A single phase 480 volt 8,000 watt heater will draw 16.67 amps, and three such heaters connected in delta to a 3 phase 480 volt supply will draw 28.9 amps.
If one phase of the supply is lost, then the current on the other two phases will be 25.00 amps, and the total power used will be 12,000 watts.
Presume that phase C is lost.
This will result in the heater connected between phases A and B working as normal and drawing 16.66 amps (8,000 watts)
The other two heaters are now in series between phases A and B , therefore each heater will receive only half its design voltage, 240 volts instead of 480.
The current drawn will therefore be halved to 8.33 amps (4,000 watts in total, or 2,000 watts in each of the 2 heaters)
The total of the full current drawn by one heater, and the half current drawn by the series pair, is of course 25 amps, 12,000 watts.
This presumes that the resistance of the heaters stays the same when worked at a lower voltage, and therefore a lower temperature, than was intended.
In practice the lower temperature would probably result in a slightly lower resistance of the series pair.
This would result in the series pair drawing slightly more than the calculated 8.33 amps. The difference should be small in practice since the alloys used for the manufacture of heating elements are chosen for a minimum variation of resistance with temperature.
If this is a question of actual practical importance, rather than theoretical calculation, then accurate measurements under working conditions would be advisable.
