Calculating the net force on fluid flowing through a pipe bend

in this problem we’re looking at the flow of a chemical we have 200 kilograms per second of a chemical flowing through a pipe and the flow comes in horizontally 2 meters per second and then it swerves upwards through the bend of the pipe and exits vertically at 3 meters per second now what we’re looking for is the net force that’s applying on the fluid as it transits through the pipe to answer this question we need to work with the momentum balance equation and the momentum balance equation gives us the force, the net force like so if we can solve a rather long and torturous equation we say I’m gonna write here the way I like to write this equation is try to not remember every detail of it but remember the general structure and the general structure says that we have a change in time of an integral over the whole volume like so over the whole volume and to this we add an integral over the whole surface of the control volume and this is an integral with dA like so and the contents of those integrals change from equation to equation and so I’d like to not remember those in particular detail I just look them up in the formula sheet every time and so for net force we have here Rho times vector V like so and we have in this equation this surface integral here we have Rho V again the momentum per unit volume times this curious and rather annoying term which is the V relative; so the relative velocity, relative to the control surface dot the i vector which is a unit vector which is always pointing outwards like so and this integral here on the right is done over the control surface that’s a surface integral now this integral is very useful it’s very general you can use it in all kinds of cases actually I don’t know of any case where you cannot use this equation in fluid mechanics but it’s overkill for our problem and the equation we want to use for this particular case where is it there’s only one Inlet and one outlet and the flow is steady is it much simpler and it’s written like so in this case where we are if we have F net like so as a vector here is the mass flow multiplied by v2 the outlet velocity as a vector minus v1 the inlet velocity as a vector like so and we directly use this equation and start solving but I’d like to show you how to go from the top equation which is general and true for all cases down to this equation which is true for are particularly useful case here today but may not be generally true and so to do this let’s take a little time to go through this and let’s make the little space next so you push this guy down like so and let’s try to get from the top to the bottom of this equation see what we cross out in this big equation so the first thing we’re gonna cross out is this whole first term here here this is because this is the change in time of the momentum inside the control volume the momentum inside the control volume is not zero for sure but it’s change in time is because we have at the bottom here we have a flow that is completely steady the inlet and the outlet do not change at all the pipe is not expanding it’s not contracting you take ten pictures of this flow you will get ten times the same picture and so this whole component there let’s read it in light blue like so this whole component here this is equal to zero so this adds up to vector zero like so and then we have this a surface integral here and the surface integral is made over the whole control volume which in this case would circle around the pipe down here so something like this this would this would be my control volume let’s make a box like so yes would be my control volume over this whole area we do with this surface integral and so this comes down down to just two components one for internet and one for outlet let’s have a look at two things in there the first thing is this annoying term here Vrel dot n this is called the orthogonal and the orthogonal is going to be positive when the flow is outgoing and negative when it’s incoming these are the conventions we use in from mechanics so it’s just right here positive outgoing negative in like so and we’re going to leave the rest intact so let’s now write the rest of this equation here we have here minus the integral in coming here of Rho times the velocity vector here times here the absolute value of the orthogonal because V orthogonal here is negative when I put an absolute value here I have minus over there yes and this is for ta integral da this is for the inlet and for the outlet I’m gonna have exactly the same thing is there except it’s a plus and so this writes as integral out of again Rho vector V times here and then da like so okay so let’s try to work this some more we have the integral here when we look at the integral that we have we’re integrating with respect to area but when we look at the inlet of our control volume all the way down here then in there we have only one velocity vector a meaning in this simplified problem all the incoming vectors are all parallel and identical one to the other the velocity is uniformly distributed at the inlet this means that when we integrate anything with respect to area at the inlet they’re all going to stay constant with respect to area so this is really no need for an area for any real integral so this here adds up as being minus Rho vector V times the absolute value of V orthogonal here a and all of those terms are for the inlet so I add here the indices 1 1 1 in like so this is for the inlet and for the outlets here I have the same thing Row 2 vector V 2 like so then the absolute value of the orthogonal to and then e to myself and now here we notice that every time we have Rho times V orthogonal times a and again Rho the orthogonal a and this sums up here every time - of course mass flow so that we have I’m sorry I forgot the - here’s the - the