On May 28, 5:38*pm, "
wrote:
On Sat, 28 May 2011 15:21:47 -0500, Jim Yanik wrote:
"RBM" wrote :

"Mark" wrote in message
....
On May 27, 8:47 pm, Home Guy wrote:
RBM used improper usenet message composition style by full-quoting:

3) There are (on average) 30.4 days in a month, and therefore 730
hours in a month. If I take the above wattage calculation (120 x
total_current) and multiply it by 730, then divide by 1000, I
should get a quantity energy measurement (KWh) that should match
(or approximate) my bill from the local utility - assuming that
the load in use at the time of the readings are representative of
daily or continuous use. If this method of obtaining a
representative monthly KWh measurement is not correct (or needs
more refinement) then please state what, why or how.

**You're not going to be the least bit accurate trying to calculate
that way.

I want to establish several use-case situations, primarily a
"worst-case" KWh monthly usage by assuming that all the devices that are
normally on during a week-day 9-am to 5-pm work day and turned off at
all other times are instead left on continuously 24/7.

I can also get current readings for other use-case situations (evenings
and week-ends) that should give me a more closer-to-reality current
reading and factor in their time-of use over the course of a month.

This is a small office - not a home. There are fewer variable involved.

There is no reason that you shouldn't be able to look at the
electric meter.

The meter is in a locked cabinet. The only time I get to see it is when
the meter-reader guy comes around once a month to read it.

I suspect the meter is in a locked cabinet to prevent tampering / bypass
(the meter is inside the utility / furnace room of the building and is
not accessible from the outside).

And besides, having the ability to lay my eyes on the meter won't tell
me anything about the accuracy of the meter, or the real-time current
consumption.

All I want to know is - should the current readings from a clamp-on
meter (when extrapolated across the 730 hours of a typical month) jive
with the accumulated KWh reading as measured by a typical billing
meter? Do I have to do any "special" math to the wattage I calculate
with the meter to arrive at what the billing meter is measuring?

yes you are correct you can get a rough reading this way..
you need to measure and sum only the 3 black cables, do not include
the white striped cable in the sum.

Add up the 3 currents and multiply by 120 and this is your VAs and the
actual Watts will be equal to or less then the VAs.

Multiply by hours and you have Watt Hours.

Divide by 1000 and you have kWh which is how you are billed.

Mark

This is a commercial service and metering equipment. How is he going to
guestimate demand?

since the white return wire returns the current from all 3 phases,wouldn't
that give you the sum of all the currents?

Only the *difference* of the three phases is returned in the neutral. *With a
balanced load there will be zero current in the neutral.


YOU ARE ALL SO FULL OF HOOEY.....THE ABOVE STATEMENT IS INCORRECT.
ON SAID NEUTRAL THERE WILL BE A POTENTIAL TO GROUND EQUAL TO THE
LOAD....
KRW YOU ARE ONE DUMBASS ENDANGERING MOTHER****ER.

I HAVE COME TO TNE CONCLUSION THAT THOUGH SOME OF YOU ARE VERSED AND
EDUCATED YOUR OBSERVATIONS AND KNOWLEDGE IS WANDERING...RENDERING
THESE GROUPS USELESS, MISGUIDING AND NOT WORTH THE EFFORT PUT INTO
THEM...

PATECUM
TGITM