REF Protection Useful

Need for REF Protection for Transformers - Dr G. Pradeep Kumar, PEARL

Restricted earth fault protection is a common protection given for most power transformers. Better sensitivity and more coverage for the star winding are generally given 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. The rating and connection details of the transformer are, Rating Ratio Vector group Neutral Grounding Resistor

: : : :

5MVA 33/6.6kV Dyn11 76.2 Ω

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 neutral current 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 winding 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 of the star winding. The transformer leakage impedance is given as 6.73%, which translates to 6.73%*6.62/5 = 0.586Ω on the 6.6kV side. Compared to the neutral grounding resistance value of 76.2Ω, 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.

©2008 Protection Engineering And Research Laboratories

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Need for REF Protection for Transformers

Author : Dr. G. Pradeep Kumar

Thus the maximum fault current on the low voltage side of the transformer for a fault at the winding terminal can be calculated as,

I f max =

V LV / 3 6600 / 3 = = 50 A 76.2 R

This current on the secondary of the CT will appear as,

I f max_ sec =

50 = 0.1A = 10% I nom 500

(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 current will 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,

I n = I f max = 50 A 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

I HC = − I HA = I f max ×

Author : Dr. G. Pradeep Kumar

6.6 / 3 33

(2)

6.6 / 3 = 50 × = 5.77 A 33 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

I HC − max = − I HA− max

5.77 = 0.0577 A = 5.77% I nom 100

(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 also be 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

If =

xV LV / 3 = xI f max R

(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 transformer’s 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,

x6.6 / 3 x6.6 / 3 = xI f max × 33 33 6.6 / 3 = x 2 50 × = x 2 5.77 A 33

I HC = − I HA = I f ×

(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 fault current on the HV side varies as a square of fault position “x”. With value of x being less 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 be seen from the comparative graph, for the same fault position, measurement on the LV side yields more current.

©2008 Protection Engineering And Research Laboratories

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Need for REF Protection for Transformers

Author : Dr. G. Pradeep Kumar

Secondary Fault Current Seen by RELAY

0.12

LV HV 0.1

Current in Amps

0.08

0.06

0.04

0.02

0

0

0.1

0.2

0.3

0.4 0.5 0.6 Fault position "x"

0.7

0.8

0.9

1

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 the HV 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 (HV side fault current)

(6)

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 effective required operating current can be calculated as Ieff-op = IS+m*IBIAS For the biased differential relay to operate, the measured differential current should be 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 which the 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 secondary of 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 the phase current mismatches, CT error and spill due to OLTCs. Performance of REF Relay The REF relay would measure the difference between the residual current measured on 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 the differential current flowing through it exceeds the setting. Let us say that for the present example the LV REF relay was set to 5% (which would translate 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 the corresponding 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 the secondary 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 to 95% of the star LV winding. Summary This analysis brings out the importance of REF protection for a transformer. This also shows how a biased differential relay can at times not provide any protection against LV earthfaults. ©2008 Protection Engineering And Research Laboratories

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