Tutorial 2 Stability

Tutorial Tutorial #2: Power System Stability

Question # 1: A four-pole, 60-Hz synchronous generator has a rating of 200 MVA, 0.8 power factor lagging. The moment of inertia of the rotor is 45,100 kgm 2. Determine  M and H . Solution:

Question # 2: A two-pole, 60-Hz synchronous generator has a rating of 250 MVA, 0.8 power factor lagging. The kinetic energy of the machine at synchronous speed is 1080 MJ. The machine is running steadily at synchronous speed and delivering 60 MW to a load at a power angle of 8 electrical degrees. The load is suddenly removed. Determine the acceleration of the rotor. If the acceleration computed for the generator is constant for a period of 12 cycles, determine the value of the power angle and the rpm at the end of this time. Solution:

,

,

, or Integrating the above expression, we have

, Speed at the end of 0.2 second is Question # 3: 1

Determine the kinetic energy stored by a 250-MVA, 60-Hz, two-pole synchronous generator with an inertia constant  H of 5.4 MJ/MVA. Assume the machine is running steadily at synchronous speed with a shaft input of 331,100 hp. The electrical power developed suddenly changes from its normal value to a value of 200 MW. Determine the acceleration or deceleration of the rotor. If the acceleration computed for the generator is constant for a period of 9 cycles, determine the change in the power angle in that period and the rpm at the end of 9 cycles. Solution:

or

Integrating the above expression, we have

or

Speed at the end of 0.15 second is

2

Question # 4: The swing equations of two interconnected synchronous machines are written as

Solution:

Denote the relative power angle between the two machines by    =   1 -   2. Obtain a swing equation equivalent to that of a single machine in terms of   , and show that

where

Subtracting the second swing equation from the first, we obtain

or

or

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Question # 5: Two synchronous generators represented by a constant voltage behind transient reactance are connected by a pure reactance  X 0.3 per unit, as shown in Figure Q5. The generator inertia constants are  H 1 4.0 MJ/MVA and  H 2 6 MJ/MVA, and the transient reactances are X 1' = 0.16 and X 2' = 0.20 per unit. The system is operating in the steady state with  E 1' =1.2, Pm1 = 1.5 and  E 2' =1.1, Pm2 = 1.0 per unit. Denote the relative power angle  between the two machines by    =   1 -   2. Referring to Question 4, reduce the twomachine system to an equivalent one-machine against an infinite bus. Find the inertia constant of the equivalent machine, the mechanical input power, and the amplitude of its  power angle curve, and obtain the equivalent swing equation in terms of   . =

=

=

Fig. Q5

Solution:

Referring to Question.4, the equivalent parameters are

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Question # 6: A 60-Hz synchronous generator has a transient reactance of 0.2 per unit and an inertia constant of 5.66 MJ/MVA. The generator is connected to an infinite bus through a transformer and a double circuit transmission line, as shown in Fig. Q6. Resistances are neglected and reactances are expressed on a common MVA base and are marked on the diagram. The generator is delivering a real power of 0.77 per unit to bus bar 1. Voltage magnitude at bus 1 is 1.1. The infinite bus voltage V   1.00 per unit. Determine the generator excitation voltage and obtain the swing equation.

Fig. Q6 Solution:

or

The total reactance is X  = 0.2+0.158+0.4 = 0.758, and the generator excitation voltage is

the swing equation with ± in radians is

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Question # 7: A three-phase fault occurs on the system of Question 6 at the sending end of the transmission lines. The fault occurs through an impedance of 0.082 per unit. Assume the generator excitation voltage remains constant at E' = 1.25 per unit. Obtain the swing equation during the fault. The impedance network with fault at bus 1, and with with  Z  f  = j0.082 is shown in Fig. Q7. .

Fig. Q7 Solution:

Transforming the Y-connected circuit in FigQ7 into an equivalent , the transfer reactance between  E ' and V  is

Therefore, the swing equation during fault with ± in radians is

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Question # 8: The machine in the power system of Question 6 has a per unit damping coefficient of  D = 0.15. The generator excitation voltage is E'  = 1.25 per unit and the generator is delivering a real power of 0.77  per unit to the infinite bus at a voltage of V   = 1.0 per unit. Write the linearized swing equation for this power system. Find the equations describing the motion of the rotor angle and the generator frequency for a small disturbance of  = 15o. Solution:

The initial operating angle is given by

The synchronizing power coefficient given by

The undamped angular frequency of oscillation and damping ratio are

The linearized force-free equation which determines the mode of oscillation given  by with    in radian is

The damped angular frequency of oscillation is

corresponding to a damped oscillation frequency of

the motion of rotor relative to the synchronously revolving field in electrical degrees and the frequency excursion in Hz are given by equations

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Question # 9: Write the linearized swing equation of Question 8 in state variable form.    AX   BU   X  . Y   CX   DU  Solution:

Question # 10: The generator of Question 8 is operating in the steady state at   o = 27.835o when the input power is increased by a small amount P  = 0.15 per unit. The generator excitation and the infinite bus voltage are the same as before. Find the equations describing the motion of the rotor angle and the generator frequency for a small disturbance of P = 0.15 per unit. Solution:

Substituting for  H , we get

  o,

 and

 n evaluated

in Question 9 and expressing the power angle in degree,

or

Also,

or

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