 Gasoline is flowing through a horizontal pipeline at 20 degrees Celsius. The distance and pressure drop between pumping stations are 13 kilometers and 1.4 megapascals respectively. The pipe is 0.6 meters in diameter, that's the inside of the pipe, and is made out of galvanized iron. Determine the flow rate of gasoline through the pipe in cubic meters per second. So again, I will start with a very good diagram, pumping station and pumping station. Not to be confused with just squares. They are of course not squares. I'm going to call state 1 this side of this pumping station, state 2 this side of this pumping station. The distance between them is 13 kilometers. The pressure drop across that distance is going to be 1.4 megapascals. And then I'm going to set up a conservation of energy in between state 1 and 2. And my next step is going to be to eliminate terms that aren't relevant to this situation. First of all, I have a horizontal pipe, so Z1 and Z2 are the same. And then I have a constant diameter across my pipe and a mass flow rate that is flowing steadily. Since I have incompressible flow, or the density is constant, and the same mass flow rate, that means that the volumetric flow rate at 1 must be the same as the volumetric flow rate at state 2. Since my volumetric flow rate is the average velocity times cross-sectional area, and the cross-sectional area is the same, that then means V1 has to be the same as V2. I was told the change in pressure, which means that I can leave that term. I'm going to neglect turbine head and pump head. Now I know what you're thinking, but John, the problem specifically mentions a pump. And that's true. Basically we have two ways to approach this. We can either start our control volume here and end it here, at which point we would have no pressure drop, but we would have a pump head, or we start a control volume here and end it here, at which point the pump doesn't appear, but a pressure drop does. Since we have a pressure drop already, we know that number. We might as well draw our control volume as not including either of the pumps. Then I can also plug in the definition of my friction head, since we're only using major losses for now. It's going to be F times L over D times V squared over 2 times gravity. Therefore, my conservation of energy equation simplifies to P1 minus P2 divided by density times gravity is equal to F times L over D times V squared over 2 times gravity. We know the length and we know the diameter. Our goal is going to be to calculate a volumetric flow rate. You could think of this as solving for the velocity in this equation and then using the velocity to calculate a volumetric flow rate. Or we can just write our velocity in terms of volumetric flow rate. You guys know I'm not here to do math, so I'm going to substitute V as 4 times volumetric flow rate divided by pi times diameter squared. Then because I'm squaring the velocity, that's actually going to be 4 squared times volumetric flow rate squared divided by pi squared times diameter to the fourth power. I'll make that substitution, 4 squared times volumetric flow rate squared divided by 2 times pi squared times diameter to the fourth power because it's squared already and then we're squaring it again. Then I have gravity at the end. So in this relationship I know P1 minus P2. I can look up the density of gasoline at 20 degrees Celsius. I can assume standard gravity. I know the length, I know the diameters, I know the constants leaving me with just volumetric flow rate. So volumetric flow rate is going to be the square root of P1 minus P2 times 2 times pi squared times d to the fifth power times gravity divided by f times 4 squared times l times density times gravity. Gravity cancels gravity and nothing else immediately simplifies. In order to calculate volumetric flow rate I'm going to need to plug in something for f and for my f value I have two different equations. One for laminar flow, one for turbulent flow. Which means that to be able to continue I have to know whether I have laminar flow or turbulent flow. Under most circumstances we would calculate a Reynolds number. We would compare our Reynolds number to the critical point Reynolds number and then make a deduction from that comparison. Since we have internal flow through a pipe if our Reynolds number is below 2300 we're treating it as laminar if it's above 2300 that means we're treating it as turbulent. However, we don't know the velocity. We don't have that. Which means the only way that we can proceed is to make an assumption about our Reynolds number use that assumption to calculate a velocity use that velocity to calculate a new Reynolds number use that new Reynolds number to calculate a new velocity and to repeat until we converge on a solution. If we had a computer we could run MATLAB and have it do the guess and check process for us but since all we have is digital paper and a digital pen and a disgruntled calculator we're going to have to do this by hand. So my guess and check process is going to go as follows. One, I'm going to guess an f. Two, I'm going to use