slides electromechanical

AV-222Electromechanical Systems

Dr Salman Aslam

Wing Commander, PAFAssociate Professor

Avionics DepartmentCollege of Aeronautical Engineering

PAF Academy Risalpur

,


readme1. Introduction2. Theory

(a) Electromagnetics(b) Power(c) Drive electronics(d) Mechanics(e) Transformers(f) Motors & Generators

- DC- Linear- Brushed

- AC- Synchronous- Induction

- Other- Universal motor- Reluctance motor- Hysteresis motor- Stepper motor- BLDC motor- Servo motor

3. Applications4. Labs5. Problems

,

Textbook

These slides are under construction. Should be done bythe end of the semester around Aug 2015.

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Textbook3 / 412








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Famous scientistsAndre Marie Ampere

http://en.wikipedia.org/wiki/Andre-Marie_Ampere

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• 1775-1836, France

• Started teaching himself advanced math at the age of 12

• Ampere showed that two parallel wires carrying electric currentsattract or repel each other, depending on whether the currentsflow in the same or opposite directions, respectively - this laid thefoundation of electrodynamics

http://en.wikipedia.org/wiki/Andre-Marie_Ampere








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Famous scientistsMichael Faraday

http://en.wikipedia.org/wiki/Michael_Faraday

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• 1791-1867, England

• Discovered benzene and electromagnetic induction

• When asked by the British government to advise on theproduction of chemical weapons for use in the Crimean War(1853-1856), Faraday refused to participate citing ethical reasons

http://en.wikipedia.org/wiki/Michael_Faraday








,

Course OverviewMagnetic field creation and 3 applications

• This course is about transformers, motors and generators

• Magnetic fields are the fundamental mechanism by which energyis converted from one form to another in all these devices

• Create a magnetic field: This is the first step.(Creation, Ampere’s Law): A current carrying wire produces amagnetic field in the area around it. Now that a magnetic fieldhas been generated, one of the following 3 are possible if you havea conductor placed in a magnetic field:

1 Change a magnetic field to create a voltage(transformer action, Faraday’s Law): A time-changingmagnetic field induces a voltage in a coil of wire if it passesthrough that coil

2 Put a current-carrying wire in the magnetic field(motor action, Lorentz Law): A current-carrying wire inthe presence of a magnetic field has a force induced on it

3 Put a moving wire in the magnetic field(generator action, Faraday’s Law): A moving wire in thepresence of a magnetic field has a voltage induced on it

Chapman 5th ed, pg 8

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,

Voltage, Current and ResistanceAn overview

http://www.build-electronic-circuits.com/wp-content/uploads/2014/09/

Ohms-law-cartoon-by_unknown.jpg

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http://www.build-electronic-circuits.com/wp-content/uploads/2014/09/Ohms-law-cartoon-by_unknown.jpg

http://www.build-electronic-circuits.com/wp-content/uploads/2014/09/Ohms-law-cartoon-by_unknown.jpg

Maxwell’s equationsSummary

• E and H are the electric and magnetic field intensities measured in V/m and A/m respectively.• D and B are the electric and magnetic field densities respectively, measured in coulombs and

teslas respectively.• D = εE, where ε is permittivity. The permittivity of free space is ε0 = 8.854x10−12 F/m.• B = µH, where µ is permeability. The permeability of free space is µ0 = 4πx10−7 H/m.• J is current density measured in A/m2.• φ is flux measured in Webers.

Ampere’s Law I =∮

LH.d` =

∫A

(J + ∂D∂t ).dA ∇×H =J + ∂D

∂t

Faraday’s Law V =∮

LE.d` = −

∫A∂B∂t .dA = − dφ

dt ∇× E =−∂B∂t

Gauss’s Law∮

AB.dA =0 ∇.B =0

Gauss’s Law∮

AD.dA =ρ ∇.D =ρ

• Maxwell introduced 2 new things:

• The induced voltage∫

A∂B∂t.dA

• The displacement current∫

A∂D∂t.dA

• The conduction current density is J = σE (Ohm’s Law) while the displacement current density isJD = ∂D

∂t. Therefore, conduction current I =

∫A J.dA and displacement current ID =

∫A JD.dA.

The displacement current is a result of the time-varying electric field, eg, current through acapacitor when a time-varying voltage is applied to its plates.

• For the time invariant form, ∂B∂t

= ∂D∂t

= 0. This means that the divergence equations remain thesame and only the curl equations change.

,





LH.d` =

∫A

(J + ∂D∂t ).dA ∇×H =J + ∂D

∂t


LE.d` = −


dt ∇× E =−∂B∂t

Gauss’s Law∮

AB.dA =0 ∇.B =0

Gauss’s Law∮

AD.dA =ρ ∇.D =ρ

• We see that• ∫

A B.dA = φ (from Faraday’s Law)• ∮

A B.dA = 0 (from Gauss’ Law for a closed surface, meaning that no monopole exists)

• Also, notice• ∫

A(J + ∂D∂t

).dA = σ∫

A E.dA + ε∫

A∂E∂t.dA = I (from Ampere’s Law)

• ∫A B.dA = µ

∫A H.dA = φ (from Faraday’s Law)

• Now, notice parallels between• E and H (intensities)• B and J,D (densities)• I and φ (what flows in circuits)• µ and σ, ε (material constants) ,





LH.d` =

∫A

(J + ∂D∂t ).dA ∇×H =J + ∂D

∂t


LE.d` = −


dt ∇× E =−∂B∂t

Gauss’s Law∮

AB.dA =0 ∇.B =0

Gauss’s Law∮

AD.dA =ρ ∇.D =ρ

• For a conductor of length ` meters in a uniform magnetic flux density B,

• Motor action: If the conductor carries current i , then the force on it is F = i(`× B)• Generator action: If the conductor moves with velocity v, the voltage induced in it is

e = (v × B).`

,





LH.d` =

∫A

(J + ∂D∂t ).dA ∇×H =J + ∂D

∂t


LE.d` = −


dt ∇× E =−∂B∂t

Gauss’s Law∮

AB.dA =0 ∇.B =0

Gauss’s Law∮

AD.dA =ρ ∇.D =ρ

• For an inductor, the voltage that is induced by the time variations in the current of a circuit iscalled the electromotive force (emf) of self-induction, and is expressed in terms of theself-inductance L by

e = N dφdt

= L dIdt

⇒ Nφ = LI

⇒ L = NφI

Inductance is therefore the flux linkage per ampere

,

From Current to Induced VoltageAn overview

electric charges

separation motion

Electric field Magnetic field

current(amperes)

Ampere's Law

magnetic fieldintensity

magnetic flux density

magneticflux

if changing

"magnetic current" Faraday's Law(induced voltage)

In a magnetic circuit, such as a transformer core,

where,

,

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,

Ampere’s Law (1/4)

Ampere’s circuit law states that the line integral of thetangential component of H around a closed path is thesame as the net current Ienc enclosed by the path

∮H.d` = Ienc

H is the magnetic field intensity measured inampere-turns/m

Chapman, pg 8Elements of Electromagnetics, Sadiku pg 273

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,


Example 1: wire ∮H.d` = Ienc

⇒∮

Bµ .d` = Ienc

⇒2π∮0

Brdθ = µIenc

⇒ B = µ2π

Iencr

- http://www.physics.upenn.edu/courses/gladney- also see Biot-Savart Law

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Example 2: wire wound on core

• We have a core with a winding of N turns of wire wrapped aboutone leg of the core

• If the core is made of ferromagnetic material, then all themagnetic field produced by the current will remain inside the core

• Therefore, the path of integration in Ampere’s Law is the meanpath length of the core, `c

Chapman, pg 8

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,


Example 2: wire wound on core cont.∮H.d` = Ienc

⇒ H`c = Ni

⇒ Bµ `c = Ni

⇒ B = Ni`cµ

(B = µH)

⇒ φ = Ni`cµA

(φ = BA)

⇒ = NiR (R = `c

µA )

• Ni is the mmf (magnetomotive force, F), equivalent to voltage

• B is the magnetic flux density measured in webers/m2, or teslas

• φ is the total flux measured in webers and is equivalent to current

• The reluctance R is equivalent to resistance

Note- H is linearly related to F (think voltage)

- B is linearly related to φ (think current)

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,

Faraday’s Law (1/5)

If a flux passes through a turn of a coil of a wire, avoltage will be induced in the turn of wire that isdirectly proportional to the rate of change in the fluxwith respect to time

eind = −dφ

dt

where eind is the voltage induced in the turn of the coiland φ is the flux passing through the turn.

The minus sign in the equation is an expression ofLenz’s Law

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,


If a coil has N turns and if the same flux passes throughall of them, then the voltage induced across the wholecoil is given by

eind = −Ndφ

dt

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,


Determine polarity of eind using Lenz’s Law

Lenz’s Law states that the direction of voltage buildupin the coil in Faraday’s Law is such that if the coil endswere short-circuited, it would produce current thatwould cause a flux opposing the original flux change

To see this clearly, consider the example on the nextslide

Chapman, pg 30

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,


Determine polarity of eind using Lenz’s Law

• In the left figure below, φ is increasing and willtherefore induce a voltage eind in the coil

• In the right figure below, a current i flowing asshown would produce a flux in the oppositedirection of φ

• The polarity of the voltage will be such that itcould drive the current i in an external circuit

Chapman, pg 30

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Determine polarity of eind using Lenz’s Law cont.

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,

Analogy between electric and magneticcircuits (1/3)

Conductivity σ Permeability µField intensity E Field intensity HCurrent I =

∫J.dA Magnetic flux φ =

∫B.dA

Current density J = IA

= σE Flux density B = φA

= µH

Electromotive force (emf) V Electromotive force (mmf) FResistance R Reluctance RConductance G = 1/R Permeance P = 1/R

• Permeability is the measure of the ability of a material to supportthe formation of a magnetic field within itself. Hence, it is thedegree of magnetization that a material obtains in response to anapplied magnetic field.

• In SI units, permeability is measured in henries per meter.

• A good magnetic core material must have high permeability.

Elements of Electromagnetics, Sadiku, pg 348

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,


Chapman, pg 11

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,


Determine polarity of mmf in magnetic circuit

Chapman, pg 12

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,

Hysteresis

https://www.kjmagnetics.com/blog.asp?p=magnet-grade

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https://www.kjmagnetics.com/blog.asp?p=magnet-grade








,

Magnetization curve (1/2)

Chapman, pg 22

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,

Magnetization curve (2/2)

Chapman, pg 26

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,

AC circuitsPowers

Voltage V = V∠αCurrent I = I∠βPhase lag θ = α− β (θ is negative for inductive circuit)

Power factor PF = cos θ

PowerReal P = V I cos θ (equal to average power)

Reactive Q = V I sin θ

Complex S = P + jQ= V I cos θ + jV I sin θ= V I∠θ= V I∠(α− β)= V∠αI∠−β= VI∗

Apparent S = V I= |S|

Instantaneous p(t) =√

2V cos(ωt)√

2I cos(ωt − θ) (assume α = 0)

= 2V I cosωt cos(ωt − θ)= V I cos θ(1 + cos 2ωt) + V I sin θ sin 2ωt= P + P cos(2ωt) + Q sin(2ωt)

Chapman 5th ed, pg 47-51

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DC Motor Drivers (1/5)

Stepper → Driver stage → L298

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Step 1: Pick an L-298.

Connect 2 voltages (5V, 36V), 2 capacitors (100 nF), 2 resistors (RSA,RSB ), and ground it.








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Step 2: Study the circuit.

Notice that we have 2 similar circuits which are totally independent ofeach other. The left circuit is controlled by EnA while the right circuitis controlled by EnB.








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Step 3: Let’s focus on only one side of the circuit. Theother side works exactly the same way.

Let’s use the left side. Connect a coil (motor winding) as shown.








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Step 4a: Current flow.

Let EnA=1n1=5V. This causes current to flow through the coil.








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Step 4b: Current flow.

Let EnA=1n2=5V. This causes current to flow through the coil in theopposite direction.








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MotionDisplacement, velocity, acceleration

• Displacement• Linear: r• Angular: θ (radians)

• Velocity• Linear: v = dr/dt• Angular: ω = dθ/dt

• ωm: radians/sec• fm: revs/sec• nm: revs/min

• Acceleration• Linear: a = dv/dt• Angular: α = dω/dt


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MotionForce, torque, work, power

• Force: F• Torque: τ = rF sin θ

• Work: W =∫

Fdr• Work: W =

∫τdθ (rotational motion)

• Power: P = dW /dt = d(Fr)/dt = Fdr/dt = Fv• Power: P = dW /dt = d(τθ)/dt = τdθ/dt = τω (rotational motion)


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,

Modeling (1/4)

Constant acceleration model

s(t) = a

t∫t0

s(τ)dτ =t∫

t0

a dτ

s(τ)|tt0= a τ |tt0

s(t)− s(t0) = at − at0 Notice this is vf = vi + at

t∫t0

s(τ)dτ −t∫

t0

s(t0)dτ =t∫

t0

aτdτ −t∫

t0

at0dτ

s(τ)|tt0− s(t0)τ |tt0

= 12

a τ2∣∣tt0− at0τ |tt0

s(t)− s(t0)− s(t0)t + s(t0)t0 = 12

at2 − 12

at02 − at0t + at0

2

let initial time t0 = 0, initial distance s(t0) = s i = 0, and some initialvelocity s(t0) = vi , to get the familiar equation,

s(t) = vi t +1

2at2

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,

Modeling (2/4)

Constant acceleration model• The equations s = s i + vi t + 1

2at2 and vf = vi + at

can be written in discrete time with sampling time T as,[svf

]=

[1 T0 1

] [si

vi

]+

[12

T 2

T

]a

and writing in terms of states x , we get,

xkT =

[xkT

xkT

]=

[1 T0 1

] [xkT−1

xkT−1

]+

[12

T 2

T

]a

• For simplicity, let T = 1,

xk =

[xk

xk

]=

[1 10 1

] [xk−1

xk−1

]+

[121

]a

• It may be noted that the following subsitution may be used sincef = ma and using f seems more logical to use as input. Keep inmind that both formulations are equivalent.

[12

T 2

T

]a =

12

T 2

m

Tm

f

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,

Modeling (3/4)

Classical mechanics

Description Symbol Formula Units

radius r - m

angular velocity ω dθdt

rad/sec

1 linear momentum p mv kg m/sec2 force F ma kg m/sec2 = N3 angular momentum L r × p = Iω kg m2/sec4 torque τ r × F kg m2/sec2 = N m5 moment of inertia I mr2 kg m2

First, focus only on blue, then focus only on green

http://en.wikipedia.org/wiki/Torque

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Modeling (4/4)

Damping

Applied force

displacement

damping coefficient,in this case, wall friction b

spring constant k

Oscillatory force(Hooke's Law)

Damping force

Net force

3constants

k, b, M

Mass M

Unitsk: N/m = kg/s2

b: N s/m=kg/s M: kg

Dorf pg 45, http://en.wikipedia.org/wiki/Damping

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,

Transformers (1/12)

Magnetic circuit

• A Transformer is a device that changes AC electric power at onevoltage level to AC electric power at another voltage level throughthe action of a magnetic field.

• It consists of two or more coils of wire wrapped around a commonferromagnetic core. These coils are not directly connected. Theonly connection between the coils is the common magnetic fluxpresent within the core.

Chapman, pg 18








,

Transformers (2/12)

Turn ratios

Vp

Vs= Is

Ip=

Np

Ns= a

Vp/IpVs/Ip

= a

⇒ Vp/IpVs/(Is/a) = a

⇒ Zp

Zs= a2

Chapman, pg 89

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,

Transformers (3/12)

Equivalent circuit

• The losses that occur in real transformers have to be accountedfor in any accurate model of transformer behavior.

• The major items to be considered in the construction of such a

model are:

• Windings: Copper I 2R losses• Windings: Leakage flux• Core: Eddy current losses• Core: Hysteresis losses

• It is possible to construct an equivalent circuit that takes intoaccount all the major imperfections in real transformers.

