Harmonic Current in Single Phase

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drMpower

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Hye there,

I am doing works on harmonic current analysis for my thesis. and my scope in for single phase wiring system.

I've got several data here on selected nonlinear loads which I connect on 240/415 system:

PC (1) THD = 199.2%, W = 300W, Drawn Current = 1.25A
TV set (1) THD = 167.8%, W = 300, Drawn Current = 1.25A
Flourescent Lamp THD = 8%, W = 120W, Drawn Current = 0.5A
Hair Dryer THD = 40.5%, W = 1875, Drawn Current = 7.813A

All are connected to 4 pieces of 13 Amp socket and as far as I know, those sockets connected in parallel to the MainBreaker.

Can i just confirm this, that my total Drawn Current = 1.25+1.25+0.5+7.813A = 10.813A ??

and that, the total THD of the line leading to my MainBreaker CANNOT be calculated by just adding those THD of the equipments?

I am doing consultation thesis on cable sizing.
 
drMpower said:
Can i just confirm this, that my total Drawn Current = 1.25+1.25+0.5+7.813A = 10.813A ??

and that, the total THD of the line leading to my MainBreaker CANNOT be calculated by just adding those THD of the equipments?
Click to expand...
No the total drawn current is not the same as the mathematical sum. First, the fundamental currents may have different phase angles (different displacement power factors). Second, some of the harmonic currents may cancel each other out.

You are correct, the total THD is not the sum of the individual THD's. Think of this: If all circuits were identical, the THD of each would be the same and the THD of the total would be the same as each circuit.
 
drMpower said:
PC (1) THD = 199.2%, W = 300W, Drawn Current = 1.25A
TV set (1) THD = 167.8%, W = 300, Drawn Current = 1.25A
Fluorescent Lamp THD = 8%, W = 120W, Drawn Current = 0.5A
Hair Dryer THD = 40.5%, W = 1875, Drawn Current = 7.813A.

Can i just confirm this, that my total Drawn Current = 1.25+1.25+0.5+7.813A = 10.813A ??.
Click to expand...
Not necessarily, but in this case the total will be about 10.81 amps since the harmonic load is small.
The total RMS current = sqrt( I1? + I2? + I3? + ...In?)
A true RMS amp meter will give you the correct measurement.

and that, the total THD of the line leading to my MainBreaker CANNOT be calculated by just adding those THD of the equipments?
Click to expand...
THD % fundamental = [(I2 + I3 + ...In)/I fundamental] x 100

The problem with your example is that the major portion of the load is the dryer which will have little or no harmonic content since it is all resistive except the motor. Residential loads usually have small harmonic content. Perhaps a small commercial load might be better.
 
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I am sorry Bob, I wasnt aware the reply you wrote. I was pretty occupied this weekend. was out of town for 2 days. Thanks for your reply though.

the second person answering my question said that I cannot just plainly total my breaker current by adding those each equipments' intake current. How is that possible? I mean, all the wirings are all paralleled, right? so, going back to theoretic laws of current, I can just add all of those? all i have got to take into account is the harmonic current which come with the nonlinear loads. right?

both of you are right. need to find one true RMS meter.

I am working on the cable sizing method which I take harmonic current into consideration. And the important variable is that I notice harmonic existence do increase the loss of the conductor which up until now, I cant exactly know to which degree the loss extended. And as I said before, harmonic do increase the loss of the conductor, and subsequently reduce the ampicity capacity of the cable.

Anyway, thank you so much for your replies. I really appreciate it. THank you again.

ps- since you guys helped me, surely i'll return the favour by participating in the threads. thanks~
 
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