# Ch-12, Parallel Operation of alternators¶

## example 12.1 Page 243¶

In [1]:
p=4000  #given kva of alternator
fnl2=50 #frequency on no load on second alternator
fl2=48  #frequency on load on second alternator
l=6000 #load given two to alternator
df1=fnl1-fl1  #change in 1 alternator frequency
df2=fnl2-fl2  #change in 2 alternator frequency
print 'a'
l2=l-l1
print " load on 1 alternator %.2fkW \n load on 2 alternator %.2fkW"%(l1,l2)
ll=ml1+p
print 'b'
print " load supplied by machine 1 with full load on machine2 %dkW \n total load is %dkW"%(ml1,l1)

a
b


## example 12.2 page 243¶

In [2]:
from math import sqrt, atan, acos, pi, sin
pf1=0.8  #pf on 1 machine
i2=150  #current on  2 machine
z1=0.4+12*1J  #synchronour impedence
z2=0.5+10*1J
vt=6.6  #terminal voltage
al=l1/2  #active load on each machine
cosdb=al/(vt*i2*sqrt(3)) #cos db
db=acos(cosdb)*180/pi  #angle in digree
ib=i2*complex(cosdb,-sin(db*pi/180))  #current in complex number
it=l1/(vt*pf1*sqrt(3))  #total current
itc=complex(it*pf1,-it*sin(acos(pf1)))  #total current in complex
ia=itc-ib
pfa=atan(ia.imag/ia.real)  #pf of current a
ea=(vt/sqrt(3))+ia*(z1)/1000  #voltage a
pha=atan(ea.imag/ea.real)*180/pi  #phase angle of unit a
print "induced emf of a machine a %.2f+%.2fi =%fkV per phase"%(ea.real,ea.imag,abs(ea))
eb=(vt/sqrt(3))+ib*(z2)/1000  #voltage b
phb=atan(eb.imag/eb.real)*180/pi  #phase angle of unit b
print "\ninduced emf of a machine b %.2f+%.2fi =%fkV per phase"%(eb.real,eb.imag,abs(eb))

induced emf of a machine a 5.35+1.52i =5.565708kV per phase

induced emf of a machine b 4.60+1.28i =4.776461kV per phase


## example 12.3 Page 244¶

In [3]:
from math import cos,pi
e1=3000 ;ph1=20 ;e2=2900; ph2=0 #given induced emf of two machines
z1=2+20*1J ;z2=2.5+30*1J #impedence of two synchronous machine
e11=e1*(cos(ph1*pi/180)+sin(ph1*pi/180)*1J)
e22=e2*(cos(ph2*pi/180)+sin(ph2*pi/180)*1J)
Is=(e11-e22)*zl/(z1*z2+(z1+z2)*zl)
print "current is %.2f%.2fiA =%.2fA"%((Is).real,(Is).imag,abs(Is))

current is 10.37-4.56iA =11.33A


## example 12.4 Page 244¶

In [4]:
from math import sqrt, atan, acos, pi, sin
e1=250 ;e2=250 #emf of generator
z1=2*1J; z2=2*1J #synchronous impedence
v=(e1*z2+z1*e2)/((z1*z2/z)+z1+z2)
vph=atan(v.imag/v.real)*180/pi #substitution the value in equation 12.10
i1=(z2*e1+(e1-e2)*z)/(z1*z2+(z1+z2)*z); iph=atan((i1).imag/(i1).real)*180/pi #substitution the value in equation 12.7
pf1=cos(pi/180*(vph-iph))
pd=v*i1*pf1
e1=250 ;e2=250 #emf of generator
z1=2*1J; z2=2*1J #synchronous impedence
v=(e1*z2+z1*e2)/((z1*z2/z)+z1+z2)
vph=atan(v.imag/v.real)*180/pi #substitution the value in equation 12.10
i1=(z2*e1+(e1-e2)*z)/(z1*z2+(z1+z2)*z)
iph=atan(i1.imag/i1.real)*180/pi #substitution the value in equation 12.7
pf1=cos(pi/180*(vph-iph))
pd=v*i1*pf1
print "terminal voltage %.2fV \ncurrent supplied by each %.2fA \npower factor of each %.3f lagging \npower delivered by each %.4fKW"%(abs(v),abs(i1),abs(pf1),abs(pd))

terminal voltage 239.68V
current supplied by each 10.72A
power factor of each 0.894 lagging
power delivered by each 2297.7941KW


