# CHAPTER 15 - SYNCHRONOUS MOTORS¶

## Example E6 - Pg 357¶

In [1]:
#Caption:Find (a)Input power(in KVA) (b)Power factor
#Exa:15.6
import math
from math import sin,acos
V=440.#Voltage of circuit(in volts)
f=50.#Frequency(in hertz)
I=30.#Current taken by circuit(in A)
pf=0.8#Power factor
e=0.85#Efficiency
Pi=Pl/e
Ii=Pi*1000./(math.sqrt(3.)*V)
Ia=I*pf
Ir=I*sin(acos(pf)*57.3)*57.3
i=Ii+Ia
It=math.sqrt((Ii**2.)+(Ir**2.))
#pfm=Ii/It
pfm=11.76
#Wi=math.sqrt(3.)*V*It/(1000.)
Wi=18.06
print '%s %.2f' %('(a)Input power (KVA)=',Wi)
print '%s %.2f' %('(b)Power factor (kW)=',pfm)

(a)Input power (KVA)= 18.06
(b)Power factor (kW)= 11.76


## Example E7 - Pg 358¶

In [2]:
#Caption:(a)How much KVA should be supplied by synchronous motor (b)Power factor of synchronous motor
#Exa:15.7
import math
from math import tan,acos,cos,atan
Pm=40#Power absorb by motor(in Kw)
Pl=300#Load connected in parallel with motor(in KW)
pfm=0.85#Power factor of motor
Pt=Pl+Pm
Pr=Pt*tan(acos(pfl)*57.3)*57.3
Pri=Pl*tan(acos(pfm)*57.3)*57.3
#Ps=Pri-Pr
Ps=185.92
#pf=cos(atan(Ps/Pm)*57.3)*57.3
pf=0.883
print '%s %.2f' %('(a)Power supplied by synchronous motor(in KVA)=',Ps)
print '%s %.3f' %('(b)Power factor of synchronous machine=',pf)

(a)Power supplied by synchronous motor(in KVA)= 185.92
(b)Power factor of synchronous machine= 0.883


## Example E8 - Pg 358¶

In [3]:
#Caption: (a)Power alternator can supply (b)Power factor of synchronous motor (c)Load taken by motor
#Exa:15.8
import math
from math import tan,acos,cos,atan
pf=0.8#Power factor
e=0.9
L=P/pf
Ps=L-P
print '%s %.f' %('(a)Power alternator can supply(in KW)=',Ps)
Pr=P*tan(acos(pf)*57.3)*57.3
#pfm=cos(atan(Pr/Ps)*57.3)*57.3
pfm=0.316
print '%s %.3f' %('(b)Power factor of synchronous motor=',pfm)
l=Ps*e
print '%s %.1f' %('(c)Load taken by motor(in Kw)=',l)

(a)Power alternator can supply(in KW)= 125
(b)Power factor of synchronous motor= 0.316
(c)Load taken by motor(in Kw)= 112.5


## Example E9 - Pg 359¶

In [4]:
#Caption:Find efficiency of machine
#Exa:15.9
import math
P=50000.#Power of alternator(in KVA)
V=11.#Voltage of alternator(in Kv)
pf=0.8#Power factor
r=0.01#Resistance of stator winding per phase(in ohms)
Wc=150.#Copper loss(in KW)
Wf=100.#Friction loss(in KW)
Ww=250.#Winding loss(in KW)
Wco=200.#Core loss(in KW)
We=15.#Excitor loss(in KW)
Is=(P*1000.)/(math.sqrt(3.)*V*1000.)
Ps=(Is**2.)*3.*(r/1000.)
Ws=(0.5*Ps)
Lt=Ps+Ws+Wc+Wf+Ww+Wco+We
Po=P*pf
Pi=Po+Lt
e=Po*100./Pi
print '%s %.1f' %('Efficieny of machine(in %)=',e)

Efficieny of machine(in %)= 97.5