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',,"v British Columbia Transmission,-7 fjo-\.(::l:> -77<11.)

Maintenance StandardProtection and ControlCORPORATION

Subject SEL 387Title: SEL 387 Relay .~.0 -$ -(,

Supplementary Training Notes Number 42.87.732-8

1. Introduction

The SEL 387 relay provides four-input variable-percentage current differential protection andovercurrent protection associated with each input. The relay includes

• three restrained and unrestrained differential elements. The restrained differentialelement has two settable percentage restraint slope characteristics and can be blocked bysecond or fifth harmonic blocking elements. The unrestrained element can be used toquickly clear high magnitude internal faults.

• eleven overcurrent elements for each of the 4 three-phase current inputs. Each group iscomprised of nine torque-controlled elements (one instantaneous, definite time, andinverse-time for phase, negative sequence, and residual currents), and two phasesegregated elements for targeting of level sensing.

• four combined overcurrent elements which operate on the sum of phase currents from tworelay terminals.

2. Overcurrent Protection

Phase overcurrent elements operate on the maximum phase current of A-,B-, or C-phase currents atany time.

Negative sequence elements respond to 312current, where 312= Ia + Ib(1I240) + Ic(1/120).

Residual current elements respond to 310current, where 310= Ia + Ib + Ie.

3. Differential Protection

3.1 Differential Element Logic

Figure 1 illustrates how the SEL 387 relay compares input winding currents IWICI throughIWIC4. These quantities are the result of input currents scaled by the value of their associated TAPsetting and then compensated for the given transformer / CT connection configuration (see 3.2below). Resultant quantities are then added

vectorially (phasor sum) to provide lOP

and

magnitudinally (scalar sum) and divided by two to provide IRT

See Figure 1, following

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~ritish ColumbiaTransmissionCORPORATION

Maintenance Standard".•.

Protection and ControlSubject: SEL 387Title: SEL 387 Relay

Supplementary Training Notes Number 42.87.732-8

E87WlIIWICI ~o--

E87W2IIW2Cl -if 0-

E87W3 ~ ..II W3C1 ~C:r--t-t+--\ru I ~ II w.ci] 1-------- IOPl

E87W4

llW4Cl ~o- ~ I II W 1C 1 I 1---------,.

YI IIW2C111-------7( ~~IIIW3C11r----~

'---~ I II W4C 1111----~

+2 t---lRTl

IOP= IIW1La + IW2L~ + IW3L8 + IW3L~ I if a = ~+ 180°and 8 = ~ = ~then lOP = IIW11 -IIW21-1 IW31-1 IW41

IRT=(IIW11 + IIW21 + IIW31 + IIW41 )/2

Figure 1 - Differential Element Operate (lOP) and Restraint (IRT) Quantities

3.2 Transformer and CT Connections (Relay Compensation)

The SEL 387 relay utilizes CT Connection Compensation Matrices for each winding to adjust fortransformer and CT connection phase shifts.

Connecting a transformer in a delta-wye or wye-delta configuration shifts secondary voltages andcurrents 30° with respect to comparable, primary quantities. Conventionally, connecting associatedCTs in an opposite configuration compensates for this phase shift, i.e. CTs are connected delta-wyefor a wye-delta connected power transformer.

See Figure 2, following

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Title:

British Columbia TransmissionCGRPORATION .

Maintenance StandardProtection and Control

SEL 387 Relay Subject: SEL 387

Supplementary Training Notes

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Number 42.87.732-8

Recommended By: PDS Issue Date: 04 Jan 2007

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~:e,"'British Columbia Transmission

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Maintenance Standard..,.

Protection and Control

\'\

Subject SEL 387Title: SEL 387 Relay

Supplementary Training Notes Number 42.87.732-8

With reference to Figure 2, note that the SEL 387 relay provides the option of using wye-wyeconnected CTs with wye-delta or delta-wye connected power transformers.

The compensation matrix required to compensate for this phase shift is specified by settingWnCTC=m,

where n = winding number, 1-4, and

m = 0, 1, 2, ... , 12, where m = the number of increments of 30 degrees that a balanced, ABCrotating system is rotated in a counter-clockwise direction.

