How to calculate maximum possible demand in a network?

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Prasanna M

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Nov 4, 2014, 8:36:46 AM11/4/14
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Hi All,

          I have simulated a small network consisting of a reservoir, pump, tank and a distribution line. The tank is 0.2m in height, 0.252m dia and consists of inlet and outlet valves(FCV). The size of this pipeline is 20mm dia. The outlet is connected to a distribution network of pipe dia 4mm. When I simulate this network, I can give only a maximum demand of 0.022 LPS irrespective of setting any flow to the inlet valve of the tank. If I exceed this limit, the outlet pipe of the tank is closed and negative pressure is generated. Why is it so?? How to calculate the  maximum possible demand in a network?? Do I need to change any other parameters in order to get higher demands??
PFA : network model
Any Idea regarding this problem is much appreciated :-)

Regards,
Prasanna M
WDN subntwrk.net

Daniel Antonio Colocho Arévalo

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Nov 4, 2014, 9:23:59 AM11/4/14
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Hello Prasanna: Before you do the model of a WDS, you shall do some previous calculous and know some engineering specifications, those previous values it give you the guide to asign measures to: diameters (mimimun at beginning), velocities [0.60 to 3.00 m/s], service pressure (10 mca at least in the more elevated points in the net), an so on. To do first a random model  without guidelines is to lose your time.

I advice you, read, first the EPANET user´s manual (2 or 3 hours is enough to do it), in the www.epanet.es you find some elemental books you can read online or buy them, those are books of Santiago Arnalich that guide you throug the design process. 

Remember, EPANET is not a software to do design, is for simulate. 

The maximum possible demand in a network is a value you know trough of engineering parameters asociated to the net design, is the consequence of the net users demand proyected to a time called "period´s design". This concept get the actual users, asign a water quantity and make a projection of this in the period´s design. 

Really, try to explain all the concepts you need is a big work that require a lot of time. Read the books of Santiago and you´ll be an expert in network models.

Your model it has small diameters, those are smallest that the minimum required in water networks. If you are using Hazen &Williams coeficient, this diameters are not apropriated, I thing this is the reason tou have those results.

Over, I advice you, read the user´s manual an the Santiago books.

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Prasanna M

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Nov 4, 2014, 11:54:05 PM11/4/14
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Thank you very much for your kind suggestions. 

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Laurent Wismer

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Nov 5, 2014, 11:13:51 PM11/5/14
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Hi Prasanna,

Just remove the FCV valves (that force your 1 lps in the first part and 0.22 in the distiripbution part) better not to use any flow regulation valve and add a "pump curve" for your pump and you will have a system that will work...
Hope it helps...

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Prasanna M

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Nov 6, 2014, 2:38:31 AM11/6/14
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Hi Laurent,

                Thanks for spending some time for me. I know it works but am in need of those flow control valves which I will be controlling using some motorized actuators. So is there any other possibilities of meeting higher demands without exceeding the pipe diameter of 20 mm?? Thanks again for your suggestion....

Regards,
Prasanna M 

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Lucas Vasconcelos

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Nov 6, 2014, 4:18:12 PM11/6/14
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Hi,

With diameters this low, you should use DW headloss equation. Although Hazen Williams' is widespread, with diameters this small the difference between the equations will most certainly effect the result. EPANET's manual provides some values, as you're using high HW coefficients, you should go for the smaller values for DW valid intervals. :-)

Now, you have a flow control valve (FCV) set to 0.022Lps in the outlet of your tank. Epanet does not mimic these valves actual behavior (as a matter of fact, it does not mimic any of the valves actual behavior). No matter how high the pressure is, the flow through this valve won't be higher than its setting. It doesn't matter, as well, the network's demand.

Epanet's hierarchy will follow some rules:
- The set demands (base or multiplied) will be fulfilled. If there is not enough flow, if the demanding nodes are disconnected (due to an tank's outlet closing because the tank reached minimum level, for instance) or even production in the system to met the demands, Epanet  will create water with ridiculously low negative pressures and will met the demands;
- The set values for valves will limit the system, and the valve will behave as if it wasn't if the values are below the setting point. No flow above the setting point will go through a FCV, even if it means demands won't be met. Epanet will obey the FCV setting AND create flow. :-)
- The pre-defined valves diameters are not used. They won't be used to calculate an pre defined local headloss. Epanet also won't automatically set local headloss coefficients for junctions and splits in the pipings. :-(

That said, as for answering your question, there is no demand high enough in a node that Epanet won't fulfill. If it is unstable or unable to find a solution, just allow for more attempts in options>hydraulic.
The maximum demand of each node that won't cause negative absolute pressures are tricky to find. By lowering a node's demand, you can raise all others'. And so it goes on. If you need something in that line, you'll naturally need some limiting parameters, and resort to a coded solution. Even in the simplest network this might be a hard problem.

Now, there is an approach used to assess the maximum flow for calculating tank's differential pressure. I do assume that either it will be caused by the piping's breakage in the lowest point, or it the tank's base. Orifice flow will follow an equation:
Q = Cd.S.squareroot(2.P.g), I use .61*area*suareroot(2.Pressure.9.81). You can manipulate it to match epanet's noozle, Q=C.squareroot(Pressure). But units for pressure and flow will be the ones set for the rest of the system. In the other, it is set for cubic meters per second and meters. C is the emitter coefficient for each node.

In your system, set as it is, it will be 0.022Lps because of the tank's outlet FCV. :-)

Prasanna M

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Nov 11, 2014, 1:13:16 AM11/11/14
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Hi Lucas,

              Thank you very much for your detailed description on how EPAnet works. It gave me some ideas to proceed further. I came to know that the limitation in providing base demand is because of not providing the initial level in the tank. If we did not provide any initial value to the tank, then as the simulation runs, as soon as the tank starts filling it gets empty. This forces EPAnet to close the outlet of the tank. Again the tank starts filling and again it starts emptying. Hence several such time steps takes place and forces Epanet to end at closed link thereby producing ridiculously high  negative pressure to meet the demands.

            Therefore I provided a very very minimum initial level (say 0.001m) and increased the base demands at nodes and the program runs successfully .  

            But now I have another problem to over come :-( The pressure at the demand points are in the range of 2.55~2.99 m ( Which is Approx 0.2467~0.2893 atm ) In order to provide flow, the pressure at the demand points must be atleast greater than 1 atm isn't?? So my next doubt is, does the output of EPAnet shows the pressure in Absolute scale or Gauge scale?? If it is in gauge scale, then there ll be no problem. It ll be very much helpful if someone knows about it.....

Regards,
Prasanna M

Lucas Vasconcelos

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Nov 11, 2014, 5:50:04 PM11/11/14
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Hi,
Epanet will run the model as if the pressure 0 is the ambient pressure, so, as long as the pressure is positive, flow will leave a node, and as long as it is above vapour pressure (actually, vapour-ambient, the value will roam around -7m and -10m), there will be flow (although there might be cavitation).

Epanet won't assess these effects, so you should always be cautious of negative pressures.

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