# Chapter 09 : Insulated Cables¶

## Example 9.1, Page No 196¶

In [1]:
import math
#initialisation of variables
V=85.0     # working voltage (kV)
gmax=65.0  # dielectric strength of insulating material (kV/cm)

#Calculations
r=V/gmax
d=2*r
D=2.6*math.e

#Results
print("Diameter of the sheath =%.2f cm\n" %D)

Diameter of the sheath =7.07 cm



## Example 9.2, Page No 200¶

In [2]:
import math
#initialisation of variables
e1=4.0
e2=4.0
e3=2.5
g1max=50.0
g2max=40.0
g3max=30.0

#Calculations
r1=r*e1*g1max/(e2*g2max)
r2=r1*e2*g2max/(e3*g3max)
V=66.0
lnc=(V-((r*g1max*math.log(r1/r))+(r1*g2max*math.log(r2/r1))))
m=lnc/(r2*g3max)
R=r2*(math.e**m)
D=2*R

#Results
print("minimum internal diameter of the lead sheath,D=%.2f cms " %D)

minimum internal diameter of the lead sheath,D=7.52 cms


## Example 9.3, Page No 202¶

In [3]:
import math
#initialisation of variables
g1max=34.0
er=5.0
r1=1.0
R=7.0/2    #external dia(cm)

#Calculations
g2max=(r*g1max)/(er*r1)
V=((r*g1max*math.log(r1/r))+(r1*g2max*math.log(R/r1)))
V=V/(math.sqrt(2.0))

#Results
print("Maximum safe working volltage ,V =%.2f kV r.m.s\n" %V)

Maximum safe working volltage ,V =11.34 kV r.m.s



## Example 9.4 Page No 202¶

In [4]:
import math
#initialisation of variables
r=0.9
r1=1.25

#Calculations
r2=r1+.35
r3=r2+.35    # radius of outermost layer
Vd=20.0        # voltage difference (kV)
g1max=Vd/(r*math.log(r1/r))
g2max=Vd/(r1*math.log(r2/r1))
g3max=(66-40)/(r2*math.log(r3/r2))

#Results
print("g1max =%.1f kV/cm" %g1max)
print("g2max =%.2f kV/cm" %g2max)
print("g3max =%.0f kV/cm" %g3max)

g1max =67.6 kV/cm
g2max =64.81 kV/cm
g3max =82 kV/cm


## Example 9.5, Page No 206¶

In [5]:
import math
#initialisation of variables
V=20.0     #voltage (kV)
w=314.0

#Calculations
KVA=V*V*w*C*1000.0

#Results
print("3-phase kVA required =%.0f kVA" %KVA) 	#Answer don't match due to difference in rounding off of digits

3-phase kVA required =764 kVA


## Example 9.6, Page No 206¶

In [6]:
import math
#initialisation of variables
C1=0.208
C2=0.096
Cx=3.0*C1
w=314.0
V=10.0

#Calculations
Cy=(C1+ 2*C2)
Co=((1.5*Cy)-(Cx/6))
C=Co/2.0

#Results
print("(i)Capacitance between any two conductors=%.3f micro-Farad/km" %C)
c=((2*C2 + ((2/3)*C1)))
print("(ii)Capacitance between any two bunched conductors and the third conductor=%.2f micro-Farad/km" %c)
I=V*w*Co*1000*(10**-6)/math.sqrt(3)
print("(iii)the charging current per phase per km =%.3f A" %I)

(i)Capacitance between any two conductors=0.248 micro-Farad/km
(ii)Capacitance between any two bunched conductors and the third conductor=0.19 micro-Farad/km
(iii)the charging current per phase per km =0.899 A


## Example 9.7 Page No 213¶

In [7]:
import math
#initialisation of variables
rm=(2.28/2)-(.152/2)# mean radius of sheath (cm)
d=5.08
a=d/rm
w=314.0

#Calculations
Xm=2*(10**-7)*math.log(a)     # mutual inductance (H/m)
Xm2=2000*Xm
V=w*Xm2*400

#Results
print("Voltage induced =%.2f volts \n" %V)#Answer don't match exactly due to difference in rounding off of digits i between calculations

Voltage induced =78.54 volts



## Example 9.8, Page No 214¶

In [8]:
import math
#initialisation of variables
R=2*0.1625
Rs=2*2.14
M=314.0

#Calculations
w=6.268*10**-4
r=Rs*M*M*w*w/(R*((Rs**2)+(M*M*w*w)))

#Results
print("ratio=%.4f " %r)

ratio=0.0278