Naked Science Forum
Non Life Sciences => Technology => Topic started by: Ben Bleiman on 15/04/2009 12:30:02
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Ben Bleiman asked the Naked Scientists:
I have a question about power factor. I know that compact fluorescent bulbs are more efficient that normal bulbs, but I also know that they have a power factor of .5 or .6. What does this mean? Am I only getting billed for .5 times the power it is actually using? Does this mean it is not as efficient as it appears but you save money anyway?
Thanks
Ben
What do you think?
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You pay for the actual power you use. The power factor only affects the peak current and volts needed to transmit it.
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I have a question about power factor. I know that compact fluorescent bulbs are more efficient that normal bulbs, but I also know that they have a power factor of .5 or .6. What does this mean? Am I only getting billed for .5 times the power it is actually using? Does this mean it is not as efficient as it appears but you save money anyway?
A simple resistive load such as a heating element obeys Ohms law I=V/R instantaneously, ie throughout the AC voltage waveform the current is always proportional to the instantaneous voltage. This will also hold true for a conventional lightbulb. The power factor is 1.
Any load which contains an inductor (such as an electric motor) or capacitor on the mains will draw current "out of phase" with the voltage. A pure capacitance will draw current proportional to the rate of change of voltage, so the current will be maximised as the voltage crosses zero (in the sinewave this is when it's changing fastest). Summing up the instantaneous power (I.V) you would find that over one cycle, no "power" is drawn by the capacitive load (and no power would be metered by a domestic meter). However, you have still been pulling current through the wires which means that there will be real energy losses (costs incurred) in the distribution network. Capacitive loads draw current "ahead" of the voltage whereas inductive loads draw current "behind" the voltage - consequently an inductive load cna be balenced by applying a capacitor to it.
Modern devices with switch-mode power supplies including computers and low-energy bulbs typically feed the mains straight into a bridge rectifier with a capacitor on the DC side. This means that they draw essentially zero current for most of the AC cycle, and just a brief burst of current at the peak of the voltage waveform. This means that the instantaneous peak current is much higher than the average current of a resistive load of the same nominal power. If a large fraction of the load on the grid (or even just one premesis) starts loading the network in this fashion it can distort the waveform of the mains and cause other issues (I'm not too sure what exactly). It's not really very easy to correct or compensate for loads which only draw currents at the peak of the waveform.
See also http://en.wikipedia.org/wiki/Power_factor
One consideration is that mains electricity is distributed as three phases 120degrees apart. Assuming all loads are resistive, the current in the neutral wire to the substation will sum to zero (both instantaneously and over time). Consequently no power is dissipated in the neutral wire, and it can be relatively light gauge.
If all the loads only drew current at the peak of the voltage waveform (so there's no instantaneous overlap of loads) the average absolute current in the neutral would be the sum of the currents in the phases, and the neutral cable would need to have three times the cross-section of the phases (for power-loss / cable self-heating considerations). This is an extreme example, but illustrates problems that can arise.
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Try the new LED lighting systems if you are looking for energy-efficiency. you will definetly see a difference!
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A power factor of 0.50 means that the lamp draws twice the amount of power that it consumes; it uses ½ of it and returns ½ of it to the power company. A power factor of 0.25 means that the lamp draws four times the amount of power that it consumes; it uses ¼ of it and returns ¾ of it to the power company. As far as I know, power companies monitor the power factor of large industrial and commercial customers, but not residential customers. So, don’t get all worried about your fluorescent lamps with a power factor of 0.5 to 0.6.
Good point Dave. A minor clarification:
A power factor of 0.25 means that the lamp is drawing four times the amount of current that it actually needs. The lamp never really draws any more power than it needs.
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Dave - A power factor of 0.25 means the current supplied is four times greater than it would be at a power factor of 1.
According to Basic Electrical Engineering - Shepherd, Morton & Spence:
"the power factor when the current and voltage are sinusoidal is cos a, i.e. the power factor is the cosine of the angle of phase difference between the current and the voltage."
"power absorbed (by the load) = V.I.cos a"
Therefore, for a given power absorbed at a constant voltage, the current (I) is proportional to 1/cos a
If the power factor is 0.25, cos a is 0.25 and therefore I is four times greater than it would be for a PF of 1.
No extra "power" was ever delivered to the lamp, however, a large amount of unnecessary current alternated through the transmission equipment, and that can waste power and overload the transmission equipment.
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More from Shepherd, Morton & Spence.
"The disadvantage of low power factor is that the current required for a given power is increased. This leads to a large kVA for a given power.
Alternators, switchgear, transformers and cables are all limited in their current carrying capacity by the permissible temperature rise (proportional to I^2). Thus they may be fully loaded with respect to kVA (all a.c. equipment is rated on a kVA basis because of the temperature rise limitation) without delivering their full power." etc. etc.
