 Consider a spout on a Roman style water service reservoir as shown in the figure below. If the spout length is relatively short, determine the flow rate of water out of the spout. So when it says spout length is relatively short, what that's indicating to us is that we can neglect the frictional losses due to the straight section of spout. Therefore, the only frictional losses we'll account for are the minor losses associated with the round entrance to the pipe here. To relate how that loss will affect our flow rate, we can set up a conservation of energy from state 1 to state 2. Next, I can simplify the problem by neglecting some things that aren't relevant to us. First of all, I recognize that state 1 and state 2 are probably going to be both at atmospheric pressure. And if p1 and p2 are approximately the same, I can neglect both terms on both sides of the equation. Next, I'm going to be assuming that the cross-sectional area at state 1 is very large relative to state 2. If the cross-sectional area at state 1 is very large, that means that for the same mass flow rate, the velocity will be relatively small. When we consider the effect of squaring that small velocity, we can say v1 is approximately zero. We can't neglect the changes in potential energy as that is what is actually driving our flow rate. I can get rid of the turbine and pump head, and I also will treat the major losses as being essentially zero. Next, I have to consider whether or not I have laminar or turbulent flow. I don't have enough information yet to be able to calculate a Reynolds number, so my best chance at getting through the problem is going to be to make an assumption about whether or not it's laminar or turbulent, and then check that assumption at the end. In general rule of thumb, if I have liquids flowing and the cross-sectional area isn't extremely small, I probably have turbulent flow. So if I have turbulent flow, that means my alpha value is essentially 1, which means that I can write z1 is equal to alpha 2, which is 1, times v2 squared over 2 times gravity, plus z2, plus the minor losses. Next, I will make my substitution for my calculation for the minor losses. So the velocity in that equation is the velocity through that interface itself. Here, I'm going to call that velocity essentially the same as v2, because at the point of the round transition, the velocity is relatively close to the velocity at state 2. Because v2 squared over 2 times gravity appears in both the kinetic energy term and the minor loss calculation, I can factor that out. At this point, I have enough information to look up k, I know gravity, I know the difference in elevation, I can calculate v2. But I don't actually care about v2, what I really care about is the volumetric flow rate. So I will make that substitution directly to limit how many calculations I have to do to calculate the volumetric flow rate. Then I can solve for the volumetric flow rate. Now I know the diameter at state 2, I know the elevation change, I can assume gravity. All I need to do is look up k. To determine that k value, I'm going to go to section 6.9 of my textbook and I will look around for minor losses that are relevant to this entrance. Note again that that's a rounded entrance to a pipe. For entrances, I'm going to be using 6.21. On 6.21, specifically 6.21b, I can see that for a rounded inlet, I'm going to have a loss coefficient that is a function of the proportion of the radius over the diameter. On figure 6.21b, I can see that for a rounded inlet, I'm going to have a loss coefficient that is a function of the bend radius over the diameter of the pipe. Which means I have to go calculate that value. I have a bend radius of 0.5cm and the diameter of my pipe is 2.5cm. So if I pop up my calculator, I can take 0.5 divided by 2.5 and I get 0.2. For 0.2, I'm going to be looking at the intersection of that lowest blue line with the far right side of my graph. I'm going to eyeball this as about half and maybe this is about a quarter. So I would say about a quarter of the way between 0 and 0.2. Which means that I'm looking at a loss coefficient of about 0.05. You could probably say that that's maybe a fifth of the way up. In which case it would be about 0.04. But whatever the case, I have an approximate K value. Which means that I can continue on in my problem. So right now I'm calculating a quantity in meters to the sixth per second squared. And I can take that quantity and square root it and I'll be left with cubic meters per second. So I start with pi squared times 2 times 9.81 times 2.5 to the fourth power times 1.5 divided by 4 squared times 1.05 times 100 to the fourth power. The reason I brought that out as a separate calculation is because I'm ending up with a very small number. Which means that my intermediate step is also going to be a very small number. So I have 0.000007 meters to the sixth per second squared. Then I take the square root of that value and I have my actual volumetric flow rate. So the last thing to do now is to double check my assumption of turbulent flow. To do that I will calculate a Reynolds number. And because I have water I'm going to use the kinematic form to limit how many things I have to look up. Again, since I have volumetric flow rate not velocity I will make the same substitution that I did earlier. 4 times volumetric flow rate over pi times diameter squared. One of the diameters cancels and I'm left with 4 times the volumetric flow rate. Divided by pi times diameter times the kinematic viscosity of water at 20 degrees Celsius. For that I will go into table A1. Water at 20 degrees Celsius has a kinematic viscosity of 1.005 times 10 to the negative sixth. So 4 times 0.00255 excuse me 0.002599 cubic meters per second divided by pi times two and a half centimeters times 1.005 times 10 to the negative sixth meters cubed excuse me squared per second. Then I want this to be a unitless proportion which means I have to get rid of the centimeters 100 centimeters in a meter. Then I have square meters times meters cancels cubic meters seconds cancel seconds and I'm left with a unitless proportion. Specifically I'm left with 1.317 E6 which is going to be about 1,317,000 which is greater than 2300. Therefore my turbulent assumption was accurate. Just for the sake of being complete let me formalize my list of assumptions. The incompressible flow flowing steadily at a short spout meaning that I could neglect major losses. I assumed P1 was pretty close to P2. I assumed D1 was much much greater than D2. Therefore V1 is approximately zero. Then I had assumed turbulent flow until it checked out. And that's the first form of this example problem.