VFD Line Reactor Losses

Tom Jenkins

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In another thread Dick DV mentioned voltage drop through line reactors, which of course means there is a power loss as well.

In the wastewater industry I routinely encounter engineers whou specify both line and load side reactors, usually at the highest rating optionally available. (For example, if hte manufacturer's standard input line reactor is 3% but a 5% is available, they will spec the 5%). This is partly because they aren't paying for the equipment, and partly because they don't realize that the improvement (sometimes marginal at best) in transient protection and harmonics comes with an energy penalty.

I have never seen specific power losses for line reactors. Since these are inductors I assume VxA isn't accurate. Some suppliers have suggested that a 3% reactor will result in a power loss of 3% of the motor power, but this seems too simple. Making the analysis more complicated is the fact that much of the time the motor load is significantly less than nameplate. Finally, most VFDs claim a power factor of 95% or better - what is the impact of line reacors on the power factor?

I'm sure the answer is not simple, but any guidance would be appreciated.

Tom
 
Tom

Am responding with what little I know in part to help and to learn.

Re Power Factor. In many cases this term is used to include both power factor and harmonics.
I think
1. power factor should be power factor ie displacement between amps and volts (same ole PF that has been that way for eons). VFDs generally have PF close to 1.0.
2. Harmonics should be described and use the term Total Harmonic Distortion (or ???).

FILTERS I have come to conclusion that every time you add a goldangit there is inefficiency associated thus loss. The question in my mind what is the benefit and how does that compare to loss?

Since the filters are inductors Xl = 2 pi FL (as I know you know well) but it these are a choke type confguration I am now lost with the air gap consideration. SIMPLISTIC - what if you measured temperature on the filter surface and used that as rough / wild estimate of heat lost from its surface then get power (3.4 BTU per watt) then divide the "power loss" by power in?

Dan Bentler
 
I've never really considered the power losses in reactors but there certainly are some. I've always focussed on the filtering capability and the voltage drop across them.

It's that voltage drop part that is conveniently ignored, that is, until mysteriously, the drive runs out of voltage somewhere around 55hz and the motor becomes voltage-starved above that frequency.

I think the power issue would best be answered by MTE, TCI, or Hammond. I'd look forward to learning something in that regard. I do know that a simple drive isolation transformer is nominally 3% of load kw for losses.
 
Tom, one of my pet peeves is a spec written "all drives must include 5% line reactor in supply leads" or equivalent. What about those manufacturers that routinely include 3% reactors in their standard drives? Do they really want a total of 8% reactance.

Let's see now. We have a nice stiff 480V source so, at full rated current the reactor drop is 480 x .08 = 38.4V leaving 441.6volts for the drive input.

Oh, yes, they also spec a 5% motor lead reactor so that's 5% more drop in the motor leads which is 23 volts at full load leaving 418.6 volts max to the motor.

And then, at commissioning, everyone is mystified as to why the motor amps are so high above 52 hz when the pump curve is clearly below the motor hp!

It makes my brain hurt to just remember those startups and the uproar that always ensued.
 
This is a representative example from the MTE website:

http://www.mtecorp.com/lineinf4.html

Just picking the RL-5503 used in a 480 VAC system as a representative example, the reactor will eat 71 watts, or 0.15% of the energy sent into it at full load. I don't think that is it's continuous power dissipation capacity. That has got to assume a pure sine wave power input and represents the resistive losses. As the harmonic content increases the energy dissipated in the reactor will go up.

Keith
 
Kieth, that is excellent information.

Nowadays each kW in a 24/7 application typically costs between $500 and $1,000 a year. (Example $.09/kWh x 8760 hrs/year = $800) If you take a 400 A unit for example, then the full load losses will run to about $240 per year (.3 kW at $800/yr/kW). Not too obnoxious, but why spend it if the harmonics reduction isn't significant?

I agree with you, Dick, about the A&E firm's shelf specs. These guys seem to never have heard of the Law of Unintended Consequences, Murph's Law, or even the First Law of Thermodynamics! I frankly had not even thpought about the voltage drop, and now it seems like a much bigger potential problem than the power loss or even the harmonics!
 

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