 A water nozzle is to be attached to a vertical pipe as shown in the figure below. The engineer designing the piping system needs to know the force exerted on the coupling by the nozzle, including the weight of the nozzle itself and the force of the water pressure, so that it can be designed effectively. The nozzle inlet diameter is 3 inches, the outlet diameter is 1 inch, the static pressure at the inlet is 19.2 PSI-G, the weight of the empty nozzle, which is fabricated out of stainless steel, is 10 pounds, and it has an internal volume of 150 cubic inches. Determine the horizontal and vertical forces acting on the coupling by the nozzle, being sure to indicate direction. Let's start with our assumptions. First up, we have incompressible flow, then steady state, and I am assuming all of our velocities are uniform velocities, that's an average velocity, uniform flow at 1 and 2, and I'm going to be neglecting any body forces in the x direction, and then let's assume that the water is at standard temperature and pressure, for which the density is 1.937. That value comes from table A1 at standard temperature and pressure, which is 1 atmosphere and 68 degrees Celsius for pressure and temperature respectively, 137 is our density for water. Next up, do I care about the x direction or the y direction? The answer is yes, I care about both actually. I have to perform a conservation momentum in the x direction to determine the x force, and a conservation momentum in the y direction to determine the y force. I can do them independently, I just have to do both. So let's start with the x direction, and this should be becoming old hat, so I will go through it a little bit more quickly. For surface forces, I have our reaction force, which I'm drawing as rx into the right direction, so I'm going to plug that in here as a positive value. I have a gauge pressure at state 1, but the surface force exerted by that gauge pressure appears in the y direction. Therefore, I don't have a surface force exerted by that pressure in the x direction. Then there are no body forces, because I've neglected any body forces in the x direction because the only body force we're considering is gravitational acceleration. Then I have steady state, which means this entire term is also 0, and I have one opportunity for water to cross the boundary in the x direction, and that is state 2. So my control surface integral is going to be the integral across state 2 of density 2, u2 times velocity vector 2, times da2. So rx is equal to rho 2 comes out of the integral, u2 comes out of the integral, and I've assumed uniform flow, so the integral of velocity vector 2 with respect to a2 is going to collapse down to magnitudes of average velocity times area 2, and you know it's coming. Is that a positive or negative value? Well, our velocity vector and area vector are in the same direction, which means that it's going to be a positive value. Then I'm going to try to rewrite u2 in terms of average velocity 2, and for that I recognize that I have an x component of velocity here, but I can describe by using a cosine. So cosine of 30 degrees is equal to that is the proportion u2 over v bar 2. Therefore I'm writing u2 as v bar 2 times cosine of 30 degrees, which means that I'm actually writing this as density of water. It's incompressible, so the density is the same everywhere, so it's just density times cosine of 30 degrees times the average velocity at state 2 squared times a2. And this point, I know a2 in terms of diameter, I know rho, I know the cosine of 30 degrees, but I don't know the velocity at state 2. So I can rewrite the velocity at state 2 in terms of the velocity at state 1 because I know both cross-sectional areas by using our conservation of mass. So just like last time, it's going to simplify down to the velocity at state 1 times the area at state 1 is equal to the velocity at state 2 times the area at state 2. And just because I want to show it a little bit differently, I'm going to calculate that numerically and plug it in as opposed to doing what I normally do, which is calculating a value symbolically. It doesn't really matter. You do. I'm going to say v bar 2 a2 is equal to v bar 1 a1. Therefore, v2 is equal to v bar 1 times the quantity a1 over a2. And that would be v bar 1 times pi over 4 times diameter 1 squared over pi over 4 times diameter 2 squared. Which is equal to average velocity at state 1 times the proportion d1 over d2 squared. My velocity at state 1 is given as 6 feet per second. So if I were to take 6 multiplied by the proportion 3 over 1 squared, I would end up with 54. So my velocity at state 2 is 54 feet per second. I'm going to make that a little bit bolder. Not because it's important, but just because when we refer to this later, I don't want to have to just scrub around the page going, where is it? I know I calculated it. You cannot find it. Oh no. 54. Okie doke. So then rx is going to be 1.937 slugs per cubic foot times cosine of 30 degrees times our velocity at state 2, which is 54 feet per second, and then I square everything and then our area at state 2 is pi over 4 times the diameter at state 2, which I believe is 1 inch. Yup. 1 inch. And then I square everything. And I'm looking for a result in pounds. Yes, pounds of force. So I will start at my destination and work backwards. A pound of force is 1 slug times 1 feet per second squared. Slugs cancel slugs. Second squared cancels second squared. And I have square feet times square inches in the numerator and cubic feet times feet in the denominator. So I need 12 inches in 1 foot. I square everything. 