Wattage consumption

I think the missing factor is that the amount of water pumped is much less when you add head height. Therefore, much less work is being done then if there was no head.

 

You already answered your own statement. Wattage consumed is by water moved. More head, or resistance to flow, less flow, less water moved, less watts consumed. Throttling a pump with a valve and adding head is exactly the same thing. Lowered power consumption is perfectly measurable in each case with and without.

And as was stated to the OP.... the powerhead is designed to move water which is what it is doing. You are not harming it in any way. If you have enough pressure for the head you have then it is working fine.

 

Head is a static thing, even if the water is moving. It does take work to move water up... that is the force over distance, but it does not consume watts to overcome head. At cutoff hight, no water will flow and not much power will be consumed.

Now there are losses.... power factor, heat in windings, fluid friction... so those are consuming power, but you can't get around those. Power consumption will be at it's lowest at cutoff height with no flow.

 

Thanks for the replies.

Makes sense, thank you.

Would be nice to see those results. Just out of curiosity.

 

that would be fun blackraven! if you think of it send me a message and let me know if you make a thread or post on this thread? I'd like to see the results!

 

I think the other side of the missing coin is WHY is the pump moving less water at the higher head? Because it reaches it's max power (watts used) and can't draw more power to lift water faster (volume) at that head (resistance). Therefore at the higher head... when the volume drops... the pump is drawing electricity at max to try to move as much water as possible against that extra resistance/head.

I'll be looking for the test results too! LOL
twas a good question!

 

I'm pretty sure that's not the way it works, greybeard. What I think happens is that the magnetic push of the motor pushes at the same strength regardless as to head, but the amount of flow is determined by the resistance. Any extra energy that gets taken due to resistance in the pipes causes more heat in the water, and no energy savings.

Like I said though, I'm gonna test this out sometime soon, since it keeps coming up. I need to find a bigger pump than the one I was thinking of using, though, as I think it's gonna be too small of a draw to get a reliable reading.

 

So what you are telling me is this:

I have a reservoir attached to a centrifugal pump, attached to 200 feet of vertical piping. I turn on my pump and water is drawn from the reservoir and into the piping to the maximum head height of the pump, let us say in this example 175 feet.

Once the water is at this height no more energy is consumed by the pump therefore no work is being done on the water. Since the water is held in place for free (no energy) I can reasonably assume that if I turn off the pump the water column in the pipe will stay at 175 feet and not run back into the reservoir?

 

Gravity - it's always doing work. So until that stops you always have to account for it's effects. Once the maximum head height is reached - the effect of gravity is equal to the full strength of the pump moving the water and you are at a standstill.

 

i figured it out not by consumption but by the increase of impeller speed with head height and some manufactures will show an increase in rpm with head height to me increased rpm tells me less load = less amp draw

i can listen to iwaki and as i slow the flow down the rpm increases due to cavatation of the pump it bigins to recerculate the water within the impeller