Power systems miscellaneous
- 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.
-
View Hint View Answer Discuss in Forum
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.
- 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?
-
View Hint View Answer Discuss in Forum
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
- A 1 – φ load of 200 kVA is delivered at 2500 V over a transmission line having R = 1.4 Ω, X = 0.8 Ω. What is the power at the sending end when the power factor of the load is 0.8 leading?
-
View Hint View Answer Discuss in Forum
I = 200 × 1000 = 80A 2500
Z = R + jX = 1.4 + j0.8 = 1.6 |29.74°
cos φ = 0.8 (lagging)
φr = 36.86° (lagging)
Ir = Ir |+φr = 80 |+36.86°
Vs = Vr + ZIr
= 2500 + 1.6 × |29.74° × (80 × |36.86°)
= 2500 + 128 |66.61°
= 2500 + 128 (cos 66.61° + jsin 66.61°)
= 2500 + 50.8 + j117.84
= (2550.8 + j117.84)= 2553.5 tan-1 
117.84 
2550.8
= 2553.5 |2.63°
φs = 2.63° + 36.86° = 34.23°
cos φs = cos 34.23° = 0.82 (leading)Correct Option: B
I = 200 × 1000 = 80A 2500
Z = R + jX = 1.4 + j0.8 = 1.6 |29.74°
cos φ = 0.8 (lagging)
φr = 36.86° (lagging)
Ir = Ir |+φr = 80 |+36.86°
Vs = Vr + ZIr
= 2500 + 1.6 × |29.74° × (80 × |36.86°)
= 2500 + 128 |66.61°
= 2500 + 128 (cos 66.61° + jsin 66.61°)
= 2500 + 50.8 + j117.84
= (2550.8 + j117.84)= 2553.5 tan-1 117.84 2550.8
= 2553.5 |2.63°
φs = 2.63° + 36.86° = 34.23°
cos φs = cos 34.23° = 0.82 (leading)
- A single phase load of 200 kVA, is delivered at 2500 V over a tranmissi on line having a R = 1.4Ω, X = 0.8Ω. What is the voltage at the sending end, when the power factor of the load is 0.8 lagging?
-
View Hint View Answer Discuss in Forum
I = 200 × 1000 = 80A 2500
Z = R + jX = 1.4 + j0.8 = 1.6 |29.74°
Let Vr is the reference phasor. Therefore
Vr = Vr + j0 = 2500 + j0
cos φ = 0.8 (lagging)
φr = 36.86° (lagging)
Ir = Ir |-φr = 80 |-36.86°
Vs = Vr + ZIr
= 2500 + (1.6 |29.74° (80 |36.86°)
= 2500 + 128 |7.12°
= 2500 + 128 (cos 7.12° – j sin 7.12°)
= 2500 + 127 – j 15.87
= [(2627)2 + (15.87)2]1/2
= 2627.06 V
= 2.627 kVCorrect Option: A
I = 200 × 1000 = 80A 2500
Z = R + jX = 1.4 + j0.8 = 1.6 |29.74°
Let Vr is the reference phasor. Therefore
Vr = Vr + j0 = 2500 + j0
cos φ = 0.8 (lagging)
φr = 36.86° (lagging)
Ir = Ir |-φr = 80 |-36.86°
Vs = Vr + ZIr
= 2500 + (1.6 |29.74° (80 |36.86°)
= 2500 + 128 |7.12°
= 2500 + 128 (cos 7.12° – j sin 7.12°)
= 2500 + 127 – j 15.87
= [(2627)2 + (15.87)2]1/2
= 2627.06 V
= 2.627 kV
- A three phase synchronous generator delivers 10 mvA, at a voltage of 10.5 kV. The line impedance is 5Ω. What is the voltage drop in the line in per unit and in volts ? Use the reference base as 12 mVA at 11 kV.
-
View Hint View Answer Discuss in Forum
S3φ = √3 VL. IL VA
∴ IL = S3φ √3VL
Here, S3φ = 10 mVA = 10 × 106 VA,
VL = 10.5 kV = 10.5 × 10³V∴ IL = 10 × 106 = 549.36 amp √3 10.5 × 10³
Iph = IL = 549.36 A
Voltage drop in the line per base = Zph. Iph
= 5 × 549.36 = 2.749 k V.Correct Option: C
S3φ = √3 VL. IL VA
∴ IL = S3φ √3VL
Here, S3φ = 10 mVA = 10 × 106 VA,
VL = 10.5 kV = 10.5 × 10³V∴ IL = 10 × 106 = 549.36 amp √3 10.5 × 10³
Iph = IL = 549.36 A
Voltage drop in the line per base = Zph. Iph
= 5 × 549.36 = 2.749 k V.
