 Consider a piping system routing incompressible water at 20 degrees Celsius between two reservoirs as shown in the figure below. If the relative roughness of the piping is 0.001 and the required flow rate of water is 12 cubic feet per minute, determine the power of the pump required. So take a minute to appreciate this excellent drawing. We have a piping system that is 400 feet of length in pipe. Each of those pipes is 2 inches in diameter. I have a couple of fittings which are interacting with my flow. I can account for the minor losses associated with those fittings by looking at the minor loss coefficients corresponding to an open globe valve, a screwed regular 90 degree elbow, not to be confused with Elbu. Look at that. Like it never happened. A 12 inch bend and then a half open globe valve also screwed. In addition to those fittings, I'm going to have to account for the minor losses associated with the entrance and exit of the piping system into and out of the reservoirs. And I could in theory account for the minor losses associated with the pump itself, but I don't have any data for that. And depending on the type of pump, it could be a lot of different types of losses. So let's work with what we've got. And let's start this off with a conservation of energy analysis from the first reservoir to the second reservoir. Next, I can neglect terms that aren't relevant to this situation. First of all, I will recognize that P1 and P2 are probably both pretty close to atmospheric pressure. I'm assuming the top of the reservoirs is exposed to atmosphere, and I'm assuming that the elevation difference given is actually between the tops of the fluids in the reservoirs. Next, I can neglect turbine head because I don't have any turbines. And the next, I'm going to cancel the kinetic energy terms. So take a minute, think on why I did that. And if you said, while John, back in Chapter 3, we recognize that if we have steady state analysis, that means m.1 is equal to m.2, 5.1 in Latin, 1 outlet. And since mass flow rate can be represented as density times volumetric flow rate, and if I have a constant density, i.e. an incompressible flow, that means volumetric flow rate at 1 has to be the same at volumetric flow rate at 2. Then volumetric flow rate can be written as average velocity times cross sectional area. If the cross sectional area doesn't change, that means the average velocity must not change. You are exactly correct. I have to leave my potential energy terms, I have to leave both major and minor losses, and because I'm calculating the work of the pump, I have to leave the pump. So so far my simplification goes, z1 is equal to z2 plus my friction losses, both major and minor, minus the pump head. So for my major losses, I will use v squared over 2 times gravity times f times l over d. And for my minor losses, I will use v squared over 2 times gravity times the sum of my loss coefficients. The reason I can factor out my velocity squared over 2 times gravity is because the velocity at all of the opportunities for minor losses are all the same, because the velocity isn't changing because I have steady state flow of incompressible fluid. Furthermore, when I combine my major and minor losses together, I can actually write that as v squared over 2 times gravity times f times l over d plus the sum of all the minor losses. The v squared over 2 times gravity term comes all the way out. Now I have z1 is equal to z2 plus v squared divided by 2 times gravity times the quantity f times l over d plus the sum of k minus the pump head. And then you'll remember from chapter 3 that my pump head can represent the specific work of the pump divided by gravity, so I can make that substitution. What I'm actually looking for is the power of the pump. And the power of the pump represents the mass flow rate of fluid through the pump times the specific work of the pump. And specific work of the pump can be written as pump head times gravity. And then I can write mass flow rate in terms of volumetric flow rate to limit how many secondary calculations I have. And I can do the same substitution for volumetric flow rate that I have the last couple of examples, meaning that I'm going to use volumetric flow rate is equal to average velocity times cross-sectional area. Since the cross-sectional area is the same across the entire pipe, I don't have to write that out in terms of state points. So I will just write the velocity can be written as 4 times volumetric flow rate divided by pi times diameter squared. So now I have z1 is equal to z2 plus 4 squared volumetric flow rate squared. Look at that, I remember that the first time divided by pi squared, let's see if I remember the 4 times diameter to the fourth power. Hey look, I did it guys. Times gravity times 2. And then multiplied by the quantity f times l over d plus the sum of k. And then my pump head can be written as 4 of the pump divided by volumetric flow rate times density times gravity. So then I will write minus power of the pump divided by density times gravity times volumetric flow rate. So in this relationship, I know the difference in elevation. I know the volumetric flow rate is 12 cubic feet per minute. I know the diameter of the pipe. I can assume gravity. I know the length of all the straight piping. I know the diameter of the straight piping. I can determine f. I can determine the minor loss coefficients. I know the density of water or rather I can look it up. I know gravity. I know the volumetric flow rate. So the only step left here is going to be to look up f, to look up the density of water at 20 degrees Celsius and to look up all of minor loss coefficients. So let me rewrite this as the power of the pump. So I will start with my major loss coefficient. I need to determine the friction factor for a situation where the relative roughness is 0.001, I believe, 0.001. Yep. And I can relate the Reynolds number to the volumetric flow rate by recognizing that I can write it as velocity times diameter divided by kinematic viscosity and my velocity I just substituted for in terms of volumetric flow rate by recognizing that average velocity is going to be four times volumetric flow rate divided by pi times diameter squared, which means that I can write my Reynolds number as four times the volumetric flow rate divided by pi times diameter times kinematic viscosity. So kinematic viscosity for water at 20 degrees Celsius is going to come from table A1 using 20 degrees Celsius and then grabbing the imperial kinematic viscosity because remember the rows of this table spread across both the imperial and metric unit system. 