AC0458EN2.pdf

Protection

and control

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Sepam range

Sepam 100 LD

High impedance

differential protection

n Merlin Gerin n Square D n Telemecanique

presentation

contents

page

presentation

protection

specifying the sensors

surge limiter

connections

characteristics

installation

ordering information

Sepam 100 is a group of modules that may

be used:

c separately to perform a function,

c in combination with Sepam 1000

or 2000.

Each module has been designed to perform

a complete function. It includes all the

elements required, such as:

c tripping output relays,

c annunciation, settings,

c connections.

description

Sepam 100 LD comes in 4 versions:

c single-phase for restricted earth

protection,

c three-phase for busbar and machine

protection.

c 50 or 60 Hz

50 Hz single-phase: 100 LD X 51

50 Hz three-phase: 100 LD X 53

60 Hz single-phase: 100 LD X 61

60 Hz three-phase: 100 LD X 63

The front of Sepam 100 LD includes:

c 2 signal lamps:

v power “on” indicator,

v latching “trip” indicator indicating output

relay tripping,

c protection setting dial,

c “reset” button for acknowledging output

relays and the “trip” indicator.

When the button is activated, the “trip”

indicator undergoes a lamp test.

The back of Sepam 100 LD includes:

c input/output connectors:

v an 8-pin connector for toroid inputs

and remote acknowledgment,

v an 8-pin connector for “tripping” outputs

and power supply,

v a 4-pin connector for “tripping” outputs,

c a microswitch used to configure the relay

“with” or “without” latching.

Sepam 100 LD has:

c 1 or 3 current inputs with a common point

according to whether it is a single-phase

or three-phase version,

c a logic input (isolated) for remote

acknowledgment,

c “tripping” output relay with 5 contacts

(3 normally open contacts and 2 normally

closed contacts).

Sepam 100 LD operates in 5 voltage ranges

(please specify when ordering):

c 24-30 Vdc,

c 48-125 Vdc,

c 220-250 Vdc,

c 100-127 Vac,

c 220-240 Vac.

Sepam 100 LD is associated with a

stabilization plate (or 3 plates) with variable

resistance, enabling operation with 1A or 5A

transformers.

applications

Sepam 100 LD is a high impedance

differential relay.

It provides restricted earth fault, busbar

and machine protection.

advantages

c stability with respect to external faults,

c sensitivity to internal faults,

c speed (typical response time:

15 ms to 5 x Is),

c outputs with or without latching,

c local and remote acknowledgment,

c high level of immunity to electromagnetic

interference.

trip

30 35

% In

reset

sepam 100

S01 LD

Sepam 100 LD

protection

operation - setting curve

t (ms)

100

l/ls

settings (1)

setting values

setting current Is

5 to 40 % In by steps of 5 % In

40 to 80 % In by steps of 10 % In

W to 68 W

stabilizing resistor plate

Rs = 0

P = 280 W

W to 150 W

Rs = 0

P = 280 W

Rs = 0

to 270

P = 280 W

W to 470 W

Rs = 0

P = 180 W

Rs = 0

to 680

P = 180 W

accuracy / performance

setting

pickup (%)

93% ± 5%

response time

i 10 ms for I u 10 x Is

i 16 ms for I u 5 x Is

i 25 ms for I u 2 x Is

memory time

i 30 ms

(1) The dial on the front of the device is used for setting.

parameter setting

Microswitch SW1, accesible on the back of

Sepam 100 LD , is used to choose “with“ or

“without“ latching.

without latching:

SW1

with latching :

SW1

Sepam 100 LD

specifying the sensors

To ensure the stability and sensitivity of

Sepam 100 LD , the stabilization resistor

and characteristics of the CTs are calculated

as follows.

Protection sensitivity

The CTs consume magnetizing current and

the surge limiter, when installed, creates

fault current. The minimum residual primary

current detected by the protection is

therefore:

Id = n x (i m 1 + …i m p + if + is)

with

v i m 1, …i m p are read on the CT

magnetization curves at V = Rs x is.

v if is the total earth leakage current of the

surge limiter for Vs = Rs x is, i.e. the sum of

the earth leakage currents of the N limiter

units installed in parallel: if = N x ib

(see: surge limiter).

