Symmetrical faults

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Short Circuit Capacity(SCC) / Fault Level of Bus SCC is defined by the product of the magnitude of pre-fault voltage and the post fault current

For a solid fault, Zf= 0

Say, pre-fault voltage, Vf Fault current, |If |= |Vf |/ XT ; XT = Total impedance in the faulted

circuit

SCC= |Vf | *|If |= |Vf | * |Vf |/ XT = |Vf |2

/ XT For, Vf ≈ 1 p.u.; SCC= 1/ XT p.u. (If all the values are given in p.u)

SCC= 1/ XT MVA (If all the values are given in actual values)

Strength of a Bus A symmetrical fault occurs at bus-3. Say, the pre-fault voltage at bus-3 is 1.0 p.u. and just after the fault this voltage will be reduced to almost zero. The reduction in voltage of various buses indicate the strength of the network. So, the strength of a bus is the ability of the bus to maintain its voltage when a fault takes place at other bus Strength of a bus related to SCC. • The higher the SCC of the bus, the more it will be able to maintain its voltage in the case of fault on any other bus • If XT =0 then SCC= ∞. For this case the bus is known as Infinite Strong Bus

Symmetrical faults

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Example-8 A small generating station has a bus-bar divided into three sections. Each section is through a reactor rated at 5MVA, 0.1 p.u. A generator of 8MVA. 0.15p.u. is connected to each section of the bas bar. Determine the SCC of the bus if a 3-Φ fault takes place on one of the section of the sections of bus-bar.

Note: The bus is chosen arbitrarily. The answer will be same regardless which is selected as the faulty bus. Let the base: 8MVA in the generator circuit.

Xg= 0.15p.u.

So the reactors reactance at 8MVA, XT = 0.1*8/5 = 0.16p.u.

So the diagram becomes as follows: j0.15 and j(0.16+0.155) is parallel to each with respect to the fault bus bar.

SCC = 1/Xeq =1/j0.1016 = -j9.86 p.u. Considering only the magnitude SCC = 9.86p.u. = 9.86p.u. * 8MVA = 78.73MVA.

Example-9 Two generating station having 1200MVA and 800MVA respectively and operating at 11kV are linked with an inter-connected cable having reactance of 0.5Ω per phase. Determine SCC of each station. [Xg1=Xg2= 0.1 p.u] Let the base: 1200MVA, 11kV in the generator circuit. Xg1= 0.1p.u. Xg2= 0.1*(1200/800)=0.15p.u. Base impedance= (11)2 /1200 Cable = 0.5/[(11)2 /1200] = 4.96p.u

Symmetrical faults

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Say fault at the terminal of generator-1:

Say fault at the terminal of generator-2:

SCC= 1/Xeq =1/j0.098 = -j10.2 Considering only the magnitude; SCC = 10.2p.u. = 10.2*1200 = 12,234MVA

SCC= 1/Xeq =1/j0.085 = -j11.73 Considering only the magnitude; SCC = 11.73p.u. = 11.73*1200=14,072 MVA

Symmetrical faults

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Symmetrical faults

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