a. For a bolted balanced three-phase fault at location C, calculate the magnitude and phase of the...

Question

a. For a bolted balanced three-phase fault at location C, calculate the magnitude and phase of the transmission line currents flowing away from bus B and line currents flowing towards the transformer for both transformers. Use the given line to neutral voltage at bus 1 as the reference. b. For single phase fault at location C, calculate the magnitude and phase of the transmission line currents flowing away from bus B and line currents flowing towards the transformer for both transformers. Use the given line to neutral voltage at bus 1 as the reference. Assume phase `a' is faulted fault resistance is RI c. For phase to phase fault at location C, calculate the magnitude and phase of the transmission line currents flowing away from bus B and line currents flowing towards the transformer for both transformers. Use the given line to neutral voltage at bus 1 as the reference. Assume phases `b' and c' are faulted and fault resistance is zero. d. For phase to phase to ground fault at location C, calculate the magnitude and phase of the transmission line currents flowing away from bus B and line currents flowing towards the transformer for both transformers. Use the given line to neutral voltage at bus 1 as the reference. Assume phases `b' and c' are faulted and fault resistance is zero.
Expressions for currents i1, i2, and i3 and values for the resistance will be given via the StudyDesk. Piecewise linear magnetising curves for the CTs and the piecewise linear voltage-current characteristic of resistance R will also be provided.
a. Prove that even in the presence of CT saturation of one of the relays, the protection system is secure against a through fault. Use the given values of resistances, line currents under the close-in through fault, CT ratios and relay setting. b. Explain what happens if R was chosen to be linear? Use the given values of resistances, line currents under the bus fault, CT ratios and relay setting. c. Explain what happens if during normal operation (that is if there are no primary side faults), the secondary terminals of one of the CTs are shorted. Use the given values of resistances, line currents at maximum demand, CT ratios and relay setting.
Use the data in the row corresponding to the last two digits of your USQ student number. Those are in the first column of the table (labelled YOUR ID).
Note : Bus A nominal voltage =110 KV ; Bus B nominal voltage = 11kV
TPFL= 3 phase fault level (kA) at bus A; Purely reactive fault current assumed
SPFL= single phase to ground fault level (kA) at bus A; Purely reactive fault current assumed
MVA1 = transformer 1 MVA rating; MVA2= transformer 2 MVA rating;
X₁ = pu leakage reactance of transformer 1; X₂ = pu leakage reactance of transformer 2.

TPFL (kA)	SPFL (kA)	MVA1	MVA2	X₁ (%)	X2 (%)	R1 (ohm)	R2 (ohm)	Z⁺ (ohm)	Z⁰ (ohm)	R_f (ohm)
8	8	80	80	6.0	4.5	2	2	5+j6	5+j11	50

005_e78sajo-5smw2lvp.doc

Robert · Accepted Answer

Solution 
a) TPFL= 3 phase fault level (kA) at bus A = 8kA 
Base Voltage = 110 kV at Bus Bar A side 
Base Voltage = 11 kV at Bus bar B side 
Base MVA = 80
Base Current =  80000 KVA / 110 kV * √3 = 419.9 A = 420 A
TPFL in p.u = 8000/420 = 19 p.u
Considering that the Generators are supplying power to Bus bar A, The TPFL = 1/Zp.u
19 = 1/Zp.u    => Zp.u = 1/19 = 0.052 pu (This is positive sequence impedance of circuit at the left 
of Busbar A
X1 = 0.06 pu and X2 = 0.045 pu
 For 3 Phase fault only positive sequence impedance will appear hence converting positive 
sequence impedance of transmission line in pu 
=  5+j6 * (80000/(1000*11*11) = 3.3 + 4j pu
Total Xpu =  0.052j + (0.06 II 0.045) + 4j = 0.052j +0.026j+ 4j  = 4.07j  pu with angle 90
o 
Total R pu = 3.3 pu 
Total Zpu  = 5.23 pu and 51
o phase difference 
Fault Current = base Voltage / Net Impedance = 1/5.23  = 0.19 pu and 51o phase difference 
Base Current at busbar B = 4200 A => Fault Current = 803 A. 
Current Flows through Busbar B = 803 A and 51o phase difference 
Current through T1 at the side of bus bar B= 344 A 
Z1 
X1 
X2 
Z+ 
Current through T2 = 459 A at the side of Bus bar B 
b) Negative sequence network will remain same as positive sequence network
Z+   =  Z-   = 5+6J = 3.3 +4j  pu 
Zo  = 5+11j  = 3.3 + 7.3 j pu
Rf  = 50 ohm = 33 pu 
R1 = R2 = 2 ohm = 1.3 pu
single phase to ground fault level (kA) at bus A = 8kA 
Hence Z1 = 0.052j pu
Net positive sequence reactance = (0.06 II 0.045)j + 4j = 0.026+4 = 4.026j 
Net positive sequence resistance = 3.3 pu 
Net positive sequence impedance = 3.3+4.03j = 5.2 pu with 51o phase difference 
 = negative sequence impedance
Total zero sequence reactance = (0.06 II 0.045)j + 7.3j = 7.326j 
Net zero sequence resistance = 3.3 pu 
Net zero sequence impedance = 3.3+7.33j = 8 pu with 65.7o phase difference
 Fault Current 3/(3.3+4.026j +3.3+4.026j +3.3+7.33j +3*33+0.052j) = 3/(108.

a. For a bolted balanced three-phase fault at location C, calculate the magnitude and phase of the transmission line currents flowing away from bus B and line currents flowing towards the transformer...

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