 incompressible liquid water is diverted through an elbow as shown in the figure below. I want us to determine the horizontal and vertical forces acting on the coupling by the nozzle, being sure to indicate direction. So here again, we are talking about a situation where the nozzle is attached to some sort of interface where a fluid upstream is being provided, maybe by a pipe. So we are trying to figure out what the force on that pipe is going to be, or how the force is going to affect the requirements of the coupling to the nozzle. I recognize that because I'm looking for a force in the x-direction and a force in the y-direction, I'm going to have to perform a conservation of momentum analysis in both the x-direction and y-direction. And just to get back into the swing of things here, we can set up a list of assumptions and probably populate quite a few already. First of all, we have incompressible flow that isn't technically an assumption because we were told we had incompressible water, so I'm going to leave that off from now. Or maybe I'll write no indication of an incompressible flow, so that's not assumption number one, because it was explicitly told to us. We don't have to make that assumption. Then our first real assumption is going to be steady state. I'm going to assume that because we were given no indication as to the size of the thing, I don't know if this is super long or super short. I don't know what it's made out of. I don't know enough information to describe the weight of the nozzle itself, nor the weight of the water inside the nozzle. So I'm just going to be neglecting gravity here. Neglecting gravity, because we have no other way of proceeding without making big sweeping assumptions as to the size of the thing and what it's made out of. Assumption number three, I'm going to assume that our water is at standard temperature and pressure, which from table A1 or A3, I can look up a density of water, and that would be 998 kilograms per cubic meter. Our x-force is going to come from the conservation of momentum in the x-direction. Our y-force is going to come from the conservation of momentum in the y-direction, and both of those conservation of momentum analyses are going to require that I have velocity at state one and state two. So before we get into conservation of momentum, let's talk about mass. I can describe the mass flow rate at state one as a function of the velocity in the cross-sectional area, and I can describe the mass flow rate at state two as a function of velocity in cross-sectional area. And since I was told a volumetric flow rate, perhaps it would be easier to consider the volumetric flow rate as the average velocity times the cross-sectional area, at which point I could say q, which is volumetric flow rate, which is the volume d time here, remember that that is average velocity at state one times area one, and also average velocity at state two times a2, because for there to be steady flow rate of mass, i.e. m dot in is equal to m dot out, and incompressible flow, i.e. the density doesn't change, the volumetric flow rate at the inlet must be the same as the volumetric flow rate at the outlet. So I could say that our velocity at state one is the volumetric flow rate divided by cross-sectional area at state one. This is a circular nozzle, so we have a diameter. So I could say four times our volumetric flow rate divided by pi times diameter at state one squared, and velocity two is volumetric flow rate divided by cross-sectional area two, which would be four times volumetric flow rate divided by pi times diameter two squared. Now again, we could proceed symbolically or we could plug in numbers and get an actual velocity. Let's see, last time we had calculated a number for velocity the previous time we had done it symbolically, so it seems like the next rotation here would be to handle it symbolically. So let's see how that goes, and if we need to calculate a number for algebra convenience, we can cross that bridge when we get there. I will begin somewhat arbitrarily with my x momentum, and then we will calculate an x-force, and then we can go into y momentum and calculate a y-force. My x-momentum equation is going to begin with our surface forces in the x direction, and then our body forces, which we are neglecting because the only body force we consider is gravitational acceleration, and we are neglecting gravity or rather the effects of gravity. Then I have d dt integral velocity in the x direction times density integrated with respect to volume across our control volume. This entire term is also zero because of the steady state assumption, and then I'm adding to that the integral across my control surface of our x component of velocity times density times velocity vector d area vector. I have two opportunities for mass to cross the boundary of my system in the x direction, that would be state one, wherein our average velocity of state one is the same as u one because the velocity is entirely in the x direction, and state two, where u two and squiggly v two can be described relative to our average velocity at state two, and