Making sense of the 3D linear momentum balance equation (Fluid Dynamics with Olivier Cleynen)

the momentum balance in fluid mechanics is super useful because it allows us to simply draw a control volume wherever we want in the flow and then compute the net force that’s applying to that fluid inside based just on measurements at the outlet and at the inlet however it’s also very intimidating and so let’s take a look at the different terms inside this equation we have here three different terms on the left is Fnet is what we want to calculate this is the net force the sum of all the forces due to a propeller inside or a compressor or due to shear inside the flow or due to pressure inside the flow due to gravity the net vector sum of all of those as one force this is what we want to calculate and this is equated to two things the first part is the change in time of the momentum inside the control volume and this is the sloshing back and forth this is, if you have a control volume with no inlet and no outlet but the fluid inside is going back and forth it still has movement and that distribution is changing with time then this will result in a net force on the fluid to this we add the net flow of momentum through the boundaries of the control volume this is the total sum negative in and positive out of all the flows carrying momentum with them inside the control volume so instead of trying to prove to you this equation let me try to show what it’s good for and what is not good for let’s take a case where we have only one Inlet and no it has only one Inlet and two outlets now like this what is this Vrel and this dot n vector in this equation so those terms here what are those for this is to designate the inlet or outlet velocity incoming outgoing velocity through the control surface through the surface of the control volume and so let’s imagine now that you have through this control volume this control volume in blue could be expanding or contracting could be moving the velocity of the fluid relative to the surface here V rel at the inlet one would be like this could be uniform for example what is V orthogonal or what is Vrel dot n what we’re looking for is the amount of velocity, the component of Vrel that is perpendicular to the area this is V orthogonal here and this Vorthogonal is following the direction of the flow so with it’ll be incoming when the flow is incoming it will be outgoing when the flow is outgoing and in this equation here Vrel dot n, the dot product of this relative velocity vector and a unit vector that’s pointing outwards every time, is the length it’s the length of V orthogonal and by convention in fluid mechanics this is very upsetting for thermodynamicists the orthogonal will be negative inwards and positive outwards okay let’s take a case where there’s only one Inlet and one outlet and let’s have a look at this equation the net force here is equal to the change in time here of the momentum inside the control volume plus whatever is going out here with density the component of velocity perpendicular to the surface and the vector velocity v2 as we do this integral here over the whole area 2 to make sure that we catch any non-uniformities in the velocity distribution and then minus the same thing for the incoming velocity. so in this what could create a net force? what would result in net force? well let’s take a look first at the first term what could be this term here this term here is not zero or not the vector zero if you want if the momentum inside the control volume changes. And it changes if the distribution of velocities is changing yes or if the distribution of mass following the velocity is changing so if the velocity vectors are all the same but the mass is traveling along those velocity (the density is not uniform) then you would have a change in this term let me give you an example you’re on the cruise boat and cruise boat rocks back and forth and because it rocks back and forth the swimming pool inside the cruise boat -I’m told this is a thing- the swimming pool inside the boat sees the water slosh back and forth and there is a force exerting on this water it’s equivalent to a force exerting on this water if it’s pushing it back and forth that’s a net force here will be computed by calculating the distribution of momentum inside the control volume, and if that changes in time because water is sometimes left sometimes right, then you have a net force that’s appearing okay so that’s so much for the first term now what about the other two terms what could cause if we say this is zero because nothing is changing in time the fluid flow is steady then what could cause a non zero force what could cause a force the with two remaining terms well there are many possible reasons why it could be nonzero the first could be that the mass flow is different so for example you have a tank with no inlet and the tank is leaking and flow is rushing out of this, the outgoing velocity and the going mass flow is greater than the (zero) incoming mass flow and so because of this even if you have zero velocity Inlet it will have a net force applying on the flow you could have velocity vectors with different lengths so great output velocity with small input velocity this would be a jet engine for example admits air with low velocity shoots it out the back with high velocity different lengths this is equivalent to a net force you could have different directions you have the same velocity incoming and the same velocity outgoing but they have different directions so these are different vectors v1 and v2 and if so you have a net force that is resulting from this, and you could have also which is very tricky different distributions of velocity and I’ll take an example of this later on in the chapter so examples practical examples we’ve seen this before this is an example where you have low velocity incoming into the pipe here and because the cross section of the pipe changes and you have now hollow cylinder of water shooting out the nozzle then you have a non zero force and the net force on the water is pushing this way so out of the boat and so of course the this results in the force opposite on the pipe and this is what these people here are trying to compensate [for] trying to exert the opposite force on the pipe. A helicopter would have here low incoming diversity and high outgoing velocity below even if the cross-sections are different and so this results in a net force this is what keeps the helicopter up in the air that’s what you can see on the bottom is this when the stream of air hits the water it creates mist on the surface of the water very pretty a rocket here’s a typical example of a machine where if you draw a control volume that surrounds a rocket completely move along with the rocket then