Example 2

Busbar (3-phase relay)

switchboard with 10 cubicles incomer,

feeder, without coupling

isc = 30 kA

n : 2000/5 A

R = 1.76 W

CTs situated 15 m from the relays,

connected by 2.5 mm 2 wiring (copper):

Rf = 0.02 x

CT1

CT2

protected zone

(2 x 15)

2.5

= 0.24

Number of sensors in parallel: 10 (same)

Setting: is = 50% In, is = 2.5 A

c CT knee-point voltage

isc = 30000 x 5/2000 = 75 A

(R + Rf) x isc = (1.76 + 0.24) x 75 = 150 V

Vk > 2 x 150 = 300 V

e.g.: Vk = 320 V.

c stabilizing resistor

Rf1

Rf2

Example 1

Restricted earth (single-phase relay)

Isc primary = 8 kA

n: 400/1 A

R = 2.4

100 LD

150

2.5

320

(2 x 2.5)

< Rs i

CTs situated 60 m from the relays,

connected by 6 mm 2 wiring (copper)

< Rs i 64

The resistance is adjustable from

0 to 68 W . It is set to 64

W .

n :

CT transformation ratio

(2 x 60)

Rf = 0,020 x = 0,4 W

Number of sensors in parallel: 4 (same)

Setting: Is = 20% In, is = 0.2 A

c CT knee-point voltage

isc = 8000/400 = 20 A

(R + Rf) x isc = (2.4 + 0.4) x 20 = 56 V

Vk > 2 x 56 = 112 V

e.g.: Vk = 140 V.

c stabilizing resistor

c surge limiter ?

V = 2

p :

Number of CTs

r x 320 (75 (1.76 + 0.24 + 64) - 320)

V = 3443 V > 3 kV

It’s necessary to install a surge limiter

in parallel.

c minimum primary default current

detected by the protection

for V = Rs x is = 160 V

The CT magnetization curve indicates

im = 20 mA. The leakage current curve

of the surge limiter indicates i l = 4 mA.

The number of surge limiters is:

Rf1, Rf2 :

Wiring resistance on either side

of Rs

Rf = max (Rf1, Rf2)

R1, …Rp :

CT secondary

resistances

R = max (R1, …Rp)

Rs :

Stabilizing resistor

Rl :

Surge limiter

isc :

Maximum external short-circuit

current in CT secondary winding

0.2

140

(2 x 0.2)

is :

Protection setting (A)

< Rs i

if :

Current in Rl

2000

= 2 hence

i m 1, i m p :

CT magnetizing currents

280 < Rs i 350

The resistance is adjustable from

0 to 470 W . It is set to 300

Vk1, Vkp :

CT knee-point voltages

Vk = min (Vk1,…Vkp)

Id =

(10 x 0.02 + 2 x 0.004 + 2.5) = 1083 A

W .

c surge limiter ?

V = 2

Choice of current transformers

c all the CTs must have the same

transformation ratio n.

c the knee-point voltages are chosen

so that:

Vk > 2 x (R + Rf) x isc

r x 140 (20 (2.4 + 0.4 + 300) - 140)

V = 2574 V < 3 kV

It’s not necessary to install a surge limiter

in parallel.

c minimum primary default current

detected by the protection

for V = Rs x is = 60 V

The CT magnetization curve indicates

im = 10 mA

Id = 400 (4 x 0.01 + 0.2) = 96 A

Choice of stabilizing resistor

2 x is

R + Rf

x isc < Rs i

Surge limiter

The approximate voltage developed

by a C T in the event of an internal fault is :

V = 2

r x Vk x (isc x (R + Rf + Rs) - Vk)

If the value exceeds 3 kV, it is necessary to

add an Rl surge limiter in parallel with the

relay and stabilizing resistor in order to

protect the CTs.(see: surge limiter).

Sepam 100 LD

surge limiter

If the calculations have shown that it is

necessary to install a surge limiter in parallel

with the relay and Rs to protect the CTs, it is

determined as follows.

choice

Standard references

c the surge limiters on offer consist of limiter

blocks which are independent of each other.

Each block accepts a maximum current

of 40 A rms for 1 s. By installing the blocks

in parallel, it is possible to obtain the

appropriate limiter for the application.

Examples

c Examples 1 and 2 :

Max. RMS short-circuit current in the primary

winding of a CT with a ratio of 1000: 60 kA .

v calculation of the number of blocks

per phase:

60000

(1000 x 40)

= 1.5

Hence N = 2

v modules to be ordered:

2 triple modules for a 3-phase relay.

2 single modules for a single-phase relay.

c there are two standard references:

v a single module, comprising one block,

v a triple module, comprising three

independent blocks which are aligned.

c Example 3 :

Max. RMS short-circuit current in the primary

winding of a CT with a ratio of 400: 50 kA .

Single-phase relay.

v calculation of the number of blocks

per phase:

Calculation of the number of blocks

per phase

According to i, max. RMS short-circuit

current in the secondary winding of a CT,

the number of blocks required per phase

is calculated:

50000

(400 x 40)

= 3,1

Hence N = 4

v modules to be ordered:

1 triple module + 1 single module to have

4 blocks in all.

c for a 3-phase relay, N triple modules

should be ordered,

c for a single-phase relay, N blocks,

made up of triple and single modules.

earth leakage current

c a limiter block accepts a max. steady state voltage of 325 V rms and presents an earth fault

current ib:

1000

U (V ms)

100

0.001

0.01

Ib (A rms)

0.1

Sepam 100 LD

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