either sine or cosine of 20 degrees. So u two is going to be cosine of 20 degrees times average velocity at state two, and squiggly v two is going to be sine of 20 degrees times the average velocity of state two, where u two and squiggly v two are the x component of velocity and the y component of velocity respectively. So back over here I will split my control surface integral into the integral across our state one control surface, wherein we are going to bring out u one and rho one, and because I'm assuming uniform flow, which by the way I should have written down, uniform flow, which really doesn't actually matter if it's an assumption or not because we are using the average velocity here for all of our calculations, that's, that is a uniform flow because it's average, but writing down the assumption is a good measure here anyway. Then I'm going to describe my u one density. I'm going to describe the simplification of the integral, collapsing the vector into a magnitude notation, and then I'm going to write average velocity one times area one, and then I'm going to ask you if I have to add a negative or not. And you should be good at this by now. Our velocity and area vectors are in opposite directions, which means that there's a negative out front. And then because the average velocity state one is you one, they're the same thing because the velocity is entirely in the x direction, I can write this as negative density times average velocity state one times area one. Then our integral at the state two exit point here is going to simplify down to u two times density times average velocity state two times area two. And do I add a negative or not? I do not because the velocity in the area vectors are in the same direction. And remember that I'm writing you two in terms of the average velocity state two by plugging in cosine of 20 degrees. So I have density times the cosine of 20 degrees multiplied by average velocity at state two multiplied by cross sectional area at state two. And this is going to equal all of our surface forces in the x direction. I have a gauge pressure or rather I have a difference in pressure acting in the right direction. But since I have absolute pressures here, I will account for them both by writing f x plus p one a one minus p two a two times cosine of 20 degrees. Just like with the velocities here, I have to account for the fact that the force exerted on the control volume by the pressure is in this direction. So the x component of that force is going to be pressure times area multiplied by cosine of 20 degrees. Furthermore, f x here is just the name of that force that could just as well be anything else, including but not limited to Brian or epsilon or whatever you like. This is the thing that we're indicating. I know it looks a lot like f s x and f b x, but it is just a place holder variable for that reaction force required. So I will populate this down just to be consistent. And then we can start to consider what we actually know about this problem. I know p one and a one. Excuse me. I know p two and p one. I know diameters at state one and two, so I can describe the areas as a function of diameter. I know cosine of 20 degrees, I know density of water. I know cosine of 20 degrees, I can write the average velocity in terms of volumetric flow rate, which I know. And diameter, I can write the area as a function of diameter, I can write the velocity as a function of volumetric flow rate and diameter in the area as a function of diameter. So I will first write this as f x is equal to that would be p two a two times cosine of 20 degrees minus p one a one minus density times average velocity state one times a one plus density times cosine of 20 degrees times average velocity state two times a two. And it looks like I didn't square those terms. Truly, I said that when I was writing it. I just didn't write the two. Because remember, you want is the same as average velocity. So I'm combining the two together. And you two is average velocity times cosine of 20 degrees. So I'm combining the two together. So squares here as well. There we go. Now I will write my areas as a function of diameter and I will write my velocities as a function of volumetric flow rate. And this is going to be symbolically heavy. So I'm going to scoot it over first preemptively. I'm actually going to go all the way over just to be optimistic. So p two is a number that I know. Multiplying by pi over four times diameter to squared times cosine of 20 degrees minus p one times pi over four times diameter one squared minus density times our volumetric flow rate times four divided by pi times diameter one squared and I have to square that entire term oh boy so that's going to be four squared times q squared times a one oh no not times a one okay we'll come back to that four squared times q squared divided by pi squared times diameter one to the fourth power and then we are multiplying by a one which was pi over four times diameter one squared. So four in the numerator and four in the denominator are going to cancel leaving me with density times four times q squared then the