Power systems miscellaneous
- A three phase line has conductor of 5 mm diameter placed at the corner of an equilateral triangle of 1.5 m side. The capacitive reactance to netural per phase per km will be
-
View Hint View Answer Discuss in Forum
Here, d = 5 mm, r = 2.5 mm = 2.5 × 10-3m, D = 1.5 m
∴ Deq = (DAB. DBC. DCA)1/3
= (1.5 × 1.5 × 1.5)1/3 = 1.5 mCn = 2πε0 = 1 In(D/r) 18 × 109 × In 
1.5 
2.5 × 10-3 = 1 18 × 109 × In 1.5 × 10-3 2.5
= 8.68 × 10-12F/m
= 8.68 × 10-9F/km
= 8.68 × 10-3µF/km
Capacitive reactance,XC = 1 2πƒ(Cn) = 1 2π × 50 × 8.68 × 10-9
= 3.66 × 105 Ω/kmCorrect Option: D
Here, d = 5 mm, r = 2.5 mm = 2.5 × 10-3m, D = 1.5 m
∴ Deq = (DAB. DBC. DCA)1/3
= (1.5 × 1.5 × 1.5)1/3 = 1.5 mCn = 2πε0 = 1 In(D/r) 18 × 109 × In 1.5 2.5 × 10-3 = 1 18 × 109 × In 1.5 × 10-3 2.5
= 8.68 × 10-12F/m
= 8.68 × 10-9F/km
= 8.68 × 10-3µF/km
Capacitive reactance,XC = 1 2πƒ(Cn) = 1 2π × 50 × 8.68 × 10-9
= 3.66 × 105 Ω/km
- A two conductor 1 – φ line operates at 50 Hz. The diameter of each conductor is 4cm and are spaced 6 m apart. What is the capacitive susceptance to neutral per km?
-
View Hint View Answer Discuss in Forum
Here, D = 6m, r = 2 cm = 2 × 10-2m
∴ Cn = 2πε0 = 1 In(D/r) 18 × 109In 6 2 × 10-2 = 1 18 × 109In 3 0.01
= 9.74 × 10-9F/k m
Capacitive susceptance to neturalbc = 1 = 2π. ƒ. Cn Xc
= 2π × 50 × 9.74 × 10-9 s/km
= 3.06 × 10-9s/k mCorrect Option: D
Here, D = 6m, r = 2 cm = 2 × 10-2m
∴ Cn = 2πε0 = 1 In(D/r) 18 × 109In 6 2 × 10-2 = 1 18 × 109In 3 0.01
= 9.74 × 10-9F/k m
Capacitive susceptance to neturalbc = 1 = 2π. ƒ. Cn Xc
= 2π × 50 × 9.74 × 10-9 s/km
= 3.06 × 10-9s/k m
- A two conductor 1 – φ line operates at 50 Hz. The diameter of each conductor is 2 cm and are spaced 3m apart. What is the line to line capacitance?
-
View Hint View Answer Discuss in Forum
Here, D = 3m, r = 2/2 = 1 cm = 10–2 m
∴ Cn = 2πε0 = 1 In(D/r) 18 × 109In 3 2
= 9.74 × 10-12 F/m = 9.74 × 109 F/km
Line to line capacitance,CL = Cn = 1 × 9.74 × 10-9 2 2
= 4.87 × 10-9 F/kmCorrect Option: B
Here, D = 3m, r = 2/2 = 1 cm = 10–2 m
∴ Cn = 2πε0 = 1 In(D/r) 18 × 109In 3 2
= 9.74 × 10-12 F/m = 9.74 × 109 F/km
Line to line capacitance,CL = Cn = 1 × 9.74 × 10-9 2 2
= 4.87 × 10-9 F/km
- What is the time of operation of a relay of rating 5 amp 2.2 sec, IDMT and having a relay setting of 125% Tms = 0.6 ? It is connected to a supply circuit through a CT, 400 ratio. The fault current is 4000 amp.
-
View Hint View Answer Discuss in Forum
The pick up value of the relay is 5 amp but since the relay setting is 125%,
Operating current of the relay
= 5 × 1.25 = 6.25 Amp.PSM = Secondary current Relay current setting = Primary current (fault current) Relay current setting × CT ratio = 4000 = 8 6.25 × 80
From the standard 2.2 sec curve, the operating time for PSM = 8 is 3.2 sec.
Since the TMS is 0.6, the actual operating time of the relay is 1.92 sec.Correct Option: B
The pick up value of the relay is 5 amp but since the relay setting is 125%,
Operating current of the relay
= 5 × 1.25 = 6.25 Amp.PSM = Secondary current Relay current setting = Primary current (fault current) Relay current setting × CT ratio = 4000 = 8 6.25 × 80
From the standard 2.2 sec curve, the operating time for PSM = 8 is 3.2 sec.
Since the TMS is 0.6, the actual operating time of the relay is 1.92 sec.
- What is the capacitance of neutral/km of a 1 – φ phase line composed of number 2 single strand conductor radius is 0.328 cm, spaced 3 m apart and 7.5 m above the ground ? Neglecting earth effect.
-
View Hint View Answer Discuss in Forum
Neglecting the presence of earth, we have
Cn = 0.0242 F/km log D r = 0.0242 = 0.00817 F/km log 300 0.328 Correct Option: B
Neglecting the presence of earth, we have
Cn = 0.0242 F/km log D r = 0.0242 = 0.00817 F/km log 300 0.328
