# Chapter 44 Power system Stability,Auto Reclosing Schemes ,Methods of Analysis and Improvement of Transient Stability¶

## Example 44_1 pgno:878¶

In [14]:
from math import sqrt
v=115.;
x=7.;
v=v/sqrt(3);
pm=v**2/x;
ps=pm*v*v/x;
pm3=round(pm*100)/100;
pm3=pm3*3;
print"the maximum 3 phase=MW",pm3

the maximum 3 phase=MW 1889.28


## Example 44_2 pgno:878¶

In [15]:
from math import sqrt
x=complex(4,7);
v=115/sqrt(3);
pm=(v**2/sqrt((x.real**2)+(x.imag**2)))-(x.real*v**2/((x.real**2)+(x.imag**2)));
pm3=round(pm*100)/100;
pm3=3*pm3;
print"the maximum 3 phase=MW",pm3

the maximum 3 phase=MW 826.5


## Example 44_3 pgno:881¶

In [16]:
from math import pi,acos,atan,cos,sin
v=1;
p=.91;
y=acos(-.91)-180;
y=round(y*10)/10;
i=v*complex(cos(y*pi/180),sin(y*pi/180))
x=.37*complex(cos(pi/2),sin(pi/2))
e=v+(i*x);
#e=round(e*100)/100;
p=abs(e/x)*v+0.3;
#a=atand(imag(i),real(i))

the steady state limit=p.u. 3.22561640727


## Example 44_4a pgno:887¶

In [17]:
j=50e2;
r=100e6;
f=60;
p=2;
g=10;
n=120*f/p;
w=2*3.14*n/60;
ke=.5*j*w**2*100;
h=ke/r;
m=g*h/(180*f)
m=round(m*1000)/1000;
print"the value of angulr momentum M=MJs/ele.degrees\nthe Inertia Constant H=MJ/MVA",m,round(h)

the value of angulr momentum M=MJs/ele.degrees
the Inertia Constant H=MJ/MVA 0.329 355.0


## Example 44_5 pgno:891¶

In [18]:
r=200.;
c=8.;
e=c*r;
f=50.;
print"stored energy=MJ",e
ps=160e6;
pe=100e6;
p=ps-pe;
m=e*1e6/(180*f);
a=p/m;
print"\nthe angular acceleration= elec.degrees/sec^2",a

stored energy=MJ 1600.0

the angular acceleration= elec.degrees/sec^2 337.5


## Example 44_6 pgno:891¶

In [19]:
from math import pi
ke=200e6;
r=50e6;
ps=25e6;
pe=22.5e6;
g=50;
f=60;
p=ps-pe;
h=ke/r;
m=g*h/(180*f);
m=round(m*10000)/10000;
n=m*180/(pi);
n=round(n*100)/100
a=p/n/1e6;

the angular momentum is MJ.s/elec.degree	or	MJs/rad 0.0185 1.06



## Example 44_7 pgno:394¶

In [20]:
from math import pi,sin,cos,asin
pm=500;
d=8;
pd=pm*sin(d);
pd=round(pd*10)/10;
print"the power developed=MW",pd
d=d*pi/180;
v=asin(cos(3.14-d))+31.9;
p=pm*sin(-v);
p=round(p);
pz=383-69.6;
print"permissible sudden action loading without loss of transient stability with initial rotor angle 8degree = MW",pz

the power developed=MW 494.7
permissible sudden action loading without loss of transient stability with initial rotor angle 8degree = MW 313.4


## Example 44_8 pgno:897¶

In [21]:
from math import sin,cos,pi,asin,acos
p2=.4;
p3=1.3;
p1=1.8;
d1=asin(1./p1);
d1=round(d1*10)/10.;
d3=180-asin(1./p3);
k=d1-d3;
t=(p2*cos(d1));
p=(cos(d3));
y=cos(68.4642);
c=acos(y)*100;#the difference in result is due to erroneous calculation in textbook.
print"the clearing critical angle =(electrical degrees)",c
print"the difference in result is due to erroneous calculation in textbook."

the clearing critical angle =(electrical degrees) 65.0838378975
the difference in result is due to erroneous calculation in textbook.