# Chapter 18: Unsymmetrical Fault Calculations¶

### Example 18.2, Page Number: 429¶

In [2]:
from __future__ import division
from pylab import *
import cmath
import math

#Variable declaration:
Ir = cmath.rect(100,30*math.pi/180)          #current in phase R(A)
Iy = cmath.rect(50,300*math.pi/180)          #current in phase Y(A)
Ib = cmath.rect(30,180*math.pi/180)          #current in phase B(A)

#Calculation:
a = cmath.rect(1,120*math.pi/180)
I0 = 1/3*(Ir+Iy+Ib)                    #A
I1 = 1/3*(Ir+a*Iy+a**2*Ib)             #A
I2 = 1/3*(Ir+a**2*Iy+a*Ib)             #A
In = Ir+Iy+Ib                          #A

#Result:
print "The positive, negative and zero sequence currents in the R-line are"
print "Return current in the neutral wire is"

The positive, negative and zero sequence currents in the R-line are
I0 = ( 27.29 4.69 ) A
I1 = ( 57.98 43.3 ) A
I2 = ( 18.97 24.96 ) A
Return current in the neutral wire is
In =( 81.88 4.7 ) A


### Example 18.3, Page Number: 430¶

In [3]:
from __future__ import division
from pylab import *
import cmath
import math

#Variable declaration:
Ir = 12+6j                  #current in phase A(A)
Iy = 12-12j                  #current in phase Y(A)
Ib = -15+10j                  #current in phase B(A)

#Calculation:
a = cmath.rect(1,120*math.pi/180)
#Red phase
Ir0 = 1/3*(Ir+Iy+Ib)            #A
Ir1 = 1/3*(Ir+a*Iy+a**2*Ib)     #A
Ir2 = 1/3*(Ir+a**2*Iy+a*Ib)     #A

#Yellow phase:
Iy0 = Ir0                         #A
Iy1 = a**2*Ir1                    #A
Iy2 = a*Ir2                       #A

#Blue phase:
Ib0 = Ir0                        #A
Ib1 = a*Ir1                   #A
Ib2 = a**2*Ir2                   #A

#Result:
print "The sequence currents of red phase are:"
print "Ir0 =",Ir0.real+round(Ir0.imag,2)*1j,"A"
print "Ir1 =",round(Ir1.real,2)+round(Ir1.imag,2)*1j,"A"
print "Ir2 =",round(Ir2.real,2)+round(Ir2.imag,2)*1j,"A"
print "\nThe sequence currents of red phase are:"
print "Iy0 =",Iy0.real+round(Iy0.imag,2)*1j,"A"
print "Iy1 =",round(Iy1.real,2)+round(Iy1.imag,1)*1j,"A"
print "Iy2 =",round(Iy2.real,1)+round(Iy2.imag,2)*1j,"A"
print "\nThe sequence currents of red phase are:"
print "Ib0 =",Ib0.real+round(Ib0.imag,2)*1j,"A"
print "Ib1 =",round(Ib1.real,2)+round(Ib1.imag,2)*1j,"A"
print "Ib2 =",round(Ib2.real,2)+round(Ib2.imag,2)*1j,"A"

The sequence currents of red phase are:
Ir0 = (3+1.33j) A
Ir1 = (10.85+10.13j) A
Ir2 = (-1.85-5.46j) A

The sequence currents of red phase are:
Iy0 = (3+1.33j) A
Iy1 = (3.35-14.5j) A
Iy2 = (5.7+1.13j) A

The sequence currents of red phase are:
Ib0 = (3+1.33j) A
Ib1 = (-14.2+4.33j) A
Ib2 = (-3.8+4.33j) A


### Example 18.4, Page Number: 430¶

In [4]:
from __future__ import division
from pylab import *
import cmath
import math

#Variable declaration:
Er0 = 100                        #zero sequence voltage of phase R(V)
Er1 = 200-100j                        #positive sequence voltage of phase R(V)
Er2 = -100                        #negative sequence voltage of phase R(V)

