Chapter 05 : High Voltage DC Transmission

Example 5.1, Page No 107

In [1]:
import math
#initialisation of variables
Vo=3*math.sqrt(2)*110/math.pi

#Calculations 
Vd=Vo*(math.cos(math.radians(0))+math.cos(math.radians(15)))/2.0 
Vd1=Vo*(math.cos(math.radians(30)) + math.cos(math.radians(45)))/2.0 
Vd2=Vo*(math.cos(math.radians(45)) + math.cos(math.radians(60)))/2.0 

#Results
print("(a)For a=0, Vd=%.2f kV" %Vd) 
print("(b)For a=30,Vd=%.2f kV" %Vd1) 
print("(c)For a=45,Vd=%.2f kV" %Vd2) 

(a)For a=0, Vd=146.02 kV
(b)For a=30,Vd=116.85 kV
(c)For a=45,Vd=89.66 kV

Example 5.2, Page No 107

In [2]:
import math
#initialisation of variables

#Calculations
VL=(100.0*2*math.pi)/(3*math.sqrt(2.0)*(math.cos(math.radians(30)) + math.cos(math.radians(45)))) 
print("VL=%.2f kV" %VL)  #Answers don't match due to difference in rounding off of digits
Tr=VL/110.0

#Results 
print("tap ratio=%.2f " %Tr) 

VL=94.14 kV
tap ratio=0.86

Example 5.3, Page No 109

In [3]:
import math
#initialisation of variablesVd=100000
Id=800.0  # current

#Calculations
X=((3*math.sqrt(2.0)*94.115*0.866*1000.0/math.pi)-Vd)*math.pi/(3.0*Id)

#Results
print("effective reactance per phase , X=%.2f ohm\n" %X)	#Answer don't match due to difference in rounding off of digits

effective reactance per phase , X=143.89 ohm

Example 5.4 Page No 112

In [4]:
import math
#initialisation of variables
a=15.0
d0=10.0
y=15.0
X=15.0
R=10.0

#Calculations
Id=(3*math.sqrt(2)*120*(math.cos(math.radians(a))-math.cos(math.radians(d0+y)))*1000.0)/((R+(3.0*2*X)/math.pi)*math.pi)

#Results
print("Id=%.2f amp." %Id)

Id=249.99 amp.