Power systems miscellaneous
- A conductor is composed of seven identical copper strands, each having a radius r. The self GMD of the conductors will be
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Self GMD of the seven strand conductor is 49th root of the distances. Thus
Ds = [(r')7 × (D²12 × D²26 × D14 × D17)6 × (2)6]1/49
= [(0.7788 r)7 × (2²r² × 3 × 2²r² × 2²r × 2r)6]1/49= 2r[3(0.7788)]1/7 = 2.177 r 61/49 Correct Option: A
Self GMD of the seven strand conductor is 49th root of the distances. Thus
Ds = [(r')7 × (D²12 × D²26 × D14 × D17)6 × (2)6]1/49
= [(0.7788 r)7 × (2²r² × 3 × 2²r² × 2²r × 2r)6]1/49= 2r[3(0.7788)]1/7 = 2.177 r 61/49
- A three phase circuit line consists of 7/4.75 mm hard drawn copper conductors. The arrangement is shown in the figure. The line is completely transposed. What is the inductance, per phase per km of the system?
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Diameter of each conductor = 3 × 4.75 = 14.25 mm.
r = 7.125 × 10-3 m.
Dab = √3² + 1.5² = 3.354 m
Dbc = Da'b' = Db'c' = 3.354 m
Dab' = √3² + 7.5² = 8.077 m
Dbc' = Dba' = Db'c = Dab' = 8.077 m
Daa' = √6² + 6² = 6√2 m
Dcc' = 6√2 m
Dm = (Dab.Dbc.Dca.Dab'.Db'c.Dc'a.Da'b'.Db'c'.Dc'a'.)1/2
= (3.354 × 3.354 × 6 × 8.076 × 8.076 × 6 × 8.076 × 8.076 × 6 × 3.354 × 3.54 × 6)1/12
= 5.4576 m
DsL = (r'³ × Daa' × Dbb' × Dcc')1/6
= [(0.7788 × 7.125 × 10-3)³ × 6√2 × 9 × 6√2]1/6
= 0.21913 m= 2 × 10-7 In 
Dm 
DsL = 2 × 10-7 In 5.4576 
H/m 0.21913
= 6.43 × 10-7 H/m = 6.43 × 10-4 H/km = 0.643 mH/k m.Correct Option: D
Diameter of each conductor = 3 × 4.75 = 14.25 mm.
r = 7.125 × 10-3 m.
Dab = √3² + 1.5² = 3.354 m
Dbc = Da'b' = Db'c' = 3.354 m
Dab' = √3² + 7.5² = 8.077 m
Dbc' = Dba' = Db'c = Dab' = 8.077 m
Daa' = √6² + 6² = 6√2 m
Dcc' = 6√2 m
Dm = (Dab.Dbc.Dca.Dab'.Db'c.Dc'a.Da'b'.Db'c'.Dc'a'.)1/2
= (3.354 × 3.354 × 6 × 8.076 × 8.076 × 6 × 8.076 × 8.076 × 6 × 3.354 × 3.54 × 6)1/12
= 5.4576 m
DsL = (r'³ × Daa' × Dbb' × Dcc')1/6
= [(0.7788 × 7.125 × 10-3)³ × 6√2 × 9 × 6√2]1/6
= 0.21913 m= 2 × 10-7 In Dm DsL = 2 × 10-7 In 5.4576 H/m 0.21913
= 6.43 × 10-7 H/m = 6.43 × 10-4 H/km = 0.643 mH/k m.
- What is the inductance per km per phase of a single cir cuit 460 kV line using two bundle conductors per phase as shown in the figure, if diameter of each conductor is 5 cm?
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Let effect of transposition is negligibly small.
Ds = (0.025 × 0.8 × 0.7788)1/2 = 0.124 m
Dm = (8.5 × 17 × 8.5)1/3 = 10.7 m
Inductance per km/phase= 2 × 10-7 × 103 In 10.7 mH/km/ph 0.124
= 2 × 10-4 × 4.45 mH/km/phase
= 8.90 × 10-4 mH/k m/phaseCorrect Option: C
Let effect of transposition is negligibly small.
Ds = (0.025 × 0.8 × 0.7788)1/2 = 0.124 m
Dm = (8.5 × 17 × 8.5)1/3 = 10.7 m
Inductance per km/phase= 2 × 10-7 × 103 In 10.7 mH/km/ph 0.124
= 2 × 10-4 × 4.45 mH/km/phase
= 8.90 × 10-4 mH/k m/phase
- If diameter of the conductor is 0.1 cm, then what is the inductance of a 3φ line operating at 50 Hz and conductors are arranged as follows ?
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r' = 0.7788 × 0.10 = 3.894 × 10-4 m. 2 × 100
Deq = (3.2 × 3.2 × 6.4)1/3 m = 4.031 m
Inductance per km= 2 × 10-7 × 103 In 4.031 3.894 × 10-4
= 1.848 mH/kmCorrect Option: D
r' = 0.7788 × 0.10 = 3.894 × 10-4 m. 2 × 100
Deq = (3.2 × 3.2 × 6.4)1/3 m = 4.031 m
Inductance per km= 2 × 10-7 × 103 In 4.031 3.894 × 10-4
= 1.848 mH/km
- A three phase 50Hz line consists of three conductors each of diameter 21 mm. The spacing between conductors is as shown. The inductance per phase per km of the line will be
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r' = 0.7788 r = 0.7788 × 21 × 10-3 m. 2
Deq = (3 × 5 × 3.6)1/3 = 3.78 mL = 2 × 10-7 In Deq r' 2 × 10-7 
In (3.78) 
(0.7788 × 10.5 × 10-3)
= 2 × 10-7 ln (46.22)
= 2 × 10-7 × (6.136)
= 12.2 × 10-7 H/m
= 12.2 × 10-4 H /kmCorrect Option: A
r' = 0.7788 r = 0.7788 × 21 × 10-3 m. 2
Deq = (3 × 5 × 3.6)1/3 = 3.78 mL = 2 × 10-7 In Deq r' 2 × 10-7 In (3.78) (0.7788 × 10.5 × 10-3)
= 2 × 10-7 ln (46.22)
= 2 × 10-7 × (6.136)
= 12.2 × 10-7 H/m
= 12.2 × 10-4 H /km
