Are you poor?

=A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0Dan

Thanks for all those suggestions, they are greatly appreciated!

Currently I have a floating chlorinator that I fill with 3" dia. "pucks" every 10 days or so. For July and August I also add an algicide. I used to have an automatic chlorinator, also with pucks, but it was very difficult to keep the chlorine level properly adjusted.

Spraying part of the returned pool water through the air to absorb oxygen can be done with a small pump while the main pump is shut off. Running the main pump 4 to 6 hours per day sounds REALLY appealing!

Ozonation is another item to look into. I seem to recall that this method is also used for drinking water treatment in some areas?

All good stuff. Thanks again.

Wolfgang

Currently I have a floating chlorinator that I fill with 3" dia. "pucks" every 10 days or so. For July and August I also add an algicide. I used to have an automatic chlorinator, also with pucks, but it was very difficult to keep the chlorine level properly adjusted.

Wolfgang

Check your stabilizer with a cyanuric acid tester, if you haven't already. Stabilizer (cyanuric acid) keeps chlorine from offgassing from the surface. About $20, and you need to test it every six to twelve months. Very simple test that even a girl could do. If that is your problem, you will have an instant reduction in the amount of chlorine you use, and stabilization in levels. Best damn thing I ever found for pools.

Steve

(...)

Keep us posted, willya?

--Winston

I did that for a neighbor's fish pond to control the algae. It worked great, till someone talked her into some plants that were supposed to do the same thing with less water loss. They ripped out what I'd built and made their changes. It was completely green again in a couple weeks. I had built a small waterfall out of rocks, about 24" high and a foot wide.

=A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0Dan

Pool pump efficiency stinks!

Yesterday I measured the current draw of my 3/4 HP pool pump. Meter read 8.8 amps, and the name plate states 13.8 amps full load current. Doing some calcs I determined that at that operating point of the motor the electrical efficiency is 35%! Something clearly needs to be done. (knowledgable electric motor guys may wish to chime in on this, please). I think that maybe a 1/2 HP pump would suffice? The suction line is 1 1/2" dia. with 3 elbows and about 40 ft in length.

I was perusing a pool maintenance website and the moderator recommended that the pump be run at least 6 hrs per day.

Since it is too late in the season to do much with the pool (I close and winterize it on the Labour Day weekend) I have put the timer back in the motor circuit and set it such that it cycles 4 hours on and 4 hours off; thus cutting the power consumption by almost 50 %.

For next year I will do the same while investigating better electric motor efficiencies and how to achieve this. Perhaps by using algecide earlier and more regularly in the season the algae can be avoided with cycled pump times.

I'd be interested in learning much more about motor efficiencies and how to achieve this; any recommended readings are greatly appreciated! Shit, one can buy electric model motors running at tens of thousands of RPM with 90 % efficiency, dirt cheap.

Wolfgang

Then fill in the pool and do without.

That simply means that the pump isn't fully loaded. The load depends on the 'head', or how high the pump has to lift the water. It also allows the pump to start faster and with less mechanical strain.

A smaller pump would have to run at full load, and would wear out a lot faster.

Try one and see how long it will last on a pump. Have a fire extinguisher handy.

You can not just measure the current and calculate the motor efficiency. You need to know the power factor.

Dan

=A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0Dan

Michael,

The point I was trying to make is this, that if you can have an inexpensive electric motor of 1/2 HP, that fits into the palm of your hand and turns at tens of thousands of RPM with an electrical efficiency of 90%, then it ought to be possible to produce a 3/4 HP motor the size of a breadbox and running at 3450 RPM at say 10x or 20x that price.

Dan,

Here is how I guesstimated the motor efficiency at that operating point:

Name plate rating: 3/4 HP, 13.8 amps full load current Measured motor current is 8.8 amps.

Now I do know about power factor, but I thought that the following calcs would not be out too far.

8.8 running amps / 13.8 full load amps =3D .638. I conclude that the motor is producing .638 x 3/4 HP =3D .48 HP.