you have a zero incoming velocity and all the velocity is outgoing and so you have a net force as exerting as a result of this and so finally instead of showing you more pictures and more math equations let me try to show you how this equation works in practice for this we are looking in the case of a blast deflector this is what you put on a aircraft carrier when aircraft are taking off and the reason you put this is to shield the aircraft behind. so this aircraft is about to take off the jet engines will blast some very hot very fast air it’s incredible how annoyed people get when you just destroy their 30 million dollar fighter jet with your hot blast of air from the engines of the previous jet and so to avoid this then you put this deflector plate and it doesn’t help the airplane in any way this is just to protect the following airplane so you have a hot jet coming in bumping onto this plate and being pushed upwards so I’m gonna try to simplify the situation we’re gonna say this is a deflector here and the flow is coming from the left with a uniform velocity and it’s leaving towards the top with a uniform velocity and the length of those vectors aren’t the same it’s just being deflected upwards like this I would say there is 10 kilograms per second coming in at 20 meters per second it’s a uniform like uniform and my question is what is the net force exerted on the fluid as it passes along and is deflected along with this deflector let’s take a look first thing we do is we build a control volume we draw a potato around the flow and we make sure that we cut the inlet and the outlet at nice 90-degree angles so that our math is gonna be a lot easier once you’ve done this this is a really cool thing about integral analysis once you’ve done this you can just remove all the rest from your mind and so what is resulting now from the physical and mathematical point of view is that you have a control volume you have incoming flow you have flow and nothing else matters nothing inside nothing outside matters all you’re computing is a difference in momentum between the outlet and the inlet so what is going on here let’s write this equation and let me try to switch the slide so we can see them all together like so yes no not yet let me see like so yes all right here we go so we have a little picture here to just remind us what we’re looking at and we have a general equation which is here for the net force and the first thing we’re gonna do is we say this is a steady flow it doesn’t change with time and so whatever this amount is inside it’s not changing with time this goes to zero or vector zero to be precise and this part here we’ll split into two one for the inlet and one for the outlet of our control volume and so now it looks like so we have this part is split into Inlet and outlet like this and we replace now this term the relative velocity dot the n vector by the length of this orthogonal vector here and the length of this vector V orthogonal this length is positive outwards and negative inwards yes so that this is gonna be a negative number because we’re in the incoming integral and this is gonna be a positive number in the outgoing integral so now if I rewrite again the same equation and I try to replace this velocity here with the absolute velocity so the length in absolute terms positive terms I have to put a minus at the inlet and have to keep the plus at the outlet okay the velocities are uniform so I can drop the integrals it means that the integral of whatever with respect to area is going to be just the area multiplied by whatever’s here and so I can just drop the integral and I’m left with this air at the inlet Rho density V orthogonal times a multiplied by the velocity vector in and rho V orthogonal times a multiplied by V out.

now rho with 11:40 - v orthogonal a this is the mass flow and so I can rewrite this as so I can say this is the mass flow here m dot in multiplied by the in velocity and m dot out V out and I can also simplify all this because it’s the same mass for incoming and outgoing in our case and so this just turns out to be net force here as a mass flow multiplied by the difference in velocity as a vector, difference in velocity between outlet and inlet okay so this is kind of nice and dot times v2 minus v1 how does this compute now with math we split into components this is three equations one for x one for y one for z but in our problem we only have two dimensions I think it’s plotted below here let me try this switch perhaps the picture yes in our problem we only have two dimensions of interest which are x and y X being the horizontal part and Y being the vertical one so I split into F net x and f net y and every time I have to take the X&Y components of V 2 and V 1 taking those components now may be a little bit tricky because V 2 X and V 2 y may be positive or negative depending on where V 2 is pointing and depending on where positive is overall for this I need to look at the diagram very carefully and picture where V 2 X is and so when I have these numbers which are plotted in blue here I have to pay attention to the sign positive or negative depending on the coordinate system what is the coordinate system in this case well if you look below in this little picture here we have v2 is going let’s say v2 is going in this way like so and so the X component of v2 is pointing this way and the X component of v2 is pointing away from positive x which is in this direction so v2 X here will be a negative number and so it will be here minus 20 cosine of 40 degrees as I’m taking the 20 degrees sorry I’m taking the 20 m/s of v2 which are like this and I’m taking the projection of this length down onto the x axis like this so this becomes minus 20 cos 40 yes the minus here is from above and then I had a length of V1 V 1 X, V 1 is purely horizontal and it’s pointing in that way and that way turns out to be in the opposite direction of positive x so the length of V one is minus 20 and then do the same for Y and it turns out that V 2 y is in the positive y direction and V 1 Y is just 0 so I end up with those numbers and I get those two things F net as a vector is two components one is positive in X and the other is positive in Y positive 46 positive 128 so if you plot this on the diagram it would go in this direction here I have a little difficulty with the mirroring of the video so you’re pointing in this direction positive x and positive Y this is the net force that is applied to the fluid and of course the net force applied by the fluid on the deflector is exactly the opposite and so would be pointing down into the corner of the deflector so this is how you make sense of the momentum balance equation for simple cases where you have one incoming and one outgoing vector with different directions. .