# Fluid Mechanics Flow Rate Problem

I am working on a project at an internship this summer regarding flow rates of a certain system that is used to test flow rates on tools, and it requires a bit of a hefty explanation.

I need to increase the current flow rate (1232 GPM) to the desired flow rate (1800 GPM) because some of the tools require testing at flow rates that high.

The pump should be able to handle the flow rates with 180 ft of pressure head no problem. There is about 80 ft of piping from the pump to the tool and then another 80 feet back with a few 90 degree bends. I know that this will add friction head and I have found the equivalent head in feet to help find the total pressure drops.

Where I am at a loss is here:

The majority of the system is 6" ductile iron but once it goes into the building it goes to a 4" flex hose for 15 ft then into the tool (about 2" diameter) then back into another 15 ft of 4" flex hose before heading back to the tank through the 6" ductile iron.

I know that the 30 ft of 4" hoses are greatly decreasing the flow rate, because I ran the pump without going through the tool and already increase from 1232 to 1476 GPM and the tool is only about 5 ft long. I have not had a fluid mechanics course yet but have been trying to do research online to teach it to myself.

I don't expect a solution, but I would greatly appreciate guidance on what the best equations to use would be and how to go about it. I think that completely removing any 4" components and knocking up the rpm's on the pump slightly will allow us to achieve the 1800 GPM that is necessary but I need to have it in calculations before I can put in the request for the 6" fittings and flex hoses.

Thank You

You're in a tough situation because you have not had a FM course. You are going to have to learn how to use a Moody diagram which will enable you determine the head loss at different velocities for each part of the system. A Moody diagram provides friction factors based on Reynolds Numbers and relative roughness. The friction factor is then related to head losses.

I have moody diagram but I am having trouble finding my Reynolds number for the ductile iron. All of the 6" is ductile iron in the system. I found my relative roughness for it (.017) and the hydraulic diameter is 6.0125 inches.

I have found that Reynolds is (Velocity * hydraulic diameter) / kinematic viscosity for pipes. The kinematic viscosity I found for the water is about .926 ft2/s * 10^-5. But I am at a loss on how to find the velocity of the water flowing through the pipe. Is it as easy as just doing the Q=v*A so I would do 1232 GPM or 147.7 cubic feet per minute / (pi *.5 ft ^2) which gets you 3.495 ft per second.

Do I need to be doing this also for the 4" part of the system (the flex hoses) and the 2" part (the tool).

The solution is found by trial and error. You know the pressure at the pump. The pump should have a pressure volume curve so you can use it. What you do is go to the pump curve and pick a point on it giving you a discharge pressure and volumetric flow rate. Based on the flow rate you can compute the velocity and Reynolds number for all portions of the pipe system. Then go the Moody diagram and knowing roughness ratio, you can get the friction factor. Knowing the friction factor you can compute the head loss for that section of pipe. Subtract all these losses from the intial pressure you picked from the point on the curve and see if you get the pressure at the respervoir which, I suspect, is atmospheric. If the pressure you compute is wrong, pick another point from the pump curve and continue until you get the pressure in the reservoir.

I have the pressure volume curve of the pump. There are a couple different curves for it on the chart. Will I just be picking a random point on the curve to start or is there a method to it i.e. should I be picking a point that is of my desired flow rate (1800-2000), or should I start with the point that is at the flow rate that I am at now (1232).

And once I find the friction factor, I then just use the total lengths of the sections for pressure drops? Also for the 90 degree bends in the 6" pipe, will I just use the equivalent lengths and add that to the total length to compute the pressure drop?

And lastly, once I do this and figure out where/how much all of the pressure drops are. How can use that to show how much faster the system would flow with the 6" hoses instead of the 4". Or is there a way to just computer it with the 6" hoses in place of the 4 in the first place to get an estimate on how much greater the flow rate would be?

I flowed the system without the few feet of the tool just going from the 6 to 4 and back to 6 and the flow rate already increased from 1232 to 1476 so I really feel like taking the 30 ft of 4" flex hose out and making it 6" it will greatly increase the flow rate.

Thanks again for the help, this has really helped me to streamline my thinking on the subject matter, I have already learned more and made more progress than a week of trying to look over websites on this stuff.

You can start at the desired flow rate. With it you acquire the pump outlet pressure. Then determine the pressure drop for each section of pipe. You can adjust the lengths to accommodate the elbows. You should see that with a large flow rate and the pump curve starting pressure and small pipe diameters, you will end up with a negative exit pressure which is impossible because the reservoir is probably vented to the atmosphere for safety reasons. That means the pressure drop is too great to sustain the velocities. For your next trial, pick a lower flow rate (higher pump exit pressure) and see what you wind up with at the reservoir. Also, you can change the 4 inch sections to 6 inch sections to see how it affects the downstream pressures.

