🚀 go-pugleaf

In article <3272@ihlpm.ATT.COM> wrv@ihlpm.ATT.COM (Vogel) writes:
>This really belongs in sci.electricity, but since there isn't one, here goes:
>
>I have a swimming pool, which has a filter pump that runs all summer.
>Over the weekend I had a conversation with the local pool supply
>company service person.  We got on the subject of pumps, and he asked
>me if my pump was ok, and how it was connected.  I told him it was connected
>to 110V (by previous owner), but it could also run at 220.  He said I
>should convert it, that this would give a great savings in the power bill.
>
>Well, I went home wondering how this could be true.  I examined the
>pump motor, its rated:
>
>	9  amps @ 220v
>	18 amps @ 110v
>
>So, in my mild knowledge of electronics and Ohm's law, Power = E x I
>1980 watts in both cases.
>
>So somebody please tell me, is the pool guy whacko, or am I missing
>something?
>
>	-Bill [waiting for summer so I can swim......] Vogel, ihlpm!wrv

Like, the guy is out to lunch.  First of all, the voltages you get at the
meter are 120 and 240 by power company tariff.

Second, we're not talking about motor efficiency, but circuit efficiency --
the motor doesn't run any differently at the 240VAC connection.

Third, in an AC circuit with reactance (the motor inductance) the voltage
and current are out of phase and cannot be multiplied to get watts -- you
get volt-amperes.  To get watts, you must multiply by the power factor.

However, let us assume that your filter pump requires a wire run of 100 feet
from the breaker box to the pump motor, for a total conductor length of 200
feet.  Assume also that the wire gauge used is #10AWG, which has an
approximate unit resistance of 1 Ohm/1000ft.  Thus, the circuit resistance
will be 0.2 Ohm. (It's possible that #12 wire was used, at 1.26 Ohm/1000ft,
but the following calculations will not be significantly changed.) Finally,
let's assume your energy costs are 9.5 cents per KWHr, about what they are
here, that you leave the filter on continuously for 100 days, and that the
motor is actually running at full load (this depends on the water head).

Now, for the high-voltage connection, the wire (I^2R) losses are:

	9A * 9A * 0.2 Ohm = 16.2 watts.

For 100 days at 24Hrs/day, your energy cost is:

	16.2W * 24Hr/day * 100 Day / 1000W per KWhr * $0.095/Kwhr = $3.69
	
Doing the same for the present 18-Amp load brings it up to four times that,
or a whopping $14.77 for the season, so you will save $11.08 over the season.
Play with the numbers, if you like, and insert your own.

This is the kind of superficial analysis that the partially educated (the
pool folks, not you) do, if they do it at all.  Being a well-trained (I
think) electrical engineer, however, I will give you the benefit of a true
engineering analysis:

What's your time worth?  Got the number?  Let's proceed...

You will have to purchase and install a two-pole circuit breaker (if you can
free up two circuits) in the breaker box.  If not, you will need an
auxiliary box.  (Fuse boxes are even worse!)  You will then have to replace
the switch at the pump (you DO have one there for safety, don't you?) with a
240-Volt unit.  Now, assuming that you do all of the labor yourself, the job
will cost around $50.00 plus your labor plus a building permit plus any fee
for a certified inspector, if required by your municipality.  When you
figure out the economic payback period, remember the time-value of money
before you attempt simple division.  Today's dollars are not tomorrow's.

The above was rather cute, I realize, but the analysis and numbers are
correct.

NOW, LET'S GET REALLY SERIOUS...

Finally, the other issues -- Do you really want 240 volts running around
your pumphouse and pool?  Have you provided a ground-fault interrupter on
ALL of the outdoor circuits?

SAFETY FIRST!  PLEASE, IF YOU HAVE ANY DOUBTS, DO NOT ATTEMPT THE CONVERSION
YOURSELF.  MILLIAMPS KILL!


Rich
-- 
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Ideas have consequences.                                    RSD@sei.cmu.edu
Richard Weaver
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