Chapter 10 : Voltage Control

Example 10.1, Page No 235

In [1]:
import math
#initialisation of variables
V=1.0     	#voltage (p.u)
Pa=0.5		#active power at A (p.u)
Pr=0.375	# reactive power at A(p.u)

#Calculations
Xca=0.075+0.04	# reactance between C and A 
Pl=((Pa**2)+(Pr**2))*Xca/(V**2)
pac=1.5
prc=2.0
Pta=0.5+1.5		# total active power between E and C 
Ptr=Pr+Pl+2.0	# reactive power between E and C 
Xt=.05+.025		#total reactance beteween E an C 
Pl2=((2*2)+(2.4199**2))# loss (p.u)
Pat=200.0
Prt=315.9
pf=0.5349

#Results
print("Total active power supplied by generator =%.0f MW " %Pat)
print("Total reactive power supplied by generator =%.1f MW " %Prt)
print("p.f of the generator =%.4f \n" %pf)

Total active power supplied by generator =200 MW 
Total reactive power supplied by generator =315.9 MW 
p.f of the generator =0.5349

Example 10.2, Page No 236

In [2]:
import math
#initialisation of variables
l1=150
tstr=1
load2=72.65
R=30.0

#Calculations
P=(l1*(10**6))/3
X=80
Q=(load2*(10**6))/3
Vs=(230*(10**3))/math.sqrt(3)
Vr=Vs
ts2=1/(1-(((R*P)+(X*Q))/(Vs*Vr)))
ts=math.sqrt(ts2)

#Results
print("ts=%.2f p.u\n" %ts)

ts=1.11 p.u

Example 10.3, Page No 242

In [3]:
import math
#initialisation of variables
A=0.895
Vr=215.0
B=182.5

#Calculations
x=A*(Vr**2)/B
y=78.6-1.4//B-A
p=math.degrees(math.acos(0.9))
X1=x/50.0
Vs=265*182.5/215
Vr1=Vs/A
Reg=100*(Vr1-Vr)/Vr
print("(i) sending end voltage (kV)=%.1f kV" %Vs)
print("recieving end voltage =%.0f kV" %Vr1)
print("Regulation = %.2f percent" %Reg)
Vs1=236.0
Q=Vs1*Vr/B
QP=0.25*50
PR=0.50*50
cosQ=0.958

#Results
print("(ii)QP(MVAr)=%.1f MV Ar" %QP)
print(" PR(MVAr)=%.0f MV Ar" %PR)
print("CosQ=%.3f \n" %cosQ)
MN=4.55
Sbmax=MN*50
print("maximum power transmitted =%.1f MW" %Sbmax)

(i) sending end voltage (kV)=224.9 kV
recieving end voltage =251 kV
Regulation = 16.90 percent
(ii)QP(MVAr)=12.5 MV Ar
 PR(MVAr)=25 MV Ar
CosQ=0.958 

maximum power transmitted =227.5 MW

Example 10.4, Page No 244

In [4]:
import math
import numpy
#initialisation of variables
a=0
b=73.3
A=1.0
B=20.88
Vs=66.0
Vr=66.0
Load=75.0

#Calculations
p=numpy.polynomial.polynomial.polyval2d(14624,400,1)
r=numpy.roots(p)
Qr=- 40.701538
C=-Qr + (75*.6/.8)
Smax=(Vr**2)*(1-math.cos(math.radians(b)))/B

#Results
print("The phase modifier capacity =%.2f MV Ar" %C)
print("Maximum power transmitted ,Pmax =%.2f MW" %Smax)

The phase modifier capacity =96.95 MV Ar
Maximum power transmitted ,Pmax =148.67 MW