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I have an APC BX1500M UPS rated at 900 W and 1500 VA.

I want to attach three devices for a combined wattage of 816 Watts and a combined amperage of 8.1 Amps.

Note: This total was determined from the specs of the attached devices.

Given that the UPS has Automatic Voltage Regulation, let's assume it always outputs 120 volts under normal fluctuation conditions, so that we do not need to be concerned about possible output voltage fluctuations.

The total load falls within the Watt Rating and VA Rating. However, if we divide the Watt Rating of 900 W by the assumed output voltage of 120 V this leads me to believe that the UPS is further limited to only be able to supply 7.5 Amps total output without detriment.

Therefore, I believe I cannot hook up all three devices without come kind of drawback.

If I were to connect the three devices:

  1. How would the UPS internally deal with this issue of an 8.1 Amp load?

  2. What consequent drawbacks or other effects would occur over the short, medium, and or long term upon the UPS, it's batteries, and the appliances attached to the UPS due to this load?

  3. How would the results differ from the scenario where two of the three devices connected to the UPS for a total of 576 Max Watt draw and 4.8 Max Amp draw?

  4. Is Max Amps (the 7.5 Amps) another requirement when planning UPS that was not mentioned in the tutorials on how to "size" a UPS?

Edit: My current understanding is that the Watt Rating represents the true power available, and so it represents a robust limit that I can use to determine the capacity of a UPS, due to input A/C phase shifting. I'm not aware of how the input VA could be relied on to supply all of its amps into the output power of a UPS, so my "safe assumption" is that the amps are limited to what the Watt Rating provides. If my assumptions are wrong, explanation would be most helpful.

Update:

I talked to someone at APC and they gave me a lower number than 7.5 Amps.

They mentioned the 80 - 90 % Utilization Rule and gave me a number that was 1 full amp lower than the calculated theoretical max amperage of the UPS -- they told me 6.51 Amps.

Given that they are representing the UPS brand, I'm inclined to take this as the authoritative answer from APC.

This number will give me a roughly 4 minute UPS runtime and will ensure that no device connected will ever attempt to draw more electricity than the UPS can push out, so all the equipment should last as long as possible!

I may consider a 3rd UPS ; we'll see what I can manage

Any good thoughts on this outcome are appreciated and may also count as an answer if they help explain the situation

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  • How would the UPS deal with this? It will, but probably not for long.
    – Massimo
    Apr 3 at 21:29

3 Answers 3

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As we can see, my total load falls within the Watt Rating and VA Rating ; however, if we divide the Watt Rating of 900 W by the assumed output voltage of 120 V this leads me to believe that the UPS is further limited to only be able to supply 7.5 Amps total output without any detriments or side-effects of any kind.

Don't do that. divide the max VA instead, that's what it's for. (I get 12.5 for max amps)

If you want to know more on this subject you can look up "VAR" and "Complex Power" for the details of why. (College 100 level physics)

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Almost certainly the only difference will be run time. Loaded that heavily, your UPS will have a very short run time, considerably less than half of what it would have with the lighter load (576 watts, 4.8A, two devices), as batteries can provide less energy when loaded more heavily. That said, ratings on name plates are usually maximum rather than average power, so in normal operation your three devices should draw much less than the 8.1A / 816 watts that they would need at peak.

The simplistic calculation you're using to determine current actually is somewhat flawed by the fact that the waveform your UPS produces is not a true sine wave; instead, it's a stepped approximation, probably only two voltage levels each side of 0. The significant amounts of higher-frequency harmonics in the power waveform will cause reactive currents in the devices that will affect power calculations, making the simple E x I calculation turn into something much more complicated. I don't know, but I would suspect that this approximation for the sine wave would be the reason that output amperage limits are not explicitly listed, and generally not considered when sizing a UPS. I would actually expect that particular device to be fused at 10A and to have its driver transistors and transformer sized to match that.

As a side note: Do not put power filtering on the output side of a stepped-sine UPS. That causes fires. Surge suppression is usually all right, if unnecessary because of surge suppression built into the UPS, but the really good power filters will burn and explode when the UPS kicks in.

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How would the UPS internally deal with this issue of an 8.1 Amp load?

The maximum current load can be derived from the UPS's maximum impedance load - the VA value. 1500VA ÷ 120 V = 12.5A maximum current.

Accordingly, 8.1A is not a problem.

What consequent drawbacks or other effects would occur over the short, medium, and or long term upon the UPS, it's batteries, and the appliances attached to the UPS due to this load?

Running a UPS with 2/3 of the maximum load is likely to put significant stress on the battery on failure, in combination with a rather low runtime. I'd estimate that you'd need to replace the battery after 2-4 power failures (or more, depending on how quickly the stress is relieved by load shutdown or power return).

How would the results differ from the scenario where two of the three devices connected to the UPS for a total of 576 Max Watt draw and 4.8 Max Amp draw?

If you exceed either maximum effective load (wattage) or impedance load (VA) of the UPS then it'll shut down on power failure, without being able to provide backup power.

For the battery runtime, the effective load is the relevant figure. Unless your UPS has additional, external battery packs, I'd advise not to exceed 1/3 of the maximum load for a somewhat decent runtime and battery life.

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