pi in the numerator and one of the pies in the denominator are going to cancel leaving me with d one excuse me and d one squared is going to cancel two of the d ones in the denominator so this four is gone this pi is gone. Two of these are gone. Pi times d1 squared. And then I can do the same thing for our velocity at state 2. It's going to be density times cosine of 20 degrees times 4 squared times q squared divided by pi squared times diameter 2 to the fourth power times pi over 4 times d2 squared. So 2 pi pi leaving me with density times cosine of 20 degrees times 4 times q squared divided by pi times diameter 2 squared 4 4 pi pi diameter squared row q squared cool. And at this point I can probably make my life easier by factoring out some stuff. Let me populate this down and see what factors out if anything. So I have diameter appearing in each term but they're not all the same. I have q appearing in the last two terms at pi appearing in the last two terms at 4 and row both appearing in the last two terms. So I will write this as pi over 4 times the quantity p2 d2 squared and I will write that p a little better. So it's easier to see that it's not a d times cosine of 20 degrees minus p1 times d1 squared. Then I'm going to subtract 4 times row times q squared divided by pi multiplied by 1 over d1 squared plus cosine of 20 degrees over d2 squared. And now I can begin to populate numbers. So I will say pi over 4 times pressure 2 which was 70 kilopascals multiplied by diameter 2 which was 0.05 meters. So 0.05 squared meters squared times cosine of 20 degrees. And I'm going to eventually want an answer in Newton's. So I'm going to make it my goal to get this into Newton's. And I can do that by breaking the kilopascal apart into pascals and then a pascale into Newton's per square meter. And kilopascals will cancel kilopascals, square meters will cancel square meters, pascals will cancel pascals leaving me with Newton's. And then I'm subtracting p1 which was 150 kilopascals. Squish this down. 150 kilopascals multiplied by d1 squared. d1 was 0.1 meters. And then I'm going to have the exact same unit conversion appearing. So I will actually factor that out. I'm bringing this calculation here out. That way I hopefully can limit how many things I have to write in my calculator. I'll make the calculators life easier and hopefully ensure a better chance of success when typing in numbers. That's out front. And I multiply by the quantity 70 times 0.05 times cosine of 20 minus 150 times 0.1 squared. So that will yield this first group over here in Newton's. And then I'm going to subtract from that with a big horizontal line here. The density was 998 kilograms per cubic meter. And then I'm multiplying by 4, which honestly I should write out front. Don't want to be consistent. 4. Then our volumetric flow rate was 0.025. 0.025 and that will be squared. And then I was in cubic meters per second. So cubic meters squared is going to be meters to the sixth second squared. And then I am dividing by pi, which I will write out here. And then I am multiplying by 1 over d1 squared. d1 was 0.1 meters. So 0.1 squared meters squared. And then I'm going to add to that cosine of 20 degrees divided by d2 squared, which is 0.05 squared meters squared. And then let's look at the units. I have meters to the sixth in the numerator. I have meters squared times cubic meters in the denominator. That's going to yield kilogram meter per second squared, which is a Newton. So I will just add that back here. A Newton is defined as a kilogram meter per second squared. And that meter has entirely too many hops. Try that again. Meter. Much more better. And meter is written meter. And just because I want this to kind of match visually, I'm going to bring this to the front side here and write this here. Beautiful. Apologies to everyone who's trying to write while I'm writing and write it in the exact same way. Okay. There we go. That will yield an answer in Newtons. So I can open up my calculator. And let's get rid of all this nonsense. Starkly. Here we go. We have no ability to scroll up and hold it there. So I will back out a little bit. First up pi over four pi over four times 1000 times the quantity 70 times 0.05 squared times the cosine of 20 degrees. My calculator is already in degrees minus 150 times 0.1 squared. And then fingers crossed and the number of parentheses minus four over pi times 998 times 0.025 squared times nothing in particular. And then I multiply that by the quantity. I'll go with double parentheses here just to hopefully make it easier to keep this straight. And you know, give me the opportunity to completely forget about that in half a second and have the wrong number of parentheses on the tail end cosine of 20 degrees. I'm already in degrees. So I am going to divide by 0.05 squared. And two parentheses. Fire in the hole. Calculator yields negative 1426. So let's go through here and double check that this all seems to match what we had written down. Power four times 1000 times the 70 times 0.05 squared times cosine of 20 minus 150 times 0.1 squared minus four over pi times 998 times 