# CHAPTER 1 - Breakdown Mechanism of Gases Liquid and Solid Materials¶

## EXAMPLE 1.1 - PG NO.51¶

In [1]:
#Chapter 1,Example 1.1 Page 51
import math
I = 600. # micor amps
x = 0.5 # distance in cm
V = 10. # kV
I2 = 60. # micro amps
x2 = 0.1 # distance in cm
#Calculation 600 = I0*exp(0.5*alpha) and 60 = I0*exp(0.1*alpha)
alpha =math.log(600./60.)/(0.5-0.1)
print'%s %.3f %s' %("Townsends first ionising coefficient = ",alpha," ionizing collisions/cm")

#Answers may vary due to round of error

Townsends first ionising coefficient =  5.756  ionizing collisions/cm


## EXAMPLE 1.2 - PG NO.52¶

In [2]:
#Chapter 1,Example 1.2 Page 52
import math
# Refering the table in example 1.2
# slope between any two points (math.log(I/I0)/x)
# taking the gap between 2 and 2.5 mm
I1= 1.5*10**-12
I2= 5.6*10**-12
I0 = 6*10**-14
gi1 = math.log(I1/I0) # gradual increase when gap is 2
gi2 = math.log(I2/I0) # gradual increase when gap is 2.5 #claculation in text is wrong
slope = (gi1-gi2)/0.05
print'%s %.3f %s' %("Slope = ", -slope,'\n')
#evaluvating ghama
e1 = math.exp(-slope*0.5)
e2 = math.exp(-slope*0.5) # -1 is ignored due to the large magnitude
ghama = (7*10**7-6*e1)/(e2*7*10**7)
print'%s %.3f %s' %("Ghama for set 1= ", ghama*100000,"*10^-5 /cm \n ")
#Gap between the slope for set 2
alpha = math.log(12./8.)/0.05
print'%s %.1f %s' %("Alpha = ", alpha," collosions/cm \n")
e1 = math.exp(alpha*0.5)
e2 = math.exp(alpha*0.5) # -1 is ignored due to the large magnitude
ghama = (2*10**5-e1)/(e2*2*10**5)
print'%s %.1f %s' %("Ghama for set 2=", ghama*100,"*10^-2 colissions/cm \n")

#Answers may vary due to round of error

Slope =  26.346

Ghama for set 1=  0.182 *10^-5 /cm

Alpha =  8.1  collosions/cm

Ghama for set 2= 1.7 *10^-2 colissions/cm



## EXAMPLE 1.3 - PG NO.53¶

In [3]:
#Chapter 1,Example 1.3 Page 53

#employing equation Vb = K*d**n
#88 = K*4**n --- eq(1)  165 = K*8**n ---eq(2)
#dividing eq(2)/q(1)
Vb1 = 88.
Vb2 = 165.
n1 = 0.6286/0.693
K1 = Vb1/4**n1
#135 = K*6**n --- eq(1)  212 = K*10**n ---eq(2)
#dividing eq(2)/q(1)
Vb1 = 135.
Vb2 = 212.
n2 = 0.4513/0.5128
K2 = Vb1/6.**n2
n = (n1+n2)/2.
K = (K1+K2)/2.
print'%s %.2f %s %.2f' % ("n =",n,"K = ",K,)

#Answer may vary due to round of error

n = 0.89 K =  26.46


## EXAMPLE 1.4 - PG NO.53¶

In [4]:
#Chapter 1,Example 1.4 Page 53
# Determine (pd)min Vbmin
import math
A = 12.
B = 365.
e = 2.718
ghama = 0.02
K = 51.
pd = (e/A)*math.log(1.+(1./ghama))
Vbmin = (B/A)*e*math.log(K)
print'%s %.2f %s %d %s' % ("(pd)min = ",pd," Vbmin = ",Vbmin,"Volts")

#Answers may vary due to round of error

(pd)min =  0.89  Vbmin =  325 Volts