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#41
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Power Factor & kWH?
wrote in message ... : chocolatemalt wrote: : : ... If the power company is feeding you 1000W on a straight V*A basis, : and your motor is seeing just 700W of useful work on a PF*V*A basis, : there are 300W of energy that have "disappeared". I guess the question : is so simple that the answer is obvious: The energy is consumed within : the motor as non-useful heat. : : No. If your motor used 700 W with a power factor of 1, 700/120 = 5.83 amps : would flow in your wiring. If the wiring resistance were 0.1 ohms, it would : dissipate 5.83^2x0.1 = 3.4 watts of heat, and you would pay for 703.4 watts. : : With a 0.7 PF, 5.83/0.7 = 8.33 amps flow, so the wiring would dissipate : 8.33^2x0.1 = 6.9 watts, and you pay for 706.9 watts. Your only penalty : is the difference, 6.9-3.4 = 3.5 watts. The power company is less happy : because they lose more power in their wiring, but they usually don't : complain, in the case of houses. : : Nick : Yeah, I"ve since come across some info that indicates a " very bad" PF would be in the order of single-digit percentages, as in a few % or so. I was surprised it could get that low, but I guess it can. Besides, it's not jsut heat lost to the wires & system; it's the fact that the voltage and current never reach peaks at the same point in time and thus IE never represents true power without including a PF. Apparently most of the loss is in the energy required to build the flux fields, interference to it, and then offset by the collapsing field, which is creating voltage by trying to increase current, etc etc etc.. Everybody seems to be missing the phase relationship between the voltage and current thru a reactive load. |
#42
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Power Factor & kWH?
chocolatemalt writes:
I guess the question is so simple that the answer is obvious: The energy is consumed within the motor as non-useful heat. No. |
#43
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Power Factor & kWH?
chocolatemalt writes:
Perhaps with superconducting transmission lines someday (assuming it's achievable), they will no longer care if your PF deviates since their own energy expenditure will be equal to your real wattage, and reactance will be irrelevant. No, there is still an economic penalty to the current capability of the generation plant being used up for no delivered power. Capital is tied up without producing power for the customer. This may be more costly than the line losses today. It is not much different than the peak demand charges that even small commerical customers now endure, and which ought to be paid by everyone. The true cost of electric power is not just fuel and transimission, but the capital investment for peak capacity. |
#44
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Power Factor & kWH?
chocolatemalt wrote:
I understand the theory of PF, but the question is much simpler: If the power company is feeding you 1000W on a straight V*A basis, and your motor is seeing just 700W of useful work on a PF*V*A basis, there are 300W of energy that have "disappeared". I guess the question is so simple that the answer is obvious: The energy is consumed within the motor as non-useful heat. In a motor. the current creates a magnetic field which varies in magnitude and direction during the cycle. Some energy is stored in the magnetic field for part of the cycle and sent back toward the utility on another part of the cycle. The 300 watts of reactive power is from energy being stored and returned 120 times per second and is not power that disappears. Because of the properties an inductor the current is shifted and the current flow sine wave is later than the voltage sinewave and the reactive power is 90 degrees out of phase with the real power. A lightly loaded motor will have a lower PF than fully loaded motor because the real power of the fully loaded motor will be higher and will be a higher percentage of the total power. My memory of dim teachings is that the reactive power doesn't change much with loading. The higher current caused by the magnetic field interchange does cause real I squared R losses in wire as detailed in other posts. In a capacitor energy is stored in an electric field and is also stored and released 120 times per cycle. But in a capacitor the current leads the voltage. Power factor correction capacitors cancel the effects of inductance. Over correction - too much capacitance to balance the inductance can cause resonance (if I remember right), which can cause the voltage to rise as in the post by Beachcomber. This can be made worse by harmonics which can be generated by VFDs, power supplies, dimmers. bud-- |
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