FAQ

How do I figure size requirements for new UPS unit?

So you need an uninterruptible power supply (UPS) unit, but you're having trouble sizing it? Some systems are rated in kilowatts (kW) and others in kilo-volt-amperes (kVA). How do you figure it out? First, let's review some confusing terminology and facts:
  • 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) supplies our buildings and equipment. AC is more efficient for power companies to deliver, but when it hits the equipment's transformers, 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 PF 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 UPS systems continue to be designed that way, even though the majority of technology today has power factors of 0.95 – 0.98.

    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 UPS systems on the market that are PF-corrected so that the kW and kVA ratings are the same.

    Nameplate data on UPS systems

    The biggest problem when figuring size requirements for UPS units is determining their actual load. Many data hardware manufacturers still provide very inadequate or erroneous power data on their equipment. Bigger manufacturers are usually linked to or have a configurator on their websites. These tend to give quite accurate information.

    Beware of using the nameplate. This is a legality rating, and will usually give a much higher volt-ampere rating than the unit will ever draw. For example, consider a unit with a nameplate that reads 90 – 240 volts at 4 – 8 amps with 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.

    ABOUT THE AUTHOR: Robert McFarlane is a pioneer in the field of building cabling design. He has been asked to speak at countless seminars on building infrastructure for electronic communications, evolving technologies and the requirements of trading floor and data center design. Mr. McFarlane served for twelve years as President of Interport Financial, Inc., a firm specializing in designs for financial trading floors and critical data centers.

    This was first published in February 2007

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