# Three phase heaters and amps

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.

• Who decided on the value of 25A for the thermal limit swtch? If the switch is never tripped it should live a reasonable time. If it is often operated at 96% of rated load I would not expect it to last very long if it was intended as a safety switch and not a thermostat. Jan 22 '17 at 21:48

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.

• It's all a question of where the design margins are allocated. If a manufacturer of 25A switches had them failing regularly at 24A, he probably wouldn't stay in business very long. Therefore, he's going to make sure his switches have an actual capacity in excess of 25A to begin with, in order to cover the statistical distribution from manufacturing tolerances, test equipment calibration, aging, etc. So it comes down to how much you trust the manufacturer and how much additional margin you want to build into your product. Nov 12 '13 at 18:42
• @Dave Tweed A 25 Amp switch is rated to switch 25 Ampere under specified conditions. If it doesn't, it's a bad switch. That does not mean it is rated to carry 25 Ampere under any other particular condition: that's not the way high current switches are normally described. It's very unlikely you can get specific information for any particular situation other than the exact switch specification. For all other loads, you should start with a margin. Nov 13 '13 at 8:50

Is 208V your line voltage or phase to neutral voltage?

If it is phase to neutral voltage, then each (5kW) heater has across it 360V RMS ($208 V\times\sqrt3$) and this means a current of 13.88A per phase.

If 208V is your line voltage, then the current per phase will be 24A.

Let _L = line and _Ph = phase. For star or delta connections total power = 3 * phase power. For a delta connection line voltage = Phase Voltage. Therefore : Total Power (Pt) = 3*V_L*I_Ph. Rearrange to get: I_Ph = Pt/(3*V_L) I_Ph = 15000W/(3*208) = 24.04A

So 25A is spot on.

• Welcome to EE.SE. Notice that the question was asked in 2013. Aug 1 '19 at 14:16

There is no need to reconcile the two approaches. You calculated the line current and your coworker calculated the phase current. (By the term phase current I mean the current in one of the delta-connected elements.) They are different because each line is connected to two phases. The two phase currents have equal magnitude, but are 120 degrees out of phase. The difference between the two phase currents is the line current, which is √3 times the phase current (and shifted by 30 degrees).

I like to think of Delta as sort of parallel and I like to think of wye as sort of series.

The correct location for your limit switch in a Delta configuration would be 1 limit switch each in series with only one element. that would require 3 limit switches. As a practicing component-level troubleshooting technician after 40 years I can tell you that you want 15 to 20% over rating for those switch contacts, wire and terminations, Unless you like to sell parts. Many manufacturers thrive on repeat parts sales.

Also most manufacturers would put that limit switch in the contactor control circuit where it has little current passing through it and lasts a long time, you only need one and it can be a 1 amp limit switch, that is if you're limiting temperature. If you're limiting current then you would use a current limiting relay to monitor the current in the contactor with 1 contact set used to interrupt the contactor coil.

It all depends on the application requirements.