Power Equations 3

[Gav316]

I have a problem.
We have an unbalanced 3 phase load from 2 distribution boards fed from a 60kVa ups.
Does anyone know of the equation for working out kVa of an unbalanced load from the 3 different line currents. The voltage is 415v (U.K.)

Regards
Gavin

 

[maximka]

if your 3-phase load is not ballanced and not grounded, you will need to know the impedances of each phase, and all 3 currents. if your load has a ground (3-phase, 4 -wire), sum up the currents and multiply by 415V. What is your load?

 

[piterpol]

Apparent power (KVA): S(3ph)=P+jQ=VaIa*+VbIb*+VcIc*, of course, each of these phase voltages and currents are phasors and (*) is the conjugate of...

 

[jghrist]

Or if you don't know about conjugates:

P(3ph)=Va*Ia*cos(a)+Vb*Ib*cos(b)+Vc*Ic*cos(c)

Q(3ph)=Va*Ia*sin(a)+Vb*Ib*sin(b)+Vc*Ic*sin(c)

Where a is the angle between Va and Ia, b is the angle between Vb and Ib, and c is the angle between Vc and Ic.

 

[DanDel]

For kVA, simply multiply each phase current by the appropriate phase voltage(L-N, or L-L Voltage/1.73) and sum. For circuits with similar PF values, this will be reasonably accurate.

 

[jbartos]

Suggestion: The Fortescu symmetrical component approach also may be applied. Visit

http://www.eng-tips.com/faqs.cfm?spid=238&newpid=238&sfid=314

http://grouper.ieee.org/groups/sag/IEEEP1564_00_08.doc

etc. for more info