1 squared is boring. Feet squared. And feet squared cancel feet and cubic feet. Square inches cancels square inches, leaving me with a result in pounds of force. So, calculator, if you would help us out here. And that would be, oh no, screen went away. Okay, we made it through the void. So 1 point, 1, come on calculator. YouTube, 1.937 times the cosine of 30 degrees times 54 squared. Looks like I dropped a trilling presidency there. Times 54 squared times high over 4 times 1 over 12 squared. And I get 43.25, oh, no, pi over 4, why are you appearing there? What happened there? Oh, it's what was that a device on? I don't know what's happening to my calculator. It's all the calculator's fault. That's what's important. We get 26.67 26.67 pounds of force and you know what's going to happen next. Is that force on the nozzle to the left or to the right? Well, we calculated a reaction force. So the force required to hold the nozzle in place has to be to the right. We have to counteract the force on the nozzle by providing 26.68 pounds of force to the right. That means the actual force in the nozzle is to the left. I'm going to say 26.67 the x component of force is force on nozzle is to the left. Because the force required to hold the nozzle in place the reaction force is to the right. And we know that because our rx value is to the right. We got a positive value. Therefore, the reaction force is to the right. So one dimension down now we can do our y momentum equation and for that I will open up a new sheet. And the y momentum equation is going to have the same general terms as our x momentum equation. The only real difference here is that we are talking about the y components of velocity which we indicate with an italics v. Now, here's the thing. I know my handwriting is difficult to discern if I'm writing an italics v or not. So instead of writing an italics v I'm basically going to be writing the greek letter nu and that's just to try to exaggerate it by writing it like this. So hopefully you guys are able to interpret what I'm trying to indicate here. Nu for our purposes here is just going to indicate the y component of velocity. I'm going to write fs y plus fby is equal to the integral of some stuff that isn't going to matter because of steady state and that is density times nu times dv plus the integral across our control surface of density times v times velocity vector dA. Then we can plug in and cancel terms I'll get rid of my calculator here. I have a pressure at state one that is higher than atmospheric pressure which means I have a gauge pressure which is exerting a force on the control volume which is going to appear in the negative direction because I'm defining my y axis as being up. Furthermore I have a surface force in the form of ry which is up so it's going to appear as a positive value. So my surface forces here are going to be ry minus p1 times a1. And again if you work the problem in gauge pressure then I only have to care about state one's pressure if you were to try to model this with absolute pressure you would have to account for the pressure at state two as well. And it's again really just the pressure difference that matters anyway so because p2 is at atmospheric pressure the difference between the two is gauge pressure so p1a1 is the surface force that's relevant as a result of the pressure differential exerting a force on the control volume. Then we have body forces this time because we are talking about the y axis and gravity is down in the y axis presumably actually you know just for good measure here let's just say g is down now it's not just presumably it's assumably we have weight in the form of the weight on the nozzle itself we also have the weight of the water in the nozzle because we were told that the internal volume is 150 cubic inches so I'm going to subtract the weight of our nozzle and I'm going to subtract the weight of our water that's everything on the left for the right I can get rid of our volume term because we are at steady state and we have two opportunities for water across the boundary in the y direction that's at state one in the downward direction and to state two in the downward direction as well so I am going to write for my first integral here I have the y component of velocity at state one times density because they come out of the integral and then I'm taking the integral of the velocity vector with respect to the area vector which because of the uniform flow assumption is going to simplify down to a magnitude so that's v bar one times a one and do I make that a negative or not well the area vector is going to be acting upward because it's always defined in the outward direction and my velocity vector is in the downward direction which means that they are in opposite directions which means I add a negative and then at state two I don't have quite enough room so I will make a little bit of room here so I am going to write that as plus the y component of velocity at state two times the density at state two times and again, okay let's take this down a line hopefully that's not confusing then I'm going to bring out my velocity vector as a magnitude because we are assuming a compressible flow and do I make that a negative or not no I do not because the velocity and area vectors are in the same direction now we can go through and consider these terms one by one r y is what we are looking for p one we know we can determine as a function of d one which we know the way to the nozzle we know the way to the water we can calculate because we know the volume so that would be the mass of the water times gravity which is the volume of the water times density times gravity then the y component of velocity at state one is going to be negative v bar one you follow because the average velocity at state one is in the downward direction which means the y component of velocity is a negative value because we define our y axis is going up this is going in the downward direction and density one is the same as density v bar one we know a one we can determine in terms of d one the y