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I'm very confused. I don't know of any power company that bills residential customers in a way where power factor matters. They bill you by the KWH - which is often termed "real power
From my understanding they do bill you by the KWH however a true watt is not simply volts x amps when referring to ac current. Volts x amps is VA or KVA. Power factor is what percentage of KVA is KW so 100 KVA with a .8 power factor is 80 KW. I believe most residential homes operate in the >.95 range so KVA and KW are usually pretty close however some devices have lower power factors so you can't just do volts x amps to get watts very actually on all devices.
If you are a large business customer, they will further penalize you for poor power factor - but that's usually some kind of multiple added to the per-KWH rate - it doesn't change how much power you consumed - that remains in KWH.
I believe commercial accounts get billed for KWH usage but also a demand charge which is a rate multiplied by the peak KVA so having a lower power factor will increase their demand charge as they'll have higher KVA but not their actual usage charge as the KWH isn't changed. A 10 horse motor consumes roughly 7680 watts so a motor with a .95 pf will need 8074 VA to run but a 10hp motor with a .85 pf will need 9035 VA to run however they'll both use that same wattage and cost the same to run except for the demand charge and maybe some slight heat loss in the wires from the extra current.
Maybe I'm wrong but that's how I understand it.
This begs the question on what is being measured. When the hall sensor - or the shunt - gives me an amperage rating, and i multiply that by the voltage - am i measuring apparent power or real power? I believe - and i may be wrong - but these sensors are not going to show reactive power - since they don't monitor the power consumption at the sub-jiffy level - which is where reactive power is observed.
Regardless, KW is real power - if you have a proper KW reading for in and out, efficiency is simply OUT / IN. Worth mentioning reactive power is not observed on DC circuits, so if your kill-a-watt is showing KW, then that plus the A and V on the DC bus is enough.
A demand charge is a peak wattage (note, real-power). AKA, you may only use an average of 100KW over a month, but you might peak to 1MW of power (aka when your heavy equipment is operating). You would be charged for 1MW of "demand". Because at any moment the utility must be able to handle the power that you "demand".
This begs the question on what is being measured. When the hall sensor - or the shunt - gives me an amperage rating, and i multiply that by the voltage - am i measuring apparent power or real power?
That would be apparent power. Real power is apparent power multiplied by pf.
Regardless, KW is real power - if you have a proper KW reading for in and out, efficiency is simply OUT / IN. Worth mentioning reactive power is not observed on DC circuits, so if your kill-a-watt is showing KW, then that plus the A and V on the DC bus is enough.
Yes that's how I ran the efficiency numbers. The Kill-A-Watt Metter shows the real power in watts (VA x PF). So I took the battery volts multiplied by the battery amps and divided it by what the meter claimed the in watts to be.
A demand charge is a peak wattage (note, real-power). AKA, you may only use an average of 100KW over a month, but you might peak to 1MW of power (aka when your heavy equipment is operating). You would be charged for 1MW of "demand". Because at any moment the utility must be able to handle the power that you "demand".
For the demand charge I believe they go off of the peak apparent power. Which is how a business would be penalized for poor pf.
That would be apparent power. Real power is apparent power multiplied by pf.
Yes that's how I ran the efficiency numbers. The Kill-A-Watt Metter shows the real power in watts (VA x PF). So I took the battery volts multiplied by the battery amps and divided it by what the meter claimed the in watts to be.
I need to look more into this. But i would expect an amp sensor's reading (assuming you're not looking at it in a sub-jiffy time frame) to basically average-out the apparent power, as the reactive power is both received and returned during each jiffy.
Sid could probably skool me real good on this one.
For the demand charge I believe they go off of the peak apparent power. Which is how a business would be penalized for poor pf.
That's not what the demand charge is for. The demand charge is for the availability of additional surge-capacity on the grid - because it costs more to do that. Without that additional capacity (called spinning reserve), the large consumer would cause an outage when their demand suddenly hits the grid. It is measured in KW -- true power, not KVA.
Power factor is a direct penalty - usually applied as an additional cost on every KWH billed. I've only seen it done this way, though some utilities may bill it as a fee per KVAR.
Power factor is what percentage of KVA is KW so 100 KVA with a .8 power factor is 80 KW.
A 10 horse motor consumes roughly 7680 watts so a motor with a .95 pf will need 8074 VA to run but a 10hp motor with a .85 pf will need 9035 VA to run
Umm...this kinda what's throwing me for a loop: where PF is being rather freely applied in either direction, both as a subtract and as an addition.
In power source ratings, yes, it "subtracts" from the nameplate (i.e. 100kva source with a 0.8PF load has an effective maximum load of 80kw)
But in power consumption, it "adds" to the nameplate (i.e. an 80kw load with a 0.8PF has an effective draw of 100kva).
With that in mind, and the GS inverter consuming power, I would expect the low PF to add to the "kw" rating.
I mean, I really like the idea of the GS inverter being freak efficient. But I just don't see that being electrically or mathematically possible--especially if we run the math on a power factor of 0.1.
With the inverter drawing power, the of will make the KVA higher than the KW, because KVA is always higher than KW
With that in mind, and the GS inverter consuming power, I would expect the low PF to add to the "kw" rating.
One thing I'm really not to sure on is what really happens to the unused power between the KW and KVA (Reactive Power). Everything I read says it goes back to the source but what does that really mean? The GS inverters can easily do their respective 12 and 6 ratings in KVA no problem but say you put a 6 KW load on the GS 6 but that load has a pf of .5 then what? Is the difference "recycled" in the transformer and spit back out?
One thing I'm really not to sure on is what really happens to the unused power between the KW and KVA (Reactive Power).
HEAT . Inductive load produce a lot of heat . The GS12kw is very efficient as the fan is very quiet .
The KW is the power used
The AC sign wave is a constantly changing voltage
Drawing current at a lower voltage part of the wave hat the same amps but lower watts due to the lower voltage at the time of the current
The unused part of the power didn't move
The wires will still get hit with the amps and voltage drops associated