Chapman 5th ed, Sec 2.5, pg 86-94

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,

Transformers (4/12)

Equivalent circuit # 1


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,

Transformers (5/12)



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,

Transformers (6/12)


• We will mostly be using the simplified equivalent circuit givenbelow

• The magnetizing branch has been moved to make calculationseasier


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,

Transformers (7/12)


• A very simplified equivalent circuit that will not be used much

• The magnetizing branch has been completely eliminated


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,

Transformers (8/12)

Equivalent circuit

For the magnetizing branch,

Resistance, (Ω) = Rc

Reactance, (Ω) = Xm

Impedance, (Ω) = ZE

= Rc//jXm

= jRc Xm

Rc +jXm

Conductance, (Siemens) = Gc = 1Rc

Susceptance, (Siemens) = Bm = 1Xm

Admittance, (Siemens) = YE = 1ZE

= Rc +jXm

jRc Xm

= 1Rc− j 1

Xm


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,

Transformers (9/12)

Equivalent circuit

Open Circuit Test

• One transformer winding is open-circuited and theother winding is connected to full rated line voltage

•


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Transformers (10/12)

Autotransformer

VCVSE

= ISEIC

= NCNSE

VLVH

= IHIL

= NCNSE +NC

SWSIO

= NSENSE +NC

Chapman, pg 110-113

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Autotransformer

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,


Regulation

• Because a real transformer has series impedance within it, theoutput voltage of a transformer varies with the load if the inputvoltage remains constant

• To conveniently compare transformers in this respect, it iscustomary to define a quantity called voltage regulation (VR)

• Full-load voltage regulation is a quantity that compares theoutput voltage of the transformer at no load with the outputvoltage at full load

• It is defined as

VR =VS,nl−VS,fl

VS,fl× 100%

=Vpa−VS,fl

VS,fl× 100% since Vs =

Vpa

at no load

• Usually, it is good practice to have as small a voltage regulationas possible

• For an ideal transformer, VR=0 %









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MotorsDefinition

A motor is an electrical machine that coverts electricalenergy to mechanical energy


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,

MotorsTheory

The figure below shows a conductor present in a uniformmagnetic flux density B, pointing into the page. The conductor is `meters long and contains a current of i amperes.

The force induced on the conductor is given by,

F = i(`× B)


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MotorsTheory cont.

The direction of ` defined to be in the direction of current flow

The direction of the force is given by the right hand rule

(see Example 1.7 )


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MotorsTypes

WoundRotor

SquirrelCage

ShadedPole Capacitor Split

Phase

CapacitorStart

PermanentSplit

Capacitor

Two ValveCapacitor

ReluctanceStart

WoundField

Perm.Magnet

Reluctance HysteresisMultipleSpeedPole

Switching

suonorhcnySnoitcudnI

Single/PolyphaseSingle-PhasePolyphase

MultipleSpeed

SingleSpeed

SynchronousPhase-Locked LoopSteppers

Synchronous Induction

Switched SynchronousReluctanceReluctance

ReluctancePerm.

MagnetInverter PM Assisted

SynchronousReluctance

Driven

RotorControl

StatorControl

Perm.Magnet

WoundRotor

ElectronicCommu-

tationHybrid

VariableFrequency

BrushlessDC Motor

SquareDrive

SineDrive

Series

AC-DC Split Field

ConventionalConstruction

MovingCoil

DCTorquer

dnuopmoCtengaM .mrePtnuhS

(universal)

(bru

shed

)

SMMA, The Motor & Motion Association, http://www.smma.org/technical-info.htm- The words ”universal” and ”brushed” have been added later- All these motors are rotating motors, linear DC and AC motors also exist ,

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http://www.smma.org/technical-info.htm








,

MotorsTypes cont.

• In this course, we aim to study the following six types of motors:

1 DC linear2 DC brushed

3 AC synchronous4 AC induction

5 Electronically controlled: brushless (BLDC)6 Electronically controlled: stepper

• In the next slide, we present the voltages on the rotor and statorfor these kinds of motors, followed by a uniform graphicalrepresentation of magnetic, electrical and mechanical signals

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MotorsComparison of voltages on rotor and stator

Rotor(DC voltage)

Rotor(no voltage)

Rotor(permanent magnet)

Stator (DC voltage)1. Linear DC motor(Strictly speaking,should not use theword ”rotor” heresince there is linearmotion)

- -

Stator(DC voltage appliedthrough commuta-tor)

Mechanicalcommutation2. Brushed DCmotor

- Electroniccommutation5. Brushless DC(BLDC)motor

6. Stepper motorStator(AC 3-phase) 3. Synchronous AC

motor4. Induction ACmotor

-

http://electronics.stackexchange.com/questions/93710/

how-do-dc-motors-work-with-respect-to-current-and-what-consequence-is-the-curre,

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http://electronics.stackexchange.com/questions/93710/how-do-dc-motors-work-with-respect-to-current-and-what-consequence-is-the-curre

http://electronics.stackexchange.com/questions/93710/how-do-dc-motors-work-with-respect-to-current-and-what-consequence-is-the-curre

1. Linear DC MotorElectrical, magnetic and mechanical signal flow

Electromagnet (linearly moving conductor)

Magneticflux density

Magneticfield intensity

KVLAmpere's

LawMaterial

propertiesMagnetic "current"

Magneticflux

Magnetomotiveforce (mmf)

Caterfor turns

CurrentApplied DC

voltage

+-

Inducedvoltage Lorentz

forceNewton's2nd Law

Faraday'sLaw

1

23

4

Electromagnet (stator)



KVLAmpere's

LawMaterial


Magneticflux


Caterfor turns

CurrentApplied DC

voltage

,

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2. Brushed DC MotorElectrical, magnetic and mechanical signal flow




KVLAmpere's

LawMaterial


Magneticflux


Caterfor turns

CurrentApplied DC

voltage

Electromagnet (rotor)



KVLAmpere's

LawMaterial


Magneticflux


Caterfor turns

CurrentApplied DC

voltage

+-

Inducedvoltage

Mechanicalcommutation

Lorentzforce

TorqueNewton's2nd Law

Faraday'sLaw

1

23

4

,

59 / 412

3. AC Synchronous MotorElectrical, magnetic and mechanical signal flow




KVLAmpere's

LawMaterial


Magneticflux


Caterfor turns

CurrentApplied 3-phase AC

voltage




KVLAmpere's

LawMaterial


Magneticflux


Caterfor turns

CurrentApplied DC

voltage

+-

Inducedvoltage

Slip rings(rotary joints)

Lorentzforce


Faraday'sLaw

1

23

4

rotating

,

60 / 412

4. AC Induction MotorElectrical, magnetic and mechanical signal flow




KVLAmpere's

LawMaterial


Magneticflux


Caterfor turns

CurrentApplied 3-phase AC

voltage




KVLAmpere's

LawMaterial


Magneticflux


Caterfor turns

Current

Inducedvoltage Lorentz

force


Faraday'sLaw

2

34

1

rotating

,

61 / 412

5. Brushless DC (BLDC) MotorElectrical, magnetic and mechanical signal flow




KVLAmpere's

LawMaterial


Magneticflux


Caterfor turns

CurrentApplied DC

voltage

+Electrical

commutation

- 1

Permanent magnet (rotor)


Magnetic "current"

Magneticflux


Faraday'sLaw

23

4

,

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,

MotorsFeature Comparison

http://www.nidec.com/en-NA/technology/capability/brushless/

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,

MotorsWindings

There are 2 kinds of windings in electromechanical machines:

1 Field winding: In general, this term applies to the windings that

produce the main magnetic field

• For synchronous machines, the field windings are on therotor (Chapman, pg 267)

• For DC machines, the field windings are on the stator(Chapman, pg 520)

2 Armature winding: This term applies to the windings where themain voltage is induced (Chapman, pg 267, 520)

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,

Linear DC Motor (1/12)

Overview

• A linear DC motor is the simplest and easiest-to-understand DCmotor

• Yet, it operates according to the same principles and exhibits thesame behavior as real motors


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,


Overview cont.

• A linear DC motor is shown below

• It consists of a battery and a resistance connected through aswitch to a pair of smooth, frictionless rails

• Along the bed of this ”railroad track”, is a constant,uniform-density magnetic field directed into the page

• A bar of conducting metal is lying across the tracks


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,


Overview cont.

• The behavior of the linear DC motor, like any DC motor, is

governed by four equations that come into play in the following

sequence:

1 Kirchoff’s Law i = VB−eindR

2 Lorentz Force F = i(`× B)3 Newton’s 2nd Law Fnet = ma4 Faraday’s Law eind = (v × B).`


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,


Starting at no load

To start the motor, simply close the switch. After this, thefollowing sequence of events happens:

1 Kirchoff’s Law: compute current

• A current flows in the bar which is given by i = VB−eindR

• Since the bar is initially at rest, eind = 0 and so i = VBR• The current flows down through the bar across the tracks

2 Lorentz Force: compute force

• A current flowing through a wire in the presence of amagnetic field induces a force on the wire

• This force is F = i`B to the right

3 Newton’s 2nd Law: compute acceleration

• The bar will accelerate to the right (due to Newton’s Law)• The velocity of the bar begins to increase

4 Faraday’s Law: compute induced voltage

• A voltage appears across the bar which is given byeind = vB`

• This voltage reduces the current in the bar due toKirchoff’s Law (back to step 1!)


68 / 412








,


Starting at no load cont.

Given below is the linear DC motor under starting conditions and noload.


69 / 412








,



• The result of this action is that the bar will eventually reach aconstant steady-state speed where the net force on the bar is zero

• This will occur when eind has risen all the way up to equal thevoltage VB

• At this time, the bar will be moving at a speed given by

VB = eind = vss B`, and so vss = VBB`

• The bar will continue to coast along at this no-load speed foreverunless some external force disturbs it (Newton’s first law ofmotion)

• This is precisely the behavior observed in real motors on starting

• On the next slide, we show the velocity v , induced voltage eind

and induced force Find , from when the motor is started till itstarts running at no-load steady-state


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,




71 / 412








,


Applying an external load

• Assume that the linear DC motor is initially running at theno-load steady-state conditions described previously

• What will happen to this motor if an external load is applied to it?

• Examine the figure below where the load is applied to the baropposite to the direction of motion

• Since the bar was initially moving with steady state velocity,application of the force Fload will result in a net force on the bar inthe direction opposite the direction of motion (Fnet = Fload −Find )

• The effect of this force will be to slow the bar


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,


Applying an external load cont.

• But just as soon as the bar begins to slow down, the inducedvoltage on the bar drops

• As the induced voltage decreases, the current flow in the bar rises

• Therefore the induced force rises too

• The overall result of this chain of events is that the induced forcerises until it is equal and opposite to the load force, and the baragain travels in steady state, but at a slower speed

• On the next slide, we show the velocity v , induced voltage eind

and induced force Find , from when a load is attached to a motorrunning at steady state, and compare with starting at no load


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,




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,



• A question that can come to mind is, why is the steady statevelocity slower than before?

• Remember that the force that the motor must supply hasincreased, and since power P is a product of induced force Find

and velocity v , the velocity must decrease

• The power consumed by the bar is eind i

• This power is converted to Find v

• Therefore, Pconv = eind i = Find v


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,


Construction

https://www.youtube.com/watch?v=o_VjkUTZQXg

76 / 412

https://www.youtube.com/watch?v=o_VjkUTZQXg








,

Linear DC generator (1/2)

Operation

• Once again, consider the Linear DC machine initially running atno-load steady-state conditions

• Now, what will happen if we apply a force in the direction ofmotion to it?

• See the figure below

• Fapp is applied to the bar in the direction of motion

Chapman, pg 41

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,

Linear DC generator (2/2)

Operation cont.

1 Increasing velocity and voltage Since the bar was initially atsteady state, application of the force Fapp will result in a net forceon the bar in the direction of motion Fnet = Fapp − Find . Theeffect of this force will be to speed up the bar causing the inducedvoltage eind to increase and become more than VB .

2 Increasing reverse current and force As the induced voltageincreases, the current i starts to increase in the reverse direction.This creates an increasing induced force to the left.

New steady state (faster constant velocity) The overall result of this

chain of events is that the induced force increases till it is equal and

opposite to the applied force and the bar again travels in steady state,

but at a faster speed

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,

Brushed DC Motor (1/86)

Introduction

• A very simple motor can be made from two permanent magnets,one static, one able to rotate, and the interaction of thesemagnets creates rotation

• But there is a problem here, the rotating magnet will not rotate ifits north pole is aligned with the stationary magnet’s south pole

• So, we need to keep changing polarities of the rotating magnet, aprocess called commutation

commutation

commutation

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,


Introduction cont.

• How to change polarities, i.e, how to do commutation?

• Well, first of all, make the rotating magnet an electromagnet sowe have control over its polarities

• Now, there are 2 ways of changing polarities of the electromagnet

1 Mechanical commutation: This gives us a brushed DCmotor

2 Electrical commutation: This gives us a brushless DC motor(BLDC)

• This gives us the simplest DC motor

• Simplest DC motor: consists of one permanent magnet and one

electromagnet

• The permanent magnet produces a uniform magnetic field• The electromagnet is made from a simple DC current

carrying loop• Let us see a couple of animations of this before getting into

the mathematics and explanation

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,


Definitions

• Mechanical

• Rotor: The rotating part of the motor.• Stator: The stationary part of the motor.

• Electrical

• Armature: The power-producing component of the motor.The armature can be on either the rotor or the stator.

• Field: The magnetic field component of the motor. Thefield can be on either the rotor or the stator and can beeither an electromagnet or a permanent magnet.

• For a brushed DC motor, the armature is on the rotor and thefield is on the stator

• The armature circuit is represented by an ideal voltage source EA

(also written as eind ) and a resistor RA.

• This representation is really the Thevenin equivalent of the entirerotor structure, including rotor coils, interpoles, and compensatingwindings, if present.

http://en.wikipedia.org/wiki/Armature_(electrical_engineering)


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http://en.wikipedia.org/wiki/Armature_(electrical_engineering)








,


Definitions cont.

• The distortion of the flux in a machine as the load is increased iscalled armature reaction.

• To take care of this, compensating windings are connected inseries with the rotor windings, so that whenever the load changesin the rotor, the current in the compensating windings changes,too

Chapman 5th ed, pg 433 (armature reaction), 443 (compensating windings)

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Single rotating loop in uniform magnetic field (1/15)

http://web.ncf.ca/ch865/englishdescr/DCElectricMotor.html ,

83 / 412

http://web.ncf.ca/ch865/englishdescr/DCElectricMotor.html








,



• On the previous animation, the method of connecting the wire tothe commutator is not shown

• This is done through brushes

• On the next slide, we look at another animation to get a betterfeel for how a DC current carrying loop placed in a magnetic fieldworks

• This animation clearly shows brushes

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,



https:

//nationalmaglab.org/education/magnet-academy/watch-play/interactive/dc-motor

85 / 412

https://nationalmaglab.org/education/magnet-academy/watch-play/interactive/dc-motor

https://nationalmaglab.org/education/magnet-academy/watch-play/interactive/dc-motor








,



• The Lorentz force is given by F = i(`× B)

• The direction of ` defined to be in the direction of current flow

• The direction of the force is given by the right hand rule

• Note that there is zero force on the wire sides that are parallel tothe magnetic flux B

• When the loop is in the horizontal position, current flow isstopped and it tips over using its momentum


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,



• The figure below shows a simple DC motor consisting of a largestationary magnet producing an essentially constant and uniformmagnetic field B and a DC current carrying loop of wire abcdplaced within that field.

• The rotating part of the motor, the loop, is called the rotor.

• The stationary part of the machine, the stationary magnet, iscalled the stator.


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,



• The magnetic field B always points to the right and is in theplane of the paper

• Segments ab and cd are always out of the plane of the page andare perpendicular to B

• Segments bc and da are always in the plane of the page and arecontinuously changing angles with B


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,



Segment ab

• Lorentz force F = i(`× B)

• The angle between ` and B is always 90 deg• The induced force is Fab = i`B down

• Torque τ = r × F

• The angle between r and F changes between 0 and 90 deg• The induced torque τab = ri`B sin(θab) clockwise


89 / 412








,



Segment bc


• In this segment, the angle between ` and B changesbetween 0 and 180 deg

• The induced force is Fbc = i`B into the page


• The angle between r and F is always 0 deg• The induced torque τbc = 0


90 / 412








,



Segment cd


• The induced force is Fcd = i`B up.• The angle between ` and B is always 90 deg


• The angle between r and F changes between 0 and 90 deg• The induced torque τcd = ri`B sin(θcd ) clockwise


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,



Segment da


• In this segment, the angle between ` and B changesbetween 0 and 180 deg

• The induced force is Fda = i`B out of the page.


• The angle between r and F is always 0 deg• The induced torque τda = 0


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,



• Torque is only produced by segments ab and cd

• θab = θcd = θ

• The total induced torque is τind = 2ri`B sin θ

• Notice that the torque is maximum when the plane of the loop isparallel to the magnetic field, and the torque is 0 when the planeof the loop is perpendicular to the magnetic field

• Given below is the variation of torque as the loop rotates


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,



Define Bloop = µiG, G depends on the geometry of the loop

⇒ i =Bloop G

µ

τind = 2r i`Bs sin θ B=Bs (s for stator) to distinguish from Bloop

= 2rBloop G

µ`Bs sin θ Substitute i =

Bloop G

µ

= AGµ

BloopBs sin θ Substitute A = 2r` is the area of the loop

= kBloopBs sin θ k depends on the construction of the machine

= kBloop × Bs

• θab=θcd =θ is also the angle between Bloop and Bs


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,



τind = kBloop × Bs

• This produces a torque vector into the page, indicating that thetorque is clockwise, with the magnitude given by kBloopBs sin θ

• Thus, the torque produced in the loop is proportional to

• The strength of the loop’s magnetic field• The strength of the external magnetic field• The sine of the angle between them• A constant representing the construction of the machine

(geometry, etc.)


