Good evening,

I have a 35W water pump at 220 VAC, I would like to reduce its consumption, I thought of two methods:

1- use a transformer to reduce the voltage from 220V to 110V

2- use two power resistors in series of 470 ohm each and which resist a thermal dissipation of 20w each, this method will cause a voltage drop of 150V (the goal is to make the voltage drop as much as possible), so it will remain 70V for the pump.

Can the pump work if I use one of these two methods or not?

Thanks for your help.

 

35W and you need to reduce consumption??
Throttle down the output.!

 

35W and you need to reduce consumption??
Throttle down the output.!

Yes, because I use it to cool a solar panel, and thus improve its efficiency, so this cooling will allow me to gain a few more watts which must not be less than the pump's consumption.

 

Your not going to use less if anything you'll use more.
You need something like PWM and current limting

 

Yes, because I use it to cool a solar panel, and thus improve its efficiency, so this cooling will allow me to gain a few more watts which must not be less than the pump's consumption.

Any reduction in RPM is going to reduce the flow, if you can reduce the dia of the output pipe and still maintain the required cooling, it will run at reduced current.

 

35W is when it's operating at maximum load.
The motors power used is basically determined by its load, so if the pump load stays the same then the power will stay the same since typical AC motors run at nearly a constant speed, independent of load .
For an AC motor, all that reducing the voltage will do is increase the current to maintain the same power (unless you reduce the power sufficiently to lower the motor speed, in which case the efficiency will likely drop.)

 

If you reduce the voltage, you are likely to stall the motor and burn it.
The current drawn by the motor is proportional to the flow of water. Put a valve at the output and close it till the power drawn reduces to the level you want. Remember that it cannot be zero.
Current = [a fixed value] + [ proportional to water flow]

 

Thank you all for your answers

Any reduction in RPM is going to reduce the flow, if you can reduce the dia of the output pipe and still maintain the required cooling, it will run at reduced current.

Yes, reducing the flow rate will not cause me any problems, because the maximum flow rate that the pump can deliver is much higher than the flow rate I would like to use.

If you reduce the voltage, you are likely to stall the motor and burn it.
The current drawn by the motor is proportional to the flow of water. Put a valve at the output and close it till the power drawn reduces to the level you want.

So, as I understand it, reducing the cross-sectional area at the pump outlet will reduce the current and thus the consumption of the pump. So the power consumed by the pump is a function of the flow it delivers. A good and simple idea, I will apply it.

Current = [a fixed value] + [ proportional to water flow]

Is there a formula or an experimental method to determine this fixed value of current?

Thank you once again

 

The easiest (and safest) way to accurately measure the power used by an AC motor is to use a power meter, such as the Kill A Watt.

 

Yes, reducing the flow rate will not cause me any problems, because the maximum flow rate that the pump can deliver is much higher than the flow rate I would like to use.
So, as I understand it, reducing the cross-sectional area at the pump outlet will reduce the current and thus the consumption of the pump. So the power consumed by the pump is a function of the flow it delivers. A good and simple idea, I will apply it.

Reducing the output discharge dia will reduce current as per #2 & #5 !

 

The easiest (and safest) way to accurately measure the power used by an AC motor is to use a power meter, such as the Kill A Watt.

Can a simple measurement of current and voltage with an ammeter and a voltmeter determine the power using this equation P= U.I ?

 

Reducing the output discharge dia will reduce current as per #2 & #5 !

OK, I understand, Thank you

 

Does the pump have a sticker so we can see it's ratings, or is the 35W @ 220V (which is 160mA) it's rating? It's a motor- current charges the magnets, and the voltage controls how reactive the magnets are. Not providing the power the pump is rated for can cause it to run badly, run hot, or not operate correctly potentially.

 

Does the pump have a sticker so we can see it's ratings, or is the 35W @ 220V (which is 160mA) it's rating? It's a motor- current charges the magnets, and the voltage controls how reactive the magnets are. Not providing the power the pump is rated for can cause it to run badly, run hot, or not operate correctly potentially.

yes on the pump sticker it says, 35W - 220V - 21 L/min and h=1.2m

 

Is there a formula or an experimental method to determine this fixed value of current?

Close the output of the pump.
Run the pump.
Measure the current.

 

So the power consumed by the pump is a function of the flow it delivers.

Yes.

 

Can a simple measurement of current and voltage with an ammeter and a voltmeter determine the power using this equation P= U.I ?

Not accurately, since it doesn't include the AC power factor of the motor (phase-shift between voltage and current).
That calculation gives the apparent power, which is a combination of real power and reactive power.
But it may be good enough for your purposes.

 

Not accurately, since it doesn't include the AC power factor of the motor (phase-shift between voltage and current).
That calculation gives the apparent power, which is a combination of real power and reactive power.
But it may be good enough for your purposes.

Not sure. As the pump flow increases, the change may be only an improvement in the PF, the value of I being the same.

 

As the pump flow increases, the change may be only an improvement in the PF, the value of I being the same.

Yes, the PF will improve since the reactive current is the magnetizing current and the real current is from the load.
But there certainly will be an increase in current with load, since the magnetizing current is not much affected by load.

 

ok
thanks again for the information.