 Consider a flow of water being directed through a flanged joint as shown in the figure below. The velocity of the fluid at the joint inlet is 4.25 feet per second, and a static pressure gauge next to the inlet indicates a fluid gauge pressure of 2.28 psi of gauge pressure. The joint redirects the fluid by 30 degrees and decreases the diameter of water pipe from 12.5 inches to 6.25 inches. Determine the horizontal force in the flanged joint and be sure to indicate direction that is tension or compression. So we have an x component of momentum because we are looking for this force in the x direction. That fx value is again going to be confusing because I'm throwing out fsx and fbx and fx here is just a name like anything else so I'm going to call it something let's call it f. So f is the placeholder name for that force in the left direction and I will describe my x-axis as going to the right so it is positive in the right hand direction. Next I can consider the assumptions that I'm going to be making and since we are a little bit more practiced at this we can probably get further out in our assumptions before we start. I will start by recognizing that I have steady state operation of incompressible flow I'm going to neglect any body forces in the x direction because remember the only body force that we care about for our purposes is gravitational acceleration. I mean I guess there could be other types of acceleration if we were analyzing I don't know. The box of a pickup truck being used as a pool by putting a tarp in the back of the pickup truck and filling it with full of water having people sit back there and we wanted to try to figure out the forces acting as the truck took tight cookies in a parking lot and exerted some acceleration in the outward direction I mean those are all things that we could consider but gravitational acceleration is the only body force for the scope of this class. Anything else we want to assume well I was told the velocity of the fluid at the joint inlet and I'm going to be determining a velocity at the outlet and let's just say that those are average velocities doesn't give us enough information to be able to describe a velocity profile here so we would have to assume something anyway and if we're going to be making an assumption about that velocity profile we might as well assume that the average velocity is 4.25 v per second and that's what is given. So I'm treating state 1 and state 2 as having uniform flow which again just is saying that the velocity doesn't change with respect to area so the velocity at the top is the same as the velocity in the middle which is the same as the velocity at the bottom etc. Uniform flow at 1 and 2. Okay I think that's enough to get started on our x momentum equation. First question I will ask do we have any surface forces and yes we do we have a force acting on the surface in the form of the forces acting on this joint so like you can basically think of this as like imagine this is bolted together each of these pipes has a flange on the end of it and there's some sort of bassiner connecting these together so what the problem is saying is determine how much force is required to hold that joint in place does that make sense so that force F that we are solving for is going to be the force at that joint so at the flange how much force is required either in the left direction or the right direction okay so I'm going to write a surface force in the form of F and that is to the left I define my x-axis as being positive to the right that means I write minus F and F is just a placeholder you can use epsilon you can use brian you can use whatever you like next question do I have any other surface forces yes I do I do because at state two the pressure of the fluid is atmospheric pressure but at state one the pressure of the fluid is going to be higher than atmospheric pressure that means that there's going to be a net force in the right direction as a result of the pressure that pressure is acting on the control volume because the gauge pressure at state one is higher than zero in the previous problems that we had analyzed we'd always had the same pressures at the inlet and outlet so there was no difference in pressure driving any sort of force but in this problem we do we have an unbalanced pressure which is going to result in a force acting on our control volume and I will count for that force as being in the right direction specifically in the positive x direction because the pressure is acting on the control volume so it's acting like this so the force exerted by the pressure at state one is going to be to the right so I will leave it as a positive quantity over here so my surface forces are negative F plus E1A1 because that's how I account for the force exerted by a fluid in terms of its pressure and area those are my surface forces my body forces I have already neglected and then I have steady state operations so this entire term disappears and I am left with the integral across each of the control surfaces that cross the boundary in the x direction I have two state points that cross the boundary in the x direction that's at least at least up here in the x direction it's not as though state two doesn't count because it's not perfectly in the x direction there's an x component of velocity at state two and there's a y component of velocity at state two so I have an integral at state one and an integral at state two I'm splitting this control surface integral across each of the state points let's row one times u1 times velocity vector one times da1 that would mean the wrong place to work the one and then plus integral at state two let's draw that better of row two u2 velocity vector two da2 okay I have uniform flow at states one and two which means