20 degrees Celsius is equivalent to 68 degrees Fahrenheit or their amounts and the kinematic viscosity is going to be 1.082 times 10 to the negative fifth square feet per second. So from table A1 kinematic viscosity of water at 20 degrees Celsius is 1.082 times 10 to the negative fifth square feet per second. So my Reynolds number is going to be four times, let's draw a big horizontal line, four times 12 cubic feet per minute divided by pi divided by the diameter of my pipe which I believe was two inches, yeah divided by two inches and then divided by 1.082 times 10 to the negative fifth square feet per second. In order to get the units to cancel I will have to convert inches to feet once and seconds to minutes. So 60 seconds in one minute and 12 inches in a foot, inches cancel inches, seconds cancel seconds, minutes cancel minutes, cubic feet cancels, square feet and feet leaving me with a unitless proportion. Come on calculator, let's see if you can't do this. I've always believed in you, don't let anyone tell you, otherwise four times 12 times 12 divided by pi times two times 1.082 times 10 to the negative fifth times 60. We get a Reynolds number of 141,210, 141,210 and for the Moody chart we're going to want that in an exponential format so I'll take 1.4 times 10 to the 1, 2, 3, 4, 5 times 10 to the fifth. I will find the intersection of this relative roughness and this Reynolds number, oh by the way 141,210 is greater than 2,300 which is the critical Reynolds number for internal flow through a round pipe therefore I have turbulent flow. So Moody chart here we come, 1.4 times 10 to the fifth, oh I will draw a bigger line so that hopefully you can see it. That's appropriate. I'll find 1.4 times 10 to the fifth so here's 1 times 10 to the fifth, here's 1.5 times 10 to the fifth, 2 times 10 to the fifth and then 2.5, 3, 3.54, 4.55, 6, 7, 8, 9. So I want, oh let's go with like here, yeah right about there. That is exactly 1.4 times 10 to the fifth plus or minus 20% and then I need to find the intersection of that line with my relative roughness. That relative roughness was 0.001 which is here so the intersection is going to be here which means over I'm on this line here I'll draw a horizontal line so that hopefully you can see that a little bit better. That corresponds to 0.02 to 1, 2, 2, 3, 2, 4, 2, 5 so that corresponds to 0.022 therefore my friction factor for the frictional losses between the fluid flowing through this pipe and the pipe itself is going to be 0.022. Next up I have to tabulate the minor losses so let's build a table draw a whole bunch of lines. I'm going to have the following minor losses let's keep track of these on our fingers. First up we have the entrance to the pipe, second we have the open globe valve, third we have the 12 inch bend, fourth we have the screwed regular 90 degree elbow, fourth we have the excuse me fifth we have the half open globe valve and sixth we have the exit back into the reservoir. I'm going to assume that my entrance is sharp because it appears to be a sharp entrance on this crude drawing so first up we have a sharp entrance and we'll go into section 6.9 and I will look for entrances. I see that my entrances are going to be on figure 6.21B. First of all I note that it doesn't matter what my type of exit is while I'm here because exits are one for all shapes I can write that down while I'm here let's figure 6.21B exit as a k value of one and then sharp edged means that I'm going to be here which is a k value of 0.5 and five. Next up I need to find globe valves for two and five those globe valves we have butterfly valves we have partially opened valves we have a table of valves fully open diagram of our globe valve and for partially open valves we can find our value from this figure so for fully open we can use the table or the partially open we can use the figure so that's 6.18B figure 6.18B for the fully open we can use the table which was 6.5 so fully open globe valve it is of the screwed variety it is two inches nominal diameter so it's going to be a loss coefficient of 6.9 and then for the partially open valve we're talking about a fractional opening which we can see a little bit more clearly on figure 6.17 so globe valve is variety B which means that I'm talking about the proportion of height to diameter so I am assuming that half open is referring to that fractional opening in which case I'm going to be reading off a value from this intersection of our line and 0.5 so this might be hard to see on the video but the globe valve is a slightly darker blue than the disc valve which means that I'm looking at the top line not the bottom line so I'm looking at a loss coefficient of about 6 as opposed to about 3 I'm just going to call that 6 and then we have a 12 inch 90 degree bend so I will go looking for bends hey look it's bends so figure 6.2 is going to be my smooth walled 90 45 degree and 180 degree bends so we are using a value that was determined at a reynolds number of 200 000 that's not our two that's not our reynolds number but it's the best we got so we're talking about the big radius over the small diameter so the radius of the bend was 12 inches calculator 12 inches 12 divided by the diameter of the pipe which is two inches so the proportion is 6 so I'm reading off the intersection of this line here and then maybe yeah that looks all right two three five yeah right about here so I'm going to call that a k value of uh 0.3 0.2 0.24 maybe being a little bit arbitrary and that was figure 6.2 figure 6.2 got to get out of the way and that k value was again exactly approximately 0.24 lastly I have the elbow which is a 90 degree regular screwed elbow back to 6.9 I can find my 90 