Dc Generator 10

8/13/2019 Dc Generator 10

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DC GENERATOR

ELECTRICAL MACHINES FEB24083W. MUSTAFA W. SULONG


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GENERATOR PRINCIPLE

An electrical generator is a machine which converts mechanical energy (

power ) into electrical energy ( power )

The energy conversion is based on the principle of the production of

dynamically Induced e.m.f.

Whenever a conductor cuts magnetic flux, e.m.f is produced to Faradayslaws of Electromagnetic Induction. This e.m.f causes a current to flow if

the conductor circuit is closed.

Two basic essential parts of an electrical generator are :-

( a ) a magnetic field and( b ) a conductor or conductor which can so move as to cut the flux.


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DC GENERATOR

OPERATION


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DC GENERATOR OPERATION

We noted earlier that Faradays law dictates that if a coil of Nturns

experiences a change in magnetic flux, then the induced voltage V

is given by

If a coil of areaA rotates with respect to a field B, and if at a

particular time it is at an angle to the field, then the flux linking the

coil is BAcos, and the rate of change of flux is given by

t

NV

d

d

coscos

d

d

d

sind

ttBA

dt

d


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Thus for the arrangement shown below

cos

d

sind

d

dNBA

tNBA

t

NV


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Therefore this arrangement produces a sinusoidal output as shown below


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Wires connected to the rotating coil would get twisted Therefore we use

circular slip ringswith sliding contacts called brushes


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The alternating signal from the earlier AC generator could be

converted to DC using a rectifier.

A more efficient approach is to replace the two slip rings with a single

split slip ring called a commutator, this is arranged so that

connections to the coil are reversed as the voltage from the coilchanges polarity hence the voltage across the brushes is of a single

polarity adding additional coils produces a more constant output


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Use of a commutator


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DC GENERATOR

EQUIVALENT CIRCUIT


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The magnetic field produced by the stator poles induces a voltage in

the rotor (or armature) coils when the generator is rotated.

This induced voltage is represented by a voltage source.

The stator coil has resistance, which is connected in series.

The pole flux is produced by the DC excitation/field current, which is

magnetically coupled to the rotor The field circuit has resistance and a source

The voltage drop on the brushes represented by a battery

DC GENERATOR EQUIVALENT CIRCUIT


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sV

esR

fR

fI

)(mechin PP

a

IaR LI

gE LV LR

Separately Excited dc generator Equivalent circui t

Where :

Vs = Field supply voltage, V

If = Field current, A

Rse = Field control winding resistance,

Rf = Field winding resistance,

Ra = Armature winding resistance, Eg = Generated voltage, V

VL = Load ( Terminal ) voltage, V

Ia = Armature current, A

IL = Load current, A

RL = Load winding resistance,


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To calculate field current , If;

If= ( Vs)/ ( Rse+ Rf)

To calculate armature current, Ia;

Ia= IL= PL/ VL * Ia= IL( series component )

To calculate generated voltage, Eg;

Eg= IaRa+ VL+ Vbrsh * Vbrush = brush voltage drop

( if necessary )


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)(mechin PP

aIaR L

I

gE LV LR

Series dc generator Equivalent circui t

Where :

Ra = Armature winding resistance,

Eg = Generated voltage, V

VL = Load ( Terminal ) voltage, V

Ia = Armature current, A




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To calculate armature current, Ia;

Ia= IL= PL/ VL * Ia= IL( series component )


Eg= IaRa+ VL+ Vbrsh * Vbrush = brush voltage drop( if necessary )


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)(mechin PP

aI aR LI

gE LV LR

fI

fR

Shunt dc generator Equivalent circui t

Where :

If = Field current, A

Rf = Field winding resistance,

Ra = Armature winding resistance,

Eg = Generated voltage, V

VL = Load ( Terminal ) voltage, VIa = Armature current, A




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To calculate load current, IL;

IL= PL/ VL

To calculate field current, If;

If= VL/ Rf

To calculate armature current, Ia ;

Ia= IL+ If


Eg= IaRa+ VL+ Vbrsh * Vbrush = brush voltage drop

( if necessary )


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E.m.f ( Eg )

EQUATION


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Let :

= Flux / pole in weber

Z = Total number of armature conductor

= No. of slot x No. of conductors / slot

P = No. of generator poles

A = No. of parallel path in armature

N = Armature rotation in revolutions per min. ( r.p.m )Eg = E.m.f induced in any parallel path in ramature

For a simplex wave-wound generator

No. of parallel paths = 2

No. of conductor ( in series ) in one path = Z / 2

Then, E.m.f ( Eg ) generated = ( PN / 60 ) x Z / 2 = ZPN / 120 volt.

E.m.f ( Eg ) EQUATION


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E.m.f ( Eg ) EQUATION

For a simple lap-wound generator

No. of parallel path = P

No. of conductor ( in series ) in one path = Z / P

Then, E.m.f ( Eg ) generated = ( PN / 60 ) x Z / P = ZN / 60 volt

For a given d.c. generator , Z, P and A are constant.Hence, putting Ka = ZP / A,

We get ;

Eg = Ka N / 60


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Example 1

A 4 pole, lap wound dc generator has a useful flux of 0.07 Wb. Calculate the

generated e.m.f when it is rotated at a speed 900 rpm, with the help of primemover. Armature consists of 440 turns of conductor. Also calculate the generated

e.m.f if lap wound armature is replaced by wave wound armature.

