Powering the IT infrastructure and your BladeSystems can be a challenge. Derek Cockerton helps to explain the physics, mathematics, and theory behind a question that came from a customer:
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We have had a question regarding Maximum Phase shift on 3 phased Power from a large customer.
Apparently the customer has been faced with a requirement from IBM and have asked us, whether we have the same requirement.
IBM have stated that at a solution with two 3 phased connections from two different sources A, B - no more than 7 degrees phase shift are allowed.
Situation is that the Customer has a number of blade enclosures with 3 phase input,
Apparently the two UPS system on the campus are not synchronized could create a difference in phase shift.
? DO WE HAVE THE SAME LIMITATION AS IBM ?
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Good stuff from Derek:
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Well, we have a constantly lagging power factor which means we are not as constrained as them – there is always a small element of phase shift as electricity is rerouted through breakers etc., but we are certainly able to tolerate the 7 Degrees IBM have stated is their limitation – we can certainly deal with more but as our power factor is typically 0.95 to 0.97 at load we HELP the customer by working within a 3-5 Deg range because the energy transfer is more effective (imagine fewer bends in a straighter pipe if that helps with a mental picture)
More importantly – because IBM’s BladeCenter power supply designs are not quite as efficient as ours means there are times when the Power Factor can become leading, not constantly lagging, and it means that IBM BladeCenter can never be connected to a Radial UPS, (see this source for Radial UPS info: http://www.wbdg.org/ccb/NAVFAC/OPER/mo201.pdf) whereas HP’s BladeSystem can be connected to both radial UPS’ and more conventional battery and generator designs without issue! Lagging power factors are almost always assisted by a power factor correction capacitor at the breaker or contactor, so not for us to supply or worry about (they are typically needed anyway), and extra protection can be added in the form of an anti-single phase relay, but ignore that piece for now.
This may help explain the relationship between two issues of phase shift and power factor for you anyway, and why they really matter!
When the need arises to correct for poor power factor in an AC power system, you probably won't have the luxury of knowing the load's exact inductance to use for your calculations. You may be fortunate enough to have an instrument called a power factor meter to tell you what the power factor is (a number between 0 and 1), and the apparent power (which can be figured by taking a voltmeter reading in volts and multiplying by an ammeter reading in amps). In less favourable circumstances you may have to use an oscilloscope to compare voltage and current waveforms, measuring phase shift in degrees and calculating power factor by the cosine of that phase shift.
Most likely, you will have access to a wattmeter for measuring true power, whose reading you can compare against a calculation of apparent power (from multiplying total voltage and total current measurements). From the values of true and apparent power, you can determine reactive power and power factor. Let's do an example problem to see how this works: (Figure below)
Wattmeter reads true power; product of voltmeter and ammeter readings yields apparent power.
First, we need to calculate the apparent power in kVA. We can do this by multiplying load voltage by load current:
As we can see, 2.308 kVA is a much larger figure than 1.5 kW, which tells us that the power factor in this circuit is rather poor (substantially less than 1). Now, we figure the power factor of this load by dividing the true power by the apparent power:
Using this value for power factor, we can draw a power triangle, and from that determine the reactive power of this load: (Figure below)
Reactive power may be calculated from true power and apparent power.
To determine the unknown (reactive power) triangle quantity, we use the Pythagorean Theorem “backwards,” given the length of the hypotenuse (apparent power) and the length of the adjacent side (true power):
If this load is an electric motor, or most any other industrial AC load, it will have a lagging (inductive) power factor, which means that we'll have to correct for it with a capacitor of appropriate size, wired in parallel. Now that we know the amount of reactive power (1.754 kVAR), we can calculate the size of capacitor needed to counteract its effects:
Rounding this answer off to 80 µF, we can place that size of capacitor in the circuit and calculate the results: (Figure below)
Parallel capacitor corrects lagging (inductive) load.
An 80 µF capacitor will have a capacitive reactance of 33.157 Ω, giving a current of 7.238 amps, and a corresponding reactive power of 1.737 kVAR (for the capacitor only). Since the capacitor's current is 180o out of phase from the load's inductive contribution to current draw, the capacitor's reactive power will directly subtract from the load's reactive power, resulting in:
This correction, of course, will not change the amount of true power consumed by the load, but it will result in a substantial reduction of apparent power, and of the total current drawn from the 240 Volt source: (Figure below)
Power triangle before and after capacitor correction.
The new apparent power can be found from the true and new reactive power values, using the standard form of the Pythagorean Theorem:
This gives a corrected power factor of (1.5kW / 1.5009 kVA), or 0.99994, and a new total current of (1.50009 kVA / 240 Volts), or 6.25 amps, a substantial improvement over the uncorrected value of 9.615 amps! This lower total current will translate to less heat losses in the circuit wiring, meaning greater system efficiency (less power wasted).
Hope this helps a little – apologies to those of you who may already have this info
A long but thorough explaination.
