a. Three phase (3ph), phase to phase, and single line to ground (SLG) at 138kV bus. b. 3ph, and SLG at 34.5kV bus. 138/34.5 transformers have the same Z and are in parallel. c. 3ph, and SLG at 34.5kV at 34.5/12.47 substation high side. d. 3ph, and SLG at 12.47kV bus e. 3ph, and SLG at 12.47kV tap f. 3ph, and SLG at 12.47kV end of line (EOL)

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1) For the utility radial distribution system shown on figure below (only generator is at 138kV). All impedances are in per unit on a 100MVA base and nominal voltage shown. All transformer connected taps equal nominal voltage shown. All breakers and switches are closed unless marked with N.O. (Normally Open). Calculate fault current magnitudes in rms symmetrical amps at the following locations (faults are bolted, one at a time, not simultaneous.)
a. Three phase (3ph), phase to phase, and single line to ground (SLG) at 138kV bus.
b. 3ph, and SLG at 34.5kV bus. 138/34.5 transformers have the same Z and are in parallel.
c. 3ph, and SLG at 34.5kV at 34.5/12.47 substation high side.
d. 3ph, and SLG at 12.47kV bus
e. 3ph, and SLG at 12.47kV tap
f. 3ph, and SLG at 12.47kV end of line (EOL)
g. State the 138kV source (generator) amps for the 3ph at 12.47kV EOL fault.

 

This is a typical Ameren Missouri distribution system.
Eq.
Source
Symmetrical
Components
Distribution
Example
138 kV
j2.382
34.5 kv
3.99 mi
12.47KV
L
کے سلت
f
Z₁ = 2.266+; 1.549
Z₁ = 3.586 +5 4565 pu
|Z₁ = 0.0379 +0. 2036 pu
|Z₁ = 0.1465 +0.9036 pu
Да
15,000 ft
1/0 AC SR
2000 ft
ри
Z₁ = 0.00167 + 0.01483,
0.00831+; 0.03284 pu
Zo
utu
Single &
tap
1/0 ACSR
j2.38.22
*
b
N.O.
N.O.
=
e
2.6mi
10,000 ft
556.5 A.A.2,
³12,-0.0041+ 0.1468
2
B
ch
A
NO
Z₁ = 0.0041 +0.1468
Z₁ = 0.025+j0. 1318 pu
Z.=0.1071 +0.5949pu
DA
-A
HE
Z₁=0.0338+0.5823 AV
Z.=0.0338+0.5823;
0.250+ j 0.7716 pu
Z₁₂= 1.087+; 2,608 pu
Z₁ = 0.302 1 + 0.2065
ри
Z₁ = 0.4782 +0.6086 pu
Transcribed Image Text:This is a typical Ameren Missouri distribution system. Eq. Source Symmetrical Components Distribution Example 138 kV j2.382 34.5 kv 3.99 mi 12.47KV L کے سلت f Z₁ = 2.266+; 1.549 Z₁ = 3.586 +5 4565 pu |Z₁ = 0.0379 +0. 2036 pu |Z₁ = 0.1465 +0.9036 pu Да 15,000 ft 1/0 AC SR 2000 ft ри Z₁ = 0.00167 + 0.01483, 0.00831+; 0.03284 pu Zo utu Single & tap 1/0 ACSR j2.38.22 * b N.O. N.O. = e 2.6mi 10,000 ft 556.5 A.A.2, ³12,-0.0041+ 0.1468 2 B ch A NO Z₁ = 0.0041 +0.1468 Z₁ = 0.025+j0. 1318 pu Z.=0.1071 +0.5949pu DA -A HE Z₁=0.0338+0.5823 AV Z.=0.0338+0.5823; 0.250+ j 0.7716 pu Z₁₂= 1.087+; 2,608 pu Z₁ = 0.302 1 + 0.2065 ри Z₁ = 0.4782 +0.6086 pu
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