# Chapter 06 : Corona¶

## Example 6.1, Page No 142¶

In [1]:
import math
#initialisation of variables
t=21.0     # air temperature
b=73.6     # air pressure
do=3.92*73.6/(273+t)
m=0.85
r=0.52
d=250.0

#Calculations
Vd=21.1*m *do*r*math.log(250/.52)
vd=math.sqrt(3)*Vd
m=0.7
vv=21.1*m*do*r*(1+(0.3/math.sqrt(r*do)))*math.log(250/0.52)
Vv=vv*math.sqrt(3)
Vvg=Vv*0.8/0.7

#Results
print("critical disruptive line to line voltage=%.2f kV " %vd)
print("visual critical voltage for local corona=%.2f kV " %vv)
print("visual critical voltage for general corona=%.2f kV " %Vvg)

critical disruptive line to line voltage=97.89 kV
visual critical voltage for local corona=66.09 kV
visual critical voltage for general corona=130.83 kV


## Example 6.2, Page No 142¶

In [2]:
import math
#initialisation of variables
d=2.5
di=3.0  # internal diameter
do=9.0  # external diameter
ro=do/2.0  # external diameter

#Calculations
g1max=20/(1.25*math.log(ri/(d/2))+0.208*1.5*math.log(ro/ri))

#Results
print("g1max=%.0f kV/cm" %g1max)
print("Since the gradient exceeds 21.1/kV/cm , corona will be present.")

g1max=35 kV/cm
Since the gradient exceeds 21.1/kV/cm , corona will be present.


## Example 6.3, Page No 143¶

In [3]:
import math
#initialisation of variables
m=1.07
r=0.625

#Calculations
V=21*m *r*math.log(305.0/0.625)
Vl=V*math.sqrt(3.0)

#Results
print("critical disruptive voltage=%.0f kV " %V)
print("since operating voltage is 110 kV , corona loss= 0 ")

critical disruptive voltage=87 kV
since operating voltage is 110 kV , corona loss= 0


## Example 6.4 Page No 143¶

In [4]:
import math
#initialisation of variables
r=0.5

#Calculations
V=21*r*math.log(100.0/0.5)

#Results
print("critical disruptive voltage=%.1f kV" %V)

critical disruptive voltage=55.6 kV


## Example 6.5, Page No 146¶

In [5]:
import math
#initialisation of variables
D=1.036 	       # conductor diameter(cm)
d=2.44	           #delta spacing(m)

#Calculations
Ratio=d*100.0/r
j=r/(d*100.0)
Rat2=math.sqrt(j)
t=26.67     	   #temperature
b=73.15            # barometric pressure
mv=0.72
V=63.5
f=50.0	           #frequency
do=3.92*b/(273+t)        #do=dell
vd=21.1*.85*do*r*math.log(Ratio)
print("critical disruptive voltage=%.2f kV" %vd)
Vv=21.1*mv*do*r*(1+(0.3/math.sqrt(r*do)))*math.log(Ratio)
Pl=241*(10**-5)*(f+25)*Rat2*((V-vd)**2)/do        #power loss
Vd=0.8*vd
Pl2=241*(10**-5)*(f+25)*Rat2*((V-Vd)**2)*160/do        #loss per phase /km
Total=3.0*Pl2

#Results
print("visual critical voltage=%.0f kV" %Vv)
print("Power loss=%.3f kW/phase/km" %Pl)
print("under foul weather condition ,")
print("critical disruptive voltage=%.2f kV" %Vd)
print("Total loss=%.0f kW\n"%Total)

critical disruptive voltage=54.72 kV
visual critical voltage=66 kV
Power loss=0.672 kW/phase/km
under foul weather condition ,
critical disruptive voltage=43.77 kV
Total loss=1626 kW