# Chapter 6: Power Factor Improvement¶

### Example 6.1, Page Number: 108¶

In :
#Variable declaration:
kW1 = 300
pf1 = 0.6                  #power factor
pf2 = 1

#Calculation:
kVA = kW1/pf1
kW2 = kVA*pf2
kW = kW2-kW1

#Result:
print "Increased power supplied by the alternator is",kW,"kW"

Increased power supplied by the alternator is 200.0 kW


### Example 6.2, Page Number: 109¶

In :
import math

#Variable declaration:
V = 400                     #rated voltage(V)
Im = 31.7                    #motor curtrent(A)
pf1 = 0.7                    #initial power factor
pf2 = 0.9                    #raised power factor
f = 50                       #Hz

#Calculation:
#Referring to the phasor diagram,
Ima = Im*pf1                #Active component of Im(A)
I = Ima/pf2                 #total current(A)
Imr = Im*math.sqrt(1-pf1**2)     #Reactive component of Im(A)
Ir = I*math.sqrt(1-pf2**2)       #Reactive component of I(A)
Ic = Imr-Ir                 #A

C = Ic/(V*2*math.pi*f)

#Result:
print "Required value of capacitor is",round(C,7),"uF"

Required value of capacitor is 9.46e-05 uF


### Example 6.3, Page Number: 110¶

In :
from __future__ import division
import math

#Variable declaration:
#(i):
kW1 = 20
pf1 = 1

#(ii):
kW2 = 100
pf2 = 0.707

#(iii):
kW3 = 50
pf3 = 0.9

#Calculation:
kVA1 = kW1/pf1
kVA2 = kW2/pf2
kVA3 = kW3/pf3

kVAR1 = kVA1*0
kVAR2 = -kVA2*math.sin(math.acos(0.707))
kVAR3 = kVA3*math.sin(math.acos(0.9))

kW = kW1+kW2+kW3
kVAR = kVAR1+kVAR2+kVAR3
kVA = math.sqrt(kW**2+kVAR**2)
pf = kW/kVA

#Result:
print "Total kW is",kW
print "Total kVA is",round(kVA)
print "Power factor is",round(pf,3)

Total kW is 170
Total kVA is 186.0
Power factor is 0.913


### Example 6.4, Page Number: 111¶

In :
from __future__ import division
import math

#Variable declaration:
pf1 = 0.75                #Original p.f lag
P = 5                     #motor input(kW)
pf2 = 0.9                  #final p.f lag

#Calculation:
kVAR = P*(math.tan(math.acos(pf1))-math.tan(math.acos(pf2)))
R = kVAR/3

#Result:
print "Rating of capacitors is",round(R,3),"kVAR"

Rating of capacitors is 0.663 kVAR


### Example 6.5, Page Number: 111¶

In :
from __future__ import division
import math
from sympy import *

#Variable declaration:
Vph = 400                  #rated voltage(V)
n = .93                  #efficiency
P = 74.6                  #power output(kW)
pf1 = 0.75                #power faactor
pf2 = 0.95               #raised power factor
n1 = 4                  #no. of capacitors in each branch
V1 = 100                 #rating of capacitors(V)
f = 50                   #frequency(Hz)

#Calculation:
Pi = round(P/n)                    #motor input(kW)
phy1 = math.acos(pf1)
phy2 = math.acos(pf2)
#Leading kVAR taken by the condenser bank
kVAR = round(Pi*(math.tan(phy1)-math.tan(phy2)),2)
kVAR1 = round(kVAR/3,2)             #Leading kVAR taken by each of three sets

#Fig. above shows the delta* connected condenser bank.
C = symbols('C')
Icp = 2*math.pi*f*C*Vph           #Phase current of capacitor(A)

c = solve(Vph*Icp/1000-kVAR1,C)  #capacitance(F)

#Result:
print "Capacitance of each capacitor is",round(c*4*10**6,1),"uF"

