Tricky Power Systems Questions and Answers

This set of Tricky Power Systems Questions and Answers focuses on “Switchgear and Protection – 3”.

1. A power system network is connected as shown in the figure below.

Sd1=15+j5 pu
Sd2=25+j15 pu
Zcable = j0.05pu
|V1|=|V2|=1 pu. The torque angle for the system will be _______
a) 14.4
b) 22.1
c) 16.2
d) 18.2
View Answer

Answer: a
Explanation: As the resistance is zero, losses will be zero.
P_G1=PD1+PD2=40 pu
For the equal sharing of load at the station
P_G1=P_G2=20pu
Real power flow from bus 1 to 2

Real power flow from bus 1 to 2 in given figure

δ = 14.4^o.

2. A single phase distributor of 1 km long has resistance and reactance per conductor of 0.1Ω and 0.15 Ω respectively. If the far end voltage Vb=200V and current is at 100A at 0.8 lag. At the midpoint a current of 100A is tapped at a pf of 0.6 pf with ref to voltage Vm at mid point. The voltage magnitude at M is __________
a) 218V
b) 200V
c) 232V
d) 220V
View Answer

Answer: a
Explanation: Drop in MB= (100ʟ-36.67)(0.1+j0.15)= 18.027ʟ19.44
V_M= 200+18.027 = 218.027 V.

3. A single phase motor is connected to 400V, 50Hz supply. The motor draws a current of 31.7A at a power factor 0.7 lag. The capacitance required in parallel with motor to raise the power factor of 0.9 lag (in micro farads) is ___________
a) 94.62
b) 282.81
c) 108.24
d) 46.87
View Answer

Answer: a
Explanation: Active power drawn by the motor=VIcosФ = 400*31.7*0.7 = 8876 W
Reactive power = VIsinФ=400*31.7*sin(45.57.29) = 9055.3 VAR
New power factor=cosθ2 = 0.9
θ2=cos^-1(0.9)
Q2=8876*tan(25.84) = 4298.855 VAR
Change in reactive power=9055.3-4298.855 = 4756.4 VAR
Qc = V²/Xc = V²*2πfC
C=4756.4/(400²*2π*50) = 94.62μF.

4. A single phase motor is connected to 400V, 50Hz supply. The motor draws a current of 31.7A at a power factor 0.7 lag. The additional reactive power (in VAR) to be supplied by the capacitor bank will be ___________
a) 4756
b) 4873
c) 4299
d) 9055.3
View Answer

5. A 275 kV TL has following line constants A=0.85ʟ5o, B=200ʟ75o. The active power received if the voltage to be maintained is 275kV will be ______________
a) 117.63
b) 220
c) 120
d) 115.25
View Answer

Answer: a
Explanation: |Vs|=|Vr|=275 kV
α= 5^o , β= 75^o

The active power received if the voltage to be maintained is 275kV will be 117.63

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6. A 275 kV TL has following line constants A=0.85ʟ5o, B=200ʟ75o. The active power angle such that the voltage to be maintained at the other end will be 275 kV ______________
a) 22
b) 16
c) 18
d) 24
View Answer

Answer: a
Explanation: |Vs|=|Vr|=275 kV
α= 5^o , β= 75^o

The active power angle such that voltage to be maintained at other end will be 275 kV 22

7. A power system has a maximum load of 15 MW. Annual load factor is 50%. The reserve capacity of plant is _____ if the plant capacity factor is 40%.
a) 3.75 MW
b) 4.75 MW
c) 18.75 MW
d) 5.75 MW
View Answer

Answer: a

The reserve capacity of plant is 3.75 MW if the plant capacity factor is 40%

Plant capacity= (0.5/0.4)*15 = 18.75 MW
Reserve Capacity = 18.75-15 = 3.75 MW.

8. A 100 MVA synchronous generator operates on full load at a frequency of 50 Hz. The load is suddenly reduced to 50 MW. Due to time lag in governor system, the steam valve begins to close after 0.4s. The change in the frequency (H=5 kW-s/KVA) is ___________
a) 1
b) 0.5
c) -1.5
d) 0.8
View Answer

Answer: a
Explanation: Energy transferred in 0.4 sec = 50*0.4= 20 J

The change in the frequency (H=5 kW-s/KVA) is 1 due to time lag in governor system

9. A 50 Hz four pole turbo-generator rated 100 MVA, 11 kV has an inertia constant of 8 MJ/MVA. If the mechanical input is suddenly raised to 80 MW for an electrical load of 50MW, then the rotor acceleration is __________
a) 337.5
b) 3.375
c) 457.5
d) 4.57
View Answer

Answer: a
Explanation: Energy stored = 100*8 =800MJ
Accelerating power = Md²δ/dt²
M=GH/180f = 800/(180*50) = 4/45 MJs/elect. Deg

Rotor acceleration is 337.5 if input is raised to 80 MW for electrical load of 50MW

α= 337.5 elect deg/s².

10. A single phase TL has copper conductor of 0.775 cm2 cross section through which 200 kW at UPF at 330 V is to be maintained. If the efficiency of transmission line is 90%, then the minimum length of TL is _________(in km and take specific resistance to be 1.785 μΩ/cm).
a) 13.6 km
b) 14 km
c) 136 km
d) 16.4 km
View Answer

Answer: a
Explanation: Pr=200 kW, efficiency=0.9
Ps= 200/0.9=222.22 kW
Losses=22.22 kW
Current, I=200000/3300 = 60.60 A
Line losses=2I²R (for a 2 wire line)
R=22.22/(2*60.602)=3.02 Ω
R=ρl/a
Length, l = (3.025*0.775)/(1.785*10-6) = 13.6 km.

11. An isolated generator connected to a turbine with its continuous maximum power of 20 MW, 50 Hz. Generator connected with two loads of 8 MW, each operate at 50 Hz.. Generator has 5% droop characteristic. If an additional load of 6 MW is added then frequency will be ___________
a) 50.25
b) 40.75
c) 49.75
d) 48.75
View Answer

Answer: a
Explanation: