Question

Consider the system Shown in the single-line diagram of Figure 4. The required sequence reactances in per unit to the same base are as follows.

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Consider the system shown in the single-line diagram of Figure 4. The required sequence reactances in per unit to the same base are as follows:

G1 X1 = X2 = 0.12 

G2 X1 = X2 = 0.24 

G3 X1 = X2 = 0.12 

Transformers X11=X12= 0.12 

Lines: Positive and X13=X12= 0.20 Negative Sequence 

a. Draw the positive-, and negative- sequence reactance diagrams. [6 Points] 

b- Determine the The'venin's equivalent of each sequence network as viewed from a fault in the middle of line 1-3. [7 Points) 

c-A line to line fault on phases B and C takes place in the middle of line 1-3. Determine the fault currents in phases B and C in per unit. [7 Points) 

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Answer #1

Griven G1 X, X = 0.12 G2 X,=X2 = o24 G13 XT, Tran sformers 0-12 Lines Posidive and negalive Seguun Ce 13- x- O 20 only PositiNegative Sequmce readam ca diogram Thuu wil no voltoge Teactan & diagram in nig ative Sequen Source For GrenerraJm For ger 3

Since taul line occevrs im the middle Cuvant tous im 1-3 NO Current tlows in Fault 1-a line line 1-3 Thevenins Equivalent Tm

For negative Seajuemce nat work jo12 ul jo 12 jo 20 O12 gjo ay 0 2 Thavemins equivalrit Tmpedume is nagatiue bun (0-12+0-12 +A L.L Faut in middle d line 1-3 Fault current Phare Band c in phasus B and c. A C raut Cwwn Fauit Duving Curren zf TAI NR2 poF,-1TA E Z,+ tZf TFault Cwrrwd +a TA2 t (a- a) TA, (-0-5-j0866) - C-05 +jo 866) -jJ IA 2, +22 t z , Fautt undome in positiue

IB =16.037/3

IB=-IC=5.345 Amps

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