Dc Machines

Microsoft PowerPoint Presentation Graphics for

EE 315: Basic Electrical Engineering IIIPrepared by Brian Manhire, Ph.D.Professor of Electrical Engineering

Stocker Center, home of Ohio University’sRuss College of Engineering & Technology

12:21 PM Ohio University’s Russ College of Engineering & TechnologyOhio University’s Russ College of Engineering & TechnologyOhio University’s Russ College of Engineering & Technology 2

Microsoft PowerPoint Presentation Graphics© Copyright 1998 Brian Manhire

For Part 3 of

Introduction to ElectricalEngineering, 2/e

by C.R. Paul, S.A. Nasarand L.E. Unnewehr

1992, 1992, McGraw-Hill, Inc.

Ohio University’s Russ College of Engineering & TechnologyOhio University’s Russ College of Engineering & TechnologyOhio University’s Russ College of Engineering & Technology

IntroductionIntroduction

13.113.1 The Faraday Disk and Faraday’s LawThe Faraday Disk and Faraday’s Law 13.2 13.2 The Heteropolar, or Conventional, DC MachineThe Heteropolar, or Conventional, DC Machine 13.3 13.3 Constructional DetailsConstructional Details

13.4 13.4 Classification According to Forms of ExcitationClassification According to Forms of Excitation13.513.5 Performance EquationsPerformance Equations

13.613.6 Effects of SaturationEffects of SaturationBuidup of Voltage in a Shunt GeneratorBuidup of Voltage in a Shunt Generator13.713.7 Generator CharacteristicsGenerator Characteristics13.813.8 Motor CharacteristicsMotor Characteristics

13.913.9 Starting of DC MotorsStarting of DC Motors13.1013.10 Speed Control of DC MotorsSpeed Control of DC Motors13.1113.11 Losses and EfficiencyLosses and Efficiency

Chapter 13:Chapter 13: DC MachinesDC Machines


Introduction

Mechanical energy = Teωm∆t

Ideal (lossless)Electric Machine

Motor

GeneratorElectrical energy = vi∆t

Teωm

v

i→

Pelec. = vi = Teωm = Pmech.

+

−

The machine’s magnetic field is themedium of (energy) conversion

Electromechanical Energy Conversion


Section 13.1: The Faraday Disk and Faraday’s Law

Flux-cutting Rule:Voltage e = Bluis induced in themoving conductorascending thestationary rails

B (flux density) is in T., the moving conductor’s lengthand speed (l and u) are in m. and m./sec. respectively


Section 13.1: The Faraday Disk and Faraday’s Law cont.

From Example 13.1 (pp. 494-495): v = 0.5 ω B r12 = 6.28 V. for B = 0.4 T.,

ω = 1200 RPM and r1 = 0.5 m.which for the machine’s physical size andspeed, isn’t very much. Moral: The homopolar scheme isn’t practical


Section 13.2 The Heteropolar, or Conventional, DC Machine

AC Generator

v(t) = Emsin(t), where: Em = 2BNlrωω


Section 13.2 The Heteropolar, or Conventional, DC Machine cont.

DC Generator

Commutator-brush assembly mechanicallyrectifies the coil’s AC voltage so that DCvoltage appears across the brushes


Section 13.2 The Heteropolar, or Conventional, DC Machine cont.

DC Motor

Conductor aunder N pole

Conductor aunder S pole

A unidirectional (CCW) torque acts on the coil per text Fig. 13.7,p. 499 (left-hand rule for forces acting on the coil’s conductors)


Section 13.3 Construction Details

←← DC Machine Parts

Armature Lamination →→


Section 13.3 Construction Details cont.

Commutator Assembly Details



←← Lap Winding

Wave Winding →→Odd- and even-numbered conductors are atthe top and bottom of the slots respectively



← Lap winding (single bar)

Multi-turn wavewinding →

← Coil in slot

Slot details →(layered coils)


Section 13.4 Classification According to Forms of Excitation

Separately excited (armature & field separated)

Shunt excited (arma-ture & field in parallel)

Series excited (arma-ture & field in series)


Section 13.4 Classification According to Forms of Excitation cont.

