Quote Originally Posted by Evap0rat0r View Post
Formulas:

Airflow (CFM) = Air Velocity (fpm) * Cross-sectional Area of Opening (ft sq).

You need to ensure to keep units consistent or account for them using appropriate conversion factors.

1 ton ref = 12000 btu/hr = 3.516 kW.

The specs on your unit is typically the nominal capacity as determined by the manufacturer. To determine the theoretical performance of the system given the temp and humidity conditions you mentioned, you would need to do the following:

Measure the cross-sectional opening of the duct and convert your result to feet squared; multiply this value by your fpm air velocity reading; the product will give you your air flow in CFM.

You know your room supply temp and the room RH%. You need to find out the entering (outside) air temp and RH%. Now, it gets a bit confusing for a novice: you need to consult your psychrometric chart and plot the state point for the OA conditions and the SA conditions. Get the delta h value between these two state points. Then, apply the formula for total cooling capacity:

Qtotal = 4.5 X CFM X delta h

you can convert your answer to kW and then divide this result by the nominal provided by the manufacturer, multiplied by 100, you get the percentage utilization of the system capacity.

Of course, this is a fairly simplistic view of the processes taking place here, but you asked for some formulas to curb your enthusiam so there you go.
Brilliant stuff.This is what i am after something straight forward and quick calculations.Actually i use a software to apply the measurments.The folmulaes i need just to understand what is what.Btw can you upload a psycrometric example of what you explained.When i start adding these values i am not 100% sure whether the state point is for the room temp. or outdoor temp. and of course is it room RH% i put or ambient RH%.
Thanx