#Calculation:
a = cmath.rect(1,120*math.pi/180)
Er = Er0+Er1+Er2                 #V
Ey = Er0+a**2*Er1+a*Er2          #V
Eb = Er0+a*Er1+a**2*Er2          #V

#Result:
print "The phase voltages are:"

The phase voltages are:
Er = (223.61, -26.57) V
Ey = (213.0, -99.9) V
Eb = (338.59, 66.2) V


### Example 18.5, Page Number: 431¶

In [5]:
from __future__ import division
from pylab import *
import cmath
import math

#Variable declaration:
Er0 = 0.5-0.866j                  #zero sequence components of red phase(V)
Er1 = 2+0j                      #positive sequence components of red phase(V)
Er = 3+0j                         #phase R voltage(V)

#Calculation:
a = cmath.rect(1,120*math.pi/180)
Er2 = Er-(Er0+Er1)                #negative sequence voltage of red phase(V)
Ey = Er0+a**2*Er1+a*Er2            #phase Y voltage(V)
Eb = Er0+a*Er1+a**2*Er2            #phase B voltage(V)

#Result:
print "The negative sequence component of red phase is"
print "\nThe phase voltages are:"

The negative sequence component of red phase is
Ir2 = ( 1.0 60.0 ) V

The phase voltages are:
Iy = ( 3.0 -120.0 ) V
Ib = ( 0.0 120.0 ) V


### Example 18.6, Page Number: 431¶

In [6]:
from __future__ import division

#Variable declaration:
In = 12                      #current from neutral to ground(A)

#Calculation:
I0 = In/3                       #A

#Result:
print "The zero phase sequence components in phases are",I0,"A  each"

The zero phase sequence components in phases are 4.0 A  each


### Example 18.7, Page Number: 432¶

In [7]:
from __future__ import division
import cmath
import math

#Variable declaration:
#(i) Before removal of fuses.
Iri = cmath.rect(90,0)                  #phase R current(A)
Iyi = cmath.rect(90,math.pi/180*240)         #phase Y current(A)
Ibi = cmath.rect(90,math.pi/180*120)         #phase B current(A)

#(ii) After removal of fuses.
Irii = 90+0j                          #A
Iyii = 0                               #A
Ibii = 0                               #A

#Calculation:
a = cmath.rect(1,120*math.pi/180)
#(i) Before removal of fuses.
#Since the system is balanced, it will have only positive sequence currents.
Ir1i = Iri                          #A
Iy1i = Iyi                          #A
Ib1i = Ibi                          #A

#(ii) After removal of fuses.
#The sequence currents in the three lines will be:
Ir0ii = 1/3*(Irii+Iyii+Ibii)         #A
Iy0ii = Ir0ii                         #A
Ib0ii = Ir0ii                        #A

Ir1ii = 1/3*(Irii+Iyii+Ibii)          #A
Iy1ii = a**2*Ir1ii                    #A
Ib1ii = a*Ir1ii                       #A

Ir2ii = 1/3*(Irii+a**2*Iyii+a*Ibii)    #A
Iy2ii = a*Ir2ii                        #A
Ib2ii = a**2*Ir2ii                     #A

#Result:
print "(i) Before removal of fuses, the symmetrical components are:"
print "\n(ii) After the removal of fuses, the symmetrical components are:"

(i) Before removal of fuses, the symmetrical components are:
Ir1 = ( 90.0 0.0 ) A
Iy1 = ( 90.0 240.0 ) A
Ib1 = ( 90.0 120.0 ) A

(ii) After the removal of fuses, the symmetrical components are:
Ir0 = ( 30.0 0.0 ) A
Ir1 = ( 30.0 0.0 ) A
Ir2 = ( 30.0 0.0 ) A

