First off, I'm an IT guy and not an engineer. I've been tasked with figuring out and monitoring the efficiency of some pumps. Thing is, I don't have a lot of data to work with. I don't know the total head for any of the pumps. One or maybe two have a voltmeter. Heck, I don't even know the size of the wells they are pumping from. About the only data I know I have for certain is:

  • Pump Start/Stop times
  • Pump Start/Stop set points
  • Water level

So I need help with the logic here, not the math.

I know I'm not going to be able to calculate the hydraulic efficiency because I don't know the model of the pumps being used let alone things like impeller sizes or shaft rotational speeds. But I think I can calculate a quasi volumetric efficiency. For that, I only need to know how much stuff gets moved in a certain period of time. Then I compare today's flow rate to yesterday's flow rate to determine if it is more or less efficient than it was yesterday/last week/last month under the assumption that the only thing that is changing is wear and tear on the pump.

Since the pumps don't move, any efficiency lost due to working against pressure/gravity/head should be roughly the same as they were yesterday so I can effectively ignore that, right? And the size of the well isn't changing so I can just as easily say it is 1. Unit doesn't really matter since I am comparing it to yesterday's calculation. It could be 1 cubic meter, 1 gallon, 1 swimming pool. The unit of measure (UOM) is only useful for comparing it to other calculations in other systems. Here I'm just comparing this pump's flow rate with its flow rate yesterday, so the UOM could be 1 Tim for all we care.

So my thought is to calculate the fill time of the well by looking at the elapsed time between the pump stopping and starting, and calculate the draw down time of the pump by looking at the time that lapsed between the pump starting and stopping. I know I need the fill time because the well will be filling while the pump is pumping. That means it not only pumped out "1 Tim" of water, it also pumped out a little more. This will give me the "Tim's per minute". If I compare today's number with yesterday's number, that should tell me if a pump is starting to fail, no? Even if the pump starts to fail mechanically, I would see this in my calculation because it would take longer to pump the same amount as it pumped yesterday.

Does this seem like relatively solid logic? Is there a way I can get a better efficiency calculation with the very limited information I have available? (The pumps are at various farms and travelling to them is not really feasible, nor is there any kind of budget to add meters or additional equipment.)


1 Answer 1


Efficiency = work out divided by energy in. What is anyone doing to actually do with thi number? It seems to me a more useful number is KW/flow rate or something like that. I worked in a plant with a ton of pumps (and wells) and I can't remember anyone caring how efficient any of them were.

Anyway, assuming that efficiency is what you're after, first you need the work out. You don't know the water level at the bottom of the well, so you will not be able to figure out the work out. Water tables change, so the work required to pump will be different day to day, vertianly month to month. The "Work out" is how much mass is moved to what height over time. Power in is easy with a clamp-on ammeter, though for a motor you really need to know the power factor (but let's leave this out for now, I doubt a power factor analyzer is in the budget).

Well people can tell you what the water level is in the well. They figure this out when they do a draw-down test. If you know the aquifer you're drawing from, there might also be local resources who know. This might be useful for right now, but the water table will change over time. So, if you're looking for efficiency changes over time (again, why?) you will need to be able to determine the water level in the well. if you can't figure out the water level in the well, determining the efficiency will be impossible.

So, you can figure out the work done in a minute (volumetric flow times height times mass/volume times g), and get the watt-hours used in a minute from the ammeter. Get the units the same and you have efficiency. Yes, the units matter. They have to be the same to give you efficiency. Swimming pools per fortnight won't be useful to you. Someone needs to measure the flow and tell you what it is, and someone needs to measure the current and tell you what it is as well. There is literally no way to do this without being there.

Now, if all this is to figure out if a pump is about to fail I think you will be disappointed. You might get a little more current draw from bearings going bad, but most of the energy is likely lifting the water, and, as discussed earlier, that will be governed by the water level.

The "measure the time" method you mention has nothing to do with efficiency. It might be useful for a flow rate chart, but unless you live where water tables don't change that won't tell you much. I'm not clear what you're filling up, but expect a pump to run longer when you're pulling water out of whatever tank you're filling.

If you want to predict failures, put in vibration sensors the next time you take the pumps out. Anything else will be guesswork.

  • $\begingroup$ Maybe efficiency isn't the word/measurement I'm looking for. The owner wants to know when a pump's performance has degraded. I believe the intent is to proactively prepare for a pump's failure rather than reacting. $\endgroup$
    – Tim
    Sep 21, 2022 at 12:53
  • $\begingroup$ if all you can get is run time then that's all you can get, but I don't tihnk you'll be happy with the results. $\endgroup$
    – Tiger Guy
    Sep 21, 2022 at 14:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.