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How to figure size requirements for new UPS unit

We need a new UPS, but we're confused by how to figure the size. Some ratings are in kW and kVA, others in only kVA or kW. Some have the same kW and kVA ratings, but for most the numbers are different. Then we have equipment rated in watts, VA or kVA, and some with just volts and amps ? often with a range of values. We don't want to over- or under-buy. What does it all mean?

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The difference between watts (W) and volt-amperes (VA) confuses everyone, so don't feel alone. (kW and kVA simply mean 1,000 watts or 1,000 volt-amperes – the "kilo" prefix being used for larger numbers.) The basic rule of physics that "watts = volts x amps" is based on direct current (DC) circuits. Alternating current (AC), which supplies our buildings and equipment, is more efficient for the power company to deliver, but when it hits the transformers (coils of wire or inductors) inside most of our technology, it exhibits a characteristic known as reactance. Reactance reduces the useful power (watts) available from the apparent power (volt-amperes). The ratio of these two numbers is called the power factor (pf). Therefore, the actual power formula for AC circuits is "watts = volts x amps x power factor." Unfortunately, the power factor is rarely stated for most equipment, but it is always a number of 1.0 or less, and about the only thing with a 1.0 pf is a light bulb.

For years, large UPS Systems have been designed based on a pf of 0.8, which means that a 100 kVA UPS will only support 80 kW of "real" power load. Most UPSes continue to be designed that way, even though the majority of technology today has power factors of 0.95 – 0.98. (This is mainly because Europe won't allow anything with a pf less than 0.95 to be sold.) Neither the kW nor the kVA capacity of the UPS can be exceeded, but because of the higher pf numbers, it is usually the kW rating that governs today. There are, however, some UPSes on the market that are power factor corrected so that the kW and kVA ratings are the same.

The biggest problem is determining actual load, since many data hardware manufacturers still provide very inadequate or erroneous power data on their equipment. For the bigger manufacturers, investigate configurator web sites, which tend to give quite accurate information. But if you're forced to use the nameplate data on the devices, beware! This is a legality rating, and will usually give a much higher volt-ampere rating than the unit will ever draw. Consider a unit with a nameplate that reads 90 – 240 volts at 4 – 8 amps, and has a 500 watt power supply. First, the numbers are backwards. The larger amperage goes with the lower voltage. If you assume a nominal 120 volts at 8 amps, you get 960 VA. A pf of 0.95 would yield 912 watts. No power supply is that inefficient, and a power supply almost never runs at full power. Therefore, it is highly unlikely that this device will ever draw more than 500 watts of power, but if you want to be really conservative, multiply by 1.1 and figure 550 watts of input power. And don't get trapped by dual-corded equipment. The power supplies share the load and either one is supposed to be able to carry the full load. Therefore, a unit with two 500 watt supplies should still be figured as if it had only one.

This was first published in January 2005

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