# Chapter 8: Overvoltage phenomenon and Insulation coordination in electric power systems¶

## Example 1: pg 350¶

In :
#example 8.1
#calculation of surge impedance,velocity and time taken by the surge to travel to the other end
from math import sqrt
#given data
L=1.26*10**-3#inductance(in H/km)
C=0.009*10**-6#capacitance(in F/km)
l=400.#length(in km) of the transmission line

#calculation
v=1/sqrt(L*C)
Xs=sqrt(L/C)
t=l/v
#results
print 'The value of surge impedance is (ohm) = ',round(Xs,1)
print '%s %.1e %s' %('\nThe value of velocity is ',v,' km/s')
print '\nThe time taken by the surge to travel to the other end is (ms) = ',round(t*10**3,2)

The value of surge impedance is (ohm) =  374.2

The value of velocity is  3.0e+05  km/s

The time taken by the surge to travel to the other end is (ms) =  1.35


## Example 2: pg 350¶

In :
#example 8.2
#calculation of the voltage build up at the junction

#given data
Z1=500.#surge impedance(in ohm) of transmission line
Z2=60.#surge impedance(in ohm) of cable
e=500.#value of surge(in kV)

#calculation
tau=(Z1-Z2)/(Z2+Z1)#coefficient of reflection
Vj=(1+tau)*e
#results
print 'The value of the voltage build up at the junction is (kV) = ',round(Vj)

The value of the voltage build up at the junction is (kV) =  893.0


## Example 5: pg 352¶

In :
#example 8.5
#calculation of the transmitted,reflected voltage and current waves
from math import sqrt
#given data
L1=0.189*10**-3#inductance(in H/km) of the cable
C1=0.3*10**-6#capacitance(in Farad/km) of the cable
L2=1.26*10**-3#inductance(in H/km) of the overhead line
C2=0.009*10**-6#capacitance(in Farad/km) of the overhead line
e=200.*10**3#surge volatge(in kV)

#calculation
Z1=sqrt(L1/C1)#surge impedance of the cable
Z2=sqrt(L2/C2)#surge impedance of the line
tau=(Z2-Z1)/(Z2+Z1)#when wave travels along the cable
edash=tau*e#reflected wave
edashdash=(1+tau)*e#transmitted wave
Idash=edash/Z1#reflected current wave
Idashdash=edashdash/Z2#transmitted current wave
Z2n=Z1
Z1n=Z2
taun=(Z2n-Z1n)/(Z2n+Z1n)#when wave travels along the line
edashn=taun*e#reflected wave
edashdashn=(1+taun)*e#transmitted wave
Idashdashn=edashdashn/Z2n#transmitted current wave
Idashn=edashn/Z1n#reflected current wave
#results
print 'When wave travels along the cable,the transmitted voltage is (kV) = ',round(edashdash*10**-3,2)
print '\nWhen wave travels along the cable,the reflected voltage is (kV) = ',round(edash*10**-3,2)
print '\nWhen wave travels along the cable,the transmitted current is (kA) = ',round(Idashdash*10**-3,3)
print '\nWhen wave travels along the cable,the reflected current is (kA) = ',round(Idash*10**-3,2)
print '\nWhen wave travels along the line,the transmitted voltage is (kV) = ',round(edashdashn*10**-3,2)
print '\nWhen wave travels along the line,the reflected voltage is (kV) = ',round(edashn*10**-3,2)
print '\nWhen wave travels along the line,the transmitted current is (kA) = ',round(Idashdashn*10**-3,3)
print '\nWhen wave travels along the line,the reflected current is ',round(abs(Idashn*10**-3),3),' kA or ',round(abs(Idashn)),'A'

When wave travels along the cable,the transmitted voltage is (kV) =  374.85

When wave travels along the cable,the reflected voltage is (kV) =  174.85

When wave travels along the cable,the transmitted current is (kA) =  1.002

When wave travels along the cable,the reflected current is (kA) =  6.97

When wave travels along the line,the transmitted voltage is (kV) =  25.15

When wave travels along the line,the reflected voltage is (kV) =  -174.85

When wave travels along the line,the transmitted current is (kA) =  1.002

When wave travels along the line,the reflected current is  0.467  kA or  467.0 A


## Example 6: pg 353¶

In :
#example 8.6
#calculation of value of voltage at the receiving end in Bewley lattice diagram

