LE260 : Electrical Machines 1

the principles and analysis of electromechanical systems. Students will develop analytical techniques for predicting device and system interaction characteristics as well as learn to design major classes of electric machines. Problems used in the course are intended to strengthen understanding of the phenomena and interactions in electromechanics, and include examples.

1. FUNDAMENTAL ELECTRICITY AND MAGNETISM

1.1 Conventional and electron current flow

           In the battery,there are positive and negative poles. To make the battery works, we have to connect the positive and negative to make the potential difference by the electric current flow from high potential (positive charge) to low potential (negative charge).

      Conventional current flow is the direction of the current flow in electric power technology which is from positive charge to negative charge. For the electron flow, it will flow in the opposite direction to the conventional current flow.  

1.2 Distinction between source and load

   Every electrical device will have a source which stand for the delivery of electrical power and a load which is the one that absorb the power.

   

     

  The way to classify which one is source and load is that we have to see the conventional current flow. For the source, the current will flow out from the positive terminal.On the other hands, for the load, the current will flow into positive terminal.

1.3-1.6 Sign notation

     For electricity, the positive and negative signs use to tell the direction. If you set one direction to be positive, the opposite direction will be negative.

    

 For voltage, there are double subscript notation to tell the potential difference and the polarities.

    For example, potential from A to B is +50V , it means that A is positive respect to B because the current flows from A to B as you see from the potential difference. If the value of potential difference is negative, it means that at node A has lower potential than node B.

 

    The same for current, it also use positive and negative sign to show the flow direction.It you fix one direction to be positive, you will know that the opposite direction will be represent in negative sign.

1.7 Graph of an alternating voltage

  The graph of an alternating voltage is the alternating voltage represent by the means of a graph. It means that you use the voltage value to draw the graph which x-axis is the time and y-axis is the voltage. From the graph, the negative region means that it has negative voltage ( changing in the direction of the positive and negative terminals).

1.8-1.9  Sinusoidal Voltage

  For ac voltage generation, the graph is nearly a perfect sine wave which can represent in the formula 

e=Emcos(2πft+θ)

                                         e= instantaneous voltage (v)                                 t=time(s)

                                      Em= amplitude (peak value of voltage)            θ=fixed angle (rad)

                                      f= frequency (Hz)

      

  To convert cosine function to sine function is to add right angle to the sine function because sine function and cosine function have phase difference 90 degree.

     sine function and cosine function have phase difference 90 degree.

   To convert cosine function to sine function is to add right angle to the sine function.

1.10 Effective value of an ac voltage

For ac voltage, it is better to use effective voltage value for voltage that varies sinusoidally. 


   The rms or effective voltage or current is the value that converts the same energy as does a dc value.

  In other words, the effective value is an equivalent DC value which tells you how many volts or amps of DC that a time-varying sinusoidal waveform is equal to in terms of its ability to produce the same power. 

The relationship between the effective value and maximum value are shown in the form of formula:

 

Eeff = Em/√2             Ieff = Im/ √2

 

      Sometimes, effective voltage value called RMS or root mean square. It also use to measure of heating effect compare to dc voltage. Additional, most of the electrical equipment such as ammeters and voltmeters measure both current and voltage in effective value.

 

1.11 Phasor representation

We will not concern much with the instantaneous value because voltage is measured in the effective value and also phase angles rather than the maximum voltage value.

            The phasor diagram is to show the magnitude and phase angles between voltages and currents. Phasor is similar to a vector with the arrow because it shows both magnitude and also the direction in the form of angle. The angle between 2 phasors is equal to the electrical phase angle between the quantities.

There are some rules for the phasor :

  • Phasors in phase means that they are parallel and go in the same direction with no phase angle.
  • Phasors out of phase are when phasors point in the different directions and there are some phase angle.
  • Phasor rotate clockwise is said to be a lead phasor. On the other hands, the phasor will be a lag phasor if it rotate counterclockwise.
  • Phasors do not have to start from a common origin to show their magnitudes and phase relationship.
  • If one phasor has lag angle to another phasor when it rotates clockwise to be in phase with another phasor but it will be still some lag angle between them.

1.12 Harmonics

     Harmonics are the signal that come to distort the main signal and make the form of the graph change from the original. This distortion can be produced by magnetic saturation in the core of transformer or by the switching action of thyristors in the electronic drives.

 

   For example of 2 sources in the same series circuit with different voltage, it will result the terminal voltage wave in the form of flat-topped wave which is the non-sinusoidal waveform whose degree of distortion depends on the magnitude of the harmonic it contains.

    A square wave is composed of a fundamental wave and an infinite number of harmonics. The higher harmonics have smaller and smaller amplitude. However, these high-frequency harmonics produce the steep sides and pointy corners of the square wave. In practical way, square waves are not produced by adding sine waves but any wave shape can be built up from a fundamental wave and the number of harmonics.

 

   In ac circuit, the fundamental current and fundamental voltage together produce fundamental power which is the useful power that cause power to rotate. The product of a harmonic current also produces a harmonic power. Additional, the product of a fundamental voltage and a harmonic current yields zero net power.

1.13 Energy in an inductor

The energy store in the coil in its magnetic field when it has a current I is given by  the formula: 

where 

W = energy storing (J)

L= inductance of the coil (H)

I = current (A)

1.14 Energy in the capacitor

 A capacitor will store the voltage when a voltage appears across its terminals. The  energy is given by

 

where

W= energy stored in the capacitor (J)

C= capacitance of the capacitor (F)

E= voltage(V)

1.15 Some useful equations

The image above is the set of the equations that use to analyst the common AC circuit which will be useful to compose the impedances to be easy for calculation.

1.16 Magnetic field intensity H and flux density B

 The presence of magnetic flux intensity is given by the formula

 

H=U/I

Where

H= magnetic field intensity (A/m)

U = magnetomotive force acting on the component (A)

l = length of the component (m)

The formula for the flux density is

B=Φ/A

Where

B = flux density (T)

A= cross section of the component (m2)

Φ= flux in the component (Wb)

   There are the relationship between the flux density (B) and the magnetic field intensity is represented graphically by B-H curve of the material.

1.17 B-H curve of vacuum

   The B-H curve of vacuum is a straight line. A vacuum never saturates, no matter how great the flux density may be. For nonmagnetic material such as copper, paper, rubber,and air have B-H curves almost identical to the vacuum.

In vacuum, the magnetic flux density B is directly proportional to the magnetic field intensity H, and it shows in the equation 

 

B=μoH

Where

B= flux density (T)

H= magnetic field intensity (A/m) 

μo = magnetic constant =(4∏ x10-7)

1.18 B-H curve of the magnetic material

 The flux density of the magnetic material depends on the magnetic field intensity  and its value is given by

B=μoμrH

where

B= flux density (T)

H = magnetic field intensity (A/m)

μo= magnetic constant

μr= relative permeability of the material 

  Different magnetic material will have different typical saturated curve  because the value of μr is not constant so it makes the graph not linear.

 

 

 

1.19 Determining the relative permeability

 The relative permeability of the material is the ratio of the flux density in the material to the flux density that would be produced in vacuum under the same magnetic field intensity (H).

There are the approximate equation to find the relative permeability 

μo = 80,000B/H

where

B= flux density in the magnetic material (T)

H= corresponding magnetic field intensity (A/m)

  As the magnetic field intensity increases, the magnetic material saturated more and more and all B-H curves follow the B-H curve of the vacuum.

1.20 Faraday’s law of electromagnetic induction

   Faraday’s law of electromagnetic induction is the law of electromagnetism that represent the relationship between the voltage and flux in the circuit.

Faraday’s law states:

  • If the flux linking a loop (or turn) varies as a function of time, a voltage is induced between its terminals.
  • The value of the induced voltage is proportional to the rate of the change of the flux.

Faraday’s law states:

  • If the flux linking a loop (or turn) varies as a function of time, a voltage is induced between its terminals.
  • The value of the induced voltage is proportional to the rate of the change of the flux.

  By the definition, when the flux inside the loop varies at the rate of 1 Wb/sec, a voltage 1V is induced between its terminals. If the flux varies inside a coil of N turns, the voltage induced is given by:

E = N(∆Φ/∆t)

where

E= induced voltage (V)

N= number of turns in the coil

∆Φ= change interval during which the flux change (s)

∆t = time interval during which the flux change (s)

  

Faraday's law of electromagnetic induction is the basis of operation of transformers, generators ,and alternating current motors.


1.21 Voltage induced in the conductor

    In many motors, the coils move with respect to a flux that is fixed in space. The relative motion produces a change in the flux linking the coils and a voltage induced according to Faraday’s law. 

 

 

There are the special case to calculate the induced voltage with reference to the conductors rather than the coil itself. The value of the induced voltage is given by

E= B/v

where

E= induced voltage (V)

B= flux density (T)

I = active length of the conductor in the magnetic field (m)

v= relative speed of the conductor (m/s)

1.22 Lorentz force on a conductor

     Lorentz force or electromotive force is when a current caring conductor is placed in a magnetic field and cause this force to happen. This force is of fundamental importance because is the basic structure of the operation of the motors and generators of many electrical instrument.

 

  The magnitude of the force is depends on the orientation of the conductor respect to the field.The force is greatest when the conductor is perpendicular to the field and zero when it is parallel to it. Between these two extremes, the force has intermediate values.

  

 The maximum force acting on the straight conductor is given by:

F=BlI

where 

F= force acting on the conductor (N)

B= flux density of the field (T)

l= active length of the conductor (m)

I= current in the conductor (A)

1.23 Direction of the force acting on a straight conductor

  Whenever a conductor carries a current. It is surrounded by a magnetic field. For a current flow in to the page, the circular lines of force have the direction. The figure below shows the magnetic field created between the N,S poles of a powerful permanent magnet.

However, the magnetic field does not. Because as you can see from the figure because lines of force never cross each other.

    The lines of force created by the conductor and the permanent magnet act in the same direction above the conductor and in the opposite directions below it. These make the magnetic field change the shape. Additional, the line of flux act like stretched elastic bands. It tends to push it downward.

1.24 Residual flux density and coercive force

Residual flux density and coercive force

   the coil surround a magnetic material formed in the shape of the ring. A current source, connected to the coil will form a current whose value and direction can be change.

   This picture shows the method of determining B-H properties of magnetic material.

If you see from B-H curve,when the current increase B and H will be increase until it reach Bm and Hm which are maximum magnetic flux density and magnetic field strength.

  If current is gradually reduced to zero the graph will not follow the same line as it increase because as we reduce the magnetic field intensity, the magnetic strength is influenced by the field Hm to remain original orientation. This phenomenon  is called hysteresis.H is reduced to zero but the substantial flux density remain is called residual induction. 

  If we want to eliminate the residual flux, we have to reverse the current in the coil and gradually increase H in the opposite direction. When you reverse it until magnetic flux density become zero , this magnetic field intensity called coercive force.

  In reducing the flux density from residual induction to zero, we will gain a furnish energy. This energy is used to overcome the frictional resistance of the magnetic domains as they oppose the change in orientation. This energy is in the form of heat, it will make the ring temperature rise.

1.25 Hysteresis loop

The Hysteresis Loop and Magnetic Properties

  A great deal of information can be learned about the magnetic properties of a material by studying its hysteresis loop. A hysteresis loop shows the relationship between the induced magnetic flux density (B) and the magnetizing force (H). It is often referred to as the B-H loop. 

  Transformers and most electric motors operate on the alternating current. In the devices, the flux in the iron changes continuously both in value and direction. It will have both positive and negative residual induction and coercivity which we called this BH curve as hysteresis loop.

1.26 Hysteresis loss

    The work was done by the magnetising force against the internal friction of the molecules of the magnet, produces heat. This energy which is wasted in the form of heat due to hysteresis is called Hysteresis Loss. When in the magnetic material magnetisation force is applied, the molecules of the magnetic material are aligned in one particular direction, and when this magnetic force is reversed in the opposite direction, the internal friction of the molecular magnets opposes the reversal of magnetism resulting in Magnetic Hysteresis.

   To wipe out or overcome this internal friction or in other words known as residual magnetism, a part of the magnetising force is used. This work, done by the magnetising force produces heat; this wastage of energy in the form of heat is termed as hysteresis loss.



 

1.27 Hysteresis losses caused by rotation

  Hysteresis losses are also produced when a piece of iron rotates in a constant magnetic field.

 

  An armature AB, made of iron that revolves in  a field produced by permanent magnets N and S. The magnetic domains in the armature tend to line up with the magnetic field. As the armature rotates, the N poles of the domains point first toward A and then toward B. Hysteresis losses are produced just as they are in an ac magnetic field.

1.28 Eddy current

  There are an ac flux that link a rectangular-shaped conductor.According to Faraday's law, an ac voltage E1 is induced across its terminals. If the conductor is short-circuited, the alternating current will flow, it makes the conductor temperature increase.

 

 

  If the second conductor is placed inside the first, a smaller voltage is induced because it links a smaller flux. As you can see from the picture, the short- circuit current I2 is less than I1 which is the power dissipated in this loop. A currents are progressively smaller as the area of the loops surrounding the flux decreases.

  The ac flux passes through a solid metal plate is basically equivalent to a densely packed set of rectangular conductors touching each other. There are currents swirl back and forth inside the plate which is called eddy currents.

  Consequently, a metal plate that is penetrated by an ac flux can become very hot.so, you should take care of the transformers not to get overheat. 

  Due to the Lenz's law, the eddy currents flow in such a way as to oppose the increasing flux.

1.29 Eddy currents in a stationary iron core

    For all ac motors and transformers, there are a problem to eddy current when iron has to carry an ac flux. Because of the carrying of ac flux in the solid iron, the core can become overheat due to the eddy current losses. We can reduce the losses by splitting the core in two along its length because the voltage induced in each section is one half of what it was before. These will make the eddy current losses considerably reduced. The more you divide the iron core, the more losses decreasing.

 

    In practice, in the core, there are the stacked lamination which is a small amount of silicon. It will alloy with steel to increase the resistivity to reduce the losses of eddy current. Therefore, we can conclude that eddy current losses decrease in proportion to the square of the number of laminations.

1.30 Eddy-current losses in a revolving core

The stationary field in direct-current motor and generator produce dc flux. This flux induces eddy currents in the revolving armature.

  From the figure, there are the cylindrical iron core between the magnet. As it turns, the core cut the flux lines. From Faraday's law, a voltage is induced along its length having a polarities. It makes large eddy currents flow in the core because the resistance of the core is very low. These eddy currents produce large power loss in the form of heat. This power is proportional to the square of the speed and the square of flux density.

