electric engineering ii ee 326 lecture 2 &...

Electric Engineering II “Dr. Ahmed El-Shenawy”

Electric Engineering IIEE 326

Lecture 2 & 3

<Dr Ahmed El-Shenawy>


DC Machines

The direct current (dc) machine can be used as a motor or as a generator.

DC Machine is most often used for a motor.

The major advantages of dc machines are the easy speed and torque regulation.

However, their application is limited to mills, mines and trains. As examples, trolleys and underground subway cars may use dc motors.

Automobiles are equipped with dc dynamos to charge their batteries.

Most DC machines are similar to AC machines: i.e. they have AC voltages and current within them. DC machines have DC outputs just because they have a mechanism converting AC voltages to DC voltages at their terminals. This mechanism is called a commutator; therefore, DC machines are also called commutating machines.

Electric Engineering II “Dr. Ahmed El-Shenawy”DC Generator

In a generator, moving a conductor through a stationary magneticfield generates voltage. If a coil is rotated through a magnetic field as shown, an alternating voltage will be produced.


DC Generator

For the external circuit to produce DC voltage, it is necessary to reverse the polarity of the external leads at the same time the voltage in the coil is reversed. This is accomplished by segmenting a slip ring to form what is called a commutator



Improving the DC output waveform:

To improve the DC signal 4 coils and 4 commutators are introduced. The voltage between the bruches is more uniform.

Electric Engineering II “Dr. Ahmed El-Shenawy”DC Machines Construction

The construction is similar for both d.c generators and d.c motors

1. Stator

• Provides mechanical support for the machine

• consists of the yoke and poles

Yoke provides high permeable path for

magnetic flux (thin laminated steel sheets)

Poles are designed to accommodate field

windings

Electric Engineering II “Dr. Ahmed El-Shenawy”DC Machines ConstructionStator

Electric Engineering II “Dr. Ahmed El-Shenawy”DC Machines Construction

2. Field Winding

The coils are wound in such a way that the poles

alternate in their polarity

a)Shunt Field Winding: many turns of fine wire

b)Series Field Winding: few turns of heavy wire,

which are placed in series with armature winding

3. Armatureo The rotating part of the dc machine

o Has Circular Cross section and is made of thin

laminations that have axial slots to house the

armature coils


DC Machines ConstructionArmature


DC Machines ConstructionArmature

Electric Engineering II “Dr. Ahmed El-Shenawy”DC Machines ConstructionArmature


DC Generators

Electric Equivalent Circuit of a DC Generator

E+-Ra

ia

RL

Field Winding

Field winding is responsible of setting up a flux


Types of DC Generators

1. Permanent Magnet Generators

This type is characterized by being smaller, lighter and more efficient

compared to wound generators

Ia = IL

E+-Ra

Ia

RL

IL

φ Constant Flux


2. Electro-magnet Generators

E+-

RaIa

RLIL

RfIf

VF

a. Separately Excited

Ia = IL

b. Self Excited

Rf

E+-

RaIa

RLIL

If

Ia = If + IL


Induced E.M.F Equation

e.m.f “EC” is generated in armature windings

EC = BLVB: Flux density (T)

L: length of conductor (m)

V: velocity of conduction motion

N

S

L

rV =Wr

B =φ / A = φ/ L Ap = φ/ [L(πD/P)]

V = Wr = (2πN/60)r (N: rev/min)

Then

Machine e.m.f “Ea” = EC (total no. of conductors/no. of parallel paths)

Where Ka = ZP/2 π a ……………….. Machine constant

And Z = 2 C NC (C:no. of coils, NC : no. of turns)