component of velocity at state two we can describe relative to our 30 degree angle so sine of 30 degrees is going to be y component of velocity divided by average velocity which means this is going to be negative cosine of 30 excuse me sine of 30 times v bar two because again that's in the downward direction density two is the same as density and a two we can determine as a function of diameter so at this point we have everything that we need just a matter of plugging stuff in so I'm going to do something dangerous I'm going to try to solve for our y and substitute in the same step so we begin with p one a one and then we add our weight of our nozzle and then we add density times water times gravity and then we subtract oh excuse me it's negative v bar one times negative v bar one which is going to yield a positive quantity density times v bar one squared times pi over four times diameter one squared and then at state two we only have one negative so it doesn't cancel so we're left with minus density times sine of 30 degrees times v bar two squared times pi over four times diameter two squared and if we had left our v bar two symbolic we could make that substitution now and I'm sure we could write this in terms of all the diameters ever but for now this is perhaps a little bit more straightforward so p one was given as 16.2 psi g and we are multiplying by a one which is going to be pi over four times diameter one squared I really should have made that substitution up here in order to be consistent with my other areas. Diameter one was three inches and then I square everything and I'm looking for pounds of force and a psi is a pound of force per square inch which means I'm already in pounds of force then we add the weight of our nozzle which was 10 pounds of force oops then density of water was 1.937 slugs per cubic foot and then we multiply by our volume of water which was 150 cubic inches if I am not mistaken and I am not mistaken then we multiply by gravity which is presumably 32.2 feet per second squared I guess I should add that to our assumption per second squared and then I'm going to want that in pounds of force so I will start at our destination and work backwards upon a force is one slug times one foot per second squared slugs cancel slugs and then 12 inches are in one feet and I cube everything one cube is boring second squared cancel second squared cancel inches cubed feet and cubic feet cancel cubic feet and feet leaving me with pounds of force so next up is going to be the density of water times pi over 4 times the average velocity is day 1 times d1 squared so 1.937 slugs per cubic foot times pi over 4 times velocity 1 squared which was 6 feet per second yeah 6 and then 54 6 feet per second and then I square everything and then multiply by the diameter at day 1 which was 3 inches 3 squared inches squared and I want pounds of force so I will start at my destination and work backwards out of force is a slug times a feet per second 12 inches in one foot I square everything one squared is boring slugs cancel slugs feet and cubic feet cancel square feet and square feet leaving me with pounds of force and then we are subtracting what appears to be just a part of a row we are subtracting 1.937 and you know what just to be a little bit more efficient I am going to copy and paste this line because like 80% of it is the same density times sine of 30 degrees so I should go to this all over to the right sine of 30 degrees as pi over 4 times the velocity is 2 squared which was 54 and then we are multiplying by the diameter at state 2 squared which is 1 which looks right so density times sine of 30 times pi over 4 times velocity 2 squared times diameter at 2 squared I will point out here that if you had plugged in the velocity at state 2 symbolically in terms of velocity at state 1 and a bunch of diameters it would probably be easier to factor out a bunch of stuff and write it in terms of a unit conversion that only occurs once with the ingress and egress point but this works just as well now I can calculate for days calculator you are needed so I am going to take 19.2 and then I am going to multiply by pi over 4 which is going to be a theme in this calculation I really wish there was a pi over 4 button on my calculator times 3 squared and then we add to that quantity 10 then we add to that quantity 1.937 times 150 times 32.2 and then we multiply by 1 divided by 12 u and then we add to that 1.937 times pi over 4 times 6 squared times 3 squared times 1 over 12 squared and then we subtract 1.937 times the sine of 30 degrees my calculator is already in degrees times pi over 4 times 54 squared times 1 squared which is boring times 1 divided by 12 squared ok it's 2 carats that's entirely too many carats and let me sanity check this calculation 19.2 times pi over 4 times 3 squared plus 10 plus 1.937 times 150 times 32.2 times 1 over 12 cubed plus 1.937 times pi over 4 times 6 squared times 3 squared times 1 over 12 squared and then we are subtracting 1.37 it's 1.937 not 1.37 calculator you made a mistake again 937 times sine of 30 times pi over 4 times 54 squared times 1 over 12 squared I'm just going to double check that last term make sure there's nothing doesn't seem quite right yep looks good to me so our answer is going to be 139.15 and then it wants to know the direction again for some nozzle is we got an Ry value that was positive which implies that the reaction force is up so in order to hold the nozzle in place you have to supply a force in the upper direction of 139 pounds therefore the actual force on the nozzle is going to be in the downward direction or rather the force on the pipe guess that's what I really meant the force on the pipe is downward meaning the pipe would be in tension I'll clarify that over here it's not the force on the nozzle it's the force on the pipe