  In conclusion, to reduce the eddy current losses, we laminate the armature suing thin circular laminations that are insulated from each other. The lamination will stack with the flat side running parallel to the flux line.

1.31 Current in an inductor

In inductive circuit, the voltage and current are related by the equation:

e= L∆i/∆t

Where

e =  instantaneous voltage induced in the circuit (V)

L = inductance of the circuit (H)

∆i/∆t =  rate of change of current (A/s)

This equation enables to calculate the instantaneous voltage when we know the rate of change of current.

   However, when we want to calculate the current and we know the instantaneous voltage, it hards to calculate as you can see from the formula, it needs advanced mathematics to solve. We can use graphical method called volt- second method to solve. graphical method will show the trend of increasing and decreasing of the current with time in the response of voltage.

 

From the figures, the voltage applied across the inductance. If the inductance has the current I1 at time t1 and we want to determine the current in the short time.

If you want to calculate the current at time t2 when t2 is many change of time from t1. We have to increase the incremental changes in current during a long period (t2-t1).

 

The net area A after time interval T is equal to (A1-A2) volt-seconds.

To generalize, the current after interval always given by the formula

I = I1+A/L

where 

I1 = current at start interval T

I = current after time interval T

A= net area under voltage curve during time T

L = inductance

Consider the inductance having negligible resistance connected to the source which voltage is according to the curve in the figure below.

  If the initial current is zero, the value at t1 equal to I=A1/L. As time goes by the current increase in the same direction as the area under curve increase.However, when the current reaches the maximum at t2 the area cannot increase anymore so it starts to be negative. At t3 the area will include A1 and A2 also A3 in the negative way. So at t3, I=(A1+A2-A3)/L. At the same time, at t4 it will include area 4 in the negative sign same as A3.

  From another graph, you will see that the graph is charged between t1 to t2 and discharged from t2 to t4 which this characteristic is similar to the capacitor.

If the current at the beginning of interval T is not zero, we simply add the initial value to all the ampere values calculated by the volt-second method.

1.32 Kirchhoff's voltage law

  Kirchhoff's voltage law states that that the algebraic sum of the voltages around a close loop is zero. In the closed circuit, Kirchhoff's voltage law means that the sum of the voltage rises is equal to the sum of the voltage drops.

  

  Voltages can be expressed in either double-subscript or sign notation. If the current go from positive sign to negative sign, the voltage will represent in positive sign but if it goes in the opposite direction, voltage will represent in negative sign.

1.33  Kirchhoff's voltage law and double-subscript notation

  Because voltage is an "across" variable and exist between 2 points, the double-subscript notation defines differences in potential.

  The double- subscript notation Vab specifies point as a high potential. If this is not the case, the negative sign must be associated with the magnitude of Vab. 

Note that Vab is the voltage at point A respect to point B.

  From the figure below, there are six circuit elements connect in the close loop. In going around the circuit loop, we can start with any node and move either clockwise or counterclockwise direction until we come back to the starting point.

  The set of voltages designated by the KVL equations may be ac or dc. If they are in ac, the voltage will represent in the form of phasors, having certain magnitudes and phase angles.

1.34 Kirchhoff's current law

   Kirchhoff's current law states that the algebraic sum of currents that arrive at a point is equal to zero. It means that the sum of the current that flow into a terminal is equal to the sum of the currents that leave it.

1.35 Currents, impedances,and associated voltages

  Consider the impedance Z carrying a current I, connect between 1 and 2. A voltage E12, having a magnitude IZ will appear across the impedance. However, there are the question about voltage across E12 equal to IZ or -IZ.

  From the rule, you know that when moving across an impedance Z in the same direction as a current flow I, the voltage IZ will be positive sign. In the opposite way, if it goes against the current flow, the voltage IZ will represent in the negative sign. The current can be both dc or ac,and impedance can be resistive(R), inductive(jX1), or capacitive (-jXc).

 

  In the most circuits, it is impossible to predict the actual direction of current flow in the various circuit elements. For example below, there are 2 known voltage sources E13 and E24 are connected to 4 impedances. Because the actual directions of current flows are presently unknown, we will assume arbitrary directions.

1.36 Kirchhoff's law and ac circuits

  The same basic rules of writing double-subscript equations can be applied to ac circuits, including 3-phase circuits The only difference is that the resistive elements in dc circuits are replaced by resistive, inductive, or capacitive elements. Also, the voltages and currents are expressed as phasors which having both magnitude and angle.

1.37 KVL and sign notation

  Voltage in ac and dc circuits are frequently indicated with sign notation and designed by symbols such as E1, Ea,em,and so on. As we cruise around the loop, we observe the polarity of the first terminal of every voltage we meet. If only the positive terminal of the voltage source is marked, the unmarked terminal will be negative. The polarity will be use to write KVL equation.

1.38 Solving ac and dc circuits with sign notation

   In circuits using sign notation, we treat the IZ voltages in the same way as in circuits using double-subscript notation. The IZ voltage across an impedance Z is preceded by a positive sign whenever we move across the impedance in the direction of current flow and it will be negative if it goes the the opposite direction as the current flow.

1.39 Circuits and hybrid notation

In some circuits, it is useful to employ both sign notation and double-subscript notation.

 

Exercise 1

  • In phase means that one phasor is the part of another phasor
  • out of phase means two phasors go in the different direction.
  • Lead phase can be checked by looking at the magnitude of the phasor.
  • Lag phase can be checked by rotate clockwise and in phase with another phasor.
Which one is correct?

Exercise 2

  • Conventional current flow is the direction of current flow in electric power technology.
  • You can tell that it is the source by seeing the current flow out of the positive terminal.
  • The difference in potential caused by the electron at the negative terminal compare to the positive terminal.
  • Electron and electricity flow in the same direction in the electrical circuit.
Which one is wrong?

Exercise 3

  • increase magnetic flux density
  • reverse current
  • increase resistance
  • increase magnetic field intensity in the opposite direction
  • reduce the flux density
  • increase induced voltage
State 2 ways to eliminate the residual flux

Exercise 4

Hint: there are 2 ways to reduce the eddy current in the iron core.

  • Make the conductor become laminates.
  • increase the size of the iron core
  • mix silicon and steel to the iron core
  • reduce the air gap between the core
  • increase the cross-sectional area of the core
  • increase the volume of the core
State how to reduce eddy current ( There will be more than 1 way to reduce eddy current).

Exercise 5

  • The sum of current in the close loop equal to zero.
  • current that pass every resistance in the circuit will be equal.
  • Voltage around the close loop equal to zero.
  • current in the parallel circuit will equal to the current in series circuit.
What is Kirchhoff's current law ?

Exercise 6

Hint: use the formula of voltage induced in a conductor.

  • 10V
  • 20V
  • 30V
  • 40V

    A conductor 2m long moves at a speed of 60km/h through a magnetic field having a flux density of 0.6 T. Calculate the induced voltage.

Solution E.6

Exercise 7

  • 1.8V
  • 1.25V
  • 1.35V
  • 1.67V

    A coil having 200 turns links a flux of 3mWb, produced by a permanent magnet. The magnet is moved,and the flux linking the coil falls to 1.2m Wb in 0.2s. Calculate the average voltage induced.

Solution E.7

Exercise 8

  • 600N
  • 200N
  • 100N
  • 300N

    A conductor 3 m long carrying a current of 200A is placed in a magnetic field whose density is 0.5T. Calculate the force acting on the conductor if it is perpendicular to the lines of force.

Solution E.8

Exercise 9

  • a. 6A b. 200 V c. 1660W d. 3124 W
  • a. 12A b. 169.7 V c. 1440 W d. 2880 W
  • a. 8A b. 170.65 V c. 2567 W d. 1367 W
  • a. 5A b. 143.29V c. 1329 W d. 1236 W

 A sinusoidal voltage of 120 V is applied to a resistor 10 ohms. Calculate

  • a. the effective current in the resistor
  • b. the peak voltage across the resistor
  • c. the power dissipated by the resistor
  • d. the peak power dissipated by the resistor

 

Solution E.9

Exercise 10

  • I1=3A I2=2A I3=5A
  • I1=1.9A I2=-2.7A I3=4.87A
  • I1=1.43A I2=4A I3=-5.43A
  • I1=-1.25A I2=2.32A I3=-7.87A

Calculate the currents  in this circuit.

 

Solution E.10

2. MAGNETIC CIRCUIT

2.1 Magnetic Circuit

In electrical machines, the magnetic circuits may be formed by ferromag-netic materials only (as in transformers) or by ferromagnetic materials inconjunction with an air medium (as in rotating machines). In most electricalmachines, except permanent magnet machines, the magnetic field (or flux)is produced by passing an electrical current through coils wound on ferro-magnetic materials.

i-H RELATION 

first study how the current in a coil is related to the magnetic field intensity (or flux) it produces. When a conductor carries current a magnetic field is produced around it,

The equation        

                           ∮ Hdl = ∑i = i1 + i2  - i3               

                                                                                                   ∮ H dl cosθ = ∑i

                                                                       when θ = 0,     ∮ Hdl   =  i              

                                                                                                        H2πr  = i

 

B-H RELATION

The magnetic field intensity produces magnetic flux density everywhereit exists. These quantities are functionally related by      

                                                              B = μweber/m²      or   Tesla

                                                              B = μᵣμ₀/Wb/m²        or   

For free space or electrical conductors (such as aluminum or copper) or insulators, the value of μᵣ is unity. However, for ferromagnetic materialssuch as iron, cobalt, and nickel, the value of pr varies from several hundredto several thousand. For materials used in electrical machines, μᵣ varies in the range of 2000 to 6000.Alarge value of μᵣ implies that a small current can produce large flux density in the machine.

MAGNETIC EQUIVALENT CIRCUIT

a simple magnetic circuit having a ring-shaped magneticcore, called toroid, and a coil that extends around the entire circumference.

When current i flows through the coil of N turns, magnetic flux is mostlyconfined in the core material. The flux outside the toroid, called leakage flux,is so small that for all practical purposes it can be neglected.

∮ Hdl = Ni              

H2πr  = Ni

                                          Hl  = Ni  =  F

                                          H   = (N ⁄ l )i   At ⁄ m

                                          B    = (μNi ⁄ l )T

 

If we assume that all the fluxes are confined in the toroid, that is, there is no magnetic leakage, the flux crossing the cross section of the toroid is 

Φ = ∫B dA = BA Wb

where B is the average flux density in the core and A is the area of cross section of the toroid. The average flux density may correspond to the path at the mean radius of the toroid. If H is the magnetic intensity for this path, 

Φ = (μNi  ⁄ l)A  = (Ni  ⁄ l  ⁄ μA)  =  Ni  ⁄ ℛ  = F  ⁄ ℛ

ℛ = l /μA = 1/P

 

MAGNETIZATION CURVE 

If the magnetic intensity in the core is increased by increasing current, the flux density in the core changes. The flux density B increases almost linearly in the region of low values of the magnetic intensity H. However, at higher values of H, the change of B is nonlinear. The magnetic material shows the effect of saturation.

The B-H curve  is called the magnetization curve. The reluctance of the magnetic path is dependent on the flux density. It is low when B is ow, high when B is high. The magnetic circuit differs from the electric circuit in this respect; resistance is normally independent of current in an electric circuit, whereas reluctance depends on the flux density in the mag- netic circuit.  

MAGNETIC CIRCUIT WITH AIR GAP 

         In electric machines, the rotor is physically isolated from the stator by the air gap. Practically the same flux is present in the poles (made of magnetic core) and the air gap. To maintain the same flux density, the air gap will require much more mmf than the core.

         If the flux density is high, the core portion of the magnetic circuit may exhibit a saturation effect. However, the air gap re- mains unsaturated, since the B-H curve for the air medium is linear 

      

 

INDUCTANCE

A coil wound on a magnetic core,  isfrequently used in electric circuits. This coil may be represented by an idealcircuit element, called inductance, which is defined as the flux linkage ofthe coil per ampere of its current.

2.2 Hysteresis

HYSTERESIS LOSS 

The hysteresis loops obtained by slowly varying the current i of the coil  over a cycle. When i is varied through a cycle, during some interval of time energy flows from the source to the coil-core assembly and during some other interval of time energy returns to the source. How- ever, the energy flowing in is greater than the energy returned. Therefore, during a cycle of variation of i (hence H), there is a net energy flow from the source to the coil-core assembly

 

EDDY CURRENT LOSS 

Another power loss occurs in a magnetic core when the flux density changes rapidly in the core. The cross section of a core through which the flux density B is rapidly changing .Consider a path in this cross sectmn. Voltage will be induced in the path because of the time variation of flux enc osed by the path. Consequently, a current iₑ , known as an eddy current will flow around the path. Because core material has resistance, a power loss i R will be caused by the eddy current and will appear as heat in the core. 

The eddy current loss can be reduced in two ways. 

  1. A high-resistivity core material may be used. Addition of a few percent of silicon (say 4%) to iron will increase the resistivity significantly. 

  2. A laminated core may be used. The thin laminations are insulated from each other. The lamination is made in the plane of the flux. In transform- ers and electric machines, the parts that are made of magnetic core and carry time-varying flux are normally laminated. 

     The eddy current loss in a magnetic core subjected to a time-varying flux is 

                                       Pₑ  = Kₑ(B²max)f²

CORE LOSS 

The hysteresis loss and the eddy current loss are lumped together as the core loss of the coil-core assembly

ₕₕIf the current in the coil of is varied very slowly, the eddy currents induced in the core will be negligibly small. The B-H loop for this slowly varying magnetic intensity is called the hysteresis loop or static loop.

If, however, the current through the coil changes rapidly, the B-H loop becomes broader because of the pronounced effect of eddy currents induced in the core. This enlarged loop is called a hystero-eddy current loop or dynamic loop.    

 

 

   

2.3 Sinusoidal Excitation

In ac electric machines as well as many other applications, the voltages and fluxes vary sinusoidally with time 

EXCITING CURRENT      

No Hysteresis                                                                                     first consider a B-H characteristic with no hysteresis loop.The B-Hcurve can be rescaled (Φ = BA,                  i = Hl / N. Note that the exciting current  is nonsinusoidal, but it is in phase with theflux wave and is symmetrical with respect to voltage e. 

