Chapter 6: Generation of High voltages and Currents

Example 1: pg 196

In [1]:
#example 6.1
#calculation of percentage ripple,the regulation and the optimum number of stages for minimum regulation in Cockcroft-Walton type voltage multiplier
from math import sqrt
#given data
C=0.05*10**-6#value of capacitance(in F)
Vmax=125.*10**3#value of supply transformer secondary voltage(in V)
f=150.#frequency(in Hz)
I=5.*10**-3#load current(in A)
nst=8.#number of stages

#calculation
n=nst*2#number of capacitors
#from equation of ripple voltage
deltaV=(I/(f*C))*(n*(n+1)/2)
perripple=(deltaV*100)/(16*Vmax)
deltaVn=(I/(f*C))*(((2*nst**3)/3)+(nst*nst/2)-(nst/6))#voltage drop...here n = nst = number of stages
reg=deltaVn/(2*nst*Vmax)#regulation
nopt=round(sqrt(Vmax*f*C/I))#optimum number of stages
#results
print 'the value of percentage ripple is (percentage) = ',round(perripple,2)
print '\nthe value of the regulation is (percentage) = ',round(reg*100,1)
print '\nthe optimum number of stages for minimum regulation is ',round(nopt)

the value of percentage ripple is (percentage) =  4.53

the value of the regulation is (percentage) =  12.4

the optimum number of stages for minimum regulation is  14.0

Example 2: pg 197

In [3]:
#example 6.2
#calculation of series inductance and input voltage to transformer
from math import pi
#given data
kva=100.*10**3#value of volt-ampere of transformer(in VA)
V=250.*10**3#value of transformer secondary voltage(in V)
Vi=400.#value of transformer primary voltage(in V)
Vc=500.*10**3#voltage(in V)
Ic=0.4#charging current(in A)
perX=8.#percentage leakage reactance
f=50.#value of frequency(in Hz)
perR1=2.#percentage resistance
perR2=2.#percentage resistance of inductor


#calculation
I=kva/V#maximum value of current that can be supplied
Xc=Vc/Ic#reactance of cable
Xl=(perX*V)/(100*I)#leakage reactance
adrec=Xc-Xl#additional reactance
Xadrec=adrec/(2*pi*f)
perR=perR1+perR2#total resistance
R=(perR*V)/(100*I)
VE2=I*R#excitation at secondary
VE1=VE2*Vi/V#primary voltage
IkW=(VE1/Vi)*100#input kW
#results
print 'The value of series inductance is (H) = ',round(Xadrec)
print '\nThe value of input voltage to the transformer is (V) = ',round(IkW)

The value of series inductance is (H) =  3820.0

The value of input voltage to the transformer is (V) =  4.0

Example 3: pg 198

In [7]:
#example 6.3
#calculation of series resistance ,damping resistance and maximum output voltage of the generator
from math import exp
#given data
n=8.#number of stages
C=0.16*10**-6#value of condenser(in farad)
Cl=1000.*10**-12#value of load capacitor (in farad)
t1=1.2*10**-6#time to front(in second)
t2=50.*10**-6#time to tail(in second)
Vc=120.*10**3#charging voltage(in V)

#calculation
C1=C/n#generator capacitance
C2=Cl#load capacitance
R1=(t1*(C1+C2))/(3*C1*C2)
R2=(t2/(0.7*(C1+C2)))-R1
V=n*Vc#dc charging voltage for n stages
alpha=1/(R1*C2)
betaa=1/(R2*C1)
Vmax=(V*(exp(-alpha*t1)-exp(-betaa*t1)))/(R1*C2*(alpha-betaa))
#results
print 'The value of series resistance is (ohm) = ',round(R1)
print '\nThe value of damping resistance is (ohm) = ',round(R2)
print '\nThe value of maximum output voltage of the generator is (kV) = ',round(-Vmax*10**-3,2)
print 'The answer is a bit different due to rounding off error in textbook'

