calculating kVA from 277 lighting load

[biofueler]

in a 480/277 delta circuit, should I use the 1.73 factor when calculating kVA from 277 lighting currents?

In other words when calculating KVA for a 480V motor load, the formula is V x A x 1.73 /1000

if I have 3 1A lights on three legs of a 480/277, is the kVA calculated by:

3 x 1 x 277 x 1.73 /1000 = 1.43 kVA

effectively this is the same as a 1A motor load, correct?

so why wouldnt it be

1 x 477 x 1.73 /1000 = 0.82 kVA

thanks as always

 

[biofueler]

oops, that obviously should be 480 in that last equation - friday afternoon in effect

 

[rbulsara]

Review thread238-101450 and you should be able to draw your own conclusions. Please post your understanding here.


Hint: Total 3 phase power = total of power in each phase.

The 3 phase formula is for a balanced 3 phase system and uses Line to line voltage.

For a single phase power use single phase voltage (L-N).

 

[biofueler]

ok, so after reading that, I think I got it. Lets say I have a lighting load spread out as evenly as possible amongst the three phases. The single-phase currents are 3A, 2.5A, and 3.5A

calculate kVA per phase:
3 x 277/1000 = 0.83 kVA
3.5 x 277/1000 = 0.97 kVA
2.5 x 277/1000 = 0.69 kVA

total kVA for the three-phase circuit = 0.83+0.97+0.69= 2.48 kVA

neutral current is vector sum of currents:

=sqrt((3 cos(0) + 3.5cos(120) + 2.5cos(240))^2 + (3 sin(0) + 3.5sin(120) + 2.5sin(240))^2)

=sqrt((3*1 + 3.5*(-0.5) + 2.5*(-0.5))^2 + (3*0 + 3.5*.866 + 2.5*(-.866))^2)

=sqrt((0)^2 + (.866)^2)

=0.866 A

 

[rbulsara]

Biofuelr:

Yes, good work. Very good in fact.

The check is, consider this being a balanced circuit, the current in each phase being the average of 3, 2.5 and 3.5 which is 3. The 3 phase kva = sqrt(3)*480V*3A/1000= 2.5kVA. Note 480 is the line to line V in this formula.