The fundamental component of the exciting current lags the voltage e by 90°. Therefore nopower loss is involved. This was expected, because the hysteresis loop,represents power loss, was neglected. The excitation current is therefore apurely lagging current and the exciting winding can be represented by apure inductance                                                              


With Hysteresis

consider the hysteresis loop of the core,The wave form of the exciting current from the sinusoidal flux waveform and the multivalued Φ-icharacteristic of the core. The exciting current is nonsinusoidal as well as nonsymmetrical with respect to the voltage waveform. 

The exciting current can be split into two components,one (ic in phase with voltage accounting for the core loss and the other(im) in phase with and symmetrical with respect to e, accounting  for the magnetization of the core. This magnetizing component im is the same as the exciting current if the hysteresis loop is neglected.

2.4 Permanent Magnet

MAGNETIZATION OF PERMANENT MAGNETS

the magnetmaterial is initially unmagnetized.           large mmf is applied, and on its removalthe flux density will remain at the residual value Bᵣ .

If a reversed magnetic field intensity of magnitude Hᵢ is now applied to the hard iron, the operating point moves to pointb. If Hᵢ is removed and reapplied, the B-H locus follows a minor loop.

The minor loop is narrow and for all practical purposescan be represented by the straight line bc, known as the recoil line. The slope of the recoil line is called the recoil permeability.

the reversed magnetic field intensity does not exceed H₁ , themagnet may be considered reasonably permanent. If a negative magneticfield intensity greater than H₁, is applied, such as Hthe flux density of thepermanent magnet will decrease to the value B₂ . If H₂ is removed, theoperation will move along a new recoil line de.

APPROXIMATE DESIGN OF PERMANENT MAGNETS 

Let the permanent magnet be magnetized to the residual flux density. If the small soft iron keeper is removed. The air gap will become the active region for most applications.

In order to determine the resultant flux density in the magnet and in the atr gap, let us make the following assumptions.            

1. There is no leakage or fringing flux.

2. No mmf is required for the soft iron. 

PERMANENT MAGNET MATERIALS

A family of alloys called alnico (aluminum-nickel-cobalt) has been usedfor permanent magnets since the 1930s. Alnico has a high residual flux density.

Since 960 new class of permanent magnets known as rare-earth perma-nent magnets has been developed. The rare-earth permanent magnet materi-als combine the relatively high residual flux density of alnico-type materialswith greater coercivity than the ferrites. These materials are compounds ofiron, nickel, and cobalt with one or more of the rare-earth elements. Acommonly used combination is samarium—cobalt. Another rare-earth magnetmaterial that has come into use recently is neodymium-iron-boron.

The residual flux density and coercivity are both greater than those for samarium-cobalt.It is expected that this neodymium—iron—boron will be used extensivelyin permanent magnet applications.


Exercise 1

  • conductance
  • conductivity
  • resistance
  • flux density
Permeability in a magnetic circuit corresponds to ........ in an electric circuit

Exercise 2

  • increase the resistivity
  • decrease the resistivity
  • use a laminated core
  • increase the energy density in the core
How to reduce the eddy current loss

Exercise 3

  • magnetising components in phase with flux and symmetrical with respect to e
  • energy components in phase with voltage e accounting for the core loss
  • electric components in phase with flux and symmetrical with respect to e
  • power components in phase with voltage e accounting for the core loss
The exciting current can be split into two components are

Exercise 4

  • reluctivity
  • magnetomotive force
  • permeance
  • reluctance
The property of a material which oppose the creation of magnetic flux in it is know as

Exercise 5

  • increase their cost of manufacture
  • lower their power efficiency
  • leads to their increased weight
  • produces fringing
Point out the wrong statement. Magnetic leakage is undesirable in electric machines because is

Exercise 6

  • 10 V.
  • 20 V.
  • 30 V.
  • 40 v.
A conductor 2 m long moves at a speed of 60 km/h through a magnetic  field having a flux density of 0.6 T. Calculate the induced voltage.

solution E.6

Exercise 7

  • 3840 Ampere turns
  • 3920 Ampere turns
  • 4100 Ampere turns
  • 5350 Ampere turns
We want to produce a flux density of 0.6T in an air gap having a length of 8 mm. Calculate the mmf required.

solution E.7

 

Exercise 8

  • 65.73 A
  • 69.71 A
  • 70.71 A
  • 71.71 A
A sinusoidal current has an effective value of 50  A. Calculate the peak value of current.

solution E.8

Exercise 9

  • 2400 W, -800 W
  • 2600 W, -800 W
  • 2700 W, -1000 W
  • 2800, -1000 W
The voltage applied to an ac magnet is given by the expression E = 160 sin Φ and the current is I = 20 sin(Φ-60°), all angles begin express in degree. Calculate the peak positive power and the peak negative power in the circuit.

solution E.9

Exercise 10

  • 1.42 ms
  • 1.55 ms
  • 1.61 ms
  • 1.67 ms
The current in a 60 Hz single-phase motor lags 36 degrees behind the voltage. Calculate the time interval between the positive peaks of the voltage and current.

solution E.10

3. TRANSFORMER

3.1 Ideal Transformer

A transformer with two windings, a primary winding of N₁, turns and a secondary winding of N₂ turns.In a schematic diagram will show the two windings in the two legs of the core, although in an actual transformer the windings are interleaved.

 An ideal transformer properties

  1. The winding resistances are negligible. 

  2. All fluxes are confined to the core and link both windings; that is, no leakage fluxes are present. Core losses are assumed to be negligible. 

  3. Permeability of the core is infinite . Therefore, the exciting current required to establish flux in the core is negligible; that is, the net mmf required to establish a flux in the core is zero. 

IMPEDANCE TRANSFER

Consider the case of a sinusoidal applied voltage and secondary impedance Z₂.An impedance Z₂ connected in the secondary will appear as an impedanceZ₂ looking from the primary.

Impedance can be transferred from secondary to primary if its value is multiplied by the square of the turns ratio. An imped-ance from the primary side can also be transferred to the secondary side.



POLARITY 

Windings on transformers or other electrical machines are marked to indi- cate terminals of like polarity. 

           If the windings can be visually seen in a machine, the polarities can be determined. However, usually only the terminals of the windings are brought outside the machine. Nevertheless, it is possible to determine the polarities of the windings experimentally. 

          Polarities of windings must be known if transformers are connected in parallel to share a common load. 

 

 

 

 

the parallel connection of two single-phase ( 1 ⌀) transformers. This is the correct connection because secondary voltages  e₁₂ and e₂₂ oppose each other internally.  

PRACTICAL TRANSFORMER 

            In a practical transformer the windings have resistances, not all windings link the same flux, permeability of the core material is not infinite, and core losses occur when the core material is subjected to time-varying flux. In the analysis of a practical transformer, all these imperfections must be considered. 

           Two methods of analysis can be used to account for the departures from the ideal transformer
1. An equivalent circuit model based on physical reasoning.
2. A mathematical model based on the classical theory of magnetically coupled circuits.
Both methods will provide the same performance characteristics for the practical transformer. 

3.2 Practical Transformer

           In a practical transformer the windings have resistances, not all windings link the same flux, permeability of the core material is not infinite, and core losses occur when the core material is subjected to time-varying flux. In the analysis of a practical transformer, all these imperfections must be considered. 

           Two methods of analysis can be used to account for the departures from the ideal transformer
1. An equivalent circuit model based on physical reasoning.
2. A mathematical model based on the classical theory of magnetically coupled circuits.
Both methods will provide the same performance characteristics for the practical transformer.

1. REFERRED EQUIVALENT CIRCUITS

The ideal transformer can be moved to the right or left byreferring all quantities to the primary or secondary side.The referred quantities are indicated with primes. By analyze equivalent circuit the referred quantities can be evaluated, and theactual quantities can be determined from them if the turns ratio is known.

  • Approximate Equivalent Circuits

This approximateequivalent circuit simplifies computation of currents, because both the excit-ing branch impedance and the load branch impedance are directly connectedacross the supply voltage. Besides, the winding resistances and leakagereactances can be lumped together.This equivalent circuit is frequently used to determine the performance characteristics of a practical transformer.

In a transformer, the exciting current  is a small percentage of the rated current of the transformer (less than 5%). A further approximation of theequivalent circuit can be made by removing the excitation branch.

 

2.DETERMINATION OF EQUIVALENTCIRCUIT PARAMETERS

If the complete design data of a transformer are available, these parame-ters can be calculated from the dimensions and properties of the materialsused.

The magnetizing inductances Lm can be calculated from thenumber of turns of the winding and the reluctance of the magnetic path.The calculation of the leakage inductance will involve accounting forpartial flux linkages and is therefore complicated.

These parameters can be directly and more easily determined by per-forming tests that involve little power consumption. Two tests, a no-loadtest (or open-circuit test) and a short-circuit test, will provide informationfor determining the parameters of the equivalent circuit of a transformer.

  • Transformer Rating    

The kilovolt-ampere (kVA) rating and voltage ratings of a transformer aremarked on its nameplate.These voltages are proportional to their respective numbers of turns,and therefore the voltage ratio also represents the turns ratio.

The winding that is connectedto the supply (called the primary winding) will carry an additional compo-nent of current (excitation current), which is very small compared to therated current of the winding.

 

  • No-Load Test (or Open-Circuit Test)
    This test is performed by applying a voltage  to either the high-voltage sideor low-voltage side, whichever is convenient.

 

  • Short-Circuit Test

This test is performed by short-circuiting one winding and applying ratedcurrent to the other winding.

3.3 Voltage Regulation

Most loads connected to the secondary of a transformer are designed tooperate at essentially constant voltage. As the current is drawnthrough the transformer, the load terminal voltage changes because of 
volt-age drop in the internal impedance of the transformer.

3.4 EFFICIENCY

Equipment is desired to operate at a high efficiency. Fortunately, losses in transformers are small. Because the transformer is a static device, there are no rotational losses such as windage and friction losses in a rotating machine. In a well-designed transformer the efficiency can be as high as 99%. 

MAXIMUM EFFICIENCY

3.5 Autotransformer

 

 

This is a special connection of the transformer from which a variable ac voltage can be obtained at the secondary.

3.6 Three-phase Transformers

A three-phase system is used to generate and transmit bulk electrical energy.

  • Three-phase transformers are required to step up or step down voltages inthe various stages of power transmission. 
  • A three-phase transformer can bebuilt in one of two ways: by suitably connecting bank of three single-phasetransformers or by constructing a three-phase transformer on a commonmagnetic structure.

3.6.1. BANK OF THREE SINGLE-PHASE TRANSFORMERS(THREE-PHASE TRANSFORMER BANK)

A set of three similar single-phase transformers may be connected to forma three-phase transformer. The primary and secondary windings may beconnected in either wye (Y) or delta (Δ) configurations. There are thereforefour possible connections for a three-phase transformer: Y-Δ, Δ-Y, Δ-Δ,

Phase Shift

Some of the three-phase transformer connections will result in a phaseshift between the primary and secondary line-to-line voltages.

Y-Δ

This connection is commonly used to step down a high voltage to a lower voltage. The neutral point on the high-voltage side can begrounded, which is desirable in most cases.

Δ-Y

This connection is commonly used to
 step up voltage.

Δ

This connection has the advantage that one transformer can be removed for repair and the remaining two can continue to deliverthree-phase power at a reduced rating of 58% of that of the originalbank. This is known as the open-delta or V connection.

This connection is rarely used because of problems with the exciting current and induced voltages. 

Y-Y

Single-Phase Equivalent Circuit

If the three transformers are practically identical and the source and load are balanced, then the voltages and currents on both primary and secondary sides are balanced. The voltages and currents in one phase are the same as those in other phases, except that there is a phase displacement of 120°. Therefore, analysis of one phase is sufficient to determine the variables on the two sides of the transformer. A single-phase equivalent circuit can be conveniently obtained if all sources, transformer windings, and load imped- ances are considered to be Y-connected. 

The load can be obtained for theload by the well-known Y-A transformation.

The equivalent Y representation of the actual circuit,in which the primary and secondary line currents and line-to-line voltages are identical to those of the actual circuit

V Connection

It was stated earlier that in the Δ-Δ connection of three single-phase trans- formers, one transformer can be removed and the system can still deliver three-phase power to a three-phase load. This configuration is known as an open-delta or V connection. 

Transformer windings ab and be deliver power    

 

 

 

 

 

 

 

 

 

 

3.6.2 THREE-PHASE TRANSFORMER ON A COMMON MAGNETIC CORE (THREE-PHASE UNIT TRANSFORMER) 

           A three-phase transformer can be constructed by having three primary and three secondary windings on a common magnetic core. If balanced three-phase sinusoidal volt- ages are applied to the windings, the fluxes Φₐ , Φb,Φc and will also be sinusoidal and balanced.
           If the three legs carrying these fluxes are merged, the net flux in the merged leg is zero. Both primary and secondary windings of a phase are placed on the same leg. 
           a three-phase transformer of this type. Such a transformer weighs less, costs less, and requires less space than a three-phase transformer bank of the same rating. The disadvantage is that if one phase breaks down, the whole transformer must be removed for repair. 

3.7 Harmonics in Three-phase Transformer Bank

              If a transformer is operated at a higher flux density, it will require less magnetic material. Therefore, from an economic point of view, a transformer is designed to operate in the saturating region of the magnetic core. This makes the exciting current non sinusoidal,the exciting current will contain the fundamental and all odd harmonics. How ever, the third harmonic is the predominant one, and for all practical purposes harmonics higher than third can be neglected.

               At rated voltage the third harmonic in the exciting current can be 5 to 10% of the fundamental. At 150% rated voltage, the third harmonic current can be as high as 30 to 40% of the fundamental. In this section we will study how these harmonics are generated in various connections of the three-phase transformers and ways to limit their effects. 

  • Switch SW₁ Closed and Switch SW₂ Open

Because SW2 is open, no current flows in the secondary windings.The currents flowing in the primary are the exciting currents.  

 

 

 

 

Both SW, and SW2 Open

In this case the third-harmonic currents cannot flow in the primary windings. Therefore the primary currents are essentially sinusoidal. If the exciting current is sinusoidal, the flux is nonsinusoidal because of nonlinear B-H characteristics of the magnetic core, and it contains third-harmonic components. This will induce third-harmonic voltage in the windings. The phase voltages are therefore nonsinusoidal, containing fundamental and third- harmonic voltages. 

Switch SW₁ Open and Switch SW₂ Closed

If switch SWis closed, the voltage v𝝙will drive a third-harmonic currentaround the secondary delta. This will provide the missing third-harmoniccomponent of the primary exciting current and consequently the flux andinduced voltage will be essentially sinusoidal.

