Power systems miscellaneous
- Transmission line has a total series reactance of 0.2 p.u. Reactive power compensation is applied at the midpoint of the line and it is controlled such that midpoint voltage of the transmission line is always maintained at 0.98 pu. If voltage at both ends of the line are maintanined at 1.0 pu, then steady state power transfer limit of the transmission line is _________ pu
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P = EsEr sinδ Xtr
Steady state power transfer limit= 1 × 1 = 5 pu. 0.2 Correct Option: B
P = EsEr sinδ Xtr
Steady state power transfer limit= 1 × 1 = 5 pu. 0.2
- A 500 MVA, 11 kV sychronous generator has 0.2 p.u. synchr onous r eact ance. The p.u. synchronous reactance on the base values of 100 MVA and 22 kV is __________
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NA
Correct Option: A
NA
- A surge of 15 kV magnitude travels along a cable towards its junction with a overhead line. If inductance and capacitance of the cable and overhead line are respectively 0.3 mH, 0.4 µH and 1.5 mH, 0.012 µH per km, the voltage rise at the junction due to the surge, will be ________ kV
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The surge travels from the cablet owards the overhead line and hence there will be positive voltage reflection at the junctions.
Natural impedance of the cable= 0.5 √ 0.3 × 10-3 = 27.38Ω 0.4 × 10-6
Natural impedance of the line= √ 1.5 × 10-3 0.12 × 10-6 = √ 1.5 × 10-3 = 35.3Ω 0.12 × 10-7
The voltage rise at the junction is the voltage transmitted into the over head line as the voltage is zero before the surge reaches the junction.E" = 2 × 35.3 × 15 (35.3 + 2.7) = 2 × 35.3 × 15 = 27.87kV (35.3 + 2.7) Correct Option: A
The surge travels from the cablet owards the overhead line and hence there will be positive voltage reflection at the junctions.
Natural impedance of the cable= 0.5 √ 0.3 × 10-3 = 27.38Ω 0.4 × 10-6
Natural impedance of the line= √ 1.5 × 10-3 0.12 × 10-6 = √ 1.5 × 10-3 = 35.3Ω 0.12 × 10-7
The voltage rise at the junction is the voltage transmitted into the over head line as the voltage is zero before the surge reaches the junction.E" = 2 × 35.3 × 15 (35.3 + 2.7) = 2 × 35.3 × 15 = 27.87kV (35.3 + 2.7)
- If two pole 50 Hz, 50 MVA turbogenerator has a moment of inertia of 8.6 × 10³ kg-m², then inretia constants H in MJ/MVA will be __________
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Kinetic energy stored = 1 jω² 2 = 1 × (8.6 × 10³) × 
2π × 3000 
² = 424.4 MJ. 2 60 H = Kinetic energy stored in MJ MVA rating = 424.4 = 8.48 MJ/MVA. 50 Correct Option: C
Kinetic energy stored = 1 jω² 2 = 1 × (8.6 × 10³) × 2π × 3000 ² = 424.4 MJ. 2 60 H = Kinetic energy stored in MJ MVA rating = 424.4 = 8.48 MJ/MVA. 50
- If two pole 50 Hz, 50 MVA turbogenerator has a moment of inertia of 9 × 10³ kg-m², then kinetic energy in MJ at rated speed will be _________ MJ
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N = 120 × ƒ = 120 × 50 = 3000 rpm P 2 Kinetic energy stored = 1 jω² = 1 (J). 2π × N ² 2 2 60 = 1 × 9 × 10³ × 2π × 3000 ² = 444 MJ. 2 60 Correct Option: D
N = 120 × ƒ = 120 × 50 = 3000 rpm P 2 Kinetic energy stored = 1 jω² = 1 (J). 2π × N ² 2 2 60 = 1 × 9 × 10³ × 2π × 3000 ² = 444 MJ. 2 60
