Power systems miscellaneous
- 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 charging kVAr taken by a 10 km length of the cable when connected to an 11 kV, 50 Hz supply? The capacitance measured between any two cores of a three phase belted cable is 0.5 µF/km.
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Let CL be the capacitance between any two cores.
CL = C0 2
C0 = 2 CL = 2 × 0.5 × 10 = 10µFIC = Vp.ω.C0 = VL ω.2CL √3
Total charging kVAr
= √3 (VL).(LC) × 10-3= √3 × VL × VL × 2CL × 10-3 √3
= 2 (VL)². ω. CL × 10-3= 2 × (11000)² × 2π × 50 × 10 × 10-3 × 10-6 2
= 380 kVAr.Correct Option: A
Let CL be the capacitance between any two cores.
CL = C0 2
C0 = 2 CL = 2 × 0.5 × 10 = 10µFIC = Vp.ω.C0 = VL ω.2CL √3
Total charging kVAr
= √3 (VL).(LC) × 10-3= √3 × VL × VL × 2CL × 10-3 √3
= 2 (VL)². ω. CL × 10-3= 2 × (11000)² × 2π × 50 × 10 × 10-3 × 10-6 2
= 380 kVAr.
- 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.
- For a 500 Hz frequency excitation, a 50 km long power line will be modelled as
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NA
Correct Option: C
NA
- Bundled conductors are employed to
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NA
Correct Option: C
NA