Y-Y System with Tertiary (𝝙) Winding

For high voltages on both sides, it may be desirable to connect both primaryand secondary windings in Y,  third set of windings, called a tertiary winding, connected in is normally fittedon the core so that the required third-harmonic component of the excitingcurrent can be supplied. This tertiary winding can also supply an auxiliaryload if necessary.

3.8 Per-unit (PU) System

Computations using the actual values of parameters and variables may be lengthy and time-consuming. However, if the quantities are expressed in a per-unit (pu) system, computations are much simplified. The pu quantity is defined as follows: 

There are two major advantages in using a per-unit system:

  1.  The parameters and variables fall in a narrow numerical range when expressed in a per- unit system; this simplifies computations and makes it possible to quickly check the correctness of the computed values. 
  2.  One need not refer circuit quantities from one side to another; therefore a common source of mistakes is removed. 

 

3.8.1 TRANSFORMER EQUIVALENT CIRCUIT IN PER-UNIT FORM 

3.8.3 FULL-LOAD COPPER LOSS

Exercise 1

  • maximum resistance on primary side and least resistance on secondary side
  • minimum resistance on primary side and maximum resistance on secondary side
  • equal resistance on primary and secondary side
  • no ohmic resistance on either side
For an ideal transformer the winding should have

Exercise 2

  • primary = 22.7 A, secondary = 602.4 A
  • primary = 602 A, secondary = 22.7 A
  • primary = 301 A, secondary = 22.7 A
  • primary = 11.4 A, secondary = 301 A
A single phase transformer has specifications as 250 KVA, 11000 V/415 V, 50 Hz. What are the values of primary and secondary currents?

Solution E.2

 

Exercise 3

  • load current
  • maximum flux density
  • supply frequency
  • both 2 and 3
In a transformer, hysteresis and eddy current losses depend upon

Exercise 4

  • high frequency supply
  • direct current
  • rated transformer voltage
  • rated transformer current
Open circuit in a transformer is prefered with

Exercise 5

  • 3200 W and 2500 W
  • 3200 W and 5200 W
  • 1600 W and 1250 W
  • 1600 W and 5000 W

The full load copper loss and iron loss of transformer are 6400W and 5000W respectively. The copper loss and iron loss at half load will be respectively?

Solution E.5

Exercise 6

  • Pi/Pc
  • (Pi/Pc)²
  • √(Pi/Pc)
  • Pc/Pi
If Pi and Pc represent core and full load copper loss respectively, the maximum KVA delivered to load corresponding to maximum efficiency is equal to rated KVA is

Solution E.6

Exercise 7

  • ratings
  • per unit impedance
  • leakage reactance
  • efficiency
Two transformers operating in parallel will share the load depending on their

Exercise 8

  • Conduction process only
  • Induction process only
  • Both 1 and 2
  • None of the above
n an auto-transformer, power is transferred through

Exercise 9

  • iron loss
  • copper loss
  • rotational loss
  • all of the above
In a transformer which of the following losses are zero ?

Exercise 10

  • Coulombs law.
  • Amperes law.
  • Faradays law of electromagnetic induction.
  • Newtons law.
The working principle of transformer depends on

4. ELECTROMECHANICAL ENERGY CONVERSION

4.1 Energy conversion process

There are various methods for calculating the force or torque developed in an energy conversion device. The method used here is based on the principle of conservation of energy, which states that energy can neither be created nor destroyed; it can only be changed from one form to another. An electromechanical converter system has three essential parts : ( 1 ) an electric system, (2) a mechanical system, and (3) a coupling field 

The electrical energy loss is the heating loss due to current flowing in the winding of the energy converter. This loss is known as the i²R loss in the resistance ( R ) of the winding. The field loss is the core loss due to changing magnetic field in the magnetic core. The mechanical loss is the friction and windage loss due to the motion of the moving components. All these losses are converted to heat.

The energy balance equation can therefore be written as 

          Electrical energy                                =       mechanical energy          +      increase in stored   
input from source - resistance loss                  output + friction                 field energy + core loss 
                                                                                       and windage loss 

Consider a differential time interval dt during which an increment of electrical energy dWₑ (excluding the i²R loss) flows to the system.
          Let   dWf  the energy supplied to the field (either stored or lost, or part stored and part lost) 
                   dWmₘₘₘ the energy converted to mechanical form (as loss, or part useful and part as loss)

dWe = dWm + dWf 

 

4.2 FIELD ENERGY

Let  assume that the movable part is held stationary at some air gap and the current is increased from zero to a value i. Flux will be established in the magnetic system. 

dWm  =  0 
dWₑ =  dWf 

If core loss is neglected, all the incremental electrical energy input is stored as incremental field energy. Now,

e = d λ / dt

                                   dWₑ =  ei dt 

When the flux linkage is increased from zero to λ , the energy stored in the field is

Wf  =   ∫ i dλ

Hc = magneticintensityinthecore
Hg = magnetic intensity in the air gap
lc   = length of the  magnetic core material
 Ig  = length of  the air gap 

 

   

4.3 MECHANICAL FORCE IN THE ELECTROMAGNETIC SYSTEM

Let the movable part move from one position (say x=x1)to another position (x =x2) sothatattheendofthe movement the air gap decreases. 

If the movable part has moved slowly, the current has remained essentially constant during the motion. The operating point has therefore moved up- ward from point a to b. During the motion, 

LINEAR SYSTEM 

If the reluctance of the magnetic core path is negligible compared to that of the air gap path, the A-i relation becomes linear. For this idealized system 

λ =L(x)i 

where L(x) is the inductance of the coil, whose value depends on the air gap length. The field energy is

Wf  =  ∫i dλ  = 1/2 L(x)i²

fₘm  = 1/2(i²) [dL(x)/dx]

For a linear system ,                                               Wf  =  W'f  =  1/2 L(x)i²

if the reluctance of the magnetic core path is neglected, 

                                                                                      Ni   =  Hg2g   =  Bg/μ₀ (2g )

            Wf  =  (Bg²/2μ₀) x volume of air gab    =  (Bg²/2μ₀) x (Ag2g)

fm  =  (Bg²/2μ₀) x 2Ag

The total cross-sectional area of the air gap is 2Ag . Hence, the force per unit area of air gap, called magnetic pressure Fm , is

Fm  =  Bg²/2μ₀ (2Ag) N/m²

 

4.4 ROTATING MACHINES

                    The production of translational motion in an electromagnetic system has been discussed in previous sections. However, most of the energy converters, particularly the higher-power ones, produce rotational motion.  
                    The fixed part of the magnetic system is called the stator, and the moving part is called the rotor. The latter is mounted on a shaft and is free to rotate between the poles of the stator. 

                    The current can be fed into the rotor circuit through fixed brushes and rotor-mounted slip rings. The stored field energy Wf of the system can be evaluated by establishing the currents is and z r in the windings keeping the system static, that is, with no mechanical output. 

4.5 CYLINDRICAL MACHINES

Consider the following cases. 

  1. ωr =0, α=0 ,ωm = ωs. The rotor current is a direct current and the machine rotates at the synchronous speed. 

If the machine is brought up to its synchronous speed (ωm = ωs), it will develop an average unidirectional torque and continuous energy conversion can take place at the synchronous speed. This is the basic principle of operation of a synchronous machine, which normally has dc excitation in the rotor and ac excitation in the stator. 

at ω = 0, the machine does not develop an average torque and therefore the machine is not self-starting.    With one winding on the stator, the machine is called a single-phase synchronous machine. Although it develops an average torque, the instantaneous torque is pulsating.

The pulsating torque may result in vibration, speed fluctua- tion, noise, and waste of energy. This may be acceptable in smaller machines but not in larger ones. The pulsating torque can be avoided in a polyphase machine, and all large machines are polyphase machines. 

  

2.  ωm  =  ωs  -  ωr, Both stator and rotor windings carry ac currents at different frequencies and the motor runs at an asynchronous speed 

This is the basic principle of operation of an induction machine, in which the stator winding is excited by an ac current and ac current is induced in the rotor winding. 

Note that the single-phase induction machine is also not self-starting, because at ωm = 0 no average torque is developed.The machine is brought up to the speed  ωm=ωs -ωr so that it can produce an average torque. To eliminate pulsating torque, polyphase induction machines are used for high-power applications. 

 

Exercise 1

  • increase the co-energy at constant flux
  • increase the co-energy at constant mmf
  • decrease the stored energy at constant mmf
  • decrease the stored energy at constant flux

The developed electromagnetic force and/or torque in electromechanical energy conversion systems, acts in a direction that tends to.....

Which of the above statements are correct?

Exercise 2

  • electrical system in order to extract energy from mechanical system
  • mechanical system in order to extract energy from mechanical system
  • electrical system in order to extract energy from electrical system
  • electrical or mechanical system for electro-mechanical energy conversion
A coupling magnetic field must react with

Exercise 3

  • electrical side is associated with emf and current
  • electrical side is associated with torque and speed
  • mechanical side is associated with emf and current
  • mechanical side is associated with torque and speed
 In an electro-mechanical energy conversion device, the coupling field on the

Exercise 4

  • Electrical energy to mechanical energy only
  • Mechanical energy to electrical energy only
  • All of the mentioned
  • None of the mentioned
An electro-mechanical energy conversion device is one which converts.......

Exercise 5

  • Magnetic field
  • Electric field
  • Magnetic field or Electric field
  • None of the mentioned
What is the coupling field used between the electrical and mechanical systems in an energy conversion devices?

Exercise 6

  • stored in the magnetic field
  • stored in the electric field
  • divided equally between the magnetic and electric fields
  • zero
A physical system of electromechanical energy conversion, consists of a stationary part creating a magnetic field with electric energy input, and a moving part giving mechanical energy output. If the movable part is kept fixed, the entire electrical energy input will be.....

Exercise 7

  • increase the stored energy at constant mmf
  • decrease the stored energy at constant mmf
  • decrease the co-energy at constant mmf
  • increase the stored energy at constant flux
The developed electromagnetic force and/or torque in electromechanical energy conversion system, acts in such a direction that tends to.....

Exercise 8

  • 9.75 V
  • 10.46 V
  • 11.73 V
  • 14.50 V

The coil has 400 turns and a resistance of 5 ohms. The reluctance of the magnetic material is negligible. The magnetic core has a square cross section of 5 cm by 5 cm. When the sheet of steel is fitted to the electromagnet ,air gap, each of length = 1 mm, separate them. An average force of 550 newtons is required to lift the sheet of steel.

Determine the dc source voltage.

Solution E.8

Exercise 9

  • 78.33°
  • 79.33°
  • 80.33°
  • 81.33°

When current flows through the curved solenoid coil, a curved ferromagnetic rod is pulled into the solenoid against the torque of a restraining spring. The inductance of the coil is L = 4.5 + 180 µH, where θ is angle of deflection in radius. The spring constant is 0.65 mN•m/rad.

Determine the deflection in degree for a current of 10 amperes (rms).

Solution E.9

Exercise 10

  • 2.18 mm
  • 2.63 mm
  • 3.05 mm
  • 3.14 mm

 The plunger has a cross-sectional area 0.0016 m². The coil has 2500 turns and a resistance of 10 ohm. A voltage of 15 V (dc) is applied to the coil terminals. Assume that the magnetic material is ideal.

Determine the air gap g in mm for which the flux density in the air gap is 1.5 T. Determine the stored energy for this condition.

 

Solution E.10

5. LINEAR MACHINE

5.1 The Linear DC Machine

  The linear dc machine is like a rotating dc motor,generates mechanical force by the interaction of current in conductors and magnetic flux provided by permanent magnets.

  The structure of linear dc machine consists of a battery and a resistance connected to the switch to a pair of smooth,frictionless rails. Along the bed of this "railroad track" is a constant and there are uniform-density magnetic field directed into the page. A bar of conducting metal is lying across the tracks.

The behavior of the device can be determined from 4 equations to the machine :

1. The equation for the force on a wire in the presence of a magnetic field :

where

F= force on wire

i = magnitude of current in wire

l = length of wire, with direction of l defined to be in the direction of             current flow

B = magnetic flux density vector

 

2. The equation for the voltage induced on a wire moving in a magnetic field :

where

e(ind) = voltage induced in wire

v = velocity of the wire

B = magnetic flux density vector

l = length of conductor in the magnetic field

 

3. Kirchhoff's voltage law for this machine :

 

4. Newton's law for the bar across the tracks :

 

 

5.2 Starting the Linear DC Machine

  From the figure , it shows the dc machine under starting conditions. The circuit will start when you connect the switch. It will make the current flow in the bar which can determine by Kirchhoff's voltage law:

  When the bar is initially at rest, voltage induced in the wire equal to zero, so i = VB/R. The current flows down through  the bar across the tracks. But from the equation F=i(l×B), a current flowing through a wire in the presence of a magnetic field induce the force on the wire.

  Therefore, the bar will accelerate to the right which come from Newton's law. However, when the velocity of the bar begins to  increase, a voltage appears across the bar. The voltage induced will reduce to

 The voltage now reduces the current flowing in the bar, since by Kirchhoff's voltage law. As voltage induced increase, the current i decreases

  The result of this action is that eventually the bar will reach a constant steady-state speed where the net force on the bar is zero. This will occur when voltage induced has risen all the way up to equal the voltage VB. At that time, the bar will move at a speed given

  The bar will continue to coast along all this no-load speed forever unless some external force disturbs it. When the motor is started, the velocity induced voltage, current,and induced force.

 

To summarize, at starting, the linear dc behave as follow:

1. Closing the switch produces a current flow i = VB/R

2. The current flow produces a force on the bar which come from the formula F= ilB

3. The bar accelerates to the right, producing an induced voltage as it speeds up.

4. This induced voltage reduce the current flow

5. The induced force decreases until F=0. At that point, the induce voltage equal to VB, current flow equal to zero and the bar moves at a constant no-load speed.

 

5.3 The Linear DC Machine as a Motor

   For the motor, as you can see from the figure, a load force is applied to the bar opposite to the direction of motion. Since the bar is initially at steady state, the applied force of the load will result in a net force on the bar in the direction opposite to the direction of motion ( F(net)= F(load)-F(ind)). The effect of this force will slow down the bar.

When the bar slow down,the induced voltage on the bar drops. As the induced voltage decreases, the current flow in the bar rises.

  Therefore, because of the current flow in the bar increase, the induced force rises too. The induced force rises until it is equal and opposite to the load force, and the bar travel in steady state in lower speed. This assumption are represent in the form of graph below.