Cumulative compound(field windings aid one-another)

Differential compound(field windings opposeone-another)


Section 13.4 Classification According to Forms of Excitation cont.

Long-shunt compound(armature in series withseries field winding)

Short-shunt compound(armature in parallelwith shunt field winding)


Section 13.5: Performance Equations

emf equationE = φφnZ(p/a)/60torque equation

Te = φφZ(p/a)Ia/(2π)Where:

φ = Flux per pole (Wb.) n = Armature speed in RPM Z = Number of active armature conductors p = Number of field poles = a for lap winding a = Number of parallel paths in the armature winding = 2 for wave winding Ia = Armature current in Amperes


Section 13.5: Performance Equations cont.

Example 13.2Determine the voltage induced in the armature of a DC machinerunning at 1750 RPM and having four poles. The flux per pole is25 mWb. and the armature is lap-wound with 728 conductors

Since the armature is lap-wound, p = a. So the emf equation yields:

E = φnZ(p/a)/60

= 25×10-3Wb.×1750 RPM×728×(p/a = 1)/60

= 530.8 V.



Example 13.3A lap-wound armature has 576 conductors and carries a current of123.5 A. If the flux per pole is 20 mWb., calculate theelectromagnetic torque developed by the armature.

The armature is lap-wound so p = a. Then the torque equationyields:

Te= φZ(p/a)Ia/(2π)

= 20×10-3Wb.×576×(p/a = 1)×123.5A./(2π)

= 226.4 N.m



Example 13.4If the armature of Example 13.3 rotates at 123.5 radians per second,what is the induced emf in the armature?

Assuming the armature has no resistance (so it’s lossless), then itselectrical power equals its mechanical power, ergo:

E×Ia = Te×ω

∴ Ε × 123.5 Α. = 226.4 N.m × 123.5 rad./sec.

which has solution E = 226.4 V.



E

+

+

−−+ Vf −

↓ If

Shaft

n

Ia

→

Vt

RaIa

+

−Ra

Back emf voltage →

Separately Excited Motor

Supply +−

Armature

FieldRf

E = Vt - RaIa (KVL) and E = k1nφ so that …

n = (Vt - RaIa )/(k1φ ) which is the so-called speed equation

If the field isn’t saturated, then φ = kfIf so that …n = (Vt - RaIa )/(kIf) where k = k1kf



E

+

+

−− ↓ If

Ia

→

Vt = 250 V.

RaIa

+

−

Shunt Excited Motor

Supply +− Rf = 125 Ω

Ra = 0.25 Ω

I→

Example 13.5A 250 V. shunt motor has an armature resistance of 0.25 Ωand a field resistance of 125 Ω . At no load (1200 RPM), themotor draws a current of 5 A. At full load, the motor draws52 A. What’s the motor’s full-load speed (in RPM)?



E

+

+

−− ↓ If

Ia

→

Vt = 250 V.

RaIa

+

−

Shunt Excited Motor


Ra = 0.25 Ω

I→

Example 13.5 cont.The field current is If = Vt/Rf = 250V./125Ω = 2A. so (by KCL),the armature current at no-load is: Ia = I - If = 5A.- 2A. = 3A.

By KVL, ENL = Vt - RaIa = 250V. - 0.25Ω×3Α. = 249.25V.

and the corresponding (no-load) speed is nNL = 1200 RPM



E

+

+

−− ↓ If

Ia

→

Vt = 250 V.

RaIa

+

−

Shunt Excited Motor


Ra = 0.25 Ω

I→

Example 13.5 cont.The field current is If = Vt/Rf = 250V./125Ω = 2A. so (by KCL),the armature current at full load is: Ia = I - If = 52A. - 2A. = 50A.

