Easy First Step: 'Stiffness' and Electrical Systems

During the Q&A portion of a seminar I recently gave to a regional power quality group on PQ standards, a discussion of IEEE 1547, Standard for Interconnection and Interoperability of Distributed Energy Resources with Associated Electric Power Systems Interfaces, raised the concern on how distributed energy resources (DERs) are able to contribute to the “stiffness” of the electrical distribution system.

We had to backtrack for some of the attendees to explain what “stiffness” means in the context of electrical systems. It is one of those abstract terms we use (or misuse) and assume everyone knows what we mean. It is also one of the first items we should check in a power quality investigation, but we often overlook it because we assume the electrical supply source is stiff.

As we use the term, stiffness reflects the electrical supply’s source impedance. The stiffer it is, the less impact changing loads will have on the voltage. This is true both at the power frequency (60 hertz) and at the harmonic frequencies. Like most PQ conditions, there isn’t an exact number that makes it good or bad. Exact numbers are not what is needed for most power quality problems. It would involve a lot of complex mathematical equations to get an exact number, and too many unknown values are in the equations. But we know, when it is bad, problems get amplified and are generally more numerous. Rules of thumb based on those same two laws that we keep bringing up—Ohm’s and Kirchhoff’s Laws—along with a variation on a theme, the power superposition theorem, are all that’s needed here.

The ideal situation is that the loads use all of the power generated. Due to impedances along the path, as well as the generator’s source impedance, that can’t happen in the real world. The goal is then to maximize the transfer of power within economical limits. A good rule of thumb is 100:1—or 99 percent transfer. To do that, the source impedance (all of the impedances summed together when looking back from the point of common coupling toward the generator) are 1 percent of the value of the load impedances (all the impedances of the load’s combined looking toward them from the point of common coupling).

For example, if the equivalent source impedance is 0.1 ohm, and the equivalent load impedance is 10 ohms, the ratio is 10/0.1 or 100:1—a good, stiff system.

While the electric utility distribution system’s source impedance is often in the 0.1–0.5-ohm range, what about the DERs? Do most solar photovoltaic systems or wind turbines do as well? We know from years of data collection and analysis that many local motor-generator sets used for backup power are generally not as stiff as the local utility. In addition, they lack the “electrical inertia” a large hydro plant or nuclear power plant has. When a large load turns on while running off a M-G set, the voltage often drops as does the fundamental power frequency for a while. Is that true with solar photovoltaics and wind turbines when islanded? Even more complex is what happens when the DER is feeding the grid and the power flow changes direction back and forth as loads change or clouds and wind changes.

We need more field data to understand this better. The required calculations aren’t complex, just a couple arithmetic functions. To calculate the approximate load impedance, take the root mean square (rms) voltage and rms current at the same point in time, and divide volts (V) by amperes (A). Do this over a dozen or more data points, and find the average. If using a spreadsheet, you can do hundreds and throw away some of the outliers. That will give us the approximate load impedance.

The approximate source impedance calculation is a bit trickier. Instead of V/I, its ∆V divided by ∆I, or the change in instantaneous voltage between two points in time divided by the change in current at the same points in time. To do that, I like to use data with a significant difference in the voltage at two points in time. If the voltage is only 1/10 volt difference, there are too many factors making the accuracy of the answer suspect. But, for example, if one cycle has the voltage at 121.4V and current at 10.4A, while another has 118.5V and 35.1A, subtract 118.5 from 121.4, and subtract 10.4 from 35.1. Then divide the second result from first. (121.4-118.5) / (35.1 – 10.4) = 0.12 ohms.

Repeat that for a dozen or so points in time, using the same set of numbers for one of the points. Then, compute the average of those numbers. In this example, the ratio turned out to about 90:1, with a source impedance 0.12 ohm and a load impedance of 10.8—a stiff source by most people’s standards.

If you have access to PQ monitor data with DERs, try to do some of the calculations, and send along the results. Perhaps when I speak to the group next time, I can give them a better answer than “I don’t know.”

About the Author

Richard P. Bingham

Power Quality Columnist

Richard P. Bingham, a contributing editor for power quality, can be reached at 732.248.4393.

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