Load Loss Factor is expressed as heat, yes, but the way it gets there is important.
When AC electric current flows in a purely resistive load, say, a light bulb, as the voltage rises on the half-cycle, the current also begins to flow at the same time, and in proportion to the rising voltage. The voltage and current are said to be "in-phase" because the rise and fall together.
This isn't true for capacitive or inductive loads; these devices are "springy" (if you will): the current lags the voltage or leads the voltage depending whether it is a net capacitive or inductive load.
Keeping to your question: What is Load Loss Factor?, the load loss factor is an expression of how much power is lost to the internal "springyness" of the circuit. It is possible to create a circuit that does nothing but generate heat, that is, it just oscillates back and forth in your circuit and does no useful work, and so is to be avoided. Kinda like pounding sand...
One reason you personally would want to avoid it is because your power meter charges you for the total energy consumed, even that portion lost to heat due to Load Loss Factor. Load Loss Factor is also called phase-shift (the phase relationship between voltage and current) or Power Factor.
Fluorescent lights, motors are inductive devices and cause the current to lag the voltage; this can be corrected with capacitor banks which have the opposite effect. If you have a whole bunch of motors/inductive devices, sometimes the power company will REQUIRE you to install capacitor banks to correct the Power Factor.
This subject is either advanced high school or college electrical course material. Below is a link explaining Load Load Factor in some detail. Examples and references to standards of the electric formula are given in the second link.
2006-07-06 13:52:27
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answer #1
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answered by jimdempster 4
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Loss Load Factor
2016-12-18 16:02:14
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answer #2
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answered by Anonymous
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You don't mention what type of motor this is, which is important information. Has the load dropped? Power factor will drop with load. You call it a loss which I don't understand. Depending on your power consumption and agreement with you supplier, you may not be penalized for power factor and even if you are, its KVAr's you pay for so you have to factor in the power.
2016-03-16 21:49:22
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answer #3
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answered by Virginia 4
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i am assuming you are trying to minimize losses in a utility distribution system...when i was a system engineer at a utility..the quickest payback items were 1)correcting power factor as close to unity as possible with switched capacitor banks 2)have a policy of sizing distribution transfomers properly in relation to the load (i.e not oversized) to miminimize NL (no load) losses and not undersized too much either to minimize LL (load losses). 3)size distribution feeder conductors to miminize both NL and LL losses. The correct point at which you achieve max savings has a good bit to do with your %rate of capital as well as your cost of power both KW and KWH rates.
2016-04-08 21:07:35
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answer #4
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answered by ? 4
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Heat.
2006-07-06 12:32:28
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answer #5
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answered by normy in garden city 6
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