WKaZPWE paa

2

22)(PWWDL

PDLE c


Example

Calculate the voltage generated by a six pole dc machine if its windings are: 1)

lap wound 2) Wave wound, if the flux per pole is 0.05 wb. and the generator

speed is 120 rev/min, 200 armature coils with 15 turns each

Solution

Z = 2CNC = 2×200×15 = 6000 conductors

1)Lap wound

a = P = 6

Ea = Ka φp w = [6000×6/2π ×6]×0.05×[2π×120/60] = 600 V

2)Wave wound

a=2

Ea = Ka φp w = [6000×6/2π×2]×0.05×[2π×120/60] = 1800 V

Electric Engineering II “Dr. Ahmed El-Shenawy”Torque Equation

P = Ea Ia = Ka φp w Ia = T × w

Then the developed torque is given by: Td = P/w = Ka φp Ia

ExampleA 24 slot, 2 pole Dc machine has 18 turns per coil, the average flux density per

pole is 1T. The effective length is 20 cm and the armature radius is 10 cm. the magnetic poles are designed to cover 80 % of the armature periphery. If the armature velocity is 183.2 rad/sec and the armature current is 25 A, determine:• the current in each conductor• the developed torque• the developed power

Solution•As the machine is a 2 pole,

•then IC = Ia / 2 = 25/2 = 12.5 A

• Ap = 2πrL/p = 2×3.14×0.1×0.2/2 = 0.063 m2,

•as Ae = 0.8 Ap ,

•then Ae = 0.8×0.063 = 0.05 m2,


•then φp = B Ae = 1×0.05 = 0.05 wb

•and Ka = (Zp/2πa) = (2×24×18)×2/2П×2 = 137.51

•Td = Ka Ia φp = 137.51×25×0.05 = 171.89 N.m

• Pd = Td ×w = 171.89×183.2 = 31490 W


Magnetization characteristics of DC machine

As Ea α φp w, now if the armature circuit is left open, and armature is rotated at rated

speed, then Ea = K1 φp = Ka wm φp and since φp = Kf If (flux per pole depend on the

m.m.f provided by current If ) then Ea = K1 Kf If

Field Current If

Indu

ced

E.M

.F a

t no

load Air gap line

Magnetic circuitAir gap

m.m.f

From previous operation of the machine Er

Since Ea is an indirect measure of the

flux per pole and since If is a measure

of the applied m.m.f (Nf If ), then the

curve is similar to a B-H curve

m.m.f required for magnetic material is almost negligible at small values of flux density


DC Machines

Armature Circuit Model

The equivalent circuit model for the armature of a dc machine is below. The induced armature voltage, EA is represented by a voltage source, connected via 2 brushes to the rest of the circuit. The armature winding resistance is RA and terminal voltage is

VT.

The armature circuit behavior is dependent on the flux in machine, which is traditionally controlled by a field winding. We will consider three types of wound field DC machine. 1) Separately Excited 2) Shunt Excited 3) Series Excited


DC Machines

Separately Excited

Self Excited

Series woundShunt wound

Electric Engineering II “Dr. Ahmed El-Shenawy”Separately Excited Dc Generator

Ea = Vt + Ia Ra

IL = Ia

Vf = If Rf E+-

RaIa

RL IL

RfIf

VF

Vt+-

Ea = VN L

Vt

Drop due to armature resistance and reaction

Speed


Shunt Generator

If = Vt /Rf

Ia = IL +If

Vt = Ea – IaRa =ILRLE

+-

RaIa

RLIL

Rf

If

Ia

- Vt +

Term

inal

Vol

tage

If

Field re

sistance lin

e

Magnetization curveVNL

Er

•Er induces an e.m.f in armature winding (Er) and as the field is parallel then a small Ifflows giving a m.m.f that sets up a flux giving aid to the residual flux.•An increase in the flux per pole will increase the induced e.m.f thus increasing If•The shunt generator continues to build up voltage until the point of intersection of the field resistance line and the magnetization saturation curve


Note :

A decrease in the field circuit resistance will cause the shunt generator to build up

faster to a higher voltage, the generator will not build up if the field resistance is

greater than or equal the critical resistanceTe

rmin

al V

olta

ge

If

VNL

Er

RF1

RFRF2

RC

RC >RF2 >RF >RF1


Series Generator

Ea

+

-

Ra

Ia

RL

Rf-

V t+

TS

wm

ILIL = Vt /RL

Ia = IL = If

Vt = Ea – Ia Ra –Ia Rf

•Series generators are used to supply a constant load current

•Rf should be small to reduce voltage drop (so it is made of thick wire)


