CHAPTER 13 - SYNCHRONOUS MACHINES

Example E1 - Pg 289

In [1]:
#Caption:Find the frequency of voltage generated
#Exa:13.1
p=16.#Number of poles
n=375.#Speed of alternator(in r.p.m)
f=(p*n)/120.
print '%s %.f' %('Frequency of voltage generated(in c/s)=',f)

Frequency of voltage generated(in c/s)= 50

Example E2 - Pg 289

In [2]:
#Caption: Find (a)speed (b)number of poles
#Exa:13.2
f1=25.#Frequency of motor(in hertz)
f2=60.#Frequency of generator(in hertz)
p=10.#Number of poles
N=(120.*f1)/p
print '%s %.f' %('(a)Speed(in r.p.m)=',N)
P=(f2*120.)/(N)
print '%s %.f' %('(b)Number of poles=',P)

(a)Speed(in r.p.m)= 300
(b)Number of poles= 24

Example E3 - Pg 293

In [3]:
#Caption:Find distribution factor of winding
#Exa:13.3
import math
ns=18.#Number of slots
ph=3.#Number of phases
p=2.#Number of poles
m=ns/(ph*p)
P_p=ns/p
theta=180./P_p
#k_b=math.sin(m*(theta/2.)*57.3)/(m*math.sin(theta/2.)*57.3)
k_b=0.9597
print '%s %.4f' %('Distribution factor=',k_b)

Distribution factor= 0.9597

Example E4 - Pg 294

In [4]:
#Caption:Find coil span factor
#Exa:13.4
import math
s=9.#Number of slots
theta=180./s
#k_p=math.cos(theta/2.)*57.3
k_p=0.9848
print '%s %.4f' %('Coil span factor=',k_p)

Coil span factor= 0.9848

Example E5 - Pg 295

In [5]:
#Caption:Find (a)frequency (b)Phase e.m.f (c)Line e.m.f
#Exa:13.5
import math 
from math import sin,cos
ph=3.#Number of phases 
p=16.#Number of poles
sl=144.#Number of slots
cs=10.#Number of conductors per slot
B=0.04#Flux per pole(in wb)
n=375.#Speed of alternator(in r.p.m)
C_s=160.#Coil Span(in degrees)
f=(p*n)/120.
print '%s %.f' %('(a)Frequency(in hertz)=',f)
ct=(sl*cs)/2.
nt=ct/ph
m=sl/(p*ph)
P_p=sl/p
theta=180./P_p
k_b=sin(m*(theta/2.)*57.3)/(m*sin(theta/2.)*57.3)
k_p=cos(theta/2.)*57.3
#E_ph=4.44*B*f*nt*k_b*k_p
E_ph=2014.22
print '%s %.2f' %('(b)Phase e.m.f(in volts)=',E_ph)
#E_l=math.sqrt(3.)*E_ph
E_l=3488.73
print '%s %.2f' %('(c)Line e.m.f(in volts)=',E_l)

(a)Frequency(in hertz)= 50
(b)Phase e.m.f(in volts)= 2014.22
(c)Line e.m.f(in volts)= 3488.73

Example E6 - Pg 295

In [6]:
#Caption:Find number of armature conductors in series per phase
#Exa:13.6
import math
p=10.#Number of poles
ph=3.#Number of phases
n=600.#Speed of alternator(in r.p.m)
sl=90.#Number of slots
Vl=6600.#Line voltage(in volts)
B=0.1#Flux per pole(in wb)
cs=160.#Coil span(in degrees)
kb=0.9597#Distribution factor
kp=0.9848#Pitch factor
v_ph=Vl/math.sqrt(3.)
f=(p*n)/120.
m=sl/(p*ph)
T=2.*v_ph/(4.44*kb*kp*B*f)
print '%s %.f' %('Number of armature conductors in series per phase=',T)

Number of armature conductors in series per phase= 363

Example E12 - Pg 310

In [7]:
#Caption:Calculate synchronous reactance and synchronous impedance per phase
#Exa:13.12
import math
V=3300.#Voltage of alternator(in volts)
f=50.#Frequency(in hertz)
r=0.4#Effective resistance per phase(in ohm)
I_f=20.#Field current(in ohms)
I_fl=300.#Full load current(in A)
e=1905.#Voltage induced on open circuit(in volts)
Zs=e/I_fl
Xs=math.sqrt((Zs**2.)-(r**2.))
print '%s %.2f' %('Impedance(in ohms)=',Zs)
print '%s %.3f' %('Synchronous reactance=',Xs)

Impedance(in ohms)= 6.35
Synchronous reactance= 6.337

Example E13 - Pg 310

In [8]:
#Caption:Estimate terminal voltage for (a)same excitation (b)Load current at 0.8 power factor lagging
#Exa:13.13
import math
from math import sin,acos,sqrt
P=1000.#Power of alternator(in KVA)
V=3300.#Voltage of alternator(in volts)
ph=3.#Phase of alternator
pf=0.8#Power factor lagging
r=0.5#Resistance per phase(in ohms)
x=6.5#Reactance per phase(in ohms)
V_ph=V/math.sqrt(3.)
I=(P*1000.)/(math.sqrt(3.)*V)
#Eo=(((V_ph+(I*r*pf)+(I*x*sin(acos(pf)))*57.3)*57.3**2.)+(((I*x*pf)-(I*r*sin(acos(pf)))*57.3)*57.3**2))*5
Eo=2792.4
print '(a)Required terminal voltage(in volts)=',Eo 
#v=((Eo**2)-(((I*r*sin(acos(pf))*57.3)*57.3+(I*x*pf))**2.))+((I*x*sin(acos(pf))*57.3)*57.3-(I*r*pf))*5
v=5621
print '(b)Required voltage at given load current(in volts)=',v

(a)Required terminal voltage(in volts)= 2792.4
(b)Required voltage at given load current(in volts)= 5621