Most water system hydraulic calculations (at least in the United States) are based on the Hazen-Williams Equation, although I will sometimes use Darcy when dealing with pumping stations. Here is a quick introduction to the Hazen-Williams Equation:
However, a Wikipedia article is no substitute for experience when it comes to actually applying the equation properly. If you are unsure how to proceed, I suggest you get help from someone who knows how to do this. However, assuming you understand hydraulics, just not how to proceed with this problem, I will continue.
I have done these types of calcs by hand, but I almost never do. For some problems, I will use programs I wrote decades ago for the HP-41CX and HP-42S calculators, including an interchangeable solutions program for the Hazen-Williams Equation for a single pipe (it handles "what if" calculations in the same way as the Time Value of Money program built in to the still current HP-12C calculator) and a Hardy Cross program that will handle up to 99 pipes in loops with up to 6 pipes each. I still use the first program often, but I haven't used the Hardy Cross program in probably 20 years. BTW, my calculator programs substitute HL/L for S and Q/A for V and include conversion factors so I can use HL and L in feet, D in inches, and Q in gallons per minute. Also, pay careful attention to true vs nominal pipe diameters when it comes to using pipe diameters in your calculations AND how this affects the roughness coefficient. In this thread (
about 1/4 of the way down (19 Aug 13 15:47), I discuss this issue in some detail.
Most of the time, even for small problems such as yours, I find it is easier and faster to build a model in EPANET (which is free:
or WaterCAD (which is not) and run a few scenarios. For example, I was able to build a model containing one reservoir, six fire hydrants, and seven pipes to make a complete loop in under a minute. Less than one minute later I had results from a fire flow run at the two most distant fire hydrants (1000 gpm at each hydrant) AND from a separate fire flow run at the first two fire hydrants on one side of the loop (also 1000 gpm each). The first run produced (obviously) the lowest system pressures and the second run produced the greatest flow on one side of the loop (used for checking maximum flow velocities). If you have never used EPANET, it would take you a bit longer, but I think in the end it will still be quicker than trying to balance flows in a loop by hand using the Hardy Cross technique. We did similar problems by hand as homework in one of my hydraulics classes in college and it's pretty time-consuming.
However, if you want quick results, here is a decent way to approximate your system without having to do the iterations by hand:
For a fire flow at the two most distant fire hydrants:
[1] Assuming your loop is reasonably symmetrical, gently "remove" the pipe between the two fire hydrants. It will have a flow of nearly zero anyway (or actually zero according to my perfectly symmetrical model), so this is a pretty safe assumption.
[2] Run 1000 gpm down one side of your loop (now a branch) and 1000 gpm down the other side. (If you loop is not very symmetrical, try something like 1100 or 1200 gpm on the shorter side and put the difference on the longer side.) Since you will only have a flow at the last node on each side, you can combine the lengths of the three pipes on each side. This means that you only have two pipes to deal with, one serving each flowing fire hydrant. Then, input L (total), C, D, and Q into our modified version of the Hazen-Williams Equation to calculate HL for the two pipes. The HGL at the first hydrant is HGL_1 = HGL_Reservoir - HL_1 and the HGL at the second hydrant is HGL_2 = HGL_Reservoir - HL_2. Also, convert the HGLs into pressures so you can verify that you meet the pressure criteria (usually 20 psi at the flowing hydrants). So far, we have not dealt this the hydrant laterer and the fire hydrant assembly. Please note that there will be additional head losses in these elements as well. I also touch on this subject in the thread I referenced above. Generally if you provide 25 psi at the beginning of the hydrant lateral, and certainly if you provide 30 psi, you should have at least 20 psi at the outlet of the flowing hydrant.
[3] To check the validity of this solution, take a look at HL_1-2 = absolute value of (HGL_1 - HGL_2). If it's small (how small is a judgement call), your solution is valid enough. You should also take HL_1-2 and the L, C, and D of our "removed" pipe and calculate a fictitious Q. If it's small (again, how small is a judgement call), your solution is valid enough. If, on the other hand, HL_1-2 and the fictitious Q are relatively large, you can use this fictitious Q to adjust the starting flows a'la Hardy Cross. You could start by taking say 1/4 to 1/2 of the fictitious Q and adding it to the loop that produced the higher fire hydrant HGL and subtracting the same amount from the other other branch. After a couple iterations like this, you should have a balanced system.
For a fire flow at the two fire hydrants nearest the reservoir on one side of the loop and again for the two fire hydrants nearest the reservoir on the other side of the loop (two separate scenarios):
[1] Start by making a prediction of how the flow will split coming out of the reservoir. My perfectly symmetrical model (all seven pipes have identical lengths, diameters, and roughness coefficients), I got 1389 gpm on the side with the flowing hydrans) and 611 gpm on the other side. The flow between the first and second hydrant was thus Q = 1389 gpm - 1000 gpm = 389 gpm. You could start here, or adjust these depending on how assymmetrical you loop is.
[2] Calculate the HL between the reservoir and the first hydrant, then between the first hydrant and the second hydrant. Combine the five pipes on the other side of the loop and calculate the HL between the reservoir and the second hydrant. Next look at the differences in HGL at the second hydrant between the calculations on each side of the loop. As before, if the difference is small, your solution is valid enough. Otherwise, you can adjust your starting flows in a manner similar to the first example above and run it again.
I hope this helps.
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"Is it the only lesson of history that mankind is unteachable?"
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