Efficiently control a pump using VFD

[sundaysfantasy]

I recently started to work as a PLC programmer at a local pump manufacturing company. Among other projects, the most I deal with are wastewater pumping stations that use two pumps to empty the tank.
Before working here, I did a few similar projects and the solution was simple: start one pump at a preset level, stop it when tank empty, start the other when it fills up again. Pumps were not controlled via VFD so no further control was necessary.
Now however, most of the pumping stations have VFDs controlling the pumps and the only thing I do with those VFD besides soft starting the motor to nominal frequency is testing for clogs.
I recently came across a control system for this type of pumping station that has a high efficiency mode which means, from what I managed to understand, that it varies the rpm in order to reduce power consumption while keeping the pump on it's curve. So I started researching ways to do this, but the math is killing me.
I have the following information: a graph of the pump curve and efficiency, level of the tank and the out flow of the station.
My question is, how can I use the graph and flow in order to find a suitable control method for my pump using the VFD in order to reduce power consumption? Is there a math formula to calculate a parameter that needs to stay constant?



here's the graph: https://ibb.co/Y3jZSkv

 

[drbitboy]

Set the wayback machine, Sherman, we're going to see if I can remember some thermodynamics ...

Fluid work is

dW = d(PV)
dw = dW / η(Q) = d(PV) / η(Q)​

where

W = Specific work applied to the fluid; "specific" means per unit quantity (mass) of fluid; N.B. ignore sign conventions as we are interested in magnitude
P = Pressure
V = Specific Volume i.e. Volume per unit quantity
w = External Specific Work (energy) applied
Q = flow rate, dV/dt
η(Q) = efficiency as a function of flow rate
​

So

dw = d(PV) / η = (dP V + P dV) / η​

since water is incompressible

dV = 0
dw = dP V / η
Δw = ΔP V / η
​

Power is the time rate of work

power = ΔP dV/dt / η = ΔP Q / η​

Although OP wants to minimize power convention, every quantity of wastewater that comes into the pumping station is going to be pumped out, so the Δw specific work form is good enough, because it tells us how much work is done per unit of wastewater, and minimizing that will minimize power consumption.

ΔP is the pressure rise across the pump, and will be equal to the head difference between the suction of the pump and the destination, plus any frictional pressure drop from flow:

ΔP = (Pdest - Psuction) + ΔPfriction = ΔPhead + ΔPfriction​

ΔPhead is a constant, and to first order, ΔP friction is proportional to the square of flow rate:

ΔP = ΔPhead + k1 Q**2
​

where k1 is a constant and a property of the piping between the pumping station and the destination. So

Δw = ΔP V / η(Q) = (ΔPhead + k1 Q**2) / η(Q)
​

So we have two terms, a constant plus flow-squared term in the numerator, and a flow-dependent efficiency term in the denominator.

As a thought experiment/estimate, using a power law model for efficiency,

η(Q) = k2 Q**k3; k2, k3 are constants
​

based on the supplied pump curve, my calibrated eyeballs say k3 < 1

η(Q=120) / η(Q=60) ~ 60 / 40 ~ 1.5
(Q=120) / (Q=60) = 120 / 60 = 2
∴ k3 ~ 1.5 / 2 < 1
​

So, when ΔPhead is small relative to [k1 Q**2] (high flow rate), the numerator of the Δw equation, with a power of 2, would dominate the specific work (and power) required.

The way to minimize power consumption, in that situation i.e. high average flow rate, is to minimize flow rate, which would mean maintaining a constant level. There is however a lower limit to the flow rates where this is true; pumping below that minimum uses more work per unit of water because the efficiency drops too much i.e. faster than ΔP. If the level is still dropping at that minimum flow, because the incoming flow is less, then the level should be allowed to drop to the low limit float switch and the pump turned off.

The remaining pieces of the puzzle required are

• the system curve i.e. the constants ΔPhead and k1
• a better model for η(Q); if the power law were actually implemented, then k3 would be a function of Q, and eventually it would drop to 0 and go negative with increasing Q.

Both of those are required to determine the minimum Q.

Conclusion

That was a first-principles approach to the problem. It may be far too oversimplified; it may also contain fundamental errors. That said, it might provide a starting point for further analysis.

Notes

Inalt. de pompare: High.(?) pumping
randament hidraulic: hydraulic efficiency
randament total: total yield (efficiency?)
valori npsh: npsh values
turatie: speed
domeniul de utilizare: field (domain?) of use

 

[sundaysfantasy]

Thanks for the detailed information. I, however, am not looking to maintain a constant level, that wouldn't even be possible because of the low intake flow on these tanks.

What I want to do is empty the tank as usual, but controlling the speed of the pump in such a way that it pumps all the water out while using the least amount of energy.