Capacitance of each capacitor is 1173.8 uF


### Example 6.6, Page Number: 112¶

In :
from __future__ import division
import math

#Variable declaration:
pf1 = 0.8                #initial power factor
pf2 = 0.9                #improved power factor
t = 3000                 #working hours
c1 = 100                 #1st part tariff(Rs/kVA)
c2 = 20                  #2nd part tariff(paise/kWh)
c3 = 60                  #cost of capacitors(Rs/kVAR)
n = 10                  #interest & depreciation on capacitors(%)

#Calculation:
phy1 = math.acos(pf1)
phy2 = math.acos(pf2)
#Leading kVAR taken by the capacitors:
kVAR = P*(math.tan(phy1)-math.tan(phy2))

#Annual cost before p.f. correction:
kVAm1 = P/pf1                #max demand
T1 = kVAm1*c1+P*t*c2/100      #total annual cost(Rs)

#Annual cost after p.f. correction:
kVAm2 = P/pf2               #max demand
T2 = c1*kVAm2+P*t*c2/100+n*c3*kVAR/100

T = T1-T2                #annual saving(Rs)

#Result:
print "Annual saving is Rs",round(T)

Annual saving is Rs 9836.0


### Example 6.7, Page Number: 112¶

In :
from __future__ import division
from sympy import *
import math

#Variable declaration:
pf1 = 0.85               #initial power factor
pf2 = 0.9                #improved power factor
t = 2500                 #working hours
c1 = 150                 #1st part tariff(Rs/kVA)
c2 = 5                   #2nd part tariff(paise/kWh)
c3 = 420                 #cost of capacitors(Rs/kVAR)
Pcl = 100                #capacitor loss(W/kVAR)
n = 10                   #interest & depreciation on capacitors(%)

#Calculation:
phy1 = math.acos(pf1)
phy2 = math.acos(pf2)
#suppose the leading kVAR taken by the capacitors is x:
x = symbols('x')
Pcl1 = Pcl*x/1000                #capacitor loss(kW)
P2 = P1+Pcl1                     #total power(kW)
x1 = solve((P1*math.tan(phy1)-P2*math.tan(phy2))-x,x)

#Annual cost before p.f. correction:
kVAm1 = P1/pf1                #max demand
T1 = kVAm1*c1+P1*t*c2/100      #total annual cost(Rs)

#Annual cost after p.f. correction:
kVAm2 = P1/pf2               #max demand
T2 = c1*kVAm2+P1*t*c2/100+n*c3*x1/100+round(Pcl*x1*t*c2/100000)

T = T1-T2                #annual saving(Rs)

#Result:
print "Annual saving is Rs",round(T)

Annual saving is Rs 553.0


### Example 6.8, Page Number: 113¶

In :
from __future__ import division
import math

#Variable Declaration:
pf = 0.8                           #power factor
kVA = 750                          # monthly demand
c1 = 8.50                          #monthly power rate(Rs/kVA per month)
R = 250                            #rating of capacitors(kVA)
C2 = 20000                        #installed cost of equipment(Rs)
r = 10                            #fixed charge rate(%)

#Calculation:
phy = math.acos(pf)
kW = kVA*math.cos(phy)             #kW component of demand
kVAR = kVA*math.sin(phy)           #kVAR component of demand
kVAR1 = kVAR-R                     #Leading kVAR taken by the capacitors
kVA1 = math.sqrt(kVAR1**2+kW**2)        #kVA after p.f. improvement
kVA11 = round(kVA-kVA1,1)                   #Reduction in kVA
T1 = c1*kVA11*12                   #Yearly saving on kVA charges
T2 = r*C2/100                      #Fixed charges/year(Rs)
T = T1-T2                          #net saaving(Rs)

#Result:
print "The annual saving is Rs",T

The annual saving is Rs 9985.0


### Example 6.9, Page Number: 113¶

In :
from __future__ import division
import math

#Variable declaration:
pf1 = 0.8                    #initial power factor
pf2 = 0.9                     #improved power factor

