Courses In
Electrical
Engineering
Volume IV
COURSE ON ELECTRICAL MACHINES
FOR TECHNICAL COLLEGE
By
Jean-Paul NGOUNE
DIPET I (Electrotechnics), DIPET II (Electrotechnics) M.Sc. (Electrical Engineering)
To Jesus-Christ,
My Lord and Saviour.
FOREWORD
This is the first chapter of a Course on Electrical Machines for Technical College that I am writing. The concern of this chapter is the study of Direct Current Machines. That is, DC Generators and DC Motors. The chapter begins with the study of DC generators and end with that of DC motors. Many examples are treated in order to enable the reader to assimilate easily the course.
My aim is to bring my humble contribution for the improvement of Technical Education in my country (Cameroon, Central Africa) and to help anyone to whom this document may be useful.
This Document and many other pedagogical resources produced by me are available
and freely downloadable at the following link: www.scribd.com/jngoune. Any
suggestion or critic is warmly received; send me a mail at the following address:
[email protected]. Stay blessed.
NGOUNE Jean-Paul, 03 August 2012, 12:18 Kumbo, Cameroon.
CONTENTS
1.0 Specific objectives………. 6
1.1 Introduction………. 6
1.2 Structure of Direct Current machines………. 8
1.3 DC generators……… 11
1.3.1 Working theory of DC generators………... 11
1.3.2 Equation of induced emf……….. 13
1.3.3 Classification of DC machines……… 14
1.3.4 Equations of Separately Excited DC generators………. 16
1.3.5 Equations of Self Excited DC generators………. 17
1.3.6 Armature Reaction……… 22
1.3.7 Compensating windings………... 23
1.3.8 Commutating poles (Interpoles windings)………. 23
1.3.9 Losses in DC generators………. 24
1.3.10 Powers and efficiencies……… 26
1.3.11 DC generators characteristics………. 29
1.4 DC motors……….. 47
1.4.1 Principle of DC motors………. 47
1.4.2 Back emf………. 48
1.4.3 Power relationship in DC motors……… 49
1.4.4 Types of DC motors………. 49
1.4.5 Conditions for maximum power……….. 54
1.4.6 Torque………. 55
1.4.7 Power flow and efficiency………. 56
1.4.8 Speed control………. 66
1.4.9 Speed regulation……… 67
1.4.10 DC motors characteristics……… 68
1.4.11 Starting and breaking methods of DC motors……….. 75
REVIEW QUESTIONS………. 84
References/ Acknowledgements……… 93
Courses In
Electrical
Engineering
Volume IV
ELECTRICAL MACHINES FOR TECHNICAL COLLEGE
CHAPTER ONE: DIRECT CURRENT MACHINES
By
Jean-Paul NGOUNE
DIPET I (Electrotechnics), DIPET II (Electrotechnics) M.Sc. (Electrical Engineering)
Chapter One
DIRECT CURRENT
MACHINES
1.0 Specific objectives:
At the end of this chapter, the student will be able to:
- Define electrical machine;
- Describe the structure of a direct current machine;
- Differentiate a direct current motor from a direct current generator;
- Establish the fundamental equations for the various types of direct current motors an generators;
- Plot interpret the characteristic curves of direct current motors and generators
for different types of test (open circuit characteristics, external characteristics);
- Give specific applications for each type of direct current motors and
generators;
1.1 Introduction:
Electrical machines are converters that are used to continuously translate electrical input to mechanical output or vice versa. This process of translation is known as an electromechanical energy conversion. If the conversion is from mechanical energy to electrical energy, the electrical machine is said to be a
generator; if it is from electrical energy to mechanical energy, the electrical machine is amotor. The electromechanical conversion results basically from the two following electromagnetic phenomena:
When a conductor moves in a magnetic field, voltage is induced across the
When a current carrying conductor is placed in a magnetic field, the conductor
experiences a mechanical force (Motor action).
