ref protection useful
TRANSCRIPT
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Need for REF Protection for Transformers- Dr G. Pradeep Kumar, PEARL
Restricted earth fault protection is a common protection given for most powertransformers. Better sensitivity and more coverage for the star winding are generallygiven as the reasons for providing REF protection even in situations where
differential protection is provided. Through this document we try to explain these
reasons with a numerical example.
Let us take the case of a typical step down transformer, feeding a radial system. Therating and connection details of the transformer are,
Rating : 5MVA
Ratio : 33/6.6kV
Vector group : Dyn11
Neutral Grounding Resistor : 76.2
Figure 1 shows the single line diagram of the transformer with the high voltage, low
voltage and neutral currents marked. Under normal load condition the neutralcurrent IN will be zero.
Fig. 1: Single line diagram of the transformer
Fault Current Distribution in HV and LV for Ground Fault in the LV w inding
Let us now consider a single phase to ground fault on the low voltage star winding.
The worst case of single phase to ground fault is when the fault is at the terminal ofthe star winding. The transformer leakage impedance is given as 6.73%, which
translates to 6.73%*6.62 /5 = 0.586 on the 6.6kV side. Compared to the neutral
grounding resistance value of 76.2, we can safely ignore the transformer leakage
impedance in our calculation without introducing much error.
Figure 2 shows the connection diagram for this transformer and the current
distribution for a C phase to ground fault on the LV star winding.
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Thus the maximum fault current on the low voltage side of the transformer for a
fault at the winding terminal can be calculated as,
AR
VI LVf 50
2.76
3/66003/max
===
This current on the secondary of the CT will appear as,
nomf IAI %101.0500
50secmax_
=== (1)
Fig. 2: Fault current distribution for C phase to ground fault on the star winding
As can be seen from the Figure 2, for a fault in the LV star winding, the fault currentwill flow in the faulted LV phase winding to ground and flow back from ground
through the neutral (grounded through the resistor). Thus the neutral current during
this fault will be,
AII fn 50max ==
In the HV side this current will reflect in two phases. With Dyn11 vector group the
current will appear in the A and C phases.
The three phase has its HV winding connected in Delta and LV winding in Star. Thus
to get a voltage ratio of 33/6.6kV, the individual phase turns ratio should be 33/
(6.6/3)
Therefore the fault current in the phase A and C on the HV side for a ground fault
on the LV winding will be,
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Need for REF Protection for Transformers Author : Dr. G. Pradeep Kumar
A
IIIfHAHC
77.533
3/6.650
33
3/6.6max
==
==
(2)
The maximum fault current will be when x=1, and that value is
IHC-max = - IHA-max = 5.77A
On the secondary of the HV side CT, this current will reflect as
nomHAHC IAII %77.50577.0100
77.5maxmax
=== (3)
Comparing (1) and (3), it can be seen that for the same fault, when measured on
the LV side the current is fault current is 10% of the CT rated current (which will alsobe the relay nominal current), whereas when measured on the HV side, the fault
current appears as only 5.77% of rated current of CT (and relay).
When the fault position is anywhere within the winding, the expression for the fault
current can be written as
max
3/f
LV
f xIR
xVI == (4)
Where x is the per unit fault distance from the neutral. For a terminal fault, x=1
and a fault at the neutral point, x=0.
In a transformer the ampere-turns on either winding of a phase always balances.When fault occurs in the transformers LV winding, the fault current flows only in a
portion of the LV winding. If Ns is the total LV winding turns, then the effective
turns becomes xNs. Thus we can re-write expression (2) as,
Axx
xxI
xIII ffHAHC
77.533
3/6.650
33
3/6.6
33
3/6.6
22
max
==
===
(5)
Varying the value of x from 0 to 1, we can find the fault current on the LV and HV
side (for ground faults at different locations on the star winding). The reflected faultcurrent on the HV side varies as a square of fault position x. With value of x beingless than or equal to 1, the reflected current on HV side rapidly diminishes for faults
closer to neutral.
Figure 3 shows the fault current seen by the relay when measured on the LV neutral
and HV phase CT for faults at different positions in the star LV winding. As can beseen from the comparative graph, for the same fault position, measurement on the
LV side yields more current.
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Need for REF Protection for Transformers Author : Dr. G. Pradeep Kumar
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10
0.02
0.04
0.06
0.08
0.1
0.12Secondary Fault Current Seen by RELAY
Fault position "x"
CurrentinAmps
LV
HV
Fig. 3: HV and LV fault current for different fault position on the star LV winding
Performance of Biased Differential Relay
A phase differential relay would measure the difference between the current on theHV and LV side for each phase individually. For this case there is no current on the
LV phase side, so the differential current seen by the phase differential relay will be
(we will look at the C phase element),
IPH-DIFF- C = IHC ILC = IHC-0 = IHC (6)(HV side fault current)
The bias current measured by the relay will be
IBIAS = (|IHC| + |ILC|)/2 = |IHC|/2 (7)
Let us take the differential setting in the relay as IS and the slope as m. The effectiverequired operating current can be calculated as
Ieff-op = IS+m*IBIAS
For the biased differential relay to operate, the measured differential current shouldbe more than the effective required operating current. That is,
IDIFF > IS+m*IBIAS (8)
Usually the differential relay pick-up setting is set to 10% or above (0.1A for a 1A
relay). The minimum slope is also typically 20%. Thus we can write Equation (8) as,
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Need for REF Protection for Transformers Author : Dr. G. Pradeep Kumar
IDIFF > 0.1+0.2*IBIAS (9)
Substituting the expression from Equations (6) and (7) in (9) we get,
IHC > 0.1+0.2*IHC/2
The above expression can be re-written to find the minimum value of IHC for whichthe relay will operate. Thus we get the condition to operate as,
IHC - 0.2*IHC/2 > 0.1
or
IHC > 0.1/0.9 = 0.11A
From the Figure (3) (and equation 4) we can see that the maximum fault current (for
fault on the LV winding terminal) that flows in the HV side (as seen on the secondaryof the HV CT) is only 0.0577A. This means that the phase differential relay does not
provide any protection for the LV winding ground faults.
Reducing the pick-up setting or slope is not an option as they are governed by thephase current mismatches, CT error and spill due to OLTCs.
Performance of REF Relay
The REF relay would measure the difference between the residual current measuredon the LV phase side and the LV neutral current. Again as the LV side phase currents
are absent for this case, the differential current seen by the REF relay will be,
IREF-DIFF = In (ILa+ILb+ILc) = In - (0+0+0) =In.
The REF relay has a fixed threshold setting and the relay will operate when thedifferential current flowing through it exceeds the setting.
Let us say that for the present example the LV REF relay was set to 5% (which wouldtranslate to 0.05A). For this REF relay to operate the differential current should
exceed 0.05A.
In the present example, the LV side fault current will be more than 0.05A for faults
beyond 50% of the winding (refer Figure 3, for LV current of 0.05A thecorresponding value of x is 0.5).
Here again there is a possibility of improving the amount of secondary winding
protected by selecting a different CT ratio for REF protection. Say if you select a CT
ratio of 50/1 for REF protection (three phase CTs and one neutral CT), then thesecondary fault current on the LV will increase to 10 times of that shown in Figure
(3). Thus with a setting of 5%, the REF protection will provide coverage of up to95% of the star LV winding.
Summary
This analysis brings out the importance of REF protection for a transformer. This alsoshows how a biased differential relay can at times not provide any protection against
LV earthfaults.
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