#Calculation:
phy1 = math.acos(pf1)
phy2 = math.acos(pf2)
#In Fig. above, Δ OAB is the power triangle for load,
#Δ ODC for combined load and Δ BEC for the motor.
#(i):
kVAR = L1*math.tan(phy1)-L*math.tan(phy2)   #Leading kVAR taken by the motor
#(ii):
kVArat = math.sqrt(L2**2+kVAR**2)          #kVA rating of motor
#(iii):
pf = L2/kVArat                        #p.f. of motor

#Result:
print "The leading kVAR taken by the motor is",round(kVAR,2)
print "kVA rating of the motor is",round(kVArat,2)
print "Power factor at which the motor operates is",round(pf,3),"leading"

The leading kVAR taken by the motor is 14.39
kVA rating of the motor is 81.28
Power factor at which the motor operates is 0.984 leading


### Example 6.10, Page Number: 114¶

In :
from __future__ import division
import math

#Variable declaration:
#for induction motor:
kW1 = 37.3
pf1 = 0.8                           #lagging
n1 = 0.85

#for synchronous motor:
kW2 = 18.65
n2 = 0.9

kW3 = 10
pf3 = 1
c1 = 60                          #Rs/kVA of max demand
c2 = 5                           #paie per kWh
t = 2000                         #working hours

#Calculation:
P1 = kW1/n1                     #power input to induction motor(kW)
Q1 = P1*math.tan(math.acos(pf1)) #Lagging kVAR taken by induction motor

P2 = kW2/n2                      #Input power to synchronous motor(kW)
Q2 = P2**math.tan(math.acos(pf1))  #Leading kVAR taken by synchronous motor

#Since lighting load works at unity p.f., its lagging kVAR = 0.
Q3 = 0

kVAR = Q1-Q2                    #net kVAR
kW = round(P1+P2+kW3,1)                      #total active power
kVA = round(math.sqrt(kW**2+kVAR**2))

T = c1*kVA+c2*kW*t/100

#Result:
print "The annual electical charges is Rs",round(T)

The annual electical charges is Rs 12140.0


### Example 6.11, Page Number: 114¶

In :
from __future__ import division
import math

#Variable declaration:
P1 = 500                        #kW

P2 = 400                         #kW
pf2 = 0.707                      #power factor lagging

P3 = 800                        #kW
pf3 = 0.8                       #power factor leading

P4 = 500                        #kW
pf4 = 0.6                       #power factor lagging

#(v)a synchronous motor driving a d.c. generator:
P5 = 540                        #power rated at generator(kW)
n = 0.9                         #overall efficiency

#Calculation:
#Total lagging kVAR taken by loads (ii) and (iv)
kVAR24 = P2*math.tan(math.acos(pf2))+P4*math.tan(math.acos(pf4))

kVAR3 = P3*math.tan(math.acos(pf3))

#Leading kVAR to be taken by synchronous motor
kVAR5 = kVAR24-kVAR3
Pi = P5/n                       #motor input(kW)
phy = math.atan(kVAR5/Pi)       #phase angle of synchronous motor
pf = math.cos(phy)

#Result:
print "The power factor of synchronous motor is",round(pf,2,),"leading"

The power factor of synchronous motor is 0.79 leading


### Example 6.12, Page Number: 115¶

In :
from __future__ import division
import math

#Variable declaration:
P1 = 100                        #power of synchronous motor(hp)
P2 = 200                        #power aggregated by induction motor(hp)
pf2 = 0.707                     #lagging power factor for induction motor
n2 = 0.82                       #efficiency of induction motor
c1 = 100                        #1st part tariff(Rs per kVA per annum)
c2 = 0.06                       #2nd part tariff Rs per kWh
pfa = 0.8                       #power factor factor initially(lagging)
na = 0.93                       #efficiency of synchronous motor initially
pfb = 0.8                       #power factor factor changed later(leading)
nb = 0.93                       #efficiency of synchronous motor changed later

