Chapter 5 : Methods for Calculating Regulation of Alternator

Example 5.1 page no : 12

In [3]:
import math 

# Variables
P = 1000.*10**3 			#load power
phi = math.acos(math.radians(0.8)) 			#power factor lagging angle
V_L = 11.*10**3 			#rated terminal voltae
R_a = 0.4 			#armature resistance per phase
X_s = 3.			#synchronous reactance per phase

# Calculations
I_L = P/(math.sqrt(3)*V_L*math.cos(math.radians(phi)))
I_aph = I_L 			#for star connected load
I_a = I_L			#current through armature
V_ph = V_L/math.sqrt(3) 			#rated terminal volatge phase value

E_ph =  math.sqrt(  (V_ph*math.cos(math.radians(phi))+I_a*R_a)**2+(V_ph*math.sin(math.radians(phi))+I_a*X_s)**2   )  			#emf generated phase value
E_line = E_ph*math.sqrt(3) 			#line value of emf generated
regulation = 100*(E_ph-V_ph)/V_ph  			#pecentage regulation

# Results
print 'Line value of e.m.f generated is %.2f kV  \nRegulation is %.3f percent'%(E_line*10**-3,regulation)

# note : book answer is wrong.

Line value of e.m.f generated is 11.05 kV  
Regulation is 0.428 percent

Example 5.2 Page no : 14

In [5]:
import math 

# Variables
VA = 1200.*10**3
V_L = 6600.
R_a = 0.25 			#armature resistance per phase
X_s = 5.			#synchronous reactance per phase

# Calculations and Results
I_L = VA/(math.sqrt(3)*V_L)
I_aph = I_L 			#for star connected load
I_a = I_L
V_ph = V_L/math.sqrt(3)

#Part(i)
phi1 = math.acos(0.8)			#and lagging
E_ph1 =  math.sqrt(  (V_ph*math.cos(phi1)+I_a*R_a)**2+(V_ph*math.sin(phi1)+I_a*X_s)**2   )
regulation = 100*(E_ph1-V_ph)/V_ph  			#percentage regulation
print 'i)Regulation at 0.8 lagging pf is %.2f percent'%(regulation)
#Part(ii)
phi2 = math.acos(0.8)			#and leading
E_ph2 =  math.sqrt(  (V_ph*math.cos(phi2)+I_a*R_a)**2+(V_ph*math.sin(phi2)-I_a*X_s)**2   )
regulation2 = 100*(E_ph2-V_ph)/V_ph 			#percentage regulation
print 'ii)Regulation at 0.8 leading pf is %.f percent'%(regulation2)

i)Regulation at 0.8 lagging pf is 9.33 percent
ii)Regulation at 0.8 leading pf is -7 percent

Example 5.3 Page no : 17

In [6]:
import math 

# Variables
#full load
V_L_FL = 1100.
V_ph_FL = V_L_FL/math.sqrt(3)

# Calculations
#no load
V_L_NL = 1266
E_line = V_L_NL
E_ph = E_line/math.sqrt(3)
regulation = 100*(E_ph-V_ph_FL)/V_ph_FL

# Results
print 'Regulation at full load is %.2f percent'%(regulation)

Regulation at full load is 15.09 percent

Example 5.4 Page no : 23

In [7]:
import math 

# Variables
V_L = 866.
VA = 100.*10**3
I_L = VA/(math.sqrt(3)*V_L)    			#because VA = math.sqrt(3)*V_L*I_L
I_aph = I_L			#full load and star connected alternator
V_ph = V_L/math.sqrt(3)

# Calculations
#Graph is plotted and V_oc_ph and I_asc_Ph is obtained for 
#SCC for I_asc = 66.67 A
I_f = 2.4  # A
#OCC for I_f = 2.4 A
V_oc_ph = 240    # V

#for measruemnt of impedance
V_oc_ph = 240 			#for I_f = 2.4..From o.c.c graph
I_asc_ph = 66.67 			#for I_f = 2.4...From s.c.c graph
Z_s = V_oc_ph/I_asc_ph
R_a = 0.15
X_s = math.sqrt( Z_s**2-R_a**2 )

V_ph_FL = 500.
phi = math.acos(0.8)  			#lagging pf
E_ph = math.sqrt((V_ph_FL*math.cos(phi)+I_aph*R_a)**2+(V_ph_FL*math.sin(phi)+I_aph*X_s)**2)
regulation = 100*(E_ph-V_ph)/V_ph

# Results
print 'Full-load regulation at 0.8 lagging pf is %.2f percent '%(regulation )

Full-load regulation at 0.8 lagging pf is 35.57 percent

Example 5.5 Page no : 25

In [8]:
import math 

# Variables
V_OC_line = 230.
I_asc = 12.5  			# when I_f = 0.38
V_OC_ph = V_OC_line/math.sqrt(3)
Z_s = V_OC_ph/I_asc 

R_a = 1.8/2  			#1.8 is between terminals..0.9 is per phase
X_s = math.sqrt(Z_s**2-R_a**2)

I_a = 10.			# when regulation is needed
V_L = 230.
V_ph = V_L/math.sqrt(3)

