# Chapter 1: Synchronous Machines(Additional problems)¶

## Problem 1, Page 120¶

In [1]:
import math

#Variable declaration
phase = 3  #three phase winding
ns = 36.    #number of slots
np = 4     #number of poles

#Calculations
nsp = ns/phase  #number of slots per phase
npp = nsp/np    #number of slots pole/phase
alpha = 180*np/ns #slot angle
Kd = math.sin(npp*alpha/2*math.pi/180)/(3*math.sin(alpha/2*math.pi/180))

#Results
print "The distribution factor is %.2f"%Kd

The distribution factor is 0.96


## Problem 2, Page 120¶

In [2]:
import math

#Variable declaration
np = 3    #number of phases
npl = 4   #number of poles
ns = 24   #number of slots

#calculaions
nsp = ns/npl    #number of slot/pole
alpha = 180/nsp #slot pitch(degrees)
m = ns/(npl*np)
Kd = math.sin(m*alpha/2*math.pi/180)/(m*math.sin(alpha/2*math.pi/180))
Kc = math.cos(alpha/2*math.pi/180)

#Results
print "Distribution factor = %.2f"%Kd
print "Pitch factor = %.3f degrees"%Kc

Distribution factor = 0.97
Pitch factor = 0.966 degrees


## Problem 3, Page 120¶

In [3]:
import math

#Variable declaration
Vl= 11000     #V
N = 1500      #rpm
f = 50        #Hz

#Calculations
P = (120*f)/N
Vp = Vl/math.sqrt(3)

#Results
print "Number of poles = %d"%P
print "Voltage per phase of alternator = %d V"%Vp

Number of poles = 4
Voltage per phase of alternator = 6350 V


## Problem 4, Page 120¶

In [4]:
import math

#Variable declaration
P = 4    #number of poles
ph = 3   #number of phases
s = 36   #number of slots
cs = 8   #coil span

#calculations
nsp = s/P    #number of slots/pole
alpha = 180/nsp   #slot pitch
m = s/(P*ph)
Kd = math.sin(m*alpha/2*math.pi/180)/(m*math.sin(alpha/2*math.pi/180))
ns = nsp-cs   #no. of slots by which coil is short pitched
B = 1*alpha
Kc= math.cos(B/2*math.pi/180)

#Results
print "Distribution factor = %.3f"%Kd
print "Pitch factor = %.3f degrees"%Kc

Distribution factor = 0.960
Pitch factor = 0.985 degrees


## Problem 5, Page 120¶

In [5]:
import math

#Variable declaration
P= 16.    #number of poles
N = 375  #rpm
s = 144.  #number of slots
c = 10   #number of conductors
phase = 3
phi = 0.035  #flux per pole

#Calculations
f = (P*N)/120  #Hz
ns = s/P         #slot/pole
m = ns/phase
alpha = 180/ns   #slot angle
Kd = math.sin(m*alpha/2*math.pi/180)/(m*math.sin(alpha/2*math.pi/180))
Tc = s*c         #total number of conductor
Tcp = Tc/phase   #total number of conductors/phase
Ncp = Tcp/2      #number of turns/phase
emf = 4.44*Kd*phi*f*Ncp
Vl = math.sqrt(3)*emf

#Results
print "Frequency = %d Hz"%f
print "E.M.F = %d V"%emf
print "Line voltage = %d V"%Vl

Frequency = 50 Hz
E.M.F = 1789 V
Line voltage = 3100 V


## Problem 6, Page 120¶

In [6]:
import math

#Variable declaration
P = 4   #no. of poles
s = 36  #no. of slots
ph = 3  #no. of phases
cs = 8  #coil span

#Calculations
ns = s/P  #no. of slots/pole
alpha = 180/ns  #slot pitch
m = s/(P*ph)
Kd = math.sin(m*alpha/2*math.pi/180)/(m*math.sin(alpha/2*math.pi/180))
Ns = ns-cs
Kc = math.cos(alpha/2*math.pi/180)

