redsnapper wrote: A diversion, I think.
So the electrical output leaves the power plant via the grid, and the thermal output via the cooling towers. One does not include the other, and they both came from the fission reaction. Neglecting any other significant losses, the total energy generated by the fission reaction is the sum of the two (1st law of thermodynamics). Efficiency is defined as useful energy output (electrical) divided by thermal input to the cycle, so you must do it the way Brian H has described: eff=w(elec)/(w(elec)+w(waste)).
… ah, but as external observers we don’t have the temperature or the watts of the thermal input to the cycle… so doesn’t that imply that we can’t get an actual efficiency rating for the plant using external observation and the standard equation?
(The aardnanas move into flanking positions…)