power usage and apparent power

Re: power usage and apparent power

I went through some of the math for AC power and motors in school for engineering--it hurt my head then, and I have forgotten most of it by now... And unless you are really into the math--the details for our needs here don't really matter.

So, basically, imagine you are pulling a cart on tracks. If you stand in front of the cart (0 degrees) and pull--100% of your efforts go into moving the cart.

If you stand 90 degrees to the cart an pull--0% of your efforts go into pulling the cart forwards... So the math is:

Force@cart = force @you * Cos (off-center-angle to cart)

And the energy part of the equation, if we use pulling up hill would just include the distance pulled:

Work/Energy = Dist*Force@cart = Dist*force @you * Cos (off-center-angle to cart)

The AC electrical equation is basically the same thing except it is the Voltage and Current Sine Waves and their "angle" to each other...

Watts = Volt*Amps * Cos (angle)
Watts = Volt*Amps * PF
PF = Watts/Volt*Amps = Cos (angle)

For a resistor, PF=1, for a motor under load, the PF=0.6-1.0 (roughly, and depending on several design issues)

So:

2.5 amps
260 VA (not watts--sometimes called VAR, I think)
150 Watt

V=Volt*Amps/Amps=260 VA / 2.5 A = 96 Volts AC (sounds a bit low)

PF=150W/260VA=0.58

Angle=Cos^-1 (0.58) = 55 degrees (angle offset between voltage and current)

So, the inverter will require, from the battery:

BATTamp=(Watts/BATTvolt) * 1/Inv eff
Bamp = (150W/11.5 volts) * 1/0.85 inv eff = 15.3 amps at 11.5 volts input current

Notice (within a first level approximation), that the VA rating does not affect inverter input current because the Inverter is a Power Converter... It takes power from the battery (plus a bit for 15% losses in above example) and converts it into POWER at 120 VAC.

However, the output of the inverter will have to conform to the requirements of the load--if it is a motor, because the current and voltage are "out of phase", more current will need to be supplied by the inverter... But because this is not work (see cart above), the battery does not have to supply any more current to the inverter. So, the inverter will have to output 2.5 amps, even though, from a watt meter reading, would would appear the current should be less.

The offset between the current and the voltage can also be thought of as pushing a spring against the wall--the amount of energy from the distance you push on the spring for 1/2 the cycle, is given back to you and the 2nd half of the cycle. So, overall you still moved things (and needed to eat food), but you did no "work" because the spring is back to the same place it was before.

The inverter needs to be designed based both on Watts and VA load requirements. The higher VA Current is "real", and it generates heat losses (P=I^2 * R) so it has to be accounted for too. And of course Watts because the input from the battery needs to be designed too.

Lastly, with non-linear wave forms and loads, they create non-sine waves. The math is much more complex, but is simplified into a discussion of Power Factor... Modified Square Wave Inverters driving motors will waste ~20% or so as heat (because it is not a simple sine wave, but looks like a bunch of sine wave of higher frequencies added together--those higher frequencies which don't help the motor to turn--become heat). You can read more here.

For inductive and capacitive loads, the PF will vary 0-1.0

For resistive loads, the PF=1 because a resistor cannot offset the voltage and current wave forms.

Inverters are constant power devices... So if your battery voltage is low, they will draw more current. So, when you design your system, you will have to understand how much power you will need, the inverter efficiency, and how low the battery voltage will go (including voltage drop in your wiring).

For the AC side, most inverters should be able to work with a PF=0.6 ... And I have seen some inverters that list their VA rating in their model name, and you have to read their specification for the watt rating (VA rating is bigger, and looks better in the marketing documentation).

-Bill

Re: power usage and apparent power

Regarding battery Amp*Hours--Basically, the standard battery is rated at a 20 hour rate--or if you can discharge a battery at 5 amps for 20 hours, then the rating would be:

5a*20h=100 Amp*Hours.

If you multiply the Amp*Hours times the average battery voltage, you would get the capacity of the battery in Watt*Hours:

100 Amp*Hours * 12 volts = 1,200 Watt*Hour or 1.2 kWhrs

Some other issues... Batteries, the more current you take out of them the less Amp*Hours they appear to have. So, if you take current out at a 15 amp rate, the battery may have a 75 Amp*Hour rating at a 5 Hour Rate.

Also, you should not drain a lead acid by more than 50% for long life...

So, the 1,200 Whr battery would have ~600 WH of usable capacity (or 50 Amp*Hours).

So, now assume you have your 150 watt load (the 260 VA does not matter here)... Assuming 85% efficient inverter:

600 Watt*Hours * 0.85 * 1/150 Watts = 3.4 hours

But, remember that for our fictional battery, it only had 75 amp*hours at a 5 hour rate...

Our DC Inverter load would be:

150 Watts * 1/0.85 * 1/12 volts = 14.7 amps load on battery

And our 100 AH battery (20 hour rate) is rated at 75 AH (5 hour rate), and remember that we only want a maximum 50% state of discharge:

75 AH * 0.50 / 14.7 A = 2.6 hours of recommended operation

Remember too that batteries when they are cold also lose AH/WH capacity.

-Bill