M o to r G e n e ra to r In p u t O u tp u t In p u t O u tp u t E le c tr ic a l e n e r g y M e h a n ic a l e n e r g y M e c h a n ic a l e n e r g y E le c tr ic a l e n e r g y
Figure 1.1: Energy conversion for motor and generator
M
Field
Armature
Figure 1.2: Symbol of a DC machine
The two electromagnetic phenomena mentioned above occur simultaneously
whenever energy conversion takes place from electrical to mechanical or vice versa. In motoring action, the electrical system makes current flow through conductors that are placed in the magnetic field. A force is produced on each conductor. If the conductors are placed on a structure free to rotate, an electromagnetic torque will be produced, making the rotating structure (rotor) to rotate at a given speed. If the conductors rotate in a magnetic field, a voltage will also be induced across each of
them. In generating action, the rotating structure is driven by a prime mover such as
a steam turbine or a diesel engine. A voltage is induced in the conductors that are rotating with the rotor. If an electrical load is connected to the windings formed by those conductors, and electrical power will be produced to supply it. Moreover, the current flowing in the conductors will interact with the magnetic field to produce a
1.2 Structure of direct current machine:
The following figure presents the main parts of a direct current machine.
Figure 1.3: Structure of a direct current machine.
The essential parts of a DC generator consist of:
Magnetic frame or yoke,
Pole coils or field coils,
Armature windings,
Brushes and bearings,
Pole cores and pole shoes,
Armature core,
Commutator.
Of these, the pole core, the armature core, the yoke and the air gaps between
the poles and the armature cores form the magnetic circuit whereas the rest form the
electrical circuit. Yoke:
The yoke is the outer frame of the machine. It provides mechanical support for the poles and also canalizes the magnetic flux produced by the poles. It acts as a protective covering for the whole machine. The pole is made of cast iron or cast steel.
Pole cores and pole shoes:
The pole shoes help to support the field coils and also to spread out the magnetic flux in the air gap. The pole cores and pole shoes are made up of steel plates laminated and retrieved together. They are bolted to the yoke. The purpose of
laminating the core is to reduceEddy current losses.
Pole coils or field coils:
The field coils consist of insulated copper wires wound round the pole cores.
When current is flowing through these wires, they magnetise the poles which produce
Armature core:
The armature core houses the armature conductors in the slots. It is built up of laminated steel disc mounted on a shaft.
Figure 1.6: Photographic view of an armature of DC machine
Armature windings:
Two types of windings are mostly employed for the armatures of DC
machines; they are:Lap Winding and Wave Winding. The difference between the two
is merely due to the different arrangement of the connections at the front or commutator end of the armature. The following rules apply to the types of windings:
- In a lap winding, the number of parallel paths is always equal to the number of
poles and also the number of brushes.
- In wave windings, the number of parallel paths is always two and there may be
two or more brush positions.
The lap winding is suitable for high current, low voltage machines like welding
plants. The wave winding is suitable for high voltage, low current machines, like DC
generators used for lighting. Brushes:
The brushes are used to collect current from the commutator in the case of generator. In the motor, they lead current into the armature windings through the commutator. The brushes are made of carbon.
Commutator:
The function of the commutator in a generator is to convert the alternating current induced in the armature conductor into direct current in the external load circuit. The commutator has a cylindrical structure and is built up of insulated copper segments.
Figure 1.7: Photographic view of commutator.
1.3 DC generators.
When a conductor is moving in a magnetic field, an emf e is induced across its
terminals. That emf is given by the Faraday’s law as follows:
Where
DC generators are built using this basic principle.
1.31 Working theory of DC generators:
DC generator is actually an ac machine which is furnished with a special device: the commutator. The commutator is therefore a rectifying unit in a dc
generator as it converts the alternating current generated in the armature to a direct
BLv dt
d
e e = induced emf in volts (V)
B = magnetic flux density (T)
L = Length of conductor cut by the flux (m) v = Speed of the conductor (m/s)
Generators are driven by some mechanical means like petrol or diesel engine, hydropower, steam engine etc.
The brushes always keep the same polarity despite the fact that the emf induced in the armature windings changes his polarity twice per period.
Figure 1.8: Simple loop generator.
0 0 0 t t t Ua Ub Uc Emf induced
in the armature windings
Emf received on the brushes
Direct voltage
after filtering and stabilisation
1.3.2 Equation of induced emf.
Let us assume that:
Flux per pole;
Z = Total number of armature conductors; P = Number of pairs of poles;
N = Armature speed in rps;
E = emf induced in one of the parallel path;
A = Number of parallel path (A = 2 for wave winding, A = 2P for lap winding). According to the Faraday’s law of electromagnetic induction, we have:
dt d e .