By KVL, EFL = Vt - RaIa = 250V. - 0.25Ω×50Α. = 237.5V.

and the corresponding (full-load) speed nFL is unknown.



E

+

+

−− ↓ If

Ia

→

Vt = 250 V.

RaIa

+

−

Shunt Excited Motor


Ra = 0.25 Ω

I→

Example 13.5 cont.Then ENL = k1φnNL and EFL = k1φnFL with constant φ so that

nFL = (EFL/ENL)× nNL = (237.5V./249.25V.)×1200 RPM, ∴

nFL ≈ 1143 RPM


Section 13.6: Effects of SaturationBuildup of Voltage in a Shunt Generator

E

+

+

− −↓ If

Ia

→

Vt

RaIa = 0V.−

+

Unloaded Shunt Excited Generator

Rf

Ra = 0 Ω

I = 0 A.→

Given that the machine isn’t loaded and that Ra is negligible, Ia =If and E(If) = RfIf where E(If) is a non-linear function of the fieldcurrent If. The implications of this are shown in the next slide.


Section 13.6: Effects of SaturationBuildup of Voltage in a Shunt Generator cont.

Voltage buildup depends on residual magnetism and Rf’s value

E(If) = RfIf

If

E(If), RfIf

Residual magnetism →

Slope = Rf

E(If)

RfIf

E++

− −

↓ If

Ia

→

Rf If

Unloaded Shunt Excited Generator

I = 0 A.→


Section 13.7: Generator Characteristics


Section 13.7: Generator Characteristics cont.Example 13.6

A 50 KW, 250 V., short-shunt compound generator has thefollowing date: Ra = 0.06 Ω, Rser. = 0.04 Ω and Rf = 125 Ω.Calculate the induced armature emf at rated load and terminalvoltage. Take 2 V. as the total brush-contact voltage drop.

I = 50KW/250 V. = 200A.

←

E

++

−−

↓ If

Ia

→

Vt = 250 V.

RaIa

−

+Rf = 125 Ω

Ra = 0.06 Ω

Rser. = 0.04 Ω

50 KW


Section 13.7: Generator Characteristics cont.Example 13.6 cont.

I = 200A.←

KVL: Vf = Vt + Rser.I = 250V. + .04Ω×200A. = 258 V.

If = Vf/Rf = 258V./125Ω = 2.064 A. and by KCL:

Ia = I + If = 200 A. + 2.064 A. = 202.064 A.

KVL: E’ = Vf + RaIa = 258 V. + .06Ω×202.064A. = 270.12 V.

KVL: E = E’ + Vbrush = 270.12V. + 2V. = 272.12 V.

Note that Vbrush isn’t shown on the schematic

E

++

−−

↓ If

Ia

→

Vt = 250 V.

RaIa

−

+Rf = 125 Ω

Ra = 0.06 Ω

Rser. = 0.04 Ω

50 KWVf

+

−

+ Rser.I −


Section 13.8: Motor Characteristics


Section 13.9: Starting of DC Motors

E

+

+

−− ↓ If

Ia

→

Vt = 230 V.

RaIa

+

−

50 hp Shunt Excited Motor

Supply +−Rf

Ra = 0.05 Ω

I→

On starting, E = k1nφ = 0 V. (since n = 0 RPM)By KVL: Ia = (Vt - E)/Ra = 230 V./.05 Ω = 4.6 KA. andPa = (Ia)2Ra = 1.058 MW = 1.42 Khp >> 50 hp = rated hpWhere is this power going?

In practice, high starting currents are limited by way of astarting mechanism which places resistance in series withthe armature during startup


Section 13.9: Starting of DC Motors cont.

E

+

+

−− ↓ If

Ia

→

Vt = 230 V.