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,



• Now, mapping our newly created Bloop onto segments ab and cd ,shown in the left and right figures below


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,



• τind = kBloop × Bs

• τind is directed into the plane of the paper, i.e., the torqueis clockwise

• The torque induced in the loop is proportional to thestrength of the loop’s magnetic field, the strength of theexternal magnetic field, and the sine of the angle betweenthem

• This equation also shows that if there are 2 magnetic fieldspresent in a machine, a torque will be created that will tendto line up the magnetic fields

• The torque therefore depends on

1 Rotor magnetic field2 Stator magnetic field3 Sine of the angle between them4 A constant representing the construction of the machine

(geometry etc.)


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,


Single rotating loop in magnetic field generated bycurved pole faces(1/3)

• The loop of rotor wire lies in a slot carved in a ferromagnetic core• The iron rotor, together with the curved shape of the pole faces,

provides a constant-width air gap between the rotor and stator• The reluctance of air is much higher than the reluctance of the

iron in the machine


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,



• To minimize the reluctance of the flux path through the machine,the magnetic flux must take the shortes t possible path throughthe alr between the pole face and the rotor surface

• Since the magnetic flux must take the shortest path through theair, it is perpendicular to the rotor surface everywhere under thepole faces

• Also, since the air gap is of uniform width, the reluctance is thesame everywhere under the pole faces

• The uniform reluctance means that the magnetic flux density isconstant everywhere under the pole faces


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,



• As before, the torque is τind = 2ri`B sin θ = 2ri`B, since θ = 90o

• Since there are two poles, the area of the rotor under each pole(ignoring the small gaps between poles) is Ap = πrl

• Therefore, φ = BAp

• We can therefore rewrite τind = 2π

Ap iB = 2πφi

• Thus, the torque produced in the machine is the product of theflux in the machine and the current in the machine, times somequantity representing the mechanical construction of themachine (the percentage of the rotor covered by pole faces)

• In general, the torque in any real machine will depend on th e

same three factors:

1 The flux in the machine2 The current in the machine3 A constant representing the construction of the machine


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,


Working

http://www.learnengineering.org/2014/09/DC-motor-Working.html

101 / 412









,


Types

1 Separately excited (pg 468)

• Field circuit is supplied from a separate constant-voltagepower supply

2 Shunt (parallel) (pg 469)

• Field circuit gets its power directly across the armatureterminals

3 Series (pg 493)

• Field windings consist of a relatively few turns connected inseries with the armature circuit

4 Compound (pg 500)

• A motor with both a shunt and series field

5 Permanent magnet (pg 491)

• Field comes from a permanent magnet rather than a circuit


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,


Type # 1: Separately excited

• The equivalent circuit of a DC motor is given below• In this figure, the armature circuit is represented by an ideal

voltage source EA and a resistor RA

• The brush voltage drop is represented by a small battery Vbrush

opposing the direction of current flow in the circuit• The field coils, which produce the magnetic flux, are represented

by inductor LF and resistor RF

• The separate resistor Radj represents an external variable resistorused to control the amount of current in the field circuit


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,


Type # 1: Separately excited cont.

• There are a few variations and simplifications of the basicequivalent circuit

• The brush drop voltage is often small, and therefore in caseswhere it is not too critical, the brush drop voltage may be left outor approximately included in the value of RA

• Also, the internal resistance of the field coils is sometimes lumpedtogether with the variable resistor, and the total is called RF


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,



So, there are 4 equations required to analyze a DC motor:

1 KVL, IA = VT−EARA

2 The induced torque τind = KφIA

3 The internally generated voltage EA = Kφω

4 The magnetization curve relates EA with the field current IF

1.

2.

Armature

4. Magnetization curve

Relation betweenfield circuit andarmature circuit

3.


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ampere-turns

webers

Magnetization curve of a ferromagnetic material Magnetization curve of a DC motor

Chapman 5th ed, pg 467-469 ,

106 / 412



ampere-turns

webers

Magnetization curve of a ferromagnetic material Magnetization curve of a DC motor (3D)


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,



So, how can we use the magnetization curve?

• IF → φ

• If I change my field current IF by a certain ratio, the ratiowith which the resulting flux φ changes is linear up to acertain point before saturation sets in

• Using the magnetization curve, if I know the ratio withwhich IF changes, I can find the ratio with which the flux φchanges despite the non-linearity due to saturation

• So, for IF 1 and IF 2, read the corresponding EA1 and EA2

from the magnetization curve• Remember that the magnetization curve is given for a fixed

value of ω• Then,

EA1EA2

= Kφ1ωKφ2ω

⇒ φ1φ2

=EA1EA2

• This idea is used in Example 8.3


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,


Type # 2: Shunt

• In a separately excited motor, two power supplies are used,

1 VF to supply the field circuit2 VT to supply the armature circuit

• If only one power supply is used for both field and armaturecircuits, we get a shunt DC motor

Therefore,a shunt DC motor is equivalent to aseparately excited DC motor,as long as VF = VT


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,


Type # 2: Shunt cont.


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Motor winding on left and terminal characteristics on right

+

-

+ -

http://www.learnengineering.org/2014/09/DC-motor-Working.html ,

111 / 412









,



• The voltage supplied by the user, VT , which is constant in most

cases and is parallel to VF , is used for the generation of 2 kinds of

currents:

1 Stator: Field current IF which generates a magnetic fieldφF .

2 Rotor: Armature current IA which generates a magneticfield whose interaction with φF causes the rotor to rotate,in turn inducing a voltage EA

• Therefore, the current supplied by the user, the load current, canbe given by IL = IF + IA


112 / 412








,



• How does a shunt dc motor respond to a load?

• Suppose that the load on the shaft of a shunt motor is increased

• Step 2: Then, the load torque τload will exceed induced torqueτind = KφIA

• Step 3: The motor will start to slow down

• Step 4: When the motor slows down, its internal generatedvoltage EA = Kφω drops

• Step 1: This causes the armature current to increase, sinceVT = EA + IARA

• Step 2: As the armature current increases, so does the inducedtorque until it equals the load torque at a lower mechanical speedof rotation


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,



• For a motor, the output quantities are shaft torqueand speed

• Therefore, the terminal characteristic of a motor is a plot of itsoutput torque versus speed

VT = EA + IARA

= Kφωm + τindKφ

RA

⇒ ωm = VTKφ

− RA(Kφ)2 τind

• This equation is just a straight line with a negative slope


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,



Speed control can be achieved by

1 Adjusting the field resistance RF and thus the field flux

2 Adjusting the terminal voltage applied to the armature

3 Inserting a resistor in series with the armature circuit (lesscommon)


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,


Type # 3: Series

• A series DC motor is a DC motor whose field windings consist ofa relatively few turns connected in series with the armature circuit

• The equivalent circuit is shown below

• Armature current, field current and line current are the same

• KVL isVT = EA + IA(RA + RS )


116 / 412


Type # 3: Series cont.

Motor winding on left and terminal characteristics on right

http://www.learnengineering.org/2014/09/DC-motor-Working.html ,

117 / 412









,



• The terminal characteristics of a series DC motor is very differentfrom that of the shunt motor

• The basic behavior of a series DC motor is due to the fact thatthe field flux is directly proportional to the armature current(φ ∝ IA), at least until saturation is reached

• As the load on the motor increases, its armature current increases,and so does the field flux

• An increase in flux decreases the speed of the motor

• So we have a ”double drop” in velocity

• Therefore, a series DC motor has a sharply drooping torque-speedcharacteristic


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,



• The equations are

τind = KφIAφ = cIA

⇒ τind = KcIA2

• Since torque is directly proportional to the armature currentsquared, the series DC motor gives more torque per ampere thanany other DC motor

• It is therefore used in applications requiring very high torque

• Examples of such applications are the starter motors in cars,elevator motors, and tractor motors in locomotives


119 / 412



• To determine the terminal characteristics of a series DC motor, an analysis will becarried out based on the assumption of a linear magnetization curve

• In a magnetization curve, we plot φ vs IF , but since IA = IF , it is a plot of φ vs IA, implyingthat φ = cIA

• As shown earlier,τind = KcIA

2 (but IA = φc)

= Kcφ2

⇒ φ =√

cK

√τind

• The KVL equation is,

VT = EA + IA(RA + RS )

= Kφω +√τindKc

(RA + RS )

= K√

cK

√τindω +

√τindKc

(RA + RS )

VT −√τindKc

(RA + RS ) =√

Kc√τindω

⇒ ω = VT√Kc√τind− RA+RS

Kc

• A problem here is that if τind = 0, then its speed goes to ∞


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,



• In practice, the torque can never go to zero because of themechanical, core and stray losses that must be overcome

• However, if no other load is connected to the motor, it can turnfast enough to seriously damage itself

• Never completely unload a series motor, and never connect one toa load by a belt or other mechanism that could break

• If that were to happen, and the motor were to become unloadedwhile running, the results could be serious


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,



• Unlike with the shunt DC motor, there is only one efficient way tochange the speed of a series DC motor

• This method is to change the terminal voltage of the motor

• If the terminal voltage is increased, the first term in

ω = VT√Kc√τind− RA+RS

Kcincreases, resulting in a higher speed for

any given torque

• Until the last 40 years or so, there was no convenient way tochange VT , so the only method of speed control available was thewasteful series resistance method

• That has all changed today with the introduction of solid-statecontrol circuits


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,


Type # 4: Compound

• A compounded DC motor has both a shunt (parallel) and a seriesfield

• There are 2 ways to connect this motor, long shunt and shortshunt

• So, there are 2 field coils and one armature coil

• If the mmf of the shunt field coil enhances the mmf of the seriesfield coil, the situation is called cumulative compounding

• If the mmf of the shunt field coil diminshes the mmf of the seriesfield coil, the situation is called differential compounding

• The advantage of this motor is that it combines the speedregulation of a shunt motor with the high starting torque of aseries motor


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,


Type # 4: Compound: long shunt

Chapman 5th ed, pg 500-505http://www.electrical4u.com/compound-wound-dc-motor-or-dc-compound-motor/

124 / 412

http://www.electrical4u.com/compound-wound-dc-motor-or-dc-compound-motor/








,


Type # 4: Compound: short shunt

Chapman 5th ed, pg 500-505http://www.electrical4u.com/compound-wound-dc-motor-or-dc-compound-motor/

125 / 412

http://www.electrical4u.com/compound-wound-dc-motor-or-dc-compound-motor/








,


Type # 5: Permanent magnet

http://autosystempro.com/tag/motor/

126 / 412

http://autosystempro.com/tag/motor/


Comparison of equivalent circuits

3. SERIES

2. SHUNT1. SEPARATELY EXCITED

5a. COMPOUNDED(cumulatively)

5b. COMPOUNDED(differentially)

t

4. PERMANENT MAGNET

,

127 / 412








,


Power flow and losses


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,


Efficiency


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Modeling

Dorf pg 63-65 ,

130 / 412


Modeling cont.

Laplace Domain

1

2

plug I(s) from eqn 1 into eqn 2+

-

+

-

angular velocity (rad/sec) multiply by 60/2pi to go to rpm

angular distance (rad)

Typical values are:R: electric resistance 1 OhmL: electric inductance 0.5 HJ: moment of inertia of the rotor 0.01 kg.m^2b: motor viscous friction constant 0.1 N.m.sKb: electromotive force constant 0.01 V/rad/secKm: motor torque constant 0.01 N.m/AmpGiving:

1

2motor torqueload torque

back emfarmature voltage

Time Domain

1

2

differential equations

state space

Dorf uses va, ia, Ra, La, while we use v, i, R, L for armature values

Dorf pg 63-65, http://ctms.engin.umich.edu/CTMS/index.php ,

131 / 412

http://ctms.engin.umich.edu/CTMS/index.php


Modeling cont.

Laplace Domain

1

2

plug I(s) from eqn 1 into eqn 2+

-

+

-

angular velocity (rad/sec) multiply by 60/2pi to go to rpm

angular distance (rad)

Typical values are:R: electric resistance 1 OhmL: electric inductance 0.5 HJ: moment of inertia of the rotor 0.01 kg.m^2b: motor viscous friction constant 0.1 N.m.sKb: electromotive force constant 0.01 V/rad/secKm: motor torque constant 0.01 N.m/AmpGiving:

1

2motor torqueload torque

back emfarmature voltage

Time Domain

1

2

differential equations

state space

Dorf uses va, ia, Ra, La, while we use v, i, R, L for armature values

Dorf pg 63-65, http://ctms.engin.umich.edu/CTMS/index.php ,

132 / 412

http://ctms.engin.umich.edu/CTMS/index.php








,


Modeling cont.

In z domain, the open loop transfer function of a DC motoris given by,

G(z) = Z

G0(s)Gp(s)

= Z(

1−e−sT

s

)(2

s2+12s+20.02

)

= (1− z−1)Z

2s3+12s2+20.02s

= (1− z−1)Z

0.0999s− 0.1249

s+2.0025+ 0.025

s+9.9975

= (1− z−1)

0.09991−z−1 − 0.1249

1−e−2.0025T z−1 + 0.0251−e−9.9975T z−1

= 0.0999− 0.1249(1−z−1)

1−e−2.0025T z−1 + 0.025(1−z−1)

1−e−9.9975T z−1

133 / 412








,


Modeling cont.

[x1

x2

]=

[−R/L −Kb/LKm/J −b/J

] [x1

x2

]+

[1/L

0

]v

⇒[

x1

x2

]=

[−2 −0.021 −10

] [x1

x2

]+

[20

]v

y =[0 1

] [x1

x2

]

134 / 412








,


Modeling cont.

C(sI− A)−1B =[0 1

] [s + 2 0.02−1 s + 10

]−1 [20

]

=[0 1

]s + 10 1−0.02 s + 2

T

(s+2)(s+10)−(0.02)(−1)

[20

]

=[0 1

]s + 10 −0.02

1 s + 2

s2+12s+20.02

[20

]

=

[1 s − 2

]20

s2+12s+20.02

= 2s2+12s+20.02

135 / 412


Modeling cont.

G1(s) =θ(s)

V(s)=

1

s

Km

[(Ls + R)(Js + b) + KbKm]

Gp(s) =θ(s)

V (s)=

Km

[(Ls + R)(Js + b) + KbKm]

Note that we have set Td (s) = 0 to compute G1(s) and Gp (s).

Dorf pg 64 ,

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,


Modeling cont.

A motor can be represented simply as an integrator. Avoltage applied to the motor will cause rotation. Whenthe applied voltage is removed, the motor will stop andremain at its present output position. Since it does notreturn to its initial position, we have an angulardisplacement output without an input to the motor.See Nise pg 381 for a discussion on finding the transferfunction of a motor.See Nise pg 451 for a nice motor transfer functiondiagram.

Nise pg 343

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,


Computing parameters

• A brushed DC motor has 6 parameters, but we have to measure 5:

1 Armature resistance Ra

2 Armature inductance La

3 Moment of inertia of the rotor J4 Viscous friction coefficient B5 Back emf constant Kb = Torque constant KT

• The first 4 parameters can be seen in the figure of the armaturebelow:

• The equation is given by ea = iaRa + Ladiadt

+ eb

• The input voltage is ea, the resultant current is ia and eb is theback EMF

https://www.coursehero.com/file/1801696/ge320Lab2/

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,


Computing parameters cont.

Step 1. Find Ra

1 Original equation: ea = iaRa + Ladiadt

+ eb

2 Rotation: no, therefore eb = 0 since eb ∝ ω3 Response: steady state, therefore dia

dt= 0

4 Extra steps: none

5 New equation: ea = iaRa

6 Measure: ea, ia

Step 2. Find La


+ eb

2 Rotation: no, therefore eb = 0 since eb ∝ ω3 Response: transient

4 Extra steps: put a resistor Rs in series with the motor so that wecan then measure the voltage drop Vs across Rs to graphicallyobtain τ

5 New equation: ea = ia(Ra + Rs ) + Ladiadt⇒ τ = La

Ra+Rs

6 Measure: Vs (to get ia = Vs/Rs ) with time to get τ


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,



Step 3. Find Kb (KT = Kb)


+ eb

2 Rotation: yes

3 Response: steady state, therefore diadt

= 0

4 Extra steps: none

5 New equation: ea = iaRa + Kbω

6 Measure: ea, ia, ω

Step 4. Find B

1 Original equation: Tm = KT ia = J dωdt

+ Bω

2 Rotation: yes

3 Response: steady state, so dωdt

= 0

4 Extra steps: none

5 New equation: KT ia = Bω

6 Measure: ia, ω


140 / 412









,



Step 5. Find J

1 Original equation: Tm = KT ia = J dωdt

+ Bω

2 Rotation: yes

3 Response: transient

4 Extra steps: cut current so that ia = 0

5 New equation: 0 = J dωdt

+ Bω ⇒ τ = JB

6 Measure: ω with time to get τ


141 / 412



Construction: brushes but no commutator

https://www.youtube.com/watch?v=WKklyuzghQg ,

142 / 412

https://www.youtube.com/watch?v=WKklyuzghQg








,


Construction in IE workshop at CAE

• Basic structure

143 / 412








,


Construction in IE workshop at CAE

• Basic structure

144 / 412








,


Construction (Porter Cable 690 Router motor)

• This is the serviceable portion of the brush assembly.