#Calculation:
#(a) When synchronous motor runs at p.f. 0·8 lagging.
kW1 = round(P1*735.5/(na*1000))        #input to synchronous motor(kW)
kVAR1 = round(kW1*math.tan(math.acos(pfa)),2)    #Lagging kVAR taken by the synchronous motor(kVAR)
kW2 = round(P2*735.5/(n2*1000),1)        #input to induction motor(kW)
kVAR2 = kW2*math.tan(math.acos(pf2))    #Lagging kVAR taken by the induction motor(kVAR)
#Since lighting load works at unity p.f., its lagging kVAR is zero.
kVARt = kVAR1+kVAR2                 #Total lagging kVAR
kWt = kW1+kW2+P3                  #Total active power(kW)
kVAt = round(math.sqrt(kWt**2+kVARt**2),1)          #total kVA
C1 = kVAt*c1                       #annual kVA demand charges
E1 = kWt*8760                     #Energy consumed/year(kWh)
C2 = round(0.06*E1)                      #annual energy charges(Rs)
T1 = C1+C2                        #total Annual bill(Rs)

#(b) When synchronous motor runs at p.f. 0·8 leading.
kVARn = kVAR2-kVAR1            #Net lagging kVAR
kWtt = kWt                      #total active power(kW)
kVAtt = math.sqrt(kVARn**2+kWtt**2)     #total kVA
C11 = c1*kVAtt                #annual kVA charges(Rs)
C22 = C2                        #Annual energy charges(Rs)
T2 = C11+C22                   #total annual bill(Rs)
Ts = T1-T2                    #annual saving(Rs)

#Result:
print "The annual saving is  Rs",round(Ts/100)*100

The annual saving is  Rs 6200.0


### Example 6.13, Page Number: 118¶

In :
from __future__ import division
import math

#Variable declaration:
pf1 = 0.75                        #Power factor of the factory(lagging)
x = 72                            #Max. demand charges(Rs per kVA per annum)
d = 10                           #interest & depreciation of capital investment(%)
M = 175                         #max. demand(kW)
c2 = 120                          #cost of phase advancing equipment(Rs/kVAR)

#Calculation:
y = c2*d/100                    #Expenditure on phase advancing equipment(Rs/kVAR/annum)
pf2 = math.sqrt(1-(y/x)**2)           #power factor

#Result:
print "Most economical p.f. at which factory should operate is",round(pf2,3),"lagging"

Most economical p.f. at which factory should operate is 0.986 lagging


### Example 6.14, Page Number: 119¶

In :
from __future__ import division
import math

#Variable declaration:
L = 400                         #avg. demand(kW)
pf1 = 0.8                        #original power factor(lagging)
LF = 0.5                       #annual load factor
c1 = 50                        #first part tariff(Rs/kVA per annum)
c2 = 0.05                      #second part tariff(Rs/kWh)
pf2 = 0.95                     #improved power factor
c3 = 100                       #phase advancement equipment cost(Rs/kVAR)
d = 10                        #annual interest and depreciation(%)

#Calculation:
M = L/LF                     #maximum demand(kW)
kVAR1 = M*(math.tan(math.acos(pf1))-math.tan(math.acos(pf2)))
y = d*c3/100               #Expenditure on phase advancing equipment(Rs/kVAR/annum)
kVA1 = M/pf1                #Max. kVA demand at pf1 power fctor
kVA2 = round(M/pf2)                #Max. kVA demand at pf2 power fctor
Cs = c1*(kVA1-kVA2)           #Annual saving in maximum demand charges
Ce = y*kVAR1                #Annual expenditure on phase advancing equipment
NS = Cs-Ce                  #net annual saving

#Result:
print "(i) The capacity of the phase advancing equipment is",round(kVAR1),"kVAR"
print "(ii)The annual saving is  Rs",round(NS)

