Direct Current Motors[1]

8/2/2019 Direct Current Motors[1]

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Direct Current Motors


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Introduction

Most of the power is generated, transmittedand distributed in AC form.

A dc motor is a machine that converts dcpower into mechanical power.

The horsepower is used to represent the unitof a dc motor.


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The applications of dc motor are aeroplanes,

computers, printers, robots, toys, video andDVD players, mining industry, printingpresses, etc.

Its operation is based on electromagneticforce. This force plays an important role toproduce a powerful torque. Its magnitude is

given by,

F = B I L


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Working Principle

The armature terminal of a 2-pole directcurrent (dc) motor are connected to a dcvoltage source. As a result, dc current flows

through the motor armature which creates thealternate magnetic poles.

Armature conductors under N-pole areassumed to carry current upward (dots) andthose under S-poles to carry currentdownwards (crosses).


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Fig : 4.1


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It is found that due to lorentz force producesby each conductor under the poles produces

a torque which tends to rotate the armature ina clockwise direction.

All these forces add together to produce a

powerful torque, which sets the armaturerotating.

The direction of the current will be reversedwhen the conductor passes from one pole toanother pole. This process helps to develop acontinuous and unidirectional torque of themotor.


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Back or Counter EMF

When the armature of the motor rotates, theconductors also rotate in the magnetic field.According to laws of electromagneticinduction, an emf will be induced in the

conductors. By Lenzs Law, the direction of this induced

emf is opposite to applied voltage.

This induced voltage is known as back emf

(Eb).Vnet = Ia Ra = Vt - Eb


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Necessity of Back EMF

The back emf plays an important role tocontrol the additional load in a dc motor.

At the time of motor starting, the back emf is

zero. At this stage, the armature current isvery high and it can damage the fuses or canaccelerates the armature rapidly.

The back emf increases with the speed of themotor, hence decreasing the net voltage andin result armature current of the motor.


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Classification of DC Motor

Shunt Motor

In a shunt motor, the field winding isconnected in parallel with the armature

circuit.

Vt = Eb + Ia Ra

IL = Ia + Ish


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Fig: 4.3


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Series Motor

In a series motor, the field winding is

connected in series with the armature circuit.

Ia = IL

Vt = Eb + Ia (Rse + Ra)


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Fig 4.4


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Compound Motor

There are two field windings in the compound

motor. When the field is connected in parallelwith the armature then it is called short shuntcompound motor. When the field isconnected in series with the armature then itis called long shunt compound motor.


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Fig 4.5


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Mechanical Power of a DC Motor

The voltage equation of the dc motor,

Vt = Eb + Ia Ra

The power equation,

Vt Ia = Eb Ia + Ia Ra

Vt Ia = electric power supplied

Eb

Ia

= Pmd

= mechanical power developed byarmature

Ia Ra = armature copper loss

2

2


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The power developed by the motor dependson the armature current.

For the maximum power, the differentiation ofthe mechanical power developed with respectto armature current should be zero.

d / dIa (EbIa) = Vt 2IaRa = 0

IaRa = Vt/ 2

Vt = Eb + Vt/ 2Eb = Vt/ 2


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Example : 4.1

A shunt motor takes a total current of 20Afrom 250V supply. The shunt field andarmature resistances are 200 and 0.3

respectively. Determine the (i) value of backemf and (ii) mechanical power developed inthe motor armature.


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Example 4.2

A shunt motor has an armature circuitresistance of 0.7 and applied voltage 240V.The no-load and full load armature current is

5A and 30A respectively. Determine thechange in back emf from no load to full load,


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Practice Problem 4.1

A 230V shunt motor takes a total current of25A. The shunt field and armatureresistances are 55 and 0.5 respectively.

Find the (i) value of back emf and (ii)mechanical power developed in the motorarmature,


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A 4-pole shunt motor is energized by 450Vsupply and takes an armature current of 55A.The resistance of the armature circuit is 0.4.

