The KVAR unit
A KVAR unit can save
you 8% to 15% of the typical
bill. It will cost $350 for the
typical 1400 sq. ft. house with A/C, so,
if your power bill runs $325.00 Jun. - Sep. and about $200 a month for
the rest of
the
year, the 1st years savings alone would pay for the unit.
The KVAR unit is made with sealed capacitors and safety
bleed down circuits included to discharge any residual high voltages
and an 'on' indicator.
These
capacitors are used to put capacitative loads on AC
circuits to correct typical power factors
toward a '1.0' power factor
from the typical
customer's
.85 kvar power
factor. (.85 is a utility industry
accepted typical value) If a you do NOT have such a capacitor,
you will
always use additional wattage to overcome this .85 power factor. This
is wattage you have to pay the
utility for, but it does no real
work
for you. In the typical .85 power
factor example it will be 15% of the $ costs
shown on your bill.
** If you are
in the world of Pacific
Gas & Electric's surcharges for
high use,
it can exceed the 15% of the $ cost by as much
as an
additional 25% for a total
of 40
% of the bill.
OUCH!!
This is what got me
started investigating power factor correction, as my 2002
PG&E bills totaled $4,400
with 4 months over $750. I was a design engineer working as a
systems
analyst at the time, and had done power factor correction work for
mainframe computer power supplies on mainframe computers for
NCR, so I was aware of the concept of power factor correction, unlike
most people, even electricians and architects. |
What causes
this .85 power factor? All
inductive
devices
create a lagging power factor load ( +0.25)
because of when wire is wound
around a core, current changes will induce a magnetic field,
but not all of
your power loads are inductive loads, so yours will vary between 1.00
and .75 usually. Pure resistive loads have a power factor
of 1.00, so they pull the combined power factors toward the ideal 1.00.
However, capacitative
loads (like the KVAR unit) create a offsetting power factor load ( -0.25),
so they can be used to cancel out equalalent
inductive loads.
I
do recommend replacing as many incandescent bulbs as practical with
their lumen equivalent fluorescent bulbs. Although these
equivalent fluorescent bulbs do create an inductive load,
this load is very small, typically about 5 watts for each bulb and
would be easily compensated for with this KVAR unit, thereby reducing
the effective wattage of the 60 W bulbs replaced to only 18 W.
| This
means if you replaced 10 of your 60 W. bulbs with equivalent
fluorescent bulbs and if they were turned on 1.5 hours a day, you would
save 70% of the annual power cost. This is either $206.96 @ $0.09 / kWh
(national average) or for suffering Pacific Gas & Electric
customers
(like me) $347.00 @ $0.1509 / kWh.
(my 2002 average rate) |
|
What are some inductive devices:
1. Any motor in air
conditioners, refrigerators,
freezers, washers,
dryers, any pumps, fans transformers, fluorescent light
ballast's, power
supplies, fans and microwave ovens.
What are NOT
an inductive devices: (These do not cause
power factor losses but may quite inefficient)
1. Regular incandescent
light bulbs, Stove burners & regular ovens, portable heaters,
electric water heaters, baseboard heaters, electric furnaces and
electric clothes & hair dryer heating elements. Sorry, I
can't help you here! I did put my electric water heaters (2 - 50
gal.) on timers so they don't heat during the peak load period.