For one revolution, each conductor is cut by a flux under the north pole and the and the south pole (2P). Therefore, the change in flux is
P d 2
The average emf induced per conductor is
dt P dt
d
e 2
For one revolution, the time taken is
n dt 1 Hence P n n P dt d 2 1 2 .
This is the emf generated across each armature conductor. Per path, the number of conductors is given as the total number of armature conductors divided by the number of parallel paths, that is
A Z
. Therefore, the emf generated per path is given by: A Z N P A Z n P E 60 2 2
For wave wound machine, A = 2, hence
A P ZN E 2 60
For lap wound machine, A = 2P, hence
Application exercise.
A 6 pole DC generator has 1530 armature conductors. The machine is nun at 2000 rpm and the useful flux per pole is 10 mWb. Calculate the emf generated in the following cases:
- The machine is wave wound;
- The machine is lap wound.
Solution to the application exercise.
Data 3; 2000; 1530 2 6 ; 10 mWb P N Z
Case 1: wave winding
V P ZN E 1530 60 3 10 10 2000 1530 60 3
Case 2: Lap winding
V ZN E 510 60 10 10 2000 1530 60 3 1.3.3 Classification of DC machines.
The field circuit and the armature circuit of a DC machine can be interconnected in various ways to provide a wide variety of performance characteristics and outstanding advantage of DC machines. Also, the field poles can be excited by two field windings, a shunt field winding and a series field winding as shown in the following figure.
Figure 1.10: Shunt and series field windings for a DC generator.
Zn ZN E 60
The shunt winding has a large number of turn and takes only a small current
(less than 5% of the rated armature current). The resistance of the shunt winding is far greater than that of series winding. The shunt winding is connected in parallel with the armature. The series winding has fewer turns but carries a large current. It is connected in series with the armature
In short we can say that there are two main groups of DC machine: separately excited DC machines and self-excited DC machines.
Separately excited DC machines.
In the separately excited DC machine, the field winding is excited from a separate source (Figure 1.11a).
Self-excited DC machines.
For this type of DC machine, the field winding can be connected in three different ways:
- In series with the armature resulting in a series DC machine;
- In parallel with the armature resulting in shunt DC machine;
- Both shunt and series windings may be used, resulting in compound machine.
There are two types of compound DC machines: long shunt and short shunt. The figure 1.11 below show the classification of DC machines.
A rheostat is normally included in the circuit of the shunt winding to control the
field current and thereby to vary the field mmf. Field excitation may also be provided
by permanents. This maybe considered as a form of separately excited machine, the
permanent magnet providinga separate but constant excitation.
1.3.4 Equations of separately excited DC generators.
Let us consider the figure below presenting the electrical diagram of a separately excited DC generator.
RA E
If
I
U
Uf
Rf
Rh
Let us assume the following notations: If = Field current;
Rf = Resistance of the field winding; Uf = Field circuit voltage;
E = Emf generated; U = Terminal voltage;
Ra = resistance of the armature winding; Ia = Armature current;
I = line or load current; Rh = Rheostat.
We can deduce the following formula: ; h f f R R U If I Ra E Ia Ra E U . .
1.3.5 Equations of self-excited DC generators.
The field winding of a self excited generator is energized by the current
produced by the generator itself. In this machine, residual magnetism must be
present in the machine iron to get the self excitation process started. When the armature is rotated, some emf is produced which goes through the field coils and strengthens the residual magnetism.
a. Shunt excited DC generator.
RA E I U Rh Ish Ia Application exercise.
A 4 pole lap wound DC shunt generator has a useful flux per pole of 0.07Wb. The armature winding consists of 440 conductors having a total resistance of 0.055Ω. Calculate the terminal voltage when running at 900rpm if the armature current is 50A.