• The unit consists of a graphite brush and integral spring assembly.

• Observe the curvature of the brush where it mates with themotor’s commutator.

• Also notice there is plenty of length remaining in this brush, somany more years of service may be expected from this brush.

http://www.ncwoodworker.net/forums/content.php?r=

33-Brush-Inspection-and-Maintenance-for-Universal-Motors

145 / 412

http://www.ncwoodworker.net/forums/content.php?r=33-Brush-Inspection-and-Maintenance-for-Universal-Motors









,



• Notice the smooth face where the brush mates with the motor’scommutator.

• A little bit of wear along the trailing edge can be seen, but this istypical of normal wear.

• A brush in good condition will look much like this brush – smoothfaces, plenty of length, and no signs of abnormal wear, arcing orpitting



146 / 412










,



• The photo below shows the brush housing (brass housing at left),the graphite brush (center, just visible between brush housing andmotor commutator), and the motor commutator (the circulararray of copper conduction strips).



147 / 412










,



• Notice how intimately the brush and commutator mate with oneanother, indicative of a well seated brush.

• Also notice no obvious damage, pitting, or overheating in thecommutator (the copper strips).



148 / 412










,



• The blackening is normal and is residue from the graphite brush –it also provides lubrication between the brush and commutator.



149 / 412










,



• A view from another angle of brush housing, brush, andcommutator.

• Also visible in the background are the motor windings.



150 / 412










,


Construction: rotor windings

• Three ways to classify1 connection (need a better word!)

• progressive• retrogressive

2 plex• simplex• duplex• triplex• multiplex

3 sequence• lap• wave• frog-leg

Chapman, pg 492

151 / 412








,


Construction: Wave winding

http://www.sciencedirect.com/science/article/pii/S0736584512000828

152 / 412








,


Construction: Wave winding

http://www.gotwind.org/forum/viewtopic.php?t=3545

153 / 412








,


Construction: Lap vs Wave winding

http://www.tpub.com/neets/book5/15g.htm

154 / 412








,


Construction: Lap vs Wave winding

• Lap winding• Advantage: If high current is required, it can be

split among several paths, so the size of individualrotor conductors remains reasonable

• Disadvantage: A very tiny imbalance among thevoltages in the parallel paths will cause largecirculating currents through the brushes andpotentially serious heating problems

• Wave (series) winding• Advantage: Can be used to build high-voltage DC

machines• Disadvantage:

Chapman, pg 493

155 / 412








,


Construction: Winding table

http://forumrc.alexba.eu/nutpol e.htm

156 / 412








,


Construction: Commutator

http://encyclopedia2.thefreedictionary.com/Commutation

157 / 412








,



http://www.daviddarling.info/encyclopedia/C/commutator.html

158 / 412








,


Construction: Commutator (clean)

http://homerecording.com/bbs/general-discussions/analog-only/reel-motors-tascam-34b-grind-halt-296058/

159 / 412








,


Construction: Commutator (dirty)

http://homerecording.com/bbs/general-discussions/analog-only/reel-motors-tascam-34b-grind-halt-296058/

160 / 412








,



http://mrmackenzie.co.uk/category/standard-grade/using-electricity/

161 / 412








,



http://www.rctech.net/forum/rookie-zone/522906-boosted.html

162 / 412








,



http://www.answers.com/topic/commutator

163 / 412








,

Brushed DC Generator (1/10)


• This is the simplest possible machine that produces a sinusoidal

ac voltage (and dc voltage with a commutator installed)

• This case is not representative of real ac machines, sincethe flux in real ac machines is not constant in eithermagnitude or direction

• However, the factors that control the voltage and torque onthe loop will be the same as the factors that control thevoltage and torque in real ac machines

Chapman, pg 230-238

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,



• The figure below shows a simple generator consisting of a largestationary magnet producing an essentially constant and uniformmagnetic field and a rotating loop of wire within that field.

• The rotating part of the machine is called the rotor.

• The stationary part of the machine is called the stator.


165 / 412








,



• e = (v × B).`

166 / 412








,



• (v × B)

• The magnetic field B always points to the right and is inthe plane of the paper

• The velocity v takes on every possible direction incounter-clockwise direction for all segments and is always inthe plane of the page

• v × B is therefore always out of the plane of the page

• (v × B).`

• Segments ab and cd are always out of the plane of the pageand so voltage is induced in them

• Segments bc and da are always in the plane of the page andso voltage is not induced in them

167 / 412








,



1 Induced voltage for segments in the plane of the page

1 Segment ab: eba = vB` sin(θab) into the page2 Segment cd: edc = vB` sin(180o − θcd ) =vB` sin(θcd ) out of the

page

2 Induced voltage for segments out of the plane of the page

1 Segment bc: ebc = 02 Segment da: eda = 0

168 / 412








,



• Since both induced emfs reinforce each other, the total inducedvoltage eind = 2vB` sin θ

169 / 412








,



• eind = 2vB` sin θ

• θ = ωt• v = rω (r is the radius of rotation)

• eind = 2rωB` sin(ωt)

• A = 2r` (area of the loop)

• eind = ABω sin(ωt)

• φmax = AB (maximum flux)

• eind = φmaxω sin(ωt)

• Therefore, the induced voltage is sinusoidal, and dependson the flux and the speed of rotation

170 / 412








,


Single rotating loop in magnetic field generated bycurved pole faces (1/3)

• Single loop of wire rotating about a fixed axis

• If the rotor is rotated, a voltage will be induced in the wire loop

given by Faraday’s Law


171 / 412








,



http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/motorac.html

172 / 412









,



This is the same as a single rotating loop in a uniformmagnetic field for the DC motor case except that the commutator isreplaced with slip rings



173 / 412









,

3φ Synchronous & Induction AC motorsRotating magnetic field (1/7)

• In a current carrying loop placed in a magnetic field,τind = kBloop × Bs

• This equation shows that if there are 2 magnetic fields present ina machine, a torque will be created that will tend to line up themagnetic fields

• If one magnetic field is produced by the stator of an ac machine,and the other one is produced by the rotor of the machine, then atorque will be induced in the rotor which will cause the rotor toturn and align itself with the stator magnetic field


174 / 412174 / 412








,


• If there were some way to make the stator magnetic field rotate,then the induced torque in the rotor would cause it to constantly”chase” the stator magnetic field around in a circle

• This in a nutshell, is the basic principle of all ac motor operation

• How can the stator magnetic field be made to rotate?

• The fundamental principle of ac machine operation is that if athree-phase set of currents, each of equal magnitude and differingin phase by 120o , flows in a three phase winding, then it willproduce a rotating magnetic field of constant magnitude

• The three-phase winding consists of three separate windingsspaced 120 electrical degrees apart around the surface of themachine


175 / 412175 / 412








,


• The rotating magnetic field concept is illustrated inthe simplest case by an empty stator containing justthree coils, each 120o apart

• Since such a winding effectively produces only one north and onesouth pole, it is a two pole winding

• In the figure below, current• iaa(t) in coil aa flows into the a end and out of the a′ end• ibb(t) in coil bb flows into the b end and out of the b′ end• icc (t) in coil cc flows into the c end and out of the c ′ end


176 / 412176 / 412


• The orientation of the three coils in the previous figure can be visualized better in this figure• The three coils are placed 120o apart• Notice the resultant rotating magnetic field, shown by the green letters N and S

http://www.learnengineering.org/2013/08/

three-phase-induction-motor-working-squirrel-cage.html ,

177 / 412177 / 412

http://www.learnengineering.org/2013/08/three-phase-induction-motor-working-squirrel-cage.html

http://www.learnengineering.org/2013/08/three-phase-induction-motor-working-squirrel-cage.html








,


• The currents in the three coils are given by

iaa′ (t) = IM sin(ωt) Aibb′ (t) = IM sin(ωt − 120o ) Aicc′ (t) = IM sin(ωt − 240o ) A

• The magnetic field intensites are given as follows. The angles arespatial angles.

Haa′ (t) = HM sin(ωt)∠0o A.turns/mHbb′ (t) = HM sin(ωt − 120o )∠120o A.turns/mHcc′ (t) = HM sin(ωt − 240o )∠240o A.turns/m

Baa′ (t) = BM sin(ωt)∠0o TBbb′ (t) = BM sin(ωt − 120o )∠120o TBcc′ (t) = BM sin(ωt − 240o )∠240o T

• The magnetic flux densities are given by

Baa′ (t) = BM sin(ωt)∠0o TBbb′ (t) = BM sin(ωt − 120o )∠120o TBcc′ (t) = BM sin(ωt − 240o )∠240o T


178 / 412178 / 412


• We now compute the net magnetic flux density

Bnet (t) = Baa′ + Bbb′ + Bcc′

= BM sin(ωt)∠0o + BM sin(ωt − 120o )∠120o + BM sin(ωt − 240o )∠240o

= BM sin(ωt)x−0.5BM sin(ωt − 120o )x +

√3

2BM sin(ωt − 120o )y−

0.5BM sin(ωt − 240o )x−√

32

BM sin(ωt − 240o )y

=

(BM sin(ωt)− 0.5BM sin(ωt − 120o )− 0.5BM sin(ωt − 240o )

)x+(√

32

BM sin(ωt − 120o )−√

32

BM sin(ωt − 240o )

)y

= 1.5BM sin(ωt)x− 1.5BM cos(ωt)y

• The magnetic flux density is a constant 1.5BM and the angle changes continually in acounterclockwise direction at angular velocity ω.


179 / 412179 / 412








,


• The rotating magnetic field of the stator can be represented as anorth pole and a south pole


180 / 412180 / 412








,

Simplest AC Motor (1/1)

Single rotating loop in uniform magnetic field

This is the same as a single rotating loop in a uniformmagnetic field for the DC motor case except that the commutator isreplaced with slip rings



181 / 412









,

Synchronous MotorsIntroduction

• Stator: Rotating magnetic field (created by 3-phase AC currents)

• Rotor: Fixed magnetic field (created by DC current)

Synchronous vs Induction motorStator sameRotor DC field vs no DC field on the rotor

• The basic principle of synchronous motor operation is that therotor ”chases” the rotating stator magnetic field around in acircle, never quite catching up with it

Chapman, pg 346

182 / 412








,

Synchronous MotorsOperation

• Given below is a 2-pole synchronous motor• The current in the stator produces a rotating

magnetic field BS• The current in the rotor produces magnetic field BR• Therefore, there are two magnetic fields present in the motor and

the rotor field will tend to line up with the stator field, just as twobar magnets will tend to line up if placed near each other

• Since the stator magnetic field is rotating, the rotor magnetic field(and the rotor itself) will constantly try to catch up


183 / 412








,

Synchronous MotorsStarting

• The net starting torque is 0!


184 / 412








,

Synchronous MotorsStarting

Three starting methods:

• Reduce the speed of the stator magnetic field.

• Use an external prime mover.

• Use amortisseur windings. This is by far the most popular way tostart a synchronous motor. Amortisseur windings are special barslaid into notches carved in the face of a synchronous motor’s rotorand then shorted out on each end by a large shorting ring.

Chapman 5th ed, pg 290-297http://en.wikipedia.org/wiki/Squirrel-cage_rotor

185 / 412

http://en.wikipedia.org/wiki/Squirrel-cage_rotor








,

Synchronous MotorsStarting: amortisseur windings (squirrel cage)

• Squirrel cage for an actual motor


186 / 412









,



187 / 412








,


• The induced torque is sometimes counterclockwise and sometimes0 but it is always in the same direction

• Since there is a net torque in a single direction, the motor’s rotorspeeds up

• Although the motor’s rotor will speed up, it can never quite reachsynchronous speed

• Once the motor starts up, the rotor DC current is restored andthe motor locks into synchronous speed


188 / 412








,

Synchronous MotorsEquivalent circuit (for each phase)


189 / 412








,

Induction MotorsIntroduction

• Same as a synchronous motor with amortisseur windings

• In other words, there is no DC current in the rotor

• The stator is the same as the synchronous motor

Chapman, pg 346

190 / 412








,

Synchronous generatorY-connection

Chapman, pg 278

191 / 412








,

Synchronous generator∆-connection

Chapman, pg 278

192 / 412








,

Synchronous generatorEquivalent circuit (for each phase)

Chapman, pg 279

193 / 412








,

Synchronous generatorPower losses

Chapman, pg 281

194 / 412








,

Synchronous generatorOCC (open circuit characteristic)

1 Field current IF is set to 0

2 Terminals are disconnected from all loads, and therefore IA = 0and EA = Vφ

3 Then IF is increased gradually in steps, and VT is measured

4 Therefore, a plot of IF vs VT , the OCC, can be constructed

5 However, note that,

• In a Y-connection, VT =√

3Vφ• In a ∆-connection, VT = Vφ• Therefore, a plot of IF vs Vφ can also be constructed, and

it is also called the OCC

6 Also note that since we have an open circuit, EA = Vφ andtherefore a plot of IF vs EA, again also called the OCC, can beconstructed

7 Also worth noting is that Vφ or EA cannot be measured directlywhile VT can

• Although called OCC, the name is only because open circuit isused to make the plots

• Otherwise, you can see what field current IF is needed to createwhat induced EA even when a load is connected

Chapman, pg 283

195 / 412

Synchronous generatorOperating alone: changing load conditions

• Terminal voltage

• In DC machines, denoted by VT

• In AC machines, denoted by Vφ since it’s on a phase by phase basis

• Phasor directions

1 Vφ: Reference phasor direction is fixed at 0o

2 IA:

• Inductive load: lags Vφ (not necessarily by 90o )• Resistive load: has the same direction as Vφ• Capactive load: leads Vφ (not necessarily by 90o )

3 jXs IA: leads IA by 90o

4 EA: has constant magnitude, i.e., vector tip moves along a circle

Chapman, pg 290 ,








,

Induction Motors3 phase induction motor energy efficiencies over the years

US Department of Energy, Advanced Manufacturing Office, Premium Efficiency Motor Selectionand Application Guide,http://energy.gov/sites/prod/files/2014/04/f15/amo_motors_handbook_web.pdf

197 / 412

http://energy.gov/sites/prod/files/2014/04/f15/amo_motors_handbook_web.pdf








,

Induction MotorsEquivalent circuit (for each phase)

Chapman, pg 394

198 / 412

Stepper motor (1/11)

Sequence

http://wineyardstudents.blogspot.com/2011/05/stepper-motor.html ,

199 / 412








,


4 phase

http://cr4.globalspec.com/blogentry/1749/Making-a-Telescope-Part-3-The-Mount

200 / 412








,


4 phase

• Like other types of electric motors that produce a rotating force astepper motor consists of two primary components a stator whichis the stationary part of the motor and a rotor which is the partthat rotates and is used to drive whatever it is connected to.

• In our case we have a four phase 9 stepper motor but there are afew other things that you may need to use:

• Definitions• Phases: This is the number of separate coils that make up

the system.• Step Angle: This is the angle that the motor steps through

every time the next coil in sequence is energized.• Holding Torque: This is the amount of force that is needed

to cause the rotor to turn while being locked in position byan energized coil.

• Driving or Dynamic Torque: This is the amount of torquethe motor can supply as it steps from one step to the next.

• Voltage: This is fairly obvious and is the voltage that isneeded to operate the motor.

• Holding Current: This is the current that the coils drawwhen in the locked position.

• Dynamic or Peak Current: This is the current the motordraws as it steps from one step to the next.

201 / 412








,


4 phase

• The Rotor

• It is the part of the motor that rotates.• Consists of a cylindrically shaped permanent magnet that

has multiple magnetic pole pairs arranged in a radialmanner around its axis.

• As the rotor turns a fixed point adjacent to itscircumference will see a sequence of alternating magneticpoles.

• The rotor can be though of as a series of horse shoemagnets arranged so their poles form a circle with equallyspaced around its circumference

• In this case we have 10 pole pairs labeled a-j andrepresented by the Grey U shapes arranged at 36increments with their polarity shown by the RED N for thenorth poles and BLUE S for the south poles.