that f value to calculate a volumetric flow rate. Calculate the volumetric flow rate using this equation that we just built. Three, I'm going to use that volumetric flow rate to calculate a Reynolds number. Four, I'm going to use equation that's either the Colbrook equation if we have turbulent flow or 64 divided by the Reynolds number if we have laminar flow or the chart to look up a new f at which point we repeat the process. So my two equations for steps two and three are going to be this volumetric flow rate equation and then the Reynolds number equation which is going to be velocity times diameter divided by kinematic viscosity or velocity times diameter times density divided by dynamic viscosity. We're going to want to write our volumetric flow rate instead of a velocity so that we can eliminate one of the steps of the iteration process. So I'm going to use the same substitution we made earlier. This one here, down here as well. At that point we have four times volumetric flow rate times diameter and then I'll use the kinematic equation, excuse me, the kinematic viscosity form of the equation to reduce the number of lookups. Actually, we should probably go look and see what we have for gasoline because we might not have the option. Into my tables I have table A3 which is going to give me properties of gasoline and I see for gasoline I have a density, a dynamic viscosity and surely we'll have a kinematic viscosity, oh no we don't. That's unfortunate. So we don't even have the option. We have to use dynamic viscosity so I should have just left my first arrow. So from A3 at 20 degrees Celsius we're going to use a density of gasoline and a dynamic viscosity of gasoline of 680 and 2.92 times 10 to the negative fourth respectively. That's kilograms per meter second. So I have four times volumetric flow rate times density times diameter divided by dynamic viscosity times pi times diameter squared. Then diameter cancels one of the diameters leaving me with four times rho times volumetric flow rate divided by mu times pi. I'll write that so that we don't get confused when we are iterating. So this is the equation for step two. This is the equation for step three. Step four is either going to be the laminar form of the F equation, the turbulent form of the F equation, which is the Colbrook equation, or the Moody chart. And then I will split this into columns for our different iterations. And I'll just start with four iterations. We'll see how this looks. So iteration one. I have to start with an F value. And that F value is mostly arbitrary. We have to pick a number and then apply it. I could guess a good value and reduce how many iterations I have to do, or I could guess a bad value and repeat lots of iterations for the purposes of educating people watching this video on how the iteration process looks. To inform my guess, I'm going to go to my Moody chart. I don't know any information about the type of pipe. It's galvanized iron. Okay, I know a little bit of information about the type of pipe. If I go into table 6.1, I can see that for galvanized iron, we have an epsilon value of 0.15. And in our pipe, we're told we had a diameter of 0.6 meters. So if I take 0.15 millimeters, and I divide by, and I divide by calculator, 0.6 meters, I get a relative roughness of 0.00025. So in my Moody chart, that means I'm using a line that is a quarter of the way between here and here. So I'm going to attempt to draw what that line would look like. Yeah, that's good. Good drawing. So I know that my F value is going to be somewhere between here and here. Those correspond to Reynolds numbers between three-ish times 10 to the fourth and 10 to the eighth where it becomes fully rough. So my F value should start somewhere on either end. I could say 0.025 and then work my way down, or I could use 0.015 and work my way up. That's the logic. Oh, and by the way, that's only for turbulent flow. If I have laminar flow, it's going to be over here, which is from the 64 divided by the Reynolds number equation. Generally speaking, if I'm using the chart, and I'm guessing anyway, it's usually good to guess over here. Everything to the right of this dashed line is what we call full turbulence or full roughness. If I have a fully rough condition, that means that the F value doesn't really change with respect to Reynolds number anymore. You'll notice it's almost a horizontal line. So if I were to pick a Reynolds number of, oh, 10 to the eighth, that would give me an F value that's a quarter of the way between 0.0002 and 0.0004. That would be as good a place as any to start, and then I can work my way left, and if I get far enough to the left, I can jump all the way to this line. Does that logic make sense? We are starting our guess by treating it as a turbulent process, and more than that, we're treating it as very turbulent. And then we're backing off the amount of turbulence until we get it right. We could just as easily start with an F value that's crazy high, and then work our way down. But I'm going to start by treating it as fully rough. And if it's fully rough, that means my F value is, again, about a quarter of the way between this value here and this value here. The quarter of the way up is maybe here-ish. I will use the fact that I can draw perfectly horizontal lines to help me with that process. So about there, and that F value is going to be 0.0143-ish. 0.0143-ish. That's my first guess. If you were doing this with the Darcy Colbrook equation, you could get to approximately that same result by letting the calculator start with our Reynolds number that was very large. But just in the interest of showing both methods here, 1 over x to the one-half power is equal to negative sign, John. 2 times log base 10 of the relative roughness, which was 0.00025 divided by 3.7 plus calculator. 