 Incompressible water at 20 degrees celsius is flowing at 15 kilograms per second through a 0.1 meter inside diameter horizontal concrete pipe. How much does the pressure of the water drop over a 100 meter length of pipe? I will start with a very accurate system diagram. We have a horizontal pipe. That is 100 meters long. I'm going to define the inlet and outlet as state points. 1 and 2 respectively. I have water at 20 degrees celsius which means that I can look up its properties in table A1. I know an inner diameter. I know that that diameter is applicable across the entire pipe both states 1 and state 2. That also allows me to deduce things like if it's flowing steadily that means the mass flow rate has to be the same at 1 and 2. If it's incompressible that means the density has to be the same at 1 and 2. Therefore the volumetric flow rate has to be the same at 1 and 2. If the cross-sectional area is the same that means that the average velocity also has to be the same. So whatever the velocity is at state 1 it has to be the same at state 2. I can relate that velocity to the mass flow rate at which point I can relate density diameter and velocity to the mass flow rate. Since I can look up density and I know the diameter and I know the mass flow rate I can calculate the average velocity which is going to be useful in my Reynolds number calculation. Now I know what you're thinking. You're thinking why does the Reynolds number have anything to do with this? Well when we're calculating the head loss across our pipe we have to use either the laminar equation or the turbulent equation. Which equation we plug in is based on what our Reynolds number is. Since we have internal flow through a pipe we have to compare our Reynolds number to 2300. If it's laminar we can use the laminar equation. If it's turbulent we can either use the Colmbrook equation or the Moody chart. Both of those are going to be useful in plugging into the conservation of energy. So actually why don't we start there? Next we can neglect the terms that aren't relevant to this setup. I have a horizontal pipe so there's no change in elevation. I have no turbine. I have no pump. I can neglect the changes in kinetic energy. Because again if my mass flow rate is constant and I have incompressible flow that means my volumetric flow rate has to be constant. And if my diameter doesn't change that means the average velocity must be the same. So whatever the velocity is at one it has to be the same at two. Therefore there is no change. Therefore I can subtract that term from both sides. I have to leave the pressure terms because I'm determining a pressure drop which is going to be how much bigger p1 is than p2. So I want to write that as p1 minus p2 eventually. Furthermore I have to leave my friction head which I can calculate by taking the friction factor times the length of the pipe which is 100 meters divided by diameter times the velocity through the pipe squared divided by two times gravity. I know length. I know diameter. I can calculate velocity. I can assume gravity. So the real question between us and answering the question is what is f? If I have a laminar flow f is just 64 divided by the Reynolds number. If I have turbulent flow I have the Darcy-Kohlberg equation and which equation to use is based off of our Reynolds number. I calculate Reynolds number I compared to the critical Reynolds number which for this case is 2300. If it's less than 2300 that means I'm treating it as laminar flow. If it's greater than 2300 that means I'm treating it as turbulent flow. So the next thing we should do is calculate a Reynolds number. For convenience I'm going to write this in terms of kinematic viscosity so I only have to look up one property. I have incompressible water at 20 degrees Celsius which means I'm going to go into table A1. On table A1 I can read off the density dynamic viscosity and kinematic viscosity as a function of temperature for water. At 20 degrees Celsius the kinematic viscosity is 1.005 times 10 to the negative sixth. So I'll indicate that here A1 and that was in units of meters squared per second. Now I could calculate the velocity and plug that in or I could plug it in symbolically. I'm here to do as little math as possible so I'm actually going to plug in this relationship in place of velocity and at that point my Reynolds number becomes velocity which is mass flow rate times four times diameter divided by density times pi times diameter squared times kinematic viscosity. So in this case the kinematic viscosity didn't actually help us much because we still have to look up a density. Furthermore if we had just looked up dynamic viscosity in the first place the densities would have canceled at which point I would have only had to