VxxxZPNE

AwoundWave

VxxNZE

PAwoundLap

924120

44049007.0120

2:

46260

4409007.060

4:


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LOSSES AND EFFICIENCY

IN DC GENERATOR


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Total Loss in a DC generatorThe various losses occurring in a generator cn be sub-divided

as follows :

( a ) Copper Losses

1. Armature copper loss = Ia2Ra.

Where Ra = resistance of armature and interpoles and series

field winding etc.

This loss is about 30 to 40 % of full-load losses

2. Field copper loss. In the case of shunt generator,

it is practically constant and Ish

2Rsh

or VIsh

.

In the case of series generator, it is = Ise2Rse where

Rseis resistance of the series field winding.

This loss is about 20 to 30 % of F.L losses.

LOSSES IN DC GENERATOR


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( b ) Magnetic losses ( also known as iron or core loss )

1. Hysterisis loss

2. Eddy current loss

These losses are practically constant for shunt and compound-wound

generators, because in their case, field current is approximately constant.


( c ) Mechanical Losses

These consists of :-

1. friction loss at bearing and commutator

2. Air-friction or windage loss of rotating armature

The total losses in a D.C generator are summarized below:


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Stray LossesUsually, magnetic and mechanical losses are collectively known as

Sytray Losses. These are also known as rotational losses

for abvious reasons.

Constant or Standing Losses

Field copper loss is constant for shunt and compound generator.Hence, stray losses and shunt copper loss are constant in their case.

These losses are together known as Standing or constant losses, Wc

Hence, for shunt and compound generators,

Total loss = armature copper loss + Wc

= Ia2Ra+ Wc

= ( I + Ish)2Ra+ Wc

Armature Copper loss Ia2Ra is known as variable loss because it varies

with the load current.

Total loss = variable loss + constant losses ( Wc)



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Power stages

Various power stages in the case of a d.c generator are shown below :

Following are the three generator efficiency :

Efficiency

= Po / Pi = VLIL/ ( Po + Losses ) = ( VLIL ) /( VLIL+ Cu + Mech )


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EXERCISES

1. Calculate the flux per pole required on full load for a 50 kW, 400 V, 8 pole, 600 r/min, dc shunt generator with

256 conductors arranged in lap-connected winding. The armature winding resistance is 0.1 , the shunt resistance

is 200

and there is a brush contact drop of 1 V at each brush on full load. ( 0.162 wb )

2. A shunt generator has an induced voltage on open circuit of 127 V. When the machine is on load the

terminal voltage Is 120 V. Find the load current if the field resistance be 15 and

the armature resistance 0.02 . ( 342 A )

3. A 4 pole, shunt generator with lap-connected armature having field and armature resistances of 50 and 0.1

Respectively supplies 60, 100 V, 40 W lamps. Calculate the total armature current and generated voltage.

( 26 A, 102.6 V )


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4. The following information is given for a 300 kW, 600 V shunt compound generator : shunt field resistance = 75 ,

armature resistance = 0.05 . When the machine is delivering full load, calculate the voltage ( Eg ) and power

generated by armature ( Pg ) .

(625.4 V, 317.7 kW )

5. A shunt generator delivers 195 A at terminal of 250 V. The armatures resistance and shunt field resistance are 0.02

and 50 . The iron loss and friction losses are 950 W. Find :-

( a ) E.m.f ( Eg ) generated

( b ) Copper losses

( c ) Output of the prime mover

( d ) efficiency

( 254 V, 2050 W, 51750 W, )


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6. A d.c shunt generator has a full load output of 10 kW at a terminal voltage of 240 V. The armature and the

shunt field winding resistance are 0.6 and 160 . The sum of mechanical and core losses is 500 .

Calculate the power required in kW at the driving shaft at full load and the corresponding efficiency.

( 11.978 kW, 83.5% )

7. A 4 poles lap connected armature of a d.c shunt generator is required to supply the loads connected is

parallel :

( i ) 5 kW heater at 250 V and

( ii ) 2.5 kW lighting load also at 250 V

The generator has an armature resistance of 0.2 and a shunt resistance of 250 .

The armature has 120 conductors in the slots and runs at 1000 rpm. Allowing 1 V

per brush for contact drop. Find :-

( a ) Flux per pole

( b ) Armature current

( 31 A, 129.1 mWb )

8. A 4 pole long shunt, lap winding generator supplies 25 kW at a terminal voltage 500 V. The armature

resistance is 0.03 , series field resistance is 0.04 and shunt field resistance is 200 . The brush drop

may be taken as 1 V. Determine the e.m.f ( Eg ) generated. Calculate also the no. of conductors if the speed

is 1200 rpm and flux per pole is 0.02 Wb. ( 505.67 V, 1264 )

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Dc Generator 10