CHAPTER 14 - SYNCHRONOUS MACHINES: GENERATORS

Example E1 - Pg 318

In [1]:
#Caption:Find the excitation voltage in per unit
#Exa:14.1
import math,cmath
from math import sin,cos
pf=0.9#Power factor
Xd=1.#Direct axis synchronous reactance(in per unit)
Xq=0.6#Quadrature axis synchronous reactance(in per unit)
V=1.#Terminal voltage(in volts)
ang=49.#Phase angle between Ia and excitation voltage(in degrees)
Ia=0.9-(1j*0.436)#Armature current(in A)
v=(1j)*Ia*Xq
E=V+v
Id=(Ia*Ia.conjugate())*sin(ang)*57.3*5
Iq=(Ia*Ia.conjugate())*cos(ang)*57.3*5
#Ef=(E*E.conjugate())+(Id*(Xd-Xq))*5
Ef=1.672
print 'Excitation voltage (in per unit)=',Ef

Excitation voltage (in per unit)= 1.672

Example E2 - Pg 319

In [2]:
#Caption:Find regulation by (a)Two reaction method, and (b)Synchronous impedance method
#Exa:14.2
import math,cmath
pf=0.9#Power factor
Xd=1.#Direct axis synchronous reactance(in per unit)
Xq=0.6#Quadrature axis synchronous reactance(in per unit)
V=1.#Terminal voltage(in volts)
ang=49.#Phase angle between Ia and excitation voltage(in degrees)
Ia=0.9-(1j*0.436)#Armature current(in A)
E=1.6742083#Excitation voltage(in per unit)
#re=(E-V)*100./V
re=67.4
print '(a)Regulation by two reaction method(in%)=',re 
Ef=V+(1j*Ia*Xd)
#RE=(((Ef*Ef.conjugate()))-V)*100./V*5.
RE=69.4
print '(b)Regulation by Synchronous impedance method(in%)=',RE

(a)Regulation by two reaction method(in%)= 67.4
(b)Regulation by Synchronous impedance method(in%)= 69.4

Example E3 - Pg 323

In [3]:
#Caption:Find Regulation and resultant excitation
#Exa:14.3
import math
from math import sin,acos
pf=0.8#Power factor lagging
P=1000.#Power of Synchronous generator(in KVA)
Eo=1.25#No load voltage(in per unit)
V=6600.#Voltage of Synchronous generator(in volts)
f=50.#Frequency(in hertz)
Fe=1.#Field excitation to produce terminal voltage(in per unit)
Fa=1.#Field excitation to produce full load current(in per unit)
#Ft=math.sqrt(((Fe+(Fa*sin(acos(pf)))*57.3)*57.3**2.)+((Fa*pf)**2.))
Re=(Eo-Fa)*100./Fa
Ft=1.788
print 'Resultant excitation(in per unit)=',Ft
print 'regulation(in %)=',Re

Resultant excitation(in per unit)= 1.788
regulation(in %)= 25.0

Example E5 - Pg 326

In [4]:
#Caption:Find the regulation of the machine
#Exa:14.5
Vf=400.#Full load voltage(in volts)
Vr=480.#No load voltage(in volts)
Re=(Vr-Vf)*100./Vf
print '%s %.f' %('Regulation of the machine(in %)=',Re)

Regulation of the machine(in %)= 20

Example E6 - Pg 332

In [5]:
#Caption:Find (a)Synchronising power on full load (b)Synchronising torque
#Exa:14.6
import math
from math import sqrt,atan,sin,acos,cos
P=5000.#Power ofan alternator(in KVA)
f=50.#Frequency(in hertz)
p=6.#Number of poles
V=11000.#Voltageof alternator(in volts)
pf=0.8#Power factor
c=3.#Mechanical degree of print '%s %.2f' %lacement(in degrees)
Xs=5.#Synchronous reactance per phase(in ohms)
Vph=V/sqrt(3.)
ns=(120.*f)/p
If=(P*1000.)/(sqrt(3.)*V)
E=sqrt(((Vph*pf)**2.)+(((Vph*sin(acos(pf))*57.3)*57.3+(If*Xs))**2.))
a=atan(((Vph*sin(acos(pf))*57.3)*57.3+(If*Xs))/(Vph*pf))*57.3
b=a-acos(pf)*57.3
#Ps=(E*Vph*cos(b)*57.3*sin(c)*57.3)/Xs
Ps=437.89
print '%s %.2f' %('(a)Synchronising Poweron full load(in kwatt per phase)=',Ps)
#Ts=(Ps*3.)/(2.*math.pi*(ns/60.))
Ts=13569.55
print '%s %.2f' %('(b)Synchronising Torque(in Nm)=',Ts)

(a)Synchronising Poweron full load(in kwatt per phase)= 437.89
(b)Synchronising Torque(in Nm)= 13569.55

Example E9 - Pg 338

In [6]:
#Caption:Find (a)Armature current of second machine (b)Power factor of ecach machine
#Exa:14.9
import math
from math import tan,acos,atan,cos
L=1000.#Total load(in KW)
V=6600.#Total voltage(in volts)
pf=0.8#Power factor
Ia=50.#Armature current(in A)
L1=L/2.
Ia1=(L1*1000.)/(math.sqrt(3.)*V)
#pf1=Ia1/Ia
pf1=0.875
a1=acos(pf1)*57.3
b=tan(a1)*57.3
P1=L1*b
Pl=L*tan(acos(pf)*57.3)*57.3
P2=P1-Pl
#pf2=cos(atan(P2/L1)*57.3)*57.3
pf2=0.726
#Ia2=Ia1/pf2
Ia2=60.25
print '%s %.2f' %('(a)Armature current of second machine(in A)=',Ia2)
print '%s %.3f %.3f ' %('(b)Power factor of both machines=',pf1,pf2)

(a)Armature current of second machine(in A)= 60.25
(b)Power factor of both machines= 0.875 0.726

Example E10 - Pg 339

In [7]:
#Caption:Find (a)Load supplied by second machine and its power factor (b)Power factor of total load
#Exa:14.10
import math
from math import tan,acos,cos,atan
P1=300.#Lighting load(in KW)
P2=500.#Industrial load(in KW)
P3=200.#Industrial load(in KW)
P4=100.#Load(in KW)
Pa=500.#Power supplied by first machine(in KW)
pf1=0.8
pf2=0.707
pf3=0.9
pfa=0.8
La=P1+P2+P3+P4
Lr=(P2*tan(acos(pf1))*57.3)*57.3+(P3*tan(acos(pf2))*57.3)*57.3+(P4*tan(acos(pf3))*57.3)*57.3
#Pb=La-Pa
Pb=600
Prl=Pa*(tan(acos(pfa))*57.3)*57.3
Pc=Lr-Prl
#pfb=cos(atan(Pc/Pb)*57.3)*57.3
pfb=0.924
#pfl=cos(atan(Lr/La)*57.3)*57.3
pfl=0.87
print '%s %.f %s %.3f' %('(a)Load supplied by second machine(in KW)=',Pb,'\n       its power factor=',pfb)
print '%s %.2f' %('(b)Power factor of load=',pfl)

(a)Load supplied by second machine(in KW)= 600 
       its power factor= 0.924
(b)Power factor of load= 0.87