# Calculations and Results
#Part(i)
phi1 = math.acos(0.8) 			#and lagging
E_ph1 = math.sqrt((V_ph*math.cos(phi1)+I_a*R_a)**2+(V_ph*math.sin(phi1)+I_a*X_s)**2)
regulation1 = 100*(E_ph1-V_ph)/V_ph
print 'Regulation for 10 A at 0.8 lagging pf is %.2f percent'%(regulation1)
#Part(ii)
phi2 = math.acos(0.8) 			#and leading
E_ph2 = math.sqrt((V_ph*math.cos(phi2)+I_a*R_a)**2+(V_ph*math.sin(phi2)-I_a*X_s)**2)
regulation2 = 100*(E_ph2-V_ph)/V_ph
print 'Regulation for 10 A at 0.8 leading pf is %.2f percent'%(regulation2)

Regulation for 10 A at 0.8 lagging pf is 64.47 percent
Regulation for 10 A at 0.8 leading pf is -11.01 percent

Example 5.6 Page no : 31

In [9]:
import math 

# Variables
phi = math.acos(0.8)
VA = 1000.*10**3
V_L = 1905.
V_ph = V_L/math.sqrt(3)
R_a = 0.2			#Armature reactance per phase

# Calculations and Results
#Part(i)
#Ampere-turn method
I_L = VA/(math.sqrt(3)*V_L)
I_aph = I_L
V_dash = V_ph+I_aph*R_a*math.cos(phi)			#V_dash is a dummy quantity and has no significance..it's used only for mapping correcponding current
F_o = 32 			#F_o corresponds to voltage V_dash = 1148.5 from O.C.C graph
F_AR = 27.5 			#Field current required to circulate full-load short circuit current of 303.07A.From SCC F_AR = 27.5
F_R  =  math.sqrt(    F_o**2 + F_AR**2-2*F_o*F_AR*math.cos(phi+math.pi/2) )			#using Comath.sine rule

# for F_R = 53.25 E_ph = 1490 V from O.C.C
E_ph = 1490.
regulation1 = 100*(E_ph-V_ph)/V_ph
print 'Regulation on full-load by ampere-turn method is %.2f percent'%(regulation1)

#Part (ii)
#Synchronous Impedance method

I_sc = I_L
I_aph2 = I_sc
I_f = 27.5

V_OC_ph = 1060. 			#corresponding to I-f = 27.5 in the graph
Z_s = V_OC_ph/I_aph2
X_s = math.sqrt(Z_s**2-R_a**2)

E_ph2 =  math.sqrt( (V_ph*math.cos(phi)+I_aph2*R_a)**2+(V_ph*math.sin(phi)+I_aph2*X_s)**2 )   			#from phasor diagram
regulation2 = 100*(E_ph2-V_ph)/V_ph
print 'Regulation on full-load by synchronous impedance method is %.2f percent'%(regulation2)

Regulation on full-load by ampere-turn method is 35.47 percent
Regulation on full-load by synchronous impedance method is 78.09 percent

Example 5.7 Page no : 39

In [10]:
import math 

# Variables
#case(i)
V_L = 440.
V_ph = V_L/math.sqrt(3)
phi = math.acos(0.8)

#armature resistance drop from the graph
#RS = 1.1 cm and scale  = 50 V/cm
arm_leak_resis =  1.1*50    			#armature leakage resistance

OB = V_ph*math.cos(phi)
AB = V_ph*math.sin(phi) + arm_leak_resis
E_1ph =  math.sqrt( OB**2+AB**2   )

F_f1 = 6.1  			#corresponding value from OCC
F_AR = 3.1*1

# Calculations and Results
F_R =  math.sqrt( F_f1**2 + F_AR**2 -2*F_f1*F_AR*math.cos(math.radians(90+math.acos(math.radians(0.8)))))
E_ph = 328. 			#voltage corresponding to F_R = 8.33 A from OCC graph
regulation1 =  100*(E_ph - V_ph)/V_ph
print 'i)Regulation for 0.8 pf lagging is %.2f percent '%(regulation1)

#case(ii)

OC = V_ph*math.cos(phi)
BC = V_ph*math.sin(phi) - arm_leak_resis
E_1ph =  math.sqrt( OC**2+BC**2   )

F_f1 = 6.1  			#corresponding value from OCC
F_R =  math.sqrt(   F_f1**2 + F_AR**2 -2*F_f1*F_AR*math.cos(math.radians(90-math.acos(math.radians(0.8)))))
E_ph = 90 			#volatge corresponding to F_R = 3.34 A from OCC graph
regulation2 =  100*(E_ph - V_ph)/V_ph
print 'ii)Regulation for 0.8 pf leading is %.2f percent '%(regulation2)
print 'The answer in part ii doesnt match with textbook because of calculation mistake done in last step in the textbook'

i)Regulation for 0.8 pf lagging is 29.12 percent 
ii)Regulation for 0.8 pf leading is -64.57 percent 
The answer in part ii doesnt match with textbook because of calculation mistake done in last step in the textbook

Example 5.8 Page no : 44

In [11]:
import math 

# Variables
P = 1200.*10**3
V_line = 12000.
R_a = 2.
X_s = 35.			#armature resistance and synchronous reactance
phi = math.acos(0.8)

