Question

6. (i) For the system shown in FIGURE 4, determine the fault level for a fault on the cable at the 3.3 kV unit shown when the bus section switch is closed. The cable impedance is 90 microhms per metre. Neglect cable resistance and use a base of 10 MVA ii) What would be the effect on the fault level of replacing the unit shown in FIGURE 4 with a running motor? 8 MW08 pf G ( G ) = 20% Supply authority 100 MVA 11 kV 4MVA 8% 33 kV 20 m Unit FIG. 4

Answer 1

100 x 10 524 8 63 88 A Baa Current at nkv level 16 IDx to 524.8638 5248 6388lopu Fauult mnt(pu) 5 S24.8638 Souru impdnu (Xs) - nenJ) = XT(no.ogx 10 I m edanc

2 o 89 eg Yati 0 0016529 0268 322 9 pu . 37269严 2683229

Boar current at 3-3 kv lewel : 749 54A x3.3 xIo fault cwrent 6s20.3ey Ampere

ii) If a running motor is connected in place of the unit, then the fault level will increase at the fault point as the motor will also contribute to the fault the current. The increase in fault current depends on the transient reactance of the motor. The motor will contribute until the rotor of the motor is rotating. In case of synchronous motor, it will contribute to the fault current for a longer duration of time (typically 6-8 cycles of fault current) since it is separately excited and hence it will start to operate as a generator and is able to maintain the voltage at the fault point for a longer duration. In case of induction motor, since it is a single excited machine, it will contribute to the fault current only till the transient state of the fault (typically 1-4 cycles of fault current) since the rotating magnetic field would reduce to zero as the voltage at motor terminals drop due to fault.

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