Basler Open Delta

Published for the Basler Electric Power Systems GroupPublished for the Basler Electric Power Systems Group
#PC-59N01 • August, 2004#PC-59N01 • August, 2004
APPLICATIONAPPLICATION
NN
oo
tt
ee
ss
On an impedance grounded system, phase to groundOn an impedance grounded system, phase to ground
faults are detected by monitoring the zero faults are detected by monitoring the zero
sequencesequence
component, Vcomponent, V
00
, of the line voltages and tripping with, of the line voltages and tripping with
a 59N function (high Va 59N function (high V
00
). Two ways relays ). Two ways relays
determinedetermine
VV
00
are:are:
1)1)
Numerically calculated VNumerically calculated V
00
from phase-gndfrom phase-gnd
voltages.voltages.
A relay monitors all phase-gnd voltagesA relay monitors all phase-gnd voltages
from a set from a set
of of
3 VTs connected 3 VTs connected
Wye-gnd/Wye-Wye-gnd/Wye-
gnd. The relay calculates Vgnd. The relay calculates V
00
using the equationusing the equation
VV
00
= (1/3)( V= (1/3)( V
AGAG
+ V+ V
BGBG
+ V+ V
CGCG
).).
2)2)
Measured 3VMeasured 3V
00
from Wye-Gnd/Broken Deltafrom Wye-Gnd/Broken Delta
VTs.VTs.
A relay with a single voltage input monitorsA relay with a single voltage input monitors
the voltage across ththe voltage across th
e break in a set of 3 VTse break in a set of 3 VTs
connected Wye-gnd/broken Delta. The voltageconnected Wye-gnd/broken Delta. The voltage
across the broken delta is simply tacross the broken delta is simply t
he sum of systemhe sum of system
phase-gnd voltages, or 3Vphase-gnd voltages, or 3V
00
. The Wye side of the. The Wye side of the
The 59N and Broken Delta ApplicationsThe 59N and Broken Delta Applications
Figure 1: 59N Broken Delta SchematicFigure 1: 59N Broken Delta Schematic
Wye-Gnd/BrokeWye-Gnd/Broke
n delta VT n delta VT
can either be can either be
directlydirectly
connected to the high voltage terminals or to theconnected to the high voltage terminals or to the
secondary of a main stepdown VT.secondary of a main stepdown VT.
Method 1 is used by Method 1 is used by
numeric multifunction relays,numeric multifunction relays,
such as the BE1-951, -1051, -IPS, and -GPS relays.such as the BE1-951, -1051, -IPS, and -GPS relays.
Most applications of such relays need (or work bestMost applications of such relays need (or work best
with) line-gnd voltage anyway, so the inclusion of thewith) line-gnd voltage anyway, so the inclusion of the
VV
00
calculation and 59N protection requires nocalculation and 59N protection requires no
additional inputs to the relay and, hence, is easilyadditional inputs to the relay and, hence, is easily
included. Method 2 is a more historic included. Method 2 is a more historic
method thatmethod that
evolved before the era of numeric relays, but evolved before the era of numeric relays, but
it stillit still
sees substantial use, likely due to sees substantial use, likely due to
users beingusers being
comfortable with past practices, its ability to add acomfortable with past practices, its ability to add a
small zero sequence load to help stabilize the systemsmall zero sequence load to help stabilize the system
neutral, and because it allows a simple relay to neutral, and because it allows a simple relay to
bebe
used. The BE1-59N was designed for thisused. The BE1-59N was designed for this
application and is an easy relay to use and set.application and is an easy relay to use and set.

which therefore means:
Since a broken delta sums the 3 phase voltages,
From this line of reasoning, one should be able to see
that the worst case steady state fundamental frequency
voltage seen at the broke
n delta will be:
The VTR in the above equation is best thought of in
terms of the
winding turns ratio
, rather than voltage ratios.
This is highlighted because the question of whether one
should use V
LL
or V
LN
to calculate VTR is avoided if one
thinks in terms of winding turns ratio.
