Low power factor affect on PW2 capacity

[macq]

I am looking at using a PowerWall2 for essential circuit backup.

One of the circuits is a freezer which has a power factor of 0.60 and running amps of 2.0. This means energy use is 240 volt-amps (VA) but only 144 watts (120v * 2a * 0.60pf).

So, in theory should the freezer run 91.7 hours (13.2kWh / 0.144kWh) or 55 hours (13.2 / .240kVA) on a single PW2 charge? Or, are both of these wrong? Let's just focus on the difference between real (W) and apparent power (VA) due to low power factor and ignore duty cycles of the freezer, battery/inverter efficiency and the like.

I wouldn't have questioned using "real" watts until I saw this in a PW2 spec sheet:

1) Capacity is 13.2kWh of AC Energy
2) Power Factor Output Range is +/- 1.0 adjustable
3) Power Factor (full-rated power) of +/- 0.85.

Number 3 especially has me puzzled. Can anyone provide clarity? I tried Tesla sales, hot line, and their forum to no avail.

Thanks.

 

[Parzival]

I am curious about this also.

 

[wwhitney]

3) Power Factor (full-rated power) of +/- 0.85.

The spec sheet also says:

Real Power, max continuous 5 kW
Apparent Power, max continuous 5.8 kVA

And note that 5 / 5.8 = 0.85 (or so)

So I believe that means the inverter will support any total load that meets both of those constraints. In the case of a total load with a power factor less than 0.85, the limit on apparent power will control; if the power factor is greater than 0.85, the limit on real power will control.

As for how long the load will run, only real power matters (ignoring the resistive losses of the reactive current).

Cheers, Wayne

 

[macq]

As for how long the load will run, only real power matters (ignoring the resistive losses of the reactive current).

Cheers, Wayne

Thanks for clarifying that the "full rated power" spec relates to max load, not running time.

But, I just want to make sure I understand your "ignoring the resistive losses of the reactive current" comment.

Are you just saying you ignore the reactive current as its power is both dissipated and returned so nets to zero or are you saying there is an additional loss on the reactive current that isn't accounted for by the power factor? If the latter, how would I account for that?

 

[wwhitney]

Are you just saying you ignore the reactive current as its power is both dissipated and returned so nets to zero or you saying there is an additional loss on the reactive current that isn't accounted for by the power factor? If the latter, how would I account for that?

What I meant to say is that if you have a spec from an appliance that says 240 VA and 144 Watts, and you are willing to ignore the losses in your distribution wiring (which are usually small), then you can just do the simple computation of 13.5 kWh per Powerwall * 90% Depth of Discharge / 144 Watts = 84 hours.

If you do want to model the resistive losses in your distribution wiring, then you need to use the current from the apparent power figure, i.e. 240VA / 120V = 2A. If the distribution wiring is 200 feet (round trip) of #14 copper with a resistance of 0.6 ohms, then the I^R power loss is 2^2 * 0.6 = 2.4 W. A small adjustment.

On the other hand, if you have the actual appliance plugged in and measure the real power at the source end of the circuit, using an accurate meter, that measured figure will already reflect the losses in your distribution wiring, so there is no adjustment to be made. If you measure at the appliance, that measurement would not reflect the losses in your distribution wiring.

Cheers, Wayne

 

[macq]

Excellent explanation Wayne. That fully takes care of my original post.

I have a follow-up to the 90% discharge in your example. Tesla's site says 13.5kWh with depth of discharge of 100% (Tesla Powerwall). They note a 90% round-trip efficiency. Are you suggesting I should use 90% as a discount for depth of discharge instead of the quoted 100% or was that just an example?

And, note I originally said 13.2kWh was the spec but that was from a non-Tesla site so best to use 13.5kWh.

Again, many thanks.

 

[Parzival]

My understanding is that 13.5kWh is the usable capacity of the AC powerwall, with the losses factored in.