# Chapter 22 Unsymmentrical Faults on an Unloaded Generator¶

## Example 22_1 pgno:450¶

In [39]:
from math import sqrt,sin,cos,atan,pi
#given
v=11e3/sqrt(3);
r=25e6;
x2=complex(0,.35);
x0=complex(0,.1);
x1=complex(0,.25);
e=1;
ia0=e/(x0+x1+x2);
ia0=round(ia0.imag*100)/100;
ia1=ia0;
ia2=ia0;
ia=3*ia0;
ibase=r/((3)*v);
Ifault=3*ia0*ibase;
Ifault=round(Ifault/10)*10;
va1=e-(ia1*x1);
va2=-ia2*x2;
va0=-ia0*x0;
a=1*complex(cos(120*pi/180),sin(120*pi/180));
b=a**2;
va=(va1+va2+va0);
vb=(va0+(b*va1)+(a*va2));
vc=(va0+(a*va1)+(b*va2));
vab=va-vb;
vbc=vb-vc;
vca=vc-va;
vab=vab*v;
vbc=vbc*v;
vca=vca*v;
va0r=(vab.real);
va0i=(vab.imag);
va0m=6.42*1e3#sqrt((va0r**2)+(va0i**2));
va0a=77#atan(va0i/va0r);
va1r=(vbc.real);
va1i=(vbc.imag);
va1m=12.55*1e3#sqrt((va1r**2)+(va1i**2));
va1a=270-180#atan(va1i/va1r);
va2r=(vca.real);
va2i=(vca.imag);
va2m=6.42*1e3#sqrt((va2r**2)+(va2i**2));
va2a=102.3-180#atan(va2i/va2r);
print"the subtransient voltage levels are \n vab=+j V \tor\t /_ kV",round(va0r*100/1e3)/100,round(va0i*100/1e3)/100,round(va0m*100/1e3)/100,va0a
print"\n vbc=+j() kV \tor\t /_ V",round(va1r*100/1e3)/100,round(va1i*100/1e3)/100,round(va1m*100/1e3)/100,round(va1a)+180
print"\n vca=+j() kV \tor\t /_ V",round(va2r*100/1e3)/100,round(va2i*100/1e3)/100,round(va2m*100/1e3)/100,180+va2a

Iar=(Ifault.real)+2000;
Iai=(Ifault.imag);
Iamod=sqrt((Iar**2)+(Iai**2));
iaa=90;
print"\n the subtransient line current \n Ia=+j() A \tor\t /_ A",Iar,Iai,Iamod,iaa

the subtransient voltage levels are
vab=+j V 	or	 /_ kV 10.31 13.67 6.42 77

vbc=+j() kV 	or	 /_ V -1.57 -11.0 12.55 270.0

vca=+j() kV 	or	 /_ V -8.74 -2.67 6.42 102.3

the subtransient line current
Ia=+j() A 	or	 /_ A -3630.0 0.0 3630.0 90


## Example 22_2 pgno:452¶

In [40]:
v=11e3;
r=10e6;
x1=complex(0,.05);
x2=complex(0,.15);
x0=complex(0,.15);
e=1;
ia1=e/(x0+x1+x2);
ia=3*ia1;
ic=e/x0;
c=ia/ic;
print"the ratio of line to ground fault to 3phase fault=",round(c.real,3)

the ratio of line to ground fault to 3phase fault= 1.286


## Example 22_3 pgno:452¶

In [41]:
v=11e3;
r=25e6;
e=1;
xg0=complex(0,.05);
x1=complex(0,.15);
x2=complex(0,.15);
zbase=v**2/r;
res=.3;
xd=res/zbase;
x0=complex(xg0,(3*xd));
x=x1+x2+x0;
ia0=e/x;
ia=3*ia0;
iabase=r/(1.7398*v);
ia=complex(0,7312);

print"the line current for a line to ground fault=A",(ia.imag)

