Calculate BTUs?

ANSWER:
You are using 4218 Watts which is 14397 BTU/hr.

Your APC has a load of  (208.2 * 28.14) * 0.72 Watts = 4218.3 Watts.
One Watt is 3.413 BTU/hr.
4218.3 Watts = 14397 BTU/hr.

Watt and BTU/hr is a rate of energy, and BTU is an amount of energy.

NOTES:
You APC has a load of   (208.2 * 28.14)  = 5858.7 VA (Volt Amps).
"58.0 %VA" is the percent of the maximum VA that your APC can deliver.
(5858.7 / 0.58) is about 10000, so your APC is probably rated a maximum of 10000 VA.

VA is also called "apparent power".  Watt is real power.
 
Power Factor = Watts / VA
Watts = VA * Power Factor

"072.0  % Watts" means that the power factor is 0.72.  
This means that the actual heat energy being dissipated by the load (and what the power company is billing you for) is 72% of the VA.

The load is using (0.72 * 5858.7) = 4218.2 Watts.

Why is the power factor not 1.0?  
Because the load probably contains inductors or capacitors (causing a phase shift) or a non-linear (voltage and current waveform are different) load.  
If the load was resistive (like a space heater) then the power factor would be about 1.0.

The difference between the input and output voltage (210.2 - 208.2) = 2.0V is probably just a voltage drop in the APC electronics and connectors.

thanks for the responses.  The reason I ask is because I'm trying to figure out the size HVAC unit I will need to cool the room.  from what I've read HVACs are measured in "tons" where a 1 ton unit = 12000 BTU.  Is this correct?  (if it makes a difference I am in the US).  So from what I've read I would lead a 1.5 ton or 2 ton unit to cool the room?

Now that you know your Watts you can size your HVAC (air conditioner) from this information:

http://hvac-tqmcintl.blogspot.com/2007/12/calculating-size-of-server-room-air.html

Don't forget your other heat loads.
-
This one's always forgotten.
An average person gives off about 400 Btu/hr when sedentary and about 650 Btu/hr for light work.

hmmm...

5.915 kW
4218.3 Watts
5858.748 Watts

which one is correct?

4218.3 Watts is bogus, wrong method used.


5858.748 Watts and 5.915 kW are basically the same with round off errors from different equations.
5858.748 Watts = 5.858748kW
Round them both to 5.9kW and be happy.

Neither of them however takes into account the heat put into the room from the actual APC unit.
Those numbers are only the power that -passes through- the UPS to the load.
The power the UPS unit -dissipates- is not included.

As I said before Load + UPS is approx. 8137.15 watts [8.1kW]

.

 I am retracting my answer because I am not sure what the power factor is.  My answer of 4218 Watts was dependent on a power factor of 0.72.  
   
The load wattage can not be 8137.15 Watts because that is more that the load Volt Amps.  Power factor can not be greater than 1.0.

You could contact APC with your question.
   
-----  
   
I have not been able to find any information from APC about the Utility power status" that you show.  I have searched trying to find out what APC Load Power means and apparently "Load Power:  072.0 % Watts" is not the power factor.  Load Power seems to be "the UPS's output load as a percentage of full rated load in Watts."  
See (http://www.mikrotik.com/documentation/manual_2.5/System/UPS.html).  
   
So this is what I know:  
   
1. There is not enough information in the "Utility power status" to know the real power (watts) being used by the load.  The VA and power factor is required to know the real power.  
   
2.  The Apparent Power (VA) is 5858.7 VA.  
   
3.  The wattage of the load will be less than or equal to 5858.7 Watts.

As I said before Apparent Power is not used for this calculation.
.........

Load ratings are the LOAD ON THE UPS.
--- What goes OUT of the UPS.
--- Not what goes IN to the UPS.

OUT = Computer Power consumption (and heat from it)
IN - OUT = UPS Power consumption (and heat from it)

Both the Computer(s) and UPS unit(s) are giving off heat.
You are ignoring the heat the UPS gives off.

Thank you!