• Something that is worth mentioning is that while each ofthe poles is separated by 18 it is the 36 angle between thepair of poles or next pole of the same polarity that isimportant.

202 / 412








,


4 phase

• The Stator

• This consists of a series of coils that are equally spaced• In our system we have 8 coils, but they are interconnected

so that electrically there are only 4 coils A-Red, B-Green,C-Blue, and E-Purple that are spaced at 45 intervals.

• When there is no power to the stepper motor the rotor willrotate fairly freely but it will try and stop so that one of thepoles aligns with one of the coils.

• In this instance the rotor will try and settle in increments of9 or 40 separate points.

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,


4 phase

• Step 0

• In the image on the left shows what would happen whenpower was applied to the stator coils labeled A and drawnin red.

• The magnetic fields these coils produce will then cause therotor to turn till the North pole of magnet a aligns with theSouth pole of the red coil A.

• On the opposite side of the rotor the North pole of magnetf would align with the South pole of the other red A coil.

• Something worth noting is that provided the holding torqueis not exceeded the rotor will stay locked in this positionuntil the power is removed from the red A coils.

• Unlike other forms of electric motor a locked rotor will notresult in damage to the motor or burnt out coils.

204 / 412








,


4 phase

• Step 1

• If we now remove the power to the red A coils and apply itto the green B coils the rotor will rotate till the North ofrotor magnet b aligns with the South of stator green B coil.

• On the other side of the rotor the North of magnet g willalign with South of the other stator green B coil.

• The important thing to note here is that even though themagnetic field of the stator has rotated through 45 therotor has only rotated through 9.

• If we now continue to energize each of the coils red-A,green-B, blue-C and purple E in sequence the motor willstep through 9 each time a coil is energized.

205 / 412


4 phase

• Advantages

1 Simplified Feedback: There are several advantages related to feedback:

(a) Position: Since the motor can only be in one of the positions defined by the coils andpermanent magnets in the coil finding the position of the system. All you need to do isstart from a known position and then count the steps in either direction to calculatethe position.

(b) Speed: Since the speed the rotor turns at is governed by how rapidly you step fromphase to phase you don’t need a feedback mechanism to calculate the speed the motoris rotating at.

2 Locked Rotor: Unlike with other motors a locked rotor will not result in the currentthrough the coils causing them to overheat and burn out. It can also be very helpful insituations where there needs to be some sort of breaking mechanism that can hold it ina desired position.

3 Simplified Drive Electronics: Unlike other types of motor where the current and voltagebeing applied to the motor need to be controlled through a range stepper motor coilsonly need to be either on or off. This makes the driving circuit much simpler andconsequently more reliable and less expensive.

4 Reduced Maintenance: Since there is no commutator, brushes, etcetera that are proneto wear and contamination stepper motors require less maintenance and have longerlife expectancies than other DC or servo motors.

,

206 / 412


4 phase

• Disadvantages

1 Step Induced Oscillations: Because stepper motor move in a sudden jerky mannerbetween each step the steps can set up vibrations in the drive train that can bedetrimental to the process and equipment.

2 Fine Control: The staccato or jumping motion of stepper motors can be a seriousproblem. As the motors can only be in specific positions as designated by the geometryof the rotor and stator you can have problems in applications where a smooth orcontinuous drive is required. To a certain extent this can be overcome by having stepsthat are around an order of magnitude smaller than required in the application,however, it is something that engineers need to be aware of.

,

207 / 412








,


4 phase

http://www.instructables.com/id/How-to-make-an-H-bridge/step2/The-truth-about-H-bridges/

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,


4 phase

• When the coils on ”Relay 1” and ”Relay 4” are pulled high(electricity is flowing through them), then the motor will spinforwards (see ”Image 1”).

• When the coils on ”Relay 2” and ”Relay 3” are pulled high(electricity is flowing through them), then the motor will spinbackwards (see ”Image 2”).

• When the coils on ”Relay 1” and ”Relay 2” are pulled high(electricity is flowing through them), then the motor will stopspinning (see ”Image 3”).

• When the coils on ”Relay 3” and ”Relay 4” are pulled high(electricity is flowing through them), then the motor will stopspinning (see ”Image 4”).

• ********WARNING***********• You want AVOID:• ”Relay 1” and ”Relay 3” being pulled high. This is a short circuit

since there is no load for the electricity to pass through. Badthings will happen! (see ”Image 5”)

• ”Relay 2” and ”Relay 4” being pulled high. This is a short circuitsince there is no load for the electricity to pass through. Badthings will happen! (imagine ”Image 6”)

• More than 2 relays being pulled high at one time. Bad things willhappen.

209 / 412








,

Brushless DC Motor (1/28)

Analogy

•


210 / 412









,


Analogy

•


211 / 412



Construction of a simple motor

https://www.youtube.com/watch?v=ms3KOZexkmI ,

212 / 412

https://www.youtube.com/watch?v=ms3KOZexkmI


Construction of a simple motor

https://www.youtube.com/watch?v=Kudzft19coo ,

213 / 412

https://www.youtube.com/watch?v=Kudzft19coo








,


Analogy

• A humorous analogy help to remember it is to think of BLDCoperation like the story of the donkey and the carrot

• The donkey tries hard to reach the carrot, but the carrot keepsmoving out of reach

http://www.learnengineering.org/2014/10/Brushless-DC-motor.html

214 / 412

http://www.learnengineering.org/2014/10/Brushless-DC-motor.html








,

Brushless DC Motor (BLDC) (6/28)

Introduction

http://www.freescale.com/files/sensors/doc/app note/AN3461.pdf

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,


Introduction

The permanent magnet synchronous motor (PMSM)can be thought of as a cross between an AC inductionmotor and a brushless DC motor (BLDC). They haverotor structures similar to BLDC motors which containpermanent magnets. However, their stator structureresembles that of its ACIM cousin, where the windingsare constructed in such a way as to produce a sinusoidalflux density in the airgap of the machine. As a result,they perform best when driven by sinusoidal waveforms.

http://www.ti.com/lsds/ti/apps/motor/permanent magnet/overview.page

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,


Construction

http://www.bavaria-direct.co.za/models/images/Mini300 Winding Instructions.png

217 / 412








,


Phases

• A single phase BLDC motor has current passingthrough one coil only

• A twp phase BLDC motor has current passingthrough two coils simultaneously

218 / 412








,


Commutation Steps for Single Phase Machine

http://www.ti.com/motor, https://www.youtube.com/watch?v=0mQunSe2 FM

219 / 412

Let’s look at the stator

• The stator above has 6 poles, or 3 pole pairs)

• Single wire is used for a pole pair, A and A create an oppositepolarity








,




220 / 412

Step 1.

• To get that rotor to rotate, you need to commutate the statorfield such that the rotor is always chasing that magnetic field

• To do this, you turn on, in sequence, different pole pairs








,




221 / 412

Step 2.

• Turn the current off in A and A and turn the current on in C andC creating a North and South electromagnetic pole

• The rotor is then attracted to it








,




222 / 412

Step 3.

•








,




223 / 412

Step 4.

•








,




224 / 412

Step 5.

•








,




225 / 412

Step 6.

•








,




226 / 412

Back to where we started from.

•








,



• The problem with the single phase setup is that only one windingis being used to create torque

https://www.youtube.com/watch?v=ZAY5JInyHXY

227 / 412








,


Commutation Steps for Two Phase Machine

S

N

S

N

S

N

S

N

A

BS

N

A

A

B

B

C

C

Current

Torque

Vcc

228 / 412

Step 1.

•








,



S

N

S

N

S

N

S

N

A

CS

N

A

A

B

B

C

C

Current

Torque

Vcc

229 / 412

Step 2.

•








,



N

S

N

S

S

N

S

N

BC

S N

A

A

B

B

C

C

Current

Torque

Vcc

230 / 412

Step 3.

•








,



N

S

N

S

N

S

N

S

A

B

S

N

A

A

B

B

C

C

Current

Torque

Vcc

231 / 412

Step 4.

•








,



N

S

N

S

N

S

N

S

A

C

S

N

A

A

B

B

C

C

Current

TorqueVcc

232 / 412

Step 5.

•








,



S

N

S

N

N

S

N

S

BC

SN

A

A

B

B

C

C

Current

TorqueVcc

233 / 412

Step 6.

•








,



• The trick is when do you turn on that adjacent pole

• There’s no positional information in the diagram shown, you don’tknow where the rotor is so you don’t know when you’re supposedto turn on that next magnetic pole

• The turning on, the timing is extremely critical, you always wantto maximize the torque

• If you turn on that field too early or too late, you will haveperformance issues

• Therefore, typically sensors are added to the system


234 / 412








,


Mathematical Model

Stator windings:• Total flux (as a result of stator currents and rotor permanent

magnet):ψa

ψb

ψc

=

Laa Lab Lac

Lba Lbb Lbc

Lca Lcb Lcc

iaibic

+

ψam

ψbm

ψcm

=

Ls −Ms −Ms

−Ms Ls −Ms

−Ms −Ms Ls

iaibic

+

ψam

ψbm

ψcm

• Terminal voltage:

va

vb

vc

=

Rs 0 00 Rs 00 0 Rs

iaibic

+

dψadt

dψbdt

dψcdt

http://www.ti.com/motor

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,


Comparison

http://ww1.microchip.com/downloads/en/AppNotes/00885a.pdf

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,


Comparison

Also read http://www.teslamotors.com/blog/

induction-versus-dc-brushless-motors

http://ww1.microchip.com/downloads/en/AppNotes/00885a.pdf

237 / 412

http://www.teslamotors.com/blog/induction-versus-dc-brushless-motors

http://www.teslamotors.com/blog/induction-versus-dc-brushless-motors








,

ServomotorsOverview

• A servomechanism, sometimes shortened to servo, is an automaticdevice that uses error-sensing negative feedback to correct theperformance of a mechanism and is defined by its function

• A servomotor is a rotary actuator that allows for precise control of

angular position, velocity and acceleration

• As the name suggests, a servomotor is a servomechanism• More specifically, it is a closed-loop servomechanism that

uses position feedback to control its motion and finalposition

• The input to its control is some signal, either analogue ordigital, representing the position commanded for the outputshaft

• It consists of a suitable motor coupled to a sensor forposition feedback

• It also requires a relatively sophisticated controller, often adedicated module designed specifically for use withservomotors

https://en.wikipedia.org/wiki/Servomechanism

https://en.wikipedia.org/wiki/Servomotor

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https://en.wikipedia.org/wiki/Servomechanism









,

ServomotorsCan be AC or DC

• The type of motor is not critical to a servomotor and differenttypes may be used

• At the simplest, brushed permanent magnet DC motors are used,owing to their simplicity and low cost

• Small industrial servomotors are typically electronicallycommutated brushless motors

• For large industrial servomotors, AC induction motors are typicallyused, often with variable frequency drives to allow control of theirspeed

• For ultimate performance in a compact package, brushless ACmotors with permanent magnet fields are used, effectively largeversions of Brushless DC electric motors


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,

Solenoid• A solenoid is simply a specially designed electromagnet

• A solenoid usually consists of a coil and a movable ironcore called the armature.

• When current flows through a wire, a magnetic field is set uparound the wire

• If we make a coil of many turns of wire, this magnetic fieldbecomes many times stronger, flowing around the coil andthrough its center in a doughnut shape

• When the coil of the solenoid is energized with current, the coremoves to increase the flux linkage by closing the air gap betweenthe cores

http://mechatronics.mech.northwestern.edu/design_ref/actuators/solenoids.html

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,

Solenoid• The movable core is usally spring-loaded to allow the

core to retract when the current is switched off

• The force generated is approximately proportionalto the square of the current and inversely proportional to thesquare of the length of the air gap

• Solenoids are inexpensive, and their use is primarily limited toon-off applications such as latching, locking, and triggering

• They are frequently used in home appliances (e.g. washingmachine valves), office equipment (e.g. copy machines),automobiles (e.g. door latches and the starter solenoid), pinballmahines (e.g., plungers and bumpers), and factory automation


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,

Electric Vehicles (86/86)

Chevrolet FNR

http://www.extremetech.com/extreme/

203862-this-chevrolet-fnr-concept-car-is-science-fiction-made-real

242 / 412

http://www.extremetech.com/extreme/203862-this-chevrolet-fnr-concept-car-is-science-fiction-made-real

http://www.extremetech.com/extreme/203862-this-chevrolet-fnr-concept-car-is-science-fiction-made-real








,

MagnetismWorld’s strongest magnet: 27T

• Built by MagLab, largest and highest powered magnetlab in the world

• Demonstrated on 5 June 2015

https://nationalmaglab.org/news-events/news/

maglab-claims-record-with-novel-superconducting-magnet

http://nextbigfuture.com/2015/06/new-superconducting-magnet-already-at.html

243 / 412

https://nationalmaglab.org/news-events/news/maglab-claims-record-with-novel-superconducting-magnet

https://nationalmaglab.org/news-events/news/maglab-claims-record-with-novel-superconducting-magnet

http://nextbigfuture.com/2015/06/new-superconducting-magnet-already-at.html








,

Magnetism (1/12)

The Geomagnetic Field

• The magnitude of the Earth’s magnetic field (the geomagneticfield) varies over the surface of the earth from a minimum of22µT (0.22 Gauss) over S. America to a maximum of 67µT (0.67Gauss) south of Australia

• The heading of an eCompass is determined from the relativestrengths of the two horizontal geomagnetic field components andthese vary from zero at the magnetic poles to a maximum of42µT over E. Asia

• Detailed geomagnetic field maps are available from the WorldData Center for Geomagnetism athttp://wdc.kugi.kyoto-u.ac.jp/igrf/

http://cache.freescale.com/files/sensors/doc/app note/AN4247.pdf

244 / 412

http://wdc.kugi.kyoto-u.ac.jp/igrf/

http://cache.freescale.com/files/sensors/doc/app_note/AN4247.pdf








,

Magnetism (2/12)

The Geomagnetic Field cont.

• Although the Earth’s magnetic field is relatively stable over time,electric currents in the ionosphere can cause daily alterationswhich can deflect surface magnetic fields by as much as onedegree

• Normally, daily variations in field strength are on the order of0.025µT (0.25 mGuass), which would equate to about 0.03degree variation in heading

• This small change of heading is on the same order of magnitudeas the resolution of most MEMS based magnetometers, so inmost cases, the Earth’s magnetic field can be considered constantwrt time

http://www.vectornav.com/support/library/magnetometer

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,

Magnetism (3/12)

Magnetometer calibration

• Magnetic measurements will be subjected to distortions:

1 Hard iron: Created by objects that produce a magnetic field2 Soft iron: Deflections or alterations in the existing magnetic

field


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,

Magnetism (4/12)

Magnetometer calibration cont.

• A common way of visualizing and correcting hard and soft irondistortions is to plot the output of the magnetometer on a 2Dgraph

• The following plot shows measurements taken by themagnetometer as the device is slowly rotated around the Z-axis

• In the event that there are no hard or soft iron distortions present,the measurements should form a circle centered at X=0, Y=0.

• The radius of the circle equals the magnitude of the magnetic field


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,

Magnetism (5/12)


• The effect of hard iron distortions on the plot will be to shift thecenter of the circle

• As shown in the plot the center of the circle with hard irondistortions is now at X=200, Y=100

• From this we can conclude that there is 200 mGauss hard ironbias in the X-axis and 100 mGauss hard iron bias in the Y-axis

• Hard iron distortions will only shift the center of the circle awayfrom the origin

• Hard iron distortions will not distort the shape of the circle.


