MCQs on Power Systems

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01․ If a 500 MVA, 11 kV 3-phase generator at 50 Hz feeds through a transfer impedance of (0+j0.605) Ω/phase, an infinite voltage also at 11 kV, then the maximum steady state power transfer on the base of 500 MVA and 11 kV is
1 pu
0.8 pu
0.4 pu
0.5 pu

Maximum steady state power = EV/X Where, E = Sending side voltage V = Receiving side voltage X = Reactance of the line Maximum steady state power = 11*11/0.605 = 200 MW Maximum steady state power in pu = 200/MVAbase= 200/500 = 0.4 pu

02․ For 800 MJ stored energy in the rotor at synchronous speed, what is the inertia constant H for a 50 Hz, four pole turbo generator rated 100 MVA, 11 kV?

Inertia constant H = Kinetic energy / MVAbase Inertia constant H = 800/100 = 8 MJ/MVA

03․ For which one of the following types of motors, is the equal area criterion for stability applicable?
Three phase synchronous motor
Three phase induction motor
DC series motor
DC shunt motor

Equal area criterion used to transient stability analysis of synchronous machines, is consistency with transient energy function method for single machine-infinite bus system or two machines system.Equal area criteria is used more efficiently for a single generator stability analysis or single motor stability analysis and is is used to find the absolute stability of the system.

04․ An alternator having an induced emf of 1.6 pu is connected to an infinite bus of 1 pu. If the bus bar has reactance of 0.6 pu and alternator has reactance of 0.2 pu, what is the maximum power that can be transferred?
6 pu
5 pu
2 pu
2.67 pu

Maximum steady state power = EV/X Where, E = Sending side voltage V = Receiving side voltage X = Reactance of the system Maximum steady state power = 1.6*1/(0.6+0.2) = 2 pu

05․ If the inertia constant H = 8 MJ/MVA for a 50 MVA generator, the stored energy is
400 MJ
8 MJ
50 MJ
200 MJ

Inertia constant H = Energy stored / MVAbase Energy stored = H*MVAbase Energy stored = 8*50 = 400 MJ

06․ Two identical synchronous machines having same inertia constant are connected in parallel and swinging together. The effective inertia constant is 4 MJ/MVA. Then each machine has inertia constant of

When two machines swing together in parallel connection H = H1 + H2 Given that, H1 = H2 4 = 2*H1 H1 = H2 = 2 MJ/MVA

07․ Maximum power capacity of a transmission line can be increased by
increasing voltage level
decreasing reactance value
either 1 or 2
nether 1 nor 2

Maximum power Pmax ∝ V²s/Xs Maximum power transfer capability of transmission line can be increased by either increasing voltage level or reducing the line reactance value. Methods of reducing line reactance are 1. Parallel transmission lines 2. Using series capacitance 3. Using bundled conductors

08․ Which of the following is/are the disadvantages of phase modifiers?
It consumes both active and reactive power
Initial cost of synchronous phase modifier is higher
It requires more maintenance because of running equipment
all of the above

Advantages and disadvantages of phase modifiers: 1.Single unit can be used as both capacitance and inductor by adjusting the excitation 2. Smooth voltage regulation is possible by controlling excitation 3. Initial cost of synchronous phase modifier is higher 4. It consumes both active and reactive power 5. It requires more maintenance because of running equipment 6.It requires starting mechanism for bringing the synchronous motor up to synchronous speed

09․ In thermal power stations, economiser will
reduce the thermal efficiency
increase the thermal efficiency
reduce the heat
both 2 and 3

Basically, economiser is located in between exit of boiler and entry of air pre heater. When the flue gases are coming out from boiler they take away a lot of heat. Economiser utilize this heat from flue gases and these heat is used to heat the feed water which is going to feed the boiler.Economiser is used to absorbing heat from the flue gas for increasing the temperature of the water so that improving the thermal efficiency.

10․ Stock bridge dampers are used in case hill areas to over come the which problem?
unequal ice loading
increase string efficiency
increase power factor
all of the above

Due to unequal ice loading vibrations are produced in the conductors. If these vibrations are extended up to the point of support of the conductor, bolts and nuts are loosen, finally producing fault in the system. Therefore, to reduce these vibrations at the point of support, vibration or stock bridge dampers are used