I am a mechanical engineer with basic understanding of electricity. I am programming an app for ordering 3 phase heaters, and I have an argument to resolve with a coworker. Lets say I have a 208V 3-phase heater (delta connection), with a heating element on each leg L1, L2, L3. The total KW of the heater is 15 KW. My calculation for amps in each leg is 15KW*1000/[208V*1.73] = 41.7 amps. On each leg there is a thermal limit switch with 25 amp rating. Since I1 = I2 = I3 = 41.7 amps in each leg, I need to split into two branches to divide the amps into two because of the 25 amp limit switch.
A coworker is arguing the following: Each element is producing 15KW/3 = 5 KW per element. Treating each leg separately, one can treat it as a 'single phase' circuit with 208V across the element. So amps per leg is 5 KW *1000/208V = 24 amps. So no need to branch for the thermal limit switch.
Obviously one can't come up with two different answers. My thought is that even though on average each element is producing 5 KW of heat, instantaneously it might be producing more than that, as long as all three elements together are producing 15 KW at any given instant. So instantaneous amps would be 24 amps*1.73 = 41.7 amps (to reconcile the two approaches). With a Wye connection my coworker's argument would work because the voltage across each element is 120V. But intuitively I don't know why it works for Wye but not for delta connections.
Best Answer
You co-worker is right in calculating the current per element. Each element is apparently driven with 208 V accross it and is drawing 5 kW of power. 5 kW / 208 V = 24 A, as he said. (By the way, the extra 1000 in your equations is incorrect. 5 kW * 1000 is 5 MW, not 5000 W. Also, the correct units is kW (killo-Watts), not KW (Kelvin-Watts)).
You are right in that the instantaneous peak current is higher, but for sine waves it is higher by the square root of 2 from the RMS, not the square root of 3. 24 A RMS therefore has peak currents of 34 A.
However, the Amp rating of these thermal limits switches are almost certainly RMS, not peak, check the datasheet. So in theory, these 25 A limit switches can handle your 24 A load.
However, in the end I agree with you. Having a switch rated for 25 A regularly carry 24 A loads for sustained periods of time is asking for trouble. This is just bad engineering. What are you going to do, go to the customer when the inevitable failures occur and wave the datasheet saying "But it says here it should have worked"? Use beefier limit switches or split each 5 kW load into two 2.5 kW loads, which will draw only 12 A each and will be fine with the existing limit switches.