# Calculations
I_L = P/(math.sqrt(3)*V_line*math.cos(phi))
I_a = I_L
V_ph = V_line/math.sqrt(3)
E_ph = math.sqrt((V_ph*math.cos(phi)+I_a*R_a)**2+(V_ph*math.sin(phi)+I_a*X_s)**2)
regulation = 100*(E_ph-V_ph)/V_ph

# Results
print 'Regulation at 0.8 lag power factor is %.2f percent'%(regulation)

Regulation at 0.8 lag power factor is 26.66 percent

Example 5.9 Page no : 45

In [12]:
import math 

# Variables
V_L = 11000.  
V_ph =  V_L/math.sqrt(3)
VA = 1000.*1000
I_L = VA/(V_L*math.sqrt(3))

V_OC_ph = 433/math.sqrt(3)
I_asc_ph = I_L

Z_s = V_OC_ph /I_asc_ph  			#ohms per phase
R_a = 0.45 			#ohms per phase
X_s = math.sqrt(Z_s**2-R_a**2)

# Calculations and Results
#part(i)
phi = math.acos(0.8) 			#lagging
E_ph  =  math.sqrt((V_ph*math.cos(phi)+I_L*R_a)**2 +(V_ph*math.sin(phi)+ I_L*X_s)**2)
regulation = 100*(E_ph-V_ph)/V_ph
print 'Voltage regulation at 0.8 pf lagging is %f percent'%(regulation)

#part(ii)
phi = math.acos(0.8) 			#leading
E_ph2  =  math.sqrt((V_ph*math.cos(phi)+I_L*R_a)**2 +(V_ph*math.sin(phi)- I_L*X_s)**2)
regulation2 = 100*(E_ph2-V_ph)/V_ph
print 'Voltage regulation at 0.8 pf lagging is %f percent'%(regulation2)
print 'Answer mismatches due to improper approximation'

Voltage regulation at 0.8 pf lagging is 2.690067 percent
Voltage regulation at 0.8 pf lagging is -1.996182 percent
Answer mismatches due to improper approximation

Example 5.10 Page no : 46

In [13]:
import math 

# Variables
VA = 125.*10**3
V_L = 400.
V_ph = V_L/math.sqrt(3)
I_L = VA/(math.sqrt(3)*V_L)
I_aph = I_L

# Calculations
I_f = 4.
I_asc = I_aph/2 			#for half load.. refer to graph
V_OC_line = 140.
V_OC_ph = V_OC_line/math.sqrt(3)
I_asc_ph = I_asc
Z_s =  V_OC_ph/I_asc_ph
R_a = 0.1
X_s = math.sqrt(Z_s**2-R_a**2) 			#armature resistance and synchronous reactance

phi = math.acos(0.8)
E_ph  =  math.sqrt((V_ph*math.cos(phi)+I_asc*R_a)**2 +(V_ph*math.sin(phi)- I_asc*X_s)**2)
regulation = 100*(E_ph-V_ph)/V_ph

# Results
print 'Voltage regulation at 0.8 pf leading for half load is %.2f percent'%(regulation)

Voltage regulation at 0.8 pf leading for half load is -12.39 percent

Example 5.11 Page no : 48

In [15]:
import math 
from numpy import roots

# Variables
V_OC_line = 575.
V_OC_ph = V_OC_line/math.sqrt(3)
I_asc_line = 75.
I_asc_ph   = I_asc_line 
I_aph = I_asc_ph
I_L = I_aph

Z_s =  V_OC_ph/I_asc_ph
R_a = 2.16/2
X_s  =   math.sqrt(Z_s**2 - R_a**2)

# Calculations and Results
#on full load
E_ph = 6100.
phi = math.acos(0.8) 			#leading

#using E_ph  =  math.sqrt((V_ph*math.cos(phi)+I_a*R_a)**2 +(V_ph*math.sin(phi)- I_a*X_s)**2)
p = [1, -256.68, -3.71*10**7]
ans = roots(p)
V_ph = ans[0] 			#second root is ignored as its -ve
V_L = V_ph*math.sqrt(3)
print 'Rated terminal voltage between the lines is %.3f V '%(V_L)
VA_rating = math.sqrt(3)*V_L*I_L
print 'kVA rating of the alternator is %.f kVA'%(VA_rating*10**-3)

Rated terminal voltage between the lines is 10774.515 V 
kVA rating of the alternator is 1400 kVA

Example 5.12 Page no : 49

In [16]:
import math 
from numpy import roots

# Variables
V_L = 6600.
V_ph = V_L/math.sqrt(3)
VA = 1500.*10**3
I_L = VA/(math.sqrt(3)*V_L)
I_aph = I_L

# Calculations and Results
R_a = 0.5
X_s = 5			#armature resistance and synchronous reactance
phi = math.acos(0.8)
E_ph  =  math.sqrt((V_ph*math.cos(phi)+I_aph*R_a)**2 +(V_ph*math.sin(phi)+ I_aph*X_s)**2)
print 'Induced EMF per phase is %f V'%(E_ph)

#full load 
phi = math.acos(1)
#using E_ph  =  math.sqrt((V_ph*math.cos(phi)+I_a*R_a)**2 +(V_ph*math.sin(phi)- I_a*X_s)**2)
p = [1, 131.215, -1.791*10**7]
ans = roots(p)
V_ph = ans[1] 			#first root is ignored as it is -ve
print 'Terminal voltage per phase is %f V'%(V_ph)