Assume that a power transformer is connected wye-delta and that the delta connection is an "A-C"style, i.e., A0 polarity is connected to C0 non-polarity; associated CTs are connected wye-wye.

v : DACA-ph -::::=~~~l

B-ph -+----+~+--+-~

C-ph A-ph'

In Ie' Ib' la'

-C C'

\ N

<:N .A

IB 8' A'

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Maintenance StandardProtection and ControlBritish Columbia Transmission

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Title: Subject: SEL 387SEL 387 Relay

Supplementary Training Notes Number 42.87.732-8

Note that for the above example, A' lags A by 30°. Since the compensation matrices rotate phasorsin the counter-clockwise direction, aligning these vectors requires A' rotation of 30° or A rotationof 330°.

1 1 -1 0 1 1 -2 1CTC1= - 0 1 -1 CTC2= - 1 1 -2

/3 -1 0 1 3 -2 1 1

1 0 -1 1 1 -1 -1 2CTC3= - 1 0 -1 CTC4= - 2 -1 -1

/3 -1 1 0 3 -1 2 -1

1 -1 0 1 1 -2 1 1CTC5= - 1 -1 0 CTC6= - 1 -2 1

J3 0 1 -1 3 1 1 -2

1 -1 1 0 1 -1 2 -1CTC7= - 0 -1 1 CTC8= - -1 -1 2

/3 1 0 -1 3 2 -1 -1

1 0 1 -1 1 1 1 -2CTC9= - -1 0 1 CTCI0= - -2 1 1

/3 1 -1 0 3 1 -2 1

1 1 0 -1 1 2 -1 -1CTCH= - -1 1 0 CTC12 = - -1 2 -1

/3 0 -1 1 3 -1 -1 2

Figure 3: Compensation Matrices (m = 1 to 12)

As previously outlined, the compensation matrices rotate a balanced set of currents with ABC phaserotation in 30 degree increments in a counter-clockwise direction. CTC(O) creates no changes andsimply multiplies the currents by an identity matrix. Similarly, CTC(12) produces no phase shift butdoes remove zero-sequence components from the winding currents, as do all other matrices where m:;t: O.

For the above example, arbitrarily choose Winding 1 as the reference. Apply CTC(12) to Winding 1to remove zero-sequence:

1 2 -1 -1CTCI2=- -1 2 -1

3 -1 -1 2

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Maintenance StandardProtection and Control

Title: Subject: SEL 387SEL 387 Relay

Supplementary Training Notes Number: 42.87.732-8

Therefore, the relay calculates the following currents for Winding 1 using CTC(12):

Iar, = 1/3 [2Ia, - Ib, - Ic.]

Ibr, = 1/3 [-la, + 21b, - Ic.]

Icr, = 1/3 [-la, - Ib, + 2 Ic.]

where Iar, Ibr., and Icr, are internally calculated relay currents

for externally applied currents Ia, Ib, and Ic,

The 387 relay is tested on a single-phase basis to minimize the amount oftest equipment required fortesting. Therefore, ifIa, is arbitrarily applied to Winding 1 of the relay and Ib, = Ic, = 0, then

Iar, = 1/3 [2Ia, - 0 - 0]

Ibr, = 1/3 [-la, + 0 - 0]

Icr. = 1/3 [-la, - 0 + 0]

Note that Ibr, and Icr, are produced internally in the relay despite external currents Ib, = Ic, = o.

Apply compensation matrix CTC(1) to Winding 2 to effectively shift Winding 2 current in phasewith Winding 1. This results in a 30° phase shift of Winding 2 currents in the counter-clockwisedirection.

