First Time Creating a System Curve 2

[SimplyCivil]

I'm an EIT looking for advice on how to properly create the system curve for a simple water distribution network. I understand conceptually that it is the resistance the pump will "see" due to friction, fittings, etc.

I was told there is only one unique (H vs Q) curve for a given system, but I'm confused because I think that it will depend where the flow and head are measured in the distribution system.

Here's what I've done so far:
1. Input nodes, pipes, & a reservoir to EPAnet
2. Chosen a node from which to draw the demand flows
3. Vary the demand flows from 0-1250 USgpm, noting the head at the node after each run.
4. Graph the H vs Q.

But if I select a different node, I generate an entirely different curve.

Any guidance on what I might be doing wrong?

Thank you kindly!

 

[stanier]

Simply the hydraulic gradline varies with elevation and distance from the source of the head (ie the pump). The system curve is usually developed for the pump node so that a suitable pump can be selected.

“The beautiful thing about learning is that no one can take it away from you.”
---B.B. King

http://waterhammer.hopout.com.au/

 

[SimplyCivil]

Thanks for the reply. That makes much more sense that the system curve should be taken at the pump node. (I had been measuring the head at the end of the piping system, after the HGL had fallen.) I've attached a simple sketch of my system to clarify. You'll see the system of interest in red - the pump will extract water from the reservoir and provide water to a residential area uphill.

I guess what confuses me is that since I have a reservoir (underground) upstream of my pump node, every time I run my EPAnet model, the head at the pump node is always a constant too. It's like the reservoir acts like a boundary condition. This would produce a "flat" system curve, but that can't be correct because there is piping with flow-based losses downstream.

Your thoughts?

 

[stanier]

The reservoir is a system boundary and provides a constant head. Hence will provide all the flow that is necessary to meet the demand. The pump also has to provide pressure at the users system boundary. It could be that this pressure in the system is the significant proportion of the sytem loss and hence you will get a flat system curve. This suggests your piping is too large or sized for future demand.

“The beautiful thing about learning is that no one can take it away from you.”
---B.B. King

http://waterhammer.hopout.com.au/

 

[fel3]

You will often have more than one system curve. For starters, here's a response I gave in another section of this forum:

http://www.eng-tips.com/viewthread.cfm?qid=320430

(see the bottom of the thread).

For most systems I have worked on, the system curve is really a system envelope. A simple example is pumping from one tank to another. When the suction side tank is at minimum level ("empty") and the discharge side tank is at maximum level ("full"), the static lift is much greater than if the suction side tank is full and the discharge side tank is empty. In addition, demands between the tanks will fluctuate between minimum and maximum (typically a diurnal cycle, but not always) and it's even possible that the fluctuations on either side of the pump do not have the same time pattern.

If system demands are well behaved, then this type of situation requires four system curves: [1] pumping empty to full with minimum demands (maybe even zero demands to be safe), [2] pumping empty to full with peak demands, [3] pumping full to empty with minimum demands, and [4] pumping full to empty with maximum demands.

For complex systems, it may require more than four curves. In addition, you may need to model a fire flow or two. Sometimes the upper and lower bounds of the envelope are defined by curves that cross each other, so it's not just finding the one upper and one lower curve.

==========
"Is it the only lesson of history that mankind is unteachable?"
--Winston S. Churchill

 

[77JQX]

Star for fel3.

However, I notice that your sketch does not include any receiving tank. Will this be a closed loop pumping system without a tank? The concept of a system curve gets a bit tricky with closed loop systems. When you construct a traditional system curve there's a place to receive excess water, and you're finding out how much flow the system will take with varying supply pressures. With a closed loop system, the flow is always the same, regardless of the supply pressure. So the curve you construct more closely resembles a plane.

For a closed loop system, you need to find the minimum head the pump will need to develop under the worst case conditions. That usually means finding the lowest pressure delivery point in your system under maximum demands, and finding out what pressure is required at your pump node to maintain the minimum acceptible delivery pressure. The pump design will then need to include provisions to avoid over-pressuriztion and unacceptible pressure variations with other potential operating scenarios.

 

[SimplyCivil]

Yes, it is a closed system past the reservoir; the only discharges are for residential servicing and fireflow from one of two hydrants. I was starting to piece it together from fel3's different tank scenario's that I would need to do a similar thing for my closed system - just as you mentioned 77JQX that I need to look at the elevations to find the point of lowest pressure (highest elevation). Since I have Maximum Demand and then potential for 2 distinct fires (different nodes) - this gives rise to 3 system curves...

As far as modelling, I keep getting an error that says "Negative Pressures at 0hrs." My skeleton has the reservoir, a "dummy node" for the pump, and distribution piping with user demands. Since I don't have a pump in the model I think it makes sense...I believe the flow out of the reservoir must be the amount the pump needs to provide?

I also tried moving the reservoir up "theoretically" to fix the negative pressure errors. When I raised the reservoir by 120' the minimum system pressure of 20psi is met everywhere. I think that indicates that the the pump must add a minimum of 120' head to the system.

 

[fel3]

Back to your original question…

Your approach seems to be similar to what I do, but I'm not sure why you moved your varying demand to another node since your pump won't be moving node to node. Regardless, the system curve (or system envelope) is different for each node. This is to be expected because each node "sees" the system in a unique way.

Just to be sure, here's how I approach a situation like yours. I leaving a lot of details out and focusing on the key points of the analysis. BTW, this can be done in EPANET as well as other programs.

[1] Within one EPANET model I build two disconnected systems that are disconnected at my proposed pump location. One system is for the suction side of the pump and one system is for the discharge side. Each side will have one node that represents the pump. Make sure to include at least the minor losses within the pumping station since these can be significant with respect to the combined system as a whole.

[2] Set up a demand pattern to apply only to the two nodes that represent the pump. Other demand patterns can be set up for the rest of system as appropriate. For the simplest systems (e.g. reservoir-pipe-pump suction + pump discharge-pipe-reservoir…notice I did not say tank, which has varying levels unlike a reservoir) there are no demands that vary over time so the "time period simulation" implied by the demand pattern is bogus with respect to the time. The purpose of the demand pattern is to step the flows so that you get all your suction and dicharge HGL data from one run. There are two ways to do this: set base demands at the two pump nodes to 1 gpm (suction) and -1 gpm (discharge) and create demand pattern multipliers that equal the flow steps you want (e.g. 0, 250, 500, 750, etc) or set base demands equal to the first flow step (e.g. ±250 gpm) and use the multipliers 0, 1, 2, etc. The 0 multiplier gives you the static lift for zero flow.

[3] Analyze the models and extract the suction and dicharge HGLs. From here, I usually drop the data into a spreadsheet so I can plot the systems curve(s) against several pump curves. Also, if I have tanks instead of reservoirs, I can manually raise or lower the tank levels in Excel rather than complicate my EPANET model even more.

[4] Once I have selected a pump, I insert a pump where my two node are and add it's published curve. Then I run any additional simulations needed.

I hope this helps.

==========
"Is it the only lesson of history that mankind is unteachable?"
--Winston S. Churchill