Chapter 15: Voltage Control

Example 15.1, Page Number: 384

In [13]:
from __future__ import division
from sympy import *
import math

#Variable declaration:
Pl = 10000                      #load(kW)
pf = 0.8                        #power factor(lag)
Vl = 33000                      #receiving end line voltage(V)
R = 5                           #line resistance(ohm)
X = 10                          #line reactance(ohm)



#Calculation:
I2 = math.floor(Pl*1000/(3**0.5*Vl*0.8))        #load current(A)
Ip = I2*pf                        #A
Iq = I2*math.sin(math.acos(pf))     #A
V1 = round(Vl/3**0.5)                      #sending end voltage(V)
#Let Im be the current taken by the synchronous condenser.
Im = symbols('Im')                   #A
Im1 = solve((V1+Ip*R-X*(Im-Iq))**2+(Ip*X+(Im-Iq)*R)**2-V1**2,Im)[0]
C = 3*V1*round(Im1)/1000                   #kVAR


#Result:
print "Capacity of synchronous condenser is",math.floor(C),"kVAR"
Capacity of synchronous condenser is 13203.0 kVAR

Example 15.2 Page Number: 385

In [2]:
from __future__ import division
import math

#Variable declaration:
Pl = 25000                      #load(kW)
pf = 0.8                        #power factor(lag)
Vl = 33000                      #receiving end line voltage(V)
R = 5                           #line resistance(ohm)
X = 20                          #line reactance(ohm)



#Calculation:
I2 = round(Pl*1000/(3**0.5*Vl*pf),1)        #load current(A)
Ip = I2*pf                        #A
Iq = I2*math.sin(math.acos(pf))     #A
V1 = Vl/3**0.5                      #sending end voltage(V)
#Let Im be the current taken by the synchronous condenser.
Im = symbols('Im')                   #A
Im1 = solve((V1+Ip*R-X*(Im-Iq))**2+(Ip*X+(Im-Iq)*R)**2-V1**2,Im)[0]
C = 3*V1*Im1/1000                   #kVAR


#Result:
print "Capacity of synchronous condenser is",round(C/1000,2),"MVAR"
Capacity of synchronous condenser is 33.11 MVAR