Line Current on Delta Transformer with Single Phase Loads

[ProjectDelta]

Someone posted a thought provoking scenario that I can’t seem to find a consistent answer anywhere. It is as follows: suppose you connect 3 identical single phase 480v 24kW heaters to a delta connected system, one across each phase. If you put a clamp on amp meter on one of the feeder legs, what would the current reading be. My thought is that since each heater is a single phase load, the current on each winding would be 50amps, therefore since line current is 1.73 times the winding current, you’d see 86.6amps. My instructor came up with the same answer but the person who proposed the question seems to have a different answer. Thoughts?

 

[tortuga]

I'd say 86.6 amps is correct. Another way to think about it is a 72000 W 480V load.

72000 / 836.36 = 86.6

 

[Julius Right]

Total power=3*24=72 kW=72000 W
I=72000/SQRT(3)/480=86.6 A

 

[Julius Right]

See attached sketch

 

[synchro]

As can be seen on Julius' drawing above, the components of current phasors IRY and -IBR that are in-phase with each other add together to form phasor IR, which represents the line current at terminal R.

The components of IRY and -IBR that are in quadrature (i.e., at right angles) to the output phasor cancel out and do not appear in the line current. Instead, these components represent the current that flows between the loads on B-R and R-Y through their common connection at terminal R. This behavior is analogous to the current that flows between balanced L1-N and L2-N loads on a split-phase system without contributing a current on the neutral conductor.

In the case presented by the OP the current that flows between two resistive loads at their common connection in the delta configuration is sin(30°) x 50A = 0.5 x 50A = 25A. This current can be measured by arranging the wires connecting the delta to go through a clamp meter as in the drawing below, where the clamp meter is indicated by oval #2. In this case the indicated current should be 50A because two wires which each flow 25A in the same direction are encircled by the clamp's sensor.



Such a measurement would be informative about where the "missing" current is going when you use a 1.73x factor instead of a 2x factor when load currents are summed to obtain the line current in a delta. Otherwise, you probably wouldn't have a need to do this particular measurement.

The line current measured at location #1 would be 2 x cos(30°) x 50A =2 x 0.866 x 50A = 86.6A as mentioned in the posts above.