Using the formula AMPS=(KW x 1000)/(VOLTS x 1.73)  I come up with 86 amps.

Wirenut

A rule of thumb is ~1.2 amps/kVA for 480V·3ø circuits.

   ~2.8 amps/kVA for 208V·3ø
   ~2.4 amps/kVA for 240V·3ø

Is 86 Amps the total amount of current or the current in each phase.  If this is the current in each phase the wire size would be 4AWG THHN.  

86 amps is the current on each phase,  4 awg should work

Wirenut

99NEC§424-3 “…ampacity of the branch-circuit…shall not be less than 125 percent…[but] rating or setting of overcurrent protective devices shall be permitted in §240-3.”

Doesn’t that mean 110-amp overcurrent protection and 2AWG 75°C Cu conductors?

#4 is only 85 amp wire in the 75 degree C column.  It is too small for this ciruit.
Don(resqcapt19)

He did say THHN.  I show 90 degrees C with an ampacity of 95 amps.  Of course that does not take voltage drop or the fact that you would probably but a 100 amp breaker on the circuit.  More than likely he would need but a 100 amp breaker and use #2 awg.

Wirenut

I would agree with busbar, at least for a U.S. installation.  The breaker MUST be rated for 125% of the rating of the heater for electric heating.  This would be 107.5A so a 110 A breaker is the minimum allowable size.

This, in turn, would require #2 AWG, assuming that the terminations at **both ends** of the circuit are rated for 75 deg C.  But this is not a good assumption for end-use equipment 100 amps or less.

Safest approach is to assume 60 deg C terminations and use 60C insulation ampacity for all circuits up through 100A.  This would require #1 AWG.  

This is an NEC-UL centered analysis and probably would be much different in other parts of the world.  

To support dpc's reply, there is (NRTL-) listed equipment that is labeled "use 90°C conductors with 60°C ampacity.

Wirenut,

Can you please further explain your calculation. Is the 480V a line-line or phase voltage from a delta or a star. If I assume the 480V is line-line and that the 72kW heater is a delta, I get 29A.  However, if I assume it is a star, my calculation is 29A.

Which is it?

For a delta,
72000=3(VLL)(ILL)
72000=3(480)(Ip/1.73)
Ip=29A

For a star,

72000=3(Vp)(Ip)
72000=3(480/1.73)(Ip)
72000=832(Ip)
Ip=86A

Exocet

Look in any first-year electrical text.  Shouldn't be too hard to find one.  Since this is an engineering forum for engineering professionals, I assume you are an engineering professional, and already have one or more textbooks from whatever technical training course you took.

Redtrumpet,

I know the formula, I wanted to know how wirenut assumed that it was a delta? It could have been a star? I don't work with motors and nameplates on an everyday basis so that is why I am asking?

I was assuming you wanted the current from the source, which is the line current.  I doesnt matter how the load is connected, the current for the feeder breaker will be the same.  

Although I have my PE, and have sized cable for industry and Naval ships for years, I am no expert in the NEC.  Perhaps these other contributers are better sources for the complexities of the various standards and codes.  

Wirenut

exocet,

As Wirenut says, it doesn't matter how the load is connected, delta or star, if all you are interested in is the *line* current.   

480V is the line to line voltage, nothing in how the load is connected can change that.

For any three-phase balanced load:  kVA = VLL * Iline * 1.732

dpc

This could already have been assumed or is mute however I will toss it in the pot - the wire ampacity might need to be further derated depending upon how many other motor/heating circuits are in the raceway/conduit due to heating from the multiple load conductors within the same raceway. For phase angle controlled loads I will sometimes (depending upon the controller load type) further derate the conductor ampacity because of the heating from the "harmonincs".

Before we are finished we will have you running 0000....

Wirenut