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,

Magnetism (6/12)


• Soft iron distortions on the other hand distort and warp theexisting magnetic fields

• When you plot the magnetic output, soft iron distortions are easyto recognize since they will distort the circular output

• Soft iron effects warp the circle into an elliptical shape

• The center of the ellipse below is still located at X=200 mGaussand Y=100mGauss since the hard iron distortions are the same asbefore but now the major axis is aligned 30 degrees up from thebody frame X direction due to soft iron distortions


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,

Magnetism (7/12)


• It is possible to eliminate the effects of both hard and soft irondistortions on the magnetometer outputs

• VectorNavr products use the following calibration model tocorrect for hard and soft iron distortions

M =

C1 C2 C3

C4 C5 C6

C7 C8 C9

Hx − C10

Hy − C11

Hz − C12

• The above model consists of 12 hard and soft iron compensation

parameters

• The first 9 parameters correct for the soft iron while the lastthree, C10, C11, C12 parameters compensate for the hard iron

• For the previous figure, the hard and soft iron calibrationparameters would be

M =

C1 C2 C3

C4 C5 C6

C7 C8 C9

Hx − C10

Hy − C11

Hz − C12


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,

Magnetism (8/12)


• Measured magnetometer locus - no correction applied:


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,

Magnetism (9/12)


• Measured magnetometer locus - hard iron correction applied:


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,

Magnetism (10/12)


• Measured magnetometer locus - hard and soft iron correctionapplied:


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,

Magnetism (11/12)


1 Raw mode. Put magnetometer in IMU mode, i.e., it should giveraw values and not fused values. These raw values will be inGauss or Teslas. For the NV-100, they are in Gauss. Note thatthese values must vary between 0.22 and 0.67 Gauss, and if thevalues are more than these values, we have hard and/or soft irondistortions

2 Find North and South poles. Rotate magnetometer 360 degrees

and observe the following:1 Hx must reach a maximum positive value when it is aligned with the North pole

of the Earth’s magnetic field.2 Turning it CW by 90 degrees should give a value of 0 since it is now orthogonal

to the Earth’s magnetic field and pointing towards East3 Turning it CW by another 90 degrees, i.e., a total of 180 degrees should give

you a maximum negative value showing it is pointing towards the South pole4 Turning it CW by another 90 degrees, i.e., a total of 270 degrees should again

give you a value of 0 showing it is pointing West5 Finally, turning it CW by another 90 degrees, i.e., a total of 360 degrees should

again give you a maximum positive value showing you are again aligned north6 The above process can be repeated with Hy7 Finding the angle as atan(Hy/Hx) or atan(-Hy/Hx) depending on how you set it

up should give you the angle of rotation

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,

Magnetism (12/12)


3 Plot Hx and Hy points. Now, rotate magnetometer through 360degrees and plot Hx vs Hy. You should get a zero-centered circleif there are no hard or soft iron distortions. If it is a shifted circle,you have hard iron distortions. If it is an ellipse instead of a circle,you have soft iron distortions. If you have a shifted ellipse, youhave hard and soft iron distortions.

4 Finding parameters of hard and soft iron distortions. Use the

following steps for manual removal. Automatic removal will

require knowledge of transforming a shifted ellipse to a centered

circle.

1 Shift the ellipse or circle to the center.2 See what angle the ellipse makes with the y-axis and find a

rotation matrix to rotate it so that one of the axes of theellipse is y-axis aligned

3 Compress Hx or Hy so that the ellipse is now a circle

5 Using above parameters to remove hard and soft irondistortions. Now, for every measurement of Hx and Hy, applyabove parameters and check that indeed as the magnetometerrotates, the angles are more accurate than before. Also, anglesshould be evenly spaced.

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,

Tesla MotorsInduction motor in vehicles

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,

Lab # 1: Lab and area familiarization

• Get familiar with the lab environment

• Get familiar with the MES environment

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,

Lab # 2: Transformers LabMeasuring parameters

• This is the setup for an open circuit test in the lab:

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,

Lab # 2: Transformers LabMeasuring parameters cont.

• This is the setup for a short circuit test in the lab:

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,

Lab # 3: Matlab UsageNumericals and plots

• Be able to solve numericals related toelectromechanical systems in Matlab

• Be able to make plots related to electromechanicalsystems in Matlab

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,

Lab # 4: DC motor modeling, Matlab

• Go to the University of Michigan websitehttp://ctms.engin.umich.edu

• Click ”MOTOR SPEED” at the top

• Complete SYSTEM MODELING and SYSTEM ANALYSIS parts

261 / 412

http://ctms.engin.umich.edu








,

Lab # 5: DC motor modeling, Matlab(part 2)Electrical, magnetic and mechanical signal flow

• Solve Example 7.1, part (a) and (b)

• In this example, there is no notion of time, i.e., how longdoes it take the motor to reach no-load steady-state ω

• Now, instead of just using the formula to find no-load

steady-state ω, write a software loop to model the motor feedback

loop and find ω and τ in increments of 0.001 sec upto 0.5 sec

• Use moment of inertia of the rotor, J = 0.01 kg m2

• Make the following 3 plots:

1 ω against time2 τ against time3 ω against τ

• Verify that the results of this lab tally with the results ofthe solved example, as well as the results of the previous labprovided that L = 0 H and motor viscous friction constantb = 0 Nms

• Investigate the effect of varying J,R, φ


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,

Lab # 6: BKB Universal Lab MachineDC shunt motor

• Front panel, starter motor, motor, generator

263 / 412








,

Lab # 6: BKB Universal Lab MachineDC shunt motor cont.

• Load bank (bulbs), Power supply

264 / 412








,


• Front panel

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,


• Stator coils

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,


• Rotor coils

267 / 412








,


• Search coils

268 / 412








,


• Dynamometer

269 / 412








,

All examples

• Ch 1, Intro to Machinery Principles: 11 examples

• Ch 2, Transformers: 10 examples

• Ch 3, AC Machinery Fundamentals: 3 examples

• Ch 4, Synchronous Generators: 2 examples

• Ch 5, Synchronous Motors: 8 examples

• Ch 6, Induction Motors: 3 examples

• Ch 7, DC Machinery Fundamentals: 8 examples

• Ch 8, DC Motors and Generators: 4 examples

• Ch 9, Single-phase and special purpose motors: 9 examples

• Total: 58

Chapman 5th ed

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,

Example 1-1Magnetic circuits: computing flux


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,

Example 1-1 cont.

Magnetic circuits: computing flux


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,

Example 1-1 cont.



273 / 412








,

Example 1-1 cont.



274 / 412








,



275 / 412








,

Example 1-2 cont.



276 / 412








,

Example 1-2 cont.



277 / 412








,

Example 1-2 cont.



278 / 412








,



279 / 412








,

Example 1-3 cont.



280 / 412








,

Example 1-3 cont.



281 / 412








,

Example 1-3 cont.



282 / 412








,

Example 1-4Magnetic circuits: computing relative permeability


283 / 412








,

Example 1-4 cont.

Magnetic circuits: computing relative permeability


284 / 412








,

Example 1-4 cont.



285 / 412








,

Example 1-4 cont.



286 / 412








,

Example 1-5Magnetic circuits: current, relative permeability, reluctance


287 / 412








,

Example 1-5 cont.

Magnetic circuits: current, relative permeability, reluctance


288 / 412








,

Example 1-5 cont.



289 / 412








,

Example 1-5 cont.



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,

Problem 1-5Magnetic circuits: current, relative permeability, reluctance


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Problem 1-5 cont.


1 c l e a r ; c l c ; c l f ;2 %t u r n s3 N = 5 0 0 ;4 %p e r m e a b i l i t y5 mu0 = 4∗ p i ∗10E−7; % p e r m e a b i l i t y o f a i r (H/m)6 mur = 8 0 0 ; % r e l a t i v e p e r m e a b i l i t y7 %l e n g t h s8 l e n l e f t = (7.5+15+7.5) / 1 0 0 ; % l e n g t h ( me te r s )9 l e n t o p = (5+20+2.5) / 1 0 0 ; % l e n g t h ( mete r s )

10 l e n r i g h t = l e n l e f t ; % l e n g t h ( mete r s )11 l e n b o t t o m = l e n t o p ; % l e n g t h ( mete r s )12 %a r e a s13 a r e a l e f t = (10∗5) /1E4 ; % a r e a ( sq met e r s )14 a r e a t o p = (15∗5) /1E4 ; % a r e a ( sq met e r s )15 a r e a r i g h t = (5∗5) /1E4 ; % a r e a ( sq mete r s )16 a r e a b o t t o m = (15∗5) /1E4 ; % a r e a ( sq met e r s )17 %r e l u c t a n c e s18 R l e f t = l e n l e f t / (mu0 ∗ mur ∗ a r e a l e f t ) ; % r e l u c t a n c e (A t u r n s /Wb)19 R top = l e n t o p / (mu0 ∗ mur ∗ a r e a t o p ) ; % r e l u c t a n c e (A t u r n s /Wb)20 R r i g h t = l e n r i g h t / (mu0 ∗ mur ∗ a r e a r i g h t ) ; % r e l u c t a n c e (A t u r n s /Wb)21 R bottom = l e n b o t t o m / (mu0 ∗ mur ∗ a r e a b o t t o m ) ;% r e l u c t a n c e (A t u r n s /Wb)22 R = R l e f t+R top+R r i g h t+R bottom ; % t o t a l r e l u c t a n c e (A t u r n s /Wb)23 %f l u x24 p h i = 0 . 0 0 5 ; % f l u x ( webers )25 %p a r t ( a )26 I = p h i∗R/N; % c u r r e n t ( amperes )27 %p a r t ( b )28 B top = p h i / a r e a t o p ; % f l u x d e n s i t y ( t e s l a s )29 B r i g h t = p h i / a r e a r i g h t ; % f l u x d e n s i t y ( t e s l a s )

Chapman 5th ed, pg 56 ,

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,

Example 1-7Motors: Lorentz force

In this , B = 0.25T, l = 1.0m, I = 0.5A, find F.

F = ilB sin θ = (0.5A)(1.0m)(0.25T ) sin(90o) = 0.125Nto the right


293 / 412








,

Example 2-1Transformers: advantage


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,

Example 2-1 cont.

Transformers: advantage


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,

Example 2-1 cont.

Transformers: advantage

1 c l e a r ; c l c ;2 %Given3 V = 4 8 0 ; %v o l t a g e4 Z l i n e = 0 . 1 8 + 0 . 2 4 j ; %impedance ( l i n e )5 Z l o a d = 4 + 3 j ; %impedance ( l o a d )6 Z t o t a l = Z l i n e+Z l o a d ; %impedance ( t o t a l )7 a T1 = 0 . 1 ; %t u r n s r a t i o (T1)8 a T2 = 1 0 ; %t u r n s r a t i o (T2)9 % p a r t ( a )

10 %−−−−−−−−11 I l i n e = V/ Z t o t a l ; %c u r r e n t12 V l o a d = I l i n e ∗Z l o a d ; %v o l t a g e13 P l o s s e s = abs ( I l i n e ) ˆ2∗ r e a l ( Z l i n e ) ; %power14 % p a r t ( b )15 %−−−−−−−16 Z l o a d l e f t T 2 = a T2 ˆ2 ∗ Z l o a d ;17 Z l o a d l e f t T 2 l e f t T 1 = a T1 ˆ2 ∗ Z l o a d l e f t T 2 ;18 Z l i n e l e f t T 1 = a T1 ˆ2 ∗ Z l i n e ;19 Z t o t a l = Z l i n e l e f t T 1 + Z l o a d l e f t T 2 l e f t T 1 ;

21 I G = V/ Z t o t a l ;22 I l i n e = I G∗a T1 ;23 I l o a d = I l i n e ∗a T2 ;

25 V l o a d = I G∗Z l o a d ;

27 P l o s s e s = abs ( I l i n e ) ˆ2∗ r e a l ( Z l i n e )28 %Notes29 %To f i n d P l o s s e s , we used t h e magnitude o f c u r r e n t30 %and t h e r e a l p a r t o f Z l i n e


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Example 2-2Transformers: finding parameters


297 / 412








,

Example 2-2 cont.

Transformers: finding parameters

• For both the open circuit and short circuit tests, we have 8

variables:

1 P: can be measured (given in this example)2 V : can be measured (given in this example)3 I : can be measured (given in this example)4 cos θ5 Rc : transformer parameter to be computed6 Xm: transformer parameter to be computed7 Reqs : transformer parameter to be computed8 Xeqs : transformer parameter to be computed

• For the open circuit test, first find cos θ, then find Rc ,Xm.

• For the short circuit test, first find cos θ, then find Reqs ,Xeqs .


298 / 412

Example 2-2 cont.



299 / 412

Example 2-2 cont.



300 / 412








,

Example 2-2 cont.



301 / 412








,

Example 2-2 cont.


1 c l e a r ; c l c ;2 %GIVEN3 %−−−−−4 a = 8000/240; % t u r n s r a t i o5 %open c i r c u i t ( s e c o n d a r y s i d e , b e c a u s e low v o l t a g e )6 Voc = 2 4 0 ; % v o l t a g e7 I o c = 7 . 1 3 3 ; % c u r r e n t8 Poc = 4 0 0 ; % power9 %s h o r t c i r c u i t ( p r i m a r y s i d e , b e c a u s e low c u r r e n t )

10 Vsc = 4 8 9 ; % v o l t a g e11 I s c = 2 . 5 ; % c u r r e n t12 Psc = 2 4 0 ; % power13 %COMPUTATIONS14 %−−−−−−−−−−−−15 %open c i r c u i t t e s t16 PFoc = Poc /( Voc∗ I o c ) ; % power f a c t o r = cos (

→ t h e t a )17 t h e t a o c = −acos ( PFoc ) ; % n e g a t i v e s i g n f o r

→ l a g g i n g18 [ I x , I y ] = p o l 2 c a r t ( t h e t a o c , I o c ) ;19 I = I x + j∗ I y ;20 Y E = I /Voc ;21 Rc = 1/ r e a l ( Y E ) ;22 Xm = −1/imag ( Y E ) ; % Y E=(1/Rc )−j (1/Xm)23 %s h o r t c i r c u i t t e s t24 PFsc = Psc /( Vsc∗ I s c ) ; % power f a c t o r = cos (

→ t h e t a )25 t h e t a s c = −acos ( PFoc ) ; % n e g a t i v e s i g n f o r

→ l a g g i n g26 [ I x , I y ] = p o l 2 c a r t ( t h e t a s c , I s c ) ;27 I = I x + j∗ I y ;28 Z E = Vsc / I ;


302 / 412








,

Example 2-5Transformers: voltage regulation and efficiency


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,

Example 2-5 cont.

Transformers: voltage regulation and efficiency


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,

Example 2-5 cont.



305 / 412








,

Example 2-5 cont.



306 / 412








,

Example 2-5 cont.



307 / 412








,

Example 2-5 cont.



308 / 412








,

Example 2-5 cont.



309 / 412








,

Example 2-5 cont.



310 / 412








,

Example 2-5 cont.



311 / 412

Example 2-5 cont.


1 c l e a r ; c l c ; format compact ; % a l l a n g l e s a r e i n r a d i a n s

3 % ===============================================4 % GIVEN5 % ===============================================6 S rated mag = 15000; % r a t e d v a l u e , VA7 p r i V r a t e d m a g = 2 3 0 0 ; % r a t e d v a l u e , V8 s e c V r a t e d m a g = 2 3 0 ; % r a t e d v a l u e , A9 a = p r i V r a t e d m a g / s e c V r a t e d m a g ; % t u r n s r a t i o

10 sec Voc mag = 2 3 0 ; % open c i r c u i t t e s t ( low v o l t a g e s i d e )11 s e c I o c m a g = 2 . 1 ; % open c i r c u i t t e s t ( low v o l t a g e s i d e )12 sec Poc mag = 5 0 ; % open c i r c u i t t e s t ( low v o l t a g e s i d e )13 p r i V s c m a g = 4 7 ; % s h o r t c i r c u i t t e s t ( h i g h v o l t a g e s i d e )14 p r i I s c m a g = 6 ; % s h o r t c i r c u i t t e s t ( h i g h v o l t a g e s i d e )15 p r i P s c m a g = 1 6 0 ; % s h o r t c i r c u i t t e s t ( h i g h v o l t a g e s i d e )


312 / 412

Example 2-5 cont.


1 % ===============================================2 % COMPUTATIONS3 % ===============================================

5 % I . OPEN CIRCUIT TEST ( g o a l i s to f i n d Rc , Xm)6 %−−−−−−−−−−−−−−−−−−−−7 % f i n d power f a c t o r8 PFoc = sec Poc mag / ( sec Voc mag∗ s e c I o c m a g ) ; % power f a c t o r

10 % f i n d complex open c i r c u i t v o l t a g e and c u r r e n t11 s e c V o c = sec Voc mag∗(1+0 j ) ; %a r b i t r a r i l y g i v e n an a n g l e o f 0

13 I o c t h e t a = −acos ( PFoc ) ; % n e g a t i v e s i g n f o r l a g g i n g14 [ I x I y ] = p o l 2 c a r t ( I o c t h e t a , s e c I o c m a g ) ;15 s e c I o c = I x + j∗ I y ;

17 % f i n d a d m i t t a n c e18 sec YE = s e c I o c / s e c V o c ; % YE=(1/Rc )−j (1/Xm)19 s e c R c = 1/ r e a l ( sec YE ) ;20 sec Xm = −1/imag ( sec YE ) ;


313 / 412

Example 2-5 cont.