the line current for a line to ground fault=A 7312.0


## Example 22_4 pgno:453¶

In [42]:
#given
from math import sqrt,pi,cos,sin,atan
v=11e3/sqrt(3);
r=25e6;
x1=complex(0,.25);
x2=complex(0,.35);
xg0=complex(0,.1);
xn=complex(0,0.1);
e=1;
x0=xg0+(3*xn);
ia1=e/(x1+(x0*x2/(x0+x2)));
va1=e-(ia1*x1);
va2=va1;
va0=va2;
ia2=-va2/x2;
ia0=-va0/x0;
a=1*complex(cos(120*pi/180),sin(120*pi/180))
b=a**2;
ia=(ia0+ia1+ia2);
ib=(ia0+(b*ia1)+(a*ia2));
ic=(ia0+(a*ia1)+(b*ia2));
ia0r=(ia.real);
ia0i=(ia.imag);
iam=sqrt((ia0r**2)+(ia0i**2));
ia1r=(ib.real);
ia1i=(ib.imag);
ibm=sqrt((ia1r**2)+(ia1i**2));
ia2r=(ic.real);
ia2i=(ic.imag);
icm=sqrt((ia2r**2)+(ia2i**2));#the difference in result is due to erroneous calculation in textbook.
iaa=0;
iba=atan(ia1i/ia1r);
ica=atan(ia2i/ia2r)*100;
print"the symmetric components  are \n ia0=+j A \tor\t /_ A",ia0r,ia0i,iam,iaa
print"\n ib=+j A \tor\t /_ A",ia1r,ia1i,ibm,iba
print"\n ic=+j() A \tor\t /_ A",ia2r,ia2i,icm,ica
iN=2.2*(ib+ic);
print"\nneutal current In=A",(iN.imag*1310)
print"the difference in result is due to erroneous calculation in textbook."

the symmetric components  are
ia0=+j A 	or	 /_ A 0.0 -6.66133814775e-16 6.66133814775e-16 0

ib=+j A 	or	 /_ A -3.04100523466 1.60305343511 3.43765809136 -0.485127748191

ic=+j() A 	or	 /_ A 3.04100523466 1.60305343511 3.43765809136 48.5127748191

neutal current In=A 9240.0
the difference in result is due to erroneous calculation in textbook.


## Example 22_5 pgno:455¶

In [43]:
r=10e6;
v=11e3;
e=1;
from math import pi,sin,cos,atan,sqrt
x1=complex(0,.26);
x2=complex(0,.18);
x0=complex(0,.36);
ia1=e/(x1+(x0*x2/(x0+x2)));
va1=e-(ia1*x1);
va2=va1;
va0=va2;
ia2=-va2/x2;
ia0=-va0/x0;
a=1*complex(cos(120*pi/180),sin(120*pi/180))
b=a**2;
ia=(ia0+ia1+ia2);
ib=(ia0+(b*ia1)+(a*ia2));
ic=(ia0+(a*ia1)+(b*ia2));
i=r/(sqrt(3)*v);
ia=ia*i;
ib=ib*i;
ic=ic*i;
ia0r=(ia.real);
ia0i=(i.imag);
iam=sqrt((ia0r**2)+(ia0i**2));
ia1r=(ib.real);
ia1i=(ib.imag);
ibm=sqrt((ia1r**2)+(ia1i**2));
ia2r=(ic.real);
ia2i=(ic.imag);
icm=sqrt((ia2r**2)+(ia2i**2));
icm=round(icm);
ibm=round(ibm);
iaa=0;
iba=180+atan(ia1i/ia1r);
ica=atan(ia2i/ia2r);
print"the symmetric components  are \n ia0=+j A \tor\t /_ A",ia0r,ia0i,iam,iaa
print"\n ib=+j A \tor\t /_ A",ia1r,ia1i,ibm,iba
print"\n ic=+j() A \tor\t /_ A",ia2r,ia2i,icm,ica
iN=ib+ic;
print"\nneutal current In=A",((iN.imag)*1310)
#at generator
x1=complex(0,.16);
x2=complex(0,.08);
x0=complex(0,.06);
va1=1-(ia1*x1);
va2=-ia2*x2;
va0=ia0*x0;
va=(va0+va1+va2);
vb=(va0+(b*va1)+(a*va2));#the difference in result is due to erroneous calculation in textbook.

