I want to continue our discussion of flow measurement by today looking at venturi meters. A venturi meter is basically a constriction or reduction in area and it enables us to measure flow rate in pipes by observing the pressure difference at the throat. Here is the relevant segment from the reference manual which describes the variables here, and to analyze the Venturi meter we can apply the Bernoulli equation. Now, for a horizontal pipe where the elevation is different, is constant, the Bernoulli equation reduces to p over gamma plus v-squared over 2g equals constant. From which we can see that where the velocity is high, the pressure is low and vice-versa. So if I think about the pressure distribution in a Venturi like this, then the variation of pressure in the pipe upstream here, assuming there's net energy losses are negligible where the area is constant the pressure is constant. Then as the area begins to converge here, the velocity increases, therefore the pressure decreases, reaches a minimum at the throat where the area is a minimum, and then, as the pipe gradually expands again, the pressure recovers back to almost its original value. And if energy losses are negligible, which they are very small in a venturi meter like this, we should have a perfect pressure recovery, the pressure recovers back to its original value. Now we can relate the pressure difference at the throat here and the upstream pressure to the flow rate by continuing with the Bernoulli equation. So if I have station one upstream here, and station two at the throat, we have the Bernoulli equation as shown. If the meter is horizontal, then the elevations Z1 and Z2 are the same, so they can sollout leaving us with this equation. Next I invoke continuity for an incompressible fluid, the volume flow rate Q is equal to A1 V1 which is equal to A2 V2 and for round pipe, the areas are as shown. Combining these equations, we end up with this equation. But the volume flow rate Q is equal to the throat area multiplied by the square root of this expression, where (p1- p2) is the pressure difference between the upstream section and the throat. And also we recognize that the areas are proportional to diameter squared. So we can, in turn, write this equation in this form in terms of the upstream and the flow diameters. Now this equation assumes no energy losses. So let me do an example on that, so in this question we have Kerosene of specific gravity .85 flowing, we're given the pressure difference between measured on the pressure gauge is 116 kilo pascals. The density of water is given, the flow rate is most nearly which of these alternatives. So to solve this we invoke the Venturi meter equation that I gave on the previous slide as shown here, and the area is 5 by 4d squared. So continuing on I can substitute in the numbers. The diameter is .06 meters at the throat. The pressure difference is 116 kilopascals or 116 times 10 cubed pascals. The density of kerosene is the specific gravity .85 times the density of water so substituting in the answer is .05 cubic meters per second and the closest answer is C.