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Imagine that I connect a big perfect capacitor to the mains.
It has a power factor of zero, but it draws a current.
There will be points on the mains cycle where the capacitor is fully charged.
At that point I could disconnect it and discharge it into something and thereby prove that the charged capacitor had energy stored in it.
I could divide the energy stored by the time I took to transfer that energy and calculate the mean power delivered to the capacitor from the supply.
There really is power transfer, even though the power factor is zero.
Of course, if I don't disconnect the cap from the supply then the stored energy is returned to the mains.
The average power delivered to a perfect capacitor would be zero, but the instantaneous power can be large.
It is by analogy, therefore in correct to say that "No extra "power" was ever delivered to the lamp".
The power was delivered, it was stored in a cpacitor or inductor, and it was returned.
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Imagine that I connect a big perfect capacitor to the mains.
It has a power factor of zero, but it draws a current.
There will be points on the mains cycle where the capacitor is fully charged.
At that point I could disconnect it and discharge it into something and thereby prove that the charged capacitor had energy stored in it.
I could divide the energy stored by the time I took to transfer that energy and calculate the mean power delivered to the capacitor from the supply.
There really is power transfer, even though the power factor is zero.
Of course, if I don't disconnect the cap from the supply then the stored energy is returned to the mains.
The average power delivered to a perfect capacitor would be zero, but the instantaneous power can be large.
It is by analogy, therefore in correct to say that "No extra "power" was ever delivered to the lamp".
The power was delivered, it was stored in a cpacitor or inductor, and it was returned.
Sorry. That's not right either. Extra energy was delivered to the capacitor, but it was never converted into power at the capacitor.
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Right, extra power delivered to the fluorescent lamp and then returned to the power company.
If you are willing to accept that power and energy are the same thing, perhaps you would like to buy a certain 2,000 horsepower can of gasoline I happen to have. I could probably let you have it for about $1,000.
The reason I am so adamant about not using "power" in this context is because, in AC systems, volt-amperes can be very different from watts. The power factor issue is not a problem of watts, it's a problem of amps.
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Dave: I'm afraid that readers might conclude from this that the generating company has to meet the demand of the apparent power consumed. I think we need to make it clear that is not the case.
The generating company only has to meet the demand of the real power consumed, plus any power lost in the transmission system.
The difference between the apparent power and the real power in any arm of the transmission system can increase transmission losses, but that is a consequence of the increase in current, and the losses are proportional to the square of the current.
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The basic question asked was to lamp efficiency. Generally incandescent lamps are very good generators of heat, and poor at producing visible light. Halogen lamps run hotter, and more of the input power is produced as visible light. CFL and LED light sources generate light by using energy level differences in excited electron bands, and produce very little output in the form of heat, thus resulting in the assertion that they are more efficient at producing visible light.
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Dave, I'm sorry, but I don't think we are clear on a couple of points.
The utility company only has to "supply" the real power demanded by the system. It does not need additional generating capacity in the sense that more power is put into the generator to supply the VA demanded. If the generating company is charging on the basis of VAs, it might be quite happy to be paid for the delivery of apparent energy it never actually supplied. Anyway, in practice, the reactance of the load presented to the generator is compensated so that there is little difference between the power and the VA at the generator.
However, additional generating capacity is required to compensate for real power lost in those parts of the transmission system that operate at reduced power factors. These power losses are a consequence of the heating effect that results from increased current flowing to loads in the transmission system with non-unity power factors, but it is impossible to determine what they will be without detailed knowledge of the transmission system.
Reactive loads are a problem for transmission, not generation.
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The basic question asked was to lamp efficiency. Generally incandescent lamps are very good generators of heat, and poor at producing visible light. Halogen lamps run hotter, and more of the input power is produced as visible light. CFL and LED light sources generate light by using energy level differences in excited electron bands, and produce very little output in the form of heat, thus resulting in the assertion that they are more efficient at producing visible light.
Yes - the efficiency of the device is important, but we should not ignore the fact that it can have an effect on the efficiency of the total system, otherwise we could end up robbing Peter to pay Paul. That was the point that Dave raised.
For example, switched-mode power supplies are common these days. Virtually every computer on the planet uses them because they are very "efficient" (they don't generate a lot of waste heat). However, because the current demand they impose on the power distribution system is very nonlinear with respect to voltage, they can cause a lot of energy to be dissipated in the power distribution system, so, perhaps they are not so efficient after all.
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er, negatory DD.
