I am trying to develop a calculator that estimates the kWh saved by installing a more efficient heat pump. Is this possible? My approach is to divide the btuh capacity of the unit, by it’s EER rating to determine the W or kW, then multiplying by the # of heating and cooling hours for a specific area for one year (i.e. 34300btuh/20.40EER = 1.7kW, 1.7kW*3100hrs = 5212 kWh). For a unit that has an EER rating of 9.0, my estimated kWh is 11,814kWh. By increasing the efficiency of the unit, I have prevented 11814 – 5212 = 6602 kWh from being consumed. Is this logical? Any help is much appreciated!

]]>As above you can only convert apples to apples.

Thus 1kw-hour is 3412.142 BTU’s

or 1BTU = 0.0002930711 kw-hours

Be very careful don’t try to calculate cooling loads if you aren’t experienced as mistakes are costly. Get advise off an experienced data centre heating/cooling mechanical engineer or suppler

Dave Mc