2.51 divided by a large Reynolds number. Let's call that y for now. Times x to the one-half power. So x to the one-half power. 1, 2, 3, 4. We're solving for x. And we're starting with a y value that is large. Because, again, that's a Reynolds number. Let's go with, oh, 10 to the eighth. 1e8. So hopefully I remembered to speed that up for you guys. It took about 30 seconds on this end for the calculator to guess and check its way through. It got a value of 0.01438. So if we use the graphical method, we get about 0.0143, 0.144. If we use the numerical solution, we end up with 0.014384. In either case, though, we have enough to go on to begin with. So we're starting with an f value of 0.0143, let's say. And then step two is to use that f value to calculate a volumetric flow rate. So I'm going to be plugging in volumetric flow rate is equal to hard brackets. And P1 minus P2, which is 1.4 megapascals, 1.4. And we're multiplying by 2 times pi times diameter to the fifth. And then we're dividing by our shiny new f value, 0.0143, which is unitless. And then 4 squared, which was 13 kilometers. Density of gasoline, which we got, is 680 kilograms per cubic meter. And we're going to take this entire quantity raised to the one-half power, and we want cubic meters per second at the end of that, which means that we need meters to the sixth per second squared before we take it to the one-half power. So because I'm running out of room, I'm going to take advantage of the fact that I'm running this digitally and shrink it down a little bit. So my next step is going to be recognizing that a megapascal is 10 to the sixth pascals, and a pascal is a Newton per square meter. It's a kilogram meter per second squared. So my numerator, I have meters to the fifth times meters cubed times meters, which would be meters to the ninth power. And then I have meters squared and kilometers. So kilometers, 1,000 meters. Kilometer cancels kilometer. Now I have three meters in the denominator, which gets rid of three of the meters in the numerator, leaving me with meters to the sixth power. And I only have second squared after I cancel kilograms, which means that I will end up with meters to the sixth per second squared, which when I take that to the one-half power, gives me a volumetric flow rate in cubic meters per second. So since I'm typing this out on my calculator, I'm going to plug in values for f symbolically so that I can come back and change that number if I have multiple iterations. So 1.4 times 2 times pi times 0.6 squared. 0.6, excuse me, to the fifth power times 10 to the sixth power divided by f, which I'll call f, times 4 squared, times 13 times 680 times 1,000. And then we are wrapping that entire thing in parentheses and raising it to the one-half power. And then I'm going to tell my calculator, a calculator, plug in an f value of 0.0143. And it gets a volumetric flow rate of 0.58154. 0.58154 cubic meters per second. And that's iteration, excuse me, step two of the iteration. I'm going to move that over a little bit, just to try to provide some coherency here as evidenced by the very well-drawn v value. So step two done, I can calculate the Reynolds number, which is step three. For that, I'm going to use equation 3, which is 4 times density times volumetric flow rate, 4 times density, which is 680 kilograms per cubic meter, times volumetric flow rate, 0.58154 cubic meters per second. And then we are dividing that by pi times our dynamic viscosity. And our dynamic viscosity was 2.92 times 10 to the negative fourth kilograms per meter second. And we want a unitless proportion, so cubic meters canceling cubic meters, seconds canceling seconds, kilograms canceling kilograms, and being left with too many meters is a bad thing. That's why this diameter appears down here as well. And I did not write then, much more better. Multiply by 0.6 meters, of course, at the end. So 0.6 meters, meters cancels meters, and I'm left with a unitless proportion. And that will give me step three. And again, I'm going to plug this in with my volumetric flow rate as a variable, since I can on my calculator, just to make it easier to repeat iterations. So 4 times 680 times 0 point, excuse me, times V, let's call it, times nothing, divided by pi times 2.92 times 10 to the negative fourth times 0.6. And when we divide that, we get a constant times V. So we're plugging in V. V value of 0.58154. We get 2.87384 times 10 to the sixth. So we recognize that that's greater than 2,300. Therefore, this is turbulent flow. And we can use our new Reynolds number to calculate a new F value. So if I do that using the Colbrook equation, if I jump back to this step, and I plug in an F value of, not far enough, if I jump back to this step, and I plug in a Reynolds number, which I had called Y, of 2.87384 times 10 to the sixth, I can let my calculator chug and spit out a new F value. And it gets 0.014684. 