Calculating a force given a pressure distribution (Fluid Dynamics w/ Olivier Cleynen)

In this problem, we want to calculate the force exerted on the panel given the pressure distribution on both sides. What we’re looking at is a window, say a window on the side of the swimming pool, and this window has; on the right side, air; and on the left side, water. Both of those fluids are applying pressure on each side, and those pressures are distributed differently in space. On the right side we have air, and air has a uniform pressure — a pressure that’s the same everywhere in space. On the left side we have water, and water has a pressure distribution which is a little bit more complicated.

And the question we 00:37 - want to answer precisely is: what is the force coming from the right due to pressure, and what is the force coming from the left due to pressure? So let me show you how to deal with this. I’ll work it out step by step and put this [here] so we can keep the data in sight for our problem. I’m gonna start calculating. So what we have on the right side is a force due to the right side and the way we deal with this in fluid mechanics is by doing an integral. And we say the force is the sum everywhere of the pressure —the local pressure— multiplied by the area — the tiny bit of area where this pressure applies, dA, like so. And we do this integral over the whole area of the panel so it becomes the integral over the panel —over the whole window if you would like.

This force here — this integral 01:27 - here is very easy to calculate because in this case p on the right side, p is just pressure due to the atmosphere, so it’s a constant — it’s just one number — it does not change according to A. And so we can get it out of the integral and this becomes just here p multiplied by the integral over the whole panel of dA, like so. And p being here p of the atmosphere. So this is of course p atmosphere multiplied by the integral of dA, that’s just A, the area of the panel. And so this is now just a matter of putting the right numbers in. P atmosphere is one. But, be careful: one bar is not SI units we need to convert this into Pascals to have SI units and get the correct numbers in the end.

And one bar is 10 to the power 5 Pascals so it’s 1 times 10 02:27 - to the power 5 Pascal as pressure, and the area is 3 meters by 3 meters like so, so that’s 9 whoops sorry no “equal” here I should put here times 3 times 3. This is going to be Pascals, that’s gonna be meters squared and the multiplication of those is going to be Newtons and so of course this is 3 times 3 — 9 times 10 to the power 5, 9 times 10 to the power 5 Newtons as the force, the force on the right like so. We could leave it like so —whoops sorry I should move this up— We could leave it like so, but in practice in engineering we like to have Newtons, kilonewtons, or meganewtons. And so this turns out to be 900,000 Newtons, so 900 kilonewtons, yes. So this is the result that I’m going to highlight in the end.

03:25 - Just before we move on to water let’s just pause and ask ourselves how big a force this is. And the measure I like to have for this is “what’s an object that has a weight of 900 kilonewtons?” Well the answer to this is easy to calculate: 1 kilogram has a weight of approximately 10 Newtons and so 900 kilonewtons (900,000 Newtons) corresponds to about 90 tons. I’ll just put it here. And in comparison, 90 tons is the weight of 2 fully loaded trucks. That’s a lot of force and this is the force that applies on a window that has 3 meters by 3 meters. A window that has 1 meter by 1 meter you would get 10 tons.

04:08 - Yeah so it’s a lot of force that we have due to pressure of the atmosphere. Because 1 bar is a very high number and the reason for this is: that we live at the bottom of an ocean of air that’s about 100 kilometers high and so the pressure in which we live every day is very high, the atmospheric pressure. Okay so much for the right side, let’s now look at the left side let’s come back on this and have a look at this equation that we have here. We have pressure due to water is 1.2 times 10 to the power of 5 this is the starting value and then every time you go down with X here so let’s have a look at x — x goes down in this way so you start at 1.2 times 10 to the power 5 (1.2 bar yeah) and then you go down, every 1 meter you go down with x, you increase the pressure by a value of 9.

81 times 10 to the power 3 (it’s 05:04 - almost 10 to the power 4 okay). This, to us, is just magic — it’s coming from the sky— but we’ll see the reasons behind this equation — the physics behind this equation — in the chapter that’s dedicated to pressure later on. By then we’ll see probably that this looks suspiciously like gravity and this suspiciously like the density of water, but that’s a secret we keep for later yeah? At the moment to us this is just magic. And now what we want to do is calculate the force due to water on the left side with that pressure distribution so let’s take a look at that. All right. Let’s go. So the force applying on the left side is again the integral of p dA the sum over the whole panel of the local pressure force multiplied by the local amount of area.

