Electrical Engineering: Transient Analysis and Stability Homework

This document presents a comprehensive solution to an electrical engineering assignment focusing on transient analysis, stability, and surge protection. The solution begins with an analysis of synchronous generators, including equations for rotor angle and generator frequency, and calculations for critical clearing time and angle. It then provides MATLAB code for simulating system behavior, including frequency and rotor angle variations. The assignment further explores impediments to adopting renewable energy resources in existing grids, along with potential solutions. MATLAB code is used to analyze the impact of system parameters, such as inertia constant (H) and damping coefficient (D), on frequency response. The document also provides a detailed discussion on the impact of H and D on frequency and offers conclusions. Figures and graphs are included to illustrate the results, and references are provided for further reading.

Electrical Engg.
Transient analysis stability and surge protection
Student Name –
Student ID -

Solution 1 a) Synchronous Generator:
Equation describing motion of rotor angle and the generator frequency :
2 H dw/dt = J d2θ / dt2
. w = generator frequency
. θ = Rotor angle
H = inertia constant
J =Total moment of inertia
b) Critical Angle = 0 degree
Critical clearing time = 0.4 seconds
Solution 1) c)
a)
Matlab Code :
E = 1.5;
V = 1;

X = 10;
Pm = 0.6 ;
H = 9.94;
D = 0.16;
f0 = 50;
Pmax = E*V/X;
d0 = asin (Pm/Pmax);
Ps = Pmax * cos (d0);
wn = sqrt (pi*60/H*Ps);
z = D/2*sqrt(pi*60/(H*Ps));
wd = wn * sqrt (1-z^2);
fd = wd / (2*pi);
tau = wd/(2*pi);
th = acos (z);
Dd0 = 10*pi/180;
t = 0:0.01:3;
Dd = Dd0 / sqrt (1-z^2)* exp (-z * wn * t).* sin(wd*t);
d = (d0 + Dd) * 180/pi;
Dw = -wn*Dd0/sqrt(1-z^2)*exp(-z*wn*t).*sin(wd*t);
f = f0+Dw/(2*pi);
subplot (2,1,1);
plot(t,abs(d));
xlabel ('t(sec)');
ylabel('Dela(deg)');
subplot (2,1,2);

plot(t,abs(f));
xlabel ('t(sec)');
ylabel('Frequency(Hz)');
%% axis([0 3 59.85 60.1]);
Figure 1
Solution 1) c)

b)
MATLAB Code:
clear all;
clc;
E = 1.5;
V = 1;
X1 = 1;
X2 = 0.2;
X3 = 0.3;
Pm = 0.6 ;
H = 9.94;
Pe1max = E*V/X1;
Pe2max = E*V/X2;
Pe3max = E*V/X3;
delta = 0:0.01:pi;
Pe1 = Pe1max * sin (delta);
Pe2 = Pe2max * sin (delta);
Pe3 = Pe3max * sin (delta);
d0 = asin(Pm/Pe1max);
dmax = pi - asin(Pm/Pe3max);
cosdc = (Pm*(dmax-d0)+Pe3max*cos(dmax)-Pe2max*cos(d0))/(Pe3max-Pe2max);
dc = acos(cosdc);
Pmx = [0 pi-d0]*180/pi;

Pmy = [Pm Pm];
x0 = [d0 d0]*180/pi;
y0 = [0 Pm];
xc = [dc dc]*180/pi;
yc = [0 Pe3max*sin(dc)];
xm =[dmax dmax]*180/pi;
ym =[ 0 Pe3max*sin(dmax)];
d0 = d0*180/pi;
dmax = dmax * 180/ pi;
dc = dc*180/pi;
x = (abs(d0):0.1:abs(dc));
y = Pe2max*sin(x*pi/180);
y1 = Pe2max*sin(d0*pi/180);
y2 = Pe2max*sin(dc*pi/180);
x = [d0 x dc];
y = [Pm y Pm];
xx = abs(dc) :0.1:abs(dmax);
h = Pe3max*sin(xx*pi/180);
xx = [dc xx dmax];
hh = [ Pm h Pm];
delta = delta*180/pi;
if H ~=0
d0r = d0*pi/180;
dcr=dc*pi/180;
tc = sqrt(2*H*(dcr-d0r)/(pi*50*Pm));

else
end
fprintf('\n Initial power angle = %7.3f\n',d0);
fprintf('Maximum angle swing = %7.3f\n',dmax);
fprintf('\n Critical clearing angle = %7.3f\n\n',dc);
fprintf('Critical clearing time = %7.3f sec. \n\n',tc);
h = figure;
figure(h);
fill(abs(x),abs(y),'m');
hold;
fill(abs(xx),abs(hh),'c');
plot (abs(delta),abs(Pe1),'k-',abs(delta),abs(Pe2),'r-',abs(delta),abs(Pe3),'g-',abs(Pmx),
abs(Pmy),'r-',abs(x0),abs(y0),abs(xc),abs(yc),abs(xm),abs(ym));
grid;
plot (abs(Pmx),abs(Pmy),'r--',abs(x0),abs(y0),abs(xc),abs(yc),abs(xm),abs(ym));
grid;
xlabel('Power angle (degree)');
ylabel('Electrical power output (pu)');

Figure 2

Figure 3
Initial power angle = 23.578
Maximum angle swing = 173.108
Critical clearing angle = 0.000
Critical clearing time = 0.426 sec.

Solution 2 )
Impediments for adopting renewable energy resources in the existing grid in terms of
transient stability – It does not make the system stable.
Reason : The reason for this is the low amount of stability offered by such systems.
Ways to overcome such impediments to incorporate more renewable energy sources in the
grid :
They can be incorporated if the overall system can be stabilized by use of systems.
Solution 3 )
(1)
s = tf ('s');
G1 = 1/(2*3*s+1);
G2 = 1/(2*5*s+1);
G3 = 1/(2*6*s+1);
F = 1/(0.0625*(1+0.25*s)*(1+0.55*s));
T1 = feedback(G1,F);
T2 = feedback(G2,F);
T3 = feedback(G3,F);
step(T1,'b',T2,'r',T3,'g');
legend('Hs=3','Hs=5','Hs=6');

Figure 4
Solution 3)
2)
s = tf ('s');
G1 = 1/(2*5*s+0.8);
G2 = 1/(2*5*s+0.9);
G3 = 1/(2*5*s+1.2);
F = 1/(0.0625*(1+0.25*s)*(1+0.55*s));
T1 = feedback(G1,F);

T2 = feedback(G2,F);
T3 = feedback(G3,F);
step(T1,'b',T2,'r',T3,'g');
legend('D=0.8','D=0.9','D=1.2');
Figure 5
Solution 3) (3) Impact of H and D ( Rotating mass and load ) on frequency :
As the value of H increases, peak value decreases. Change in value of D does not have much
effect on response.