In the previous segment, we looked at the basic ideas of the momentum theorem. Now, we'll do some examples. So, we have a 0.3 meter diameter pipeline which terminates in a nozzle of 0.15 meters. A water jet is shooting out of the nozzle here and it's de-, it strikes a flat plate and is deflected to the side. If the water flow rate through the pipe in the nozzle is 0.25 cubic meters per second, the force required to hold the plate stationary is most nearly which of these alternatives? So the first step, so we start from our momentum theorem equation summation F. And in this case we have just on inflow here, which I'll call 1, and a radial outflow, which I'll call station 2. So the relevant form of the momentum equation is given here, where we're assuming all the flows are essentially one dimension. The next step in momentum theorem problems is to choose a control volume, and in this case I will choose a control volume as cutting through the flow in the nozzle, completely encompassing the flow as it spreads out to the side, like that. So a next useful step is to redraw that control volume as a free body diagram, which looks like this. Therefore, here's our velocity entering here and exiting radially from the sides in this direction. And I'll assume that the force exerted by the plate on the fluid is to the left. And generally speaking when you're doing a problem like this, you don't know the direction of the force ahead of time, so you just guess a direction. And if your answer turns out to be negative, it just means that your guess of the direction was wrong. So, that is my control volume, and, the next step, because this is a vector equation here, I'm going to apply this in the horizontal direction or the x direction. And now the vector equation becomes a scalar equation where now this is the summation of the forces in the x direction is equal to rho Q V2x minus V1x, where V2x and V1x are the horizontal or the x components of velocity at stations 2 and 1. Next I look at the forces, so the forces which are acting are acting on this control volume, the only force I have here is F, my assumed force F. And if I take my positive coordinate direction as being to the right, then this force is to the left, which I assumed, so that is equal to negative. Also generally speaking, we would have a pressure force, but the pressure at the output here, P, is atmospheric pressure, which is zero gauge pressure, therefore there is no pressure force acting over the nozzle exit. At the output stations here, so on the right hand side, we have rho Q and the velocity V2x here. This velocity is exiting in a radial direction, therefore, it has no horizontal component in the x direction. So, V2x is 0, and V1x, because V1 is already in the horizontal direction, V1x is just equal to V1. So our equation becomes that or the force is equal to rho Q V1. In other words, the momentum flux of the jet. So now we're ready to substitute in the numbers. First we want to find the velocity at the exit. The velocity is the volume flow rate divided by the area. Or Q over pi by 4 d squared, which is equal to 14.1 meters per second, is the jet exit velocity. And now, substituting into here, we have F is rho, which is 998 kilograms per cubic meter, times the volume flow rate 0.25 cubic meters per second, times the velocity, which gives me an answer 3,518 Newtons. Therefore, the correct, most near answer is D 3,520 Newtons. The second example is similar, except now the nozzle is forming a jet which is hitting this deflector plate and is being deflected back through an angle of 120 degrees. So the outlet diameter is 0.15 meters and the flow rate is 0.25 cubic meters per second. The force exerted by the jet on the vane is most nearly which of these answers? So, this proceeds in the same way, and first as in the last problem we need to calculate the exit velocity, which is given here. Which like the last problem is 14.1 meters per second. The basic momentum theorem for one dimensional flows, the usual starting point is here. And again I'm going to choose a control volume which encompasses the flow coming out of the nozzle, goes around the vein here, and back here. So redrawing that control volume I get this shape with velocity V1 entering here, and V2 exiting radially in a direct, at an angle of 120 degrees to the horizontal. And the force, again, I will assume is equal to F, acting to the left. So, we have one inlet and outlet, so, same as the last problem. The equation reduces to this form. Again, the re-, result is similar. The force I've assumed is to the left, so it's minus F is equal to rho Q. But now in this case, this velocity V2 has a component in the horizontal x direction. And the component of the velocity in this direction is V2 cosine theta, where theta is 120 degrees. And because that is to the left, and I'm assuming that my positive direction is to the right, this is minus V2 cosine 60 degrees, or sine 120 degrees. And the inlet velocity V1 is to the right so that is minus V1. V1x is equal to V1. So rearranging, I get this equation and now I'm ready to substitute in the numbers to find the answer, so that is equal to rho Q 998 etc. And the answer is 5,280 Newtons, so the closest answer is C, 5,300 Newtons. Another example, we have a water jet leaving a nozzle and the area is given, it's 0.06 square feet, and the velocity is ten feet per second. It strikes a, a vein or a turning vein here, and is deflected through an angle theta. The