RaIa

+

−

Shunt Excited Motor

Supply +−Rf = 75 Ω

Ra = 0.05 Ω

I→

Example 13.7The shunt motor draws 7 A. when running at 1120 RPMand 46 A. at a greater load. What’s the motor’s speed atthis greater load, and, what is its speed if the fieldresistance is increased to 100 Ω at this (46 A.) load?



E

+

+

−− ↓ If

Ia

→

Vt = 230 V.

RaIa

+

−

Shunt Excited Motor


Ra = 0.05 Ω

I→

Example 13.7 cont.At the 7 A. load, n7A. = 1120 RPM, If = 230 V. / 75 Ω = 3.07 A. and Ia = I - I f =7A. - 3.07A. = 3.93 A. so the speed equation is:

n7A. = 1120 RPM = (230V. - 0.05Ω×3.93A.)/(3.07A.×k) = (Vt - RaIa)/kIf = E/kIf

Which yields k = 0.0668 (what are k’s units?). So at the 46A. load (with R f =75Ω) Ia = I - If = 46A. - 3.07A. = 42.93 A. and the speed equation yields:

n = (230V. - 0.05Ω×42.93A.)/(3.07A.×0.0668) = 1111 RPM



E

+

+

−− ↓ If

Ia

→

Vt = 230 V.

RaIa

+

−

Shunt Excited Motor


Ra = 0.05 Ω

I→

Example 13.7 cont.If the field resistance is increased to 100 Ω , If = 230 V. / 100 Ω = 2.3 A. and Ia =I - If = 46A. - 2.3A. = 43.7 A. so the speed equation is:

n = (230V. - 0.05Ω×43.7A.)/(2.3A.×0.0668) = 1483 RPM



Example 13.8What’s the motor’s speed if it’s running as shown above (with a 0.1Ω resistor placed in the armature circuit)?

From Example 13.7, If = 3.07A., Ia = 42.93A. and k = 0.0668 so thespeed equation gives n = (230V. - 0.15Ω×42.93A.)/(3.07A.×0.0668)= 1090 RPM (Note: P0.1Ω = (42.93A.)2×0.1 Ω = 184.3 W.)

E

+

+

−− ↓ If

Ia

→

Vt = 230 V.

RaIa

+

−Supply +−Rf = 75 Ω

Ra = 0.05 Ω

I = 46A.→

0.1 Ω


Section 13.10: Speed Control of DC Motors

Switch up: Vm = Vt , Switch down: Vm = 0 V.

The average voltage (Vm) across the motor (and henceits speed) is controlled (see Figure 13.23 on text p. 516)

by the electronically controlled switch (Thyristor)

Three temporal switching schemesare shown on the next slide

Supply Vt(constant)

+

−

Ia ↓

EVm

+

−

+− Motor


Section 13.10: Speed Control of DC Motors cont.Pulse-width modulation (PWM)

Frequency modulation (FM)

FM-PWM hybrid

Vave. = αVWhere: α = Duty cycle

= ton/T


Section 13.10: Speed Control of DC Motors cont.

+

−

Ia ↓Vt

E

Supply

Vm

+

−

Motor

Ldi/dt+

−

Switch representing commutationcircuitry (needed to shut off Thyristor)

Thyristor

Freewheeling diode (provides path for motor’scurrent during Thyristor shut-off)



+

−

Ia ↓Vt

E

Supply

Vm

+

−

Motor

Ldi/dt+

−

Thyristor = SC

Reverse-biased freewheeling diode = OC

1st State: Thyristor conducting

Vm = Vt



+

−

Ia ↓Vt

E

Supply

Vm = 0 V.

+

−

Motor

Ldi/dt < 0 V.+

−

Thyristor = OC

Forward-biased freewheeling diode = SC

2nd State: Thyristor shut off

Vm = 0 V. 0 A. ↓

Ia decreasing(no supply) ∴


Example 13.9For a DC series motor, the following data are given: supply voltage = 440 V.,duty cycle = 30%, motor circuit inductance = 0.04 H., maximum allowablechange in armature current = 8 A. Determine the chopper frequency.


