Computing the net force on a fluid with non-uniform velocity distribution

in this problem we’re taking a look at cases where the flow where the velocity distribution at the inlet and at the outlet is different and as a result a net force is exerted on the floor now we’re trying to compute this the answer to how to compute the net force on a floor like this one is found again using a very powerful a very general equation which is the balance of momentum equation in fluid mechanics this allows us to compute the net force here as a function of two terms one is the change in time of the momentum inside the control volume this would be for example the change of momentum due to the sloshing back and forth of water inside the tank yes inside of the control volume and the second term is the sum the net sum of the momentum flow coming in and out of the control volume what good is non-uniform velocity distribution math for? well let’s take a look at a flow like this let’s say you are interested in the flow around this body and you would draw a control volume that’s around the boat and that moves together with the boat if you do do that and you move together with a boat on the control volume then you will get a velocity Inlet that’s pretty uniform but the velocity at the outlet that’s not at all uniform it would look like this Inlet would be the velocity of the river relative to the boat so the velocity of the boat relative to the river but at the outlet you would have a zone where you have very high velocity because the thrusters of the boat and the zone where the velocity is much lower there’s a deficit because of the drag caused by the barges and so the question becomes then in this case here with those two velocity distributions at the inlet and the outlet how do we compute the net force what is the net force applying on the water as a result of those two distributions so here’s a worked out example for the momentum equation with non-uniform flow so let’s take a look at this we’re looking in general and a family of flows and family of problems which consists in measuring the velocity at the inlet and then the velocity at the outlet the velocity at the inlet is uniform the velocity of the outlet is not uniform has some kind of change and it could be a bump could be a deficit and as a result of this we try to compute what is the net force that was exerted on the flow at the center due to the fluid flow in this particular problem we’re looking at a pipe flow and as air coming in from the left side going to the right side and I at the inlet the flow has 20 meters per second uniform distribution like so and at the outlet the flow has also 20 meters per second but this velocity distribution is changed and so you have a part of it that has only 19 this is the 19 here and as you go up towards Y is equal to 2 so as Y increases then the velocity increases and then you reach a velocity of 21 so that on average you still have 20 but the velocity distribution is changed and the question we try to answer is what is the net force that is applying on the fluid on the air as a result of this change of velocity distribution so let’s take a look well the first thing we do is we draw a control point which could be any shape we want all that matters in a control volume is to choose very carefully the inlet and the outlet of the flow so that we have a very clear cut boundary at the inlet and at the outlet once you’ve done this you can remove everything that doesn’t belong so anything that’s inside the control volume just disappears anything that’s outside of the control volume just disappears and you’re left with just the surface and it’s crossed by the inlet velocity and it’s crossed by the of the velocity on the other side and this is it so based on this how can we compute the net force on the flow well we write this equation let me try to switch the slides again so we have yes so we have the picture down there to remind ourselves now what we’re looking at and this is a general equation for the net force due to the the fluid flow and in this distribution here in particular we have the first first term here we’re going to have zero this is because this is a change in time of something and even though that’s something here in that case is not zero this is the entire amount of momentum inside the control volume it doesn’t change with time and so its change (d/dt) with time is 0 so this whole first term will be zero and then we’re gonna do what we’re gonna do is gonna split this into Inlet and outlet and so the equation looks like so here what we have and this turn here at the inlet becomes just the same thing with “in” and the same thing with “out” there then this is very annoying V rel dot n the dot product of the velocity relative to the control surface and the N vector which is a unit vector which is always by definition pointing outwards this is a number this is not a vector it’s a number and this we can write as V orthogonal so I’m gonna replace here V relative dot N every time by the orthogonal V orthogonal is a pain because by definition it is negative incoming and positive outgoing and so we need to pay real attention when we put in numbers for the orthogonal here so the hoop number one out of two is the sign of V orthogonal again the same equation as before this V orthogonal here when I’m gonna replace it with its absolute value I’m gonna have to pay attention to the sign and so I have a minus appearing in front when I put an absolute value here and when I do the same with the absolute value there at the outlet it’s a positive number and so this remains a plus like so okay hoop number two is the sign of V when we drop the vectors so again the same equation here it’s a vector equation this has x y and z components three components this is actually three equations well this equation when we drop the vectors when we only interested in X direction because the whole flow is entirely in this horizontal direction we’re going to be able to remove the vectors when we remove the vectors some components of vectors might be positive or negative and we have to pay attention so let’s remove this remove the vectors so we have a vector equation on top becomes now a