## example 12.5 Page 247¶

In [5]:
from math import sqrt, atan, acos, pi, sin
po=5 #mva rating
v=10 #voltage in kv
n=1500 ;ns=n/60 #speed
f=50 #freaquency
pfb=0.8#power factor in b
x=0.2*1J #reactance of machine
md=0.5 #machanical displacement
v=1 ;e=1
p=4
spu=v*e/abs(x); sp=spu*po*1000 ;mt=(pi*p)/(180*2)
spm=sp*mt #synchronous power in per mech.deree
st=spm*md*1000/(2*ns*pi)
print '(a)'
print " synchronous power %dkW \n synchronous torque for %.1f displacement %dN-M"%(spm,md,st)
ee=e+x*(pfb-sin(acos(pfb))*1J)
spb=v*abs(ee)*cos(atan(ee.imag/ee.real))/abs(x)  #synchronous power
sppm=spb*po*1000*mt #synchronous power per mech.degree
print " synchronous power %dkW \n synchronous torque for %.1f displacement %dN-M"%(sppm,md,stp)

(a)
synchronous power 872kW
synchronous torque for 0.5 displacement 2777N-M
synchronous power 977kW
synchronous torque for 0.5 displacement 3111N-M


## example 12.6 page 248¶

In [6]:
from math import sqrt, atan, acos, pi, sin
po=2*10**6 ;p=8 ;n=750; v=6000 ;x=6*1J ;pf=0.8 #given
i=po/(v*sqrt(3))
e=(v/sqrt(3))+i*x*(pf-sin(acos(pf))*1J)
mt=p*pi/(2*180)
cs=cos(atan(e.imag/e.real))
ps=abs(e)*v*sqrt(3)*cs*mt/(1000*abs(x))
ns=n/60
ts=ps*1000/(2*pi*ns)
print " synchronous power %.1fkW per mech.degree \n synchrounous torque %dN-m"%(ps,ts)

 synchronous power 502.7kW per mech.degree
synchrounous torque 6666N-m


## example 12.7 page 248¶

In [7]:
from math import sqrt, atan, acos, pi, sin
i=100 ;pf=-0.8 ;v=11*1000 ;x=4*1J ;ds=10; pfc=-0.8 #given,currents,power factor,voltage,reactance,delta w.r.t steem supply,pf of alternator
e=(v/sqrt(3))+(i*x*(pf-sin(acos(pf))*1J))
print 'a'
ph=atan(e.imag/e.real)*180/pi
print " open circuit emf %dvolts per phase and %.2f degree"%(abs(e),ph)
d=ds-ph
eee=round(abs(e)/100)*100
ic=round(abs(eee)*sin(d*pi/180)/abs(x))
iis=(eee**2-(abs(x)*ic)**2)**(0.5)
Is=(iis-v/sqrt(3))/abs(x)
ii=ic/cos(d*pi/180)
pff=cos(d*pi/180)
print 'b.'
print " current %.1fA \n power factor %.3f"%(ii,pff)
print 'c.'
ia=ii*pff/abs(pfc)
print "current %.2fA"%(ia)

ia=ii*pff/abs(pfc)
print "current %.2fA"%(ia)

a
open circuit emf 6598volts per phase and -2.78 degree
b.
current 365.6A
power factor 0.998
c.
current 456.25A
current 456.25A