Heating Elements in 3 Phase Configuration

Is this 3 phase with a wild leg ?

Otherwise it should be 240v from phase to phase....

No, the power supply is 208 3 phase Y , but I don't have a neutral line available.

...or you might have a weird setup like this....

amps equals voltage divided by resistance, or:

I = E/R

https://www.industrialcontrolsonline.com/training/online/electricity-101-basic-fundamentals

So yes you are correct....

https://www.watlow.com/resources-and-support/engineering-tools/3phase-delta-wye-calculator

https://www.jcalc.net/three-phase-power-calculator

That's exactly what we have in our facility, 208 v phase to phase and 120 v phase to neutral

Wait how did you calculate the ohms, because I get 1200 watts/ by 208 volts = 5.77 amps...? Because when you lower the voltage you only get 75% of the heat....

240 - 208 = 32 ....240/32= 7.5 5.77 * .75 = 4.33

The 2 lines drawing 4.33 amps for a total of 8.65 amps, then coincide at the single delta point, allowing for phase angle efficiency boost of 13.33%, that gives a total amperage at the delta point of each leg of 7.50 amps...

That's what I needed, the amps per phase so that I can specify my circuit protection and conductor size.

I should be OK with 10 amp fuses and 18 ga wiring?

You have to take into consideration the length of the run due to the resistance and voltage drop over distance, then the insulation rating of the wire, probably want to go with a thhn, the conductor stranded or solid and material cu,al or clad wire, also the wire run is bundled in a conduit or raceway...I should think a #12 stranded thhn copper conductor...

The wiring will only be a few feet and in open air.

Ba aware of the ambient temperature where the conductors must traverse.

You should terminate the heater elements with a high temperature wire,such as EPDM lead wire to exit the high temperature zone into a lower temperature area.

I agree with SE,no less than a # 12 THHN (194F) conductor.

Better safe than sorry;do it once,and do it right.

duplicate reply

There are various types of fuses - rewirable [fusing current such that according to local regs, a 10 amp rewirable fuse must feed cable rated 15 amp minimum with normal cartridge "single use" fuse], glass tube (poor fault breaking capacity, around 30 amp); high breaking capacity ceramic cartridge HBC/HRC to 38mm meter fuses breaking 50,000A upwards at 1000V. Also MCBs and other circuit breakers.

A one square millimetre PVC insulated (40 'C rise at 30 'C ambient) copper cable [18 AWG about 0.8 sq.mm] is rated 10.5 amp 3 wires in a conduit in a thermally insulated wall, but 15.5 amp clipped to a wall in a room.

Since you have not given any details, I agree with other posts #8,#9 suggesting #12 AWG.

The heaters may have integral high temperature tails which can be joined to standard cable - if not, the post #9 advice applies. Hi Temp tails should be used to enough length that standard terminals & cables are not overheated.

I guess a little more detail is required at this point of the discussion.

The tool I am designing is a punch press die. The heated part of the die is on the top half, the reciprocating portion of the die. I plan to terminate the cartridge heater leads in junction boxes mounted to the upper half of the die and insulated from the heated components. From there, I plan to use SJEOW cables, one for each phase, about 6 feet to the control cabinet containing the temperature controllers, relays, and internal circuit protection (10 amps). Power will come from a 20 amp plant circuit.

Because I want the die to be easily installed and removed from the press, I intend to use threaded pin and socket connectors between the SJEOW cables and the junction boxes. The one problem I am looking at now is how to securely connect the thermocouples (type J).

http://www.chem.cmu.edu/groups/bier/procedure_files/general/powercords.pdf

https://my.rs-online.com/web/p/thermocouples/3241844/

Your power/phase is 1200 watts @ 240V. @208 it's 208/240*1200=1040W. The line current is I=P/E√3 = 2.9A/Phase. It appears you forgot about the square-root of 3 in your calculation.

Rich, it's the heat!

All the resistors are energised line to line, at 208V.

Current = V/R; Watts = current x Volts = (V/R)* V = V²/R.

Power at V₁ = V₁²/R

Hence, Power at V1/(Power at V) = V₁²/V²

1200W*(208/240)² = 1200* 0.75111 = 901.33 = 900 watts to practical accuracy.

Alas, it is true but you have to account for the phase angles of the voltage sources and how the line current is affected. I did error in using 1040 Watts, I forgot to use the total power and used only a single leg. 3,120 watts should have been used and the line current should have been 3-times higher, about 8.7A. The current through each load resistor is voltage divided by the resistance and the power is still E^2/R but the line current isn't the same as the load current in a balanced 3-phase system, hence the square root of three in the calculation.

Well, I realized that I did a "face-plant" on this one. In calculating the ratio of the voltages I forgot to square the ratio. P1/P2 ∝ (E1/E2)^2. So, hopefully error free, P=3,600*(208/240)^2 = 2,704 watts. The line current, again hopefully sans errors, I=2,704/208/√3 = 7.5A. I do apologize for my fumbled thinking. I haven't benefited as much from aging as I had hoped! I guess math has the electronics equivalent of "the amount of equipment damaged is directly proportional to the operators experience"...