Here is the situation. Ive got two loads on the same phase. When i measure the current at the conductor that feeds the panel the current is substantially less than the sum of the loads on this particular phase when measured at the conductors feeding the actual loads. What would cause this?
strange current readings
- Thread starter wankster
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Are your loads L-N or L-L loads? If L-L and the loads are connected to two different phases from the common phase you are measuring then you cannot simply add the (2) load currents linearly you must add them vectorally which will result in a smaller current value on the feeder. I suspect that your loads are L-L loads?
- Location
- New Jersey
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- Journeyman Electrician
What is the voltage system that these loads are on?
I dont have a link off the top of my head. Somone else may have a good one.
Esentially when you add branch circuits you have to add them vectorally since there is an angle associated with each L-L current and since these angles come into play the total magnitude will always be less than the direct some of the measured branch currents.
For example sets say your single phase load is connected L-L across A-B and your three phase motor is obviously connected to all three phases. So lets say you were measuring the total current on phase A upstream. So to add all the vectors that come into play you will have current A-B @ -330deg for your L-L load plus currents A-B and A-C for you three phase motor load. The A-B and A-C currents for the motor esentially add up to the same as an A-N current which is A-N @ 0 deg. So your upstream branch current will be A-B mag @-330 deg + A mag @ 0deg.
This is why when you add loads in a panel schedule you always add the loads in kVA values and not currents.
Electric-Light
Senior Member
What are the loads? I'm guessing they're not heating elements or light bulbs.
A pure inductor that is pulling 1kVA, a pure capacitor that is pulling 1kVA, a 1kW light bulb do not add up to 3kVA. In ideal system, it adds up to 1kVA. In reality, it adds up to slightly more than 1kVA due to I^2R losses in less than perfect cancellation of inductance and capacitance.
If you put a FULLY loaded 1hp motor and an idle 1hp motor, they do not add up fully either. The fully loaded motor acts more closer to resistive load and when you put the two loads together, the average power factor is higher. so you might have something like 1.5kW load with 2kVA.
We assume the source is a constant voltage and kVA is kV * A, so they are exactly proportional. I don't know what you mean here. Things that do not add up in current do not add up in kVA either, however watts ALWAYS add up independent of each other.
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Hid outdoor lights and a combo a/c unit.
Electric-Light
Senior Member
That explains. Whats the power factor on them?
spikes2020
Member
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- Nashville, TN
Getting back into the phasors lol~! but now your talking about current lagging your voltage... This happens when you have large inductive loads such as a AC unit's compressor motor.
But also you have an unbalanced 3 phase system so 90% of the basic rules don't apply. 2 of your phases have more load than the other thus you can see odd currents on your neutral phase and such. If you have a really bad unbalance your neutral could carry more current than the phase conductors and it would be a sad day.
PF is listed at >/= .99 on the ballasts.
a/c unit has 3 motors, none of which list PF.
spikes2020
Member
- Location
- Nashville, TN
Typical Motor Power Factors
Power (hp)-Speed (rpm)-Power Factor 1/2-load 3/4 load-full load
0 - 5 ____1800 _______ 0.72 ______ 0.82_____ 0.84
5 - 20 ___1800_______ 0.74_______ 0.84_____ 0.86
20 - 100 _1800 _______ 0.79 ______0.86_____ 0.89
100 - 300 1800 _______ 0.81______ 0.88 _____0.91
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