Iy0 = ( 30.0 0.0 ) A
Iy1 = ( 30.0 240.0 ) A
Iy2 = ( 30.0 120.0 ) A

Ib0 = ( 30.0 0.0 ) A
Ib1 = ( 30.0 120.0 ) A
Ib2 = ( 30.0 240.0 ) A


### Example 18.8, Page Number: 433¶

In [8]:
from __future__ import division
import cmath
import math

#Variable declaration:
Ir1 = cmath.rect(200,0)              #positive phase sequence component of current(A)
Ir2 = cmath.rect(100,60*math.pi/180)      #negative phase sequence component of current(A)
In = cmath.rect(300,300*math.pi/180)      #current in the neutral conductor(A)

#Calculation:
a = cmath.rect(1,120*math.pi/180)
Ir0 = 1/3*In                        #Zero phase sequence current in R-line(A)
Ir = Ir0+Ir1+Ir2                    #Current in the R-line(A)
Iy = Ir0+a**2*Ir1+a*Ir2             #Current in the Y-line(A)
Ib = Ir0+a*Ir1+a**2*Ir2             #Current in the B-line(A)

#Result:
print "Current in the R-line is"
print "Current in the Y-line is"
print "Current in the B-line is"

Current in the R-line is
Ir = ( 300.0 0.0 ) A
Current in the Y-line is
Iy = ( 300.0 -120.0 ) A
Current in the B-line is
Ib = ( 0.0 180.0 ) A


### Example 18.9, Page Number: 434¶

In [9]:
from __future__ import division
import cmath
import math

#Variable declaration:
Ir = cmath.rect(10,0)                  #current in line R(A)
Iy = cmath.rect(10,180*math.pi/180)                  #current in line Y(A)
Ib = cmath.rect(0,0)                  #current in line B(A)

#Calculation:
a = cmath.rect(1,120*math.pi/180)
#For R-line:
Ir0 = 1/3*(Ir+Iy+Ib)                   #zero sequence component of Ir(A)
Ir1 = 1/3*(Ir+a*Iy+a**2*Ib)            #positive sequence component of Ir(A)
Ir2 = 1/3*(Ir+a**2*Iy+a*Ib)                   #negative sequence component of Iy(A)

#For Y-line:
Iy0 = Ir0                             #zero sequence component of Ir(A)
Iy1 = a**2*Ir1                        #positive sequence component of Ir(A)
Iy2 = a*Ir2                          #negative sequence component of Iy(A)

#For B-line:
Ib0 = Ir0                             #zero sequence component of Ir(A)
Ib1 = a*Ir1                           #positive sequence component of Ir(A)
Ib2 = a**2*Ir2                        #negative sequence component of Iy(A)

#Result:
print "The sequence components of R-line are:"
print "The sequence components of Y-line are:"
print "The sequence components of B-line are:"

The sequence components of R-line are:
Ir0 = ( 0.0 90.0 ) A
Ir0 = ( 5.77 -30.0 ) A
Ir0 = ( 5.77 30.0 ) A
The sequence components of Y-line are:
Iy0 = ( 0.0 90.0 ) A
Iy1 = ( 5.77 -150.0 ) A
Iy2 = ( 5.77 150.0 ) A
The sequence components of B-line are:
Ib0 = ( 0.0 90.0 ) A
Ib1 = ( 5.77 90.0 ) A
Ib2 = ( 5.77 -90.0 ) A


### Example 18.10, Page Number: 435¶

In [10]:
from __future__ import division
import cmath
import math

#Variable declaration:
Rrb = 10                       #resistance of branch RB(ohm)
Rry = 20                       #resistance of branch RY(ohm)
Rby = 5                       #resistance of branch BY(ohm)
Er = cmath.rect(-100,0)        #voltage across BY(V)
Ey = cmath.rect(100,60*math.pi/180)        #voltage across BR(V)
Eb = cmath.rect(100,-60*math.pi/180)        #voltage across RY(V)