   There are now an induced force in the direction of motion of the bar, and power is converted from electrical to mechanical power to keep the bar moving.

An amount of electric power equal to e(induced)i is consumed in the bar and replaced by mechanical power F(ind)v. Because power is converted to mechanical form, this bar is operating as motor.

To summarize this behavior

1. A load force is applied opposite to the direction of motion, which causes a net force opposite to the direction of motion.

2. The resulting acceleration is negative, so the bar slows down.

3. The induced force falls, and so current flow increases.

4. The induced force increase until induced force equal to load force at a lower speed.

5. An amount of electric power is converted to mechanical power, and the machine is acting as a motor.

   In a real dc motor, as a load is added to the shaft, the motor begins to slow down which reduces its internal voltage and increases the current flow. Also, the increasing in current flows induced torque, and the induced torque will equal to the load torque of the motor at a slower speed.

There are the equation the describe the relationship between converted power and torque:

where the induced torque is the rotational of the induced force and the angular velocity is the rotational of the linear velocity.

5.4 The Linear DC Machine as a Generator

The linear machine with an applied force in the direction of motion. Now the applied force will cause the bar to accelerate in the direction of motion,and the velocity of the bar will increase.

  As the velocity increases. the induced voltage increases and will be larger than the battery voltage which makes the current reverses direction that you can see from the equation

   Because this current flow up through the bar, it induces a force in the bar given by the equation which the direction is given by the right-hand rule

   Finally, the induced force will be equal and opposite to the applied force, and the bar will move in the higher speed than before. Notice that now the battery is charging. The linear machine is now become the generator which convert mechanical to electrical power.

 

To summarize this behavior

1. A force is applied in the direction of motion. The net force is in the direction of motion.

2. Acceleration is positive, so the bar speed up.

3. The induced voltage increases,and so the current flow increases.

4. the induced force increase until induced force equal to load force at a higher speed.

5. An amount of mechanical power is converted to electrical power.

   Again, for the real dc generator, a torque is applied to the shaft in the direction of motion. When the speed of the shaft increases, the internal voltage increases,and current flows out of the generator to the loads as represent in the equation

   It is interesting that the same machine acts as both motor and generator. The only difference between the two is that the applied forces are in the direction of motion for the generator and opposite direction for the motor. When induced voltage greater than battery voltage, the machine act as a generator but if in the opposite way it will act as the motor. Whether the machine is a motor or generator, both induced force(motor action) and induced voltage(generator action) are present at all times.

 

   This machine was a generator move rapidly and a motor when it moved more slowly,but whether it was a motor or generator, it always move in the same direction.

5.5 Starting Problems with the Linear Machine

  From the figure, this machine is supplied by a 250 V dc source, and the internal resistance R is 0.10 ohms ( the resistor models the internal resistance of the real dc machine).

At starting conditions, the speed of the bar is zero, so the induced voltage is equal to zero so the current at the starting is 2500 A which come from 25o V of battery voltage divided by 0.1 ohms of resistor.

This current is very high, it is in the excess of 10 times the rated current of the machine. the current is very high and it can cause damage to the motor. Both real ac and real dc machine suffer from high current on starting.

   This problem can be prevented by adding an extra resistance into the circuit during starting to limit the current flow until induced voltage build up enough to limit it.

  The same problem exists in real dc machines,and it is handled in the same fashion ( a resistor is inserted into the motor armature circuit during starting).

Exercise 1

  • Newton's law F=ma
  • Law of inertia ∑F = 0
  • Faraday's law E = N(∆Φ/∆t)
  • F = i(l x B)
  • E = l( v x B )
  • E = L( ∆i / ∆t )
  • Kirchhoff's voltage law ∑v = 0
What are the 4 basic equations for the application of dc machine?

Exercise 2

  • a c d e b
  • d e a b c
  • c d a e b
  • c d e b a

Arrange the steps of the linear dc machine behavior in order

  • a. The bar accelerates to the right,producing an induced voltage as it speeds up.
  • b. The induced force is thus decreased until eventually F=0. At that point induced voltage = source voltage and current flow = 0, and the bar moves at a constant no-load speed v = source voltage/Bl.
  • c. Closing the switch produces a current flow.
  • d. The current flow produces a force on the bar.
  • e. The induced voltage reduces the current flow.

Exercise 3

  • In a real dc motor, the increased current flow increases its induced torque.
  • The induced torque will equal to the load torque of the motor at a new slower speed.
  • For the generator, when the induced force equal and opposite to the applied force, the bar will slow down and charging the battery.
  • The power convert from electrical form to mechanical form is given by the equation P = Fv. On the other hands, the power converted from electrical to mechanical is given by the equation P=ωτ
Which one is Wrong?

Exercise 4

  • Linear dc machine use 4 basic equation for the application
  • We can use Kirchhoff's voltage law for the machine to determine the induced voltage on the bar
  • Current along the bar is in proportional to the induced voltage.
  • When induced force equal to the load force, the bar will increase its moving speed
Which one is wrong?

Exercise 5

  • Motor changes from mechanical energy to electrical energy and generator changes from electrical energy to mechanical energy.
  • For motor, load force applied to the bar in the opposite direction to the motion. For generator, load force applied to the bar in the same direction to the motion.
  • Motor has higher F net than the generator
  • Motor use higher current acting to the bar than the generator
What is the difference between linear dc machine as motor and linear dc machine as generator?

Exercise 6

  • a. 1100A b.660N c.366.67m/s
  • a. 1000A b.650N c.365.50m/s
  • a. 1200A b.670N c.370.75m/s
  • a. 1500A b.700N c.376.37m/s

A linear machine has a magnetic flux density of 0.6 Wb/m^2 directed into the page, a resistance of 0.2 Ohms, a bar length = 1m, and a battery voltage of 220V, calculate

  • a. What is the starting current flow?
  • b. What is the initial force on the bar at starting
  • c. What is the no-load steady-state speed

Solution E.6

Exercise 7

  • a. 337.87m/s b.7555W c. 7333.3 W d. motor
  • a. 422.22m/s b.7444W c. 7222.2 W d. motor
  • a. 377.78m/s b.7555W c. 7333.3 W d. generator
  • a. 222.22m/s b.7444W c. 7222.2 W d. generator

A linear machine has a magnetic flux density of 0.6 Wb/m^2 directed into the page, a resistance of 0.2 Ohms, a bar length = 1m, and a battery voltage of 220V. If a 20 N force pointing to the same direction as no-load steady state is applied to the bar, calculate

  • a. What is the steady state speed?
  • b. How much power would the bar be producing or consuming ?
  • c. How much power would the battery be producing or consuming ?
  • d. Is the machine acting as a motor or a generator?

Solution E.7

Exercise 8

  • a. 345.88 m/s b. 69.97% c. generator
  • a. 315.56 m/s b. 75.77% c. generator
  • a. 345.56 m/s b. 29.85% c. motor
  • a. 355.56 m/s b. 96.97% c. motor

  From the last question, if the bar is loaded down with a force of 20N opposite the direction of motion.

  • a. What is the new steady-state speed?
  • b. What is the efficiency of the machine under this circumstance
  • c. Is this machine acting as a motor or generator?

Solution E.8

Exercise 9

  • a. 120 m/s b. 111m/s
  • a. 100 m/s b. 109m/s
  • a. 115m/s b. 171m/s
  • a. 140 m/s b. 123m/s

   A linear machine has a magnetic flux density of 0.1T directed into the page, a resistance of 0.3 Ohms, a bar length equal to 10 m,and a battery voltage of 120 V.

  • a. What is the no-load steady state speed of the bar?
  • b. If 30 N force pointing to the left were applied to the bar. What would be the new steady state speed be? Is this machine a motor or a generator?

Solution E.9

Exercise 10

  • 160 m/s
  • 170 m/s
  • 140 m/s
  • 150 m/s

   A linear machine has a magnetic flux density of 0.1T directed into the page, a resistance of 0.3 Ohms, a bar length equal to 10 m,and a battery voltage of 120 V. Assume that the bar is unloaded and that is suddenly runs into a region where the magnetic field is weakened to 0.08 T. How fast will the bar go now?

Solution E.10

6. DC GENERATOR

6.1 Generating an ac voltage

To study of a direct current (dc) generator has to begin with a knowledge of alternating-current (ac) generator. The reason is that the voltage generated in any dc generator is inherently alternating and only become dc after it has been rectified by the commutator.


        This shows schematic diagram of an elementary ac generator turning at 1 revolution per second.
        The rotation is due to an external driving force. The coil is connected to two slip rings mounted on the shaft. The slip rings are connected to an external load by means of two stationary brushes x and y.

 

 

 

As a coil rotates. The voltage appears between the brushes and, therefore, across the load.The voltage is generated because the conductors of the coil cut across the flux produced by the N,S poles. No flux is cut when the coil is momentarily in the vertical position; consequently the voltage at these instance is zero.  Another feature of the voltage is that its polarity changes every time the coil make half a turn.The voltage can therefore be represented as a function of the angle rotation.

 

 

Voltage induced in the ac generator as a function of angle of rotation.The wave shape depend upon the shape of the N,S poles. We assume the poles were designed to generate the sinusoidal wave shown. It corresponds to an interval of one second.

 

 

 

 

We can also represent the induced voltage as function of time.

 

 

 

 

 

6.2 Direct-current generator

We would obtain a voltage of constant polarity across the load. Bush x would always be positive and bush y negative. We obtain this result  by using a commutator. A commutator in its simplest form is composed of slip ring that is cut in half,which each segment insulated from the other as well as from shaft.

This picture shown elementary dc generator is simply an ac generator equipped with a mechanical rectifier called a commutator,Due to the constant polarity between the brushes.the current in the external load always flows in the same direction. The machine called a direct-current generator or dynamo.

 

 

 the voltage between brushes x and y pulsates but never changes polarity. The alternating voltage in the coil is rectified by the commutator,which acts as a mechanical reversing switch.

 

The elementary dc generator produces a pulsating dc voltage

6.3 Difference between ac and dc generators

In each case, a coil rotates between the poles of magnet and an ac voltage is induced in the coils.The machine only differ in the way the coil are connected to the external circuit.Such machines can function simultaneously as ac and dc generators.

 


The three armatures have identical windings. Depending upon how they are connected ( to slip rings or a commutator),an ac or dc voltage is obtained.

a) dc generators require a commutator      b) ac generators carry slip rings     c) small machines which carry
                                                                                                                                                        both slip rings and commutator

 

6.4 Improving the waveshape

 We can improve the pulsating dc voltage by using focus coils and four segments.The voltage still pulsates but it never fall to zero;it is much closer to a steady dc voltage.

By increasing the number of coils and segments,can obtain a dc voltage that is very smooth.The coils and the cylinder constitute the armature of the machine.The percent ripple is the ratio of the RMS value of the ac component of the voltage to the dc component,expressed in percent.

 

For reasons of symmetry , the coils are wound so that one coil side is at the bottom of slot  and the other is at the top.The coil connections to the commutator segment are easy to follow in this simple armature. 

 

 

 

 

 

 

 

the side of a1 of coil A are now sweeping past pole tip 1 and pole tip 4 . The sides of coil C are experiencing the same flux because they are in the same flux  because they are in the same slots as coil A .

The armature winding we have just discussed is called a lap winding. It is the most common typeof winding used in direct-current generators and motors.

6.5 Induced voltage

The voltage E induced in each conductor depends upon the flux density which it cuts.The fact is based upon the equation :         E=Blv   

Because the density in the air gap varies from point to point,the value of the induced voltage per coil depends upon its instantaneous position

 

Thus, by shifting the brushes the output voltage decrease. Furthermore,in this position,the brushes continually short-circuit that generated 7 V. Large currents will flow in the short-circuit coils and brushes,and sparking will result.Thus,shifting the brushes off the neutral position reduces the voltage between  the brushes and at the same time cause sparking. When sparking occurs, there is said to be poor commutation

 

6.6 Neutral zones

Neutral zones are those place on the surface of the armature where the flux density is zero.When the generator operates at no-load, the neutral zones are located between the poles.No voltage is induced in a coil that cuts through the Neutral zones.We always try to set the brushes so they are in contact with coils that are momentarily in a Neutral zones.

This show moving the brushes off the neutral point reduces the output voltage and produces sparking.​

6.7 Value of the induced voltage

The voltage induced in dc generator having a lap winding is given the equation 
 

 

This equation shows that for a given generator the voltage is directly proportional to the flux per pole and to the speed of the rotation.The equation only holds true if the brushes are on the neutral position.If the brushes are shifted off neutral,the effect is equivalent to reducing the number of the conductors Z.

 

6.8 Generator under load : the energy conversion process

When a direct-current generator is under load,some fundamental flux and current relationship take place that are directly related to the mechanical-electrical energy conversion process.

 

The energy conversion process.The electromagnetic torque due to F must be balanced by the applied 
mechanical torque.

The current delivered by the generator also flows though all the armature conductors.the current always flows in the same direction those conductors that are momentarily under a N pole.The same is true for conductors that are momentarily under a S pole.

Because the conductors lie in magnetic field,they are subjected to a force,according to Lorentz’s law.We find that the individual force F on the conductor all acct clockwise.In effect,they produce a torque that acts opposite to the direction in which the generator is begin driven.The resulting mechanical power is converted into electrical power,which is delivered to the generator load.

6.9 Armature reaction

Assumed that the only magnetomotive force (mmf) acting in the dc generator is that due to the field. However,the current flowing in armature coils also created a powerful magnetomotive force that distorts and weakens the flux waking takes place in both motors and generators. The effect produced by the armature mmf is called armature reaction.

 

 

This field acts at right angles to the field produced by the N,S poles. The intensity of the armature flux depend upon its mmf, which in turn depend upon the current carried by the armature. Thus,contrary to the field flux, the armature flux is not constant but varies with the load.

We can immediately foresee a problem which the armature flux will produce. This picture shows that the flux in the neutral zine is no longer zero and consequently, a voltage will be induced in the coils that are short-circuited by the brushes.

The intensity of the sparking will depend upon the armature flux and hence upon the load current delivered by the generator.

 

The second problem created by the armature mmf is that it distorts the flux produced by poles.In effect,the combination of the armature mmf and field mmf produces a magnetic field.

 


It is important to note that the orientation of the armature flux
remains fixed in space;it does no rotate with the armature.

 

6.10 Shifting the brushes to improve commutation

For generators,the brushes are shifted to the new neutral zone by moving them in the direction or the rotation.For motors,the brushes are shifted against the direction of the rotation.

