At present, my country Sri Lanka is undergoing its worst crisis since its independence in 1948, with severe shortage of essential items like fuel, cooking gas, electricity, medicine, milk and food items. As a result, we are compelled to be at home, while thousands are protesting in the capital city of Colombo and other major cities.

So, during this period I spent some time on SMath, and attempted the attached example in the internet solved by Hardy Cross Method using EXCEL

Essentially it is an overdetermined nonlinear pipe network with 22 variables {pipe flows) and 23 equations (16 Nodes + 7 Loops). The solution using SMath is based on a snippet using al_nleqsolve() posted by Viacheslav N. Mezentsev (uni) in SMath Forum.

Overdetermined_Nonlinear_Pipe_Network.pdf (432 KiB) downloaded 125 time(s).
Overdetermined_Nonlinear_Pipe_Network.sm (629 KiB) downloaded 134 time(s).

Hello,ndtma
Fragments withal_nleqsolve() can be reduced to one line.
alnlegsolve(3).sm (635 KiB) downloaded 102 time(s).

Wrote

Hello,ndtma
Fragments withal_nleqsolve()can be reduced to one line.

Hello Ber7,

Thank you very much for your very valuable suggestion.
It works fine.

Hi Nugegodage. I hope that the problems in your country can be solved, for the better, and that you and your family are well.

Thank you for this excellent example.

Your worksheet with a little different setup, using tables. No one new calc actually.

alnlegsolve v2.sm (904 KiB) downloaded 108 time(s).

Best regards.
Alvaro.

Hello Nugegodage

I hope that you and your family are very well in these times of crisis in your country.

I have made your example in the Epanet program, the results are shown in the graph below.

For the network to be valid, Node A must have an input pressure equal to or greater than 129.4 mH2O,
and Node C equal to or greater than 150.28 mH2O.



Best Regards

Carlos

Hi Jean, Alvaro, Carlos and others.
Thanks for expressing your sentiments and encouragement.
We are safe and hope the situation will settle as early as possible.

Many thanks to Alvaro and Carlos for your suggestions on the example.

Thanks for your sample.
I always like your irrigation examples.
Hope crisis in your country resolves too.

Regards

Hi overlord.
Thank you for the sentiments and comments.

The subject revisited.
In my original post #1, the K values were simply computed using a given formula,
and head loss in each pipe computed accordingly.

In Method 1 of the attached example, friction factors (f) depend on the Reynold numbers (Re)
of each pipe and computed using a given formula involving Re.

In Method 2, NewtonRapson function of SMath is used for the numerical solution of the
implicit Colebrook-White equation to find f.

Many thanks to Ber7, Alvaro and Carlos for your earlier suggestions on the subject.


Overdetermined_Pipe_Network_3 LOOPS-As per CHEGUIDE_al_nleqsolve.sm (286 KiB) downloaded 109 time(s).
Overdetermined_Pipe_Network_3 LOOPS-As per CHEGUIDE_al_nleqsolve.pdf (581 KiB) downloaded 78 time(s).

Hello ndtmda

Here are the results of your example calculated
with a very old version of Pipe Flow Expert.




Best Regards

Carlos

Wrote

Hello ndtmda

Here are the results of your example calculated
with a very old version of Pipe Flow Expert.

Thank you Carlos!

Hello Carlos

Can you please explain how the marked values of gauge pressures are calculated?
Thank you.

Hello Ndtma

Wrote

Hello Carlos

Can you please explain how the marked values of gauge pressures are calculated?
Thank you.

This pressure (0.9787 bar g) is caused by the water level in the tank, which is 10 meters.


The pressure of 0.6553 bar g is obtained by subtracting from the inlet pressure
of 0.9787 bar g. the losses in the pipe of 0.3234 bar g (3.304 m hd)

The system is not correctly resolved for the pressures, since nodes 3 and 8 have
pressures below atmospheric.

Best Regards

Carlos

Thank you Carlos.

Hello Ndtma

Wrote

Thank you Carlos.

You're welcome.



Changing the diameter of tubes 9 and 10 from 0.25 m diameter to 0.5 m, the problem of negative pressures is solved.

Another way to solve it is by increasing the level below the tank from 0m to 20m. and with the liquid level at 10 m




Best Regards

Carlos

Wrote

You're welcome.



Changing the diameter of tubes 9 and 10 from 0.25 m diameter to 0.5 m, the problem of negative pressures is solved.

Another way to solve it is by increasing the level below the tank from 0m to 20m. and with the liquid level at 10 m

Thank you Carlos. Your contributions are very much appreciated.

Wrote

You're welcome.

Changing the diameter of tubes 9 and 10 from 0.25 m diameter to 0.5 m, the problem of negative pressures is solved.

Another way to solve it is by increasing the level below the tank from 0m to 20m. and with the liquid level at 10 m

Hello Carlos,

Can you please post the Pipe Data Table (with Pipe Number, Inner diameter, ..... Re, Friction Head loss)
for the case Changing the diameter of tubes 9 and 10 from 0.25 m diameter to 0.5 m..
as I get different results for this case.

Hope I am not troubling you.

Hello Ndtma

Wrote

Hello Carlos,

Can you please post the Pipe Data Table (with Pipe Number, Inner diameter, ..... Re, Friction Head loss)
for the case Changing the diameter of tubes 9 and 10 from 0.25 m diameter to 0.5 m..
as I get different results for this case.

Hope I am not troubling you.

There is no problem.

Here is the requested table




Best Regards

Carlos

Thank you very much Carlos.