I wrote already that the first message calculations were wrong since I've forgot about 3600 factor. So here are better ones (as I think): Let's use 500KW windmill, since it's this kind of a mill weighs approximately 20 tonnes). Let's assume that capacity coefficient is 0.35 (i.e. the actual % of time blades will be turning at all). Since maximum power at low wind speeds can't be reached, it is hard to make up out of thin air coefficient accounting for the lower than maximum output let it be 0.1. 500,000*3600*24*0.35*0.1=552GJ/year; So it takes about 1 year of operation to smelt 20 tonnes of steel. For more precise calculations, one needs to know windspeed distribution at a given location (i.e. # of hours/year the average windspeed is within certain limits) . Wind power density is a function of windspeed, semi-theoretical efficiency of a windmill -59%. Transform windspeed distribution into a powerdensity distribution. For crude approximation mutliply this distribution by 0.59*(bladecircle area)=maximum extractable power distribution. Integrate the area under distribution = maximum possible power extracted/year for a given windmill installed in this location. I'll try to something like that in the future. If you see flaws in the last paragraph's outline let me know.