(i) The capacity of the phase advancing equipment is 337.0 kVAR
(ii)The annual saving is  Rs 4529.0


### Example 6.15, Page Number: 119¶

In :
from __future__ import division
import math

#Variable declaration:
L = 50                         #avg. demand(kW)
pf1 = 0.75                        #original power factor(lagging)
LF = 0.5                       #annual load factor
c1 = 100                        #first part tariff(Rs/kVA of max demand per annum)
c2 = 0.05                      #second part tariff(Rs/kWh)
c3 = 600                       #cost of loss free capacitors(Rs/kVAR)
d = 10                           #depreciation & interest(%)

#Calculation:
y = c3*d/100                #Expenditure on capacitors(Rs/kVAR/annum)
pfe = math.sqrt(1-(y/c1)**2)      #Most economical power factor
M = L/LF                     #Max kW demand
kVAm1 = M/pf1                #The maximum kVA demand at pf1 power factor
kVAm2 = M/pfe                #The maximum kVA demand at pfe power factor
Cs = c1*(kVAm1-kVAm2)         #annual saving(Rs)

#Result:
print "Most economical power factor is",round(pfe,3),"lagging"
print "Annual saving is Rs",round(Cs)

Most economical power factor is 0.8 lagging
Annual saving is Rs 833.0


### Example 6.16, Page Number: 120¶

In :
from __future__ import division
import math

#Variable declaration:
L = 200                         #avg. demand(kW)
pf1 = 0.8                        #original power factor(lagging)
LF = 0.5                       #annual load factor
c1 = 100                        #first part tariff(Rs/kVA of max demand per annum)
c2 = 0.05                      #second part tariff(Rs/kWh)
c3 = 500                       #cost of phase advancing plant(Rs/kVAR)
d = 10                           #depreciation & interest(%)

#Calculation:
y = c3*d/100                #Expenditure on capacitors(Rs/kVAR/annum)
pfe = math.sqrt(1-(y/c1)**2)      #Most economical power factor
Pc = L*(math.tan(math.acos(pf1))-math.tan(math.acos(pfe))) #Capacity of phase advancing plant
E = L*5000                    #units consumed per year(kWh)
C2 = c2*E                     #annual energy charges(Rs)
C1 = y*Pc                    #Annual cost of phase advancing plant(Rs)
Cm = c1*L/pfe                #Max. demand charge(Rs)
C = C2+C1+Cm                 #annual charge(Rs)

#Result:
print "(i)  The value to which the power factor be improved is",round(pfe,3),"lagging"
print "(ii) The capacity of the phase advancing plant is",round(Pc,2),"kVAR"
print "(iii)The new bill for energy is  Rs",round(C,1)

(i)  The value to which the power factor be improved is 0.866 lagging
(ii) The capacity of the phase advancing plant is 34.53 kVAR
(iii)The new bill for energy is  Rs 74820.5


### Example 6.17, Page Number: 120¶

In :
from __future__ import division
import math

#Variable declaration:
E = 80000               #units consumed per year
kVAm1 = 500                #maximum kVA demand
pf1 = 0.707             #initial power factor(lagging)
c1 = 120                        #first part tariff(Rs/kVA of max demand per annum)
c2 = 0.025                      #second part tariff(Rs/kWh)
c3 = 50                #cost of phase advancing plant(Rs/kVAR)
pf2 = 0.9              #increased powe factor(Rs)
d = 10                  #depreciation on advancing plant(%)

#Calculation:
C1 = c1*kVAm1+c2*E         #annual cost of supply(Rs)
P = kVAm1*pf1              #max. kW demand
kVARe = P*(math.tan(math.acos(pf1))-math.tan(math.acos(pf2)))
C2 = round(kVARe*c3*d/100)        #Annual cost of phase advancing equipment(Rs)
#When p.f. is raised from 0·707 lag to 0·9 lag
kVAm2 = P/pf2               #new maximum kVA demand
kVAr = kVAm1-kVAm2          #Reduction in kVA demand
Cs = round(c1*kVAr)                #annual saving(Rs)
#As the units consumed remain the same, therefore, saving will
#be equal to saving in M.D. charges minus annual cost of
Cs2 = Cs-C2                #annual charges(Rs)

