Power systems miscellaneous
- 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
- 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
- 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
- A 1 km of a three phase metal sheathed belted cable gave a measured capacitance of 0.7 µF bet ween one conduct or and the other two conductors bunched together with the earth sheath and 1.5 µF measured between the three bunched conductors and the sheath. What is the charging current, when the cable is connected at 11 kV, 50 Hz supply?
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2Cc + Cs = 0.7
⇒ 3Cs = 1.5
⇒ Cs = 0.5
Now, 2 Cc + 0.5 = 0.7⇒ Cc = 0.2 = 0.1 F 2 CL = 1 C0 2 = 1 (3Cc + Cs) 2 = 1 (3 × 0.1 + 0.5) = 0.4 F 2
Charging current =Vp ω C0Diversity factor = VL × 2π × ƒ × 2CL √3 = 11000 × 2π × 50 × 2 × 0.4 × 10-6 √3
= 1.59 Ampere/phaseCorrect Option: C
2Cc + Cs = 0.7
⇒ 3Cs = 1.5
⇒ Cs = 0.5
Now, 2 Cc + 0.5 = 0.7⇒ Cc = 0.2 = 0.1 F 2 CL = 1 C0 2 = 1 (3Cc + Cs) 2 = 1 (3 × 0.1 + 0.5) = 0.4 F 2
Charging current =Vp ω C0Diversity factor = VL × 2π × ƒ × 2CL √3 = 11000 × 2π × 50 × 2 × 0.4 × 10-6 √3
= 1.59 Ampere/phase
- What is the corona loss of a 3 – φ line, 160 km long, conductor diameter 1.036 cm, 2.44 delta spacing, air temperature 26°, corresponding to a barometric pressure of 72 cm, operating voltage 110 kV at 50 Hz. m = 0.78?
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Here, L = 160 km; diameter = 1.036;
d = 2.44 m; r = 0.518 cm = 0.518 × 10-2mδ = 3.92 b = 3.92 × 72 = 0.943 273 + t 273 + 26
∴ Vd = 21.1 × 0.78 × 0.943 × 0.518In 2.44 × 100 = 8.039 × In 244 = 49.47 (Line to neutral) kV. 0.518 0.518
Power Loss= 241 × 10-5 ƒ + 25 × √ r (V - Vd)²kW/phase/km. δ d = 241 × 105 50 + 25 × √ 0.518 × 100(63.5 - 9.4)² 0.943 2.44
= (0.1916 × 0.046 × 198.81) = 1.75 kW/phase/km.
= 5.25 kW/km.
= 841 kW for three phases.Correct Option: D
Here, L = 160 km; diameter = 1.036;
d = 2.44 m; r = 0.518 cm = 0.518 × 10-2mδ = 3.92 b = 3.92 × 72 = 0.943 273 + t 273 + 26
∴ Vd = 21.1 × 0.78 × 0.943 × 0.518In 2.44 × 100 = 8.039 × In 244 = 49.47 (Line to neutral) kV. 0.518 0.518
Power Loss= 241 × 10-5 ƒ + 25 × √ r (V - Vd)²kW/phase/km. δ d = 241 × 105 50 + 25 × √ 0.518 × 100(63.5 - 9.4)² 0.943 2.44
= (0.1916 × 0.046 × 198.81) = 1.75 kW/phase/km.
= 5.25 kW/km.
= 841 kW for three phases.