#Calculation:
a = cmath.rect(1,120*math.pi/180)
IR = Er/Rby                    #Current in in branch BY(A)
IY = Ey/Rrb                    #Current in in branch RB(A)
IB = Eb/Rry                    #Current in in branch RY(A)

#Sequence currents in resistors:
IR0 = 1/3*(IR+IY+IB)           #Zero sequence component of IR(A)
IR1 = 1/3*(IR+a*IY+a**2*IB)    #Positive sequence component of IR(A)
IR2 = 1/3*(IR+a**2*IY+a*IB)     #Negative sequence component of IR(A)

IY0 = IR0                       #Zero sequence component of IY(A)
IY1 = a**2*IR1                    #Positive sequence component of IY(A)
IY2 = a*IR2                    #Negative sequence component of IY(A)

IB0 = IR0                    #Zero sequence component of IB(A)
IB1 = a*IR1                  #Positive sequence component of IB(A)
IB2 = a**2*IR2               #Negative sequence component of IB(A)

#Sequence currents in supply lines:
Ir = IB-IY                    #Line current in R-line(A)
Iy = IR-IB                     #Line current in Y-line(A)
Ib = IY-IR                     #Line current in R-line(A)

Ir0 = 1/3*(Ir+Iy+Ib)           #Zero sequence component of Ir(A)
Ir1 = 1/3*(Ir+a*Iy+a**2*Ib)    #Positive sequence component of Ir(A)
Ir2 = 1/3*(Ir+a**2*Iy+a*Ib)     #Negative sequence component of Ir(A)

#Result:
print "The sequence components of R-line are:"
print "\nThe sequence components of Y-line are:"
print "\nThe sequence components of B-line are:"
print "\nThe Sequence currents in supply lines are:"

The sequence components of R-line are:
IR0 = ( 4.41 160.9 ) A
IR1 = ( 11.67 180.0 ) A
IR2 = ( 4.41 -160.9 ) A

The sequence components of Y-line are:
IY0 = ( 4.41 160.9 ) A
IY1 = ( 11.67 60.0 ) A
IY2 = ( 4.4 -40.9 ) A

The sequence components of B-line are:
IB0 = ( 4.41 160.9 ) A
IB1 = ( 11.67 300.0 ) A
IB2 = ( 4.4 79.1 ) A

The Sequence currents in supply lines are:
Ir0 = ( 0.0 0.0 ) A
Ir1 = ( 20.2 -90.0 ) A
Ir2 = ( 7.64 109.1 ) A


### Example 18.11, Page Number: 437¶

In [11]:
from __future__ import division
import cmath
import math

#Variable declaration:
I = 20                       #current in each line(A)

#Calculation:
a = cmath.rect(1,120*math.pi/180)
#Let R, Y and B be the supply lines. When fuse in the
#line B is removed, the various line currents are :
Ir = cmath.rect(I,0)                     #A
Iy = cmath.rect(I,math.pi)                    #A
Ib = cmath.rect(0,0)                     #A

#For R-line:
Ir0 = 1/3*(Ir+Iy+Ib)                   #zero sequence component of Ir(A)
Ir1 = 1/3*(Ir+a*Iy+a**2*Ib)            #positive sequence component of Ir(A)
Ir2 = 1/3*(Ir+a**2*Iy+a*Ib)                   #negative sequence component of Iy(A)

#For Y-line:
Iy0 = Ir0                             #zero sequence component of Ir(A)
Iy1 = a**2*Ir1                        #positive sequence component of Ir(A)
Iy2 = a*Ir2                          #negative sequence component of Iy(A)

#For B-line:
Ib0 = Ir0                             #zero sequence component of Ir(A)
Ib1 = a*Ir1                           #positive sequence component of Ir(A)
Ib2 = a**2*Ir2                        #negative sequence component of Iy(A)

#Result:
print "The sequence components of R-line are:"
print "\nThe sequence components of Y-line are:"
print "\nThe sequence components of B-line are:"