0.014684. The other way to get to a new F value would be to plug in our new Reynolds number, which is 2.87 times 10 to the sixth, into our Moody chart. That was 2.87. So let's call iteration number two, maybe a green color. Go with green. With my shiny new green color, I can draw a vertical line. And I will move that to approximately 2.87 times, excuse me, yeah, 2.87 times 10 to the sixth. So 2 is going to be here. This would be 2.5. So 2.87 is going to be, let's call it here. So again, my goal here is to try to read up about a quarter of the way between this point and this point. So I see a quarter of the way is right about here. So I need to take my F value, add approximately that position. And I have a new F value for iteration number two, that is just a hair under 0.15. So let's overcompensate a little bit and call it 0.0149. Or even 0.015. So that would be how we would graphically get iteration number two, iteration two. So I have an F value of 0.014684 or 0.0149 if we did it with a graph. So I can plug everything in to my volumetric flow rate calculation because I had left that symbolically on my calculator. That is pretty easy to plug in. So I have an F value of 0.014684 because I happen to have the best answer possible. And I get a volumetric flow rate of 0.57384, 0.57384. Capic meters per second. And calculate a new Reynolds number with that number. And I get 2.8336 times 10 to the sixth, which I can then use to get a new F. So again, I could jump into my chart again. I was doing this with a simple calculator or by hand. And this time call iteration number three blue. Need a new vertical line. So that would be 2.82.83. So right about here. And because it's impossible for us to read any granularity down to that level, if we're doing this graphically, we can treat that as having converged. So we can say graphically, we can't get any better than a number of 0.574. Although if we were doing this graphically, we probably would have plugged in an F value of 0.0149. So just for comparison's sake, if we do that with 0.0149, we get a volumetric flow rate of 0.5697. And if we plug that in, we had a Reynolds number of 2.81, which maybe you could argue is just slightly to the left of this line, at which point we could try to look up a new F value. But that F value is going to be so close that there's no point in really trying this. So we have converged on a solution. We can call the answer to our question 0.574 or 0.573 cubic meters per second. But since I'm doing this numerically because I happen to have a fancy calculator, I could keep going if I wanted to. And just to get the extra best answer that we possibly can. If we do that, then we're using our F value, which we get out from a Reynolds number of 2.836. So I'll delete these last two steps. So we're going back into this equation. And we are plugging in a Reynolds number of 2.83602 times 10 to the sixth. And we get a new F value of 0.0147. So I'm calling that converged. Therefore, the answer to the question is 0.5738. Since we don't have a lot of confidence, we could probably best answer that question by rounding a little bit. And that is arguably as far as we should go. But I'll keep going for fun. So in conclusion, we used our conservation of energy to calculate a volumetric flow rate if we knew an F. So we don't know an F for guessing a value to calculate a volumetric flow rate. We're using that volumetric flow rate to calculate a new Reynolds number. We're using our new Reynolds number to calculate a new F. And we are repeating that process until we converge. As you can see in our four iterations here, every time we iterate, we get closer to an answer. How many iterations we use depends on what we qualify as our convergence criteria. In this case, I qualify convergence as being unable to tell the difference on the Moody chart. But since I could continue symbolically with a calculator that could do the guessing and checking for me, I can see that convergence really only took four steps to achieve within an F value of about six decimal places. We can also reduce how many iterations we have to use by starting with a good F value. And as a general rule of thumb, if we have turbulent flow, which is often the case for liquids flowing through a pipe, we should start fully rough and work our way backwards, as opposed to just arbitrarily picking an F value out of the air. We could do that too, but that would just mean that we take more iterations to get to an answer. Whatever the case though, we conclude with a volumetric flow rate of about 0.574 or 0.57384 cubic meters per second.