0.025 squared times one over 0.1 squared plus cosine of 20 degrees divided by 0.05 squared. Let me scan up to see that my algebra all makes sense. Plus looks like I canceled all those terms. So I feel pretty confident in our results. But oh, I see. Okay. When I factored out four times row times q squared over pi, I should not have written the minus sign off front because that implies that it's going to be distributed. So I should really have had a plus sign off front and then switch these two signs. I'm sorry I pulled that out incorrectly. I'm sure you guys were all screaming. I should really rename this channel too. John doesn't know how to use algebra. He's not effectively when he's working quickly. So I will switch this to a positive off front and then I will rearrange these guys or at least attempt to and let's see if that makes a difference in our calculation. We have a plus sign way out here plus and we are switching to get away with just a minus sign here, right? Yeah, that should work. Okay, I get negative 829. That seems like it makes more sense. So fx is equal to negative 829.85 Newtons. So we came up with a negative value for fx, which means that fx is acting to the left here. And since I asked for the horizontal and vertical forces acting on the coupling by the nozzle, if the nozzle needs a force to the left in order to hold it in place, that means that the nozzle is trying to actually exert a force in the right direction. Therefore, the coupling would be in tension. I can say that the force on the coupling is to the right as a result of the nozzle. Now that's our x momentum calculation out of the way. What we have to do now is a y momentum calculation. And I will begin with fsy plus fby is equal to d dt integral of our control volume of a bunch of stuff that doesn't matter. Quickly v density d volume plus the integral across the control surface of squiggly v, row, velocity vector, the area vector, we have no body forces because we are neglecting body forces. We have no change in time because we have steady analysis. So nothing can change with respect to time, eliminating that control volume term. I have a surface force in the form of fy, which was defined in the upward direction. I also have a pressure acting in the upward direction here. Because remember P2A2 acts up into the left. In order to write the vertical component of that, I'm going to have to multiply by sine of 20 degrees. So fy plus P2A2 times the sine of 20 degrees. That's up. Therefore, it's going to be in the positive direction because we define our y axis in the upward direction. And we only have to account for one surface integral because there's only one opportunity for mass across the boundary in the y direction. And it is at state two. So I will write here squiggly v2 row times and I will collapse my velocity by using the average velocity at which point I have average velocity two times area two. And do I have to add a negative out front? I do not. Because the velocity vector and the area vector are in the same direction. But if I want to try to rewrite this y component of velocity in terms of the average velocity at state two, I'm going to have to involve a sine of 20 degrees. And it's also going to be in the downward direction, which means it's going to have a negative. So this v2 is in fact negative v bar two times sine of 20 degrees, which means I have FY is equal to negative P2A2 times sine of 20 degrees minus sine of 20 degrees times v2 squared times A2. So let me just restate again that when we are talking about the x and y components of velocity, we plug in a positive or negative value based on their direction relative to the axis. So if they are in the same direction as our definition of the axis, that's going to be a positive number. If they're in the opposite direction, that's a negative number. And when I'm collapsing my integral into a magnitude for uniform flow, that's the velocity times area, the positive or negative value is only a function of are the vectors in the same direction or not the same direction. It doesn't matter if those are up or down. The positive and negative is only dependent on if they are in the same direction or not. So P2 I know A2, I can write in terms of diameter, let's go pi over four times diameter two squared times sine of 20 degrees and then I'm subtracting sine of 20 degrees times this was four squared Q squared divided by pi squared times diameter to the fourth power and then I'm multiplying pi over four times diameter two squared. So pi in the numerator cancels one of the pies in the denominator for in the denominator cancels one of the fours in the numerator two of these D2's canceled two of these D2's. So I have P2 times pi times D2 squared times sine of 20 degrees divided by four minus sine of 20 degrees times four times volumetric flow rate squared times nothing divided by pi times diameter two squared. Now let's populate some numbers. It doesn't make more sense to factor out stuff. I have sine of 20 degrees appearing in both places. I could factor out D2 squared and write D2 squared up front and then one minus one over D2 to