water density is given, the force required to hold the vane stationary is most nearly which of these? Where this is given in general for any arbitrary angle, theta. So here's our solution. The same starting point, but in this case we're not given the exit velocity. So let's consider, let me suppose that the inlet station here is station 1 and the outlet station is station two. To apply our momentum theorem we need to know what that exit velocity V2 is, which isn't given. So if I apply the Bernoulli equation between these two points, one and two, here is our Bernoulli equation, neglecting energy losses. But then I can start to simplify it, this is a free jet in the atmosphere, it's in contact with the atmosphere. So the pressure everywhere at the inlet and outlet, is atmospheric pressure, which is zero gauge, so P1 and P2 go out on both sides. If we neglect this small elevation difference here, and assume that z1 and z2 are approximately the same, that term goes out. So all we're left with is V1 squared over 2g is equal to V2 squared over 2g, in other words, the velocity's constant. V1 is equal to V2. So now, we're in a position to apply the momentum theorem, which takes on this form again, for one inlet and outlet. So in this case, I'll apply this in the horizontal direction. Here is my control volume. So redrawing my control volume over here, I have inlet velocity V1, exit velocity V2, which is the same as V1 at an angle theta. And I'll assume that the force exerted by the turning vane on the water is, again, F to the left. So, the summation of the forces is just given by this expression F is equal to rho V1 A1, V1 cosine theta. So the exit velocity, V2x here, the component of V2 in the horizontal direction is V2 cosine theta or V1 cosine theta because V1 and V2 are the same, minus V1x, but V1 is already in the horizontal direction, so V1x is just equal to V1. So taking out the constant V1 we have the force is equal to rho V1 squared A1 consine theta minus 1. And now I'm ready to plug the numbers to evaluate that. But here, because the density of the water is given in pounds mass per cubic foot, we have to be careful and divide that by the universal acceleration due to gravity, 32.2 here. And if you cancel out all the units here, you'll find that that results in a force, of pounds of force. So, evaluating that, the answer is 11.6 cosine theta minus 1 pounds force, so the correct answer is A. The second answer, by the way, B, is what you would have gotten if you had forgotten about that 32.2 factor there, which is obviously incorrect. One more example, we have water given, in, UCS units, 62.4 pounds mass per cubic foot. It's discharging at a rate of 20 cubic feet per second through a fire hose nozzle. And the diameter of the inlet is 8 inches and the outlet is 6 inches, where this is 1, this is 2 here. And the pressure of the inlet here we're given is 48 pounds per square inch. So here we have bolts which are connected through here, which is holding on the nozzle. The question is what is the force transmitted through those bolts to hold the nozzle in place? Is it most nearly which of these alternatives? So, in this case, I'm going to draw my control volume as I've shown it here, where the control volume completely encases or encompasses the nozzle and passes through the bolts at the inlet area here. So here's a redrawing of the control volume as a free body diagram. And the inlet station here is 1, outlet is 2, velocity entering here is V1, velocity exiting is V2. And I'll assume that the force in the bolts, which is of course going to be a tensile force, I'll assume that the sum of the forces in all of those bolts is T. So that is our free body diagram, and here again is our momentum theorem for one inlet and one outlet, shown here. So, applying this in the horizontal direction, which I'll call x, with positive direction to the right, is given by this expression. And now the forces which are acting, in this case we have a pressure force here we're given that the pressure P1 is 48 pounds per square inch. Therefore there is a net pressure force pushing here which is pressure times area, P1 times A1. And I'll call that pressure force F1. So the forces which are acting are F1 which is pushing to the right, T which is to the left therefore it's negative, is equal to rho Q, V2 minus V1. Where V2x and V1x are just V2 and V1 because the velocities are already in the horizontal direction. There's also potentially a pressure force up the outlook station here, F2, which is equal to P2 A2. However, because this is a free jet issuing into the atmosphere, the pressure there is atmospheric, which is zero gauge, therefore, this pressure force is zero, and we don't have to worry about it. So, rearranging for the tensile force in the bolts T, which is what we're looking for, we get T is equal to P1 A1 minus rho Q V2 minus v1. To evaluate that, we have to get the velocities. So V1, the inlet velocity, is volume flow rate over A1 Q over pi by 4d squared is 57.3 feet per second. Similarly, the velocity at the outlet is much higher because it's a smaller area, is 102 feet per second. And now substituting in we get that P1 A1 we're given as 48 pounds per square inch times the area, which is pi by 4 d squared eight inches minus rho Q. So rho is 62.4 divided by 32.2 again, multiplied by et cetera, and the answer is 680 pounds force, and the closest answer is B. And this finished our discussion of the Momentum Theorem.