# note : rounding off error.

Induced EMF per phase is 4284.243828 V
Terminal voltage per phase is 4166.921808 V

Example 5.13 Page no : 50

In [17]:
import math 

# Variables
V_ph = 2000.
R_a = 0.8
I_sc = 100.
V_OC = 500.
I_f = 2.5
Z_s = V_OC/I_sc
X_s =  math.sqrt(Z_s**2 - R_a**2 )
I_aFL = 100.
I_a = I_aFL

# Calculations and Results
#part(i)
phi = math.acos(1)
E_ph  =  math.sqrt((V_ph*math.cos(phi)+I_a*R_a)**2 +(V_ph*math.sin(phi)+ I_a*X_s)**2)
regulation = 100*(E_ph-V_ph)/V_ph
print 'i)Voltage regulation is %.2f percent'%(regulation)

#part(ii)
phi2 = math.acos(0.8)
E_ph2  =  math.sqrt((V_ph*math.cos(phi2)+I_a*R_a)**2 +(V_ph*math.sin(phi2)- I_a*X_s)**2)
regulation2 = 100*(E_ph2-V_ph)/V_ph
print 'ii)Voltage regulation is %.2f percent'%(regulation2)

#part(iii)
phi3 = math.acos(0.71)
E_ph3  =  math.sqrt((V_ph*math.cos(phi3)+I_a*R_a)**2 +(V_ph*math.sin(phi3)+ I_a*X_s)**2)
regulation3 = 100*(E_ph3-V_ph)/V_ph
print 'iii)Voltage regulation is %.2f percent'%(regulation3)

i)Voltage regulation is 6.89 percent
ii)Voltage regulation is -8.88 percent
iii)Voltage regulation is 21.11 percent

Example 5.14 Page no : 51

In [18]:
import math 

# Variables
VA = 1000.*1000
V_L = 4600. 
V_ph = V_L/math.sqrt(3)
I_L = VA/(math.sqrt(3)*V_L)
I_aph_FL = I_L
I_aph = I_aph_FL
I_sc = (150./100)* I_aph_FL
V_OC_line = 1744
V_OC_ph =  V_OC_line/math.sqrt(3)

# Calculations
Z_s = V_OC_ph / I_sc
R_a = 1
X_s = math.sqrt(Z_s**2-R_a**2)

phi = math.acos(0.8 ) 			#lagging
E_ph  =  math.sqrt((V_ph*math.cos(phi)+I_aph*R_a)**2 +(V_ph*math.sin(phi)+ I_aph*X_s)**2)
regulation = 100*(E_ph-V_ph)/V_ph

# Results
print 'Voltage regulation at full load 0.8 pf is %.2f percent'%(regulation)

Voltage regulation at full load 0.8 pf is 19.89 percent

Example 5.15 Page no : 52

In [19]:
import math 


#part(i)    Ampere turn method
# Variables
F_O = 37.5
F_AR = 20.
V_L = 6600.
V_ph = V_L/math.sqrt(3)

# Calculations and Results
#lagging
phi = math.acos(0.8) 
F_R =  math.sqrt((F_O+F_AR*math.sin(phi) )**2 + (F_AR*math.cos(phi))**2 ) 
#E_ph corresponding to F_R can be obtained by plotting open circuit characteristics
E_ph = 4350
regulation = 100*(E_ph-V_ph)/V_ph
print 'i)By Ampere-turn method or MMF methodFull-load regulation at 0.8 lagging pf is %.2f percent'%(regulation)
#leading
phi = math.acos(0.8) 
F_R =  math.sqrt((F_O-F_AR*math.sin(phi) )**2 + (F_AR*math.cos(phi))**2 ) 
#E_ph corresponding to F_R can be obtained by plotting open circuit characteristics
E_ph = 3000
regulation = 100*(E_ph-V_ph)/V_ph
print 'Full-load regulation at 0.8 leading pf is %.2f percent'%(regulation)

#EMF method
V_OC_ph = 100
V_ph = 100
I_sc =  100*(F_O/F_AR)   			#times the rated value
Z_s = V_OC_ph/I_sc
F_O =  100
F_AR =  Z_s*100

#lagging
phi = math.acos(0.8)
F_R =   math.sqrt((F_O+F_AR*math.sin(phi) )**2 + (F_AR*math.cos(phi))**2 ) 
regulation = 100*(F_R-V_ph)/V_ph
print 'iiSynchronous impedance method or EMF method'
print 'Full-load regulation at 0.8 lagging pf is %.2f percent'%(regulation)
#leading
phi = math.acos(0.8)
F_R =   math.sqrt((F_O-F_AR*math.sin(phi) )**2 + (F_AR*math.cos(phi))**2 ) 
regulation = 100*(F_R-V_ph)/V_ph
print 'Full-load regulation at 0.8 leading pf is %.2f percent'%(regulation)

i)By Ampere-turn method or MMF methodFull-load regulation at 0.8 lagging pf is 14.16 percent
Full-load regulation at 0.8 leading pf is -21.27 percent
iiSynchronous impedance method or EMF method
Full-load regulation at 0.8 lagging pf is 38.72 percent
Full-load regulation at 0.8 leading pf is -19.72 percent

