Modeling a well in WaterCAD

Question

Hello, 
 How to model a well in watercad having a known constant outflow? 
 thank you

Jesse Dringoli · Accepted Answer

If there are only fixed demands downstream then a reservoir at the well location (with no negative demand anywhere) will produce a flow equal to the sum of the downstream demands. You would not be able to use a negative demand in this case without having another boundary condition (reservoir or tank) to either accept access flow or supply a deficit.

Joeri Legierse · Answer

The easiest way would be to model it as a junction with a negative demand (enter the constant outflow as a negative number for the base demand). This however does not take into account the actual maximum pressure from the pump in the well, so please keep this in mind. For a more detailed model you could start with a reservoir with an hydraulic grade equal to the well water level, then connect it to a variable speed pump (using the pump curves of the pump that is installed in the well) and setup the well constant outflow as the variable speed pump flow target.

Jesse Dringoli · Answer

Here are some related articles to supplement Joeri's answer - 
 
 If using the negative demand approach: Modeling an inflow (negative demand) in WaterCAD or WaterGEMS 
 If your source has a chance of running dry: Modeling Fixed Inflow from a Marginal Source 
 If using a reservoir + pump approach: Accounting for well drawdown

Jesse Dringoli · Answer

You will not be able to set both the pressure and the flow. If you add a reservoir and a negative demand, it will try to solve the hydraulics around both "knowns". 
 To force 65 l/s into the system, start the system with a junction with a -65 l/s demand (inflow). Meaning, no other elements upstream. This will ensure the pipe will have 65 l/s, but the model will solve the pressure/HGL based on that flow and the other hydraulic boundaries in your system. What do you have downstream of the well? Only fixed demands? Only tanks or reservoirs? A mix?

Jesse Dringoli · Answer

An FCV will throttle to reduce the flow but that only works if a headloss causes a drop in flow. If you have a model with fixed demands, the flow from the source is essentially fixed and will always be equal to the sum of the downstream demands. Attempting to use an FCV in such a case would result in an ill-conditioned matrix . See more here: Using a Flow Control Valve (FCV)

Jesse Dringoli · Answer

With only fixed demands downstream, a reservoir + pump will produce a flow equal to the sum of the demands (which I assume is 65 l/s) and will add head so that you can simulate pressures in the network. If you have an estimate of the amount of head the pump will add, you can use the "design point" pump type and enter that head with the corresponding 65 l/s flow point. It would be best to get information from the pump manufacturer. Plus you may need information about the well water level and drawdown effects. See the third link I had provided earlier: Accounting for well drawdown

Modeling a well in WaterCAD

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