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. :-)