#Results
print "Distribution factor = %.2f"%Kd
print "Pitch factor = %.4f degrees"%Kc

Distribution factor = 0.96
Pitch factor = 0.9848 degrees


## Problem 7, Page 120¶

In [7]:
import math

#Variable declaration
El = 6600    #V
P = 4        #number of poles
s = 60       #number of slots
ph = 3       #number of phases
c = 2        #number of conductors
sn = 13      #slot number
f = 50       #Hz
T = 20       #V

#Calculations
n = s/P
m = n/ph
Zp = (s*c)/ph
B = 180/n     #degrees
Kd = math.sin(m*B/2*math.pi/180)/(m*math.sin(B/2*math.pi/180))
Cs = (sn-1)*180/n  #degrees
Kp = math.cos((180-Cs)/2*math.pi/180)
phi = El/(math.sqrt(3)*4.44*Kd*Kp*f*T)

#Results
print "The required flux pole is %.3f V"%phi

The required flux pole is 0.943 V


## Problem 8, Page 120¶

In [8]:
import math

#Varible declaration
Vl = 230    #V
Rt = 10*10**3  #VA
ph = 3      #no. of phases
Ra = 0.5    #ohms/phase
Xs = 1.2    #ohms/phase
cos_phi = 0.8  #lagging
V = 132.8      #V

#Calculations
I = Rt/(ph*V)  #A
print I
#Part(a)
Eo = math.sqrt(((V*cos_phi)+(I*Ra))**2+((V*sin_phi)+(I*Xs))**2)
Reg1 = (Eo-V)/V*100

#Part(b)
Eo = math.sqrt((V*cos_phi+I*Ra)**2+(V*sin_phi-I*Xs)**2)
Reg2 = (Eo-V)/V*100

#Results
print "Part(a): Regulation = %.2f %%"%Reg1
print "Part(b): Regulation = %.2f %%"%Reg2

25.1004016064
Part(a): Regulation = 21.81 %
Part(b): Regulation = -3.08 %


## Problem 9, Page 120¶

In [9]:
import math

#Variable declaration
El = 10000   #V
Kd = 0.96
Kp = 1.0
phi = 15*10**-2  #wb
f = 50       #Hz

#Calculations
T = El/(math.sqrt(3)*4.44*Kd*Kp*phi*f)
Zp = 2*T

#Result
print "Number of armature conductors in series/phase = %d"%Zp
#Answer differs due to rounding-off errors

Number of armature conductors in series/phase = 361


## Problem 10, Page 120¶

In [10]:
import math

#Variable declaration
s = 90   #no. of slots
P = 10   #no. of poles
ph = 3   #no. of phases
E = 11000. #E.M.F(V)
phi = 0.16 #flux(Wb)
f = 50     #Hz

#Calculations
nsp = s/(P*ph)  #slot/pole/phase
alpha = 180/(s/P) #slot angle
m = s/(P*ph)
Kd = math.sin(m*alpha/2*math.pi/180)/(m*math.sin(alpha/2*math.pi/180))
N = E/(math.sqrt(3)*4.44*Kd*phi*f)
Nc = 2*N

#Result
print "Number of conductors/phase are %d"%Nc
#Incorrect answer for N in the textbook. Hence the result differs

Number of conductors/phase are 372


## Problem 11, Page 121¶

In [11]:
import math

#Variable declaration
Vl = 400    #V
cos_phi = 1 #pf
n = 85./100 #efficiency
Zs = 10     #reactance(ohms)

#Calculations
out = Zs*735.5  #output
mi = out/n      #motor input(W)
Il = mi/(math.sqrt(3)*Vl*cos_phi)
I = Il   #since current is minimum when power factor is unity
Er = I*Zs
V = Vl/math.sqrt(3)
E = math.sqrt(V**2+Er**2)
emf = math.sqrt(3)*E

#Results
print "Minimum current = %.1f A"%I
print "Line induces e.m.f. = %d V"%emf
#Answers differ due to rounding-off errors