the fourth power that doesn't really do me any good. No, there's not much point in trying to factor stuff out it would just be a sign of 20 degrees out front. And since that has no units, it's just as easy to write it twice. So pi, excuse me. So P2 was 70 kilopascals. So I will write 70 kilopascals multiplied by pi over four multiplied by the diameter state two which was 0.05 squared meters squared times sine of 20 degrees. And I want to value in newtons so I will break the kilopascal apart and write a Pascal as a new 10 per square meter at which point kilopascals cancels kilopascals square meters cancel square meters Pascal's canceled Pascal's leaving me with a value in newtons. And then I will subtract four over pi times the volumetric flow rate which was 0.025 and then I have to be careful with those units because that's cubic meters per second which would be meters to the sixth per second squared and I have to square 0.025 and then I'm dividing by D2 which was 0.05 squared meters squared and I have a sine of 20 degrees as well. So I recognize that I'm going to be left with meters to the fourth over second squared which isn't going to be in the right dimensions I can't convert that into newtons which means that I dropped some mass along the way. Where did I drop my mass? There it is. I dropped it here. I did not carry this density down into this calculation so let's add a row out front all of those and that means I'm going to be multiplying by 998 kilograms per cubic meter. It's a good thing too because otherwise I would have implied that the force at the outlet didn't have anything to do with what fluid it was which would have been a little bit alarming. Now cubic meters and meters squared are going to cancel five of the meters leaving me with a kilogram meter per second squared which is a newton. So a newton is a kilogram meter per second squared. Now kilograms are going to cancel kilograms meters cubic meters and square meters are going to cancel all six of the meters up top second squared cancels second squared leaving me with newtons. So f y is going to be a number that we can calculate with the calculator and I'm going to take 70 multiplied by pi over four multiplied by 0.05 squared multiplied by sine of 20 degrees. My calculator is in degrees times 1000 and we subtract 4 over pi times 0.025 squared times sine of 20 degrees times 998 divided by 0.05. We are missing a parentheses because of course we are. So let's just arbitrarily place one and hope for the best. That was not the best place to put it because 0.05 was a denominator for everything which is not what we want. How about we throw it here and then let's sanity check that calculation 70 times pi over four times 0.05 squared times sine of 20 degrees times 1000 minus 4 over pi which I guess I can write it that way times 0.025 squared times sine of 20 degrees times 998 divided by 0.05 squared. So let's add that square and third time for all the potatoes we get negative 61.6418. It's not quite as large a number as I was expecting. Let's go back through and check our algebra. Hey look guys I forgot the minus sign somehow didn't propagate forward so let's add a minus and a minus and a minus and let's try that a fourth time shall we. So fourth time says negative 155.66 newtons of force which is what I was expecting to see. Negative 155.66 and that force was defined in the upward direction which means we are applying a force in the downward direction of 156 pounds in order to hold that nozzle in place. What that implies is that the nozzle is trying to move up which makes sense right because we are neglecting gravity and the nozzle is pushing water out the bottom at least partially in the downward direction which is going to apply a force in the upward direction. Therefore we are saying that the force exerted on the coupling is going to be in the upward direction because the force applied by the nozzle would be in the upward direction. It has to apply a force in the downward direction to hold it in place. That whole directionality of the reaction force conversation again and before we break let me mention one more time that the thing that catches the most people on the conservation of momentum analyses is keeping track of positive versus negative signs. The positive versus negative when you have flow in the same direction as the area vector or not the positive or negative values depending on how you define your axis and the positive or negative values just because there's so much algebra involved with these problems those positive or negative signs can only make or break you in these calculations. I mean look at that I forgot a minus sign in both my x momentum and y momentum calculation and I should be the one that has experience with this and keeping track of them. So I would encourage you to get into the habit of keeping extra special attention on your positive and negative signs and maybe periodically go back and just double check that you kept them all consistently so that hopefully you have a chance of catching them if you miss one. Remember don't do what I just did be better than me you got this.