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# Calculate BTUs?

Posted on 2009-03-30
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My APC battery system shows the following information.  How do I calculate how many Watts (and therefore BTUs) I'm currently using?

Utility power status

Input Voltage:        210.2        VAC
Input Frequency:        59.99        Hz
Maximum Line Voltage:        210.2        VAC
Minimum Line Voltage:        203.0        VAC

Output power status

Output Voltage:        208.2        VAC
Output Frequency:        59.98        Hz
Apparent Load Power:        058.0        % VA

Thank you!
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Question by:mgcIT
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Assisted Solution

oobayly earned 200 total points
Well, assuming that it's drawing 28.14A (Load Current), the power is:
P = VA = 210.2V * 28.14A = 5.915kW
In one hour, it will use 5.915kWhr (units) of electricity.

1kWhr = 3413BTU

So in one hour it's using 20.2 * 10^6 BTU

This has to be one hefty ups as the a big current it's drawing.
Though it may be using 72% of a load current of 28.14A, so (only) 14.54 * 10^6 BTU
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Expert Comment

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.

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LVL 18

Author Comment

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?
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Expert Comment

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

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LVL 26

Accepted Solution

PCBONEZ earned 300 total points
Power Factor doesn't have a thing to do with the actual heat energy being dissipated by the load.
Previous response is confusing Power Factor with Efficiency and though they affect each other they are not the same.

The Watt -> BTU conversion is based on Real Power (Watts) and Time, not Apparent Power, so forget about VA and use Watts or Kilowatts and Hours.

Using Load Current of 28.14 Amps and Load Voltage of 208.2 VAC tells you the load is up to 5858.748 Watts.

Using the -Efficiency- data (Load Power) you know the Load is 72% of the Total Power input.
Total Power = 8137.15 Watts [Max] -> This is Load + APC Unit power use.

Load + APC Unit power is what you need to determine how much heat is being disipated by ALL the equipment. [Unless your APC is in an area cooled by some other AC unit.]

1 kWhr = 3412.14 BTU/hr ... so you have 27,765 BTU/hr going on.

[yes] 1 Ton = 12,000 BTU/hr ... so you have 2.31 Tons of cooling to do.

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LVL 26

Expert Comment

-
This one's always forgotten.
An average person gives off about 400 Btu/hr when sedentary and about 650 Btu/hr for light work.
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LVL 18

Author Comment

hmmm...

5.915 kW
4218.3 Watts
5858.748 Watts

which one is correct?
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LVL 26

Expert Comment

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]

.
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Expert Comment

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.

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LVL 26

Expert Comment

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

--- 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.

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LVL 18

Author Closing Comment

Thank you!
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