The average voltage across the motor is Vave. = αVt = 0.3×440V. = 132 V.

The voltage across the motor circuit inductance is V t - Vave. = 440V. - 132V. =308 V. = 0.04H.×8A./ton = L∆i/∆t (mimicking Ldi/dt) which yields …

ton = 1.04 ms

But α = ton/T ∴ T = ton/α = 1.04 ms/0.3 = 3.46 ms so that the frequency is …

f = 1/T = 1/(3.46ms) = 289 pulses per second


ConvertersThese change AC input voltage to a controllable DC voltagewhich governs the motor’s speed. They take advantage ofnatural (line) commutation (by way of sinusoidal zeros) so nocomplex commutation circuitry is needed. See text pp. 519-522for details including performance figures of merit (i.e., inputpower, displacement and harmonic factors)



Example 13.10A 1hp, 240 V. DC motor is designed to run at 500 RPM when supplied from aDC source. The motor’s armature resistance is 7.56 Ω. Its torque and back-emf constants are 4.23 Nm/A and 4.23 V/(rad.sec.) respectively. The motor isdriven by a half-wave converter at 200 V., 50 Hz. AC and draws a 2 A.average current at a certain load. While the SCR conducts, the average motorvoltage Vm is 120 V. Determine the torque developed by the motor, themotor’s speed and the supply power factor.


The torque is τ = kIa = 4.23 Nm/A×2A. = 8.46 Nm and the back emf is

E = Vm - RaIa = 120V. - 7.56Ω×2A. = 104.88 V. = 4.23 V/(rad.sec.)×Ωm = kΩm

Which yields Ωm = 24.79 rad./sec. (which is about 237 RPM)

The voltageamperage of the motor is |S|= 200V.×2A. = 400 VA and theaverage power drawn by the motor is Vm×Ia = 120V.×2A. = 240 W so thepower factor is P/|S| = 0.6 lagging.


Ward-Leonard systemThis speed-control approach employs a DC generator supplyinga DC motor with a variable DC supply voltage (see Example13.11 on text p. 523). Speed control is obtained by varying thefield current of the generator and/or motor. The Ward-Leonardspeed-control methodology is being supplanted by theaforementioned power-electronics schemes.



Section 13.11: Losses and Efficiency

Input OutputVfIf + Tshaftωmech.

Pw&f

Pmagn.loss

Pstray

Parm. = EIa

Parm.loss = RaIa2

VtIa

Power distribution: separately excited generator


Section 13.11: Losses and Efficiency cont.

Input Output

Power distribution: separately excited motor

Parm.loss = RaIa2

Pmagn.loss

Parm. = EIa

Pstray Pw&f

VtIa + VfIf Tshaftωmech.


Section 13.11: Losses and Efficiency cont.Example 13.12

A 230 V., 10 hp shunt motor draws a full-load line current of 40 A. Thearmature and field resistances are 0.25 Ω and 230 Ω respectively. The totalbrush-contact drop is 2 V. and the core and friction losses are 380 W.(total). Calculate the efficiency of the motor assuming the stray-load lossesare 1% of the motor’s rated output.

The input power is Pin = V tIline = 230V.×40A. = 9.2 KW

The (shunt) field circuit losses are Vt2/Rf = (230V.)2/230Ω = 230 W.

The armature losses are Parm.loss = (I line - If)2Ra = (40A. - 230V./230Ω)2×0.25 Ω =380.25 W.

The core + friction losses are 380 W. and the stray losses are 0.1 hp (1% of 10 hp)which is 74.6 W.

The brush-contact losses are Vb-c×Ia = 2V.×39A. = 78 W.

The total losses are 1142.85 W. So the output power is P in - total losses = 8.05715W. and the efficiency is (Pout/Pin)×100% ≈ 87.58%

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