Solution to the application exercise
Data: 2 2 4 P ; Z 440; Ra0.055; Ia 50A; 0.07Wb; lap winding (A =2P). R A E I U R h Is h Ia
We know that U ERaIa
Let us first determine the emf E of the generator. Since the machine is lap wound, we
RaIa E U Ish I Ia Rsh U Ish
V E 462 60 07 . 0 900 440 Then we have V RaIa E U 4620.05550459.25 Exercise 1.1
An 8 pole DC shunt generator with 778 wave connected armature conductors and running at 1000 rpm supplies a load of 12.5Ω resistance at a terminal voltage of 250V. The armature resistance is 0.24Ω and the field resistance is 250Ω. Determine:
a. the armature current; b. The induced emf; c. The flux per pole.
b. Series excited DC generator.
RA E I Ia U Rs Application exercise.
An 8pole DC series generator with 778 wave wound armature conductors running at 1000 rpm supplies a load of 20Ω.The useful flux per pole amount to 4mWb. The armature and series field resistances are 0.08Ω and 0.12Ω respectively. Determine:
a. The emf generated; b. The armature current; c. The terminal voltage.
Solution to the application exercise.
Data: 4 2 8 P ; Z 778; N = 1000rpm; R = 20; Ra = 0.08Ω; Rs = 012Ω; Wave winding: A = 2. 4mWb
Ra Rs
I E U Ia I RA E I Ia U Rs a. Emf generated. V P ZN E 207.46 60 4 004 . 0 1000 778 60 b. Armature current
A R Rs Ra E Ia Ia R Rs Ra E 10.27 20 12 . 0 08 . 0 46 . 207 0 c. Terminal voltage.
Ra Rs
Ia RIa V E U . 205.4 Exercise 1.2A series generator supplies a load of 10Ω and the line current is 25A. The armature and field resistances are 0.08Ω and 0.12Ω respectively. Determine:
a. The terminal voltage; b. The emf generated.
c. Compound excited generator.
We will give the equations for long shunt connected, and then for short shunt connected DC compound generators.
i) Long shunt DC compound generator.
R A E Ia Ia R s R h U Is h I
a s
a sh a h sh R R I E U I I I R U I ii) Short shunt DC compound generator. E Rsh Rs Ra U Ia Ish I Remark
In DC machines, there is usually a voltage drop Ub at the brushes. If mentioned, it should be taken into account in the total drop.
E
Rsh
Rs
Ra
U
Ia
Ish
I
Ub
Application exerciseA 4 pole, long shunt, compound generator supplies 100A at a terminal voltage of 500V. If armature resistance is 0.02Ω, series field resistance is 0.04Ω and shunt field resistance 100Ω, find the generated emf. Take drop per brush as 1V.
Solution R A E Ia Ia R s R h U Is h I
Ra Rs
Ia Ub UE 2 (We have two brushes, so the drop due to the brushes is 2Ub) s a a sh a sh sh IR I R E U I I I R IRs U I
RaIa RsI Ub
E U 100 1 500 04 . 0 02 . 0 100 Rh V Ub V U Rs Ra A I
Ra Rs
Ia Ub U E 2 But A Rh U I Ish I Ia 105 100 500 100 Finally, Exercise 1.3A long shunt compound generator supplies a load of 10kW at 250V. The armature, series and shunt resistances are 0.08Ω, 0.03Ω and 100Ω respectively. Calculate the emf generated.
Exercise 1.4
Let a short shunt compound generator supplying 100A at a terminal voltage of 230V. The armature, series and shunt resistances are 0.02Ω, 0.05Ω and 200Ω respectively. Calculate the field current and the emf generated.
Remark: Parallel operation of shunt generators.
The connection of generators in parallel means that they are connected to the same
load, usually through a common busbar system. To connect two or more shunt
generators to the same busbar, two conditions must be observed:
- The emf of the incoming generator should be practically equal to the busbar voltage V.
- Like polarities should be connected together.
Exercise 1.5
A 4 pole, DC shunt generator with a shunt field resistance of 100Ω and an armature resistance of 1Ω has 378 wave-connected conductors in its armature. The flux per pole is 0.01Wb. If a load resistance of 10Ω is connected across the armature terminals and the generator is driven at 1000rpm, calculate the power absorbed by the load.
V1.3.6 Armature reaction (AR)
By armature reaction is meant the effect of magnetic field set up by armature current on the distribution of flux under main poles.
