Hvac electrical load

[LarryFine]

I still say you need to see the specs of what will be installed to size it right.

 

[gulkhan123]

Thanks Larry. you r right . but we dont have data sheet. and we need to estimate electrical load of the building where major contribution is from HVAC units, total 120,000 BTU/Hr or say 10 Tons( 12000 BTUH= 1 Ton) and we dont know SEER or EER. I think we need to use some thumb rule to find out the electrical KW required for this. What do u say?

 

[dbuckley]

A good rule of thumb is that the A/C consumes about 30% of the heat that it shifts. The ciurcuit needs to be rated higher due to duty cycle, inrush and other vagaries of real equipment, but the average consumption is about 30%. If you have several A/C units in one space, especially data centres, then as an average it works out well, but you still need to FLA of the individual units to actually build a panelboard to drive it.

So, convert BThU or tons to KW (this site is good for this), times it by 0.3, and thats the average load.

 

[LarryFine]

What do u say?

I say that, if you are compelled to guesstimate, you err on the side of more ampacity than less, and at their cost.

 

[gulkhan123]

Thanks Larry and Debuckley. Thats sound great as I checked with some others and say the same.i.e 30-33% rule of thumb for general AC application. However they say for data centre it is better using more reserved margin capacity as those units mostly produce large amount of heat . one spec calls for 100% conversion of BTUH to elect kw power.

 

[gulkhan123]

A good rule of thumb is that the A/C consumes about 30% of the heat that it shifts. The ciurcuit needs to be rated higher due to duty cycle, inrush and other vagaries of real equipment, but the average consumption is about 30%. If you have several A/C units in one space, especially data centres, then as an average it works out well, but you still need to FLA of the individual units to actually build a panelboard to drive it.

So, convert BThU or tons to KW (this site is good for this), times it by 0.3, and thats the average load.

Great explanation. Thanks Debuckley. Agreed.