calculating electrical load and capacity of circuit in computer room

I'm an IT system administrator and responsible for a 10' x 6' computer room. There is a two ton AC that cools this room and an exhaust unit
in the ceiling
of the computer room; there are no windows, and we're on the 14th floor. Right now, things are cool in the room, however, we'll be adding equipment to
already existing equipment so I was wondering what the best way to go about calculating the power and cooling requirements would be. I have one telco rack
and one server rack, each has its own UPS that's rated at 1980 Watts/ 2200 VA. No one ever really sits in the room, I go in there from time to time (swap out
tapes, check out patch panel, etc.).
For the power, at first I simply added up all the maximum Wattage specifications for each piece of equipment, I know this figure is highly inflated since the
equipment does not run at full capacity all of the time. After doing some research, I thought it may be better to use VA (to account for the power factor)
instead of Watts, so I ordered a Kill-a-Watt tool to measure the power in VA (still waiting for it to arrive). As it stands, the telco UPS shows its load to
be at least 17%, the server rack UPS is much higher, at least 67%. I'm trying to keep the loads on the UPS's to under 75%.
I also had the building electrician drop by a while back and explain to me how the electrical outlets are arranged so that I can load balance the circuits
that are in the computer room. From what I was told, I have a single phase 208V @ 40 amp circuit with #8 AWG. I believe the circuit breaker panel in the
computer room is a subpanel from a main panel in the hallway somewhere. At any rate, I also saw the following on the subpanel:
AMPS 225 VOLTS 120/208 PHASE 1 WIRE 3 TYPE NPA
The electrician said that some of the outlets are on phase A, some on phase B, each phase has 40 amps capacity and that I should not exceed 32 amps (80%) on
either phase. I've been trying to reconcile what the electrician said with what I saw on the subpanel in terms of the circuit's full capacity in amps so
that I'll know how much more equipment I can add. The main circuit breaker in the subpanel has 40 on it twice, I'm guessing 40 amps for each phase as I was
told.
I also came across the following equation in my research: 1 ton AC 12,000 BTU/hr = 3516 Watts
So basically that would mean I can go no higher than 3516 Watts X 2 tons = 7032 Watts, certainly lower than that to leave headroom.
This leaves me with three questions:
1)    How can I figure out the maximum amps of the circuit in the computer room?
2)    Could I just use the Kill-a-Watt tool to get a good measurement of my equipment's Wattage and then use the above equation to determine my BTU's?
3)    If I knew the maximum capacity of the circuit, could I use the Kill- a-Watt tool and just keep my load less than 80% of circuit's capacity?
Many thanks for any help in putting this all together,
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
On Wed, 05 Sep 2007 10:55:35 -0700, snipped-for-privacy@nitronet.pl wrote:

Has anyone talked to you about triplen harmonics. For that load you may need a #4 neutral.
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
The paragraph below doesn't appear to belong with your research.

It actually refers to the cooling capacity of an AC unit. This cooling capacity doesn't equate directly to the power rating of your equipment. What this rating tells you is that you could (in theory) start with your room at 20 degC and put a 7 kw heater in the room and the temperature would remain at 20 degC. A 7 Kw heater in a room that size would produce a huge temperature rise!! I'm guessing from your data below the AC unit is removing approx 0.5 to 1 kw of heat produced inside the room plus any heat introduced from outside the room.
From your data below it would appear that you have more than enough cooling capacity.
What's the current room temperature versus the temperature outside the room? How many outside and inside walls does the room have?
As you don't need accurate electrical profiles and you appear to have reasonable 'basic' knowledge the Kill-a-Watt device coupled with a little common sense will give you a good 'guide' as to what your equipment is consuming.
Newsey

Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
Thanks for your insight guys, I've decided to look into getting someone to come and evaluate what we've got in terms of power, AC capacity and current loads for heat and power. This way, there will be little doubt as to what we're looking at.
George
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload

To me this indicates single phase so balancing the phases is impossible

Try 225 Amps total load so 168.75amps for 75% or 35kw

Check the corresponding circuit breakers in the panel to the circuits feeding the equipment in the room. This will tell you the max capacity that you have at that moment in time. ie 2 x 40 amps = 80 amps total load on the room = 16.640kW You need to work out the standing load on the room as well

Seriously i have lots of experiance in IT/comms room design and your best bet is to get a decent engineer in to make sure that you comply with your countrys / states regulations. ie UK 16th edition BS 7671 and electricity at work regulations 1989
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload

Polytechforum.com is a website by engineers for engineers. It is not affiliated with any of manufacturers or vendors discussed here. All logos and trade names are the property of their respective owners.