Motor Contribution to Earth Faults

I believe that is correct...

If <...an earth fault occurs between the two transformers...on the...22kv...bus...> then it is upstream of the motors. As the motors are isolated from the <...earth fault...> by the <...22kv/6.6kv transformer...>, they cannot contribute to the fault in themselves.

However, the fault will only persist for as long as the circuit protective device(s) take(s) to disconnect it; at that point, the motors lose power and stop.

"Consider a delta/Wye 66kv/22kv transformer supplying a bus then from the bus a feeder supplying another delta/eye 22kv/6.6kv transformer."

As I read this, the output of the first transformer is a Wye connection, and the input of the second transformer is a Delta connection. Is that correct?

If all 3 phases are short-circuited then no supply voltage and the EMF in stator windings are from the rotating remaining magnetic field in rotor up to demagnetizing.

If one phase is missing the rotor still rotating in the same direction but the magnetic field is split in two: one direct rotating in the same direction with the rotor and the second is rotating in opposite direction. So we have two EMF's in all three phase windings -one positive sequence and one negative sequence.

The phase s and t connected with remaining healthy phase the direct EMF is compensated by supply voltage but the opposite does not. At transformer Tb secondary phase rs- 6.6 kV- it is magnetically connected with RN [22 kV] and so, in my opinion, the current supplied by the phase rs of the induction motor contains both components.

I take an example of 66/22 kV 20 MVA 8% ,22/6.6 10 MVA 8% and a motor 6.6 kV 670 kW [66 A rated 6.4*66=422.4 LRC ].

Vbase=22 kV

The maximum EMF produced it could be: 6.6 kV.

The motor direct impedance it is: Zm6.6kV=6.6/sqrt(3)/0.4224=9.02 ohm

At 22 kV the motor impedance will be: Zm=Zm22kV=Zm6.6kV*22^2/6.6^2=100.2 ohm

The opposite motor impedance will be the same.

Neglecting the resistances and the transformer Tb Zo the motor contribution will be:

Iscm22kV=22/sqrt(3)/(2*Zm+2*ZTb)

Iscm22kV=22/sqrt(3)/(100.2+3.872)/2=0.061025 kA

If we compare this current with the short-circuit current supply from the System[Grid] then:

if we take 3000MVA SYSTEM Short-circuit power then Zsys22kV=22^2/3000=0.161 ohm

ZTa=22^2/20*8%=1.936 ohm

ZTb=22^2/10*8%=3.872 ohm

The the short-circuit current at 22 kV supplied from System it is:

IscSys=22/sqrt(3)/(0.161+1.936)=6.057 kA

Iscm/Isc=0.061/6.57*100=1%

Er, um, the OP is talking about an <...earth fault...>, not a <...short-circuit...>.

Sorry. It is here a mistake. Instead of:

Iscm22kV=22/sqrt(3)/(2*Zm+2*ZTb)

it has to be:

Iscm22kV=3*22/sqrt(3)/(2*Zm+2*ZTb)

and then:

Iscm22kV=3*22/sqrt(3)/(100.2+3.872)/2=0.183 kA