Tam.docx

Problem 1: A Two area system with the following data: Total rated area capacity Pr1 = Pr2 = 2000 MW Normal operating load Po d 1= 1000 MW Inertia constant H = 5.0 sec Assume load frequency dependency is linear, D1=D2=1 pu MW/Hz = 8.33*10-3 pu MW/Hz Calculations: Tp1 = Tp2= (2H)/(fo*D) =20 sec Kp1 = Kp2=1/D = 120 Hz/pu MW CASES Case-1 : When there is no secondary ALFC control for the above system Case II : Considering secondary ALFC loop for the above system Case III: Tie line bias control parameter variations Case IV : Considering Generator and Turbine time constants (Tg & Tt) for the above system Assumptions: The value of Tg and Tt are equal to zero for case-1, case II & case III

Case-1 : When there is no secondary control (a) R= 4% = 2.4 Hz/pu MW (b) R= 3% = 1.8 Hz/pu MW (c) R= 5% = 3.0 Hz/pu MW for step increase of load = 20 MW = 0.01 pu

AGC control for 2 area system without secondary control

Problem 1: case I: R=2.4 Hz/pu MW; R=1.8 Hz/pu MW; R=3.0 Hz/pu MW (a)The output waveforms

From the above graphs we observe that static frequency drop at area1 and area2 and Tieline power flows are observed as R=2.4 Hz/pu MW

R=1.8 Hz/pu MW

R=3.0 Hz/pu MW

∆f1=-0.0118 Hz

∆f1=-0.0088 Hz

∆f1=-0.01478Hz

∆f2=-0.0118 Hz

∆f2=-0.0088 Hz

∆f2=-0.0148 Hz

∆P12 = -0.005 pu

∆P12 = -0.005 pu

∆P12 = -0.005 pu

Problem 2: A Two area system with the following data: AREA 1: Total rated area capacity Pr1 = 2000 MW Inertia constant H = 5.0 sec Regulation R1 = 2.40 Hz/pu MW Assume load frequency dependency is linear, D1 = 1 pu MW/Hz = 8.33*10-3 pu MW/Hz AREA 2: Total rated area capacity Pr2 = 10000 MW Inertia constant H = 5.0 sec Regulation R2 = 2.40 Hz/pu MW Assume load frequency dependency is linear, D2 = 1 pu MW/Hz = 8.33*10-3 pu MW/Hz R2 =0.48 Hz/pu MW , D2 = 0.04165 when expressed in 2GW base.

Calculations: Tp1 = Tp2= (2H)/(fo*D) =20 sec Kp1 = Kp2=1/D = 120 Hz/pu MW

Problem 2: Step load increase of 20 MW in 2GW area.

static frequency deviations in area 1 and area 2 respectively are ∆f1 and ∆f2 observed from figure as ∆f1 = ∆f2 = 0.00392 Hz Tieline power flow ∆P12 = -0.0833 pu MW.

Case II : Considering secondary ALFC loop for the above system

(Ki )critical ={ (1+Kp/R)2 }/4kp.Tp = 0.2709 for both the areas (a) (Ki ) = 0.2709 (b) (Ki ) = 1.0 (c) (Ki ) = 0.01

Observations from the graph below:  IN case (a) when Ki = (Ki )critical the system reaches steady state limit without any oscillations  IN case(b) system reaches steady state limit but with oscillations  IN case(c) system reaches stead state limit very slowly.

Case III: Tie line bias control(B) variations (a)B = β = 0.425 (b)B =0.5* β = 0.2125 (c) B =1.5* β = 0.6375

Observations from the graph below:  IN case (a) when B = β the system reaches steady state limit with acceptable variations in frequency  IN case(b) system reaches steady state but takes considerably long time  IN case(c) system reaches stead state limit but with large oscillations and high peak overshoot.

Case IV : Considering Generator and Turbine time constants (Tg & Tt) for the above system On assuming Tg = 80 ms Tt = 0.3 secs

From the above graph system attains constant frequency but with oscillations