As soon as the brushes are moved,the commutation improves, meaning there is less sparking.However,If the load fluctuates,the armature mmf rises and falls and so the neural zone shifts back and forth to obtain sparkles commutation.This procedure  is not practical and other means are used to resolve the problem.

 

6.11Commutating poles

To counter the effect of armature reaction in medium and large power dc machines, always place a set of commutating poles between the main poles.These narrow poles carry windings that are connected in series with the armature.As the load current varies,the two magnetomotive forces rise and fall together,exactly bucking each other at all times.By nullifying the armature mmf in this way,the flux in the space between the main poles is always zero and so we no longer have to shift the brushes.

 


Commutating poles produce an mmf(c) thatopposes the mmf(a) of the mature.

This picture shows how the commutating poles of a 2-poles machine are connected.The direction of the current flowing through the windings indicates that the mmf of the commutating poles acts opposite to the mmf of the armature and,therefore, neutralise its effect.

The mmf of the commutating poles is made slightly greater than the armature mmf. This creates a small flux in the neutral zone,which aids the commutation process.

 

6.12 Separately excited generator

Instead of using permanent magnets to create the magnetic field,we can use a pair of electromagnet called  field poles.

Separated excited 2-pole generator. The N,S field poles are created by the current flowing in the field winding.

When the dc field current in such a generator is supplied by an independent source,the generator is said to be separately excited . Thus,the dc source connected to terminals a and b cause an exciting current to flow. If the armature is driven by a motor or diesel engine,a voltage appears between brush terminals X and y.

 

6.13 No-load operation and saturation curve

When a separately excited dc generator runs at no-load (armature circuit open),a change in the exciting current cause causes a corresponding change in the induced voltage.

Field flux vs exciting current. 

The gradually raise the exciting current I of x so that the mmf of the field increases,which increases the flux per pole.

                         When the exciting current is relatively small,the flux is small and the iron in the machine is unsaturated. Very little mmf is needed to establish the flux through the iron,with the result that the mmf developed by the field coils is almost entirely available to drive the flux through the air gap. Because the permeability of air is constant. The flux increases in direct proportion to the exciting current.

                         However,as we continue to raise the exciting current ,the iron in the field and the armature begins to saturate.A large increase in mmf is now required to produce small increase in flux.

Induced voltage vs speed.

for given exciting current,the induced voltage increases in direct proportion to the speed. 

                         If we reverse the direction of the rotation,the polarity of the induced voltage also reverse.However,if we reverse both the exciting current and the direction of the rotation ,the polarity of the induced voltage remains the same.

 

6.14 Shunt generator

              A shunt-excited generator is a machine whose shunt-field winding is connected in parallel with the armature terminals, so that the generator can be self-excited. The principal advantage of this connection is that it eliminates the need for an external source of excitation.

 

                                a) Self-excited shunt generator      b) A shunt field is one designed to be connected
                                                                                                            in shunt  with the armature winding.

              Self-excitation achieved,When a shunt generator is started up,a small voltage is induced in the armature due to the remanent flux in the poles.This voltage produces a small exciting current in the shunt field.The resulting small mmf acts in the same direction as the remanent flux,causing  the flux per pole to increase.

6.15 Controlling the voltage of a shunt generator

 It is easy to control the induced the voltage of a shunt exciting generator. We simply vary the exciting current by means of a rheostat connected in series with the shunt field.

               To understand how the output voltage varies,suppose that E₀ is 120 v when the movable contract p is in the center of the rheostat. If we move the contract toward extremity m, The resistance R₁ between points p and b diminishes, which cause the exciting current to increase. This increase the flux and, consequently, the induced voltage E₀. On the other hand,if we move the contract toward extremity n,R₁ increase,the exciting current diminishes. The flux diminishes and so E₀ will fall.

               We can determine the no-load value of E₀ if we know the saturation curve of the generator and the total resistance of the shunt field circuit between point p and b. We draw a straight line corresponding to the slop of R₁ and superimpose it on the saturation curve.

This dotted line passes through the origin,and the point where it intersects the curve yields the induced voltage.

By changing the setting of the rheostat, The total resistance of the field circuit increase,causing E0 to decrease progressively. If we continue to resistance R1 , a  critical value will be reached where the slope of the resistance line is equal to that of the saturation curve in its unsaturated region. When this resistance is attained, the induced voltage suddenly drop to zero and will remain so for any R1 greater than this critical value.

 

6.16 Equivalent circuit

We have seen that the armature winding contain a set of identical coils, all of which posses a certain resistance. The total armature resistance R₀ is which exists between the armature terminals when that machine is stationary.

It is measured on the commutator surface between those segments that lie usually very small, often less than one-hundredth of an ohm. Its value depends mainly upon the power and voltage of the generator. To simplify the generator circuit,we can represent R₀ as if it were in series with one of the brushes. If the machine has interlopes, the resistance of these windings is induced in R₀.

 

The latter is the voltage induced in the revolving conductors.Terminal 1,2 are the external armature terminals of the machine, and F1,F2 are the field winding terminals . Using the circuit, we will now study the more common types of the direct-current generator and their behaviour under load.

 

 

6.17 Separately excited generator under load

Consider a separately excited generator that is driven at constant speed and whose field is excited by a battery

The exciting current is constant and so is the resultant flux. The induced voltage E0 is therefore fixed. When the machine operates at no-load,terminal voltage E12 equal to the induced voltage E0  because the voltage drop in the armature resistance is zero.

If we connect a load across the armature , the resulting load current produces a voltage drop across resistance R0. Terminal voltage is now less than the induced voltage As we increase the load, the terminal voltage diminishes progressively.

The graph of the terminal voltage as a function load current is called the load curve of the generator.

The induced voltage also decrease slightly with increasing load because pole-tip saturation tends to decrease the field flux. Consequently, the terminal voltage falls off more rapidly than can reattributed to armature resistance alone.

6.18 Shunt generator under load

The terminal voltage of self-excited shunt generator falls off more sharply with increasing load than that of separately excited generator. The reason is that the field current in separately excited machine remain constant,whereas in a self-excited generator the exciting current falls as the terminal voltage groups. For a self-excited generator,the drop in voltage from no-load to full-load is about 15 percent of the full-load voltage,whereas for a separately excited generator it is usually less than 10 percent. The voltage regulation is said to be 15% and 10%.

 

6.19 Compound generator

The compound generator was developed to prevent the terminal voltage of a dc generator from decreasing with the increasing load. Thus, although we can usually tolerate a reasonable drop in the terminal voltage as the load increases, this has a serious effect on lighting circuits. 

compound generator is similar to a shunt generator, except that it has additional field coils connected in series with the armature. These series field coils are composed of a few turns of heavy wire, big enough to carry the armature current. 

The total resistance of the series coils is, therefore, small  is a schematic diagram showing the shunt and series field connections.

 

                When the generator runs at no-load, the current in the series coils is zero.The shunt coils, however, carry exciting current which produces the field flux, just as in standard self-excited shunt generator. If the series coils are properly designed, the terminal voltage remains practically constant from no-load to full-load. The rise in the induced voltage compensates for the armature IR drop.

                 In some cases we have to compensate not only for the armature voltage drop, but also for the IR drop in feeder line between the generator and the load. The generator manufacturer then adds one or two extra turns on the series winding so that the terminal voltage increase as the load current rises. Such machines are called over-compound generators. If the compounding is too strong, a low resistance can be placed in parallel with the series field. If the value of the diverter resistance is equal to that of the series field,the current in the latter is reduced by half.

 

 

6.20 Differential compound generator

In differential compound generator the mmf of the series field acts opposite to the shunt field. As a result, the terminal voltage falls drastically with increasing load. We can make such a generator by simply reversing the series field of a standard compound generator. Differential compound generators we're formerly used in dc arc welders, because they tended to limit the short-circuit current and to stabilize the arc during the welding process.

6.21 Load Characteristics

 

 

6.22 Generator specifications

The nameplate of a generator indicates the power, voltage ,speed,and other details about the machine. These rating, or nominal characteristics, are the values guaranteed by the manufacture.

For example,the following information is punched on the nameplate of a 100 kW generator:

In practice,the terminal voltage is adjusted to value close to its rating of 250 V. We may draw any amount of power from there generator,as long as it does not exceed 100 kW and the current is less than 400 A. The class B designation refers to the class of insulation used in the machine.

 

Construction of direct-current generators
We have to described the basic features and properties of direct-current generators. We now look at the mechanical construction of these machines, directing our attention to the field,the armature,the commutator and the brushes.

 

6.23 Field

The field produces the magnetic flux in the machine. It is basically a stationary electromagnet compose of a set of salient poles bolted to the inside of the a circular frame.

Field coils,mounted no the poles, carry the dc exciting current. The frame is usually mad of solid cast steel,whereas the pole pieces are composed of stacked iron laminations. In some generators the flux is created by permanent magnets.

           The number of poles depend upon the physical size of the machine ; the bigger it is,the more poles it will have. By using a multipole design,we can reduce the dimensions and cost of the large machines,and also improve their performance.

The field coils of multipole machine are connected together
so that adjacent poles have opposite magnetic polarities

The shunt field coils coils are composed of several hundred turns of wire carrying a relatively small current. the coils are insulted from the pole pieces to prevent short-circuits.

               the mmc developed by the coils produces a magnetic flux that passes through the pole pieces,the frame,the armature, and that air gap. The air gap is the short space between the armature and the pole pieces,It ranges from about 1.5 to 5 mm as the generator rating increases from 1 KW to 100 kW.

               Because the armature and field are composed of magnetic materials having excellent permeability,most of the mmf produced by the field is used to drive the flux across the air gap.

               If the generator has a series field,the coils are wound no the top of the shunt-field coils. The conductor size must be large enough so that the winding does not overheat when it carries the full-load current of the generator.

 

 

6.24 Armature

the armature is the rotating part of a dc generator,consists of a commutator, an iron core,and a set of coils

The armature is keyed to a shaft and revolves between the field poles. The iron core is composed of slotted, iron laminations that are stacked to form a solid cylindrical core. The laminations are individually coated with an insulating film so that they do not come in electrical contract with each other. As a result,eddy-current losses are reduced. The slots are lined up to provide the space needed to insert the armature conductor.

The armature is keyed to a shaft and revolves between the field poles. The iron core is composed of slotted, iron laminations that are stacked to form a solid cylindrical core. The laminations are individually coated with an insulating film so that they do not come in electrical contract with each other. As a result,eddy-current losses are reduced. The slots are lined up to provide the space needed to insert the armature conductor.

 

6.25 Commutator and brushes

The commutator is composed of an assembly of tapered copper segment insulated from each other by mica sheets,and mounted on the shaft of the machine

            Great care is taken in building the commutator because any eccentricity will cause the brushes to bounce, producing unacceptable sparking. The spark burn brushes and overheat and carbonize the commutator.

            Multipole machine prossess as many brush sets as they have poles. The brush sets are composed of one or more brushes,depending upon the current that has to be carried. The brush sets are spaced at equal intervals around the commutator. They are supported by a movable brush yoke that permits the entire brushes  assembly to be rotated through an angle and then locked in the neutral position.

 

      

            The brushes are made of carbon because it has good electrical conductivity and softness does not score the commutator. To improve the conductivity, a small amount of copper is sometime mixed with the carbon. The brush pressure is set by means of adjustable springs.

 

 

 

 

6.27 The ideal commutation process

     When a generator is under load, the individual coils on the armature carry one-half the load current carried by one brush. The currents flowing in the armature winding next to a positive brush.

     If the commutator segments are moving from right to left. the coil on the right-hand side of the brush will soon be on the left-hand side. This means the current in these coils must reverse. The process whereby  the current changes direction in this brief interval called " commutation ".

     The " conductivity " between the brush and commutator is proportional to the contact area.

 

 

6.28 The practical commutation process

The current can not reverse as quickly as it should. The reason is that the armature coils have inductance and it strongly oppose a rapid change in current.

 

                 In designing dc motors and generators, every effort is made to reduce the self-inductance of the coils. One of the most effective ways is to reduce the number of turns per coil. But for a given output voltage . The number of coils must be increased and more coils implies more commutator bars.

                  Direct current generators have a large number of coils and commutator bars not so much to reduce the ripple in the output voltage but to overcome the problem of commutation.

                 Another important factor in aiding commutation is that the mmf of the commutating poles is always made slightly greater than the armature mmf.

 

Exercise 1

  • Abrasion from dust
  • Excessive spring pressure
  • Rough commutator bars
  • Double the number of poles
The rapid wear of brushes takes place due to

Exercise 2

  • Brushes should be of proper grade and size
  • Making the magnetic losses of forces parallel
  • Brushes should smoothly run in the holders
  • Smooth, concentric commutator properly undercut
Satisfactory commutation of D.C. machines requires

Exercise 3

  • Separately excited generator
  • Shunt generator
  • Series generator
  • Compound generator
An exciter for a turbo generator is a....

Exercise 4

  • 100% regulation
  • Infinite regulation
  • 50% regulation
  • 1% regulation
With a D.C. generator which of the following regulation is preferred?

Exercise 5

  • Copper lugs
  • Resistance wires
  • Insulation pads
  • Brazing
The commutator segments are connected to the armature conductors by means of

Exercise 6

  • Imperfect brush contact
  • Field resistance less than the critical resistance
  • No residual magnetism in the generator
  • Faulty shunt connections tending to reduce the residual magnetism
If a D.C. generator fails to build up the probable cause could not be

Exercise 7

  • 103 V / 288 W / 10.39 N·m
  • 103 V / 278 W / 10.12 N·m
  • 106 V / 288 W / 10.12 N·m
  • 106 V / 278 W / 10.43 N·m

A separate excited dc generator turning at 1400 r/min produces an induced voltage of 127 V. The armature resistance is 2 ohm and the machine delivers a current of 12 A.

