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O prezentare power point despre punctele trigger! Trigger POINTS!
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Given Gener ator Capacity:
25 MVA
Fr equency:
60Hz
Load:
20 MW
K.E. of Generator:
2.76 MJ/MVA at rated speed
Direct Axis Reactance Xd':
0.3 pu
Reactance of each double ckt line, X t:
0.2 pu on 25 MVA base
Vg behind transient ractance of generator Et:
1.03 pu
Soln: System Reactance: i) Pre fault X t1:
0.4 pu
ii) During fault X t2:
1.1 pu
iii) Post fault X t3:
0.5 pu
Max Power Transfer i) Pre fault P mxi:
2.575 pu
ii) During fault P mxii :
0.936 pu
iii) Post fault P mxiii:
2.06 pu
Inertia Constant
M=GH/180f
-
2.56*10
Initial Condition Power Output from generator: δo:
0.8 pu 18.1 degree
0.316 radians
Initially after the fault , angular position is not changed n
Hence, Fault power angle eq :
0.936*Sin 18.1
0.29 pu
200
150
100 Point by Point Runge Kutta 2nd Order 0.4s 50
Runge Kutta 2nd Order 0.5s Runge Kutta 2nd Order 0.6s Runge Kutta 2nd Order 0.7s
Solution by Runge Kutta Second Order Method: Initial Rotor Angle, δo
18.1 degree
0.315895 radians
Defining functions f 1(δ,ω)= dδ/dt = ω 2
2
f 2(δ,ω)= d δ/dt =( πf/H)*(Pm-PeSinδ) k1= f 1(δi,ωi)*∆t=ωi h1= f 2(δi,ωi)*∆t k2= f 1(δi+k1,ωi+h1)*∆t h2= f 2(δi+k1,ωi+h1)*∆t ∆δi= 1/2*(k1+k2) ∆ωi= 1/2*(h1+h2) δi+1= δi+∆δi ωi+1= ωi+∆ωi First iteration, i=0 ωo=0