that I can collapse that vector into a magnitude row one and u1 are both constants so they come out of the integral and I'm left with row one times u1 times average velocity state one times a1 and then do I add a negative side or not is that a positive quantity or a negative quantity right it's a negative quantity because the velocity in the area vectors are in opposite directions the area vector is always out meaning it's to the left here and our velocity vector at state one is to the right which is opposite of a1 therefore the collapse into a negative magnitude now same logic at state two we have row two times u2 times the average velocity at state two times a2 and do I add a negative sign there no I do not I don't because the velocity vector and the area vector are both in the same direction they're both appearing to the right or more accurately in the upper right direction but from the perspective of the x axis they both appear to the right okay then I recognize that state one has a velocity that is only in the x direction so the average velocity at state one and the x component of velocity at state one are both the same thing so I'll write this as negative row because I've assumed incompressible flow row one and row two are the same times average velocity at state one squared times a1 and then we are adding to that row times now I need to try to write my x component of velocity at state two in terms of the average velocity at state two I was told that I have an other offset here of 30 degrees which means the average velocity at state two is in this direction then the x component of velocity is going to be v bar two times cosine of 30 degrees and I didn't actually mean to write cosine of 30 degrees here my brain was focused on what I was about to say as opposed to what I was trying to write so the cosine of 30 degrees describes u2 over v bar two which means that I can write u2 as v bar two times cosine of 30 degrees so I'm taking row times v bar two times cosine of 30 degrees times v bar two times a2 which I can write instead as row times cosine of 30 degrees times v bar two squared times a2 and this is negative f plus e1 times a1 next let's run an inventory of what we know we have water that is incompressible and we are assuming standard temperature and pressure which I will write as an assumption water at we'll do a cool add symbol like I did last time at stp therefore the density of water is going to come from table a1 or a3 and if I were to pull that up table a1 you can see that water at standard temperature and pressure is going to have a density in embryo units of 1.937 slugs per cubic foot 1.937 slugs per cubic foot so we know density we know p1 we know p1 is a gauge pressure by the way and so we are plugging in the gauge pressure at day one if we were to use absolute pressure we would have to account for the fact that p2 is also acting on the control surface so either you can work the problem in absolute pressure at which point you would have a p1 a1 term and a p2 a2 term or you can just work the the problem in gauge pressure at which point you just have p1 a1 because the gauge pressure at state two is zero does that make sense ultimately all we only care about is the pressure difference between the two because that's what's exerting the net force anyway at state one we also know the diameter so we can calculate a1 and at state two we know the diameter as well so we can calculate a2 we were told the average velocity at state one so we know v bar one and we are looking for f so we have one equation and two unknowns right now if we had some other way of relating what was happening i could describe v2 in terms of what i know and then calculate v2 or plug it in symbolically but do i have a way of referring to the velocity at state two in terms of other stuff that i know is there some other tool that i can deploy that would help me out here you're right there is i have the conservation of mass so we could start with our reynolds transport theorem and say zero is equal to d dt of the integral of the control volume density with respect to volume plus the integral across the control surfaces of density times velocity vector times area vector we recognize that entire second term middle term is going to become zero because it's steady state therefore i'm saying zero is equal to the integral of state one's density times velocity vector integrated with respect to area plus the integral across state twos density times vector velocity vector with respect to area and the densities are going to come out because it's incompressible flow so i'm going to be left with the integral of velocity vector one with respect to a1 plus the integral of velocity vector two with respect to a2 i'm saying that that's equal to zero and then i'm collapsing that integral by writing those vectors in terms of a magnitude because i have uniform flow at which point i'm going to have v1 a1 and then i'm going to ask you is it a positive or negative quantity and you're going to say it's negative john because they're in opposite directions and i'm going to say well done so you're going to say zero is equal to negative v1 a1 plus then we have v2 a2 written in terms of uniform flow and i will ask is that positive or negative and you will say positive because they're in the same direction and i will say excellent excelsior so we have zero is equal to negative average velocity one a1 plus average velocity two a2 which is another way of saying average velocity one times a1 is equal to average velocity two times a2 does that make sense so we're saying the mass flow rate at state one is the same as the mass flow rate at state two the densities cancel so i have volumetric flow rate at state one is equal to the volumetric flow rate at state two