Calculate

a. The terminal voltage

b. The heat dissipated in the armature

c.The breaking torque exert by the armature 

        

Solution E.7

Exercise 8

  • 2600 A / 5000 A
  • 4500 A / 6400 A
  • 5000 A / 7800 A
  • 6000 A / 8200 A

Each pole of a 100 kW, 250 V flat-compound generator has a shunt field of 7 turns. If the total shunt-field resistance is 100 ohm. Calculate the mmf when the machine operates rated voltage

a. At no-load

b.At full-load

Solution E.8

Exercise 9

  • 275.78
  • 276.78
  • 277.78
  • 278.78

The generator of Fig. 4.38 revolves at 960 r/min and the flux per pole is 20 mWb. Calculate the no-load armature voltage if each armature voltage if each armature coil has 6 turns

Solution E.9

Exercise 10

  • To supply traction load
  • To supply industrial load at constant voltage
  • Voltage at the toad end of the feeder
  • For none of the above purpose
D.C. series generator is used

7. DC MOTOR

7.0 introduction to dc motor

  DC motors or direct-current motors transform electrical energy to mechanical energy. The example of the dc motor is the electrical appliances that we use in our daily life such as hoist, fan, pumps,calendars, punch-presses,and cars. These devices may have a definite torque-speed characteristic or a highly variable one. The torque-speed must be adapted to the type of the load it has to drive, and it has given rise to three basic type of motor.

 

  Direct-current motors are used in ordinary industrial applications because all electric usefulness furnish alternating current. However, for some applications such as mines and electric trains, it is sometimes advantageous to transform the alternating current to direct current in order to use dc motor because torque speed of dc motor can be varied over a wide range while retaining high efficiency. 

  

7.1 Counter-electromotive force (cemf)

   Dc motors are built the same way as generators. A dc machine can operate either as a motor or as a generator. When armature is at rest, it is connected to a dc source. The armature has a resistance and the magnetic field is created by a permanent magnet.

When the switch is closed, a large current flow in the armature because its resistance is very low. It makes the armature conductor immediately subjected to a force because they are immersed in the magnetic field created by the permanent magnets. This force will produce the powerful torque and make the armature rotate.

 

    On the other hands, when the armature begins to turn, the generator will effect. The voltage Eo is induced in the armature conductors as soon as they cut the magnetic field no matter what causes the rotation. The induced voltage Eo is therefore proportional to the speed of rotation n of the motor and to the flux per pole as a given equation

Eo=ZnΦ/60

   For a motor, the induced voltage Eo is called counter-electromotive force (cemf) because the polarity always act against the source voltage Es. It acts against the voltage in the sense that the net voltage acting in the series circuit is equal to (Es-Eo) volts.

7.2 Acceleration of the motor

The net voltage acting in the armature circuit is (Es-Eo) volts. The resulting armature current I is limited only by the armature resistance R,and so

I = (Es-Eo)/R

When the motor is at rest. the induced voltage Eo=0,so the starting current is

I=Es/R

 

 

   The starting current may be 20 to 30 times greater than the nominal full-load current of the motor. In practice, it will cause the fuses to blow or the circuit breaker to trip. However, if they are absent, the large force acting on the armature conductors produce a powerful starting torque and accelerate the armature. As the speed increases, the counter-emf Eo increases, with the result that the value of (Es-Eo).

   Although the armature current decreases, the motor continues to accelerate until it reach the maximum speed. when it is no load, this speed produces a counter-emf slightly less than the source voltage. If Eo equal to Es, the net voltage will become zero also with the current flow will also equal to zero. The driving forces would cease to act on the armature conductors,and the mechanical drag and bearing cause the motor to slow down. As the speed decreases, the net voltage increase and also the current flow increase. The speed will cease to fall as soon as the torque developed by the armature current is equal to the load torque. When the motor run at no load, the counter-emf must be less than source voltage to enable a small current to flow to produce require force.

7.3 Mechanical power and torque

   The power and torque of a dc motor are two of its most important properties. There are two equations that enable us to calculate power and torque

     The cemf induced in a lap wound armature is given by

Eo=ZnΦ/60

From this figure, the electric power Pa supplies to the armature is equal to the supply voltage multiplied by the armature current.

Pa = EsI

However, the source voltage is equal to the sum of counter-emf and voltage drop in the armature.

Es=Eo+IR

From the equation above :

Pa=EsI = (Eo+IR)I = EoI+ I(IR)

The I(IR) terms represent the heat in the armature. EoI is the electrical power that is converted into mechanical power. The mechanical power of the motor is equal to the product of the cemf multiplied by the armature current.

P=EoI

where

P= mechanical power developed by the motor.

Eo = induced voltage in the armature (cemf)

For the torque,mechanical power P is expressed in the form

P=nT/9.55

where n is the speed of the rotation

When we combine all of the equations above, we obtain

nT/9.55 = EoI= ZnΦ/60

also

T=ZnΦ/6.28

The torque developed by a lap-wound motor is given by the equation

T=ZnΦ/6.28

where 

T= torque

Z= number of conductors on the armature

Φ= effective flux per pole

I= armature current

6.28 = constant to take care of the unit ( exact value = 2π )

This equation shows that the torque of the motor either by raising the armature current or by raising the flux created by the poles.

 

 

7.4 Speed of rotation

    When dc motor drives a load between no-load and full-load, the IR drop due to armature resistance is always small compared to the supply voltage. This means that the counter-emf is very nearly equal to supply voltage.

   On the other hands, from the equation

Eo=ZnΦ/60

Replace Eo by Es, we obtain

Es=ZnΦ/60

 n = 60E/ZΦ (approximation)

where

n= speed of rotation

Es = armature voltage

Z = total number of armature conductors

This equation shows that the speed of motor is proportional to the armature supply voltage and inversely proportional to the flux per pole.

7.5 Armature speed control

   In practice, we can vary source voltage by connecting the motor armature to a separately excited variable voltage dc generator G. The field excitation of the motor is constant, but the generator excitation can be varied from zero to maximum and reversed. The generator output voltage can be varied from zero to maximum, with either positive or negative polarity. The motor speed can be varied from zero to maximum in either direction. The generator is driven by an ac motor connected to a 3 phase line. This method of speed control is called Ward-Leonard system.

   The Ward-Leonard system is more than just a simple way of applying a variable dc voltage to the armature of a dc motor. It can force the motor to develop the torque and speed required by the load. Suppose that the source voltage is adjusted to be slightly higher than the cemf of the motor, the current will flow in the direction that shows in the figure above. which makes the motor generate the positive torque then the armature of the motor will absorb power because the current flow into the positive terminal.

   If we reduce source voltage by reducing the generator excitation. When source voltage less than cemf, current flows reversely. As the result:

1. the motor torque reverses

2. the armature of the motor delivers power to generator suddenly becomes a generator and generator suddenly become the motor.

The electric power that is dc motor delivers to generator G and become a kind of kinetic energy of the rapidly decelerating armature and connected mechanical load. By reducing source voltage, the motor is suddenly forced to slow down.

  If dc power received by generator, it will operates as a motor, driving its own ac motor as an asynchronous generator. As a result, ac power is fed back into the line that normally feeds the ac motor. The fact that power can be recovered this way makes the Ward-Leonard system very efficiency.

 

 Rheostat Speed Control 

   Rheostat speed control is another way to control the speed of a dc motor is to place a rheostat in series with the armature. The current in the rheostat produces the voltage drop which subtracts from the fixed source voltage, yielding a smaller supply voltage across the armature.

  This method make us to reduce the speed below its nominal speed. It is recommend for the small motor because a lot of power and heat is wasted in the rheostat. The IR drop across the rheostat increases as the current increases. This produces a substantial drop in speed with increasing mechanical load.

7.6  Field speed control

     From equation n = 60E/ZΦ (approximation),we can vary the speed of dc motor by varying the field flux. Let keep the armature voltage constant so that the numerator is constant. Consequently, the motor speed now changes in inverse proportion to the flux so if we increase the flux, the speed will drop.This method of speed control is used when the motor has to run above its rate speed called base speed. To control the flux, we connect a rheostat in series with the field.

  To understand this method of speed control, the motor is initially running at a constant speed. The counter-emf Eo is slightly less than the armature supply voltage due to the IR drop in the armature. If we increase the resistance of the rheostat, both exciting current and the flux will diminish. The reducing of the counter-emf, causing the armature current to jump to a much higher value. The current change because the value depends on the very small difference between source voltage and counter-emf. It will accelerate until counter-emf is again equal to source voltage.

  To develop the same counter-emf with a weaker flux, the motor must turn faster. The way to increase the speed of the motor is to insert a resistance in series with the field. Broader speed ranges tend to produce instability and poor communication.

 For an abnormal conditions, the flux may drop to dangerously low values such as if the exciting current of a shunt motor is interrupted accidentally, the only flux remaining is that due to the remanent magnetism in the poles. This flux is so small that the motor has to rotate at a dangerously high speed to induce the required cemf. Safety devices are introduced to prevent such runaway conditions.

7.7 Shunt motor under load

    Consider a dc motor running at no-load. If a mechanical load is suddenly applied to the shaft, the small no load current does not produce enough torque to carry the load and the motor begins to slow down. This causes the cemf to diminish, it makes a higher current and higher torque. When the torque developed by the motor is exactly equal to the torque from mechanical load, the speed will remain constant. To sum up, as the mechanical load increases, the armature current rise and the speed drops.

   The speed of the shunt motor stays a relatively constant from no-load to full-load. In a small motors,it drops a little bit when full load is applied. In big machines, the drop is very less, due in part, to the very low armature resistance. By adjusting the field rheostat, the speed can absolutely constant as the load change.

     Typical torque- speed and torque current characteristics of the shunt motor, the speed, torque,and current are given in per unit values. The torque is directly proportional to the armature current.

7.8 Series motor

   A series motor is identical in construction to a shunt motor except for the field. The field is connected in series with the armature and carry the full armature current. This series field is composed of a few turns of wire having a cross section large enough to carry the current.

    In a shunt motor, the flux F per pole is constant at all loads because the shunt field is connected to the line. However, for a series motor, the flux per pole depends on the armature current and upon the load. When the flux is large, the current is also large. Despite these differences, they both use the same basic principle to apply in both machines.

 

   When the series motor operates at full-load, the flux per pole is the same as a shunt motor of identical power and speed. However, when the series motor starts up, the armature current is higher than normal. It follows that the starting torque of a series motor is considerably greater than that of the shunt motor. We can see the comparison between T and I curves in the graph

    On the other hand, if the motor operates at less than full-load, the armature current and the flux per pole are smaller than normal. The weaker field causes the speed to rise in the same way as it would for a shunt motor with a weak shunt field. If the load is small, the speed may rise dangerously high values. For this reason, we never permit a series motor to operate at no-load. The resulting centrifugal forces could tear the winding out of the armature and destroy the machine.

 

7.9 Series motor speed control

    When a series motor carries a load. Its speed may have to be adjusted slightly. The speed can be increased by placing a low resistance in parallel with the series field. The field current is then smaller than before which produces a drop in flux and increasing speed.

Conversely, the speed may be lowered by connecting an external resistor in series with the armature and the field. The total IR drop across the resistor and field reduces the armature supply voltage and so the speed must fall.

Typical torque-speed and torque-current characteristic is represented in the form of graph which is quite different from the shunt motor characteristic.

7.10 Applications of the series motor

   Series motors are used on equipment requiring a high starting torque. They are also used to drive devices which must run at high speed at light loads. The series motor is particularly well adapted for traction purposes such as electric trains. Acceleration is rapid because the torque is high at low speeds. The series motor automatically slow down when the train goes up and turn at the top speed on flat ground. Series motors are also used in electric cranes and hoists.

 

 

7.11 Compound motor

   Compound dc motor carries both series field and a shunt field. In the cumulative compound motor, the mmf of the two fields add. The shunt field is always stronger than the series field.

     When the motor runs at no-load, the armature current I in the winding is low and the mmf of the series field is negligible. On the other hand, the shunt field is fully excited by current Ix and the motor behaves like a shunt machine. As the load increases, the mmf of the series field increases but for the shunt field remains constant which make the total mmf and the flux per pole is greater under load than no load.    

If the series field is connected opposite the shunt field, we obtain a differential compound motor. For the differential compound motor, the total mmf decrease when the load increases with the same trend as the rising of speed. The differential compound motor has a very few application such as the dc motor in steel mill.

7.12 Reversing the direction of rotation

    To reverse the direction of rotation of a dc motor, we must reverse the armature connections or reverse both shunt and series field connections. The interpoles are considered to form part of the armature.

7.13 Starting a shunt motor

  Dc motor has such a high starting current because from the equation of induced voltage is Eo=ZnΦ/60 which at the beginning, the speed of the motor start from zero so it makes the induced voltage becomes zero. When Eo=o, Es= Eo+ IaRa makes Es=IaRa. Therefore

Ia=Es/Ra

which you can see that the source voltage is very high and voltage is proportional to the current flow, so it makes the starting current also has a high value.

If we apply full voltage to a stationary shunt motor, the starting current in the armature will be very high and it will cause some hazards such as

  • Burning out the armature.
  • Damaging the commutator and brushes
  • Overloading the feeder
  • Snapping off the shaft due to mechanical shock
  • Damaging the driven equipment due to the sudden mechanical hammer blow.

  All dc motor must limit the starting current to be reasonable values to prevent the risks. To solve the problem, we can connect a rheostat in series with the armature because the resistance is reduced as the motor accelerates and eliminate entirely when the machine has attained full speed. Today, this method uses to limit the starting current and to provide speed control.

7.14 Face-plate starter

       The diagram of a manual face-plate starter for a shunt motor start from the bare copper contacts are connected to current-limiting resistors. Conducting arm sweeps across the contact when it is pulled to the right by means of insulated handle 2. The arm touches dead copper contact M and the motor circuit will open.

     When we draw the handle to the right, the conductor will touch the fixed contact N and the supply voltage Es will immediately cause the current Ix to flow but the armature current I is limited by the resistors in the starter box. The motor begin to turn as the induced voltage build up, the armature current falls. When the motor speed increases, the arm is pulled to the next contact by removing resistor 1 from the circuit. To move to the next contact the speed has to drop down. We can see that the contact arm move depends on the speed of the motor. The arm is magnetically held by the small electromagnet which connect in the series to the shunt field.

       If the supply voltage is interrupted, the electromagnet will release the arm, allowing it to turn to its dead position under the pull spring. This will prevent the motor from restarting unexpectedly.

7.15 Stopping a motor

    To stop a dc motor that has a heavy inertia load, it has to use an hour or more to stop the system of the motor. We must apply a breaking torque to ensure a rapid stop because of the amount of time for deceleration  and the circumference of the motor. One way to brake the motor is by simple mechanical friction. A more elegant method is to calculate a reverse current in the armature to brake the motor electrically. 

There are two methods to create such an electromechanical brake which are

  • dynamic braking
  • plugging

7.16 Dynamic braking

    A shunt motor that the field is directly connected to a source voltage and an armature is connected to the same source by double throw switch. The switch connects the armature to either the line or the external resistor.

