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DC GENERATOR Vita Lystianingrum1

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REFERENCES

1. Theraja, B. L., Electrical Technology, S. Chand & Company Ltd., 1978.

2. Chapman, S. J, Electric Machinery Fundamentals, McGraw-Hill, 2005.

2

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RE1334 KONVERSI TENAGA LISTRIK I

1: Energi dan permasalahannya2: Konsep konversi energi3: Pembangkitan energi listrik konvensional & nonkonvensional4: Dasar elektromagnetik, pengenalan bahan magnetik

dan elektromekanik 5: Mesin arus searah (DC)6: Generator DC: cara kerja, klasifikasi dan persamaantegangan7: Generator DC: rugi-rugi daya dan efisiensi,efisiensi maksimum serta karakteristik 8:

9-10: Ujian Tengah Semester 3

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DC GENERATOR Emf inductionArmature reactionLoaded generator

Equivalent circuit: separately-excited; shunt; series;compoundVoltage, power, current;

Voltage control;

Energy, losses, efficiency;

Characteristics;4

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TOTAL LOSS IN A GENERATOR [1]

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Total Loss in a DC machine (1)

Total Losses

Copper Losses

Iron/Magnetic

Losses

Mechanical

Losses

Armature CuLoss

Field Cu Loss

hysteresis

Eddy current

friction

windage

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IRON LOSS IN THE ARMATURE (1)Hysteresis Loss

This loss is due to the reversal of magnetism of the armaturecore.(Every portion of the rotating core passes under N and S polealternately, thereby attaining S and N polarity respectively.

Review: Hysteresis Loss

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IRON LOSS IN THE ARMATURE (2)Eddy Current Loss

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Eddy Current Loss

.The core material is electrically conductive.

.When it rotates, it also cuts the flux.

.Hence, an emf is induced in the body of the core.

.This emf sets up current in the body of the core. .

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Eddy Current Loss

.The core material is electrically conductive.

.When it rotates, it also cuts the flux.

.Hence, an emf is induced in the body of the core.

.This emf sets up current in the body of the core. .

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Total Loss in a DC machine (1)

Total Losses

Copper Losses

Iron/Magnetic

Losses

Mechanical

Losses

Armature CuLoss

Field Cu Loss

hysteresis

Eddy current

friction

windage

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Copper Loss

Copper losses (Cu losses).Armature copper loss = I a2R a

.Field copper loss = I f 2R f

.Loss due to brush contact resistance; usually included in thearmature copper loss.

VE

Ra

Rsh

VVERaRsh

Rse

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Mechanical Loss

.Friction loss at bearings and commutator

.Air-friction or windage loss of rotating armature

Commutator

Poles

Rotor winding

Fan

Bearing

Brushes

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Total Loss in a DC machine

Total Losses

Copper Losses

Iron/MagneticLosses

MechanicalLosses

Armature CuLoss

Field Cu Loss

hysteresis

Eddy current

friction

windage

Stray Losses: usually magnetic and mechanical losses are collectivelyknown as Stray Losses.

Constant Losses: W c, consists of stray losses and shunt Cu losses.

Variable Loss: varied with the load current -> armature Cu loss.

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Generator Power Stages

MechanicalPower Input Iron and

FrictionLosses

Electrical Power developed inarmature.

EIa, Watt

Copper Losses Electrical Power Output

VI, Watts

Total Losses = Constant Losses + Variable Losses

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INTERNAL GENERATED VOLTAGE AND

INDUCED TORQUE EQUATIONSThe induced voltage in any given machine depends on 3factors:

The flux in the machineThe speed n or of the machines rotor A constant depending on the construction of the machine

E = cn

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.If T a is the torque exerted by the prime mover running at n rps, thenthe power developed is:

watt2 nT a

.and the mechanical power converted into electrical power in thearmature is:

watta I E

aa I E nT = 2then

Torque Power Conversion

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Review: HP (horse power) & Watt1 HP = 746 Watt.

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Efficiency

Mechanical Efficiency = B / A

Electrical Efficiency = C / B

Overall/Commercial Efficiency = C / A

MechanicalPower Input Iron and

FrictionLosses

Electrical Power developed inarmature.

EIa, Watt

Copper Losses Electrical Power Output

VI, Watts

A B C

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CONDITION FOR MAXIMUM

EFFICIENCYGenerator output = VI

Gen input = gen output + losses

= VI + Ia 2Ra + Wc (I = Ia + Ish)

However if Ish is negligible as compared to load current, then Ia = I

(approx.) = output / input =

Efficiency is max when denominator is minimum, when:

20

( )VI WcV IRa ++11

0=

+

VI W

V IR

dI d ca 02 = VI

W V

R caca W R I =

2

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CONDITION FOR MAXIMUM

EFFICIENCY (2)

21

ca W R I =2

Variable loss = constant loss.

The load current corresponding to max efficiency is:

a

c

RW

I =

i

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EXAMPLES21-18

21-2021-19

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