Transıent Stabılıty
12 Temmuz 2007
TRANSIENT STABILITY
Problem 3.
A 60 Hz synchoronous generator having inertia constant H=5 MJ/MVA and transient reactance pu is connected to an infinite bus through a purely reactive circuit as shown in figure 3.1. Reactances are marked on the diagram on a common system base. The generator is delivering real power of 0.8 pu and 0.074 reactive power to the infinite bus at a voltage 1.0 pu;
Figure 3.1One line diagram forProb.3.
A temporary 3-phase fault occurs at the sending end of the line at point F.When the fault is cleared,both lines are intact. Determine the critical clearing angle and the critical fault clearing time.
A 3-phase fault occurs at the middle of one of the lines,the fault is cleared,and the faulted line is isolated.Determine the critical clearing angle.
Solution:
When a generator is supplying power to an infinite bus over a parallel transmission lines, opening one of the lines may cause the generator to lose the synchronism evethough the load could be supplied over the remaining line under steady-state conditions. If a 3-phase short circuit occurs on the bus to which two parallel lines are connected,no power can be transmitted over other line.
However,if the fault is at the end of one of the lines,opening breakers at both ends of the line will isolate the fault from the system and allow power to flow through the other parallel line.
When a three-phase fault occurs at some point on a double-circuit line other than on the paralling buses or at the extreme ends of the line, there is some impedance between the paralleling buses and the fault. Therefore,some power is transmitted while the fault is still on the system.
The reactance diagram is shown in figure 3.2.
Figure 3.2 Reactance diagram for problem 3.
The series reactance between the terminal voltage and infinite bus is;
X=j0.2+j0.15=j0.35 pu
The power output of the generator is;
The output current from the generator is;
The power angle equation is;
Before the fault
The reactance diagram is shown in figure 3.3a with the fault on the point F. The effect of the fault is clearly shown by redrwing the the rectance diagram as shown figure3.3b
(a)
(b)
Figure3.3 The faulted network is shown in (a) and (b) in terms of admittance.
The of the system is;
The magnitude of the transfer admittance is 0.555;
Due to its inertia,the rotor cannot change position instantly upon occurrence of the fault. Therefore the rotor angle ? is initially 29.23 degree and electircal power output is Pe=0.5883sin(29.23)=0.287 puthe initial accelerating power Pa is 0.8-0.287=0.512 pu.
After the fault cleared, the faulted line is removed;
The net transfer admittance across the system;
The power angle equations obtained;
Before tha fault: Pmax*sin(?)=1.63*sin(?)
During the fault: r1*Pmax*sin(?)=0.5883*sin(?)
After the fault: r2*Pmax*sin(?)=1.325*sin(?)
where;
=============0============0==========0============0============0========
Problem 2.
In the system of above problem,a 3-phase fault at the middle of one line is cleared by isolating the faulted circuit simultaneously at both ends.
The fault is cleared in 0.3 second. Obtain the numerical solution of the swing equation for 1.0 second using the modified Euler method with a step size of ?t=0.01 second. From the swing curve,determine the system stability.
Solution:
The power angle equations are found in the previous example. The occurence of a fault causes a discontinuity in the accelerating power Pa,which has a zero value before the fault and nonzero value immediately following the fault. When the fault occurs, we have two values of Pa at the beginning of an interval,and we must take the average of these two values as the constant accelarating power.
The m-file is;
ro(1)=29.23;
f=60;
H=5;
delta_t=0.01; %Time variables identification part
time_initial=0; %initial time value
time_final=2.3; %final time value
time(1)=time_initial;
Pa(1)=0.5*(0.8-(0.5883*sin(ro(1)*pi/180))); %initial value of the Pa
delta_ro(1)=0;
k=180*f/H*(delta_t^2); %constant k
for i=1:1:(time_final/delta_t)
delta_ro(i+1)=delta_ro(i)+k*Pa(i);
ro(i+1)=ro(i)+delta_ro(i+1);
time(i+1)=time(i)+delta_t;
if time(i)>0.3
Pa(i+1)=0.5*(0.8-(1.325*sin(ro(i+1)*pi/180)));%after fault power angle
else
Pa(i+1)=0.5*(0.8-(0.5883*sin(ro(i+1)*pi/180)));%during the fault power angle
end
end
plot(time,ro) %plot the result
The result is;
According to equal area criteria the system is stable. Since the swing equation has sinusoidal shape. Unless it has the shape goes the infinity.
Kategori: Genel kültür