mass flow multiplied by V 1 here and then plus the mass flow multiplied by vector v2 like so and so this of course adds up and sums up let me try to reduce now the space this here is equal to the net force mass flow multiplied by the difference between the outlet and the inlet and vectors ok so so much for the general theory now let’s try to solve the actual math let me take a little space here and move this to the top and let’s try to solve this general equation and to write out what net force is going to be in this case okay this equation on top is actually three equations this is because we’re in case where we have three dimensions XYZ so whenever I the vector here it actually means the three components of that vector and so we could write it like this we could say this is F net X with F net Y with F net set these are the three components of the vector F net and this is equal to dot m the mass flow it doesn’t have three components and then a vector and the vector is this the subtraction of those two vectors V 2 and V 1 and this vector has in the X Direction V 2 X minus V 1 X in the Y Direction V 2 y minus V 1 Y and in the Z Direction V - Zed is v1z like so okay so now to be able to figure out what F net is here we want to figure out what the components of the two velocities here are and so to do this we just fill in the numbers we have here the mass flow is 200 kilograms per second 200 and now let’s put numbers here V 2 X the X component of the outlet velocity when we look at the components here we see that V 2 is purely vertical and so V 2 X will be 0 V 1 X here would be 2 meters per second but pay attention v2 is going sorry V 1 is going in this direction so towards there yes well the positive x is going in the other direction towards there and so the length V 1 X is in the negative x direction and so V 1 X here is negative so when I subtract V 1 X here I’m going to subtract the value of minus 2 meters per second okay so V 2 y 2 y is the y component of V 2 and we can see that Y component is the direction here that’s pointing outwards of the screen so it’s going out of the video it towards you and nothing is happening in this dimension everything is 0 so I can straight away fill 0 for both of those components and then V 2 Z will be the component of in Z or with V 2 and so this is 3 meters per second is going upwards Z is positive upwards and so we have here 3 I could even write plus 3 like so and then the Z component of V 1 is 0 over here because V 1 is purely horizontal has no vertical component thanks so this adds up now as being 200 x plus 4 that’s 800 I’m sorry 200 multiplied by +2 so that’s 400 in the middle I have zero and at the bottom I have 200 multiplying by 3 which is 600 like so and this is f net and immediately yeah after I get a number I always check the result of units so units of force are Newtons and so let’s square this up at the bottom squaring is very difficult I’m new with this tool so I’m not exactly clean with my straight lines okay so this is how you solve with math calculating the net force applying to a fluid as it transits along now let’s try to see what this means and what this looks like with geometry with just lines looking at just the arrow to do this let me make a little bit more space let me shift this up yeah let’s have a look at what this F net really is f net is basically v2 minus v1 yeah there will be 2 minus V 1 vectors let me try to draw a nice and straight lines let me try to put them in blue like in this assignment we have here V 2 has a vector and we have here V 1 as a vector what is V 2 minus V minus V 1 as a vector well to answer this you take V 2 like so I mean put a straight line again let me try again I’m not so good at drawing when I’m going towards the bottom of the page v2 minus v1 so let’s try again drawing okay so we have V 2 like so this is V 2 and you put then V 1 at the end of V 2 you put actually minus V 1 at the end of V 2 this is minus V 1 what’s so and this will make the sum of those two vectors is then once put it in ready here let’s put it like so it will be a vector that goes like so yes yeah this is then V 2 minus V 1 has the vectors this you multiply this V 2 minus V 1 by the mass flow m dot and then you get the net force so that if we represent now the net force that’s applying to the fluid on the bottom here it would look something like this you start at a arbitrary point you go let’s have a look at the details again that’s 400 in X in 600 in Z so you go 400 in X and then 600 in Z and so this would be then something that looks like so sorry not very good at drawing yet I will improve with time but this is a little difficult for me now yes right so this would be here the net force that’s applying to this fluid here this is how you see visually what the meaning of the numbers and this in this result are 400 and 600 this is the net force that you need to apply to the fluid to make it as it comes from here and it leaves like that to make it change velocity from this to there who applies this force it may be several things might be the pipe it might be a machine inside the pipe it might be pressure and shear due to the fluid itself we don’t know we cannot know where the information that we have here all we know is that the net force is I’m sorry mm this direction like so as represented down there lastly if you want to know what the net force applied to the fluid sorry applied by the fluid to the pipe is then you take the opposite force and so the force applied by the fluid onto the pipe would be the opposite it would be something that looks like so here this would be the force of the fluid on the pipe what would be the opposite so with this this vector here in green will have minus 400, 0 and minus 600 Newtons of dimensions so there you go this is how you calculate net force with two inlets sorry two vectors one at inlet and one at outlet .