to zero, delta z turns to zero, so that terms goes out. So, we get the very simple result again, that ps is equal to pz. So, putting this together, we get that ps = py = pz = ps, which brings us with the very important conclusion, that the pressure acts equally in all directions at a point in a stationary fluid. And this is sometimes called Pascal's law. You can also show, but I won't show here, that this also applies to a flowing fluid if there are no shear stresses acting. The second major question that we want to address is, how does pressure vary with position? In order words, both with height and with horizontal position. So to do this, again, I'll take a fluid element, although in this case, it's more convenient to take an element which is a rectangular or cubic shape. And the forces on this cubic shape are shown here. Again, the pressure in the x direction is py. In the z direction, pz, etcetera. But now, I'm assuming that the pressure varies from point to point. So, balancing up the forces in the vertical direction, we have the forces pushing upwards are p pressure force in this direction, p- dp by dz, delta z over two where dp by z is the pressure gradient of the y direction, multiplied by the area, which is delta x, delta y. This is balanced by the pressure force pushing down here, p plus dpdz, delta z over two, and in addition we have the weight of the fluid element which is gamma, the specific weight, multiplied by the volume, which is delta x, delta y, delta z. So again, simplifying this, we end up with a simple equation that dp by dz, the pressure gradient in the vertical direction is minus gamma, where gamma is the specific weight of the fluid. Next, I'll apply equilibrium in the horizontal, in the y direction. And, in the case, we have the pressure forces pushing in this direction, balanced by and equal to the pressure forces pushing in this direction. So, we have this equation p minus 2pdy delta y over 2 etcetera. And, calculating through there we get partial dp by dy is equal to zero. And I can do exactly the same thing in the x direction this way, with a similar result that partial dp by dx is equal to zero. So putting these two equations together, we see that the pressure does not vary in the x and y directions, in other words, the pressure is constant on any horizontal plane in a fluid at rest. The pressure does, however, vary vertically according to this equation here. Which I can now simplify and replace the partial derivative by an ordinary derivative because pressure only depends on height and that becomes dp by dc is equal -gamma, or minus row x g where row is the fluid density. So the important conclusion here, is that the pressure is constant on any horizontal plane. And it only depends on depth. So for example, in this strange shaped object here, if I take a horizontal line through here. The pressure at every point on this horizontal plane is exactly the same. And the shape of the fluids which sit above it is irrelevant. The only thing that matters, is the depth of this horizontal plane h, below the free surface. The shape is irrelevant. Now, we can use this to derive the equation for how the pressure varies vertically. And this is known as the hydrostatic pressure equation, which we've just derived as dp dz is minus gamma. So, if, for example, I take two locations at some depths below a free surface at stations one and two here, below the free surface and integrate that expression between two points. I get p is equal to -gamma + some constant on integration. Which, I can rewrite like this, as p over gamma + z, is also equal to a constant. And in this equation, the quantity p over gamma is known as pressure head, and the sum of these two, p over gamma + z, is known as the Piezometric head of the fluid. So the Piezometric head is constant, in the stationary fluid. If I apply this between these two points, one and two here, I get p1- p2 = gamma z2- z1. Or rearranging the pressure difference between those two elevations. p1- p2 = to gamma h, where h is the vertical height difference between those two points. So the only thing matters here is the vertical distance between the two points when we're calculating the pressure difference. So this equation gives us the so called hydrostatic pressure distribution, from which we see that the pressure increases linearly as we go downwards through the fluid. And there's a very easy way to understand this. If I consider a column of fluid here, say water which is sitting here, like a tall glass or beaker. And I balance up the forces which are acting, then the forces which are acting are gravity, which is the weight of this fluid, water, which is acting downwards. And the weight of this volume, is the specific weight, gamma, times the volume. So the weight is equal to gamma times the volume. But the volume is just the cross-sectional area of the prism, a, multiplied by the height, h. So therefore, we get this equation, that the weight is equal to gamma times hA. This must be balanced by the upwards pressure force on the bottom. So if the pressure here is p, the upward force is equal to p times A. Therefore the weight, gamma hA is equal to pA. The area A cancels on both sides, leaves us with p equals gamma h, in other words, the hydrostatic pressure distribution. So all of the hydrostatic pressure is, is the pressure that's required to support the weight of the fluid which is sitting on top of it. And that's all that hydrostatic pressure is. So in this case then, in this example, we have that the weight of the fluid is equal to, and I'll assume that the specific weight is 62.4 pounds force per cubic foot. Multiplied by the area, which is one square inch. Divide by 144 to change to square feet. Multiplied by the height of the column, which is 23.1 feet. Gives me a weight of ten pounds. This is balanced by pressure times area at the bottom, so therefore the pressure is equal to the weight divided by area which is ten over one is ten pounds per square inch. And that is an easy way to understand the hydrostatic pressure distribution.
