ELG2336 Introduction to Electric Machinesrhabash/ELG2336ElectricMachines... · 2018. 3. 15. ·...

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1 ELG2336 Introduction to Electric Machines Magnetic Circuits DC Machine –Shunt: Speed control –Series: High torque –Permanent magnet: Efficient AC Machine –Synchronous: Constant speed –Induction machine: Cheap and light weight

Transcript of ELG2336 Introduction to Electric Machinesrhabash/ELG2336ElectricMachines... · 2018. 3. 15. ·...

Page 1: ELG2336 Introduction to Electric Machinesrhabash/ELG2336ElectricMachines... · 2018. 3. 15. · Chapman: Electric Machinery 5 Machinery Principles 1. Rotation motion, Newton’s law

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ELG2336 Introduction to Electric Machines

Magnetic Circuits

DC Machine –Shunt: Speed control

–Series: High torque

–Permanent magnet: Efficient

AC Machine –Synchronous: Constant speed

–Induction machine: Cheap and light weight

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History of DC Electric Machines

1884

Frank J. Sprague

produces DC motor for

Edison systems

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3 Chapman: Electric Machinery

History of Magnetic Application

1885

William Stanley

develops commercially

practical transformer

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History of Electric Synchronous Machines

1888

Nikola Tesla presents

paper on two-phase ac

induction and

synchronous motors

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5 Chapman: Electric Machinery

Machinery Principles

1. Rotation motion, Newton’s law and power relationships

2. The magnetic field

3. Faraday’s law

4. Produce an induced force on a wire

5. Produce an induced voltage on a conductor

6. Linear dc machine examples

7. Real, reactive and apparatus power in AC circuits

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6 Chapman: Electric Machinery

Torque

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7 Chapman: Electric Machinery

Magnetic Field: Ampere’s Law

1. The magnetic field is produced by ampere’s law

2. The core is a ferromagnetic material

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8 Chapman: Electric Machinery

Example

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DC Machines

Electric Machinery

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Equivalent Circuit of a DC Machine Left: Separately Excited; Right: Self-Excited Shunt

aaat

fff

RIEV

RIV

Ia_gen

If

Vf Vt Rf

+

- Ea

+

-

Ia_mot

Ra

+

Ia

If

Vt Rf

Ea

-

IL

Ra

+

+

-

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Generated emf and Electromagnetic Torque

aaat

fff

RIEV

RIV

mdaa KE

adae IKT

meaaem TIEP

Voltage generated in the armature circuit due the flux of the stator field current

Electromagnetic torque

Ka: design constant

Motor: Vt > Ea

Generator: Vt > Ea

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Efficiency

InputPower

Losses

InputPower

LossesInputPower

InputPower

OutputPower

1

The losses are made up of rotational losses (3-15%), armature circuit copper

losses (3-6%), and shunt field copper loss (1-5%). The voltage drop between

the brush and commutator is 2V and the brush contact loss is therefore

calculated as 2Ia.

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Example

A 250V shunt motor has an armature resistance of 0.25 and a field resistance of 125. At no-load the motor

takes a line current of 5.0A while running at 1200 rpm. If the line current at full-load is 52.0A, what is

the full-load speed?

Electric Machinery

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Solution

Electric Machinery

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Solution

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Reactive Power Q and Apparatus Power S

1. Reactive power Q (VAR) is defined from instantaneous power

2. Apparatus power S (VA) is defined to represent the product of voltage and current magnitudes

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Complex Power

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Power Factor

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Synchronous Machines

Synchronous machines are AC machines that have a field

circuit supplied by an external DC source.

In a synchronous generator, a DC current is applied to the rotor winding producing

a rotor magnetic field. The rotor is then turned by external means producing a

rotating magnetic field, which induces a 3-phase voltage within the stator winding.

In a synchronous motor, a 3-phase set of stator currents produces a rotating

magnetic field causing the rotor magnetic field to align with it. The rotor magnetic

field is produced by a DC current applied to the rotor winding.

Field windings are the windings producing the main magnetic field (rotor windings

for synchronous machines); armature windings are the windings where the main

voltage is induced (stator windings for synchronous machines).

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Construction of Synchronous Machines

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Voltage Regulation

A convenient way to compare the voltage behaviour of two

generators is by their voltage regulation (VR). The VR of a

synchronous generator at a given load, power factor, and at rated

speed is defined as

%100

fl

flnl

V

VVVR

Where Vfl is the full-load terminal voltage, and Vnl (equal to Ef)

is the no-load terminal voltage (internal voltage) at rated speed

when the load is removed without changing the field current.

For lagging power factor (PF), VR is fairly positive, for unity

PF, VR is small positive and for leading PF, VR is negative.