Power draw =3D 8.8 amps x 115 VAC =3D 1012 Watts.

.48 HP x 746 =3D 358 Watts mechanical power output.

Conversion efficiency =3D 358 / 1012 x 100% =3D 35.4%.

I realize that I am making a number of assumptions here regarding AC power and power factor, which will of course introduce errors. But how large is that error in my assumptions for the above?

Thanks for any input/correction.

Wolfgang

Jim Pentagrid may comment on this. If he does pay attention as he knows what he is saying.

I looked at the W.W. Grainger catalog and found that most 120 volt,

3450 rpm, 3/4 hp motors have a full load current rating at just under 10 amps. And I know that the government has mandated that electric motors have to be more efficient than they used to be. So replacing the motor with a newer one might be worthwhile. But probably only if you can get a deal on buying a new motor. You may not be able to justify buying a new motor as your usage is fairly light.

I do not know a better way to calculate motor efficiency, but don't have much faith in your calculations being any where near accurate.

Dan

I suspect that 'step zero' would be to bring the PF up to 1.0 (Correction caps) before taking any readings.

Once we remove the effect of Power Factor, (so that Watts=VAR), for a given torque, we can take some readings. Till then, we will be reading (motor current + circulating current) and getting an incorrect number, especially under no-load conditions.

Note that induction motors are very inductively reactive when unloaded, so most the current we see is 'recirculating' and does nothing to turn the shaft.

--Winston

It is not worth the effort to correct the power factor to 1.0 and pretty much impossible to do without taking readings. Consider that if you have a real current of 5.0 amps and an imaginary current of 5.0 amps. The measured current would be 7 amps. ( The square root of 5^2

• 5^2 ). And the power factor would be 5/7 or .7 . If you correct to .9 PF. then the real current is still 5.0 amps and the imaginary current is 2.4 amps. So you have cut the imaginary current about in half and reduced the total current to 5.55.... A reduction in total current of about 1..4 amps.

To correct to 1.0 power factor you have to reduce the imaginary current by 2.4 amps and that only reduces the total current by 0.55 amps.

Dan

I meant that the power readings to calculate efficiency could only be taken after the power factor readings to correct the input current figure. I mentioned power factor correction because it is a way to lessen the stress on every component in the system and correct the OP's efficiency numbers.

If you can derive the uncorrected power factor, you can correct the current consumption reading to subtract the effect of the imaginary current. I suspect one would have to take a Power Factor reading at the torque setting used because of PF's inverse dependency on motor load. Armed with those figures you could calculate efficiency.

--Winston

I have always assumed, possibly incorrectly , that the imaginary current did not change from the unloaded value when the motor was connected to the load. With a load the real current goes up and so the power factor changes. I really ought to buy Jim Pentagrids book.

Dan

=A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 Dan

Upon reflection I concluded that the full-load current draw on my pool pump is too high for a 3/4 HP 115 VAC motor....

On my small shop air compressor I have a 3/4 HP motor with a full-load current of 10 amps.

I also have another 3/4 HP pump motor (it unfortunately doesn't fit the hayward pump) that has a name plate current of 9 amps.

On that basis the measured current draw of 8.8 amps doesn't appear too bad and the motor appears to develop almost 3/4 HP.

It turns out that the service factor on the pool pump is 1.5, and I surmise that the full load current is based on 3/4 HP x 1.5 service factor =3D 1.125 HP drawing 13.8 name plate amps.

Wolfgang

I think 'service factor' indicates that your motor can be safely used at up to (0.75 * 1.5) 1.125 HP without overload or damage. Think of it as 'headroom', not an indication of the power converted when it is loaded at 3/4 HP.

Your 8.8 A reading includes the effect of the circulating 'imaginary' current plus the current that is actually converted into mechanical motion. It is pessimistic.

--Winston

That sounds reasonable. I see citations on the web for relatively large changes in power factor (say from 0.2 unloaded to 0.8 loaded) and relatively small changes in real current (doubling from no load to full load) for inductive motors.

--Winston