1 -1 0o 1 -1-1 0 1

1CTC1=-

Iar, = 1/ -J3 [la2 - Ib2 + 0] -./3

Ibr, = 1/ -J3 [0 + Ib2 - IcJ

and

where Iar., Ibr., and Icr, are internally calculated relay currents

for externally applied currents Ia., Ib2, and Ie2

lfphase A current (Ia2) is again arbitrarily applied to Winding 2 of the relay and Ib, = Ic, = 0, then

Iar, = 1/ -J3 [Ia, - 0 + 0]

n-, = 1/-J3 [0 + 0 - 0]

Icrj= 1/-J3 [-la2+O+O]

A comparison of the internally calculated relay currents for Winding 1 and Winding 2 shows that"A" phase current appears in each of the calculated currents, Iarj, Ibr, and Icrj, of Winding 1 butonly appears in Iar-, and Icr- of Winding 2.

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Title: Subject: SEL 387SEL 387 Relay

Supplementary Training Notes Number 42.87.732-8

Therefore, in order to "match" these calculated currents for testing purposes, "-Ia2" must beconnected to "B" phase of Winding 2. Applying the same current into "A" phase, (Ia2), and out of"B" phase, (-Ia2), of Winding 2 then generates 2Ia2 in the Iar, calculation and -Ia2 in the Ibr,calculation.

Finally, to adjust scalar quantities such that Iar, = Iar., multiply Ia, by 3/2 and multiply Ia, by J3 /2.Therefore, Iar, = Iar., Ibr, = Ibr., and Icr, = Icr., and testing may proceed.

3.3 Relay Current Taps

The relay will automatically calculate current taps, TAPn, provided setting MVA-:;t.OFF. Tapsettings are limited to a range of 0.1 (IN) to 31(IN) and the tap ratio TAPmax/TAPmin:::;7.5. IN isthe nominal current rating of the relay and is equal to either 1A or 5A.

Tap settings can be manually calculated using the formula

TAPn= (MV A) (1OOO)(C)(J3 )(VWDGn)(CTRn)

Where: n = Winding number

C = 1, ifWnCT setting = Y (wye-connected cts)

C = J3 ,ifWnCT setting = D (delta-connected cts)

MV A = Maximum power transformer capacity in MV A

VWDGn = Winding line-to-line voltage setting in KV

CTRn = Current Transformer ratio setting

3.4 Percentage Restrained Differential Element

Refer to Figure 4: Percentage Restraint Differential Characteristic

Restrained, differential operation of the relay is based on a composite characteristic made up ofthree "regions". These regions are created by four different settings:

• 087P (restrained element operating current pickup set in per unit of TAP)

• SLP 1 (restraint slope 1, set in percentage)

• IRS 1 (restraint current, slope 1 limit, set in per unit of TAP)

• SLP2 (restraint slope 2, set in percentage)

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Maintenance Standard-~Protection and Control

Title: SEL 387 Relay Subject SEL 387

Supplementary Training Notes Number 42.87.732-8

Region 1 Region 2 Region 3lOP(PerUnit)

lOP = 087P lOP = (SLP2)(IRT) - (IRS1)(SLP2-SLP1)

100

Operati

087P

IRT(Per Unit)

Figure 4 Percentage Restraint Differential Characteristic

3.4.1 087P

The 087P setting is the minimum lOP (per unit operate current) level required for operation.

Therefore, region 1 operation is given by

lOP = 087P087P

forIRT~ SLPI X 100

3.4.2 SLP1

Setting SLPI provides the first restraint slope percentage ofthe relay and defines "region 2"operation. This is a linear characteristic that can be extrapolated through the origin of the lOP vs.IRT (per unit restraint current) graph. Therefore, basic algebra results in the line equation

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'1 •••

~Varitish ColumbiaTransmission- CORPORATION

Maintenance StandardProtection and Control

Subject SEL 387Title: SEL 387 Relay

Supplementary Training Notes Number 42.87.732-8

lOP = (SLP1)(IRT)100

087P -for SLP 1 X 100 ~ IRT ~ IRS 1 setting

3.4.3 IRS1

The IRS 1 setting is the IRT limit of the slope 1 setting and is the boundary current between theSLPI characteristic (region 2) and the SLP2 characteristic (region 3).