plug in one thing. So by looking up kinematic viscosity I actually ended up having to do more lookups than I would have otherwise. Glad to see that plan is working out. But it's too late to change. I already wrote it down so we're going to plug in 15 kilograms per second and a diameter of 0.1 meters. So 15 kilograms per second. Furthermore we're going to have to calculate a velocity anyway because we need it for this equation. Here but potato potato. We're multiplying by four. We're multiplying by 0.1 meters. We're dividing by the density of water which back into table A1. We have 998 kilograms per cubic meter, 998 kilograms per cubic meter and then pi and then 0.1 squared meters squared. I'm going to cancel the squares and then 1.005 e to the negative sixth meter squared per second. And at this point meters and square meters cancel cubic meters, kilograms cancels kilograms, seconds cancel seconds, leaving me with a unitless proportion which is what I want. So I'm going to fire up my most helpful calculator, type 15 times 4 and then I'm going to divide by the quantity 998 times pi times 0.1 times 1.005 times 10 to the negative sine 6 and I get 63,472. And again that's a unitless proportion. Now we're comparing that number to 2300. I see that it is 2300. Therefore I have turbulent flow. And for turbulent flow, which by the way makes sense as turbulent is in the example problem statement, I'm going to have to either use the Colbrook equation to calculate my f value or I can use the Moody chart. In the interests of character building let's actually try it both ways. But I'm going to substitute everything into the conservation of energy next and then plug in everything in terms of what I know already except for the friction head. So I will do that on another sheet of paper. So p1 minus p2, which is the pressure drop, is going to be the density of the fluid times gravity times f times l over d times v squared over two times gravity. Again just to avoid calculating the velocity honestly out of principle as much as anything else at this point. I'm going to plug in mass flow rate times four divided by the density times pi times damage squared. At which point I have density times gravity times f times l times m dot squared times four squared divided by diameter times two times gravity times the denominator terms from over here, which is density times pi times d squared and then I have to square everything that's actually pi squared density squared and d to the fourth power. Gravity cancels gravity, density in the numerator cancels one of the densities in the denominator, leaving me with four squared times l times m squared times f divided by two times pi squared times d to the fifth power times density. And I know mass flow rate, I know the density because I've now looked it up. I know diameter. I know the constants except for f. So I avoided it as long as possible. Let's calculate an f. First I will do that using the calculator method. And I can only do that because I have a fancy enough calculator that it can solve for me. It's coming up with a numerical solution by guessing and checking behind the scenes will indicate what I'm doing for people who don't watch the video. Option one equation. So I have the Reynolds number f is the thing that I'm looking for. The only other thing I need before I can actually plug everything into the calculator is the relative roughness, which is the epsilon value in this case for concrete divided by the diameter. I need an epsilon value. For that epsilon value, I'm going to have to go into table 6.1. I can read off epsilon values either in imperial units or metric units for a variety of substances. For concrete, I actually have two values, one for smoothed, one for rough. So I have to make an assumption about what type of concrete I have here. If I have a pipe through which water is flowing, especially a 100 meter long pipe and especially one that's 10 centimeters in diameter, it's more likely to be rough. So I'm going to use an epsilon value of two. I should really list that as an assumption. So I will go back and add to a list of assumptions. So using my epsilon value of 2 millimeters and a diameter of 0.1 meters. Alternatively, I could write epsilon over diameter 2 millimeters divided by 0.1 meters. And then there are 1000 millimeters in a meter. So my relative roughness is 2 divided by 100, which is 200. And now calculator, it is your time to shine. Everyone has mocked you, but you can prove them wrong. Surely. Surely you can. I believe in you. Mostly. Back to the one half power. So hopefully I remember to leave that in real time. The calculator crunched for a while and returned 0.049. And it took a long time because I mean nothing personal against this calculator. It took