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Discuss Working out Current (I) using Power / Voltage on Three phase systems in the Commercial Electrical Advice area at ElectriciansForums.net

D

DanBrown

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Hiya..
From the theory, to work out the amperage of a single phase system, you divide the power (P) by the voltage (V) as from the PVI triangle. Is this the case for a three phase calculation.

For example; 1 phase
3Kw motor = 3000w / 230v = 13.04 Amps

Does this same calculation apply to three phase??

example 3 phase;
1 - 3Kw motor = 3000w / 400v = 7.5 Amps or
2 - 3Kw motor = 3000w / 230v = 13.04 Amps

Which is the correct answer?

Thanks in advance..
 
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R

rumrunner

  • Thread Starter Thread Starter
  • #3
hi ya.
niether are correct,other factors need to be taken into consideration ,the calculation of 3 phase current is quite complex,but if you learn how to calcalate it correctly it will serve you best in the end ,i use very old methods to calculate anything electrical,by that i dont meen tripping times etc ,but electrical calculations ",real electrics" as i call it ,i cant put the formulars to your question on here as they are scientific and theres no way of typing the stuff in ,especially with worked examples,so what i propose is ill scan it into my computer and upload it as an image,obviously i cant do this to quickly ,but i feel if someones interested enough to want to learn it then ,i should be prepared to help,
i propose a simple three phase study/discussion group which i will atempt to lead ,ive got an interest in writing a book along these lines so you guys can be the first to read "real Electrics" wanna try it anyone ?
chris
ooops,,,bang
 
E

EasyFox

  • Thread Starter Thread Starter
  • #4
ive got an interest in writing a book along these lines so you guys can be the first to read "real Electrics" wanna try it anyone ?
chris
ooops,,,bang
Yes I'd be happy to Chris, so post away matey.

But this is from memory & been awhile since I was at college so might need checking :eek: ( chris sounds like a man who knows & will correct me I'm sure :))

I = P/3(Vl cosf)
Or
I = P/3(Vp cos f)

P= power
Vl= line voltage
Vp = phase voltage
f = angle
3 = root 3
I = current
 
R

rumrunner

  • Thread Starter Thread Starter
  • #5
hi everyone ,i think your formulars look right but will look them up and cheak, im very busy at present but will try and find time to post stuff on this subject as i have plenty of very valuble info ,which i wish to pass on ,before it gets forgotten
chris
 
U

Unregistered

  • Thread Starter Thread Starter
  • #7
3 phase power=square root of 3 X voltage X current X power factor

simplified

P = 1.732 X 415 X I X PF

Power factor varies depending on the type of load. For something with coils like a motor it could be 0.8. For a resistive load such as heaters it will be more or less 1.

To work out current, rearrange the formula so that:-

I = Power
--------------------- ( for a motor)
1.732 X 415 X 0.8


Generalise it so that for a typical 415 system with an inductive load:-

I = Power
------------------ so example 10kW 3 phase load will be 10000
575 -------- = 17.39 A
575

generalise it for a 10kW resistive load with a PF of 1


I = 10000 10000
------------------- = --------- = 13.9A
1.732 X 415 X 1 718
 
D

DanBrown

  • Thread Starter Thread Starter
  • #8
Thanks to you all for your replies,
 

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