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Example

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Solution

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Induction Motor

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Induction Motor Speed

The induction motor will always run at a speed lower than the synchronous speed

The difference between the motor speed and the synchronous speed is called the Slip

Where nslip= slip speed

nsync= speed of the magnetic field

nm = mechanical shaft speed of the motor

slip sync mn n n

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Equivalent Circuit of a Synchronous Motor

( )

T a a a l a ar f

s l ar

T f a a s

T f a s

V I R I jX I X E

X X X

V E I R jX

V E I Z

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Rotation Speed of Synchronous Machine

Synchronous generators produce electricity whose frequency is synchronized

with the mechanical rotational speed.

Where fe is the electrical frequency, Hz;

nm is mechanical speed of magnetic field (rotor speed for synchronous

machine), rpm;

P is the number of poles.

Steam turbines are most efficient when rotating at high speed; therefore, to

generate 60 Hz, they are usually rotating at 3600 rpm and turn 2-pole generators.

Water turbines are most efficient when rotating at low speeds (200-300 rpm);

therefore, they usually turn generators with many poles.

120

Pnf

m

e

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Selection of Induction Motors

Efficiency

Starting torque

Pull-out torque

Power factor

Starting current

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The Equivalent Circuit

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Power Flow

cos3in ssVIP

rotdevout PPP

%100in

out P

P

m

PT

dev

dev

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Example 1

A 480-V, 60 Hz, 50-hp, three phase induction motor is drawing 60A at 0.85 PF lagging. The stator copper losses are 2 kW, and the rotor copper losses are 700 W. The friction and windage losses are 600 W, the core losses are 1800 W, and the stray losses are negligible. Find the following quantities: 1. The air-gap power PAG.

2. The power converted Pconv.

3. The output power Pout.

4. The efficiency of the motor.

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Solution

1.

2.

3.

3 cos

3 480 60 0.85 42.4 kW

in L LP V I

42.4 2 1.8 38.6 kW

AG in SCL coreP P P P

70038.6 37.9 kW

1000

conv AG RCLP P P

&

60037.9 37.3 kW

1000

out conv F WP P P

37.350 hp

0.746outP

100%

37.3100 88%

42.4

out

in

P

P

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Example 2

A 460-V, 25-hp, 60 Hz, four-pole, Y-connected induction motor has the following impedances in ohms per phase referred to the stator circuit:

Rs= 0.641 Rr= 0.332

Xs= 1.106 Xr= 0.464 Xm= 26.3

The total rotational losses are 1100 W and are assumed to be constant. The core loss is lumped in with the rotational losses. For a rotor slip of 2.2 percent at the rated voltage and rated frequency, find the motor’s 1. Speed

2. Stator current

3. Power factor

4. Pconv and Pout

5. ind and load

6. Efficiency

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Solution

1.

2.

120 120 601800 rpm

4

esync

fn

P

(1 ) (1 0.022) 1800 1760 rpmm syncn s n

22 2

0.3320.464

0.022

15.09 0.464 15.1 1.76

RZ jX j

s

j

2

1 1

1/ 1/ 0.038 0.0662 1.76

112.94 31.1

0.0773 31.1

f

M

ZjX Z j

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Solution

3.

4.

0.641 1.106 12.94 31.1

11.72 7.79 14.07 33.6

tot stat fZ Z Z

j

j

1

460 0

318.88 33.6 A

14.07 33.6tot

VI

Z

cos33.6 0.833 laggingPF

3 cos 3 460 18.88 0.833 12530 Win L LP V I

2 2

1 13 3(18.88) 0.641 685 WSCLP I R

12530 685 11845 WAG in SCLP P P

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Solution

5.

6.

(1 ) (1 0.022)(11845) 11585 Wconv AGP s P

& 11585 1100 10485 W

10485= 14.1 hp

746

out conv F WP P P

1184562.8 N.m

1800260

AGind

sync

P

1048556.9 N.m

1760260

outload

m

P

10485100% 100 83.7%

12530

out

in

P

P

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Solve this Example

A 460 V, 25-hp, 60Hz, four pole, Y-connected induction motor has the

following impedances in ohms per phase referred to the stator circuit:

Rs = 0.641Ω Rr = 0.332Ω

Xs =1.106Ω Xr = 0.464Ω Xm = 26.3Ω

The total rotational losses = 110 W, rotor slip = 2.2% at rated voltage and

frequency.

Find the motor's

i) Speed, ii) Stator Current, iii) Power factor, iv) Pconv,

v) Pout vi) ind, vii) load and viii) Efficiency

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Ideas for the Course Project

https://howtomechatronics.com/tutorials/arduino/arduino-dc-motor-control-

tutorial-l298n-pwm-h-bridge/

http://www.instructables.com/id/Control-DC-and-stepper-motors-with-

L298N-Dual-Moto/

https://dronebotworkshop.com/dc-motors-l298n-h-bridge/

https://howchoo.com/g/mjg5ytzmnjh/controlling-dc-motors-using-your-

raspberry-pi

https://javatutorial.net/raspberry-pi-control-dc-motor-speed-and-direction-

java