AC Watts vs DC Watts

brulaz wrote:
1.414 is the SqRoot of 2, which, I was told on another forum, gives an "approximation" of the AC RMS Watts to DC Watts.

DC Watts = Voltage X Amperage
AC Watts expressed in terms of DC = Volts X Amps / sqrt2

RMS Watts (Root Mean Square - the sqrt2 in the AC calculation) is so you have a direct correlation to DC Watts otherwise you'd end up not comparing apples to apples.

Let me rephrase the question:

So to run a 1000W AC RMS load do I need 1414 DC watts?

And does that mean my DC cables should handle an ampacity of 1414/12=118 Amps? (assuming 100% inverter efficiency and battery V drops to 12V)

Or are inverter wattage ratings not RMS?

Thing is, I've only heard of RMS watts in discussions of HiFi equipment and speakers. Never w/r to inverters or other household items. So still confused.

No, you don't need the 1.414 factor, it's confusing you. If you know the current draw of the load mulitply it by 10 and that will tell you pretty close the DC current draw and then figure the wire size.

If you don't know the current draw but know the AC watts, 10X will get you in the ballpark for things like a toaster, coffee pot, etc. For a microwave it's current draw is higher than the watts listed, use 12 or so for the current draw.

These may be off a little, but will give you a good indication of wire size needed and ability of inverter to drive the load.

For the DC power in, do you use the voltage at the inverter terminals or the voltage at the battery posts?

( I never know what is meant by "battery voltage")

I have a MSW inverter. The voltmeter in the 120 output receptacles reads in the 90s because it is not a true RMS meter, but the watts meter on the inverter display seems to come out right for the appliance's watts rating for a resistive load.

This looks like a job for Mr Kill A Watt!!! Look, up in the sky....

BFL13 wrote:
This looks like a job for Mr Kill A Watt!!! Look, up in the sky....

Be careful using a Kill-A-Watt on a MSW inverter. Mine got very hot and blew an internal fuse.

DrewE wrote:

...

RMS vs peak and the square root of two have to do with averaging a sine wave and having the equivalent steady-state value over time--integrating over time, basically. AC voltages and currents are generally expressed as RMS values, and these are related to the peak values by the conversion factor. They're expressed as RMS values precisely because it enables one to use standard basic DC laws to compute things like power or Ohm's law and get the correct answers for resistive loads (i.e. if the power factor is 1.0). To do otherwise would basically require calculus rather than arithmetic to figure these things, as indeed often is the case with non-sinusoidal waveforms.

AC power in watts is AC Volts (RMS) times AC amps (RMS) for a resistive load. An AC watt can of course accomplish as much work in a given time as a DC watt, since they're both just watts.

...

For e.g. an electric heater, the power consumption is just the wattage, and would I suppose be RMS watts if you wanted to integrate the instantaneous power consumption over the course of one or more AC cycles. Nearly always for power we're talking about more or less steady-state operation, so such integration is assumed.

If the inverter is 100% efficient, by definition the power going in will be the same as the power going out, and your DC power consumption will be exactly the same as the AC power output.
...

OK, thanks DrewE and all others.
That clears it up for me.
I'll pass this discussion on to the fellow on the other forum.
Sounds like he could use some clarification.

Is not the following factorials as much or more important (?)

Efficiency at percentages of inverter load...

5% - 10% - 15%...etc...etc...etc...

And at what point if any in battery input voltage does an inverter fault to pure square wave? i.e. two units rated 2000 watts -- one reverts to square wave at 11.4 volts, the 2nd reverts to square wave at 11.00 volts.

Test results:

Kettle says it is 120v, 1000w

In stick house with KillaWatt, no load 117.6v, with kettle on:

115.2v, 8.02a, 924w, 924VA, PF 1.0

In RV, shore power off, MSW inverter on whole house, battery voltage knocked down from 13.6 to sit at 13.0v steady, which is full for AGMs.

KillaWatt on , no load 120.9v. With kettle on: (note that KillaWatt got the "right" voltage number despite the inverter being MSW)

117.4v, 8.14a, 939w, 962VA, PF 0.99 Inverter watts display says 1000w DC voltage 12.2v Trimetric says 12.2v, 87.1 amps.

Using the Tri numbers for DC power in, 12.2 x 87.1 = 1062.6w

1062.6/962 = 1.105 or 962/1062.6 = 90.5%
1062.6/939 = 1.132 or 939/1062.6 = 88.4%
1062.6/924 = 1.150

(The 0.8 voltage drop on 87 amps with 450AH of AGMs is mostly AFAIK from my convoluted wiring set-up. No worries--it gets the job done)

Inverters are rated in AC watts or AC VOlt amps (if it says Watts I assume it means watts)

What, you ask is the difference..... Something called Power Factor.

A motor drawing 10 amps at 120 volts is 1200 watts right.. Wrong if the power factor is 0.5 it's only 600 watts. (Power factor is between zero and 1 inclusive or may be expressed as percentage zero to 100) I've heard of a case of ZERO. only one. Motor did not turn, Lots of Amps. 440 volts. but it did not turn.. Phase angle was exactly opposite.. I know the technician who fixed it. and have seen the motor.

The guy thinking it's 1414 is apparently thinking that's the DC draw needed to support the AC peak power (which is 1.414 x the RMS power). Inverters don't work like that. They have inductors and capacitors which store the needed energy. Watts in (times efficiency) equals watts out. Conversely, watts out/efficiency = watts in.

Inverter 95% efficient? 1000 W (out) / 95% (efficiency) = 1053 W (in).

OP.. the answer is NO

you do NOT need 1441 DC watts for 1000 AC watts used

that would be terrible efficiency

upteen posts have been made on this forum, mentioning 5-10% avg loss not 40% loss

that poster on the other forum does NOT know what they are talking about