# Chapter 28: Generator Characteristics¶

## Example Number 28.13, Page Number:984¶

In :
#variable declaration
v=220#V
#emf increases by 1 V for every increase of 6 A
ra=0.02#ohm
i=96#A

#calculations
voltageincrease=i/6
vd=i*ra
voltage_rise=voltageincrease-vd
vconsumer=v+voltage_rise
power_supplied=voltage_rise*i

#result
print "voltage supplied ot consumer= ",vconsumer," V"
print "power supplied by the booster itself= ",power_supplied/1000," kW"

voltage supplied ot consumer=  234.08  V
power supplied by the booster itself=  1.35168  kW


## Example Number 28.14, Page Number:985¶

In :
#variable declaration
v=50.0#V
i=200.0#A
r=0.3#ohm
i1=200.0#A
i2=50.0#A

#calculations
vd=i*r
voltage_decrease=v-vd
feeder_drop=v*r
booster_voltage=v*v/i1
voltage_net=feeder_drop-booster_voltage

#result
print "Net decrease in voltage= ",voltage_net," V"

Net decrease in voltage=  2.5  V


## Example Number 28.15, Page Number:986¶

In :
#variable declaration
inl=5.0#A
v=440.0#V
il=6.0#A
turns=1600

#calcuations
shunt_turns1=turns*inl
shunt_turns2=turns*il
increase=shunt_turns2-shunt_turns1
n=increase/i_full#number of series turns required

#result
print "Number of series turns required= ",n," tunrs/pole"

Number of series turns required=  8.0  tunrs/pole


## Example Number 28.16, Page Number:987¶

In :
#variable declaration
n=1000#turns/pole
series_winding=4#turns/pole
r=0.05#ohm
increase_i=0.2#A
ia=80#A

#calculations
R=(current_required*r)/(ia-current_required)

#result
print "Divertor resistance= ",R," ohm"

Divertor resistance=  0.0833333333333  ohm


## Example Number 28.17, Page Number:987¶

In :
#variable declaration
v=220.0#V
i=100.0#A
ra=0.1#ohm
rsh=50.0#ohm
rse=0.06#ohm
divertor=0.14#ohm

#calculations
#short shunt
vd=i*rse
ish=v/rsh
ia=i+ish
armature_drop=ia*ra
E=v+vd+armature_drop
#long shunt
vd=ia*(ra+rse)
print vd
E2=v+vd
current_divertor=(ia*divertor)/(divertor+rse)
change=(current_divertor/ia)*100

#result
print "a)emf induced using short shunt= ",E
print "b)emf induced using long shunt= ",E2
print "c)series amp-turns are reduced to ",change," %"

16.704
a)emf induced using short shunt=  236.44
b)emf induced using long shunt=  236.704
c)series amp-turns are reduced to  70.0  %


## Example Number 28.18, Page Number:988¶

In :
#variable declaration
p=250*1000#W
v=240#V
v2=220#V
i=7#A
inl=12#A
shunt=650#turns/pole
series=4#turns/pole
rse=0.006#ohm

#calculations
shunt_increase=shunt*(inl-i)
ise=shunt_increase/series
Rd=(ise*rse)/i_d

#results
print "resistance of the series amp-turns at no-load",Rd,"ohm"

resistance of the series amp-turns at no-load 0.0212751091703 ohm


## Example Number 28.19, Page Number:988¶

In :
p=60.0*1000#W
n=1600.0#turns/pole
inl=1.25#A
vnl=125#V
il=1.75#A
vl=150.0#V

#calculations
extra_excitation=n*(il-inl)
ise=p/vl
series=extra_excitation/ise
ise2=extra_excitation/3
i_d=ise-ise2
rd=(ise2*0.02)/i_d
reg=(vnl-vl)*100/vl

#result
print "i)minimum number of series turns/pole= ",series
print "ii)divertor resistance= ",rd
print "iii)voltage regulation= ",reg,"%"

i)minimum number of series turns/pole=  2.0
ii)divertor resistance=  0.04
iii)voltage regulation=  -16.6666666667 %


## Example Number 28.20, Page Number:989¶

In :
#variable declaration
v=50.0#v
i=200.0#A
r=0.3#ohm
i1=160.0#A
i2=50.0#A

#calculations
#160 A
vd=i1*(r-(v/i))
#50 A
vd2=i2*(r-(v/i))

#result
print "voltage drop at 160 A=",vd,"V"
print "voltage drop at 50 A=",vd2,"V"

voltage drop at 160 A= 8.0 V
voltage drop at 50 A= 2.5 V

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