# Chapter 19 Applications of High Voltage Engineering in Industries¶

## Example 19.18.1 pgno:665¶

In :
#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"\nThe seperation between the particles = %.3f m"%round(2*x,3)

# Answers may vary due to round off error

The seperation between the particles = 1.633 m


## Example 19.18.2 pgno:667¶

In :
#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 = %d N/m**2"%P
# Answers may vary due to round off error

 The pumping pressure P = 900 N/m**2


## Example 19.18.4 pgno:670¶

In :
#Determine the vertical displacement of the drop
from math import pi

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

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 = %.1e m"%round(x1,4)

# Answers may vary due to round off error

 The vertical displacement of the drop = 2.9e-03 m


## Example 19.18.5 pgno672¶

In :
#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 = %.0e V/m"%Ea
print"\n Eb = %.0e V/m"%Eb
print"\n Charge density = %.2f micro C/m**2"%(psc/10**-6)

# Answers may vary due to round off error

 Ea = 9e+05 V/m

Eb = 3e+05 V/m

Charge density = 2.66 micro C/m**2


## Example 19.18.6 pgno:675¶

In :
#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 = %.2e A/m**2"%J

# Answer may vary due to round off error

 Current density = 5.30e-08 A/m**2


## Example 19.18.7 pgno:676¶

In :
#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 = %.2f mm"%round(d*1000,5)

# Answers may vary due to round off errors

 Thickness of the dust layer = 1.77 mm


## Example 19.18.8 pgno:676¶

In :
#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 = %.1e m/s"%vi
print"\n Populsion force = %.4e N"%F

# Answers may vary due to round off errors

 Ion velocity = 7.1e+04 m/s

Populsion force = 1.9715e-02 N


## Example 19.18.9 pgno:667¶

In :
#Determine the position of the particle
from math import sqrt

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

t0 = sqrt(vd/y)
# based on eq 19.51 and 19.52
dx2 = qm*V/d
x = t0**2
print"\n Velocity = %d m/s2"%dx2
print"\n Position of the particle = %.2f m"%round(x,2)

# Answer may vary due to round off error

 Velocity = 2 m/s2

Position of the particle = 0.24 m


## Example 19.18.10 pgno:679¶

In :
#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 = %.2f 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 = 11.81 kV