Voltage Regulation

Voltage regulation is a measure of the terminal voltage drop at full load

V.R % = 100×(VNL – VFL )/VFL

VNL: NO load terminal voltage

VFL : Full load terminal voltage

Losses in DC Generators

Input Mechanical Power

Rotational losses “Pr” Developed Power “Pd” = EaIaMechanical Losses:• Windage (drag on armature caused by air)• Friction (brushes-commutator , bearing-shaft)Magnetic Losses:• Hysteresis losses• Eddy losses

• Output Power “Po” =Vt IL• Copper Losses “PCU” = I2R


Power Flow Diagram

Pin = TSWm

P r(R

otatio

nal)

Pd =TdWm= EaIa

PCU = I 2R

P0 = Vt IL

Efficiency “η”η %= (P0 /Pin) ×100


Example:

The d.c Separately-excited generator shown in Fig produces an open circuit emfE of 220 V. Calculate the Terminal voltage V when generator supplies a current of 10 A. The armature resistance is 1 ohm

Vt=Ea – Ra Ia

Where Ea= 220 V Ia= 10 A Ra= 1 ohm

Vt=220 – (10)(1)=210 V


Example:

A separately-excited d.c. generator produces an open-circuit voltage of 250 V with field current of 1.5 A. If the field current is increased to 2 A, calculate

a) The new open-circuit emf.b) The terminal voltage if the generator supplies a current of 5 A. Assume

armature resistance- 0.8 ohm and the speed remains constant.

Solution:

a) Since the speed is constant b) Vt=Ea – Ra Iab)

Where E1=250 V, If1= 1.5 A and If2 = 2A

E2=250 (2/1.5)=333 V

Where Ea=333 V, Ia= 5 A and Ra = o.8 ohm

V= 333 – (5)(0.8)= 333 – 4= 329 V

2

1

2

1

f

f

II

EE


Example

A separately excited DC generator has a field resistance of 50 ohm, armature resistance of 0.125 ohm and brush drop of 2V. At no load the generated voltage is 275 V, and the full load current is 95 A. the field excitation voltage is 120 V and the friction windage and core losses are 1500 watt. Calculate :

• The rated terminal voltage and output power• the efficiency at full load

E

+-

RaIa

RL IL

RfIf

VF

Vt+-

Solution

Vt = Ea –IaRa –VB = 275-95×0.125-2 = 261.13 V

P0 = Vt Ia = 26.13×95=24.81 Kw

Pin = P0 + Plosses

= 24.81×103+1500+952×0.125+1202/50

= 27.913 Kw

η = (24.81/27.913)×100 = 88.88 %


32

DC Motor

• In a dc motor, the stator poles are supplied by dc excitation current, which produces a dc magnetic field.

• The rotor is supplied by dc current through the brushes, commutator and coils.

• The interaction of the magnetic field and rotor current generates a force that drives the motor

theory of operation


Vt

Ia

RaRF

ILIF

Ea

Generator

Vt

Ia

RaRF

ILIF

Ea

Motor

Vt = Ea – IaRaVt = Ea + IaRa

Ea = KaφW

Driving e.m.f Back e.m.f


Starting of DC motor

VS = Ea + Ia R

At Starting Ea is zero (back e.m.f = 0), so Ia = VS/R , and since R is very small then Iais very big. Consequently a DC motor should never be started at rated voltage and an external resistance must be added and acts as temporarily starting resistance that is removed as soon as the armature has attained its normal speed, thenIa = VS/(Ra+RS)

Ea

+-

Ra RF

XF

VS

Ia

RS>>>>RaSpeed Regulation (SR): is a measure of the change in speed from no load to full loadAs Ia increases, Ea decreases thus N decreases

SR % = 100(NNL -NFL)/NNL = 100(WNL - WFL )/WNL

Series Motor: SR is high (variable speed motor)Shunt Motor: SR is very low (constant speed motor)


Power Flow Diagram

Pin = VSIL

P CU=I2 a

R a+I2 F

R F

Pd =TdWm= EaIa

Pr = P

core +PM

agP0 = TS Wm

Efficiency “η”

η %= (P0 /Pin) ×100

electric engineering ii ee 326 lecture 2 &...

Documents