The sequence components of R-line are:
Ir0 = ( 0.0 90.0 ) A
Ir0 = ( 11.55 -30.0 ) A
Ir0 = ( 11.55 30.0 ) A

The sequence components of Y-line are:
Iy0 = ( 0.0 90.0 ) A
Iy1 = ( 11.55 210.0 ) A
Iy2 = ( 11.55 150.0 ) A

The sequence components of B-line are:
Ib0 = ( 0.0 90.0 ) A
Ib1 = ( 11.55 90.0 ) A
Ib2 = ( 11.55 270.0 ) A


### Example 18.12, Page Number: 438¶

In [12]:
from __future__ import division
from pylab import *
import cmath
from sympy import *

#Variable declaration:
Zr = 5-10j                       #impedance connected to line R(ohm)
Zy = 6+5j                        #impedance connected to line Y(ohm)
Zb = 3+15j                       #impedance connected to line B(ohm)
VRY = cmath.rect(3300,0)         #line RY voltage(V)

#Calculation:
a = cmath.rect(1,120*math.pi/180)
VYB = a**2*VRY                   #line YB voltage(V)

#Let VR, VY, and VB be the voltages across impedances in R, Y and B
#phases respectively and IR, IY, and IB the resulting line currents.
IR, IY,IB,IR0,IR1,IR2, = symbols('IR IY IB IR0 IR1 IR2')

#IR+IY+IB = 0
IR0 = 0; IY0 = 0; IB0 = 0;
VR = Zr*(IR0+IR1+IR2)               #V
VY = Zy*(a**2*IR1+a*IR2)            #V
VB = Zb*(a*IR1+a**2*IR2)            #V
#solving the equations:
IR11 = solve(VR-VY-VRY,IR1)[0]
VY = Zy*(a**2*IR11+a*IR2)
VB = Zb*(a*IR11+a**2*IR2)
IR22 = solve(VY-VB-VYB,IR2)[0]

VR = Zr*(IR0+IR1+IR22)
VY = Zy*(a**2*IR1+a*IR22)
IR11 = solve(VR-VY-VRY,IR1)[0]
IR = IR11+IR22                      #A

#Result:
print "The line current IR is (",round(abs(IR)),round(-rad2deg(cmath.phase(IR)),1),")  A"

The line current IR is ( 241.0 -18.5 )  A


### Example 18.13, Page Number: 440¶

In [13]:
from __future__ import division
import cmath
import math

#Variable declaration:
R = 1                             #resistance of each load(ohm)
magVRY = 200                       #line RY voltage(V)
magVBR = 400                       #line BR voltage(V)
magVYB = 346                       #line YB voltage(V)

#Calculation:
a = cmath.rect(1,120*math.pi/180)
#This is a case of a balanced star-connected load supplied
#from an unbalanced 3-phase supply.

#Now,

#  (2)**22 = (1 + 1·75 cos(theta))**2 + (1·75 sin(theta))**2
#    theta = math.acos(4-1+3*1)/3.5
theta = math.pi/2
#    alpha = math.acos(1+1.75*math.cos(theta))
alpha = 60*math.pi/180

#As the phase sequence is RYB, therefore, various line voltages are :
VRY = cmath.rect(200,math.pi)                   #V
VYB = cmath.rect(346,theta)                #V
VBR = cmath.rect(400,-alpha)               #V

#The current in any phase (or line) is equal to phase voltage
#divided by resistance in that phase.
IR = VRY/(1*3**0.5)                   #A
IY = VYB/(1*3**0.5)                   #A
IB = VBR/(1*3**0.5)                   #A

#Sequence components in red phase are :
IR0 = 1/3*(IR+IY+IB)                  #A
IR1 = 1/3*(IR+a*IY+a**2*IB)           #A
IR2 = 1/3*(IR+a**2*IY+a*IB)           #A

#Result:
print "The magnitude of current in any phase is:"