The value of series resistance is (ohm) =  420.0

The value of damping resistance is (ohm) =  2981.0

The value of maximum output voltage of the generator is (kV) =  892.02
The answer is a bit different due to rounding off error in textbook

Example 4: pg 198

In [11]:
#example 6.4
#calculation of circuit inductance and dynamic resistance

#given data
alpha=0.0535*10**6#from table
LC=65.#value of product
C=8.#value of capacitor (in microfarad)
Ip=10.#output peak current(in kA)
t1=8.#time to front(in microsecond)

#calculation
L=LC/C#inductance(in microhenry)
Rd=2*(LC*10**-6)*alpha/t1#dynamic resistance
V=Ip*14./C#charging voltage

#results
print 'The value of circuit inductance is (microhenry) = ',round(L,3)
print '\nThe value of dynamic resistance is (ohm) = ',round(Rd,4)
print '\nThe value of charging voltage is (kV) = ',round(V,1)
print 'The answer for Charging Voltage is wrong in textbook'

The value of circuit inductance is (microhenry) =  8.125

The value of dynamic resistance is (ohm) =  0.8694

The value of charging voltage is (kV) =  17.5
The answer for Charging Voltage is wrong in textbook

Example 5: pg 199

In [20]:
#example 6.5
#calculation circuit inductance and dynamic resistance
from math import pi,exp,atan
#given data
C=8.*10**-6#value of capacitor (in farad)
Ip=10.#output peak current(in kA)
t1=8.*10**-6#time to front(in second)
t2=20.*10**-6#time to first half cycle(in second)
V=25.*10**3#charging voltage
im=10.*10**3#output currennt(in A)

#calculation
omega=pi/t2
omegat1=omega*t1
alpha=omega*(1/atan(omegat1))
LC=1/((t1**2)+(alpha**2))
L=LC/C
R=2*L*alpha 
V=omega*L*10*exp(-alpha*t1)
#results
print 'The value of circuit inductance is (microhenry) = ',round(L*10**6,2)
print '\nThe value of dynamic resistance is (ohm) = ',round(R,4)
print '\nThe value of charging voltage is (kV) = ',round(V,2)
print 'The answers are a bit different due to rounding off error in textbook'

The value of circuit inductance is (microhenry) =  4.09

The value of dynamic resistance is (ohm) =  1.4302

The value of charging voltage is (kV) =  1.59
The answers are a bit different due to rounding off error in textbook

Example 6: pg 199

In [21]:
#example 6.6
#calculation of front and tail time

#given data
n=12#number of stages
C=0.126*10**-6#capacitance(in Farad)
R1=800#wavefront resistance(in ohm)
R2=5000#xavetail resistance(in ohm)
C2=1000*10**-12#load capacitance(in Farad)


#calculation
C1=C/n
t1=3*R1*(C1*C2)/(C1+C2)
t2=0.7*(R1+R2)*(C1+C2)
#results
print 'The time to front is (microsecond) = ',round(t1*10**6,2)
print '\nThe time to tail is (microsecond) = ',round(t2*10**6,1)

The time to front is (microsecond) =  2.19

The time to tail is (microsecond) =  46.7

Example 7: pg 200

In [29]:
#example 6.7
#calculation of peak value of output voltage and highest resonant frequency produced
from cmath import sqrt,pi
#given data
V=10.*10**3#voltage(in V) at primary winding
L1=10.*10**-3#inductance(in H)
L2=200.*10**-3#inductance(in H)
K=0.6#coefficient of coupling
C1=2.*10**-6#capacitance(in Farad) on primary side
C2=1.*10**-9#capacitance(in Farad) on secondary side

