How to calculate the heat load for airconditioning purposes?

I want to put airconditioner in 11m x 7m x 3m room where 34 kw heat producing equipment are installed. What should be the most appropriate tonage of airconditionar?

Answer:
Wait ! before we do a bunch of stuff lets talk about the walls and roof and floors. Don't forget those nasty doors. in a perfect world we need no A/C so.the formulae for doing heat loss through walls is Q=AxU( temp inside -temp outside)...
U= coefficient of utilization ,,,A = area of wall floor roof or whatever and temp diff= something like 88^-68^=20^(this is what you want) now this Q is in BTU and since 12K BTU = one ton you have all the facts to cool this 11x7x9 . Oh yea there are a couple of other things like air changes per hour .lighting loads human loads and of course the dreadful infiltration of air through the doors. Like i said if this was a perfect world you wouldn't need A/C anyway. Take a look at the ashrea standards for temp and humidity too those affect even the best A/C systems. Hey i know i didn't give ya an answer but i hope you see the point there is no way to cool an unknown space...however if all you have is the latent heat produced by the equipment we can offset that with 3.414btu/watt so 34000watts x 3.414=116076btu now divide that by 12000 to get tons and you have 9.673 raise that to 10 tons and you cooled the equipment....Good luck with the thermodynamics class.From the E!!

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