1 % I I . SHORT CIRCUIT TEST ( g o a l i s to f i n d Req , Xeq )2 %−−−−−−−−−−−−−−−−−−−−−−3 % f i n d power f a c t o r4 PFsc = p r i P s c m a g /( p r i V s c m a g∗ p r i I s c m a g ) ; % power f a c t o r

6 % f i n d complex s h o r t c i r c u i t v o l t a g e and c u r r e n t7 p r i V s c = p r i V s c m a g∗(1+0 j ) ; %j u s t g i v e n an a n g l e o f 0

9 I s c t h e t a = −acos ( PFsc ) ; % n e g a t i v e s i g n f o r l a g g i n g10 [ I x I y ] = p o l 2 c a r t ( I s c t h e t a , p r i I s c m a g ) ;11 p r i I s c = I x + j∗ I y ;

13 % f i n d impedance14 p r i Z E = p r i V s c / p r i I s c ;15 p r i R e q = r e a l ( p r i Z E ) ;16 p r i X e q = imag ( p r i Z E ) ;

19 % I I I . REFERRING Rc , Xm, Req , Xeq TO OPPOSITE SIDES20 %−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−21 p r i R c = ( a ˆ2)∗ s e c R c ; %r e f e r r e d to p r i m a r y22 pri Xm = ( a ˆ2)∗sec Xm ;

24 s e c R e q = (1/ a ˆ2)∗p r i R e q ; %r e f e r r e d to s e c o n d a r y25 sec Xeq = (1/ a ˆ2)∗p r i X e q ;


314 / 412

Example 2-5 cont.


1 % IV . REGULATION2 %−−−−−−−−−−−−−−3 r e g u l 1 = func POWER TRANSFORMER RegulationVsKnown ( S rated mag , s e c V r a t e d m a g , 0 . 8 , →−1, sec Req , sec Xeq )

4 r e g u l 2 = func POWER TRANSFORMER RegulationVsKnown ( S rated mag , s e c V r a t e d m a g , 1 . 0 , →0 , sec Req , sec Xeq )

5 r e g u l 3 = func POWER TRANSFORMER RegulationVsKnown ( S rated mag , s e c V r a t e d m a g , 0 . 8 , →1 , sec Req , sec Xeq )

7 % ===============================================8 % PRINT RESULTS9 % ===============================================

10 d i s p ( ’On p r i m a r y s i d e ’ )11 d i s p ( ’−−−−−−−−−−−−−−−’ )12 p r i R c13 pri Xm14 p r i R e q15 p r i X e q16 d i s p ( ’ ’ )17 d i s p ( ’On s e c o n d a r y s i d e ’ )18 d i s p ( ’−−−−−−−−−−−−−−−−−’ )19 s e c R c20 sec Xm21 s e c R e q22 sec Xeq23 d i s p ( ’ ’ )24 d i s p ( ’ R e g u l a t i o n ’ )25 d i s p ( ’−−−−−−−−−−’ )26 s p r i n t f ( ’ r e g u l a t i o n : %.2 f p e r c e n t ’ , r e g u l 1 )27 s p r i n t f ( ’ r e g u l a t i o n : %.2 f p e r c e n t ’ , r e g u l 2 )28 s p r i n t f ( ’ r e g u l a t i o n : %.2 f p e r c e n t ’ , r e g u l 3 )

Chapman 5th ed, pg 102,

315 / 412

Example 2-5 cont.


1 %Vp , Vs need to be both r e f l e c t e d on p r i m a r y s i d e , o r both on s e c o n d a r y s i d e

3 f u n c t i o n r e g u l = func POWER TRANSFORMER Regulation (Vp , Vs )

5 % use same names as book6 Vnl = abs (Vp) ;7 V f l = abs ( Vs ) ;

9 % s t e p 3 : f i n d r e g u l a t i o n10 r e g u l = ( Vnl−V f l ) / V f l ∗100;


316 / 412








,

Problem 2-7Transformers: regulation


317 / 412








,

Problem 2-7 cont.

Transformers: regulation


318 / 412

Problem 2-7 cont.


1 c l e a r ; c l c ; format compact ; % a l l a n g l e s a r e i n r a d i a n s2 % ===============================================3 % GIVEN4 % ===============================================5 Srated mag = 30000; % r a t e d v a l u e , VA6 p r i V r a t e d m a g = 8 0 0 0 ; % r a t e d v a l u e , V7 s e c V r a t e d m a g = 2 3 0 ; % r a t e d v a l u e , A

9 p r i R c = 100E3 ; % m a g n e t i z i n g branch10 pri Xm = 20E3 ; % ”11 p r i Z e q = 20+100 j ; % t r a n s f o r m e r impedance

14 a = p r i V r a t e d m a g / s e c V r a t e d m a g ; % t u r n s r a t i o


319 / 412

Problem 2-7 cont.


1 % ===============================================2 % COMPUTATIONS3 % ===============================================4 a = p r i V r a t e d m a g / s e c V r a t e d m a g ; % t u r n s r a t i o

6 %g i v e n7 p r i V p = 7967∗(1+0 j ) ;8 s e c Z L = 2+0.7 j ; %use −3j f o r p a r t ( b )

10 %c o m p u t a t i o n s on p r i m a r y s i d e11 p r i Z L = aˆ2∗ s e c Z L ;12 p r i Z t o t = p r i Z e q + p r i Z L ;13 p r i I s = p r i V p / p r i Z t o t ; %p r i I s = s e c I s /a14 p r i V s = p r i I s ∗ p r i Z L ;15 VR = func POWER TRANSFORMER Regulation ( p r i V p , abs ( p r i V s ) )

17 %c o m p u t a t i o n s on s e c o n d a r y s i d e18 sec Vp = p r i V p /a ;19 s e c Z t o t = p r i Z t o t /a ˆ 2 ;20 s e c I s = sec Vp / s e c Z t o t ;21 s e c V s = s e c I s ∗ s e c Z L ;22 VR = func POWER TRANSFORMER Regulation ( sec Vp , s e c V s )


320 / 412








,

Example 2-7Transformers: autotransformer


321 / 412








,

Example 2-7 cont.

Transformers: autotransformer


322 / 412








,

Example 2-7 cont.



323 / 412








,

Example 2-7 cont.



324 / 412

Problem 2.14Transformer

I = VGZline +Zload

= 13,20060 ∠53.1o +500 ∠36.87o = 23.66∠−38.6o A

SG = VG I∗ = 13, 200(23.66∠38.6o ) = 312, 312 ∠38.6o VA

Chapman, pg 148 ,

325 / 412

Problem 2.14 cont.

Transformer

Vline = Zline I = (60 ∠53.1o )(23.66 ∠−38.6o ) = 1, 419.6 ∠14.5o V

Sline = Vline I∗ = (1, 419.6 ∠14.5o )(23.66 ∠38.6o ) = 33, 587.74 ∠53.1o VA

= |Vline |2|Zline |2

Zline = 1,419.62

602 (60 ∠53.1o )

= |I|2Zline = (23.662)(60 ∠53.1o )

Pline = Sline cos θ = 33, 587.74 cos(53.1o ) = 20.1 kW


Rline = 1,419.62

602 ∗ 60 cos(53.1o )

= |I|2Rline = 23.662 ∗ 60 cos(53.1o )

,

326 / 412

Problem 2.14 cont.

Transformer

Vload = Zload I = (500 ∠36.87o )(23.66 ∠−38.6o ) = 11, 830 ∠−1.73o V

Sload = Vload I∗ = (11, 830 ∠−1.73o )(23.66 ∠38.6o ) = 279, 898 ∠36.87o VA

= |Vload |2|Zload |2

Zload = 11,8302

5002 (500 ∠36.87o )

= |I|2Zload = (23.662)(500 ∠36.87o )

Pload = Sload cos θ = 279, 898 cos(36.87o ) = 223.9 kW


Rload = 11,8302

5002 ∗ 500 cos(36.87o )

= |I|2Rload = 23.662 ∗ 500 cos(36.87o )

,

327 / 412

Problem 2.14 cont.

Transformer

I = VGZline +Zload

= 13,2000.6 ∠53.1o +500 ∠36.87o = 26.37∠−36.89o A

SG = VG I∗ = 13, 200(26.37∠36.89o ) = 278, 393 ∠36.89o VA

,

328 / 412

Problem 2.14 cont.

Transformer

Vline = Zline I = (0.6 ∠53.1o )(26.37 ∠−36.89o ) = 15.82 ∠16.2o V

Sline = Vline I∗ = (15.82 ∠16.2o )(26.37 ∠36.89o ) = 417.17 ∠53.1o VA


Zline = 15.822

0.62 (0.6 ∠53.1o )

= |I|2Zline = (26.372)(0.6 ∠53.1o )

Pline = Sline cos θ = 417.17 cos(53.1o ) = 250 W


Rline = 15.822

0.62 ∗ 0.6 cos(53.1o )

= |I|2Rline = 26.372 ∗ 0.6 cos(53.1o )

,

329 / 412

Problem 2.14 cont.

Transformer

Vload = Zload I = (500 ∠36.87o )(26.37 ∠−36.89o ) = 13, 185 ∠−0.02o V

Sload = Vload I∗ = (13, 185 ∠−0.02o )(26.37 ∠36.89o ) = 347, 688 ∠36.87o VA


Zload = 13,1852

5002 (500 ∠36.87o )

= |I|2Zload = (26.372)(500 ∠36.87o )

Pload = Sload cos θ = 347, 688 cos(36.87o ) = 278.15 kW


Rload = 13,1852

5002 ∗ 500 cos(36.87o )

= |I|2Rload = 26.372 ∗ 500 cos(36.87o )

,

330 / 412








,

Problem 2.14 cont.

Transformer

• Initially, the ratio of the load voltage magnitude tothe input voltage was |VL|

|VG | = 11,83013,200 = 0.896

• This increased to |VL||VG | = 13,185

13,200 = 0.9989

• Initially, the line losses were Pline = 20.1 kW

• These decreased by 80 times to Pline = 250 W

331 / 412








,


SW = 5000 VA480/120600V source to 120V load ⇒ NSE = 4NC

Chapman, pg 148

332 / 412








,

Problem 2.15 cont.

Transformer

SWSIO

= NSENSE +NC

5000SIO

= 4NC4NC +NC

⇒ SIO = 54 (5000) = 6250 VA

Ipmax = 6250600 = 10.4 A

Ismax = 6250120 = 52.1 A

SIO is 1.2 times SW

Chapman, pg 148

333 / 412








,


SW = 5000 VA480/120600V source to 480V load ⇒ NSE = 1/4NC

Chapman, pg 149

334 / 412








,

Problem 2.16 cont.

Transformer

SWSIO

= NSENSE +NC

5000SIO

= NSENSE +4NSE

⇒ SIO = 5(5000) = 25, 000 VA

Ipmax = 25,000600 = 41.67 A

Ismax = 25,000480 = 52.1 A

SIO is 5 times SW

335 / 412








,

Example 5.2Synchronous generator: changing load conditions

Chapman, pg 291

336 / 412








,

Example 5.2 cont.

Synchronous generator: changing load conditions

337 / 412

Example 5.2 cont.

Synchronous generator: changing load conditionsNote: VT = Vφ since we have a 3 phase ∆-connected machine.For part (c), use IL to find IA, use IA to find a new higher EA when the load is connected.Find IF corresponding to this new EA from the OCC curve. But the OCC curve is for VT , not EA!How can you use the OCC curve? The way to look at this is as follows: By looking at the OCC curve, you find the IF needed for opencircuit VT , but this then will drop to 480V when the load is connected.

Voltage = 480VFrequency, fe = 60HzConnection = ∆Number of poles, p = 4Synchronous reactance,Xs = 0.1ΩArmature resistance,RA = 0.015ΩFull load current, IL = 1200A, 0.8 PF lagg.

= 1200A∠−36.87o

Full load friction/windage losses = 40 kWFull load core losses = 30 kW(a) Speed of rotation, fm =?(b) Field current If if no load VT = 480V =?(c) Field current If if IL = 1200A, 0.8 PF lagging load and VT = 480V =?(d) Input power Pin, output power Pout , efficiency η =?(e) If load suddenly disconnected ,VT =?(f) Field current If if IL = 1200A, 0.8 PF leading load and VT = 480V =?

∆-connection ⇒ VT = Vφ, IL =√

3Iφ =√

3IANo load ⇒ VT = EA

,

338 / 412

Example 5.2 cont.

Synchronous generator: changing load conditions

(a) fm = fep/2

= 604/2

= 30Hz = 1800rpm

(b) If = 4.5A read directly from OCC curve (VT vs If plot)

(c) IA = Iφ = IL√3

= 1200∠−36.87o√

3= 692.8∠−36.87o

EA = Vφ + RAIA + jXs IA= 480∠0o + (0.015)(692.8∠−36.87o ) + j(0.1)(692.8∠−36.87o ) = 532∠5.3o

If = 5.7A from OCC curve

,

339 / 412

Example 5.2 cont.

Synchronous generator: changing load conditions(d) Pout =

√3VT IL cos θ

=√

3(480)(1200) cos(−36.87o )= 798 kW

Pelec. losses = 3IA2RA

= 3(692.8)2(0.015)= 21.6 kW

Pin − P stray losses − Pfric.&wind. losses − Pcore losses − PCu losses = Pout

⇒ Pin − P stray losses − Pmech. losses − Pcore losses − Pelec. losses = Pout

⇒ Pin − 0 kW − 40 kW − 30 kW − 21.6 kW = 798 kW⇒ Pin = 889.6 kW

efficiency η = PoutPin× 100% = 798 kW

889.6 kW× 100% = 89.75%

,

340 / 412








,

Example 7-1DC motors: simple rotating loop


341 / 412








,

Example 7-1 cont.

DC motors: simple rotating loopFor this motor,

• The area of the rotor under each pole is A = Ap = πrldue to the curved nature of the stator poles

• K = 2π

Chapman 5th ed, pg 413, 406 (figure)

342 / 412








,

Example 7-1 cont.

DC motors: simple rotating loop


343 / 412








,

Example 7-1 cont.



344 / 412








,

Example 7-1 cont.



345 / 412








,

Example 7-1 cont.



346 / 412








,

Example 7-1 cont.



347 / 412








,

Example 7-1 cont.



348 / 412

Example 7-1 cont. I


1 %Example 7 . 1 , Chapman 5 th ed , pg 4132 c l e a r ; c l c ; format compact

4 % =================================5 % INITIALIZATION6 % =================================7 % p h y s i c a l8 r = 0 . 5 ; % r a d i u s ( mete r s )9 l = 1 ; % l e n g t h ( mete r s )

10 A = p i∗ r∗ l ; % a r e a ( mete r s ˆ2)11 K = 2/ p i ; % machine c o n s t a n t ( no u n i t s )

13 % e l e c t r i c a l14 R = 0 . 3 ; % r e s i s t a n c e ( ohms )15 VB = 1 2 0 ; % a p p l i e d v o l t a g e ( v o l t s )

17 % magnet ic18 B = 0 . 2 5 ; % magnet ic f l u x d e n s i t y ( t e s l a s )19 p h i = B∗A ; % p h i ( webers )

21 % =================================22 % COMPUTATIONS23 % =================================24 %p a r t ( b )25 %−−−−−−−−26 I s t a r t = VB/R ; %s t a r t u p

28 e i n d = VB; %no l o a d s t e a d y s t a t e , c u r r e n t =029 w = e i n d /(K∗p h i ) ; %”

,

349 / 412

Example 7-1 cont. II


31 %p a r t ( c )32 %−−−−−−−−33 tau = 1 0 ; % l o a d t o r q u e (Nm)34 I = tau /(K∗p h i ) ; % s t e p 2 : f o r c e / t o r q u e e q u a t i o n35 e i n d = VB−I∗R ; % s t e p 1 : KVL ( motor e q u a t i o n )36 w = e i n d /(K∗p h i ) ; % s t e p 4 : Faraday

38 P mech = tau ∗ w ; % output power : m e c h a n i c a l39 P e l e c = VB ∗ I ; % i n p u t power : e l e c t r i c a l

41 %p a r t ( d )42 %−−−−−−−−43 tau = 7 . 5 ; % l o a d t o r q u e (Nm)44 I = tau /(K∗p h i ) ; % s t e p 2 : f o r c e / t o r q u e e q u a t i o n45 e i n d = VB+I∗R ; % s t e p 1 : KVL ( g e n e r a t o r e q u a t i o n )46 w = e i n d /(K∗p h i ) ; % s t e p 4 : Faraday

48 %p a r t ( e )49 %−−−−−−−−50 B = 0 . 2 ;51 p h i = B∗A ; % p h i ( webers )52 e i n d = VB; % no l o a d s t e a d y s t a t e , c u r r e n t =053 w = e i n d /(K∗p h i ) ; % ”









,

Example 7-1 cont.


• Dimensions

• r = 0.5 m• ` = 1.0 m

• Field

• B = 0.25 T = 0.25 Wb/m2

• φ = BAp = B(πr`) = (0.25 T)(π × 0.5 m× 1.0 m) =0.125π Wb

• Ap is area of rotor under pole face

• External

• VB = 120 V• R = 0.3 Ω

Chapman 5th ed, pg 413, 409 (Ap )

351 / 412








,

Example 7-1 cont.


