# Chapter 19 Applications of High Voltage Engineering in Industries¶

## Example 19_1 pgno:665¶

In [1]:
 #Chapter 19,Example 1,page 665
#Determine the sepration between the particles

# Based on the equations 19.6, 19.7, 19.8, 19.9 and 19.10
E = 8*10**5 # V/m
qm = 10*10**-6 # C/kg, qm = q/m
y = -1 # m
t = (1*2/9.8)
x = 1./2.*qm*E*t
print"\n The seperation between the particles =  m",round(2*x,3)

# Answers may vary due to round off error

 The seperation between the particles =  m 1.63265306122


## Example 19_2 pgno:667¶

In [2]:
#Chapter 19,Example 2,page 667
#Determine the pumping pressure

p0 = 30*10**-3 # C/m**3
V = 30*10**3 # V
P = p0*V
print"\n The pumping pressure P =  N/m**2",P
# Answers may vary due to round off error

 The pumping pressure P =  N/m**2 900.0


## Example 19_4 pgno:670¶

In [4]:
#Chapter 19,Example 4,page 670
#Determine the vertical displacement of the drop

d = 0.03*10**-3 # m
p = 2000 # kg/m**3
q = 100*10**-15 # C
V0 = 3500. # V
d2 = 2*10**-3 # m
L1 = 15*10**-3 # m
L2 = 12*10**-3 # m
Vz = 25. # m/s
from math import pi
m = 4./3.*pi*(1./2.*d)**3*p
t0 = L1/Vz
Vx0 = q*V0*t0/(m*d2)
x0 = 1./2.*Vx0*t0
t1 = (L1+L2)/Vz
x1 = x0+Vx0*(t1-t0)

print"\n The vertical displacement of the drop =  m",round(x1,4)

# Answers may vary due to round off error

 The vertical displacement of the drop =  m 0.0029


## Example 19_5 pgno672¶

In [5]:
#Chapter 19,Example 5,page 672
#Determine the electric stress and charge density

a = 25*10**-6 # m
b = 75*10**-6 # m
Er = 2.8
ps = 25*10**-6 # C/m**3
E0 = 8.84*10**-12

Ea = (b*ps)/(ps*E0+b*Er*E0)
Eb = (a*ps)/(ps*E0+b*Er*E0) # the negative noation is removed to obtain positive answer as in the book
psc = E0*Eb

print"\n Ea =  V/m",Ea
print"\n Eb =  V/m",Eb
print"\n Charge density =  C/m**2",psc

# Answers may vary due to round off error

 Ea =  V/m 902570.520843

Eb =  V/m 300856.840281

Charge density =  C/m**2 2.65957446809e-06


## Example 19_6 pgno:675¶

In [6]:
#Chapter 19,Example 6,page 675
#Determine the current density

E0 = 8.84*10**-12
Us = 1.5*10**-3*10**-4
V = 100
d3 = 10**-6 # d**3
J = 4*E0*Us*V**2/d3
print"\n Current density =  A/m**2",J

# Answer may vary due to round off error

 Current density =  A/m**2 5.304e-08


## Example 19_7 pgno:676¶

In [8]:
#Chapter 19,Example 7,page 676
#Determine the thickness of dust layer

Edb = 3*10**6
E0 = 8.84*10**-12
p0 = 15*10**-3
d = Edb*E0/p0
print"\n Thickness of the dust layer =  m",round(d,5)

# Answers may vary due to round off errors

 Thickness of the dust layer =  m 0.00177


## Example 19_8 pgno:676¶

In [9]:
#Chapter 19,Example 8,page 676
#Determine the velocity of the ejected ions and propolsion force
from math import sqrt
mi = 133*1.67*10**-27 # kg
qi = 1.6*10**-19 # C
Va = 3500 # V
I = 0.2 # A
vi = sqrt(2*qi*Va/mi)
F = vi*mi*I/qi
print"\n Ion velocity =  m/s",vi
print"\n Populsion force =  N",F

# Answers may vary due to round off errors

 Ion velocity =  m/s 71010.8899112

Populsion force =  N 0.0197152859477


## Example 19_9 pgno:667¶

In [11]:
#Chapter 19,Example 9,page 677
#Determine the position of the particle

V = 120*10**3 # applied voltage in V
d = 0.6 #  space b/w the plates in m
vd = 1.2 # vertical dimention in m
qm = 10*10**-6 # charge to mass C/kg
y = 4.9
from math import sqrt
t0 = sqrt(vd/y)
# based on eq 19.51 and 19.52
dx2 = qm*V/d
x = t0**2
print"\n Velocity =  m/s2",dx2
print"\n Position of the particle =  m",round(x,2)

# Answer may vary due to round off error

 Velocity =  m/s2 2.0

Position of the particle =  m 0.24


## Example 19_10 pgno:679¶

In [13]:
#Chapter 19,Example 10,page 679
#Determine the minimum voltage required for gnerating drops witha charge of 50 pC per drop
from math import pi,log
q = 50*10**-12
a = 25*10**-6
b = 750*10**-6
E0 = 8.84*10**-12
r = 50*10**-6
V = (3*q*b**2*log(b/a))/(7*pi*E0*r**3)
print"\n The minimum voltage required for gnerating drops witha charge of 50 pC per drop =  kV",round(V*10**-6,2)

# Answers may vary due to round off error

 The minimum voltage required for gnerating drops witha charge of 50 pC per drop =  kV 11.81