In working with the equations, make certain your units agree.

I have the pressure volume curve of the pump. There are a couple different curves for it on the chart. Will I just be picking a random point on the curve to start or is there a method to it i.e. should I be picking a point that is of my desired flow rate (1800-2000), or should I start with the point that is at the flow rate that I am at now (1232).

It depends on the curves you've got. Sometimes they'll put the curves for different pump speeds. Do you know how a centrifugal pump works? If you do, disregard this, if not, well in a few words: Centrifugal pumps run along curves. At a given speed (rpm) the pump can only operate along a given path, based on resistance (head) vs. capacity. The less resistance, the greater the capacity.

See HERE
and HERE for good info.

What you would do is locate the point on the curve (the curve for the speed you are running) which corresponds to your measured flow rate. Then ride down the curve to the point you want, and see what head you are aligned with. Unless you get that system head, you will not get your desired flow unless you change the pump speed.

And once I find the friction factor, I then just use the total lengths of the sections for pressure drops?
Yup. Engineeringtoolbox has a good estimator for sched 40 pipe, btw. Though, fyi, hose is notoriously tricky for doing flow calculations. It isn't rigid, the tolerances are less stringent, and there are wiggles and bends throughout.

Also for the 90 degree bends in the 6" pipe, will I just use the equivalent lengths and add that to the total length to compute the pressure drop?

Precisely.

And lastly, once I do this and figure out where/how much all of the pressure drops are. How can use that to show how much faster the system would flow with the 6" hoses instead of the 4". Or is there a way to just computer it with the 6" hoses in place of the 4 in the first place to get an estimate on how much greater the flow rate would be?

Yes, once you figure out the system losses due to the 4" hose, you can figure out the system losses for the 6" hose and compare. Have you accurately taken into consideration the static head as well (the tank discharge)?

Also, so we're clear. Is the discharge tank open to the atmosphere, or is the system closed and the water just pumps around in a loop?

I flowed the system without the few feet of the tool just going from the 6 to 4 and back to 6 and the flow rate already increased from 1232 to 1476 so I really feel like taking the 30 ft of 4" flex hose out and making it 6" it will greatly increase the flow rate.

This is less promising. Presumably your 2" inlet/outlet at the tool, and the tool itself, are creating the largest pressure drop in the system. Frictional losses are quite apparent (and with your flow and diamter hose, it's definitely a factor), but if removing a major unrecoverable pressure loss from the system (the tool) only brought it up 250 gpm, I find it extremely unlikely that you'll get another 325 gpm after mitigating friction losses.

Though, fyi, hose is notoriously tricky for doing flow calculations. It isn't rigid, the tolerances are less stringent, and there are wiggles and bends throughout.

Any particular recommendation on how to do this then? There is 15 ft of 4" hose coming from the 6" going into the tool and then another 30 ft of 4" hose going out of the tool back into the 6" pipe. The first 15 ft doesn't have much of a bend but it does have to go up about 4 feet to get to the tool and then on the otherside there is a rather large parabolic bend due to the excess of hose that was necessary.

Yes, once you figure out the system losses due to the 4" hose, you can figure out the system losses for the 6" hose and compare. Have you accurately taken into consideration the static head as well (the tank discharge)?

Also, so we're clear. Is the discharge tank open to the atmosphere, or is the system closed and the water just pumps around in a loop?

The discharge tank is at atmospheric. The static head is just the height that it has to be lifted right? Well the water in the tank is like 10 feet high, but the output and input of the tank are basically at the same level. The water comes out and into the pump about a foot higher than where it goes back into the tank.

This is less promising. Presumably your 2" inlet/outlet at the tool, and the tool itself, are creating the largest pressure drop in the system. Frictional losses are quite apparent (and with your flow and diamter hose, it's definitely a factor), but if removing a major unrecoverable pressure loss from the system (the tool) only brought it up 250 gpm, I find it extremely unlikely that you'll get another 325 gpm after mitigating friction losses.

Well I was thinking considering that the tool is 6-7 ft long of pressure drop and then the hose is 45-50 ft that that much of 4" could cause a considerable drop.

Also, the chart that i have for the pump is rate for a Pump speed of 1780 rpm. I did a test on our pump and it is running at 1730 rpm. Is that difference negligible or do I need to try to get a curve for the 1730 rpm.