The armature winding is wave-connected by650 conductors and useful flux per pole is0.025Wb, Calculate the speed of the motor.


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Torque of DC Motor

The tendency to produce the rotation of theconductor is known as torque.

T = F * r

Fig : 4.6


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Types of Torque

Armature TorqueThe torque produced by the all conductors ofthe armature is known as armature torque

(Ta

). The power is,

Pmd = Ta *

EbIa = Ta * 2N/ 60Ta = 9.55 (EbIa/N)


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Example 4.4

A shunt motor takes a current of 40A from230V supply and runs at a speed of 1100rpm.Find the torque developed by armature, if the

armature and the shunt field resistances are0.25 and 230 respectively.


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Example 4.5

A 6-pole shunt motor is energized by 230V dcsupply. The motor has 450 conductors thatare wound in lap configuration. It takes 30A

current from the supply system and developsoutput power of 5560W. The current throughthe field windings is 3A and the flux per poleis 25mWb. The armature circuit resistance is0.8. Find the speed and the shaft torque.


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A shunt motor takes a current of 45A from220V do supply and runs at a speed of1500rpm. Find the torque developed by the

armature, if the armature and the shunt fieldresistances are 0.5 and 65 respectively.


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A 6-pole shunt motor is energized by 230V dcsupply. The motor has 500 conductors thatare connected in lap wound. It takes 40A

from the supply system and develops outputpower of 5.9kW. The current through the fieldwindings is 4A and the flux per pole is0.04Wb. The armature circuit resistance is0.4 . Calculate the shaft torque.


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Speed of a DC Motor

The speed of dc motor changes slightly withchange in the load.

According to generator action of the motor,the back emf can be written as,

Eb = PZN / 60A

The expression for speed becomes,

N = 60A (Vt IaRa) = K (Vt IaRa)= KEb

PZ


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The speed of a dc motor is directlyproportional to the back emf and inversely

proportional to the flux per pole. Let N1, 1 and Eb1 be the initial values and

N2, 2 and Eb2 be the final values.

Then by combining the equations for initialand final speed, we have

N1 = Eb1 * 2

N2 Eb2 * 1 The flux in the shunt motor is constant with

the change in voltage. So, 1 = 2


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For shunt motor, equation becomes,

N1 = Eb1

N2 Eb2

For a series motor, the flux is directlyproportional to the armature current prior tosaturation.

N1 = Eb1 * Ia2N2 Eb2 * Ia1


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Example: 4.5

A dc shunt motor takes a current of 5A at noload with terminal voltage 230 V and runs at1000 rpm. The armature and Field

resistances are 0.2 and 230 respectively. Under load condition motortakes a current of 30 A. Determine the motorspeed under load condition.


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A dc shunt motor takes a current of 3A at noload with terminal voltage 240 V and runs at1200 rpm. The armature and Field

resistances are 0.25 and 135 respectively. Under load condition motortakes a current of 45 A. Determine the motorspeed under load condition.


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Example: 4.6

A dc series motor takes a current of 56A froma 230 V supply and runs at a speed of 900rpm. The armature and Field resistances are

0.2 and 0.25 respectively. Find thespeed of the motor when the line current is15 A. Consider the flux for 15A current is 40percent of the flux for the line current of 56A.


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A 220 V dc series motor takes a current of120 A and runs at a speed of 1000 rpm. Thearmature and Field resistances are 0.15

and 0.25 respectively. Find the speed ofthe motor when the line current is 35 A.Consider the flux for 35A current is 70percent of the flux for the line current of

120A.


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Speed Regulation

The speed regulation of a dc motor is thechange in speed from full load to no load andit is expressed as a percentage of the speed

at full load.

%speed reg = NL speed FL speed * 100

FL speed


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Losses in a DC Motor A small amount of power will be lost due to

I*I*R for the armature and the shunt fieldresistances. The remainder power will beconverted into mechanical power that

becomes available in armature. A small amount of the developed power in

armature will be lost due to rotational losses.The remainder becomes available at shaft asuseful work.