Minimum current = 12.5 A
Line induces e.m.f. = 454 V


## Problem 12, Page 121¶

In [12]:
import math

#Variable declaration
P = 6
phi = 25*10**-3   #wb
f = 50            #Hz
ph = 3            #no. of phases
s = 12            #no. of slots/pole
nc = 4            #no. of conductors/slot

#Calculations
Zph = nc*s*P/ph
T = Zph/2
alpha = 180*(1-5./P)
Kc = math.cos(alpha/2*math.pi/180)
m = s/ph
B = 180/s
Kd = math.sin(m*alpha/2*math.pi/180)/(m*math.sin(alpha/2*math.pi/180))
Eph = 4.44*Kc*Kd*f*phi*T
El = math.sqrt(3)*Eph

#Results
print "Line e.m.f. = %.1f V"%El
#incorrect answer for Kd in the textbook. Hence the result differs

Line e.m.f. = 372.8 V


## Problem 13, Page 121¶

In [13]:
import math

#Variable declaration
P = 1200   #kW
V = 2300   #V
Ia = 200   #A
cos_phi_a = 0.9  #lagging

#Calculations
Pa = V*Ia*cos_phi_a/10**3  #kW
phi = math.atan(cos_phi_a)
Pra = Pa*math.tan(phi)
Pr = 0    #since power factor is unity
Prb = Pr-Pra
Pb = P-Pa
tan_phi = Prb/Pb
cos_phi_b = math.cos(math.atan(tan_phi))
Ia = Pb*10**3/(V*cos_phi_b)

#Results
print "Power = %d kW"%Pb
print "Power factor = %.3f"%cos_phi_b
print "Current = %.1f A"%Ia

Power = 786 kW
Power factor = 0.904
Current = 378.2 A


## Problem 14, Page 121¶

In [14]:
import math

#Part (a) - for lighting load
P1 = 600   #kW
cos_phi =1 #power factor
tan_phi = 0
Prl = P1*tan_phi

P2 = 800    #W
cos_phi = 0.9
tan_phi = 0.4843
Pri = P2*tan_phi

#Part(c) - for capacitive power
P3 = 800
cos_phi = 0.8
tan_phi = 0.75
Prc = -P3*tan_phi

P = P1+P2+P3
Pr = Prl+Pri+Prc
Pa = 1000    #kW
cos_phi_a = 0.85
tan_phi_a = math.tan(math.acos(cos_phi_a))
Pra = Pa*tan_phi_a
Pb = P-Pa
Prb = Pr-Pra
tan_phi_2 = Prb/Pb
cos_phi_b = math.cos(math.atan(tan_phi_2))

#Results
print "Active power supplied by alternator B = %d kW"%Pb
print "Power factor of alternator B = %.4f leading"%cos_phi_b

Active power supplied by alternator B = 1200 kW
Power factor of alternator B = 0.8217 leading


## Problem 15, Page 121¶

In [17]:
import math

#Variable declaration
E = 1000   #KVa
Vl = 11000  #V
ph = 3     #no. of phases
Ra = 3.5   #armature resistance(ohms)
Xs = 40     #armature reactance(ohms)
cos_phi = 0.8

#Calculations
Ia = round(E*1000/(math.sqrt(3)*Vl),1)
V = Vl/math.sqrt(3)
phi = math.degrees(math.acos(cos_phi))
Rad = round(Ia*Ra)     #armature resistance drop/phase
Xad = round(Ia*Xs)     #armature reactance drop per phase
theta = math.degrees(math.atan(Xs/Ra))

#Part a - Unity p.f.
Eba = math.sqrt(V**2+Er**2-(2*V*Er*math.cos(theta*math.pi/180)))
Vla = Eba*math.sqrt(3)
#From triangle OAB
alpha_a = math.degrees(math.asin((Er*math.sin(theta*math.pi/180))/Eba))

