Understanding IEC 60909

Understanding

IEC60909

Terminology/definitions

Fig. 1

A fault current waveform waveform illustrating AC as well as DC decay

Withstand current

= also referred to as making currents

Interrupting Interrupting duties

= also referred to as breaking shortshort-circuit duties.

AC Decrement

= a symmetrical short-circuit short-circuit current that has a time-varying change from peak to peak during the fault (i.e. the peak to peak distance varies with time)

Near to generator

short-circuits

= fault currents which have an AC decrement

Far from generator short-circuits

= fault currents exhibiting no AC decrement

I

= initial symmetrical short -circuit  at the fault location. If there was -circuit current  at

k

to-peak no AC decay, then this current would be constant in that the peak-tovalues would not change with with time. If there is only DC offset (or decay) to-peak distances remain the same but are not symmetrical then the peak-tow.r.t. the horizontal axis. With only AC decay, the peakpeak -toto-peak distances

remain symmetrical symmetrical w.r.t. the horizontal axis but they decrease with time to a steady state value. Ik

= steady state current . This is the conventional symmetrical fault current when the AC and DC decrements have disappeared. For far from

generator faults, Ik = I k. IDC

= DC Current . The exponentially decaying DC component of the fault current waveform due to the inductance of the system. It is determined by the point on the voltage waveform at which the fault occurs. Note this is Page 1 of 3

a current that can be measured, rather it is calculated. It is a

not

mathematical

model of a decaying transient inherent in the fault current

waveform. Ip

= Peak Current . This is the maximum peak of i(t) as illustrated in Fig. 1.

The theoretical maximum is 2 2 I k as already shown. It is the half-cycle peak as express ed by equation 2.13. Ib

= Breaking Current . A current duty intended for the rating of the current interrupting device for when the contacts part. For far contributions, breaking duty equals initial symmetrical current, I k. For near contributions

special multipliers need to be applied to account for the AC decrement. Ib has no DC decrement but Ib(asym) considers both AC and DC decrement. I b is a symmetrical r.m.s. current 5 5 ms (for example) down the line from I k.

IDC

Ib2

Ib(asym)

2

(1)

2

Where IDC is the DC current calculated for the same t as Ib.

Summary The short circuit current for an RL circuit in which the pre-load current can be ignored is : i( t )

iac ( t )

2 V sin( Z

idc ( t )

t

) sin(

) e

t T

A.

(2)

Where, iac ( t )

Z

R  j

2 V Z

L

sin(

) A , idc (t )

t

R2

X2

,

arctan

2 V sin( Z

L R

arctan

) e

t T

A,

X  and R

= impedance angle = initial phase displacement (or offset) angle of the source voltage, v(t) The time constant, T, for the circuit is given by, T

L R

X R

2

X s. For a threephase fault, L f R

and R are the positive sequence inductance and resistance of the system. For an earthfault, L and R

are derived from ZTOT = RTOT + jXTOT = Z1 + Z2 + Z0 .

Page 2 of 3

The maximum possible offset occurs when

i( t )

2 V Z

sin(

t

90 )

e

t T

2 V sin( Z

= ±90°, i.e. i( t )

t 90 ) e

t T

or

A.

The asymmetrical r.m.s. equivalent of the transient current, Ir.m.s., is given by 4

Ir.m.s. ( t )

I AC 1 2 e

X

f t R

4

K t I AC A . Where K( t )

1 2 e

X

f t R

. K is called the

asymmetry factor . The asymmetrical r.m.s. fault current varies from 3·IAC when t = 0 to IAC when t is large where IAC = InitSymRMS. Note that the asymmetrical  rms current does not form part of

IEC60909. The peak current, Ip, as per IEC 60909 is given by Ip

both is when X/R approaches

giving ip(max)

2

2

Ik"

1.02 0.98 e

2 Ik"  or ip(max)

Page 3 of 3

2

2 I AC .

3R X

. The limit for