Single Phase Drum Heater

Wire them in delta (line to line).

The heaters are rated 3000W at 240V.

3000/240 = 12.5A

240/12.5 = 19.2Ω

When wired on 208V,

208/19.2 ≈ 10.83A

208 x 10.83 ≈ 2253.3W

Total output of all 3 heaters is 6760W.

(If wired in star, 6760/3 ≈ 2253.3W, which will not be enough.)

Thank for the quick answers,

I am going to wire the 3 Heaters in delta configuration

I am wondering the total amount of current by line is about 10.83A? if weird in delta.

With the Voltage and the current per line I am going to make the calculation for size of wire I need and the fuse rating.

Thank in advance

In delta configuration the phase to phase voltage from 208V AC will be 360 Volt. with 19.2 ohms heater resistance the current flow will be 18.75 amps. that is 6750 watts. kindly refer to manufacturers catalog and find out whether the heaters can handle, before connecting them phase to phase.

I think you have it backwards. A delta is a parallel connection. Voltage is the reference in a parallel circuit so coil voltage and phase to phase voltage are the same. Current on the other-hand is increased by 1.73 because it has 2 paths that you add together as vectors.

I think the OP specified that the system is 208V line-to-line. If so, this would be a 208-120 3Φ system with delta voltage of 208 and star voltage of 120.

The heater collectively have 6 leads; if all 6 are fused individually, I think 12 amps is the next size up from 10.83. Wire #14 AWG.

If only the 3 lines are fused, amps per line = 10.83 x √3 = 19.2; next size is 20 amps, and wire should be #12 AWG. In either case, if it is wired to terminals on the heaters, it must be wire with high-temperature insulation. If the heaters already have high-temperature wire leads, the connecting wires probably won't need to have such a high temperature rating.

Be sure to check local code requirements and wire manufacturer's data.

TORNADO: WELL DESCRIBED