Chapter 22 : Power System Voltage Stability

Example 22.2, Page No 725

In [1]:
import math
#initialisation of variables
Vb=500.0
Sb=1000.0

#Calculations
Zb=Vb**2/Sb
Xpu=.35*100/Zb
Zth=1000/5000.0
X=Xpu+Zth
V=1.0
Q=0
P=1.0
Eth=V+(Q*X/V)+complex(P*X/V)
Q=0.75
Eth1=V+(Q*X/V)+complex(P*X/V)

#Results
print("(i) For p.f unity , Eth= {0:.5f}+{1:.5f}i".format(Eth.real, Eth.imag))
print("(i) For p.f .8 , Eth= {0:.5f}+{1:.5f}i".format(Eth1.real, Eth1.imag))

(i) For p.f unity , Eth= 1.34000+0.00000i
(i) For p.f .8 , Eth= 1.59500+0.00000i

Example 22.3, Page No 726

In [2]:
import math
#initialisation of variables
X=.625
P=1.0
Q=0.6
V=1.0

#Calculations
Eth=V+(Q*X/V)+complex(P*X/V)
Phase_Eth=math.degrees(math.atan(Eth.imag/Eth.real))

#Results
print("Eth=%.2f at an angle %.0f degrees" %(abs(Eth),Phase_Eth))

Eth=2.00 at an angle 0 degrees

Example 22.4, Page No 732

In [3]:
import math
#initialisation of variables
P=0.5
toff=4.0

#Calculations
ton=(P*toff-0*toff)/(0.8-P)

#Results
print("Toff= 4min .")
print("ton(min.)=%.3f min." %ton)

Toff= 4min .
ton(min.)=6.667 min.

Example 22.6 Page No 739

In [4]:
import math
#initialisation of variables
V=1.0
Qload=1.0*V
Qcap=-0.75*V**2

#Calculations
Qnet=Qload+Qcap
VS=1-0.75*2*V     # voltage sensitivity

#Results
print("Voltage sensitivity=%.3f" %VS)
print("since the voltage sensitivity is negative,\nvoltage regulation by tap changing will reduce net reactive load and improive voltage stability ")

Voltage sensitivity=-0.500
since the voltage sensitivity is negative,
voltage regulation by tap changing will reduce net reactive load and improive voltage stability

Example 22.7, Page No 740

In [5]:
import math
#initialisation of variables
Y=complex(-10)
n=1+0.1

#Calculations
Y1=n*(n-1)*Y
Y2=(1-n)*Y

#Results
print("Y1= {0:.2f}+{1:.2f}i".format(Y1.real, Y1.imag))
print("Y2= {0:.2f}+{1:.2f}i".format(Y2.real, Y2.imag))
print("The shunt elements equal to a reactor of 1.1V1^2 size oin the primary side and a capacitive of sixe 1V2**2 on the secondary side")

Y1= -1.10+0.00i
Y2= 1.00+-0.00i
The shunt elements equal to a reactor of 1.1V1^2 size oin the primary side and a capacitive of sixe 1V2**2 on the secondary side

Example 22.8, Page No 745

In [6]:
import math
#initialisation of variables
P=1.0		#assuming
S1=P/.95	#For pf .95
S2=P/.8		#For pf .8

#Calculations
dMVA=(S2-S1)*100.0/P		#Increase in MVA rating 
Q1=P*math.tan(math.radians(math.degrees(math.acos(0.95))))		#Q for pf .95
Q2=P*math.tan(math.radians(math.degrees(math.acos(0.8))))		#Q for pf .8
dPc=(Q2-Q1)*100.0/Q1		#Percent additional Reactive Power Capability 

#Results
print("Percent additional Reactive Power Capability is %.2f" %dPc)

Percent additional Reactive Power Capability is 128.18