3-ph current quest

Discussion in 'General Electronics Chat' started by DYLH, Aug 13, 2013.

  1. DYLH

    Thread Starter New Member

    Aug 13, 2013
    28
    1
    Hypothetical scenario
    380/220v 3-phase wye system.
    All loads are 1.0 PF

    3-phase load @ 6582 VA
    1-phase load (line to line on phases A B) @ 1900 VA
    1-phase load (line to ground on phase A) @ 549 VA

    What is the expected current on each phase? How did you arrive at it?
     
  2. BillB3857

    Senior Member

    Feb 28, 2009
    2,400
    348
    So what have you done or what do you think would be a way to solve your homework problem?
     
  3. DYLH

    Thread Starter New Member

    Aug 13, 2013
    28
    1
    I've been out of school for nearly 15 years... don't recall a problem like this. We tended to just take the total VA and divide by the 3 phase voltage to get a total current.. with the assumption that the phases were all relatively balanced.

    Speaking w/ people I work with, there seem to be two ways this is done to simplify for a design to figure out feeder current.. so we're trying to reconcile the the rules of thumb with the theoretical application.

    Option 1
    ---------
    Divide load per phase by the line to neutral voltage
    Phase A = 3693 / 220 = 16.8 A
    Phase B = 3144 / 220 = 14.3 A
    Phase C = 2194 / 220 = 9.97 A

    Option 2
    ---------
    Determine the load per phase

    Phase A = 3693 VA
    Phase B = 3144 VA
    Phase C = 2194 VA

    6582 common across all three
    6582 / (380 * sqrt(3)) = 10.0 A for A, B, & C

    1900 common to A & B
    1900 / 380 = 5.0 A additional for A & B

    Finally, 549 only on A
    549 / 220 = 2.5

    Phase A = 10 + 5.0 + 2.5 = 17.5
    Phase B = 10 + 5.0 = 15
    Phase C = 10

    If the total VA on the phases are the same, both give the same result.

    Option 2 is more conservative, and 'works' in the scenario where there is only a 5.0 A (380V 1-ph 1900 VA) load on the 3-ph service... the load would be 1900 / 380 = 5.0 A

    Option 1 would result in the 380V 1-ph 1900 VA load reporting as 950 / 220 = 4.3 A... that doesn't seem right. How can a 5.0A current turn into 4.3 A without a transformer?

    My guess is that neither simplification is 'correct'. What is the real theory here.. how would it work out?
     
  4. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    The problem requires some attention to the phase displacements of the various currents.

    Ignoring the current due to the A line to neutral load for the moment.....

    The total VA connected directly across the A-B phases is 6582/3 + 1900 VA or 4094 VA.

    If we take the the load A-B load current as the reference phase then the A-B current due to the 4094 VA load would be 4094/380= 10.774 Amps at 0°.

    Between B-C and C-A there are equal loadings of 2194 VA equating respectively to currents of 5.774 A at -120° and 5.774 A at -240°.

    The line A to neutral current of 549/220 or 2.5 A will lag the A-B load current by 30°. So the total current draw from line A would be

    \text{I_A=I_{AB}-I_{CA}+I_{AN}=10.774\angle{0^o} - 5.774\angle{(-240^o)}+2.5\angle{(-30^o)}=17.02\angle(-21.55^o) \ Amps}

    The other currents can be found by similar reasoning.
     
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