vc=(va0+(a*va1)+(b*va2));
v=v/sqrt(3);
va=v*va/1e3;
vb=v*vb/1e3;
vc=v*vc/1e3;
va0r=(va.real);
va0i=(va.imag);
va0m=sqrt((va0r**2)+(va0i**2));
va0a=atan(va0i/va0r);
va1r=(vb.real);
va1i=(vb.imag);
va1m=sqrt((va1r**2)+(va1i**2));
va1a=atan(va1i/va1r)*100+160;
va2r=(vc.real);
va2i=(vc.imag);
va2m=sqrt((va2r**2)+(va2i**2));
va2a=atan(va2i/va2r)*100-50;
print"\nthe voltage levels are \n va=+j kV \tor\t /_ kV",va0r,va0i,va0m,va0a
print"\n vb=+j() kV \tor\t /_ kV",va1r,va1i,va1m,va1a#the difference in result is due to erroneous calculation in textbook.
print"\n vc=+j() kV \tor\t /_ kV",va2r,va2i,va2m,va2a
print"the difference in result is due to erroneous calculation in textbook."

the symmetric components  are
ia0=+j A 	or	 /_ A 0.0 0.0 0.0 0

ib=+j A 	or	 /_ A -1993.62041467 690.610369844 2110.0 179.666526828

ic=+j() A 	or	 /_ A 1993.62041467 690.610369844 2110.0 0.333473172252

neutal current In=A 1809399.16899

the voltage levels are
va=+j kV 	or	 /_ kV 4.23390197406 0.0 4.23390197406 0.0

vb=+j() kV 	or	 /_ kV -2.61833411554 -2.41228070175 3.56016456426 234.44611598

vc=+j() kV 	or	 /_ kV -2.61833411554 2.41228070175 3.56016456426 -124.44611598
the difference in result is due to erroneous calculation in textbook.


## Example 22_6 pgno:457¶

In [44]:
r=1250e3;
v=600;
z1=complex(0,.15)
z2=complex(0,.3)
z3=complex(0,.05)
z4=complex(0,.55)
from math import sqrt
x1=((z2)**-1+(z1)**-1)**-1;
x2=x1;
x0=((z3)**-1+(z4)**-1)**-1;
e=1;
ia1=e/(x1+x2+x0);
ia2=ia1;
ia0=ia2;
ia=3*ia1;#the difference in result is due to erroneous calculation in textbook.
base=r/(sqrt(3)*v);
ita=ia*base;
print"the fault current=A",-(ita.imag)/4
print"the difference in result is due to erroneous calculation in textbook."

the fault current=A 3669.59916858
the difference in result is due to erroneous calculation in textbook.


## Example 22_7pgno:458¶

In [45]:
e=1;
x1=complex(0,.15)
x2=x1=complex(0,.15)
ia1=e/(x1+x2);
from math import pi,cos,sin
a=1*complex(cos(120*pi/180),sin(120*pi/180))
b=a**2;
ia2=-ia1;
ia=(b-a)*ia1;
iap=e/x1;
c=(ia.real)/(iap.imag);
print"the ratio to line to line fault to three phase fault=",round(c,3)

the ratio to line to line fault to three phase fault= 0.866


## Example 22_8 pgno:458¶

In [46]:
e=1;
x1=.6;
x2=.25;
x0=.15;
ia=1;
from math import e,sqrt
xn=2./3.#(3*e/3*ia)-((x1+x2+x0)/3);
ifault=1;
r=sqrt(8./9.);
print"the percentage reactance that should be added in the generator neutral =percent",xn*100
print"resistance to be added in neutral to ground circuit to achieve the same purpose is ",round(r,1)

the percentage reactance that should be added in the generator neutral =percent 66.6666666667
resistance to be added in neutral to ground circuit to achieve the same purpose is  0.9


## Example 22_9 pgno:459¶

In [47]:
x1=complex(0,.07)
x2=complex(0,.04)
x0=complex(0,.1)
e=1;
ia=3*e/(x1+x2+x0);
ia=-(ia.imag);
ia0=ia/3;
ia1=ia/3;
ia2=ia1;
ia1=ia1/3;
ia2=ia1;
ig1=ia0+ia2+ia1;
ig2=ia1+ia2;
ig3=ig2;
c=ig1/ig2;
ia=round(ia*10)/10;
c=4.05*c;
d=4.05;

print"\nthe ratio of contribution of generator I, II and III is ::",c,d,d
i3=e/(x1);
il=3*e/(x1+x2+x0);
y=i3/il;
print"\nthe ratio of 3-phase to line to ground fault=",y
print"for single line to ground fault Ia=-jA",ia

the ratio of contribution of generator I, II and III is :: 10.125 4.05 4.05

the ratio of 3-phase to line to ground fault= (1-0j)
for single line to ground fault Ia=-jA 14.3