The magnitude of current in any phase is:
IR0 = ( 0.0 -90.0 ) A
IR1 = ( 176.3 -169.0 ) A
IR2 = ( 66.63 30.0 ) A


### Example 18.14, Page Number: 449¶

In [14]:
from __future__ import division

#Variable declaration:
MVAg = 10                           #MVA rating of generator
Vg = 11                             #voltage rating of generator(kV)
Zg1 = 1.2j                          #Positive sequence impedance of generator(ohm)
Zg2 = 0.9j                          #Negative sequence impedance of generator(ohm)
Zg0 = 0.4j                          #Zero sequence impedance of generator(ohm)
Zf1 = 1.0j                          #Positive sequence impedance of feeder(ohm)
Zf2 = 1.0j                          #Negative sequence impedance of feeder(ohm)
Zf0 = 3.0j                          #Zero sequence impedance of feeder(ohm)

#Calculation:
#Suppose the fault is occured on the red phase.
#Taking red phase as the reference,
ER = Vg*1000/3**0.5                  #Phase e.m.f. of R-phase(V)

#(i)The total impedance to any sequence current is given by
#the sum of  generator and feeder impedances to that sequence current.
Z1 = Zg1+Zf1                       #ohm
Z2 = Zg2+Zf2                       #ohm
Z0 = Zg0+Zf0                      #ohm

#For a line-to-ground fault,
#I1 = I2 = I0
I0 = ER/(Z1+Z2+Z0)                 #A
IR = 3*I0                        #fault current(A)

#(ii)Line-to-neutral voltage of R-phase,
VR = ER-I0*(Zg1+Zg2+Zg0)                #V

#Result:
print "(i) The magnitude of fault current is",round(IR.imag)*1j,"A"
print "(ii) Line to neutral voltage at the generator terminal is",round(abs(VR)),"V"

(i) The magnitude of fault current is (-0-2540j) A
(ii) Line to neutral voltage at the generator terminal is 4234.0 V


### Example 18.15, Page Number: 450¶

In [15]:
from __future__ import division

#Variable declaration:
Vg = 11000                            #voltage rating of the alternator(V)
MVAg = 10                            #MVA rating of generator
X0 = 0.05                            #zero sequence reactance(p.u)
X1 = 0.15                            #positive sequence reactance(p.u)
X2 = 0.15                            #negative sequence reactance(p.u)

#Calculation:
#Taking red phase as the reference, let its phase e.m.f. be
ER = 1                              #p.u

#Line-to-ground fault:
#Suppose the fault occurs on the red phase.
#I1 = I2 = I0
I0 = ER/(1j*(X0+X1+X2))
IR = 3*I0                            #fault current(A)

#Three phase fault:
#the fault current (say Ish) is limited by the positive sequence reactance only.
Ish = ER/(1j*X1)                          #fault current(A)
r = IR/Ish                             #ratio of the two fault currents

#Result:
print "The ratio of two fault currents is",round(abs(r),3)

The ratio of two fault currents is 1.286


### Example 18.16, Page Number: 450¶

In [16]:
from __future__ import division

#Variable declaration:
Vg = 11                            #voltage rating of the alternator(kV)
MVAg = 25                            #MVA rating of generator
X0 = 0.05                            #zero sequence reactance(p.u)
X1 = 0.2                            #positive sequence reactance(p.u)
X2 = 0.2                            #negative sequence reactance(p.u)
Xn = 0.3                            #neutral to ground reactance(ohm)

#Calculation:
#Assume that the fault occurs on the red phase.
#Taking red phase as the reference, let its phase e.m.f. be
ER = 1                                  #p.u.
#First of all, convert the reactance Xn into p.u.
Xnp = Xn*MVAg*1000/(Vg**2*1000)         #p.u
#For a line-to-ground fault,
#I1 = I2 = I0
I0 = ER/(X1+X2+X0+3*Xnp)                  #p.u
IR = 3*I0                              #fault current(p.u)
IRa = MVAg*10**6/(3**0.5*Vg*1000) * IR        #fault current in amperes