#given data
alpha=0.8

#calculation
Vut=2*alpha/(1+alpha**2)
#results
print 'The value of voltage at the receiving end in Bewley lattice diagram is ',round(Vut,4),'u(t) V'

The value of voltage at the receiving end in Bewley lattice diagram is  0.9756 u(t) V


## Example 7: pg 355¶

In :
#example 8.7
#calculation of sparkover voltage and the arrester current

#given data
Xs=400.#surge impedance(in ohm)
Xv=1000.#surge voltage(in kV)

#calculation
#for line terminated
Iam=2*Xv/Xs#maximum arrester current
#as Iam = 5 kA   from graph Vd = 330 kV
Vd=330.#sparkover voltage(in kV)
Vso=Vd+(Vd*5./100)
#for continuous line
Iamn=Xv/Xs#maximum arrester current
#as Iamn = 2.5 kA   from graph   Vdn = 280 kV
Vdn=280#sparkover voltage(in kV)
Vson=Vdn+(Vdn*5./100)
#results
print 'The sparkover voltage for terminated line is (kV) = ',round(Vso)
print '\nThe arrester current for terminated line is (kA) = ',round(Iam)
print '\nThe sparkover voltage for continuous line is (kV) = ',round(Vson)
print '\nThe arrester current for continuous line is (kA) = ',round(Iamn,1)
#values of sparover voltages are
#for terminated line = 346 kV
#for continuous line = 294 kV

The sparkover voltage for terminated line is (kV) =  347.0

The arrester current for terminated line is (kA) =  5.0

The sparkover voltage for continuous line is (kV) =  294.0

The arrester current for continuous line is (kA) =  2.5


## Example 8: pg 356¶

In :
#example 8.8
#calculation of rise in voltage at the other end
from cmath import pi,sqrt,cos
#given data
R=0.1#resistance(in ohm/km)
L=1.26*10**-3#inductance(in H/km)
C=0.009*10**-6#capacitance(in F/km)
l=400#length(in km) of the line
V1=230#line voltage(in kV)
f=50#frequency(in Hz)
G=0

#calculation
#Neglecting resistance of line
V1p=V1/sqrt(3)
omega=2*pi*f
Xl=complex(0,omega*L*l)
Xc=complex(0,-1/(omega*C*l))
V2=V1p*((1-(Xl/(2*Xc)))-1)

#Considering all the parameters
omegaL=complex(0,omega*L)
omegaC=complex(0,omega*C)
i=l*sqrt((R+omegaL)*(G+omegaC))
betal=i.imag*l
V2n=V1p/cos(betal)

print 'Neglecting resistance of line,the rise in voltage at the other end is (kV) = ',round(V2.real,2)
print '\nConsidering all the parameters,the rise in voltage at the other end is (kV) = ',round(V2n.real-V1p.real,2)

#By considering all the parameters the rise in voltage at the other end is 94.50 kV

Neglecting resistance of line,the rise in voltage at the other end is (kV) =  11.89

Considering all the parameters,the rise in voltage at the other end is (kV) =  94.5


## Example 9: pg 357¶

In :
#example 8.9
#working out of insulation coordination
from math import sqrt
#given data
V=220.#voltage(in kV) of substation
BIL=1050.#value of BIL(in kV)
BtoS=1.24#ratio of BIL to SIL

#calculation
Vh=245.#highest voltage(in kV)
Vg=Vh*sqrt(2.)/sqrt(3)#highest system voltage
Vs=3*Vg#expected switching voltage(in kV)
Vfw=760.#impulse sparkover voltage(in kV)
Vd1=690.#discharge voltage(in kV) for 5 kA
Vd2=615.#discharge voltage(in kV) for 2 kA
#SIL = BIL/BtoS = 846 ~ 850 kV
SIL=850.#value of SIL(in kV)
Pmlig=(BIL-Vd1)/BIL#protective margin for lightning impulses
Pmswi=(SIL-Vd2)/SIL#protective margin for switching gears
Pmspr=(BIL-Vfw)/BIL#margin when lightning arrester just sparks
#results
print 'The protective margin for lightning impulses is (percentage) = ',round(Pmlig*100,1)
print '\nThe protective margin for switching gears is (percentage) = ',round(Pmswi*100,1)
print '\nThe margin when lightning arrester just sparks is (percentage) = ',round(Pmspr*100,1)

The protective margin for lightning impulses is (percentage) =  34.3

The protective margin for switching gears is (percentage) =  27.6

The margin when lightning arrester just sparks is (percentage) =  27.6