Cooling water volume is determined mainly by the electrical output. One can say roughly that approx. 27% of the furnace output and the heat losses from the crucible wall will have to be dissipated. In rough terms, one can reckon on approx. 35% of the furnace out- put as the total loss performance that must be dissipated.
In the case of a 5-t furnace with 250 Hz and 3,000kW, this gives 1,050 kW or 903,000 kcal/h, that must be dissipated at a temperature difference of 27 K. This gives a cooling water volume of 33.5 m3/h (33.444 l/h). (kcal/h divided by the temperature difference gives the volume in litres/h).