degree regular elbow screwed nominal diameter two inches gives me a k value of 0.95 table 6.5 table 6.5 and a k value of 0.95 so again because I can factor out the velocity squared divided by the two times gravity term all I really care about right now is the sum of the k values if I had different velocities at the different fittings I would have to determine the losses individually but not in this case so the sum of k calculator adds 0.5 plus 6.9 plus 0.24 plus 0.95 plus 6 plus 1 and I get a sum of 15.59 15.59 so looking back into my equation I have change in elevation I have volumetric fluoride I have the diameter of the gravity I have f now I have the length I have the diameter of the sum of k I have gravity I have the volumetric fluoride the only unknown is the density of water so the density of water at 20 degrees Celsius in the imperial unit system is 1.937 again that's from table 8.1 so 1.937 slugs per cubic foot so now I have everything I need to calculate the power of the pump and I am going to do that by adding together three quantities in the length dimension and then multiplying by density times gravity times volumetric fluoride power of the pump we start with the change in elevation between 1 and 2 so back in my drawing z2 was 120 z1 was 20 therefore I'm using 100 feet of height difference then I'm adding to that four squared times volumetric fluoride squared so four squared times 12 squared cubic feet squared which would be feet to the sixth per minute squared and then I am dividing by pi squared times diameter to the fourth so pi squared times diameter which was two inches to the fourth power and then we're multiplying by two times gravity in the denominator so two times let's use 32.2 feet per second squared because everything else is imperial and we need second squared to cancel minutes squared and we need inches to the fourth and feet to cancel five of the feet so let's handle that unit conversion now there are 60 seconds in one minute and then I square everything and then there are 12 inches in most feet and then I take everything to the fourth power one to the fourth is one feet to the fourth and feet cancel five of the feet leaving me with feet and then inches to the fourth cancels inches to the fourth and then I should be multiplying by a unitless proportion that's going to be the friction factor times length over diameter the friction factor we determined to be 0.022 0.022 times the length of the pipe we were told there are 400 feet of pipe where there is 400 feet of pipe and then we are dividing by the diameter of the pipe which was two inches we need that to become unitless so there are 12 inches in one foot and then we are adding to that all of our loss coefficients which are 15.59 again that's a unitless proportion and then we take that entire quantity and we add excuse me multiply by density times gravity times volumetric flow rate so first up our density was 1.937 1.937 slugs our cubic foot and then we're multiplying by gravity right so 32.2 feet per second squared and then we are multiplying by our volumetric flow rate which was 12 cubic feet per minute and I'm running out of room so I will continue on the next line to handle my unit conversions I'm going to be calculating a power and because I'm in the imperial unit system that is going to be horsepower so as a general rule of thumb when you want to get to a secondary dimension I would encourage you to start at that unit conversion and work backwards into base units until things cancel so I'm going to grab power one horsepower can be written as a foot times a pound of force per second so that's one horsepower 550 feet pound force per second and then I'm going to want to cancel pounds of force and slugs so I recognize that one pound of force is one slug times one foot per second squared so now slugs are going to cancel slugs then I have seconds cubed in the numerator and second squared in minutes so let's go with a minute one minute is defined as 60 seconds that cancels the minutes and then second squared and seconds cancel cubic seconds I have feet and feet in the denominator and then cubic feet in the denominator which is feet to the fifth times but that cancels cubic feet and feet and then all of these quantities are in feet within these square brackets so that cancels the foot in that square bracket as well so that leaves me with an answer in horsepower so all I have to do now is multiply forever okay calculator you can do it I believe in you you got it 100 plus so far so good calculator four squared let's see four squared times 12 squared times one squared times 12 squared that's 12 squared there's a lot of 12 terms probably could have just no it's 12 to the fourth ah don't get distracted John and then dividing by pi squared yeah pi squared pi squared times two to the fourth power times two just by itself times 32.2 times 60 squared and then that's everything out in front of the first set of square brackets let's group that together into parentheses and we are multiplying by a second set of square brackets and that's going to be 0.022 times 400 times 12 divided by 2 and then we are adding to that 15.59 and then closing the second square bracket set let's see if my parentheses are correct so far no they're not and then we are multiplying by 1.937 times 32.2 times 12 and then dividing by 550 150 times 60 so let's see 100 yeah plus oh man you guys can't see what i'm doing that's a shame 100 plus four squared times 12 squared times 12 to the fourth divided by pi squared times two to the fourth times two just regular times 32.2 times 60 squared and then square brackets 0.022 times 400 times 12 over 2 plus 15.59 and then we're taking that entire thing multiply by 1.937 times 32.2 32.2 times 12 times 1 and then we're dividing by 550 times 60 so we get an answer in horsepower of 4.2922 4.4 excuse me 4.2922 horsepower so if you were sizing a pump for this situation you now know how much power it takes to move the fluid and you'd probably assume a pump efficiency of like 80% so you could take that divide by 0.8 you should probably buy a five and a half horsepower pump maybe six if you want a little bit of headroom