Example 5.16 Page no : 56

In [20]:
import math 

# Variables
V_L = 6000.
V_ph = V_L/math.sqrt(3)
I_ph_X_Lph  =  0.9*500  			#leakage reactance drop in volts  =  0.9 cm * 500 V/cm
phi =  math.acos(0.8) 			#lagging

# Calculations
E_1ph = math.sqrt( (V_ph*math.cos(phi))**2 + (V_ph*math.sin(phi)+I_ph_X_Lph)**2 )   			#From triangle OAB
F_f1  =  26 			#from OCC
F_AR =  2.9*5  			#2.9cm * 5 A/cm

F_R  =    math.sqrt(F_f1**2 + F_AR**2 -2*F_AR*F_f1*math.cos(phi+ (math.pi/2)) )

# Results
print 'Required field current is %.2f A'%(F_R)

Required field current is 36.59 A

Example 5.17 Page no : 58

In [21]:
import math 

# Variables
V_L = 400.
V_ph = V_L/math.sqrt(3)
VA = 40.*10**3
I_L = VA/(math.sqrt(3)*V_L) 
I_aph = I_L

# Calculations and Results
I_aph_X_Lph  =  0.65*50  			#leakage reactance drop in volts  =  2.4 cm * 500 V/cm
X_Lph =  I_aph_X_Lph/ I_aph
print 'Armature leakage reactance is %.3f ohms'%(X_Lph)
phi = math.acos(0.8)  			#lagging
E_ph  =  math.sqrt((V_ph*math.cos(phi))**2 +(V_ph*math.sin(phi)+ I_aph_X_Lph)**2)
F_f1 = 15.6 			#as obtained from OCC corresponding to this E_ph

F_AR =  2.3*3  			#2.3cm * 3 A/cm
print 'Armature reaction is %.1f '%(F_AR)
F_R  =    math.sqrt(F_f1**2 + F_AR**2 -2*F_AR*F_f1*math.cos(phi+ (math.pi/2)) )  			#math.comath.sine rule to Triangle OAB
E_ph = 267.5  			#corresponding to F_R from open circiut characteristics
regulation = 100*(E_ph-V_ph)/V_ph
print 'Voltage regulation at 0.8 pf lagging is %.1f percent'%(regulation)


#Note:This answer doesnt match with textbook as it has been reciprocated in textbook

Armature leakage reactance is 0.563 ohms
Armature reaction is 6.9 
Voltage regulation at 0.8 pf lagging is 15.8 percent

Example 5.18 Page no : 60

In [22]:
import math 

# Variables
VA = 10.*10**3
V_L = 11.*10**3
V_ph = V_L/math.sqrt(3)
I_ph_X_Lph  =  2.4*500  			#leakage reactance drop in volts  =  2.4 cm * 500 V/cm

# Calculations
I_ph_R_aph  = VA/(math.sqrt(3)*V_L)
phi = math.acos(0.8)
E_ph  =  math.sqrt((V_ph*math.cos(phi)+I_ph_R_aph)**2 +(V_ph*math.sin(phi)+ I_ph_X_Lph)**2)
F_f1 = 109 			#obtained from open circuit characteristics corresponding to calculated E_ph
F_AR =  2.8*10  			#2.8cm * 10 A/cm
F_R  =    math.sqrt(F_f1**2 + F_AR**2 -2*F_AR*F_f1*math.cos(phi+ (math.pi/2)) )  			#math.comath.sine rule to Triangle OAB
E_ph = 7700.  			#corresponding to F_R from open circiut characteristics

regulation = 100*(E_ph-V_ph)/V_ph

# Results
print 'Voltage regulation at full-load 0.8 pf lagging is %.2f percent'%(regulation)

Voltage regulation at full-load 0.8 pf lagging is 21.24 percent

Example 5.19 Page no : 63

In [23]:
import math 
from numpy import roots

# Variables
VA = 2000.*1000
V_L = 11000.
V_ph = V_L/math.sqrt(3)
R_a = 0.3
X_s = 5. 			#armature resistance and synchronous reactance

# Calculations
#case (i)
phi = math.acos(0.8) 			#lagging
I_L = VA/(math.sqrt(3)*V_L)  
I_a = I_L
E_ph  =  math.sqrt((V_ph*math.cos(phi)+I_a*R_a)**2 +(V_ph*math.sin(phi)+ I_a*X_s)**2)

#using E_ph  =  math.sqrt((V_ph*math.cos(phi)+I_aph*R_a)**2 +(V_ph*math.sin(phi)+ I_aph*X_s)**2)
#we get  V_ph**2 -579.4455 V_ph -44653301.91 = 0
p = [1, -579.4455, -44653301.91]
ans = roots(p)
V_ph = ans[0]    		#second root is ignored as its -ve
print 'Terminal voltage is %.4f V'%(V_ph)

Terminal voltage is 6978.3131 V

Example 5.20 Page no : 64

In [24]:
import math 
from numpy import roots

# Variables
R_a = 0.6
X_s = 6. 			#armature resistance and synchronous reactance per phase
E_L = 6599.
E_ph = E_L/math.sqrt(3)
I_L = 180.
I_a = I_L