scalar equation this is the length of f net and since all the vectors are in the x-direction I can just remove the arrows without changing the lengths now hoop yes that’s Vin and Vout here have positive or negative value depending on their orientation and depending on the orientation which defines where positive is and so in this case in this diagram which is right here below we have X positive in that direction and all velocities are also [pointing] in this direction so all the numbers that in this equation the Vin and Vout are going to be positive numbers and so I can just now replace Vin by the absolute value of Vin and Vout by the absolute value of Vout and I’m safe I don’t need to change the sign in front of this but this may change depending on the orientation of your vector and the orientation of where positive is yes this depends on every problem ok so let’s take this equation here on the bottom and let’s reproduce it again on the next line like so and now we’re going to integrate we’re gonna carry out this whole integration so let’s do some clean up first V in V in becomes Vin squared V out V out [becomes] V out squared like so so Vin has been condensed and dA is the area at the inlet and I split it into two components one is the component is vertical which in y dA becomes dy and then the component across the flow which is in the Z Direction becomes DZ and so this area da here becomes dy DZ and so I do the same over there what does this give us well what is happening in the Z direction the integral of everything respect to Z is nonsense, because everything that’s going on in the Z direction here is — well in the Z direction nothing is happening to formulate it better so that every component of velocity every component of every property is going to be exactly the same wherever I position myself in the Z direction so that this integral here this outer integral with respect to DZ just disappears and I just have to put here the Delta Z so this is the width of the pipe away from the screen if you want and the same thing for V out I’m gonna take the rho and put it out of the integrals and again rho out of the integrals and now I have to integrate a V in here from where to where from the start the value of y at the bottom and to the value of y at the top and this is y1 and y2 here these are just the general values and these may be different at the inlet and at the outlet yeah so pay attention in this case in this particular flow we have the same limits up and top and bottom for both but they may be different so pay a little attention to that so again taking this last equation here and reproducing it on the next slide we have this one of those two integrals will be very easy to carry out and this is the integral for Vin because Vin squared here does not change if you look at the diagram here below Vin has a uniform distribution it does not change with Y and so Vin can just get out of integral and we’re left with the integral from 0 to Y 2 of dy and this becomes just Delta Y like so delta Y In the out however has to remain like so because V out is a function it’s not it’s not a number it’s not the constant and so now I put numbers in into this thing and so for the inlet flow I have Delta z; so minus Delta z; is 1 the width across the pipe into the screen if you want is 1 meter delta Y this is the height this is 2 meters this is the density of air and this is the incoming velocity 20 squared V 1 squared and at the bottom I have again Y so delta, sorry Delta Z the density Rho and then I integrate from 0 to 2 this function for V out, V out here is 19 + y I put it to the square because there’s a square here and I integrate this with respect to y to dy I spare you the math you just plug into the result and then you get plus 885 Newtons okay so that’s so much for the math but the first time you do this surely there must be an element of surprise when you see the result how come we have + 800 Newtons of force just to change the distribution of velocity let’s take a look again at the original flow we have here in this flow an incoming velocity of 20 meters per second and an outgoing velocity but also 20 meters per second on average it’s still exactly precisely 20 meters per second we have the same mass flow incoming and outgoing and so what happened here? just because we slowed down a little bit the air on the bottom and accelerated a little bit the air on the top not changing the average velocity we have to apply force to the flow? well yes this is because momentum is not proportional to velocity it is proportional to velocity squared and even though we have not changed the average velocity we do change this average square of velocity so in other words if you slow down the air from 20 to 19 so this particle here here is now flowing with 19 meters per second and you slow it down then the air loses momentum and this amount of momentum is smaller than the amount of momentum you have to provide to accelerate the air from 20 to 21 on the top here so that overall somebody has to pay somebody has to push the fluid add momentum to the fluid to change his velocity distribution from an average of 20 meters per second to an average of 20 meters per second but with a slanted distribution like so who is doing this where is this force coming from? it could be a machine which is hidden inside the pipe it could be some moving propeller but if there’s no propeller there is no machine adding anything then typically would be pressure so there would be a higher pressure at the inlet and there is at the outlet and this change in pressure is causing or is equivalent so there is change in velocity distribution so we don’t know that so the integral analysis does not allow us to find out what is happening inside it just quantifies the net effect so in practice you could have pressure pushing the flow sorry in this direction and you have shear on the contrary which is slowing down the flow and removing momentum from the flow all we are able to calculate with Inlet and outlet is the net sum of those two effects so this is how you manipulate the net force equation, balance of momentum equation in a flow where the inlet and outlet have different velocity distributions .