(a) At t = 0, the following sequence occurs

1 Voltage and velocity: eind = 0, ω = 0

2 Current and torque: i = VB−eindR

= 1200.3

= 400 A,

τind = 2πφi = 2

π(0.125π)(400) = 100 NM

3 Velocity and voltage ↑: Motor starts to rotate, i.e., ω starts

to increase causing eind = 2πφω to increase

4 Current and torque ↓: This decreases i and therefore τind

Steady state is reached with τind = 0 and eind = VB . So, we wentfrom eind = 0 to eind = VB


352 / 412








,

Example 7-1 cont.


(b) i = 400 A (see previous part)

eind = 2πφω

120 = 2π

(0.125π)ω

⇒ ω = 480 rad/sec


353 / 412








,

Example 7-1 cont.


(c)

τind = 2πφi

10 = 2π

(0.125π)i

⇒ i = 40 A

eind = VB − iR motor

= 120 V− (40 A)(0.3Ω)

= 108 V

eind = 2πφω

108 = 2π

(0.125π)ω


Power supplied to shaft = τω = (10 NM)(432 rad/sec) = 4, 320WPower out of battery shaft = VB i = (120 V)(40A) = 4, 800W


354 / 412








,

Example 7-1 cont.


(d)

τind = 2πφi

7.5 = 2π

(0.125π)i

⇒ i = 30 A

eind = VB + iR generator

= 120 V + (30 A)(0.3Ω)

= 129 V

eind = 2πφω

129 = 2π

(0.125π)ω



355 / 412








,

Example 7-1 cont.


(e) φ = BAp = B(πr`) = (0.2 T)(π × 0.5 m× 1.0 m) = 0.1π Wb

eind = 2πφω

120 = 2π

(0.1π)ω


When the flux decreases, the speed increases !


356 / 412








,

Example 8-1DC shunt motor


357 / 412








,

Example 8-1 cont.

DC shunt motor


358 / 412








,

Example 8-1 cont.

DC shunt motor


359 / 412








,

Example 8-1 cont.

DC shunt motor


360 / 412








,

Example 8-1 cont.

DC shunt motor


361 / 412








,

Example 8-1 cont.

DC shunt motor


362 / 412








,

Example 8-1 cont.

DC shunt motor


363 / 412

Example 8-1 cont. I

DC shunt motor

1 %Example 8 . 1 , Chapman 5 th ed , pg 4722 c l e a r ; c l c ; format compact ;

4 w vec = [ ] ;5 t a u v e c = [ ] ;

7 V T = 2 5 0 ;8 R A = 0 . 0 6 ;9 R F = 5 0 ;

10 w n l = 1 2 0 0 ;

12 % p a r t ( a )13 I L = 1 0 0 ;14 I F = V T/R F ;15 I A = I L − I F ;16 E A = V T−I A∗R A ;17 w = (1200/250)∗E A ;18 tau = E A∗ I A /(w∗2∗ p i /60) ;19 w vec = [ w vec w ] ;20 t a u v e c = [ t a u v e c tau ] ;

22 % p a r t ( b )23 I L = 2 0 0 ;24 I F = V T/R F ;25 I A = I L − I F ;26 E A = V T−I A∗R A ;27 w = (1200/250)∗E A ;28 tau = E A∗ I A /(w∗2∗ p i /60) ;29 w vec = [ w vec w ] ;

,

364 / 412

Example 8-1 cont. II

DC shunt motor

30 t a u v e c = [ t a u v e c tau ] ;

32 % p a r t ( c )33 I L = 3 0 0 ;34 I F = V T/R F ;35 I A = I L − I F ;36 E A = V T−I A∗R A ;37 w = (1200/250)∗E A ;38 tau = E A∗ I A /(w∗2∗ p i /60) ;39 w vec = [ w vec w ] ;40 t a u v e c = [ t a u v e c tau ] ;

42 % p a r t ( d )43 p l o t ( t a u v e c , w vec ) ;44 ho ld on ;45 p l o t ( t a u v e c , w vec , ’ o ’ ) ;46 g r i d on ;









,

Example 8-1 cont.

DC shunt motor

150 200 250 300 350 400 450 500 550 6001110

1120

1130

1140

1150

1160

1170

1180

Torque (NM)

Ang

ular

vel

ocity

(rp

m)


366 / 412








,

Example 8-1 cont.

DC shunt motor

1 Input: constantTerminal voltage, VT = 250 V is constant.(remember that terminal voltage supplied by the user is used togenerate both field and armature currents)

2 System: unchanged

1 Stator (field): Since field resistance and VT are constant, IFis constant and so flux φ is constant

2 Rotor (armature): The input and system are unchanged,

and soEA0EA1

= ω0ω1

3 Output: changingLoad current IL increases from 100A to 200A to 300A

4 No load conditions IL = 0⇒ EA = VT

EA0=250 V

ω0 = 1200 rev/min60 sec/min

= 20 rev/sec


367 / 412








,

Example 8-1 cont.

DC shunt motor

• I need ω1, ω2, ω3 but I do not have EA1,EA2

,EA3

• All I have is IL1, IL2

, IL3

• So, let’s see how to get EA for a given IL

• We have 7 variables (3 unknown) and 3 equations:

1 VT

2 IL

3 IF : unknown4 RF

5 EA: unknown6 IA: unknown7 RA

VT = EA + IARA

VT = IF RF

IL = IF + IA


368 / 412








,

Example 8-1 cont.

DC shunt motor

• I want a relation between IL and EA. We have:

(a) IA = IL − IF (b) IF = VTRF

and therefore, IA = IL − VTRF

,

VT = EA +

(IL − VT

RF

)RA

⇒ 250 = EA +

(IL − 250

50

)0.06

⇒ EA = 250− 0.06(IL − 5)⇒ EA1

= 244.3V (IL1= 100A)

⇒ EA2= 238.3V (IL2

= 200A)⇒ EA3

= 232.3V (IL3= 300A)


369 / 412








,

Example 8-1 cont.

DC shunt motor

EA0EA1

= n0n1⇒ 250

244.3 = 1200n1

⇒ n1 = 1173 rpm

EA0EA2

= n0n2⇒ 250

238.3 = 1200n2

⇒ n2 = 1144 rpm

EA0EA3

= n0n3⇒ 250

232.3 = 1200n3

⇒ n3 = 1115 rpm


370 / 412








,

Problem 8-1DC shunt motor


371 / 412








,

Problem 8-1 cont.

DC shunt motor


372 / 412








,

Problem 8-1 cont.

DC shunt motor

In this question, we useEA1EA2

= Kφ1ω1Kφ2ω2

= ω1ω2

, where φ1 = φ2

since the field current does not change.First, the no load condition gives us the first two lines below, while KVLand the magnetization curve gives us the third and fourth lines:

IF = 0.96Aω1 = ? rpm

EA1 = 240 V (at no load)

ω1 = 1800 rpmEA1 = 241 V (acting as generator)

This question can be done in 1 step:

1 We haveEA1EA2

= ω1ω2⇒ 240

241= ω1

1800⇒ ω1 = 1793 rpm

NOTE: Just because rated velocity is 1800 rpm DOES NOT MEAN

that this is no-load velocity. It appears like that in Example 8.1(pg 472),

but over there, it is clearly mentioned that no load ω is 1200 rpm which

happens to be the same as the rated ω


373 / 412








,

Problem 8-1DC shunt motor

• Notice that IF is held constant at 0.96A, whether no load or

whether loaded

• First operationg point: For the above IF , one possibleoperating point that we get from the magnetization curve isEA = 241 V and ω = 1800, and it appears that thisoperating point is achieved IF the machine is acting as agenerator

• Second operationg point: For the above IF , anotherpossible operating point is at EA = 240 V, i.e., no load


374 / 412








,


• The distortion of the flux in a machine as the load is increased iscalled armature reaction.

• To take care of this, compensating windings are connected inseries with the rotor windings, so that whenever the load changesin the rotor, the current in the compensating windings changes,too

Chapman 5th ed, pg 486, 433 (armature reaction), 443 (compensating windings)

375 / 412








,

Example 8-3 cont.

DC shunt motor


376 / 412








,

Example 8-3 cont.

DC shunt motor


377 / 412








,

Example 8-3 cont.

DC shunt motor


378 / 412








,

Example 8-3 cont.

DC shunt motor


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,

Example 8-3 cont.

DC shunt motor


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,

Example 8-3 cont.

DC shunt motorIn this question, we use

EA1EA2

= Kφ1ω1Kφ2ω2

= φ1ω1φ2ω2

. First, simple

KVL and the given fact that EA does not change, gives usthe first two lines below, while the magnetization curve gives us thethird and fourth lines:

IF 1 = 6A IF 1 = 5A

ω1 = 1103 rpm ω2 = ? rpmEA1 = 246.4 V EA2 = 246.4 V

ω1 = 1200 rpm ω2 = 1200 rpmEA1 = 268 V EA2 = 250 V

This question can be done in 2 steps:

1 We use the magnetization curve data (third and fourth lines) asexplained in this slide to getEA1EA2

= φ1ωφ2ω⇒ 268

250= φ1

φ2⇒ φ1

φ2= 1.076

2 Now, using the KVL data (first and second lines), we getEA1EA2

= φ1ω1φ2ω2

⇒ 1 = 1.076 1103ω2⇒ ω2 = 1187 rpm

Notice that the field current ratioIF 1IF 2

= 65

= 1.2 is different from the

flux ratio φ1φ2

= 1.076 showing the non-linearity due to saturation effects


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,

Example 8-4 cont.

DC shunt motor


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,

Example 8-4 cont.

DC shunt motor


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,

Example 8-4 cont.

DC shunt motor

The important thing to note in this question is that although VA

changes, IA does not change as the load torque and flux are constant

EA1 = VA − IARA = 250− 120 ∗ 0.03 = 246.4VEA2 = VA − IARA = 200− 120 ∗ 0.03 = 196.4V

So,EA1EA2

= Kφ1ω1Kφ2ω2

246.4196.4

= 1103ω2

⇒ ω2 = 879 rpm

Therefore, if we decrease the voltage VA on the rotor, its speed

decreases


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,

Example 8-5DC series motor


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,

Example 8-5 cont.

DC series motor


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,

Example 8-5 cont.

DC series motor


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,

Example 8-5 cont.

DC series motor


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,

Example 8-5 cont.

DC series motor

In this question, we useEA1EA2

= Kφ1ω1Kφ2ω2

= ω1ω2

, where φ1 = φ2 since the

field current does not change.First, simple KVL gives us the first two lines below, while themagnetization curve gives us the third and fourth lines:

IA = 50A (NIA = 1250A)

ω1 = ? rpmEA1 = 246 V

ω1 = 1200 rpmEA1 = 80 V

This question can be done in 1 step:

1 We haveEA1EA2

= ω1ω2⇒ 246

80= ω1

1200⇒ ω1 = 3690 rpm


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,

Example 8-5 cont.

DC series motor

Pconv = EAIA = τindω

⇒ τind = EAIAω

= (246 V)(50 A)(3690 rpm)(2π rad/rev)(1 min/60 sec)

= 31.8 NM


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,

Example 8-8DC shunt motor: efficiency


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,

Example 8-8 cont.

DC shunt motor: efficiency


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,

Example 8-8 cont.

DC shunt motor: efficiency


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,

Example 8-9DC separately excited generator


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,

Example 8-9 cont.

DC separately excited generator


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,

Example 8-9 cont.



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,

Example 8-9 cont.



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,

Example 8-9 cont.



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,

Example 8-9 cont.



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,

Example 8-9 cont.



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,

Example 8-9 cont.



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,

Example 8-9 cont.

DC generator: separately excited

(a)

IF = VFRadj +RF

= 43063+20

= 5.2 A

From the magnetization curve, this corresponds to EA = 430 V at1800 rpm. However, the generator is rotating at 1600 rpm.

EA0EA

= n0n

derivation

430EA

= 18001600

⇒ EA = 430×16001800

= 382 V

Since this is no-load, VT = EA = 382 V


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,

Example 8-9 cont.


(b)IA = 360 A

EA = IARA + VT

382 = 360(0.05) + VT

⇒ VT = 364 V


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,

Example 8-9 cont.


(c) No compensating windings ⇒ armature reaction

IF = VFRadj +RF

− 450 A turns1000 turns

= 43063+20

− 0.45

= 4.75 A

From the magnetization curve, this corresponds to EA = 410 V at1800 rpm. However, the generator is rotating at 1600 rpm.

EA0EA

= n0n

derivation

410EA

= 18001600

⇒ EA = 410×16001800

= 364 V

EA = IARA + VT

364 = 360(0.05) + VT

⇒ VT = 346 V (lower than before due to armature reaction)


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,

Example 8-9 cont.


(d) To restore VT to that in part (a), we need to increase EA.For this, we need to increase IF .For this, we need to decrease Radj .


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,

Example 8-9 cont.


(e) We have compensating windings, i.e., part (b) and wewant to restore VT from 364 V (part (b)) to the no-load 382 V(part (a))

EA = IARA + VT

= (360)(0.05) + 382

= 400 V at 1600 rpm

400EA

= 16001800

⇒ EA = 450 V at 1800 rpm

From the magnetization curve, this corresponds to IF = 6.15 A at1800 rpm.

IF = VFRadj +RF

6.15 = 430Radj +20

⇒ Radj = 49.9Ω ≈ 50Ω


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,

Sample quizQuestions

1 What is the difference between field and armature?

2 Draw the equivalent circuit of a DC motor.

3 In a DC machine, torque depends on which 2 quantities?

4 In a DC machine, induced voltage depends on which 2 quantities?

5 What is meant by flux?

6 What does the magnetization curve show?

7 In a transformer, what causes the voltage from the primary toappear on the secondary?

8 Why is the startup current of a motor high?

9 If I want to develop an emf on a wire, what should i do?

10 If I want to develop a force on a wire, what should i do?

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,

Sample quiz cont.

Answers

1 What is the difference between field and armature? ”Field”windings applies to the windings that produce the main magneticfield in a machine, and the term ”armature” windings applies tothe windings where the main voltage is induced (Chapman, pg267).

2 Draw the equivalent circuit of a DC motor. see here

3 In a DC machine, torque depends on which 2 quantities?I = KφIA

4 In a DC machine, induced voltage depends on which 2 quantities?v = Kφω

5 What is meant by flux? B field passing through a surface

6 What does the magnetization curve show? Plot of flux vs themmf producing it (Chapman, pg 21), or EA vs IF for a fixed speedfor a DC machine (537)

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,

Sample quiz cont.

Answers

7 In a transformer, what causes the voltage from the primary to

appear on the secondary? Refer to Chapman, pg 78

1 Voltage ep is applied on primary coil2 Current ip flows through primary coil according to ep = ipRp

3 Flux φ is created according to Ampere’s Law which flowsthrough core

4 Since the flux is not changing, it does not induce a voltagees on the secondary coil

5 Now, change voltage ep on the primary side. This causes achanging current on the primary side, and therefore achanging flux.

6 This changing flux causes an induced voltage s on thesecondary coil according to Faraday’s Law.

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,

Sample quiz cont.

Answers

8 Why is the startup current of a motor high? Refer to Chapman,pg 573

9 If I want to develop an emf on a wire, what should i do?eind = (v × B).`, i.e. move the wire with length ` at velocity vthrough a magnetic field B

10 If I want to develop a force on a wire, what should i do?Find = i(`×B), i.e. pass a current i through wire with length ` inpresence of magnetic field B

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Sample OHTQuestions

1 A 50-hp 250-V, 1200 r/min dc shunt motor with compensating windings has an armatureresistance (including brushes, compensating windings, and interpoles) of 0.06 Ω. Its field currenthas a total resistance Radj + RF of 50 Ω, which produces a no-load speed of 1200 r/min. Thereare 1200 turns per pole on the shunt field windings. Find the speed of this motor when its inputcurrent is (a) 100A (b) 200A (c) 300A.

2 A 480-V, 60-Hz, ∆-connected, four-pole synchronous generator has the OCC shown below:

The generator has a synchronous reactance of 0.1Ω, and an armature resistance of 0.015Ω. Atfull load, the machine supplies 1200A at 0.8PF lagging. Under full-load conditions, the frictionand windage losses are 40kW, and the core losses are 30kW. Ignore any field current losses.(a) What is the speed of rotation? (b) How much IF must be supplied to the generator to makeVT = 480V at no load? (c) If the generator is now connected to a load and the load draws1200A at 0.8 PF lagging, how much IF will be required to keep VT = 480V? (d) How muchpower is the generator now supplying? (e) How much power is supplied by the prime mover?

3 Explain the operation of a synchronous generator operating at lagging power factor.4 What is the difference between a DC machine, a synchronous machine and an induction machine?

,