EE 421: Line Constants Example - College of...
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EE 421 Introduction to Power Systems Session 14; Page 1/3 Fall 2015 EE 421: Line Constants Example μ 0 4 π 10 7 H m freq 60Hz Problem 4.4 In Analysis of Faulted Power Systems by P.M. Anderson: Compute the per mile positive and negative sequence impedance for the line configuration of Figure P4.4 where the conductor is 336,400 CM, 26/7 Strand ACSR. Assume ideal transposition Define heights relative to the earth H gw 36ft 11ft H a H gw 6.5ft H a 40.5 ft H b 36ft H c H a 8ft H c 32.5 ft Define horizontal position relative to center of the tower (some are negative). X a 4 ft X b 5ft X c 6 ft Now calculate distance between conductors (be careful of negative signs) D ab H a H b 2 X b X a 2 D ab 10.062 ft D ac H a H c 2 X c X a 2 D ac 8.246 ft D bc H b H c 2 X b X c 2 D bc 11.543 ft Calculate geometric mean distance between conductors: D m D ab D bc D ac 1 3 D m 9.857 ft
Transcript of EE 421: Line Constants Example - College of...
EE 421Introduction to Power Systems
Session 14; Page 1/3Fall 2015
EE 421: Line Constants Example
μ0 4 π 107
H
m freq 60Hz
Problem 4.4 In Analysis of Faulted Power Systems by P.M. Anderson:
Compute the per mile positive and negative sequence impedance for the line configuration ofFigure P4.4 where the conductor is 336,400 CM, 26/7 Strand ACSR. Assume idealtransposition
Define heights relative to the earth
Hgw 36ft 11ft
Ha Hgw 6.5ft Ha 40.5 ft
Hb 36ft
Hc Ha 8ft Hc 32.5 ft
Define horizontal position relative to center of the tower (some are negative).
Xa 4 ft Xb 5ft Xc 6 ft
Now calculate distance between conductors (be careful of negative signs)
Dab Ha Hb 2 Xb Xa 2 Dab 10.062 ft
Dac Ha Hc 2 Xc Xa 2 Dac 8.246 ft
Dbc Hb Hc 2 Xb Xc 2 Dbc 11.543 ft
Calculate geometric mean distance between conductors:
Dm Dab Dbc Dac 1
3 Dm 9.857 ft
EE 421Introduction to Power Systems
Session 14; Page 2/3Fall 2015
GMR and Rac from table A.4 (entry for Linnet in Table A.4 in Glover, Sarma and Overbye)
GMR 0.0244ft diameter 0.721inFor comparison: r'
diameter
2
e
1
4Rac 0.273
ohm
mi at 25C and 60Hz
r' 0.023 ft
As an aside: 1 cmil is the area of conductor with diameter of 0.001in:
cmil π0.001in
2
2
1cmil 7.854 107 in
2
For this conductor with an area of:
areacmil 336400 cmil
Radiuseqareacmil
π
Radiuseq 0.29 in
Compare to the diameter of:
diameter
20.36 in the difference is due to skin effect and the
impact of stranding.
r' 0.281 in
We can calculate the positive/negative sequence impedance.
L1μ0
2 πln
Dm
GMR
L1 1.932
mH
mi
Z1 Rac j 2 π freq L1 Z1 0.273 0.728i( )ohm
mi
EE 421Introduction to Power Systems
Session 14; Page 3/3Fall 2015
Now suppose, instead of using a single conductor of 336,400 CM ACSR in each phase, with currentcarrying capacity of 530 amperes, suppose that each phase consists of a two-conductor bundle of two3/0 ACSR conductors with capacity of 300 amperes/conductor. Let the two conductors of each bundlebe seperated by 1.0ft vertically.
Rac3 0.560ohm
mi from table A4 at 25C and 60Hz
GMR3 0.006ft diameter3 0.502in
Da1a2 1.0ft
GMRbundle GMR3 Da1a2 GMRbundle 0.077 ft
Now we'll have
L1bunμ0
2 πln
Dm
GMRbundle
L1bun 1.56mH
mi
Z1bunRac3
2j 2 π freq L1bun Z1bun 0.28 0.588i( )
ohm
mi
for comparison: Z1 0.273 0.728i( )
ohm
mi
Most of change is in the inductive reactance
