Uninterruptible power supply ratings: conflict between VA and W?

[seanspotatobusiness]

My understanding is that P = VI and therefore the power rating of a device in VA should be the same as W but this doesn't seem to be the case for every uninterruptible power supply I look at where the VA rating is much higher than the rating in watts. What's that about? Here's a random example on eBay atm but they're the same everywhere.

 

[Bluejets]

To get watts in an AC system, one must apply power factor to the VA rating.

I won't go into it any further as you can find the answers on a wiki search.

 

[Ratch]

My understanding is that P = VI and therefore the power rating of a device in VA should be the same as W but this doesn't seem to be the case for every uninterruptible power supply I look at where the VA rating is much higher than the rating in watts. What's that about? Here's a random example on eBay atm but they're the same everywhere.

Your understanding of P = VI is wrong. That is only true for DC resistor circuits. When inductance and capacitance are present in AC circuits, the voltage and current become out of phase with each other due to reactance. This causes less energy to be dissipated than would be if the phase difference were zero.

The following shows one amp of current and one volt of voltage in phase with each other. This is the case if no reactance is present. The green curve shows the power in phase with both the current and voltage.


Next we have the voltage and current out of phase by 45°. This would be the case if the reactance were equal to the resistance of the circuit. Notice that the green power curve does not go as high as before and dips below zero. That shows that some of the energy is given back to the circuit.


Finally we have the voltage and current out of phase by 90°. This would be the case of the circuit consisted of a capacitor or inductor with no resistance. All the power taken is given back to the circuit when the green power curve dips below zero. No energy is dissipated if resistance is not present.


So, you see that you need to know the phase when reactance is present in order to calculate power correctly. Just multiplying voltage times current (apparent power) does not work for AC circuits.

Ratch

 

[WHONOES]

Nice description RATCH.

 

[(*steve*)]

Adding to what @Ratch said. If the phase shift is 90 degrees, the power is zero, but too might be maxing out your VA rating.

 

[kellys_eye]

Simply - most domestic electrical installations are billed based on a standard PF correction of around 0.9 to allow for losses in inductive loads in the users premises. The installed meter only measures 'real' power consumption and cannot correct for phase-loss.

If your home uses a LOT of inductive loads (motors etc) your supplier may well bill you differently to compensate for their loss i.e. calculate your consumption using a different (lower) PF value. This is common for industrial premises.

On the flip side, if your home uses NO inductive loads at all then you're being 'ripped' off to the tune of 10% as your bills are being 'adjusted' for no good reason! Most homes do, however, have inductive loads - transformers, washing machine motors, blower (heating) motors etc. Some of the worst performing items you can fit are CFL lighting where the PF can be as bad as 0.5 or worse!

It only becomes an issue if you have to power LOTS of them so the figure for VA is supplied to allow users that have 'lots' in use to compensate for the current draw by installing larger wiring i.e.

if you have 1 unit rated at 1000W (or 1100VA) it's not a problem. The difference between the current at 1000W and 1100VA is minimal and 'accepted' as a reasonable loss to the power companies. The difference in current is also negligible to your wiring.

if you have 100 units rated at 1000W (or 1100VA) then the losses to the power companies revenue (since they can't charge you for the 100 x 100 = 10,000W 'additional consumed watts') become appreciable but, NOT ONLY THAT, your wiring is carrying 10,000W 'extra' that, if you only took the 1000W basic rating at face-value (and not the 1100VA 'real' rating) to calculate your installation wiring requirement, would burn out your cabling!

 

[Ratch]

Simply - most domestic electrical installations are billed based on a standard PF correction of around 0.9 to allow for losses in inductive loads in the users premises. The installed meter only measures 'real' power consumption and cannot correct for phase-loss.

If your home uses a LOT of inductive loads (motors etc) your supplier may well bill you differently to compensate for their loss i.e. calculate your consumption using a different (lower) PF value. This is common for industrial premises.

On the flip side, if your home uses NO inductive loads at all then you're being 'ripped' off to the tune of 10% as your bills are being 'adjusted' for no good reason! Most homes do, however, have inductive loads - transformers, washing machine motors, blower (heating) motors etc. Some of the worst performing items you can fit are CFL lighting where the PF can be as bad as 0.5 or worse!

It only becomes an issue if you have to power LOTS of them so the figure for VA is supplied to allow users that have 'lots' in use to compensate for the current draw by installing larger wiring i.e.

if you have 1 unit rated at 1000W (or 1100VA) it's not a problem. The difference between the current at 1000W and 1100VA is minimal and 'accepted' as a reasonable loss to the power companies. The difference in current is also negligible to your wiring.

if you have 100 units rated at 1000W (or 1100VA) then the losses to the power companies revenue (since they can't charge you for the 100 x 100 = 10,000W 'additional consumed watts') become appreciable but, NOT ONLY THAT, your wiring is carrying 10,000W 'extra' that, if you only took the 1000W basic rating at face-value (and not the 1100VA 'real' rating) to calculate your installation wiring requirement, would burn out your cabling!

Any meter that measures real or true power automatically corrects for low power factor (PF) at the point of use. The electric supplier does not directly lose revenue on low PF loads. The problem is that they have to install higher amperage conductors to transfer the same amount of electrical power. Example: A 900 watt resistance type clothes iron could use 9 amps at 100 volts (900 VA). The PF is 1.0, so the electric company supplies 9 amps at 100 constant volts. Then it rains and the 900 watt basement sump pump with a PF of 0.9 turns on. Now, the power company has to supply 10 amps at 100 volts (1000 VA), but can only bill the customer for 900 watts (100x10x0.9), The extra amp causes IR losses in their lines and increased load on their generators. They are supplying and getting paid for the same amount of delivered power (900 watts), but it is costing more to deliver. And, the true wattmeter at the house is showing 900 watts delivered even though 1000 VA was supplied.

Ratch

 

[Bluejets]

The electric supplier does not directly lose revenue on low PF loads.

Yes, they do.
Which is why it was necessary to fit large capacitor banks to motor installations etc. with inductive loads.
The old spindle meters and the like were watt meters and are now being replaced with Kvar meters so the above will no longer be necessary.

They are supplying and getting paid for the same amount of delivered power (900 watts), but it is costing more to deliver. And, the true wattmeter at the house is showing 900 watts delivered even though 1000 VA was supplied.

Yes, which is why the first bit is codswallop.

 

[Ratch]

Yes, they do.
Which is why it was necessary to fit large capacitor banks to motor installations etc. with inductive loads.

The capacitors are for improving the PF. I said it does not cost the electric company directly to supply extra VA. That is because the extra VA power supplied during the first half the the power cycle is returned to the generator source during the second part of the cycle. The most extreme case occurs when an inductance or capacitance without any resistance is driven by a sinusoidal voltage. Although current exists to and from the reactive component, no energy is delivered to the component, because all the energy accepted by the reactive component is returned during the second part of the power cycle. The only costs to the electric company are indirect, like IR losses, and whatever other costs occur from transporting more current than would otherwise be needed.

The old spindle meters and the like were watt meters and are now being replaced with Kvar meters so the above will no longer be necessary.

I believe the meters were actually kilowatt-hour meters.

Yes, which is why the first bit is codswallop.

lf you believe that it costs the electric company extra in energy costs to supply the extra100 VA in the sump pump example I gave in my last post, then you do not quite understand how reactive power works. Reactive power causes higher current to exist, but the reactive power is recovered during each power cycle and returned back to the generator.

Ratch

 

[Bluejets]

Tell it to the supply authorities.