Reactive Loads

Our family's water well -- which was never very good, and has been downright pitiful the last few years -- died, and I had to drill a new one. (My 3rd, actually, but that's another story.) The new site was at the other end of the property, nearly 1,000 ft away, and wire (esp. the size needed to prevent an insane voltage drop) is expensive. I could've ordered separate PUD service for it, but that's also expensive, and I wanted the well tied into my existing power backup system (eventually to be full-time solar-powered with my other outbuildings). I wanted flexibility.

When it was finally time to turn the network on, I flipped the breaker, the step-up transformer hummed to life, I tested everything with and without loads on it, was gratified to see that turning on a power tool at the bottom only dropped the outlet voltage from 125.5 to 125.1, and was pleased with myself -- until I clamped an ammeter on the 240V feeder. Then I was scared.

I was shocked to find that my little starter network of 3 transformers (15kVA + 2x 5kVA) was pulling over 8A without any load (except a couple of nightlights) on it. Doing the math on that 2kW load (My transformer didn't specify core losses or idle current draw, but this was much more than the typical industry numbers I had found in my research!), I saw that I could be looking at $100 a month without even running anything!

Except that it wasn't really 2kW; it was 2kVA...  What's that, you say? Isn't VA (Volt-Ampere) just a fancy name for a Watt? Look at the chart above-- Volts * Amps = Watts!
That is true -- for DC, or purely resistive AC loads.  Ohm's law applies, and the current is always directly proportional to the voltage, i.e. if you graphed voltage & current together on an a dual-channel oscilloscope, one waveform would line up perfectly with the other on the X (time) axis, and by adjusting the Y scale, you could get them to line up perfectly; multiply the RMS values (RMS? Hmmm...) and you get power. A reactive (inductive or capacitive) load, however, messes with it, and the current ends up leading (if capacitive) or lagging behind (inductive) the voltage.

The reason is somewhat intuitive. Picture a capacitor placed across a battery or other DC power source. When power is first applied, the capacitor is empty, and it takes all the current it can get, presenting an effective short (and thus keeping the voltage near zero). As it charges, its resistance increases, and as the current flow slows the voltage can finally climb. Once charged, no current flows, and full open-circuit voltage is present.  An inductor, on the other hand, resists changes in current flow, so when you first connect it, the sudden rush of current through the relatively short length of wire quickly hits a brick wall (presenting an effectively open circuit) and it takes a while before it slowly (slowly being a fraction of a second) starts letting more and more current flow. Eventually, full DC current flows, overloading the supply and dropping the voltage to near zero.

With alternating current, before the inductor's magnetic field has a chance to settle down, the voltage has changed again, so it never gets a chance to get to that point of being a short circuit. The higher the frequency or bigger the inductor, the more this is so. Good thing, too, an ideal transformer primary would draw no power if nothing was attached to the secondary side, but the primary winding of my step-up transformer has a DC resistance of less than 0.1 Ohms -- a major short!

There's still a major inrush of current for several 60 Hz cycles  when power is first applied to the transformer, until the magnetic field stabilizes. This is the worst if power is applied right at the zero crossing; if the switch is turned on right at a peak, the "shock" of the instantaneous voltage rise will in effect trigger the inductance and stop the inrush quicker.  This inrush is why building codes allow circuit breakers feeding transformers to be oversized (esp. if there's appropriate overload protection on the secondary side or in the transformer itself), and why sometimes you need a special inrush-limiting circuit (for smaller loads a thermistor may be sufficient, perhaps with a timed relay to bypass it after 2 seconds) to protect everything. If you "right-size" a breaker, you'll be playing dice, with the breaker tripping as soon as you turn it on most of the time, and it taking several attempts to get it to "stick".  (I confess to personal practical experience.) This is not good for any of the equipment -- the breaker, the transformer, the loads -- so should be avoided.

A perfectly inductive load (a theoretical construct, but an unloaded transformer is hopefully pretty close!) will cause the current to lag by a full 90 degrees, so that when voltage is max, current is zero, and visa-versa. Anything times zero is zero, right? Starting to get the idea? In either of those conditions, there's virtually no power. Thus, a highly reactive circuit draws very little power (and also does very little work, including generating very little heat -- this is why using capacitors / inductors instead of resistors to limit A/C power to something can be a great idea in some cases.)

Thus, the real power (expressed in Watts) drawn by any reactive load is always less than its apparent power (expressed in VA).  The ratio of the two (real / apparent) is called the power factor, and is often written as a percentage.  The question I faced was: What was my network's power factor? It was obviously > 0 because (a.) this is real life and (b.) the transformers are all warm. How to answer this question?

Practical method #1: Watch your power bill.  Two weeks after turning on the transformer network (and a few days after activating the new well) I used the PUD's online tool to generate a graph of my daily usage for the last 2 months. I marked the two days that should see a power increase, and carefully examined the graph without finding any visible sign (except my 2 marks) that anything increased.

Practical method #2: Put a real power meter on it.  (Think Kill-a-Watt for 240V.) I don't have one, and don't want to buy one at the moment; I have to pay the well guy.

Nerdy method: Use a dual-trace oscilloscope to display EMF & current simultaneously! It may not provide an easy number, but measuring the phase shift should allow you to calculate the power factor -- it's just the cosine of the angular difference!

First off, I apologize for the poor pictures. No screenshots from a fancy digital storage oscilloscope (DSO) -- which might also do the integration of the waveforms through software and calculate the real power for you -- these are photos of a classic analog scope. I tossed a camera to KK7FNI and was busy teaching the kids how to use a scope, and didn't even think to address picture quality -- or even to turn up the intensity of the CRT to make it easier to see.

Given the complexity of the waveform (I was naively expecting a sinusoid), I didn't make ant attempt to measure real power. Clearly, though, it's more reasonable -- both idling and under load -- than what a glance at my digital ammeter first suggested.

• 480V property-wide electrical network: Success!

• Fodder for an educational ham article: Success!

• Adequate water supply: *sigh* still working on that.
  Time to yank everything out of the shaft and drill deeper....

Something I didn't consider until after writing the article: The current probe itself is an inductive clamp pickup w/ a circuit that converts amperage to voltage (1A --> 1 mV). From China, it's designed to be plugged into a digital VOM and doesn't advertise a use case involving a scope. The inductive pickup itself undoubtedly introduces some distortion (duh, inductive?) though my uninformed opinion is that it's likely much less than the much larger equipment being measured in this case. I would do well to test it sometime on a mostly-resistive load.