Power systems miscellaneous
- A three phase line, 20 km long, delivers 10 MW at 11kV, 50 Hz, 0.8 power factor lagging. The power losses in the line is 10 percent of the power delivered. The resistance of the each onductor will be _________ Ω.
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Power receiving end 10 MW.
Vrp = 11 × 10³ = 6350 volts √3 Irp = P 3 × Vph × cosφr cosφr = 0.8 = 10 × 106 3 × 6350 × 0.8
Power loss = 10% of power delivered= 10 × 10 × 106 = 106 watts 100
Let RP be the phase resistance of conductor.
∴ 3Ir²pRp = 106or Rr = 106 = 0.774Ω 3 × (656.16)² Correct Option: A
Power receiving end 10 MW.
Vrp = 11 × 10³ = 6350 volts √3 Irp = P 3 × Vph × cosφr cosφr = 0.8 = 10 × 106 3 × 6350 × 0.8
Power loss = 10% of power delivered= 10 × 10 × 106 = 106 watts 100
Let RP be the phase resistance of conductor.
∴ 3Ir²pRp = 106or Rr = 106 = 0.774Ω 3 × (656.16)²
- A 5kVA, 400/2000 volt, 50Hz, 1 – φ transformer has primary and secondary leakage reactance each of 2.5 Ω . The total reactance in p.u. will be _____p.u.
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Let VA on both sides of the transfomer is 500 VA, i.e.
sb = 5000 VA,
Total reactance of the transfomer referred to primaryX1C = X1 + X2 
N1 
² N2 = 2.5 + 2.5 × 400 ² 200
= 2.5 + 10 = 12.5 Ω
Base voltage on primary side,
Vb1 = 400 VXpu = XΩ Sb V²b1
Total p.u. reactance referred to primaryX1epu = X1e. Sb V²b1 = 12.5 × 5000 = 0.390 p.u. (400)² Correct Option: D
Let VA on both sides of the transfomer is 500 VA, i.e.
sb = 5000 VA,
Total reactance of the transfomer referred to primaryX1C = X1 + X2 N1 ² N2 = 2.5 + 2.5 × 400 ² 200
= 2.5 + 10 = 12.5 Ω
Base voltage on primary side,
Vb1 = 400 VXpu = XΩ Sb V²b1
Total p.u. reactance referred to primaryX1epu = X1e. Sb V²b1 = 12.5 × 5000 = 0.390 p.u. (400)²
- A two conductor 1– φ line operates at 50 Hz. The diameter of each conductor is 20 mm and the spacing between the conductors is 3m. The height of conduct or above the ground is 6m. The capacitance of the line to neutral will be ________pF/m
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Cn = 2πε0 In D - In Hm r Hs
Here, r = 10 mm = 10 × 10-3m,
D = 3 m, h = 6 m,
H1/sub> = H2 = 2h = 12 m
H12 = H21 = (D² + 4h²)1/2
= (3² + 4 × 6²)1/2 = √153
Hm = (H12 . H21)1/2 = H12 = √153m
Hs = (H1 × H2)1/2 = (2h. 2h)1/2 = 2h = 12mCn = 2πε0 In 3 - In √153 10 × 10-3 12 = 1 18 × 109 × In 3 × 12 10-3 × √153
= 9.7 × 10-12 F/m
= 9.7 pF/mCorrect Option: D
Cn = 2πε0 In D - In Hm r Hs
Here, r = 10 mm = 10 × 10-3m,
D = 3 m, h = 6 m,
H1/sub> = H2 = 2h = 12 m
H12 = H21 = (D² + 4h²)1/2
= (3² + 4 × 6²)1/2 = √153
Hm = (H12 . H21)1/2 = H12 = √153m
Hs = (H1 × H2)1/2 = (2h. 2h)1/2 = 2h = 12mCn = 2πε0 In 3 - In √153 10 × 10-3 12 = 1 18 × 109 × In 3 × 12 10-3 × √153
= 9.7 × 10-12 F/m
= 9.7 pF/m
- A three phase 50 Hz line consists of three conductors each of diameter 21 mm. The spacing between the conductors as follows: A – B = 3m, B – C = 5m, C – A = 3.6 m
If the line operates at 132 kV, the reactive voltampere generatd by the line per km is _______kVAr
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Cn = 2πε0 = 1 = 9.43 × 10-9 F/km In Dm 18 × 109 × In 3 × 5 × 3.6 1/3 DsL 2 Xc = 1 = 1 = 0.337 × 106 Ω/km 2πƒCn 2π × 50 × 9.49 × 10-3
Reactive volt ampere generated by the line= VL² = (132 × 10³)² = 51703 VAr. Xc 0.336 × 106
= 51703 VAr.Correct Option: A
Cn = 2πε0 = 1 = 9.43 × 10-9 F/km In Dm 18 × 109 × In 3 × 5 × 3.6 1/3 DsL 2 Xc = 1 = 1 = 0.337 × 106 Ω/km 2πƒCn 2π × 50 × 9.49 × 10-3
Reactive volt ampere generated by the line= VL² = (132 × 10³)² = 51703 VAr. Xc 0.336 × 106
= 51703 VAr.
- A three phase, 50 Hz transmission line consists of three equal conductors of radii r placed in a horizontal plane with a spacing of 6 m between middle and each outer conductor. If radius of each conductor is 12.5 mm, then indutance per phase per km of the transposed line will be _________ mH/km
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L = 2 × 10-7In D r'
Here Dm = (D × D × D)1/3 = 1.26
⇒ D = 1.26 × 6 = 7.56 m
r' = 0.7788 × r = 0.7788 × 12.5 × 10–3
= 9.735 × 10–3 m∴ L = 2 × 10–7 ln 7.56 9.735 × 10-3
= 13.31 × 10–7 H/m
= 1.33 × 10–3 × 10–3H/m
= 1.33 mH /k m.Correct Option: A
L = 2 × 10-7In D r'
Here Dm = (D × D × D)1/3 = 1.26
⇒ D = 1.26 × 6 = 7.56 m
r' = 0.7788 × r = 0.7788 × 12.5 × 10–3
= 9.735 × 10–3 m∴ L = 2 × 10–7 ln 7.56 9.735 × 10-3
= 13.31 × 10–7 H/m
= 1.33 × 10–3 × 10–3H/m
= 1.33 mH /k m.
