The last topic I want to discuss on flow measurement is the orifice meter. Now, we start this from the equation that we previously had for the venturi meter, which is given here. And, this equation assumes that there is no energy or head losses in the flow through the meter. But a more general form of the equation is given here, where now we've incorporated a new coefficient here, Cv, which is called the velocity coefficient or the discharge coefficient, which accounts for head losses or energy losses. And this is particularly important for example in an orifice meter. And an orifice meter is simply a small disc or an opening in a pipeline where the fluid flows, and therefore the pressure changes as shown here. And this is the relevant section from the reference handbook, and the equations can be readily derived by the same principles we used earlier. However, there's another phenomenon that occurs when you have flow through an orifice, and that is the streamlines of the flow continue to contract after passing through the orifice. So that the minimum diameter is actually smaller than the diameter of the orifice. And this effect is usually accounted for by means of a contraction coefficient. Cc which is defined as the ratio of the area of the contraction, the minimum area A2 divided by the area of the orifice. And the typical value for a sharp edged orifice is approximately .62. So, if we combine those equations then we get this, where we have a new coefficient C zero, which is given by this equation here, cvcc etc., on the bottom. And this coefficient, C zero, is called the coefficient of the meter, or the orifice coefficient in the handbook. So, here are some typical values of orifice coefficients and velocity coefficients from the handbook for various shaped orifices, and here we'll do an example for flow through a sharp edged orifice meter. And we're given that the upstream diameter is .1 meters, and the diameter of the orifice is 0.06 meters, the pressure drop is 100 kilopascals, the flow rate is most nearly which of these alternatives? So, we apply our equation, first of all, compute the orifice coefficient from this formula. And in this case we assume that the values are appropriate for a sharp edged orifice right here, so substituting in the numbers, we get that the orifice coefficient is .62. Approximately the same as the contraction coefficient. So therefore, here is our generalized orifice meter equation which becomes this. So again, substituting in the numbers we get this and the answer is 0.026 cubic meters per second or the best answer is A. We can also apply this idea to the discharge from a vessel with a free surface through an orifice. For example this tall bottle here. As the water surface decreases, the driving head decreases, and therefore the velocity of the flow which is coming out of the opening, decreases, And we can analyze this in the same manner we did previously. So, for example, if I apply Bernoulli equation from some point in the vessel out here, out to the opening, we have this expression and, as usual, the upstream velocity I can neglect here. The change in elevation for that horizontal stream line I can neglect, and the upstream velocity is also zero. So again, very simple relationship is that the exit velocity, v2, is equal to square root of 2gh where h is the height difference between the free surface and the center of the opening. So this equation was first derived by Torricelli in the 17th century, and we notice from this that the exit velocity does not depend on the diameter of the opening. It only depends on h. The volume flow rate however, does depend on the diameter of the opening because the volume flow rate Is the area times the velocity, which is therefore equal to A2 square root of 2gh, which is actually the same as the orifice equation where the pressure difference there is simply equal to the height gamma h, and the upstream diameter is very large. And you can show that those equations are the same. This applies also to devices like a siphon. So, in this case, we have the water beaker, and we're siphoning water through this tube to the opening here. Which is five feet below the bottom of the vessel which is 15 feet tall. And the question is if the tube diameter is 0.5 inches the flow rate through the tube is most nearly which of these alternatives? So in this case that same equation that we just had applies. In other words, the velocity of the water through the opening here is equal to square root of 2gh, where h is this height difference. So the only thing that matters here is the height difference between the water surface and the opening. This height which is given here, capital H, is irrelevant, it doesn't enter into it. So now we can substitute in the numbers. So that is equal to square root of 2g(z1-z2) where z1 is the elevation of the free surface and z2 or z3 is the elevation of the opening. So that is equal to that. Substituting in the exit velocity is 35.9 feet per second. And again, notice that this does not depend on the diameter of the tube. It only depends on h. So finally, to get the volume flow rate, we have Q is equal to A3 times V3, pi by 4, d3 squared, times v, substituting the numbers, the answer is .049 cubic feet per second, or c. Another example, we have water flowing through and orifice which is two centimeters diameter Through a partition which has an orifice meter. If the coefficient of contraction of the orifice is 0.83, the flow rate is most nearly which of these alternatives? So here's the solution. We start with our basic orifice equation again. And in this case, the upstream diameter here, say the diameter at this point, we can put equal to infinity, or a very large number, in which case this term, 1 minus d over d to the 4th, drops out. So the equation then is given here, or this, substituting for the pressure difference for the height difference where h is the height difference between the two water levels here and using our equation for the orifice coefficient cv times cc etc. Which is, if D is very large, this term drops out. So it's essentially equal to the velocity coefficient times the contraction coefficient or 0.98 x 0.83, which is 0.81. And now we can substitute into the equation, and the answer is 7.97 times 10 to the minus 4 meters cubic meters per second multiply that by 86,400, which is the number of seconds in a day. We get our answer 68.9 cubic meters per day. So the closest answer is A. And this completes our discussion of flow meters.