In the case of a real power distribution system, the reactance of the load presented to the generator is likely to be small. However, there may be arms in the network that have quite reactive loads and are therefore dissipating power due to the associated i^2.R losses. But the generator will not "see" a reactive load, either because capacitive branches in the network are arranged to cancel out inductive branches, or because the network includes power factor correction devices at its substations, or both.
Because the transmission company takes steps to ensure that the power factor of the network is maintained as close to unity as is practical, the generator is not affected by the PFs of the individual loads. As I may have already mentioned, load PF is not a problem for generation, it is a problem for transmission.
If you want to model the real effect, you will have to include the transmission system in your model.
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Looking at my (latest) diagram for the ideal coil, at 45°, where the voltage is Emax/√2 and the current is Imax/√2, meaning the power is Emax•Imax/2, what is providing that power? The coil?
We can't say the coil produced the power. The coil absorbed energy from the power source during one half-cycle, then returned all of it to the power source during the next half-cycle. If you want to call that power transfer, that's fine with me. I prefer not to because, to me, power implies energy conversion into work or heat. Anyway, the power source delivered no RMS power, despite the fact that a real current flowed back and forth in the circuit from the inductor.
Obviously, this circuit does not represent a practical power distribution system, because it has no resistance. If it did have resistance the phase angle could not be 90 degrees and the current flow would dissipate real power across any resistance, which would then have been dissipated as heat, despite the fact that no useful work was done. That is why the power factor of the load is so important.
The power source does have to put energy into the reactive elements in the system (including the power lines). That would seem to consume some power. I suspect the net energy stored in an actual power distribution system will tend to balance out to zero at any point in time, but that's a guess. I any event, any power required to "charge up" the system is negligible compared to the power distributed by the system.
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The basic question asked was to lamp efficiency. Generally incandescent lamps are very good generators of heat, and poor at producing visible light. Halogen lamps run hotter, and more of the input power is produced as visible light. CFL and LED light sources generate light by using energy level differences in excited electron bands, and produce very little output in the form of heat, thus resulting in the assertion that they are more efficient at producing visible light.
Yes - the efficiency of the device is important, but we should not ignore the fact that it can have an effect on the efficiency of the total system, otherwise we could end up robbing Peter to pay Paul. That was the point that Dave raised.
For example, switched-mode power supplies are common these days. Virtually every computer on the planet uses them because they are very "efficient" (they don't generate a lot of waste heat). However, because the current demand they impose on the power distribution system is very nonlinear with respect to voltage, they can cause a lot of energy to be dissipated in the power distribution system, so, perhaps they are not so efficient after all.
Initially SMPS were very poor with regards to power factor. Your first SMPS was the television set, the worst example was the Phillips G11, which drew DC current from the mains as an 11A pulse every second cycle. This caused a few substations to fail due to core saturation, and started a trend towards a unity power factor. The latest SMPS incorporate PFC to give a near unity power factor, and this active approach actually improves the efficiency by reducing heat losses internally. This is only effective above around 50W, so small units just use a resistor to reduce the current pulse on the input at the expense of a slightly worse efficiency.
The power drawn by electronics is strangely in phase with the voltage, unlike the coil, but is responsible for creating harmonics on the supply, and this causes problems for transmission and distribution, as this causes heating in lines, transformers and power factor correctors. The generation side has to provide Apparent power, not True power, as the load on the generator is only determined by the current output, nothing else. The individual plants switch in and are run up to full load current, and are kept there as the control loops that run them do not respond well to load swings, having time constants in minutes due to large rotating masses. Per se they have a unity power factor, as they are the source, but they supply VA, even though the customer is billed in Watts, with a penalty if the power factor drops below 0.7.
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SeanB:
When you say the utilities "supply VA", are you suggesting that they have to consume more energy (nuclear, wind, solar, fossil fuel, etc.) to supply those VA than they would have had to if they were only producing real power, (excluding energy lost in the transmission network due to nonlinear loads in the network)?
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Today's energy-efficient bulbs are available in wide range of colors in market and light levels you are expecting. Although prices of energy-efficient bulbs is usually higher than traditional shining.
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Up until 1965 most British TV,s had very crude power supplies that injected 0.3A DC back into the mains, I have often wondered if this had a bad effect on the transformers.
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Energy saving light bulb more efficient take more load. Difference apparent power and the real power in any arm of the transmission. Electronics is strangely in phase with the voltage. Wild range of energy bulbs is available in market. Although prices are also highly.
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The resistance of the element in tungsten lamps is very temperature sensitive (varying by a factor of ten between hot and cold) and the thermal inertia is quite low so if one draws a graph between voltage and current it is not quite what you would expect and could well have the effect of the lamps not having unity power factor.
As a higher current flows into the lamp at the start of the cycle I think that to some extent it would look like a capacitor