#Result:
print "The annual cost of supply is   Rs",C1
print "Annual saving is  Rs",Cs2

The annual cost of supply is   Rs 62000.0
Annual saving is  Rs 11955.0


### Example 6.18, Page Number: 123¶

In :
from __future__ import division

#Variable declaration:
pf = 0.7           #power factor when plant working at its max. capacity(lagging)

#Calculation:
#For calculating the cost of increasing plant capacity.
#We have, referring to Fig. 6.15, the increase in kVA capacity is BD.

#Now  OE*cos phy2 = OD*cos phy1
#or   OB*cos phy2 = OD*cos phy1        (Since OE = OB)
#or            OD = OB * cos phy2/cos phy1
#                 = OB * 0·85/0·7
#                 = 1·2143 OB

#Increase in the kVA capacity of the plant is
#BD = OD − OB = 1·2143 × OB − OB = 0·2143 OB
#Total cost of increasing the plant capacity
#                = Rs 800 × 0·2143 × OB
#                = Rs 171·44 × OB                ...(i)

cos_phy1 = 0.7;         sin_phy1 = 0.714
cos_phy1 = 0.85;         sin_phy1 = 0.527

#Leading kVAR taken by p.f. correction equipment is
#  ED = CD − CE = OD*sin phy1 − OE*sin phy2
#     = 1·2143 * OB*sin phy1 − OB*sin phy2
#     = OB*(1·2143 * 0·714 − 0·527) = 0·34 * OB
#Suppose the cost per kVAR of the equipment be Rs y.
#So, total cost of p.f. correction equipment
#                     = Rs 0·34 × OB × y         ...(ii)
#The cost per kVAR of the equipment that would justify its
#installation is when exp. (i) = exp. (ii)
#i.e.,
#                     171·44 × OB = 0·34 × OB × y
y = 171.44/0.34                     #Rs/kVAR

#If the losses in p.f. correction equipment are neglected,
#then its kVAR = kVA.

#Result:
print "The maximum cost per kVA of p.f. correction equipment"
print "that can be paid is  Rs",round(y,1)

The maximum cost per kVA of p.f. correction equipment
that can be paid is  Rs 504.2


### Example 6.19, Page Number: 124¶

In :
from __future__ import division

#Variable declaration:
pf = 0.7               #power factor when plant working at its max. capacity(lagging)

#Calculation:
#Cost of increasing plant capacity
#      BD = OD − OB
#        = OB * 0.866/0.70-OB
#        = OB*(1·237 − 1)
#        = 0·237 * OB
#∴ Annual cost of increasing the plant capacity
#        = Rs 10 * 0·237 * OB
#        = Rs. 2.37 * OB                             ...(i)

#           ED = CD − CE
#              = OD*sin phy1 − OE*sin phy2
#              = 1.237 * OB * sin phy1 − OB*sin phy2
#              = OB*(1·237 * 0·174 − 0·5) = 0·383 × OB
#Let the cost per kVAR of the equipment be Rs y.
#Annual cost of phase advancing equipment
#                  = Rs 0.1 * y * 0·383 * OB            ...(ii)
#For economical use, the two costs should be equal i.e.,
#                    exp. (i) = exp. (ii).
#or      0·1 * y * 0·383 * OB = 2·37 * OB
y = 2.37/(0.1*0.383)                 #Rs

#If the losses in the phase advancing equipment are neglected,
#then its kVAR = kVA.

#Result:
print "The maximum cost per kVA of phase advancing equipment"
print "that can be paid is  Rs",round(y,2)

The maximum cost per kVA of phase advancing equipment
that can be paid is  Rs 61.88