Answer questions Subjects : Stability and control of electric power systems Lecturer : Prof. Ir. Abraham Lomi, MT,. Ph.D Name : Supaman, ST NPM : 136060300111003 Class : Electric Power System

Assignments #1

1. The single line diagram of a generator connected through parallel transmission lines to a large metropolitan system machines is delivering 1,0 pu power and both the terminal voltage and the infinite bus voltage are 1,0 pu respectively. Number on the diagram indicate the values of the reactances on a common system base. The transient reactances of the generator is 0,2 pu as indicated, determine the power angle equation for the given system operating conditions.

Reactance diagram in the figure above is.

X=0.20+ 0.32

=0.35 pu

Because the generator has output of 1.0 pu then its angle can be obtained as below

|V t||V|X

sin ∂=(1.0x 1.0 )

0.35sin ∂=1.0

so that the voltage angle (ϑ) can be calculated as follows

∂=sin−1 0.35=19.2900

So that the terminal voltage is

V t=1.0

19.2900=0.954+ j 0.300 pu

Output current of the generator can be calculated as follows

I=1.0∠19.2900−1.0∠00

j 0.300

I=1.0+ j0.1535

I=1.012∠8.7290 pu

And stress in the transition is obtained as follows

E=(0.954+ j 0.300 )+ j (0.2 ) (1.0+ j 0.1535 )=0,23+ j 0,5≈ 1.050∠ 28,440 pu

The equation connecting the power angle transition voltage E’ and V rail voltage can be calculated from the overall reactance:

X=0.2+0.1+ 0.22

=0.4 pu

so that

Pe=1.050 x 1,0

0,5sin ∂=2,10 sin ∂ pu

2. If the system in (1) is operating under the indicated conditions when a three phase fault occurs at point F in fig. 1 determine the power angle equation for the system with fault on load and swing equation, take H=5 mj / mva

Answer:Reactance in the image above if terterjadi disorder as described in the above position and switching voltage of 1,05∠28,240 with the assumption that the constant flux in the machine

Y rel= j [−3,33 0 3,330 −5,50 2,50

3,33 2,50 −10,83]If the rail 3 will be eliminated matrix obtained as follows admintansi

[Y 11 Y 12

Y 21 Y 22]= j [−2,308 0,769

0,769 −6,923]The amount of the transfer is 0.769 so admintansi

Pmax=|E1||E2||Y 12|=(1,05 ) (1,0 ) ( 0,769 )=0,808 pu

Equations power angle disorder in the system is

Pe=0,808 sin δ pu

Assignments #2

1. A generator having H = 6.0 MJ /MVA is delivering power of 1.0 per unit to an infinite bus through a purely reactive network when the occurrence of a fault reduces the generator output power to zero. The maximum power that could be delivered is 2.5 per unit. When the fault is cleared, the original network conditions again exist. Determine the critical clearing angle and critical clearing time.

Penyelesaian:

2,5 sin δo=1.0

δo=23.580≈ 0.4115rad

δcr=cos−1 [ (π−2δo ) cosδ 0−cos δ 0 ]

δcr=cos−1 [ ( π−(2 x0.4115)) cos23.580−cos23.580 ]

δcr=cos−1 [ ( π−0.823 )cos 23.580−cos 23.580 ]

δcr=cos−1 (0.9275−0.9165 )

δcr=89.270 ≈1.560 rad

t cr=√ 4 H ( δcr−δo )ωs Pm

t cr=√ 4 x6 (1.395−0.4115 )2 π 60 x 1.0

=0.270 s

2. A 60-Hz generator is supplying 60% of Pmax to an infinite bus through a reactive network. A fault occurs which increases the reactance of the network between the generator internal voltage and the infinite bus by 400%. When the fault is cleared, the maximum power that can be delivered is 80% of the original maxi mum value. Determine the critical clearing angle for the condition described.

Penyelesaian

Pmax sin δo=60 % Pmax

δo=36.870 ,≈ 0.6435 rad

r1=0,25 r2=0.8

Pm=r2 Pmax sin δmax

Pm

Pmax

=0,6

sin δmax=0.60.8

=0.75

δmax=1800−48.590=131.410 ≈ 2.294 rad

cos δcr=0.6 (2.294−0.6435 )+( 0.8 cos131.40−0.25cos36.870 )

0.8−0.25=0.475

δ cr=cos−10.475 ≈ 61.640

Note: Answer sheets of assignment #1 and # 2 should be submitted on Tuesday, April 1, 2014.

Assignment #3

Rules: 1. Students are grouped into 3 groups and each group consists of 3 students. 2. Each group should write a paper in single column format based on group topics below. 3. Student may also refer to one or more IEEE papers (Journal) in related topics to enhance 4. the material quality of your paper. 5. Papers should be submitted by April 8, 2014 to abraham@itn.ac.id 6. The paper will be presented on Class meeting, Wednesday, April 9, 2014.

Group 1 Andrik Sunyoto, ST, David Suban Koten,ST & Suparman, ST(Class attendance list of 1-3) Topic: Improvement of Damping-Power System Stabilizer and SVS supplementary controls Group 2 (Class attendance list of 4-6) Topic: PV and QV Curves, Impact of Load and Tap-changer Dynamics Group 3 (Class attendance list of 7-9) Topic: Effect of Excitation System and AVR