How to Model FCV

Hello A.S.,

What are trying to model with this? This might help to ultimately find the best setup for the system.

One item of note is how WaterCAD works in a closed system. WaterCAD and WaterGEMS are demand-based programs, meaning the demands will always be satisfied. The demand at a location is the flow rate coming out of the system at that point.

With a closed system and only one source of flow, like a reservoir or tank, the flow that you will see out of that will be equal to the demands in the system. If you have 14 m^3/hr of demand in the system, the flow out of the source will be 14 m^3/hr as well. 

Based on this, with a closed system, an FCV may not be necessary or may actually cause issues with the results balancing. Instead, if your concern is introducing some sort of minor loss, you can include a minor loss on a pipe though the pipe properties. There is minor loss engineering library available that has some common values you can use. The following link has information on this: Adding Minor Losses to Pipes.

Regards,

Scott

It sounds as if what you have is a true closed loop with no inflow or outflow. In this case the flow is controlled by the pump and the pipe and valve losses. There are no Demands in a closed loop.

Exactly how will you control the flow? Will there be a throttling valve that is controlled by a PLC; will it be controlled by the pump and valve selection with no real time control (The usual setup) or will the pump be a variable speed pump? Don't use an FCV unless you really have one.

For a model to run, you need at least one constant head node to set the Hydraulic  grade. This depend on how the system is initially charged and how makeup water is provided to offset any leakage. This can be modeling by connecting a reservoir node to the loop with a tiny pipe such that there is negligible flow inflow/outflow. The elevation of the reservoir determines the pressure.

Hi Scott, 

thanks for your reply 

I want your advise in this i want to explain the system hoping to help me 

I have the system already designed its chilled water cooling system,  the system will fill with water for the 1st time and water will be circulated in closed loop.

What exactly im doing with watercad is trying to study the system behavior and the losses occurs, pressure losses,  flow direction.

Im facing problem with representing the elements on watercad  i have expansion tank which i model it  as hydropneumatic tank ? i have PICV which i represented as FCV , chiller as piping ......

So what i understand from you is to represent the losses from the valves by adding it into pipes without adding FCV which is good solution cause the FCV caused lots of errors

but about the demands and flow im little bit confused how the program can understand that it is a flow running through the system  not a demand to satisfy? to run the system with the right direction of the flow 

and I dont have source of water but what i did i put reservoir with minor dia and long pipe or there another way to run the system without source of water ?

Hello 

A.S.,

If you have laid out the pipes in correct direction, water will flow in right direction unless there is reverse flow occurring as per the circumstances. Unless there is demand in the system water will not be supplied from the source, so you can demand on a junction in the loop and then flow can circulate in pipes. 

This reservoir i.e. source of water would be a dummy or fake reservoir in your case to setup the HGL and pressure calculations based on that. 

Please see below wiki about modeling closed system, for reference. 

Modeling a closed loop system 

In a typical water distribution system, flows are controlled by the demands which take water out of the system. In a closed loop, flow is completely controlled by the circulating pump and the head losses (both piping and minor loss) in the loop.

In a closed loop, the head added by the pump will exactly match the head loss in the loop.  The reason you need a reservoir node is that in order to convert head loses into head, you need some starting point for the calculation. Otherwise, the pressure could be 5 kPa, or 5,000,000 kPa or anywhere in between.