I just bought an electical  water pump  (sump pump) at Home Depot
which
is rated at 1/2 HP (350Watts).

So I was curious as how much current the pump was drawing and my
electrician
put a digital amp meter (the kind you clip around of of the 110V
wires) and
to our surprise we were getting 11 AMPS, 

Then we tried another pump (similar from Sears) rated at 1/6HP (120W)
and again this  was drawing 4 AMP..

Obvisously we are off by a factor of 3-4 in what kind of amperage
350watts and 120 Watts should use.

Even when we tried to pump NO water and various combination of
hoses...the
amps were similar.

For example I am getting 5 Gallon/minute with 20foot of head which
should use
less than 50Watts of energy!



Now from my electrical engineering background I seem to remember that
motors
and inductances are bad in that the current and voltage are quite out
of
phase (could be almost 90degrees)?

Since I was drawing 11 Amps on the meter does that mean I am really
using 1200W
as far as the electrical company is concerned?

Seems quite wasteful and is there something that can be done? I seem
to also
recall that putting a capacitor in series(?)  helps

I have lots of other pumps that my plumber installed (radiant heat)
and I seem
to recall that the amperage was quite consistant with the rated Watts
of the pumps in question (I was using the same kind of meter)

---

Request for Question Clarification by sublime1-ga on 29 Aug 2003 22:23 PDT

francois777...

I'm not personally familiar with the internal construction of
sump pumps, but, electronically, a simple scenario presents
itself as a possibility. Could the 120VAC be stepped-down by
a transformer to either 30VAC, which is actually running the
AC motor, or could it be stepped down and then converted to
30 Volts DC, and used to run a DC motor?

This would explain the amperage as being consistent with
the wattage rating of the unit(s).

sublime1-ga

Hello Francois

>Since I was drawing 11 Amps on the meter does that mean I am really
>using 1200W as far as the electrical company is concerned?

No. It does not mean that. Electric company is charging you for
true RMS power, not for the product of V * A.


  The product of V * A differs from the power consumption for 
several reasons. PF - the power factor,  which measures how
much the voltage is 'out of phase' relative to current is one of
them. Typical PF is .5 to .6 with inductive load such as electric
motor.    This is the factor you mentioned. Adding of capacitor 
(not necessarily in series) to orrect PF is sometime done,
   http://www.kele.com/tech/monitor/power/trefpm8.html
 but please, read to the end before you decide to do that.    

There are  other reasons:   TDH and fluctuation of the
load. TDH measures effects due to the fact that the current 
and voltage are not perfect sine wave, assumed in RMS calculation.
Fans, pumps, cooling tower.. represent nonlinear load which lead
to high TDH. It is possible to to estimate power from the current
reading, but it is involved and typically requires calibration:
    
This report compare three methods for assessing power
 
  true RMS power measurement,
 using current as a proxy for staged loads and
 using current as a proxy for variable loads:

  http://www.pge.com/003_save_energy/003c_edu_train/pec/toolbox/tll/app_notes/proxy_for_power.shtml

   It is possible 'to condition the line' to improve the PF.
    Here is an application note  describing improvement of the  power
factor "
     www.lambda-emi.com/pdfs/application%20notes/ 93008502rA.pdf


 Here is a power supply which will correct PF
     www.astrodyne.com/astro/pdf/UPF300.pdf    

 However, you are charged for power and not for V*A and there 
is little reason to go into the extra expense in a typical situation 
of household, even when pond or sump pump is used. The saving is in
less
heat dissipated in the supply line. You do not save 33% of power, even
if you would cut current by 33%. The analogy with mechanical pendulum
can help: If there is an exessive vibration in a system, there will be
more loss of the energy due to damping, put vibrational energy of the
pendulum is not immediately lost. The 'extra current' is like extra 
motion due to the vibration.  

In the industrial environment it may be relevant and here is
detailed explanation and discussion of remedies:
     http://www.kele.com/tech/monitor/power/trefpm8.html

SEARCH TERMS

 RMS power , true RMS power
 Power factor
 watt meter, power meter