# Calculations and Results
#part(i)
# using E_ph  =  math.sqrt((V_ph*math.cos(phi)+I_a*R_a)**2 +(V_ph*math.sin(phi)+ I_a*X_s)**2) and solving for V_ph
p = [1 ,1135.83, -13338836.49]
ans = roots(p)
V_ph = ans[1]
V_L = V_ph*math.sqrt(3)
regulation = 100*(E_ph-V_ph)/V_ph

phi = math.acos(0.9)
theta = math.atan((I_a*X_s+V_ph*math.sin(phi) )/(E_ph))
delta = theta-phi
print 'i)0.9 laggingTerminal voltage is %.2f VVoltage regulation is %.2f percentLoad angle is %.2f degrees'%(V_ph*math.sqrt(3),regulation,delta*180/math.pi)

#part(ii)
phi_2 = math.acos(0.8)
p = [1, -941.53, -11399574.87]
ans = roots(p) 
V_ph = ans[0] 			#second root is ignored as its -ve
V_L = V_ph*math.sqrt(3)
regulation2 = 100*(E_ph-V_ph)/V_ph
delta_2  =  math.asin( (math.tan(phi)*(V_ph*math.cos(phi_2)+I_a*R_a) -I_a*X_s )/E_ph   )
print 'ii)0.8 leadingTerminal voltage is %.2f VVoltage regulation is %.2f percentLoad angle is %.2f degrees'%(V_L,regulation2,delta_2*180/math.pi)

i)0.9 laggingTerminal voltage is 5418.22 VVoltage regulation is 21.79 percentLoad angle is 6.83 degrees
ii)0.8 leadingTerminal voltage is 6719.93 VVoltage regulation is -1.80 percentLoad angle is 7.17 degrees

Example 5.21 Page no : 66

In [25]:
import math 

# Variables
V_ph = 2000.
R_a = 0.8
I_sc = 100.
I_a = I_sc
V_OC = 500.
I_f = 2.5
Z_s = V_OC/I_sc

X_s = math.sqrt(Z_s**2- R_a**2)
I_a_FL = 100.

# Calculations and Results
#Part(i)
phi1 = math.acos(1) 			#and lagging
E_ph1 = math.sqrt((V_ph*math.cos(phi1)+I_a*R_a)**2+(V_ph*math.sin(phi1)+I_a*X_s)**2)
regulation1 = 100*(E_ph1-V_ph)/V_ph
print 'Regulation at upf is %.2f percent'%(regulation1)

#Part(ii)
phi2 = math.acos(0.8) 
E_ph2 = math.sqrt((V_ph*math.cos(phi2)+I_a*R_a)**2+(V_ph*math.sin(phi2)-I_a*X_s)**2)
regulation2 = 100*(E_ph2-V_ph)/V_ph
print 'Regulation at 0.8 leading pf is %.2f percent'%(regulation2)

#Part(iii)
phi3 = math.acos(0.71) 
E_ph3 = math.sqrt((V_ph*math.cos(phi3)+I_a*R_a)**2+(V_ph*math.sin(phi3)+I_a*X_s)**2)
regulation3 = 100*(E_ph3-V_ph)/V_ph
print 'Regulation at 0.71 lagging pf is %.2f percent'%(regulation3)

Regulation at upf is 6.89 percent
Regulation at 0.8 leading pf is -8.88 percent
Regulation at 0.71 lagging pf is 21.11 percent

Example 5.22 Page no : 68

In [26]:
import math 

# Variables
V = 600.
VA = 60.*10**3
I_sc = 210.
V_oc = 480.
I_f = 10.
R_a = 0.2

I = VA/V        			#VA = V*I and alternator is math.single phase
I_a = I

# Calculations and Results
Z_s = V_oc/I_sc  			#Synchronous Impedance
X_s = math.sqrt(Z_s**2-R_a**2) 			#SYnchronous reactance
print 'Synchronous impedances is %f ohms and synchronous reactance is %f ohms'%(Z_s,X_s)

#PART (i)
phi1 = math.acos(0.8) 			#and lagging
E1 = math.sqrt((V*math.cos(phi1)+I_a*R_a)**2+(V*math.sin(phi1)+I_a*X_s)**2) 			#plus sign for lagging power factor
regulation1 = 100*(E1-V)/V
print 'Regulation at 0.8 lagging pf is %.2f percent '%(regulation1 )

#PART (ii)
phi2 = math.acos(1) 
E2 = math.sqrt((V*math.cos(phi2)+I_a*R_a)**2+(V*math.sin(phi2)+I_a*X_s)**2) 
regulation2 = 100*(E2-V)/V
print 'Regulation at UNITY pf is %.2f percent '%(regulation2 )

#PART (iii)
phi3 = math.acos(0.6) 			#and leading 
E3 = math.sqrt((V*math.cos(phi3)+I_a*R_a)**2+(V*math.sin(phi3)-I_a*X_s)**2) 			#minus sign for leading power factor
regulation3 = 100*(E3-V)/V
print 'Regulation at 0.6 leading pf is %.2f percent '%(regulation3 )

