Application Note

6 Fluke Corporation Troubleshooting power harmonics
The following are suggestions
of ways to address some typi-
cal harmonics problems. Before
taking any such measures you
should call a power quality
expert to analyze the problem
and design a plan tailored to
your specific situation.
In overloaded neutrals
In a three-phase, four-wire
system, the 60 Hz portion of the
neutral current can be minimized
by balancing the loads in each
phase. The triplen harmonic
neutral current can be reduced
by adding harmonic filters at the
load. If neither of these solutions
is practical, you can pull in extra
neutrals —ideally one neutral for
each phase—or you can install
an oversized neutral shared by
three phase conductors.
In new construction, under
carpet wiring and modular office
partitions wiring should be
specified with individual neutrals
and possibly an isolated ground
separate from the safety ground.
Derating transformers
One way to protect a trans-
former from harmonics is to limit
the amount of load placed on
it. This is called “derating” the
transformer. The most rigorous
derating method is described
in ANSI/IEEE standard C57.110-
1986. It is somewhat impractical
because it requires extensive
loss data from the transformer
manufacturer plus a complete
harmonic spectrum of the load
current.
The Computer & Business
Equipment Manufacturers Associ-
ation has recommended a second
method that involves several
straightforward measurements
that you can get with com-
monly available test equipment.
It appears to give reasonable
results for 208/120 V receptacle
transformers that supply low
frequency odd harmonics (third,
fifth, seventh) commonly gener-
ated by computers and office
machines operating from single-
phase branch circuits.
Derating factor
To determine the derating factor for the transformer, take the peak and true-
rms current measurements for the three phase conductors. If the phases are not
balanced, average the three measurements and plug that value into the follow-
ing formula:
HDF = Harmonic derating factor
= (1.414)(true-rms phase current)
(Instantaneous peak phase current)
This formula generates a value between 0 and 1.0, typically between 0.5
and 0.9. If the phase currents are purely sinusoidal (undistorted) the instanta-
neous peaks are 1.414 times the true-rms value and the derating factor is 1.0.
If that is the case no derating is required.
However, with harmonics present the transformer rating is the product of the
nameplate kVA rating times the HDF.
kVA derated = (HDF) x (kVA nameplate)
For example: 208/120 Y transformer rated at 225 kVA:
Conductor
name
True-rms
current amps
Instantaneous
peak current
Load currents were measured with
a Fluke Model 87 and an 80i-600 ac
current probe to produce the follow-
ing results:
01 410 A 804 A
02 445 A 892 A
03 435 A 828 A
I phase avg. = 410 + 445 + 435 = 430 A
3
I pk avg. = 804 + 892 + 828 = 841 A
3
HDF = (1.414) (430) = 72.3 %
841
The results indicate that with the level of harmonics present the transformer
should be derated to 72.3 % of its rating to prevent overheating.
Solving the problem