With no current flowing in the armature, the flux in the machine is established by the mmf produced by the field current. However, if the current flows in the armature circuit, it produces its own mmf (hence flux) that opposes the flux produces by the field current under main poles. The armature magnetic field has two effects:
- It demagnetises or weakens the main flux;
- It cross-magnetises or distorts the main flux.
a) With no armature current, the flux under pole is not distorted.
d) With current flowing in its conductors, the armature produces its own flux that distorts the flux under main poles.
1.3.7 Compensating windings
These are used for large direct current machines which are subjected to large fluctuations in load, such as rolling mill motors and turbo-generators. Their function is to neutralise the cross magnetizing effect of armature reaction. In the absence of compensating windings, the segments of the commutator may be short-circuited during rapid changes in the load of the machines.
1.3.8 Commutating poles (Interpoles windings)
To counter the effect or armature reaction in medium and large-power DC machines, a set of commutating poles (sometimes called interpoles windings) is always placed between the main poles as shown by the figure below.
Figure 1.13: Commutating poles
These narrow poles carry windings that are connected in series with the armature. The number of turns on the windings is designed so that the poles develop a
magnetomotive force mmfc equal and opposite to the magnetomotive force mmfa of
the armature. As the load current varies the two magnetomotive forces rise and fall together exactly bucking each other at all time. By nullifying the armature mmf in this way, the flux in the space between the main poles is always zero, solving then the problem of armature reaction.
Exercise 1.6
The following information is given for a 300kW, 600V, long shunt compound generator: shunt field resistance = 75Ω, armature resistance including brush resistance = 0.03Ω, commutating field winding resistance = 0.011Ω, series field resistance = 0.012Ω, divertor resistance ( in parallel with series field resistance) = 0.036Ω. When the machine is delivering full load, calculate the voltage and the power generated by the armature.
Exercise 1.7
A 4 pole, lap wound DC shunt generator has a useful flux per pole of 0.07Wb. The armature winding consists of 220 turns each of 0.004Ω resistance. Calculate the terminal voltage when running at 900 rpm if the armature current is 50A.
Exercise 1.8
A 4 pole, DC shunt generator with a shunt field resistance of 100Ω and an armature resistance of 1Ω has 378 wave-connected conductors in its armature. The flux per pole is 0.01Wb. If a load resistance of 10Ω is connected across the armature terminals and the generator is driven at 1000 rpm, calculate the power absorbed by the load.
1.3.9 Losses in DC generators
The three main losses in DC generators are copper losses, iron losses and mechanical losses.
Iron losses
These losses are due to the rotation of the iron core of the armature in the magnetic
field produced by the poles. Iron losses consist of hysteresis and Eddy Current
losses.
i) Hysteresis losses
These losses are due to the reversal of magnetism of the armature core. That is the energy required to magnetize and demagnetize the armature core as it passes through the magnetic flux of the north and the south pole.
ii) Eddy current losses
As the armature core rotates, its conductors cut the magnetic lines of flux produced by the main poles. An emf is therefore induced in those conductors and a current
known as Eddy current (In French it is called ‘courrants de Foucault’) is circulating through them. These currents cause losses called Eddy current losses. To minimize theses losses, the core should be laminated.
Copper losses
They cause heat in the machine. They are due the resistance of the windings found
in the machine. Hence, we have armature copper losses (RaIa2), shunt copper losses
(RshIsh2), series copper losses (RsIs2) Mechanical losses
Mechanical losses consist of friction losses at the bearings of the rotating armature
and also of windage losses. Mechanical losses and iron losses are known as
constant losses.
The following chart summarises the losses mentioned above.
Input Useful output Total losses Copper losses Iron losses Mechanical losses
Armature copper loss
Field copper loss
Hysterisis loss
Eddy current loss
Friction loss
Windage loss
1.3.10 Powers and efficiencies
a) Powers
Let us assume:
Po = Output power (also called useful power or electrical power);
Pin = Input power (power absorbed from the prime mover);
Pm = Mechanical power developed in the armature;
PJ = Total copper losses;
Pc= Constant losses (iron and mechanical losses).
The power distribution can be sketched as follows.
Pin
Pc Pj
Pm Po
Where
T = Shaft torque of the prime mover in (N.m). N = Rotating speed in rpm.
= Angular speed in (rad/s).
Where
E = emf generated in (V) