I was testing the power consumption of a computer (to calculate the power requirements - as in what power supply to use).
To do this, I used one of those plug in green electricity monitor thingies. It shows volts, amps, watts, power factor, etc.
Anyway, when the computer was switched off at the switch on the back of the power supply (I'm assuming that this means only the filter capacitors are connected), the electricity monitor showed that it was drawing 0.049 amps with a power factor of 0%. Am I right in thinking this means I am being charged for no electricity, while 11 watts of electricity are being dissipated in the national grid somewhere? The monitor also shows 0 watts consumption, which I assume is useful watts as charged by the electricity meter - because it then goes on to use this to show cost over a period of time. This seems very wasteful to me. Or am I miscalculating something?
No. You are absolutely right. Your 11W is actually 'volt-amps' (VA), and VA x PF = watts. As your PF is 0, your PC is not consuming any power.
However, the 0.049A is coming to your PC via the house wiring. As this will have some (hopefully very small) resistance, there will be a small 'I squared R' power loss in the wires. And, as you say, similar loss will be occurring in the National Grid wiring.
Of course, your small capacitive current might be a 'good thing'. It might be helping to compensate for an inductive load next door (well, a near neighbour on the same distribution phase as you are on), where the PF is less than 1.
s. =A0It shows volts, amps, watts, power factor, etc.
he power supply (I'm assuming that this means only the filter capacitors ar= e connected), the electricity monitor showed that it was drawing 0.049 amps= with a power factor of 0%. =A0Am I right in thinking this means I am being= charged for no electricity, while 11 watts of electricity are being dissip= ated in the national grid somewhere? =A0The monitor also shows 0 watts cons= umption, which I assume is useful watts as charged by the electricity meter= - because it then goes on to use this to show cost over a period of time. = =A0This seems very wasteful to me. =A0Or am I miscalculating something?
rive vehicles."
Power factor of 0 means the current consumed is out of phase with the mains voltage, so no power is consumed and no meter reading or cost is incurred. It will cause power waste in the grid, but nothing anywhere near 11w, it'll be under 1w.
It's not really 'out-of-phase' (as, in loose talk, that usually implies
180 degrees out of phase). It's 90 degrees OOP, with the current leading the voltage (something I still find is a weird concept!).
But not if the OP's incidental capacitive current is compensating or correcting for an equal amount of inductive load current somewhere else in the mains system.
At any point on mains and grid system, the load power factor will be the 'total' of all the individual effects from the various users downstream. My understanding is that the mains system does indeed suffer from unwanted reactive 'I squared R' losses, and this is because the total load it feeds tends to have an overall lagging power factor (due largely to all those inductive power transformers and electric motors it feeds).
Yes, quite right that its lagging overall. Even within one house the odds are that the capacitive current will be offsetting overall lagging current on the whole, so in reality it'll avoid tiny losses rather than cause them.
I've often wondered if all domestic supplies should not have (probably immediately before or after the meter) a 'typical average value power factor correction capacitor'.
Actually, I believe that large factories, companies and other establishments which consume lots of electricity are obliged to have power factor correction equipment on their supply feeds. In some cases where the load varies a lot, this is dynamic, and always tries to keep the overall power factor as close as possible to 1, regardless of how much power is being used.
No. You are drawing a current of 0.049A, but aren't actually 'consuming' it. As far as you are concerned, it's a lossless use. You are only borrowing the 11VA, then giving them back. However, there are still the transportation costs, ie the 'I squared R' losses caused by the resistances of the mains and National Grid cables and intervening power transformers. Essentially, this is completely unknown (at least to people like you and me).
How much power loss your 0.49A creates is almost impossible to calculate. It would be easy if your 230V supply came directly from a power station, just up the road. Your 0.49A would be 0.049A all the way. The resistances of the power cables and the actual generator could be measured, so the 'I squared R' power loss calculation would be dead easy. If, for example, the cables were 10 ohms, and the generator 5 ohms, the total resistance is 15 ohms. The 'I squared R' loss power loss would be 0.036015W.
However, your 0.049A won't stay at 0.049A for long. At your local substation transformer, the 230V becomes (say?) 11,000V. From there to the next upstream transformer, your 0.049A will be 0.00102A, so the 'I squared R' loss will be a lot less. Eventually, your original relatively massive 0.049A will be a relatively minuscule 0.000575A flowing in the
400kV supergrid, and the 'I squared R' losses will be very small indeed. [Note that, as the 'I squared R' losses are proportional to the square of the current, it is much more efficient to carry the electricity at as high a voltage as possible.]
But it would take someone with a complete knowledge of the specification and characteristics of the whole power system to calculate what the sum total of all the individual 'I squared R' your original 0.049A creates.
Also consider the hypothetical situation of someone on the same phase further down the road with an inductive load drawing an out of phase current matching your capacitive load. The affected current will bounce backwards and forwards between the pair of you without drawing anything from the substation and the I^2R losses only arise in the cable in your street.
Doesn't it also draw more power from the generator? Say for sake of example that you had a generator in your garden, and an infinitely thick cable running to a piece of equipment with a bad power factor. Would you be losing out? Heating in the coils of the generator?
Isn't PFC a requirement for new equipment of any type nowadays?
No, it's leading. The voltage has to "happen" at source (the power station) but the current leads the voltage measured at the load (your appliance).
Imagine this circuit (you'll need a fixed width font):
A o-------------/\/\/\/\/\/\/\------------- 10 Ohm | Resistor | --------- C Capacitor --------- D | | B o----------------------------------------
Consider the voltage measured across the capacitor between points C and D when 10 volts are applied to the input terminals A and B. Initially both plates of the capacitor are at 0V so there's a potential difference of 10V across the resistor and a current of 1A will flow through the resistor into the capacitor. As the current flows it will charge up the capacitor increasing the voltage across C-D. As the voltage of C rises the voltage across the resistor falls and the current flowing into the capacitor falls until we reach the stage when the capacitor is fully charged with 10V across C-D and no current is flowing.
So the current flowing into the capacitor leads the voltage measured across its terminals.
With an AC supply a similar thing happens in reverse for the negative half cycle and you see a current waveform which leads the voltage at the capacitor by 90 degrees.
You might find it easier to imagine the capacitative load delaying the voltage waveform seen at your end of the wire.
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