For an example, if normal system voltage was 13.8kV
LL
and 7.97kV
LG
, and a VTR of 115 were used, for a phase A
fault, the worst case broken delta voltage would be:
With a VTR of 115, the normal secondary voltage
during unfaulted conditions on each leg of the delta will be:
During a fault, secondary voltage on the two unfaulted
phases rises to 120V (=13800/115). Note that 69.3 * 3 =
207.9V, which agrees with the earlier calculations for the
maximum broken delta voltage during the fault.
If the ground fault impedance is high or the source
ground impedance is low, the voltage that will arise during a
ground fault will be something less than 3 per unit. The
calculation of what will occur can be analyzed using a set of
simultaneous equations. The circuit in Figure 2 is analyzed
in a Mathcad™ (Revision 7) document, "59N_R#.MCD"
found in the "Downloads" section of the Basler Electric
web site, www.basler.com.
VT Voltage Rating
It is important to note that, during a ground fault, two
of the VTs must withstand and reproduce full line-line
voltage. It would b
e an error to use
a VT rated only for
line-ground voltage. For instance, in the wye-broken delta
system described above, the normal system voltage is
13.8kV
LL
and 7.97kV
LG
, and secondary voltage is 69.3
under normal operating conditions but rising to 120V
during a fault. The VT in this case needs to be rated for
13.8kV/120V.
Besides the BE1-59N, Basler’s BE1-951, -1051, -IPS,
and -GPS numeric relays can also monitor a broken delta
voltage via their V
X
input. Using numeric relays gives all the
advantages thereof, such as detailed event reports and
programmable logic. The BE1-951, -1051, and -IPS relays
also have the ability to perform the 67N function and have
the user selectable option of determining direction to a fault
by a number of means, including comparing the phase
relationship of V
0
from the broken delta to I
0
or I
G
, so that
the BE1-951, -1051, and -IPS relays can effectively act as a
more modern way to do the work of the classic solid state
BE1-67N.
In method 2, it is common to place a resistor in the
broken delta as shown in Figures 1 and 2. One rationale for
the resistance is that the resistance stabilizes the measured
voltage. It does
this by a) redu
cing the risk of
ferroresonance, and b) allowing the VT to act as a very
small ground bank. The ground bank effect will help hold
the system neutral voltage closer to ground, helping to
prevent small leakage impedance to ground from causing
high neutral shifts. The ground bank effect also provides a
means for bleeding off the capacitive voltage buildup
associated with arcing ground faults on high impedance
grounded systems. The ability of the VT and resistor to act
as a ground bank and hence stabilize the neutral is fairly
limited and will be covered further below.
The issue of whether an unloaded VT is at risk of
entering into ferroresonance is difficult to answer. One
circuit for ferroresonance can be seen if the delta presents
a path for a phase to ground voltage on one phase to
energize another phase via the delta, which in turn has a
capacitance to ground, creating an LC network where the L
is saturable. If the circuit is lossy due to the resistance in the
circuit, then resonance of the LC network is less likely, but
on the other hand, leaving the delta completely open
removes the resonant path, so an argument could be made
to leave the resistance out entirely if this is the circuit of
concern. Past practice by several generations of engineers
seems to indicate that including the resistance is advisable.
This application note will not try to analyze the matter any
further than indicated below.
Voltages During a Ground Fault
Refer to the phasor diagrams in Figure 2, Page 3.
During ideal normal operation with no ground fault or line
to ground current:
During a ground fault, virtually all of the faulted
phase's voltage is impressed upon the neutral impedance.
For a phase A to ground fault, Van=0, and the voltage
across the neutral resistor is essentially the negative of the
Phase A to neutral voltage. Mathematically:
which means:

Figure 2: Phasor diagrams
during line-to-ground fault
Resistance Selection
To obtain the maximum capability of the resistor to
dampen system transients and dampen ferroresonant
circuits, a typical approach to sizing the resistor is to put in
one that can handle all the power that the VTs can supply
during a full neutral offset. Fully loading the VT will cause
some level of voltage drop and error in the secondary, but
if the VT is dedicated to the broken delta configuration, the
voltage drop that will re
sult will not cause any i
ll effects.