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Fig 4.8


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Efficiency of a DC Motor

= (Po/ Pin )*100

Pin = Po + losses

The efficiency of a dc motor will be maximumwhen variable losses are equal to constant

losses.


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Example 4.7

At no load condition, a shunt motor takes 5Acurrent from a 230 V supply. The resistancesof the armature and field circuit are 0.25

and 115 respectively. Under load conditionthe motor can carry 40A current. Calculatethe iron and friction losses and efficiency.


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Example 4.8

A shunt motor takes a current of 80 A from220 V supply and runs at 800 rpm. The shuntfield and armature resistances are 50 and

0.1 respectively. The iron and frictionlosses are 1600 W. Determine the (i) copperlosses (ii) armature torque (iii) shaft torqueand (iv) efficiency


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At no load condition, a shunt motor takes 4Acurrent from a 220 V supply. The resistancesof the armature and field circuit are 0.15

and 155 respectively. Under load conditionthe motor can carry 36A current. Calculatethe iron and friction losses and efficiency.


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A shunt motor takes a current of 75 A from220 V supply and runs at 1100 rpm. Theshunt field and armature resistances are 65

and 0.2 respectively. The iron andfriction losses are 900 W. Determine the (i)copper losses (ii) armature torque (iii) shafttorque and (iv) efficiency


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DC Motor Characteristics

1. Torque and armature current characteristic:It is the curve that is obtained from thevariation of torque with the armaturecurrent.

2. Speed and armature current characteristic:It is the curve that is obtained from thevariation of speed with the armature current.

3. Torque and speed characteristic: It is the

curve that is obtained from the variation ofspeed with the torque. It is also known asmechanical characteristic.


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Shunt Motor Characteristic

The general torque equation of shunt motor,

T Ia

As flux is constant,

T Ia


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The speed of the shunt motor can be

expressed as,N Eb/


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The speed of the motor decreases withvariation of torque.


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Series Motor Characteristics

The general torque equation of series motor,T Ia

As flux is directly proportional to current,

Ia The final expression for torque,

T Ia

After saturation current the flux remainconstant and equation of torque will become,

T Ia

2


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The torque-current characteristic of seriesmotor,

N Eb/N [Vt - Ia (Rse + Ra)] /

Since flux is proportional to current,

N [Vt - Ia (Rse + Ra)] / Ia


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Under no load condition the value of thearmature current is small.

Under loaded condition, the armature currentis high.


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The increasing armature current decreasesthe speed of the motor.


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Comparison Between Generator and

Motor

The generator is driven by mechanicalmachine and generates voltage. As a result,a current will flow through the electrical

circuit. The motor is connected across the supply.

The armature of the motor takes a current

which in turn produces a force. This forceproduces a driving torque. As a result of thistorque, the motor rotates.


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In the generator, the direction of armaturecurrent and hence direction of generatedvoltage is determined by the Flemings right

hand rule.

In the motor, the direction of force on aconductor and hence the direction of rotation

is determined by Fleming's left hand rule.


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In the generator, the generated voltage andthe armature current are in the samedirection.

In the dc motor, these are in oppositedirection.


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Applications of DC Motor

Shunt Motor

The characteristics of shunt motor is an

approximately constant speed. It has mediumstarting torque. This motor is used in thelathe machine, drill machine, shaper

machine, blower and fan, etc.


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Series Motor

This motor is known as variable speed motorand it has high starting torque. This motor is

used in the trolley car, crane, electric traction,elevators, air compressors, vacuum cleaners,hair drier and conveyor machines etc.


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Compound Motor

The differential compound motor has poortorque characteristics. As a result, this motor

is rarely used. However, cumulativecompound motor has high starting torque andit is used in the printing presses, ice

machines, air compressors and rolling milsetc.


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In the generator, the relationship betweenthe generated voltage, current and speed areconsidered as main characteristics.

In the motor, these are considered betweenthe speed, torque and current.

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Direct Current Motors[1]