Synchronous impedances is 2.285714 ohms and synchronous reactance is 2.276947 ohms
Regulation at 0.8 lagging pf is 28.60 percent 
Regulation at UNITY pf is 10.08 percent 
Regulation at 0.6 leading pf is -23.98 percent

Example 5.23 Page no : 69

In [27]:
import math 

# Variables
V_L = 3300.
V_ph = V_L/math.sqrt(3)
I_a = 100.
I_f = 5.
V_OC_line = 900. 
V_OC_ph = V_OC_line/math.sqrt(3) 
R_a = 0.8 			#armature resistance
I_aph = I_a
Z_s = V_OC_ph/I_aph
X_s = math.sqrt(Z_s**2-R_a**2) 			#synchronous reactance

# Calculations and Results
#Part(i)
phi1 = math.acos(0.8) 			#and lagging
E_ph1 = math.sqrt((V_ph*math.cos(phi1)+I_a*R_a)**2+(V_ph*math.sin(phi1)+I_a*X_s)**2)
regulation1 = 100*(E_ph1-V_ph)/V_ph
print 'Regulation at 0.8 lagging is %.2f percent'%(regulation1)

#Part(ii)
phi2 = math.acos(0.8)  			#and leading
E_ph2 = math.sqrt((V_ph*math.cos(phi2)+I_a*R_a)**2+(V_ph*math.sin(phi2)-I_a*X_s)**2)
regulation2 = 100*(E_ph2-V_ph)/V_ph
print 'Regulation at 0.8 leading pf is %.2f percent'%(regulation2)

Regulation at 0.8 lagging is 21.03 percent
Regulation at 0.8 leading pf is -9.55 percent

Example 5.24 Page no : 70

In [29]:
import math 

# Variables
V_L = 13500.
R_a = 1.5 
X_s = 30. 			#armature resistance and synchronous reactance
V_ph = V_L/math.sqrt(3)

# Calculations and Results
#CASE 1
phi1 = math.acos(0.8)
P_out = 1280*10**3
I_L =  P_out/ (math.sqrt(3)*V_L*math.cos(phi1) )    			#because P_out = math.sqrt(3)*V_L*I_L*math.cos(phi)

I_a = I_L
E_ph = math.sqrt((V_ph*math.cos(phi1)+I_a*R_a)**2+(V_ph*math.sin(phi1)+I_a*X_s)**2)
regulation = 100*(E_ph-V_ph)/V_ph
print 'Regulation at 0.8 lagging power factor is %.2f percent'%(regulation)

#Case 2
phi2 = math.acos(1)
I_L =  P_out/ (math.sqrt(3)*V_L*math.cos(phi2) )    			#because P_out = math.sqrt(3)*V_L*I_L*math.cos(phi)

I_a = I_L
E_ph = math.sqrt((V_ph*math.cos(phi2)+I_a*R_a)**2+(V_ph*math.sin(phi2)+I_a*X_s)**2)
regulation2 = 100*(E_ph-V_ph)/V_ph
print 'Regulation at unity power factor is %.2f percent'%(regulation2)

#case 3
phi3 = math.acos(0.8)
I_L =  P_out/ (math.sqrt(3)*V_L*math.cos(phi3) )    			#because P_out = math.sqrt(3)*V_L*I_L*math.cos(phi)
I_a = I_L
E_ph = math.sqrt((V_ph*math.cos(phi3)+I_a*R_a)**2+(V_ph*math.sin(phi3)-I_a*X_s)**2)			# minus sign in the second bracket beacuse of leading pf
regulation3 = 100*(E_ph-V_ph)/V_ph
print 'Regulation at 0.8 leading power factor is %.2f percent'%(regulation3)

# note : rounding off error

Regulation at 0.8 lagging power factor is 18.60 percent
Regulation at unity power factor is 3.23 percent
Regulation at 0.8 leading power factor is -11.99 percent

Example 5.26 Page no : 72

In [31]:
import math 

# Variables
V_L = 11.*10**3
VA_rating = 10.**6
R_a = 2.2 			#alternator resistance
phi = math.acos(0.8)

# Calculations
I_L = VA_rating/(math.sqrt(3)*V_L)			#VA = math.sqrt(3)V_L*I_L
I_a = I_L
V_ph = V_L/math.sqrt(3)
regulation = 24.

E_ph =  ((regulation/100)+1)*V_ph    			# because regulation = 100*(E_ph-V_ph)/V_ph
#using E_ph = math.sqrt((V_ph*math.cos(phi)+I_a*R_a)**2+(V_ph*math.sin(phi)+I_a*X_s)**2)
X_s = (math.sqrt(E_ph**2-((V_ph*math.cos(phi)+I_a*R_a)**2))-V_ph*math.sin(phi))*(1/I_a)

phi1 = math.acos(0.8)
E_ph = math.sqrt((V_ph*math.cos(phi1)+I_a*R_a)**2+(V_ph*math.sin(phi1)-I_a*X_s)**2)
regulation1 = 100*(E_ph-V_ph)/V_ph

# Results
print 'Regulation at 0.8 leading power factor is %.2f percent'%(regulation1)

Regulation at 0.8 leading power factor is -13.90 percent

Example 5.27 Page no : 73

In [32]:
import math 

# Variables
V_L = 220.
VA = 100.*10**3
R_a = 0.1 			#effective resistacne of alternator
X_a = 0.5 			#leakage reactance
X_ar = 2*X_a

