Calculate grid with Hardy-Cross
Calculate grid with Hardy-Cross
(OP)
Hi
I am trying to implement a sprinkler grid calculator.
I have implemented the Hardy Cross algorithm in c# and the i tried to
find a way to calculate a simple grid(without any luck).

We first calculate the output from a,b,c and d(the will be equal for now).
Lets say 11 l /min. If we now run Hardy Cross it will adjust the flows
in the grid so that we have
the same pressure loss no matter what way you go from a sprinkler to start.
If we now go from sprinkler a back to branch 2 calculating the
pressure drop, we now have a pressure in 2.Lets say 0,8 bar.
If this had been a tree, i would have adjusted the output of the sprinkler c, so that going back from this to node gives a pressure 0,8(+-0.005) bar in node 2.
As this is a grid i do not know how to proceed.
What if i assume the extra flow in c is going from c to 2?
We now adjust the water in sprinkler c so that when going from sprinkler c to
2 gives 0,8(+-0.005) bar. Now i calculate pressureloss back to point
1. I now have a pressure in 1 from one side.
Now i do the same for the other side. I go from sprinkler b to branch
5. Now we have a pressure i 5 from sprinkler b. Adjust flow in d som
that going from d to 5 gives the same pressure in 5 as from going from
sprinkler d.
What now. If i run the Hardy Cross now, would it ruin the balancing ?
Should i go from 5 to 2 to check if
have the same pressure in point 2 going this way ? What if pressures
are not equal at point 2 ?(based on going from sprinkler to 2 and from
branch 5 to 2).
What is we come to branch 1 with two different pressures ? What
sprinkler should we adjust ?
Hoping for help :)
I am trying to implement a sprinkler grid calculator.
I have implemented the Hardy Cross algorithm in c# and the i tried to
find a way to calculate a simple grid(without any luck).

We first calculate the output from a,b,c and d(the will be equal for now).
Lets say 11 l /min. If we now run Hardy Cross it will adjust the flows
in the grid so that we have
the same pressure loss no matter what way you go from a sprinkler to start.
If we now go from sprinkler a back to branch 2 calculating the
pressure drop, we now have a pressure in 2.Lets say 0,8 bar.
If this had been a tree, i would have adjusted the output of the sprinkler c, so that going back from this to node gives a pressure 0,8(+-0.005) bar in node 2.
As this is a grid i do not know how to proceed.
What if i assume the extra flow in c is going from c to 2?
We now adjust the water in sprinkler c so that when going from sprinkler c to
2 gives 0,8(+-0.005) bar. Now i calculate pressureloss back to point
1. I now have a pressure in 1 from one side.
Now i do the same for the other side. I go from sprinkler b to branch
5. Now we have a pressure i 5 from sprinkler b. Adjust flow in d som
that going from d to 5 gives the same pressure in 5 as from going from
sprinkler d.
What now. If i run the Hardy Cross now, would it ruin the balancing ?
Should i go from 5 to 2 to check if
have the same pressure in point 2 going this way ? What if pressures
are not equal at point 2 ?(based on going from sprinkler to 2 and from
branch 5 to 2).
What is we come to branch 1 with two different pressures ? What
sprinkler should we adjust ?
Hoping for help :)





RE: Calculate grid with Hardy-Cross
Learn from the mistakes of others. You don't have time to make them all yourself.
RE: Calculate grid with Hardy-Cross
TTFN

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RE: Calculate grid with Hardy-Cross
Is the solution that easy ?
All the examples i find using Hardy Cross or other methods, the outputs(and input) are known.
I refuse to belive that the way to solve this is to split the grid into two trees.
Tommy
RE: Calculate grid with Hardy-Cross
I guess I'm confused about your last statement, since you should know your inputs. From what I read, HC is an iterative process, but I get the impression that you are trying to solve it with a closed-form solution.
TTFN

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RE: Calculate grid with Hardy-Cross
Learn from the mistakes of others. You don't have time to make them all yourself.
RE: Calculate grid with Hardy-Cross
How can i know my input before calculation ?
Yes HC is an iterative method, and i think i have to run it several times.
RE: Calculate grid with Hardy-Cross
TTFN

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RE: Calculate grid with Hardy-Cross
https://www.google.com/url?sa=t&rct=j&q=&a...
RE: Calculate grid with Hardy-Cross
Thanks for you inputs. I have to think about it.
RE: Calculate grid with Hardy-Cross
The only viable calculation that you can do with your setup is to fix the lengths and diameters of the pipes and fix the inflow at 7 and then iterate HC until you get a stable answer for the flows out of the 4 sprinklers. If they do not meet your criteria then you can change some pipe diameters and re-run HC. But without valves or restriction orifices your chances of getting equal flows (unless your entire network is symmetrical) are close to nil because pipes come in discrete sizes.
In all the sprinkler systems I have designed I have found the most satisfactory route is using a tree design. Sometimes it is better to feed from more than 1 point, but the calculation still follows the tree method. Usually you will have several sprinkler nozzles on a single lateral, and then you have to size the lateral to have negligible pressure drop from the feed point to the last nozzle so that each nozzle sees the same pressure. "Negligible" in this case means low enough that the difference in flows from the nozzles are within the design allowance.
It may be possible to design all this into a network calculation that can come to an automatic solution, but my experience of trying to remedy installations that have been done this way by others tells me that it is not easy.
Katmar Software - AioFlo Pipe Hydraulics
http://katmarsoftware.com
"An undefined problem has an infinite number of solutions"
RE: Calculate grid with Hardy-Cross
Thanx for your answer.
No i would like that the most demanding nodes , a and b in this example to deliver minimum required flow(11 l in my setup).
Then we know that the other sprinklers(of the same type) will deliver more flow(because of higher pressure) just as in a tree system.
What would your solution be knowing that ?
What if do it this way:
I start in b, and calculate the pressure drop over sprinkler b, with sprinkler delivering 11 l/m.
Then i calculate the pressuredrop over the pipe between a and b, pressure drop over a delivering 11 l/m.
Calculate pressure drop from a to node 2. I know have pressure in 2.
I now out balance the branch 2,c,d just like i would have done for a tree(iterative so that i come back to node 2 with the same pressure as when i came from node a and b)
At this point i know the flow in and the flows out, and now i run HC ?
Best regards,
Tommy
RE: Calculate grid with Hardy-Cross
RE: Calculate grid with Hardy-Cross
Regarding my replay to Katmar, i know see that it become all wrong.
I will get 2 pressures in point 1. One from the tree, and pressure zero from 4,5,6.
Tommy,