which means that i can write v bar one a1 is equal to v bar two a2 so v bar two is equal to v bar one times the proportion a1 over a2 and you guys know how i like plugging in everything symbolically as far as possible before i get into calculating numbers so i'm going to plug that in over here and v bar two then would be v bar one squared times the proportion a1 over a2 quantity squared times a2 let's just run all that together we have negative f plus plus v1 times a1 is equal to negative rho times v bar one squared times a1 plus rho times cosine of 30 degrees times v bar one squared times a1 squared divided by a2 squared times a2 therefore i have rho times cosine of 30 degrees times v bar one squared times a1 squared over a2 is that all making sense excellent glad to hear it at that point i know everything in that equation except for f so i will solve for f and then i will simplify a little bit by factoring out where i can i could leave v1 a1 and then write density times v bar one squared times a1 times the quantity one plus cosine of 30 degrees times a1 over a2 p1 was described as being 2.28 psi g and then we are multiplying by a1 which is going to be pi over 4 times d1 the diameter it's a1 which is 12.5 inches squared and we are looking for an answer in pounds of force which means i need each of these terms that i'm adding together to be in pounds of force luckily for us a psi is a pound of force per square inch so taking 2.28 psi multiplied by quantity in square inches is going to give us pounds of force yay then we are adding to that the density of water at standard temperature and pressure which is 1.937 slugs per cubic foot multiplied by the average velocity is a1 which was 4.25 feet per second and i square that i'm just going to double check that it's 4.25 because all of a sudden i had a concern that i was reading that wrong squared squared squared multiplied by a1 which is pi over 4 times diameter one which was 12.5 inches he said hoping he was right yeah it appears to be then we are multiplying that by one plus cosine of 30 degrees times the proportion of a1 or a2 and that would be pi over 4 times diameter it's a1 which was 12.5 inches squared divided by pi over 4 times 6.25 squared inches squared it is 6.25 right yeah inches squared cancels inches squared pi over 4 cancels pi over 4 so i'm really taking 1 plus cosine of 30 degrees times the quantity 12.5 over 6.25 squared so that would be 2 squared which is 4 so i have 1 plus 4 times the cosine of 30 degrees everything inside of that parentheses is unitless so i just have to keep track of the units out front and i'm beginning to run out of space so i will scoot everything to the left here yeah take the parentheses with me that's what i want and i want pounds of force so i will start at a pound of force and work backwards i can describe a pound of force as one slug times one feet per second squared let's see how far that gets us slugs cancels slugs and second squared cancel second squared then i have feet feet feet feet in the denominator and feet squared and inches squared in the numerator which means i need 12 inches in one feet square everything that's not a square and i square everything and inches squared cancels inches squared and then i have feet feet feet and feet feet feet leaving me with pounds of force so i have everything i need to compute an answer except for the help of my calculator we can quickly remedy it looks like i can't zoom down far enough to for you guys to see that i'm just gonna grab 2.28 times the quantity pi over 4 and i'll let it scroll back down pi over 4 times 12.5 squared and then we are adding to that 1.937 times 4.25 squared times the quantity pi over 4 times 12.5 squared again times 1 over 12 squared all right as 1 over quantity 12 squared then we are multiplying by 1 plus the cosine of 30 degrees my calculator is already in degrees then i'm multiplying that by 12.5 squared over 6.25 squared i know i could just multiply by 4 but i don't trust my mental math as a general rule and we get a syntax error as is tradition so we get rid of one of our parentheses and we get 412.901 so let's just run through that again we had 2.28 times pi over 4 times 12.5 squared plus 1.937 times 4.25 squared times pi over 4 times 12.5 squared times 1 over 12 squared times the quantity 1 plus the cosine of 30 degrees times 12.5 squared over 6.25 squared cool beans and that's higher than i was expecting by must have gotten a sign wrong i'm gonna guess that happened here in this step because i was trying to keep track of the sign switches as i was solving for f in one fluid motion and as you guys know i don't trust my mental algebra there it is aha she was right here okay so when we solved for f we brought this over to the right hand side so it's negative p1a1 minus rho v1 squared times a1 plus density times cosine of 30 degrees times v bar 1 squared times a1 squared over a2 and then i negated everything but i forgot to switch this to a negative 400 pounds of force is a lot in that little joint i apologize for any confusion okay so let me try that again calculator if you would please switch that positive to a negative and we get a much more reasonable 176.5 pounds of force then the problem asked for a description of direction that is is the joint in tension or compression for that let's consider the fact that we are solving for a reaction force that's how much force is required to keep the joint in place so the force required to hold the joint in place is going to be to the left that means that the joint is going to be in tension does that make sense so that joint naturally wants to fall apart meaning in order for us to hold it together we have to provide a force in the left direction of 176 pounds of force therefore this is in tension