Z_s = complex(R_a,X_a+X_ar)

# Calculations and Results
#Part(1)
phi = math.acos(0.4)
V_ph = V_L/math.sqrt(3)
I_L = VA/(math.sqrt(3)*V_L)			#VA = math.sqrt(3)*V_L*I_L
I_a = I_L
E_ph = math.sqrt((V_ph*math.cos(phi)+I_a*R_a)**2+(V_ph*math.sin(phi)+I_a*(X_a+X_ar))**2)
print 'i)Required noload voltage is %.3f V'%(E_ph)

#Part(2)
V_ph2 = 0
E_ph2 = math.sqrt((V_ph2*math.cos(phi)+I_a*R_a)**2+(V_ph2*math.sin(phi)+I_a*(X_a+X_ar))**2)
print 'ii)Required noload voltage is %.3f V'%(E_ph2)

i)Required noload voltage is 515.848 V
ii)Required noload voltage is 394.522 V

Example 5.28 Page no : 74

In [33]:
import math 

# Variables
V_L = 2000.
V_ph = V_L/math.sqrt(3)
VA = 1000.*10**3
I_L = VA/(math.sqrt(3)*V_L)    			#because VA = math.sqrt(3)*V_L*I_L
I_aph = I_L

I_f = 28.5			#for this I_aph = 288.67513 as obtained from SCC graph
V_oc_ph = 1060.			#for I_f = 28.5 as obtained fromOCC graph
Z_s = V_oc_ph/I_aph
R_a = 0.2 			#armature effective resistance
X_s = math.sqrt( Z_s**2-R_a**2 )

# Calculations and Results
#Part(i)
phi1 = math.acos(0.8)			#lagging
E_ph1 = math.sqrt((V_ph*math.cos(phi1)+I_aph*R_a)**2+(V_ph*math.sin(phi1)+I_aph*X_s)**2)
regulation1 = 100*(E_ph1-V_ph)/V_ph
print "i)Full-load percentage regulation at 0.8 pf lagging is %.2f percent"%(regulation1)

#Part(ii)
phi2 = math.acos(0.8)			#leading
E_ph2 = math.sqrt((V_ph*math.cos(phi2)+I_aph*R_a)**2+(V_ph*math.sin(phi2)-I_aph*X_s)**2)
regulation2 = 100*(E_ph2-V_ph)/V_ph
print "ii)Full-load percentage regulation at 0.8 pf leading is %.2f percent"%(regulation2)
print 'Note that the answer mismatches because of calculation mistake done in the last step of part 1'

i)Full-load percentage regulation at 0.8 pf lagging is 73.86 percent
ii)Full-load percentage regulation at 0.8 pf leading is -9.29 percent
Note that the answer mismatches because of calculation mistake done in the last step of part 1

Example 5.29 Page no : 76

In [34]:
import math 

# Variables
V_L = 3300.
VA = 200.*10**3
R_a = 0.6
X_s = 6.			#armature resistance and synchronous reactance

# Calculations
I_L = VA/(math.sqrt(3)*V_L)			#VA = math.sqrt(3)V_L*I_L
I_a = I_L
V_ph = V_L/math.sqrt(3)
phi = math.acos(0.8)
E_ph = math.sqrt((V_ph*math.cos(phi)+I_a*R_a)**2+(V_ph*math.sin(phi)+I_a*X_s)**2)

regulation = 100*(E_ph-V_ph)/V_ph

# Results
print 'Regulation at 0.8 lagging power factor is %.3f percent'%(regulation)
print ' Note : Regulation is positive for lagging power factor loads'

Regulation at 0.8 lagging power factor is 7.802 percent
 Note : Regulation is positive for lagging power factor loads

Example 5.30 Page no : 77

In [35]:
import math 

# Variables
V_L = 2300.   
V_ph =  V_L/math.sqrt(3)
f = 50.
R_a = 0.2 			#armature resistance
I_sc = 150.
V_OC_line = 780.   
V_OC_ph = V_OC_line/math.sqrt(3)

Z_s =  V_OC_ph/I_sc
X_s  =   math.sqrt(Z_s**2 - R_a**2)
I_aph = 25 
I_aFL = I_aph

# Calculations and Results
#part(i)
phi =  math.acos(0.8)  			#lag
E_ph  =  math.sqrt((V_ph*math.cos(phi)+I_aph*R_a)**2 +(V_ph*math.sin(phi)+ I_aph*X_s)**2)
regulation = 100*(E_ph-V_ph)/V_ph
print 'Voltage regulation at 0.8 pf lagging is %.3f percent'%(regulation)

#part(ii)
phi2 =  math.acos(0.8)  			#lead
E_ph2  =  math.sqrt((V_ph*math.cos(phi2)+I_aph*R_a)**2 +(V_ph*math.sin(phi2)- I_aph*X_s)**2 )
regulation2 = 100*(E_ph2-V_ph)/V_ph
print 'Voltage regulation at 0.8 pf leading is %.3f percent'%(regulation2)

Voltage regulation at 0.8 pf lagging is 3.774 percent
Voltage regulation at 0.8 pf leading is -2.967 percent