DIYbanter - Are you poor?

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On Sep 25, 11:37*am, wolfgang wrote:

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! *****, one can buy electric model motors running at tens
of thousands of RPM with 90 % efficiency, dirt cheap.

Wolfgang

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

Dan

On Sep 25, 2:01*pm, "Michael A. Terrell"
wrote:
wolfgang wrote:

On Sep 17, 8:37 pm, "Michael A. Terrell"
wrote:
wolfgang wrote:

On Sep 16, 2:00 pm, " wrote:
On Sep 15, 8:17 pm, wolfgang wrote:

To run the pump is about $400 for 6 months. *I've tried to put the
pump on a timer but during hot weather the water doesn't like it and
turns green. *I'm afraid to cycle the pump more than once every three
hours or so as this shortens the starting switch life expectancy.

Wolfgang

A *thought. *Could you connect a lawn sprinkler to the output of the
pump and have its output come down into the pool. *The water droplets
would adsorb oxygen in the air and the agitated pool surface would
also add oxygen to the pool.

My understanding is that releasing air at the bottom of the pool does
not work so well. *Too much nitrogen gets adsorbed by the water..

* * * * * * * * * * * * * * * * * * * * * * * * * * * * * *Dan

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.

* *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.

Pool pump efficiency stinks!

* *Then fill in the pool and do without.

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.

* *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.

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.

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

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! *****, one can buy electric model motors running at tens
of thousands of RPM with 90 % efficiency, dirt cheap.

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

--
You can't have a sense of humor, if you have no sense.

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 = .638. I conclude that the
motor is producing .638 x 3/4 HP = .48 HP.

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

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

Conversion efficiency = 358 / 1012 x 100% = 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

On Sep 25, 6:30*pm, wolfgang wrote:

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 = .638. *I conclude that the
motor is producing .638 x 3/4 HP = .48 HP.

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

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

Conversion efficiency = 358 / 1012 x 100% = 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

On Sep 26, 3:27*pm, Winston wrote:

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

--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

On Sep 26, 9:23*pm, Winston wrote:

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

On Sep 26, 9:33*pm, " wrote:
On Sep 26, 9:23*pm, Winston wrote:

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

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 = 1.125 HP drawing 13.8 name plate amps.

Wolfgang

wolfgang wrote:
On Sep 26, 9:33 pm, wrote:
On Sep 26, 9:23 pm, wrote:

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

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 = 1.125 HP drawing 13.8 name plate amps.

http://www.engineeringtoolbox.com/se...tor-d_735.html

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