Today I'm going to continue our discussion of fluid mechanics and look at flow measurement. So, this is topic four and in this segment we'll look at various devices for measuring flows and velocities, particularly Pitot tubes, venturi meters, and orifice meters. And, in this segment we'll first look at Pitot tubes. So a Pitot tube is a device for measuring velocity in flowing fluids, and as illustrated in this diagram from the reference handbook, in this case we have a Pitot tube, which is basically a bent over tube, which is stuck below the water surface of a flowing river or stream. And as a result of the decrease in velocity up to the stagnation point here, P0, the pressure here is higher, so it forces the liquid column up into the tube. So it raises above the water surface by a height that we denote by h. To analyze this situation, we apply the Bernoulli equation between some point upstream here, and the stagnation point at the tip of the Pitot tube, and the Bernoulli equation is here. And this of course we can simplify right away, that if we assume that the streamline is horizontal, then the elevation Z1 and Z2 are the same, so they cancel out. The velocity of the stagnation point here 0 is 0, so the velocity there goes out. And using the notation that they used in the reference hand book, we have this equation. Next we apply hydrostatics to this point and the hydrostatic pressure in a flowing fluid with straight and parallel stream lines is the same as the hydrostatic pressure. So the pressure at that point is just the pressure due to the column of fluid, which is sitting above here, which is gamma times d where d is the depth of the tip below the water surface. Similarly, we can apply hydrostatics through the water column in the Pitot tube here from here to here, and the pressure at the tip here, the stagnation pressure P0 is equal to gamma times the height of the column of fluid, which is d times h. So, combining these three equations together which is very simple, we arrive at this equation, v the upstream velocity, is equal to square root of 2gh. Therefore, by simply observing the height of the column above the water surface, we can compute the velocity of the water. And this device is known as a Pitot tube, after Henri Pitot, who used this device in 1730 to measure velocities in the river Seine, in Paris. Notice, also in this equation, that the depth of the tip below the water surface, does not appear. In other words, if we move the Pitot tube up and down, then the height that the water column rises to here remains constant, h is constant. Now, we can also write the Bernoulli equation in this form. Ps, P0 is equal to Ps plus one half rho v squared, where P0 is the stagnation pressure in the flow, Ps is the static pressure, and the last term one half rho v squared is sometimes called the dynamic pressure. So, let's do a numerical example on that. We're given that a Pitot tube is placed in a river which is flowing at a speed of five feet per second. The tip is one foot below the water's surface. The water level in the Pitot tube above the water's surface, h, is most nearly which of these alternatives? So here's our solution, just applying the equation from the last slide. V is equal to square root of 2gh. Or rearranging for h, the water surface level is v squared over 2g. Plugging in the numbers, we have five squared divided by 2 times g is 32.2 is .39 feet or 4.7 inches. So the closest answer is C, 4.7 inches. Now, Pitot tubes of that type can't be readily used for gases. Therefore, we often use another device called a Pitot-static tube. And the Pitot-static tube shown in the photograph here and the diagram here, simultaneously measures the stagnation pressure and the static pressure in a flow. The stagnation pressure is read at the tip here, and the static pressure by the side holes here. And these holes, the signation pressure tap here, and the static pressure taps here can be seen in the photograph. By connecting those taps to a manometer, we can get a direct reading of the pressure difference between those two. So, the stagnation pressure here Is higher than the static pressure, therefore the manometer fluid here will deflect in the manner shown, and the reading h, the manometer reading is a measure of the upstream velocity. And the analysis of this is fairly simple. Here is the relevant equation from the reference handbook, and in their notation, the upstream pressure P1 is Ps, the static pressure, and the stagnation pressure, P2 in this diagram, is P0. So again, the same equation applies. If we apply Bernoulli equation from well upstream of the Pitot tube to the stagnation point, we have the same equation that we had before, and by the same analysis we end up with this equation. The unknown velocity v is the square root of 2 times the pressure difference divided by the fluid density. Or recognizing that gamma is equal to rho times g, I can write it in this form where gamma is the specific weight. And to complete this, I apply hydrostatics to the manometer fluid here, so the pressure difference is just gamma times h. And assuming that the fluid that we're measuring here is a gas, so its hydrostatic pressure is different than P0 minus Ps is equal to the specific weight of the manometer fluid times h. Combining the equations we arrive at this, v is equal to 2 gamma m, gh divided by gamma g, where gamma g is the specific weight of the gas. An example of that, we have an aircraft flying at a speed of 100 miles an hour, at an altitude of 10,000 feet. The air density is 0.057 pounds mass per cubic foot. The air pressure is given 10.10 pounds per square inch. The stagnation pressure at the tip of the Pitot tube, or at this point 2 here, is most nearly which of these alternatives? So, here's the solution. We can just apply the Bernoulli equation in the latter form that I showed it. The stagnation pressure is the static pressure plus the dynamic pressure. So plugging in the given numbers, we have the static pressure is 10.1, multiply it by 144 to change to pounds per square foot, times one half rho. And, notice here we have to divide by g, and I'll talk about that more in a minute, multiplied by the speed, 147 feet per second is 100 miles an hour. So calculating that, that is 1454 plus 19 pounds per square foot absolute, or dividing by 144 Is equal to 10.23 pounds per square inch absolute. So, the closest answer is D. And again, remember that we have to put the subscript a on there to denote that we're talking about an absolute pressure to differentiate from a gauge pressure. Now also note in this example that the density was given in pounds mass per cubic foot. To work this correctly we have to divide that by the universal acceleration due to gravity, gc, which is 32.174, and it has these units. And in that case, the density is 0.057 divided by 32.2, which is this factor here, and it's equal to this number. So in that case, this would be the density in slugs per cubic foot if we were working in those units. But generally in the FE Exam, the density in US units is given in pounds mass per cubic foot, so we have to be careful to divide by gc to get our units right. Continuing with the same example, now we're given that the pressure difference recorded or the question is, what is the pressure difference recorded by the Pitot-static tube? Is it which of these alternatives? In other words, what is the difference between the static and the dynamic pressures? The static and the stagnation pressure? So, the solution here, again, just rearranging this equation here, we have the pressure difference, P0 minus Ps, is the dynamic pressure, one half rho v squared, but we've already calculated that. It's this number right here, 19 pounds per square foot. So, dividing by 144, that is equal to 0.13 psi, and the answer is A. And in this case, we don't have to put a subscript a or g on the end, because this prefers to a pressure difference. So, this completes some early examples on the Pitot tube and I'll continue with this in the next segment.