#Result:
print " The fault current for a single line to ground fault is",round(IRa),"A"

 The fault current for a single line to ground fault is 6190.0 A


### Example 18.17, Page Number: 451¶

In [17]:
from __future__ import division

#Variable declaration:
V = 10.4                            #voltage between the lines(kV)
X0 = 0.2j                            #zero sequence reactance(p.u)
X1 = 0.6j                            #positive sequence reactance(p.u)
X2 = 0.5j                            #negative sequence reactance(p.u)

#Calculation:
#Taking red phase as the reference, its phase e.m.f. is :
ER = round(V*1000/3**0.5)                #Phase e.m.f. of R-phase(V)
IF = 3**0.5*ER/(X1+X2)              #fault current(A)

#Result:
print "The fault current is",round(abs(IF),1),"A"

The fault current is 9453.8 A


### Example 18.18, Page Number: 451¶

In [18]:
from __future__ import division

#Variable declaration:
X0 = 0.05j                            #zero sequence reactance(p.u)
X1 = 0.08j                            #positive sequence reactance(p.u)
X2 = 0.07j                            #negative sequence reactance(p.u)

#Calculation:
#Taking red phase as the reference, let its phase e.m.f. be
ER = 1                            #p.u
#For a double line-to-ground fault,
#IF = IY+IB
IF = -3*X2*ER/(X1*X2+X2*X0+X2*X0)         #fault current(A)

#Result:
print "The fault current is",round(IF.imag,1)*1j,"p.u"

The fault current is 16.7j p.u


### Example 18.19, Page Number: 452¶

In [19]:
from __future__ import division

#Variable declaration:
MVAg = 20                        #MVA rating of generator
V = 11                           #voltage rating of generator(V)
X1 = 20                          #positive sequence reactance of generator(%)
X2 = 10                         #negative sequence reactance of generator(%)
X0 = 15                        #zero sequence reactance of generator(%)
Xn = 5                       #generator 1 neutral reactance(%)

#Calculation:
ER = V*1000/3**0.5                  #V
#The % reactances in Fig. 18.19b can be converted into ohmic values as under:
x1 = X1*V**2*10/(MVAg*1000)                     #ohm
x2 = X2*V**2*10/(MVAg*1000)                     #ohm
x0 = X0*V**2*10/(MVAg*1000)                     #ohm
IR = 3*ER/(x1+x2+x0)*(-1j)                           #ohm

#Result:
print "Fault current is",round(IR.imag)*1j,"A"

Fault current is (-0-6998j) A


### Example 18.20, Page Number: 453¶

In [20]:
from __future__ import division
from sympy import *

#Variable declaration:
IF1 = 2000                       #3 phase fault current(A)
IF2 = 2600                       #line-to-line fault(A)
IF0 = 4200                       #line-toground fault(A)
MVAg = 50                        #MVA rating of the generator
V = 11                           #voltage rating of the generator(V)

#Calculation:
#Let X1 = positive sequence reactance,
#    X2 = negative sequence reactance,and
#    X0 = zero sequence reactance  of the alternator.
X1,X2,X0 = symbols('X1 X2 X0')
Eph = V*1000/3**0.5                    #phase voltage(V)
X11 = solve(Eph/X1-IF1)[0]                      #positive sequence reactance(ohm)
X22 = solve(3**0.5*Eph/(X11+X2)-IF2,X2)[0]      #negative sequence reactance(ohm)
X00 = solve(3*Eph/(X11+X22+X0)-IF0)[0]          #zero sequence reactance(ohm)

#Result:
print "The positive sequence reactance is",round(X11,3),"ohm"
print "The negative sequence reactance is",round(X22,3),"ohm"
print "The zero sequence reactance is",round(X00,3),"ohm"

The positive sequence reactance is 3.175 ohm
The negative sequence reactance is 1.055 ohm
The zero sequence reactance is 0.306 ohm