Posted by Solar Bus Driver on May 24, 2007, 1:17 am
hello,
i'm putting together a system and doing some load testing. I got my hands
on one of these handy Kill-a-watt meters. It offers readings on both watts
(defined as "active power") and VA (defined as "apparent" power, Vrms X
Arms).
I figured these numbers would be fairly close but they are not. For
example, on my "tube" computer monitor, I get 72W and 116VA. On a LCD
monitor I get 20W and 36VA. I'm getting 123 Volts.
The main question for me (besides why are they so different) is which
reading is better for designing a solar system. If they were only 10% off
from each other I wouldn't worry about it because I have a fudge factor in
my equations. But these are significant differences which would affect the
design of the system, including array size, battery size, and possibly even
inverter size.
So if anyone can tell me a bit about this I'd appreciate it. I should know,
as I've been designing systems for nearly 20 years, but I've always just
looked on the back of the appliance or looked up average figures for
appliances until now, since I got this kill-a-watt meter.
Thanks
Gary
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THE SOLAR BUS
a solar energy education project on wheels
http://solarbus.org
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Posted by nicksanspam on May 24, 2007, 6:36 am
>i'm putting together a system and doing some load testing. I got my hands
>on one of these handy Kill-a-watt meters. It offers readings on both watts
>(defined as "active power") and VA (defined as "apparent" power, Vrms X Arms).
VA^2 = W^2 + VAR^2, the latter being "reactive power."
The power factor is W/VA, which looks wrong, thinking of W = VA.
>... which reading is better for designing a solar system.
I'd say watts for battery sizing and VA for inverter sizing.
For example, your "tube" computer monitor with 72W and 116VA might
use 72 watts (eg 12 V at 6 A) of battery power and require a 116 VA
inverter or generator (at 100% efficiency.)
With some loads we can correct the power factor to one, with a reactance
Xc = V^2PF/(Psqrt(1-PF^2)). For instance, a 120V 100W motor with a measured
PF = 0.25 needs Xc = 120^2x0.25/(100sqrt(1-0.25^2)) = 37.2 ohms reactive
= 1/(2PifC) = 1/(377C) at 60 Hz, eg a parallel cap C = 1/(377x37.2) = 71
microfarads (a lot)... 120/37.2 = 3.2 amps would flow through the cap,
so the VA rating needs to be 120x3.2 = 387 VAR.
Nick
Posted by Blue Cat on May 24, 2007, 10:23 am
Watts and volt-amps are identical for dc power. For ac power you have both
real power (watts) and volt-amps (reactive power). Reactive power appears in
ac systems from two types of loads: Reactive loads such as motors and
capacitors, and non-sinusoidal loads, such as switching power supplies and
light dimmers. If you are using an inverter to supply ac power from a solar
pv system, you must calculate in volt-amps.
Posted by nicksanspam on May 24, 2007, 4:08 pm
>... For ac power you have both real power (watts) and volt-amps
>(reactive power).
People seem to measure "apparent power" in VA and "reactive power" in VAR.
>... If you are using an inverter to supply ac power from a solar pv system,
>you must calculate in volt-amps.
I'd say that's true for inverter capacity but false for battery capacity,
measured in watt-hours or DC volt-amp hours.
Nick
Posted by Solar Flaire on May 24, 2007, 7:41 pm
Basic physics Nick. Battery power may be stated in watts but it takes
VA to make VA. The VA energy doesn't magically come from thin air. It
has to be produced by the VA of the battery.
You know VA is volts times amperes from the battery.
to the OP
---------
VA is the electrical level required to produce the power needed by the
load (watts). If they are different the load is wasting it (power
factor). In DC batteries talk the watts and the VA are the same thing
but it takes VA to make VA not the watts the load takes.
The important factor for a generator source of electricity is the VA
measurements. The problem is that VA do not add up correctly but need
to be converted back to their basic components (watts and VARS =
Cartesian co-ordinates) added up and then converted back, to be
accurate.
Your Kill-a-watt meter does not display VARS (reactive power). You
will need the watts and the power factor of each load.
1) add up all the watts readings.
2) calculate all the loads to their reactive components (Vars)
- Take the VA readings and multiply them times the tan of the arccos
of the power factor tan(arccos(pf))
or
- find the VARS (rVA) by taking the squareroot of (VA^2 - watts^2)
3) add up all the VARS
4) Calculate your total VA usage by the squareroot (total watts
squared +total VARS squared)
This is the requirements of your inverter to supply these loads. Most
people add up their power (watts) and allow a bunch.
Remember: generators sources generate VA to supply the watts of the
loads. The difference is waste.
>>... For ac power you have both real power (watts) and volt-amps
>>(reactive power).
> People seem to measure "apparent power" in VA and "reactive power"
> in VAR.
>>... If you are using an inverter to supply ac power from a solar pv
>>system,
>>you must calculate in volt-amps.
> I'd say that's true for inverter capacity but false for battery
> capacity,
> measured in watt-hours or DC volt-amp hours.
> Nick
>
>on one of these handy Kill-a-watt meters. It offers readings on both watts
>(defined as "active power") and VA (defined as "apparent" power, Vrms X Arms).