And p, what is this 06:05 - time p? P is this whole big expression that we have written down here, so I’m just gonna be rigorous and just transcribe it over here. And so we have “panel” here, P becomes 1.2 times 10 to the power 5 plus 9.81 times 10 to the power 3 x, yes. This is the value of p. and then I integrate P with respect to area dA Now from a mathematical point of view this is a bit problematic because we have an x inside the integral and then we have dA which specifies according to which properties we’re doing the integral and we need to translate one into the other so let’s take a look at dA now let’s try to express dA as a function at least partly of x. I’m going to take the panel I’m just redrawing the panel that we have over here 3 meters by 3 meters and on this panel we take a little width, element of width here, and this element of width I’m gonna call it dz yeah it’s because we are looking at the distance along this direction here which is a distance z here and I’m also going to look at a tiny bit of depth depth x so I’ll call it dx yes and so I’m gonna look at a tiny bit of depth dx, a tiny bit of width dz here and the product of those this is going to be my dA yeah? So I can write that dA is equal to dx dz, a tiny piece of the window. I do not learn this by heart I find it out every time depending on the coordinates of the problem this is because sometimes you will have XYZ in different positions and you need to adapt to the coordinate coordinate system of your given problem to be able to figure out what your area is equal to in terms of infinitely small coordinates so now let’s take dA as equal to dx dz and write it in here in place of the dA yeah so we have the integral over the whole panel of 1.2 times 10 to the power of 5 plus 9.81 times 10 to the power 3 x and then we have here DX DZ now I could leave here a single —whoop! I’m sorry about this, I forgot, always forget to move up the sheet of paper so you can keep an eye on on my paper— so what I did here is it’s just dA from that equation here and I transcribe this into a dx and dz.

I could leave a single integral here and write it as the integral for the 09:01 - whole panel, but now I have two infinitely small components over here so I could try to write it as a double integral. It’s the integral of the integral of something yeah. The integral over DZ of the integral over DX of that piece here right there. Drom where to where do we evaluate these integrals? Well let’s look back at the problem. we’re gonna go in the Z direction in this direction here we’re going to go at Z is equal to zero and we’re going to start from zero and go to Z is equal to three so I write here for DZ 0 to 3 we’re gonna do the same thing for the direction X so we started at zero here and we go down to 3 over here for X so see every each time the integral from 0 to 3 over here ok this looks a little bit tricky to play with but you can see it mathematically is quite simple a double integral works as so that you do the first integral you hide the inside of the first integral and you do the first integral from 0 to 3 of everything that’s inside to dz.

So a way of representing this in 10:17 - your mind is to go from 0 to 3 of a big bracket here and this big bracket inside is the number that is the integral from 0 to 3 of ta da da da da DX like so. I’m not rewriting it I’m just in my mind conceptually replacing the inner integral by just a pair of brackets and this I’m going to then use to figure out the trick. Everything that’s inside this bracket here does not depend on that there is no z anywhere in this parenthesis dX does not depend on that either so what I can do is to say this is just the bracket that I had before multiplied by the integral from 0 to 3 of DZ like so. And so let me do it, let me just now replace the bracket by the inner integral. It is the integral from 0 to 3 of 1.2 times 10 to the power 5 plus 9.81 times 10 to the power 3 x dx, yes. And now I multiply this by the integral from 0 to 3 of DZ which I’m just right now going to call just Delta Z I could just replace it by 3 meters that’s also fine so you can keep it as an abstract representation right now and so now it’s just a matter of here carrying out the integral because all I have now is X and DX with a single integral this should be doable. So let’s let’s go with this. The integral of 1.

2 times 10 12:00 - to the power of 5 dx is going to be 1.2 times 10 to the power 5 X, so. Plus 9.81 times 10 to the power 3 X DX will integrate as ¹⁄₂ of x squared yes and I now close my brackets and I evaluate this function, this integral function, here between the limits X maximum and X minimum which in this case is 0 to 3 yes? And also multiply this by dz, by, sorry, Delta z, and so now let’s just put numbers in, write out the whole number thing before I type it into the calculator. So it’s going to be here 1.2 times 10 to the power 5 multiplied by 3 plus 9.81 times 10 to the power 3 multiplied by 0.5 x 3 squared multiplied by 3 like so. And so now just typing this into the calculator yeah I did it just previously for you I get this result here I get 1 point 4 8 2 4 times 10 to the power 6 and I remember now we had calculated the whole time coming back at the very top here, a force. What is the unit of force? Newtons. So I write down Newtons at the bottom right here this is the force on the left side.

And again in engineering we like kilonewtons, meganewtons 13:41 - and so on so forth, so I can write it here as F L is equal to 1 point 4 8 2 4 megaNewton and this is what I’m going to highlight in my answer. Yes. Yeah? So this is how you deal with a pressure distribution that is not uniform, when you want to calculate the resulting force applying on a single panel. .