As I lack hydraulics education, I fail to see what the constants and the variables are in this system, but I'm determined to learn.
A coleague tried to explain how the curve works and I though I understood it, but when I started to do the math I reached a dead end.

But taking an even more simplistic approach, based on my graph, can I draw the conclusion that for best efficiency in terms of power consumption, I must control the pump at around 1450 rpm, rather than controlling it at full power(1470 rpm)? I know, it's an insignificant difference, and maybe my graph wasn't the best, but there are pumps and situations where the difference is quite big.






Inalt. de pompareump head pressure
randament hidraulic: hydraulic efficiency
randament total: total efficiency
valori npsh: npsh values
turatie: speed
domeniul de utilizare: domain of use

 

[drbitboy]

I would say the 1450rpm is a reasonable assumption and possibly the right answer, as the speed of maximum system efficiency seems to be constant across the different pumps.

But without the system curve we cannot know where the pump will actually operate. The top plot shows the head available at different flow rates; what if the outlet piping is smaller or larger or longer or shorter than what would make that the upstream pressure from the downstream destination at that flow?

Also note that even if the efficiency from the plot is better at a given flow rate and outlet head, that gain in efficiency may not translate into a net reduction in long-term power consumption (kWh per week, not kW while running) if extra work, per quantity of wastewater, is being applied against a higher pressure.

We need the system curve.

but controlling the speed of the pump in such a way that it pumps all the water out while using the least amount of energy.

Exactly.

 

[Tom Jenkins]

First, you are missing a key piece of information to achieve your goal. Your curve is just the pump performance curve, which shows the capabilities of the pump. You cannot identify the operating point of the pump without superimposing the system curve(s) on the pump curve, as DrBitboy indicates. Then you can begin to make choices, but they will be limited.

Whether you are pumping into a long force main (mostly friction head) or uphill and then to gravity flow (mostly static head) you don't have a lot of control. The objective is to have outflow equal inflow, and this is accomplished by maintaining the wet well level. The only significant optimization you can provide is by raising the wet well level setpoint to the highest point allowable. That decreases the head across the pump and minimizes power for a given flow.

The typical wet well inflow varies 2:1 over the course of the day, plus storm surge flow increases.

Remember that the owner pays for energy, kWh, not efficiency. Pumping at a lower flow rate and minimum head reduces power, even if you are at a lower pump efficiency.

Read these articles, and then come back with more specific questions.
https://www.mswmag.com/editorial/2016/06/flow_pressure_and_pump_performance
https://www.mswmag.com/editorial/2016/07/pump_station_wet_wells_vs._dry_wells
https://www.mswmag.com/editorial/2016/08/maximize_pump_station_efficiency

 

[drbitboy]

Remember that the owner pays for energy, kWh, not efficiency. Pumping at a lower flow rate and minimum head reduces power, even if you are at a lower pump efficiency.

Cool articles, Tom!


Woo hoo, looks like I am at least close; it's been a while but apparently the fundamental laws of thermo have not changed (you can't get ahead; you can't break even; you can't get out of the game ).

Another thing to consider is that, even if a lower speed is the right direction in general, the pump+VFD will accumulate runtime more quickly and may require more frequent maintenance and/or replacement. E.g. minimizing flow so pumped outlet flow = inlet flow (constant level) might minimize kWh per cubic metre but not $ per cubic metre. That said, this might be a second-order effect.

Optimization is a rarely a trivial exercise.

 

[Peter Nachtwey]

Optimization is a rarely a trivial exercise.

Optimization/minimization is not that hard but it isn't something that is done with a PLC.

Python has some good libraries for this in their scipy package. There are many similar packages/libraries for most computer languages that would be used in engineering or scientific applications.

 

[lfe]

What I want to do is empty the tank as usual, but controlling the speed of the pump in such a way that it pumps all the water out while using the least amount of energy.

That is achieved by choosing the appropriate pump for the required flow and pressure and with a good design of the pipes.

The VFD can do little or nothing to improve efficiency

 

[Saturn_Europa]

That is achieved by choosing the appropriate pump for the required flow and pressure and with a good design of the pipes.

The VFD can do little or nothing to improve efficiency

agreed

 

[Tom Jenkins]

That is achieved by choosing the appropriate pump for the required flow and pressure and with a good design of the pipes.

The VFD can do little or nothing to improve efficiency

This is mostly correct, but ...

The important parameter isn't efficiency, it is power consumption. As you point out the efficiency at a given operating point is fixed by pump design. The design of the piping determines the system curve. An oversized pump will, as you say, probably waste power.

There is potentially one way to minimize power. Frictional losses in the piping increase as the square of the flow rate. By reducing pump speed to the minimum speed that will produce the required head you will minimize friction losses and operate at the lowest possible power draw. This will be the minimum flow and head and power available. If the head is mostly static that won't matter much.