However, if the "aux VT" approach in Figure 1 is used,
then one might need to investigate if the fully loaded aux
VT will pull down the main VT secondary voltage and
affect other relaying in the system. Ignoring the voltage
drop issue, sizing of the resistor to bring the VTs to a high
loading level results in two approaches to sizing the
resistor:
a) size the resistor so that the amps drawn are equal to
the continuous current rating of the transformer bank, or
b) if the fault will be cleared quickly, size the resistor so
that power in the resistor is equal to the full 3 phase VA of
the transformer bank, which, as seen below, will
overcurrent the bank by a factor of sqrt(3).
a) Base resistance on VT continuous current rating
This approach needs to be used if the fault might be in
place for an extended period. Assume the previous
example where V
SEC
is 69.3 normally but rises to 120V
during a fault. Assume a VT rated at 500 VA per phase and
13.8kV/120V.
During a ground fault, ignoring voltage drop in the VT
when fully loaded, the resistor will see 207.9V per the
previous example. To limit current to 4.167A,
The power in the resistor will be:

Since it is anticipated that the fault might be held for an
extended period, the resistor must be sized to handle this
heat dissipation continuously.
b) Base resistance on VT 3 phase VA rating
This approach will overload the VTs and is appropri-
ate only if the fault will be cleared before the overload can
affect the VTs. Assume, again, a VT rated at 500VA per
phase, for a total of 1500VA for all three phases. Also
ignore the voltage drop in the fully loaded VT and assume
that the full 207.9V from the previous calculations is seen
across the resistor. The resistance required to load the VT
bank to 1500VA will be:
The current drawn during the fault will be:
This current is sqrt(3) times the continuous current
rating of the VT, calculated previously. The power in the
resistor will be:
In this case, since the VT is overloaded, it is likely that
the intent is to clear the fault relatively quickly. Since the
fault will be cleared quickly, the short time rating of the
resistor might be considered, allowing a smaller resistor to
be used.
Resistor Short Time Power Rating
The power calculated in the above examples is rather
large. However, if the fault will be cleared quickly, then the
short time rating of the resistor can be used to allow a
smaller resistor watt rating. A "rule of thumb" is that a wire-
wound power resistor can handle a short time overload of:
PC-59N01 (08/04)
where t
s
= time in seconds. Solving for the required
continuous rating:
In the above equation, if t is less than 1s, use t=1, and
if t>25s, then it may be best to consider using a fully rated
resistor. If a 28.8
Ω
resistor is to be used as in the above
example, but all faults will be cleared in 5 seconds, then by
the above equation and using a x2 safety margin, a 300W
resistor might be used (1500*(5/50)* (2) = 300).
Ground Bank Effect of a VT
The ability of a VT to act as a small grounding bank,
and therefore limit neutral voltage shift, is fairly weak, but it
might be seen in cases where ground impedances are very
high. As a point of comparison, the 28.8
Ω
resistor in the
previous example reflected to the primary by VTR
2
(115
2
) is
381k
Ω
. This might help stabilize a substation bus or a short
transmission line with only minor phase to ground capaci-
tance or leakage resistance. Directly related to this concept,
the current flowing in the delta will flow through the faulted
phase VT, which will tend to push some current from the
VT back into the ground fault. If the ground fault imped-
ance is large, there can be some tendency for the VT to
sustain voltage on the faulted phase. The previously
mentioned Mathcad document that analyzes Figure 2 may
help you to determine if the ground bank effect might be
noticable in a ground fault in your system.
For More Information
For further information, visit the download section of
our website at
www.basler.com
to access product docu-
mentation on the BE1-59N, -951, -1051, -IPS, and -GPS,
the Mathcad file referenced on Page 2 of this Application
Note, and Application Notes on other topics.
To discuss your specific application, consult Basler at
the factory at [phone].
http://www.basler.com
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