The IRS 1 setting is not used if SLP2 = OFF

3.4.4 SLP2

The setting, SLP2, provides the second restraint slope percentage of the relay and defines "region3" operation. This is a linear characteristic offset from the origin of the lOP vs. IRT graph.Therefore, basic linear algebra results in the line equation

(SLP2)(IR T) - (IRS 1)(SLP2 - SLP 1)lOP =

100for IRT ;:: IRS 1

3.4.5 Determining Winding Input Currents for Testing

As outlined above, the typical restrained differential operating characteristic has three distinct"regions" .

. 087P• Region 1 occurs for IRT~ SLP1 X 100 and lOP = 087P

087P (SLP1)(IRT)• Region 2 occurs for -- X 100 ~ IRT ~ IRS 1 setting and lOP =

SLP1 100

R . 3 c. IR IRS d lOP (SLP2)(IRT) - (IRS1)(SLP2 - SLP1)• egion occurs lor T;:: 1an =100

Testing the differential characteristic has been arbitrarily set at four distinct points to verify thethree regions of the differential characteristic. These points have been chosen at

1) 95% of the region 1 boundary.

2) 125% ofthe region 1 boundary.

3) 95% of the IRS 1 setting.

4) 150% of the IRS 1 setting.

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Maintenance Standard..• '

Protection and Control'Title: SEL 387 Relay Subject· SEL 387

Number 42.87.732-8Supplementary Training Notes

3.4.5.1 Case 1: 95% of the Region 1 Boundary

lRT =(.95) (087P}100)SLPI

and lRT = (HWl] + IlW21)/2

therefore uwu = (.95{ 087P) (200) - IlW21SLPI

substituting for IlW21

IlWll = (.95)(087P) (200) - IlWll + 087PSLPI

= 087P (~+!)SLPI 2

and IlW21 = IlW11- 087P

= 087P (~-!)SLP1 2

lOP = 087P

and lOP = IlW11- IlW21

therefore IlW21 = IlW11- 087P

where lWl and lW2 are the per unit input currents into Windings 1 and 2, respectively.

Therefore, actual input winding currents for winding 1 (IWDG 1) and winding 2 (IWDG2) on asingle phase basis are

IWDGI = 087P c.:x TAP1 x l/A (where "A" is the W1CTC matrix scalar)SLP1 2

IWDG2 = ·087P (~-!) x TAP2 x liB (where"B" is the W2CTC matrix scalar)SLPI 2

and

3.4.5.2 Case 2: 125% of the Region 1 Boundary

lRT = (1.25)(087P)(100)SLP1

lOP = (lRT)(SLP1)100

and lOP = IlW11- IlW21

therefore IlW21= IlW11- (lRT{Sl~~l)

substituting for lR T

and lRT = (llW11 + IlW21)12

therefore IlW11= (1.25{ ~~;~) (200) - IlW21

substituting for IlW21

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r95Jan20

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British Columbia TransmissionCORPORATION

Maintenance StandardProtection and Control

Title: SEL 387 Relay Subject SEL 387

Number 42.87.732-8Supplementary Training Notes

IIW11 = (1.25)(~~;:)(200) - \IW1\+1.25(087P)

( 100 1)= 1.25(087P) SLP1 +'2 .

and IIW21= IIW11- 1.25(087P)

= 1.25(087P) (~-!)SLP12

IIW21= IIW11- 1.25(087P)

where IWt and IW2 are the per unit input currents into Windings 1 and 2, respectively.

Therefore, actual input winding currents for winding 1 (IWDG 1) and winding 2 (IWDG2) on asingle phase basis are

IWDGI = 1.25(087P) (~+!) x TAP1 x 1/A (where "A" is the W1CTC matrix scalar)SLPI 2

(100 1) .and IWDG2 = 1.25(087P) SLP1 -'2 x TAP2 x liB (where "B" IS the W2CTC matrix scalar)

3.4.5.3 Case 3: 95% of IRS1 Setting

IRT = .95(IRS1)

and IRT = (IIW11+ IIW2D/2

therefore IIW11= (2)(.95)(IRS1) -IIW21

substituting for IIW21

IIW11= (2)(.95)(IRS1) -\IW1\ + (.95)( Sl~~l) (IRS 1)

