 We're now going to take a look at an example problem involving the application of the Bernoulli equation. So what we have is we have a nozzle and it is flowing water through the nozzle and it is discharging to the atmosphere. So let's draw out the nozzle. So we have a nozzle. We can see that we have diameters. So it is a nozzle that is round on the inlet and on the exit. And what we are given, we're given the inlet diameter, we're given the exit diameter. We're told P2 which is equal to the atmospheric pressure and we're told Q which is the volumetric flow rate. So we have 0.1 is here, 0.2 is here. And given that we're going to apply Bernoulli's equation, what we're going to be doing is applying this along a streamline and so we will have a streamline going right down the center line of the nozzle and that is a streamline that we will apply Bernoulli's to. If we were to do flowvis of this nozzle, we'd probably have other streamlines that would be doing things like this and we could apply Bernoulli to those as well provided that they're not too close to the wall but we won't be doing that. We'll only be looking at the one going right down the middle. So that's one that we'll be taking a look at in this example. And what they are asking us to do is the following. They want us to find the upstream pressure, the gauge pressure upstream at 0.1 as a function of the volumetric flow rate Q. So in order to do this, we need to make a number of different assumptions in order to apply Bernoulli's so the assumptions will be the following. Okay, so those are the assumptions, steady, incompressible, density is constant, frictionless. We're looking at flow along a streamline. There is no change in elevation so we're assuming that it's a horizontally aligned nozzle and uniform flow at 1 and 2 so that means that it's a top hat velocity profile into and exiting. So what we're going to do, we're going to begin with the continuity equation and we have a steady flow and consequently the first term disappears and then we're left with the mass flux terms. We have to be careful about the inlet and exit. So on the inlet side we have, so this would be one, we had velocity coming in here and we know that the area vector is going to point outwards so this is going to be a negative and for the second term at the jet itself we have velocity in that direction, the area vector pointing out the dot product of those two is going to be a positive. So from this what we can do and we also know that density is a constant because we're dealing with incompressible flow so density is going to pop out of that equation. What we're left with then is V1A1 equals V2A2 so we're going to take that and set it aside. The next thing that we're going to do is we're going to look at the Bernoulli equation. Let's take a look at what Bernoulli can do for us. So that's what we get from the Bernoulli's equation. Now immediately off the bat we said that elevation changes were not significant here so that term goes with that term and what we're left with I'm going to isolate pressure on the left P1. Okay, so we get that. Now if you recall the problem statement let's take a look where was it here, go back. We want to find P1 gauge as a function of the volumetric flow rate so we want to deal with gauge pressure not absolute atmospheric plus gauge we're going to deal only with gauge pressure. So let's look at gauge pressure here. Now if you recall the way the nozzle was configured we had something like this and like this. This is point one. So here we have P1 gauge. Here we had P2 and we said that was equal to P atmospheric. So what does that tell us? That tells us that P2 gauge if P2 is P atmosphere is equal to zero because P2 is at atmospheric pressure and it turns out just so that you're aware whenever you have a free jet like we would have here the pressure out here is atmospheric. Every other free jet the pressure in there is also going to be atmospheric and that's because a free shear flow or an interface cannot sustain a pressure gradient and consequently the pressure inside of this jet is going to be equal to the pressure outside. That's just kind of an aside that is kind of important whenever you're dealing with jet flows but that's one thing that I'll point out there. But anyways what we're going to pull into the problem is this right here and that is that P2 gauge is equal to zero and consequently if we look at P1 minus P2 the difference between P1 and P2 is then going to be P1 gauge. So with that we can take this and plug it into our Bernoulli equation and I will isolate P1 minus P2 on the left and we know that that is then P1 gauge and that will be rho over 2 times the velocity difference. Okay so we're getting somewhere we have this term now and we have P1 gauge but it's not in terms of volumetric flow rate which we're after and we have a velocity at 1 and we have a velocity at 2. From continuity remember we had a relationship between velocity and 1 and velocity and 2 so let's pull that in. We had V1A1 equals V2A2 so we know from continuity V1A1 equals V2A2 so we're going to pull that into here and the way that we're going to do this we're going to say V2 is V1 times A1 over A2 and we'll make that substitution so we're going to express this in terms of V1 so we get that now we have a couple of areas on the right hand side remember we said that the cross section of this nozzle was round consequently we can say A1 can be related to the diameter and A2 we get that and as a result the ratio of A1 squared to A2 squared is then equal to the pi and the 4 are going to cancel we'll get a d1 over a d2 that's squared and squared so that is raised to the power 4 so what we're going to do let's take this and plug it into that equation okay so we get this equation now what else can we do here I remember we want to express this in terms of volumetric flow rate coming through the nozzle volumetric flow rate that will be a meters cube per second so volumetric flow rate is equal to V1A1 and it's also equal to V2A2 we're going to work with V1A1 here and so let's expand that out we then get V1 multiplied by the area at 1 and from this we can then write V1 in terms of volumetric flow rate so we get that what I'm going to do is I'm going to take this here and I'll plug it back into this equation and that will give us what we're after so that then becomes a relationship between the upstream pressure and the volumetric flow rate going through the nozzle it expressed only in terms of the density and the geometry of the nozzle itself and so that solves the problem with what they were requesting and it gives an example of how to apply Bernoulli along a streamline