 Water traveling through a pipe at an average velocity of 10 meters per second is split vertically across a t-joint as shown in the figure below. Determine how much force it would take to hold the joint in place. My first question is, is this a conservation of mass, conservation of momentum, or conservation of energy problem? We know it's conservation of momentum, not only because it says it in the category for the top, but also because there's a force involved. The way that we relate forces to information about the fluid is through the conservation of momentum equation. So we know that we're going to be using a conservation of momentum equation somewhere here. The next question is, how many directions do I care about? Do I need an x-axis, a y-axis, and a z-axis conservation of momentum equation, or can I get away with just two or maybe even one? I recognize that I'm only looking for one direction of force, fx. It implies that I'm going to need at least one conservation momentum equation in one direction, and hopefully I can get away with just one. I'm going to be defining my x-axis as being in the left and right direction here, and I'm going to define it as increasing to the right. So our x-axis is defined toward the right. That's going to be important when we start taking into account the directionality of the forces or the velocity vector. So I'm going to pull up my x-direction for the conservation of momentum, and then we can begin to simplify some of the terms. The first simplification I will make is neglecting terms that aren't relevant to my situation. I've defined a control volume in this figure, and I recognize that I have at least one surface force acting on that control volume. That is fx. Also note, because fx is acting in the left direction, and we defined our x-axis as being to the right when I plug in fx, which really should be named something else, so it's not as confusing with fsx and fbx. Let's just give it a new name. How about epsilon? You guys know how I like epsilon as a temporary variable. Epsilon is now the force that we are looking for, and epsilon is acting to the left, which means that I plug it in here as a negative value, because it's in the opposite direction as our axis. Then do I have any body forces? Well, for our purposes, the only body force we consider is gravitational acceleration, and presumably that is going in the up and down direction, not the left and right direction, so I'm going to be neglecting body forces, or rather, assuming that they aren't relevant to the x-direction. Then I can consider if I need to account for any changes in the control volume with respect to time. If I have steady state analysis, remember that that entire term disappears. Do you think it's reasonable to treat this situation as steady state? I do, at which point that entire term disappears. Then I'm going to integrate across the control surface, and note that I will write an integral here for any orifice that crosses the boundary in the x-direction, and how many ingress or egress points do I have crossing the boundary in the x-direction? That's right, I only have one. There is no x component of velocity at either state two nor state three, which means this is only going to be the integral across control surface at state one. This would be density times u1 times velocity vector one, da1. So I have negative epsilon plus zero is equal to zero plus the integral across the control surface at state one, and that integral is of the density of the fluid times the x component of the velocity times the velocity vector, integrated with respect to a, and that a has directionality as well. So I will pull out my density, and I will pull out my x component of velocity, at which point I have negative epsilon is equal to density times u1 times the integral of velocity vector da1, and then I will collapse that integral by treating the flow at state one as being uniform. I was told an average velocity at state one, so I can collapse that down to magnitudes of velocity in area. That's described here, average velocity of 10 meters per second. So when I collapse that integral into its magnitudes because I'm using uniform flow, remember that I have to keep track of if it's a positive or negative value. And if the velocity vector and area vectors are in opposite directions, that means I have to use a negative value. If they are in the same direction, I use a positive value. The area vector at state one is to the left, the velocity vector at state one is to the right, which means they are in opposite directions, which means this is going to be negative average velocity at state one times a1. My average velocity at state one is going to be entirely in the x direction, therefore I can just plug in u1. And at this point, I have negative epsilon is equal to negative density one times u1 squared times a1. And in the past, I will point out that we had been able to get rid of the density by assuming that we had incompressible flow. We still can assume incompressible flow, in fact I should probably write that down as an assumption, even though it isn't particularly relevant here. But my density doesn't disappear this time because the left hand side of the equation is not zero. So when I divide both sides of the equation by row, it doesn't disappear. I have to keep track of it. Anyway, epsilon then would just be the density of the water times the velocity that we were given at state one times the cross-sectional area at state one. I'm going to assume standard temperature and pressure for the water, which means that I'm going to be using a value of 998 kilograms per cubic meter. And that comes from table a1 or a3, the 998 kilograms per cubic meter times 10 meters per second. And then I square everything. And then I multiply by pi over four times diameter or was I just given area? I was just given area. So I just plug in 0.01 square meters. Now, what unit would you like to express an answer in? I don't believe that it asks me how it does. It asks for newtons. So let's try to hit newtons. A newton is a kilogram meter per second squared, which means kilograms cancels kilograms, second squared cancels second squared, cubic meters and meters cancels square meters and square meters, leaving me with an answer in newtons. So if my calculator would help me out here, I can take 998 times 10 squared times 0.01. And I get 998. And that was fx. So what we take away from this analysis is that all of the x component of the momentum has to be stopped by the joint. All of that momentum has to be resisted in order for the joint to not move. Therefore, all of the force associated with that moving stream of water is going to be absorbed in the x direction. So my epsilon value has to counteract that. Also note, if you are currently in thermo two and you are working through the Brayton cycle analysis, we calculate the force of thrust in those circumstances by taking mass flow rate times velocity. That mass flow rate times velocity is just as applicable here. We could have taken the mass flow rate of stream one, which would have been the density times the cross sectional area times the average velocity, and then multiplied that by the velocity at which point we would have had density times velocity squared times area, which is the same relationship we just came up with using our conservation of momentum from the Reynolds transport theorem. Cool, huh?