# Chapter 11 : Some Applications¶

## Example 11.1, Page No 622¶

In [1]:
import math
#initialisation of variables
V_s=11000.0
V_ml=math.sqrt(2)*V_s
f=50.0

#Calculations
w=2*math.pi*f
I_d=300
R_d=1
g=20 #g=gamma
L_s=.01

#Results
print("firing angle=%.3f deg" %a)
print("rectifier o/p voltage=%.1f V" %V_d)

firing angle=16.283 deg
rectifier o/p voltage=13359.3 V


## Example 11.2, Page No 623¶

In [2]:
import math
#initialisation of variables
V_d=(200.0+200)*10**3
P=1000.0*10**6

#Calculations
I_d=P/V_d
#each thristor conducts for 120deg for a periodicity of 360deg
a=0
V_d=200.0*10**3

#Results
print("rms current rating of thyristor=%.2f A" %(I_d*math.sqrt(120/360)))
print("peak reverse voltage across each thyristor=%.2f kV" %(V_ml/2/1000))

rms current rating of thyristor=0.00 A
peak reverse voltage across each thyristor=104.72 kV


## Example 11.3 Page No 627¶

In [3]:
import math
#initialisation of variables
V_m=230.0
V_s=230/math.sqrt(2)
pf=0.8
P=2000.0

#Calculations
I_m=P/(V_s*pf)
I_Tr=I_m/math.sqrt(2)
I_TA=2*I_m/math.pi
fos=2 #factor of safety
PIV=V_m*math.sqrt(2)
I_Tr=I_m/(2)
I_TA=I_m/math.pi

#Results
print("rms value of thyristor current=%.2f A" %(fos*I_Tr))
print("avg value of thyristor current=%.3f A" %(fos*I_TA))
print("voltage rating of thyristor=%.2f V" %PIV)
print("rms value of diode current=%.3f A" %(fos*I_Tr))
print("avg value of diode current=%.3f A" %(fos*I_TA))
print("voltage rating of diode=%.2f V" %PIV)

rms value of thyristor current=15.37 A
avg value of thyristor current=9.786 A
voltage rating of thyristor=325.27 V
rms value of diode current=15.372 A
avg value of diode current=9.786 A
voltage rating of diode=325.27 V


## Example 11.4, Page No 629¶

In [4]:
import math
#initialisation of variables
V=200.0
I=10.0

#Calculations
R_L=V/I
I_h=.005     #holding current
R2=V/I_h
t_c=20*10**-6
fos=2 #factor of safety
C=t_c*fos/(R_L*math.log(2))

#Results
print("value of load resistance=%.0f ohm" %R_L)
print("value of R2=%.0f kilo-ohm" %(R2/1000))
print("value of C=%.3f uF" %(C*10**6))

value of load resistance=20 ohm
value of R2=40 kilo-ohm
value of C=2.885 uF


## Example 11.5 Page No 646¶

In [5]:
import math
#initialisation of variables
u_r=10
f=10000.0 #Hz
p=4.0*10**-8 #ohm-m

#Calculations
dl=(1/(2*math.pi))*math.sqrt(p*10**7/(u_r*f))
l=0.12 #length of cylinder
t=20.0 #no of turns
I=100.0
H=t*I/l
P_s=2*math.pi*H**2*math.sqrt(u_r*f*p*10**-7)
d=.02 #diameter
P_v=4*H**2*p/(d*dl)

#Results
print("depth of heat of penetration=%.5f mm" %(dl*1000))
print("heat generated per unit cylinder surface area=%.3f W/m**2" %P_s)
print("heat generated per unit cylinder volume=%.0f W/m**3" %P_v)
#answer of P_v varies as given in book as value of d is not taken as in formulae.

depth of heat of penetration=0.31831 mm
heat generated per unit cylinder surface area=34906.585 W/m**2
heat generated per unit cylinder volume=6981317 W/m**3


## Example 11.6 Page No 646¶

In [6]:
import math
#initialisation of variables
f=3000.0

#Calculations
t_qmin=30.0*10**-6
f_r=f/(1-2*t_qmin*f)
R=0.06
L=20.0*10**-6
C=1/(L*((2*math.pi*f_r)**2+(R/(2*L))**2))

#Results
print("required capacitor size=%.4f F" %(C*10**6))
#Answers have small variations from that in the book due to difference in the rounding off of digits.

required capacitor size=94.2215 F