In [1]:

```
import math
# GIVEN DATA
kV = 345. ; # Three phase transmission line voltage in kV
Z_s = 366. ; # Surge impedance of line in Ω
a = 24.6 ; # Spacing between adjacent conductors in feet
d = 1.76 ; # Diameter of conductor in inches
# CALCULATIONS
SIL = (kV)**2/Z_s ; # Surge Impedance loading of line in MW
# DISPLAY RESULTS
print " Surge Impedance Loading of line , SIL = %.f MW "%(SIL) ;
print " NOTE: Unit of SIL is MW and surge impedance is Ω" ;
#Note :ERROR: Mistake in unit of SIL in textbook
```

In [2]:

```
import math
# GIVEN DATA
SIL = 325. ; # Surge impedance Loading in MW . From exa 3.1
kV = 345. ; # Transmission line voltage in kV . From exa 3.1
# For case (a)
t_shunt1 = 0.5 ; # shunt capacitive compensation is 50%
t_series1 = 0 ; # no series compensation
# For case (b)
t_shunt2 = 0.5 ; # shunt compensation umath.sing shunt reactors is 50%
t_series2 = 0 ; # no series capacitive compensation
# For case (c)
t_shunt3 = 0 ; # no shunt compensation
t_series3 = 0.5 ; # series capacitive compensation is 50%
# For case (d)
t_shunt4 = 0.2 ; # shunt capacitive compensation is 20%
t_series4 = 0.5; # series capacitive compensation is 50%
# CALCULATIONS
# For case (a)
SIL1 = SIL*(math.sqrt( (1-t_shunt1)/(1-t_series1) )) ; # Effective SIL in MW
# For case (b)
SIL2 = SIL*(math.sqrt( (1+t_shunt2)/(1-t_series2) )) ; # Effective SIL in MW
# For case (c)
SIL3 = SIL*(math.sqrt( (1-t_shunt3)/(1-t_series3) )) ; # Effective SIL in MW
# For case (d)
SIL4 = SIL*(math.sqrt( (1-t_shunt4)/(1-t_series4) )) ; # Effective SIL in MW
# DISPLAY RESULTS
print " a) Effective SIL , SIL_comp = %.f MW "%(SIL1) ;
print " b) Effective SIL , SIL_comp = %.f MW "%(SIL2) ;
print " c) Effective SIL , SIL_comp = %.f MW "%(SIL3) ;
print " d) Effective SIL , SIL_comp = %.f MW "%(SIL4) ;
print " NOTE: Unit of SIL is MW and surge impedance is Ω " ;
# Note : ERROR: Mistake in unit of SIL in textbook "
```

In [3]:

```
# GIVEN DATA
# For case (c)
I_normal = 1000. ; # Normal full load current in Ampere
# CALCULATIONS
# For case (a) equation is (1.5pu)*I_rated = (2 pu)*I_normal
# THEREFORE
# I_rated = (1.333pu)*I_normal ; # Rated current in terms of per unit value of the normal load current
# For case (b)
Mvar = (1.333)**2 ; # Increase in Mvar rating in per units
# For case (c)
I_rated = (1.333)*I_normal ; # Rated current value
# DISPLAY RESULTS
print " a) Rated current , I_rated = 1.333 pu*I_normal " ;
print " b) Mvar rating increase = %.2f pu "%(Mvar) ;
print " c) Rated current value , I_rated = %.f A "%(I_rated) ;
```