01․ The inertia constant of 100 MVA, 50 Hz, 4 pole generator is 10 MJ/MVA. If the mechanical input to the machine is suddenly raised from 50 MW to 75 MW, the rotor acceleration will be equal to

Rotor acceleration α = P

_{a}/M Where, P_{a}= Accelerating power M = Angular momentum = HS/(180*f) H = Inertia constant S = Rating of the machine f = Frequency M =10*100/(180*50) Rotor acceleration α = 25*180*50/1000 = 225 electrical degrees/s²02․ If a generator of 250 MVA rating has an inertia constant of 6 MJ/MVA, its inertia constant on 100 MVA base is

Inertia constant (H) in pu = H/S

_{b}Where, S_{b}= Base MVA H_{pun}= H_{puo}* (MVA)_{bo}/(MVA)_{bn}Here, "n" represents the new values "o" represents the old values H_{pun}= 6*250/100 = 15 MJ/MVA03․ If the inertia constant H = 10 MJ/MVA for a 50 MVA generator, the stored energy is

Inertia constant H = Energy stored / MVA

_{base}base Energy stored = 10 * 50 = 500 MJ04․ Equal area criterion gives the information regarding

Equal area criterion used to transient stability analysis of power system, 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.

05․ Steady state stability of a power system is improved by

Steady state stability of a power system can be increased by using
1. Parallel transmission lines
2. Bundled conductors
3. Series capacitors

06․ The sequence components of the fault current are as follows:
I

_{R1}= j1.5 pu, I_{R2}= -j0.5 pu, I_{R0}= -j1. The type of fault isFor LLG fault,
I

_{R1}+ I_{R2}+ I_{R0}= 007․ The pu parameters for a 500 MVA machine on its own base are:
Inertia, H = 20 pu; reactance X = 2 pu.
The pu values of inertia and reactance on 100 MVA common base, respectively, are

Inertia constant (H) in pu = H/S

_{b}Where, S_{b}= Base MVA H_{pun}= H_{puo}* (MVA)_{bo}/(MVA)_{bn}Here, "n" represents the new values "o" represents the old values H_{pun}= 20*500/100 = 100 pu Z_{pun}= Z_{puo}* (MVA)_{bn}/(MVA)_{bo}Z_{pun}= 2*100/500 = 0.4 pu08․ An isolated synchronous generator with transient reactance equal to 0.1 pu on a 100 MVA base is connected to the high voltage bus through a step up transformer of reactance 0.1 pu on a 100 MVA base. The fault level at the bus is

The fault level at the bus = MVA

_{base}/X_{pu}The fault level at the bus = 100/0.2 = 500 MVA09․ An over current relay, having a current setting of 125% connected to a supply circuit through a current transformer of ratio 400/5. The pick up value is

Relay pick up current = % current setting * current transformer secondary current
Relay pick up current = 1.25*5 = 6.25 A

10․ An over current relay, having a current setting of 125% connected to a supply circuit through a current transformer of ratio 500/5. The plug setting multiplier for a fault current of 5 kA is

Plug setting multiplier = Fault current/(Pick up current * CT ratio)
Relay pick up current = % current setting * current transformer secondary current
= 1.25*5
= 6.25 A
Plug setting multiplier = 5*1000/(6.25*500/5) = 8

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