#calculation
M=K*sqrt(L1*L2)
omega1=1/sqrt(L1*C1)
sigma=sqrt(1-(K**2))
omega2=1/sqrt(L2*C2)
gama2=sqrt(((omega1**2+omega2**2)/2)+sqrt(((omega1**2+omega2**2)/2)-(sigma**2*omega1**2*omega2**2)))
gama1=sqrt(((omega1**2+omega2**2)/2)-sqrt(((omega1**2+omega2**2)/2)-(sigma**2*omega1**2*omega2**2)))
fh=gama2/(2*pi)#highest frequency
V2p=(V*M)/(sigma*L1*L2*C2*(gama2**2-gama1**2))
#results
print 'The value of highest resonant frequency produced is (kHz) = ',abs(fh)*10**-3
print '\nThe peak value of output voltage is (kV) - ',abs(V2p)*10**-3

#gama1 and gama2 are imaginary numbers....Moreover their magnitudes will also be same....so peak value of output voltage from equation is zero
print 'The answers are a bit different due to rounding off error in textbook'

The value of highest resonant frequency produced is (kHz) =  8.04714659485

The peak value of output voltage is (kV) -  209.631374545
The answers are a bit different due to rounding off error in textbook

Example 8: pg 201

In [31]:
#example 6.8
#calculation of output voltage
from math import sqrt
#given data
V1=10.#voltage(in kV) at primary winding 
C1=2.*10**-6#capacitance(in Farad) on primary side
C2=1.*10**-9#capacitance(in Farad) on secondary side
pern=5.#energy efficiency(in percentage)

#calculation
n=pern/100.
V2=V1*sqrt(n*C1/C2)
#results
print 'The value of output voltage is (kV) = ',round(V2,1)
print 'correct answer is 100 kV'

The value of output voltage is (kV) =  100.0
correct answer is 100 kV

Example 9: pg 201

In [32]:
#example 6.9
#calculation of self capacitance and leakage reactance
from math import pi
#given data
Vi=350.*10**3#rating(in VA)
V=350.*10**3#secondary voltage(in V)
V1=6.6*10**3#primary voltage(in V)
perV=8.#percentage ratedd voltage
perR=1.#percentage rise
f=50.#frequency(in Hz)

#calculation
I=Vi/V
Xl=(perV*V)/(100*I)
I0=perR*V/(100*Xl)
Xc=((1+(perR/100))*V)/I0
C=1/(Xc*2*pi*f)
#results
print 'The value of self capacitance is (nF) = ',round(C*10**9,3)
print '\nThe value of leakage reactance is (kohm) = ',round(Xl*10**-3)

The value of self capacitance is (nF) =  1.126

The value of leakage reactance is (kohm) =  28.0

Example 10: pg 202

In [33]:
#example 6.10
#calculation of resistance and inductance

#given data
CR=70.6#value from table
LC=11.6#value from table
C=1#capacitance(in microfarad)
pern=98.8#percentage voltage efficiency
V=10.#rating(in kV)
LC2=65.#value from table
alpha=0.0535#value from table

#calculation
R=CR/C
L=LC/C
Vo=pern*V/100
L2=LC2/C
R2=2*L2*alpha
Ip=V*C/14.

print 'The value of resistance for 1/50 microsecond voltage is (ohm) = ',round(R,1)
print '\nThe value of inductance for 1/50 microsecond voltage is (microhenry) = ',round(L,1)
print '\nThe value of output voltage is (kV) = ',round(Vo,2)
print '\nThe value of inductance for 8/20 microsecond voltage is (microhenry) = ',round(L2)
print '\nThe value of resistance for 8/20 microsecond voltage is (ohm) = ',round(R2,3)
print '\nThe peak value of current is (A) = ',round(Ip*10**3)

The value of resistance for 1/50 microsecond voltage is (ohm) =  70.6

The value of inductance for 1/50 microsecond voltage is (microhenry) =  11.6

The value of output voltage is (kV) =  9.88

The value of inductance for 8/20 microsecond voltage is (microhenry) =  65.0

The value of resistance for 8/20 microsecond voltage is (ohm) =  6.955

The peak value of current is (A) =  714.0