 

 

When the motor is running normally, the direction of the armature current and the polarity of the cemf are shown in the figure. If we neglect the IR drop between the resistor, induced voltage will equal to voltage source.

 

 

 

 

 If we open the switch, the motor continue to turn, but the speed will drop because of the friction and windage losses. On the other hand, because of the shunt field is excited, it still has the induced voltage in the system which keep falling according to the rate of speed. Now, the motor become a generator that the armature is on open circuit.

 

 

When you close switch on another way, the armature is connected to the external resistor. Induced voltage will produce the current I2 in the opposite direction to the current I1 from the previous figure. Then , a reverse torque is developed which its magnitude depends on the value of the current flow I2 as you can see from the formula  T=ZIΦ/2π. This torque will make the machine to stop.

 

   In practice, resistor is chosen so that the initial braking current will be twice the rated motor current and it will make the initial braking torque twice the torque of the motor.

 There are a graph to compare the speed time curves for a motor equipped with dynamic braking as the graph below

   For the dynamic braking, when the motor slows down, the induced voltage will gradually decrease because of the decreasing of current flow I2. Then the braking torque will become smaller and smaller until it becomes zero when the armature turn. The speed decrease exponentially which you can see that the trend of the graph is similar to the discharge voltage of the capacitor.

 

7.17 Plugging

   Plugging or another name is reverse current braking. It is a reversing of the armature current by reversing the terminals of the sources.

The armature current is given by

I1=(Es-Eo)/Ro

     If we suddenly reverse the terminals of the sources, the net voltage will also change the direction and becomes (Eo+Es). So, the cemf of the armature is no longer counter but add to the supply voltage. From this situation, it makes the net voltage produce high amount of reverse current flow. This current will initiate an arc around the commutator, destroying segments,brushes,and supports before the line circuit breaker to open.

  To prevent a catastrophe, we must limit the reverse by adding the resistor R in series to the reversing circuit. As the dynamic braking, the resistor is designed to limit the initial braking current I2. With the plugging circuit, there will be a reverse torque even  when the armature stop, we must immediately open the armature circuit, otherwise the reverse running circuit will happen.

 

   The curve for the plugging start from the same speed as the dynamic braking with the same braking current. You will see that for the plugging method, the plugging stop the motor completely after 2To compare to the dynamic braking that still have 25 percents speed for braking. Therefore, most of the application use dynamic braking than plugging.

 

7.18 Dynamic braking and mechanical time constant

 

    For dynamic braking, the speed decreases exponentially with time when the dc motor is stopped. We can speak of a mechanical time constant in the same way as the electrical time constant of the capacitor that discharges into a resistor.

   There is a direct mathematical relationship between the conventional time constant T and the half constant To, it is given by

The mechanical time constant is given by

 

where

To=time for the motor speed to fall to one-half its previous value.

J= moment of inertia of the rotating parts, referred to the motor shaft.

n1= initial speed of the motor when braking starts.

P1 = initial power delivered by the motor of braking resistor.

  This equation comes from the assumption that the braking effect due to the energy dissipated in the braking resistor. In general, the motor is subjected to an extra braking torque due to windage and friction.

7.19 Armature reaction

Until now we have assumed that the only mmf acting in a dc motor is due to the field. However, the current flow in the armature conductors also creates a magnetomotive force that distorts and weakens the flux coming from the poles. This distortion take place in motors as well as in the generators. This magnetic action of the armature mmf is called armature reaction which is effect of armature flux on main field  flux. 


   Basically there are two windings in a dc motor which are Armature winding (on stator) and field winding. When we excite the field winding, it produces a flux which links with the armature. This causes an emf and hence a current in the armature. This current in armature produces another flux which lags the main flux. This is referred to as armature reaction. It has two effects on the machine:

1. Demagnetising effect: It reduces the strength of the main flux.
2. Cross magnetising effect: Its effect is that it distort the the main flux line along the conductor.

7.20 Flux distortion due to armature reaction

When a motor runs at no-load, the small current flowing in the armature does not affect the flux that coming from the poles at the first figure. However, when the armature has a normal current, it produces a strong magnetomotive force which would create a flux 2 in the second figure. The net flux between flux 1 and flux 2 will give a result as a flux 3.  

As you can see, the flux density increases under the left half and decrease under the right half. Because of the unequal flux density, it will cause some effects.

  • the neutral zone will shift which will make poor commutation with sparking at the brushes.
  •  the saturation sets in due to the higher flux density in pole tip.

  The increase of flux under the left hand pole is less than the decrease in the right half. Flux 3 at full load is slightly less than flux 1 at no-load. For large machine, the decreasing of the flux can cause the speed to increase with load and make it unstable. To eliminate this problem is to add series field to increase the flux under load which is called stabilized-shunt winding.

7.21 Commutating poles

  One of several small poles between the main poles of a direct-current generator or motor,which serves to neutralize the flux distortion in the neutral plane caused by armature reaction. Also known as compole  or interpole.

  To counter the effect of armature reaction and thereby improve commutation, we always place a set of commutating poles between the main poles of medium and large power dc motors. Just like the dc generator, these narrow poles develop a magnetomotive force equal and opposite to the mmf of the armature so that the respective magnetomotive forces rise and fall together as a load current.

   

   In practice, the mmf of the commutating poles is greater than the mmf of the armature. Consequently, a small flux subsists in the region of the commutating poles. The flux is used to induce in the coil undergoing commutation a voltage that is equal and opposite to the self induction voltage.

    The neutralization of the armature mmf is restricted to the narrow zone covered by the commutating poles. The flux distribution under the main poles unfortunately remains distorted. This creates no problem for motors driving ordinary loads but for the special case, it necessary to add a compensating winding.

7.22  Compensating winding

    Some dc motor in the high power range employed in steel mills perform a series of rapid, heavy-duty operations. The amount of armature current produces very sudden changes in armature reaction.

 

 

   For such motors, the commutating poles and series stabilizing winding do not neutralize the armature mmf. Torque and speed control is difficult under such transient conditions and flash-overs may occur across the commutator. To eliminate this problem, we use special compensating windings to connect in series with the armature. However, because the windings are distributed across the pole faces, the armature mmf is bucked from point to point, which eliminates the field distortion in the first figure below. Also, the field distribution remains essentially undisturbed from no-load to full-load and become the general shape in the second figure.

The addition of compensating windings effect on the design and performance of a dc motor.

  •  The short air gap can be used because we don't have to worry about the demagnetizing effect of the armature. A shorter gap means that the shunt field strength can be reduced and the coil are smaller
  • The inductance of the armature circuit is reduced which the armature current change more quickly and motor give a better response.
  • A motor equipped with compensating windings can develop 3 to 4 times its rated torque. The peak torque will be lower if the armature current is large because the effective flux in the air gap falls off rapidly with increasing current caused by armature reaction.

  We conclude that compensating windings are essential in large motors subjected to severe duty cycles.

 

 

 

7.23 Basic of variable speed control

  The most important outputs of a dc motor are its speed and torque. It is very useful to determine the limits of each as the speed is increased from zero to above base speed. The rated values of the armature current, armature voltage,and field flux must not be exceeded, although lesser values may be used.

   The figure on the left side is the ideal separately excited shunt motor that the armature resistance is negligible. All the components in the figure are expressed in per-unit values. If the rated armature voltage happens to be 240 V and the rated armature current is 600 A, they are both given a per-unit value of 1. Similarly,the rated shunt field flux also has a per-unit value of 1. The benefit of the per-unit value is that it renders the torque-speed curve universal.

The per-unit torque is given by the per-unit flux times the per-unit armature current

Also, the per-unit armature voltage is equal to the per-unit speed times the per-unit flux

 

 

 As you can see from the torque-speed curve at the left, it shows the conditions that the motor develops rated torque (T=1) at the rated speed (n=1). We called the rated speed as based speed.

    In order to reduce the armature voltage to zero, while keeping the rated value of armature current and flux to be constant at their per-unit value of 1. By apply the equation, we get T= 1x1=1 and Ea= n x 1 = n.For the ideal dc shunt motor, it can operate anywhere within the limits of torque-speed curve.

  There are the graphs that represent the state of armature voltage, armature current,and flux during motor operation, know as the constant torque mode

   In practice, the actual torque-speed curve may differ from the ideal curve. The curve indicates an upper speed limit 2 but some machine may have the limit at 3 or 4 by reducing the flux. However, the increase of the speed will make the communication problem occurs and it may be dangerous. When the motor run below base speed, the ventilation becomes poorer and the temperature tends to rise. Consequently, the armature current must be reduced that it will lead to the reducing of torque. When the speed is zero, all force ceases and even the field current must be reduced to prevent the heat. As the result, the torque may have the per-unit value just only 0.25 as it shows in the graph below

   The drastic fall-off in torque as the speed diminishes can be largely overcome by using an external blower to cool the motor. The torque-speed curve will be ideal if you deliver a constant stream of air.

 

7.24 Permanent magnet motors

   We know that the shunt field motors has the energy consumed, the hear produced and the relatively large space taken up by the field poles which are disadvantages of a dc motor. By using the permanent magnet instead of field coils can reduce the disadvantage of the motor. The result is that a small motor has a higher efficiency with the added benefit due to the field failure.

   A further advantage of using permanent magnets is that the effective air gap keep increasing because the magnets have a permeability that nearly equal to the air. As the result, the armature mmf cannot create the intense field. The field created by the magnets does not become distorted as the figure below

 

 

The armature reaction is reduced and commutation is improved as well as the overload capacity of the motor. A further advantage is that the air gap reduces the inductance of the armature to change the armature current flow.

 

 

 

   Permanent magnet motors are advantageous in capacities below 5 hp. The magnet are ceramic or rare earth-cobalt alloys to construct the PM motor. Its elongated armature ensures low inertia and fast response when use in the application. 

   The only drawback of PM motors is the relatively high cost of the magnets and the inability to obtain high speed by field weakening.

 

 

Exercise 1

  • Burning out the armature
  • Overloading the feeder
  • Damaging the driven equipment
  • cause the air gap in the core
Which one is not the risk of the high starting current in the armature ?

Exercise 2

  • Dynamic breaking stops the motor slower than Plugging
  • When a dc motor is stopped by dynamic braking, the speed decreases exponentially with time.
  • Plugging is when the armature induce voltage it suddenly open the switch and connect to the external resistor for breaking.
  • Dynamic breaking is more popular than plugging because it can break smoother than Plugging.
Which one is wrong?

Exercise 3

  • Series speed control
  • Armature speed control
  • Shunt speed control
  • Field speed control
  • Compound speed control
  • Dynamic speed control
What are two types of speed control for dc motor?

Exercise 4

  • Reverse resistance pole
  • Add more resistance
  • Reverse armature connection
  • increase the number of coil
  • Reverse the rheostat connection
  • Reverse the shunt and series field connection
State 2 ways to reverse the direction of rotation of compound motor.

Exercise 5

  • Adding series motor to the system
  • Placing the set of commutating poles
  • Creating new magnetic flux by adding permanent magnet
  • Adding rheostat
What is the way to reduce armature reaction effect?

Exercise 6

  • a. 230A b.250V,375V c. 2A, 3A
  • a. 175A b.150V,175V c. 1.85A, 4.53A
  • a. 150A b.100V,146V c. 50A, 4A
  • a. 160A b.50V,73V c. 25A, 2A

The armature of a permanent-magnet dc generator has a resistance of 1 Ohms and generates a voltage of 50V when the speed is 500 r/min. If the armature is connected to a source of 150V, calculate the following:

  • a. The starting current
  • b. The counter-emf when the motor run at 1000 r/min.At 1460 r/min.
  • c. The armature current at 1000 r/min. At 1460 r/min.

Solution E.6

Exercise 7

  • a. 225r/min, 24.7kN.m b. 225 r/min, 35.84kN.m
  • a. 228r/min, 31.8kN.m b. 228 r/min, 47.8kN.m
  • a. 220r/min, 26.8kN.m b. 220 r/min, 54.3kN.m
  • a. 178r/min, 23.65kN.m b. 178 r/min, 27.98kN.m

  A 2000 kW, 500 V, variable-speed motor is driven by a 2500 kW generator,using a Ward-Leonard control system , the total resistance of the motor and generator armature circuit is 10 mOhms. The motor turns at a nominal speed of 300 r/min, when Eo is 500 V, calculate

  • a. The motor torque and speed at Es=400V and Eo=380V.
  • b. The motor torque and speed at Es=350V and Eo=380V.

Solution E.7

Exercise 8

  • a. 50A b.120V c.5750W
  • a. 40A b.76V c.4578W
  • a. 60A b.135V c.6750W
  • a. 30A b.140V c.3250W

   A shunt motor rotating at 1500 r/min is fed by a 120V source. The line current is 51A and the shunt-field resistance is 120 Ohms. If the armature resistance is 0.1 Ohms, calculate the following:

  • a. The current in the armature
  • b. The counter-emf
  • c. The mechanical power developed by the motor

Solution E.8

Exercise 9

  • a. 5s b. 50s c. 20 min
  • a. 10s b. 55s c. 27 min
  • a. 5s b. 45s c. 26 min
  • a. 10s b. 60s c. 28 min

   A 225 kW, 250V, 1280 r/min dc motor has windage, friction,and iron losses of 8kW.It drives a large flywheel and the total moment of inertia of the flywheel and armature is 177 kg.m^2. The motor is connected to a 210 V dc source and its speed is 1280 r/min just before the armature is switched across a braking resistor of 0.2 Ohms, calculate

  • a. The mechanical time constant of the braking system
  • b. The time for the motor speed to drop to 20r/min
  • c. The time for the speed to drop to 20r/min if the only braking force is that due to the windage friction ,and iron loss

 

Solution E.9

Exercise 10

  • a. 1250 A, 150.5 kW b.50s
  • a. 1150 A, 225.5 kW b.10s
  • a. 1050 A, 220.5 kW b.20s
  • a. 1350 A, 530.5 kW b.30s

    A 225 kW, 250V, 1280 r/min dc motor has windage, friction,and iron losses of 8kW.It drives a large flywheel and the total moment of inertia of the flywheel and armature is 177 kg.m^2. The motor is connected to a 210 V dc source which it has the net voltage acting across the resistor 420V its speed is 1280 r/min just before the armature is switched across a braking resistor of 0.2 Ohms is plugged,and the braking resistor is increased to 0.4 Ohms, so that the initial braking current is the same as before.

  • a. The initial braking current and braking power
  • b. The stopping time

Solution E.10