. (SLPl)= .95(IRS 1) 1+ 200

and IIW21= IIW11- (.95{Sl~~1}IRS1)

lOP = (IRT)(SLP1)100

and lOP = IIW11- IIW21

therefore IIW21 = IIW11_ (IRT{ SLP1)100

substituting for IRT

II\v21 = IIW11- (.95) (SLP1) (IRS 1)100

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Protection and ControlSubject: SEL 387Title: SEL 387 Relay

Supplementary Training Notes Number 42.87.732-8

= .95(IRS 1)(1- SLP1)200

where IW1 and IW2 are the per unit input currents into Windings 1 and 2, respectively.

Therefore, actual input winding currents for winding 1 (IWDG 1) and winding 2 (IWDG2) on asingle phase basis are

( SLP1)IWDGI = .95(IRS1) 1+ 200 x TAP 1 x l/A (where "A" is the W1CTC matrix scalar)

( SLP1)and IWDG2 = .95(IRS 1) 1- 200 x TAP2 x liB (where "B" is the W2CTC matrix scalar)

3.4.5.4 Case 4: 150% of IRS1 Setting

IRT = 1.5(IRS1)SLP2(IR T) - IRS 1(SLP2 - SLP 1)

lOP =100

and IRT = (IIW11+ IIW21)/2 and lOP = IIW11- IIW21

SLP2(IRT) - IRS1(SLP2- SLP1)therefore IIW21= IIW11- 100

substituting for IRT1.5(IRS 1)SLP2 - IRS 1(SLP2 - SLP1)

IW2 = IIW11- 100

therefore IIW11= (2)(1.5)(IRSl) -IIW21

substituting for IIW21

and

1.5(IRS 1)SLP2 - IRS 1(SLP2 - SLP 1)IW1 = (2)(1.5)(IRS1) -IIW11 + 100

_ ( S)( SLP1 + (.5)SLP2)- 1.5 IR 1 1+ (1.5)(200)

1.5(IRS1)SLP2 - IRS 1(SLP2 - SLP1)IIW21= IIW11- 100

= 1.5(IRS0(1- SLP1 + (.5)SLP2). (1.5)(200)

where IW1 and IW2 are the per unit input currents into Windings 1 and 2, respectively.

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Maintenance StandardProtection and Control

Number 42.87.732-8

Subject SEL 387Title: SEL 387 Relay

Supplementary Training Notes

Therefore, actual input winding currents for winding 1 (IWDG 1) and winding 2 (IWDG2) on asingle phase basis are

. ( SLP1 + (.S)SLP2) . .IWDGI = l.S(IRS1) 1+ (1.S)(200) x TAP1 x l/A (where "A" IS the W1CTC matnx scalar)

and

(SLP1 + (.S)SLP2) .

IWDG2 = l.S(rRS1) 1- (1.S)(200) x TAP2 x liB (where "B" IS the W2CTC matrix scalar)

3.4.6 Harmonic Blocking

The percentage-restrained differential element can be blocked by either second and/or fifthharmonic using the PCT2 and/or PCTS settings, respectively. Set these settings to "OFF" ifblocking is not required.

Transformer magnetizing inrush current contains a significant amount of second harmonic. If theratio of second harmonic to fundamental current (IF2/IF1) is greater than PCT2 setting, blockingoccurs.

Transformer overexcitation produces odd order harmonics. A fifth harmonic componentexceeding PCTS setting enables restrained-differential blocking.

Harmonic blocking can be either "Independent" or "Common". Independent Harmonic Blocking(IHBL = Y) blocks differential operation of a particular phase if the harmonic content of thatphase exceeds the pickup level. Common Harmonic Blocking (IHBL = N) blocks all differentialelements if anyone phase exceeds the pickup level.

3.5 Instantaneous Unrestrained Differential Element

The instantaneous unrestrained differential element responds to fundamental frequency only and isnot affected by SLP1, SLP2, IRS1, PCT2, PCTS, or IHBL settings.

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