a long time because it was a complicated equation. It's okay calculator. I'm not judging you, but it had to go through and guess and check values until it decided that it was close enough. We could do that by hand to guessing a value for f and then calculating a value for f, plugging that into the equation to calculate a new value of f and repeating until we get out a value of f that we put into it in the first place. But the calculator is much faster. If you are one of my students, if you are working through this on an exam, you are welcome to use your calculator. But if you don't either want to use the calculator or if you don't have a fancy enough calculator to be able to do this, we have the moody chart instead. So we're going to take our relative roughness, which is 0.02. We're going to find that blue line, which is here. And then we can follow that blue line until we find our Reynolds number. So our Reynolds number was 63,472. I'll write that down 63,472. And since everything here is an exponential notation, that would be 6.35 times 10 to the 1, 2, 3, 4th. So we're looking for 6 times 10 to the 4th on the x-axis. A big red line that we can move around. Okay, too big. That's much more better. So 10 to the 4th is over here. It's a log axis, which means 1 times 10 to the 4th is here. This would be 1.5 times 10 to the 4th. This value is 10 to the 4th, excuse me, 2 times 10 to the 4th because it's centered on this number. This would be 2.5 times 10 to the 4th, 3 times 10 to the 4th, 3.5 times 10 to the 4th, 4 times 10 to the 4th, 4.5, 5, 6, 7, 8, and 9. And I am looking for 6.35, which means I'm not going to be here. I'm not going to be here. I'm going to be somewhere in the middle. I'm going to say right about there. And then we find the intersection of my red line, which represents a Reynolds number of 6.35 times 10 to the 4th. And the relative roughness line of 0.02, I see that they intersect right here. And I want to read off an x, excuse me, a y value corresponding to that because my y-axis is f. And I get a value that is about halfway between 0.04, 2, 4, 6, 8. It's about halfway between 0.048 and 0.05. I would call that about 0.049. So from the chart, we are reading off a value of about 0.049. With either method though, I get the same f value, which means that I have everything I need to plug into this equation. So I'm going to start plugging in numbers. We start with 4 squared divided by 2, which is going to be 8. And then I have 100 meters, as that was the length of my pipe. I have 15 kilograms per second. He said, hoping he was remembering correctly. And then I'm squaring everything because the term is squared, multiplied by my f value. I'll grab the one from my calculator just because it's more accurate. 0.04925. But of course, the value from the chart would have been just fine. I'm dividing by pi squared. Our diameter was 0.1 meters. Again, hoping that I have that right. Excuse me, 0.1 meters raised to the fifth power. And then our density, which was 998. We want a pressure difference likely in pascals or kilopascals. So I'm going to write a pascale as a Newton per square meter. And a Newton is a kilogram meter per second squared. Now I can cancel units. So kilogram squared cancels kilograms and kilograms. I have meters and meters to the fifth in the denominator. So I have square meters, square meters and meters, which means that they don't entirely cancel my missing meter, densities, kilograms per cubic meter, not square meters. Okay, let's try that again. We have meters to the fifth power times meters in the denominator, which is six, three, two, and one, meaning the meters cancel. And then second squared cancels second squared, leaving me with pascals. So to the calculator again, I have four squared multiplied by 100 multiplied by 15 squared multiplied by 0.049248 quantity divided by two times pi squared times 0.1 to the fifth power times 998. We get 89,998. Since that is so large, I'm going to convert it to kilopascals. I'll add that into my unit conversion here. A kilopascal is 1000 pascals. That number divided by 1000 at the end is going to be 89.998. And you know, just for convenience sake, let's call it 90. So the pressure drop across this rough concrete pipe with 15 kilograms per second of incompressible water at 20 degrees Celsius with a diameter of 0.1 meters is about 90 kilopascals. So if we were pumping this water, we would need to overcome that pressure drop as a result of the losses, or we would have to angle it such that there was enough kinetic energy to make up for the frictional losses. We could figure that out as well by including a potential energy term and then neglecting pressure change across the pipe and figuring out what the elevation change would have to be, etc.