#Part b - At p.f. 0.8 lagging
BOA_b = theta-phi
Ebb = math.sqrt(V**2+Er**2-(2*V*Er*math.cos(BOA_b*math.pi/180)))
Vlb = Ebb*math.sqrt(3)
alpha_b = math.degrees(math.asin(Er*math.sin(BOA_b*math.pi/180)/Ebb))

#Part c - At p.f. 0.8 leading
BOA_c = theta+phi
Ebc = math.sqrt(V**2+Er**2-(2*V*Er*math.cos(BOA_c*math.pi/180)))
Vlc = Ebc*math.sqrt(3)
alpha_c = math.degrees(math.asin(Er*math.sin(BOA_c*math.pi/180)/Ebc))

#Results
print "Induced e.m.f. and angular retardation are as below:"
print "Part a : %d V,%.2f degrees"%(Vla,alpha_a)
print "Part b : %d V,%.2f degrees"%(Vlb,alpha_b)
print "Part c : %d V,%.2f degrees"%(Vlc,alpha_c)


Induced e.m.f. and angular retardation are as below:
Part a : 11284 V,18.80 degrees
Part b : 8984 V,17.62 degrees
Part c : 13294 V,13.49 degrees


## Problem 16, Page 121¶

In [18]:
import math

#Variable declaration
Em = 6000    #V
Xs = 1.5     #ohms/phase
Il = 1000    #amps
sin_phi = 1  #power factor

#Calculations
Vt = Em-Il*Xs
Vl = round(math.sqrt(3)*Vt)   #since winding is connected in star
Pr = math.sqrt(3)*Vl*Il*sin_phi/10**6

#Results
print "Reactive power supplied to the load is %.2f MVAR"%Pr

Reactive power supplied to the load is 13.50 MVAR


## Problem 17, Page 121¶

In [19]:
import math

#Variable declaration
V = 7000   #V
I = 1400   #amps
Xs = 1.2   #ohms
f = 50     #Hz
Po = 4      #number of poles
cos_phi = 1 #power factor, since load is resistive

#Calculations
E = math.sqrt(V**2+(I*Xs)**2)  #V
P = 3*V*I*cos_phi     #W
N = 120*f/Po          #rpm
T = P/(w*9.81)

#Result
print "The required torque is %.2e kg-meter"%T

The required torque is 1.91e+04 kg-meter


## Problem 18, Page 122¶

In [20]:
#Variable declaration
Ep = 5700      #V
Xs = 1.5       #ohms

#Calculations
#Since windings are connected in star
Ip = Ep/Xs   #phase current(A)
Il = Ip

#Result
print "The current per phase is %d Amps"%Il

The current per phase is 3800 Amps


## Problem 19, Page 122¶

In [21]:
import math

#Variable declaration
Vt = 15   #KV
Ps = 100*10**6   #power supplied by generator VA
cos_phi = 0.8  #power factor(lagging)
Xs = 0.7       #ohm per phase
Vl = 15        #KV
f = 50         #Hz
Po = 2          #no. of poles

#Calculations
Vp = Vl/math.sqrt(3)   #KV
AC = Vp*cos_phi
sin_phi = 0.6
BC = Vp*sin_phi        #KV
Il = Ps/(math.sqrt(3)*Vl*10**3)  #amps
Vd = Il*Xs             #voltage drop in synchronous reactance KV
BL = 2.697
LC = BL+BC             #V
AL = math.sqrt(AC**2+LC**2)  #V
Em = AL*math.sqrt(3)          #V
Ns = (120*f)/Po           #rpm
P = Ps*cos_phi          #W
wT = 80*10**6
T = (wT*60)/(2*math.pi*Ns)
print
#Result
print "Induced e.m.f. = %.2f V"%Em
print "Torque = %.2e Nw-m"%T
print "Speed = %d rpm"%Ns
#Answers differ due to rounding-off errors

Induced e.m.f. = 18.19 V
Torque = 2.55e+05 Nw-m
Speed = 3000 rpm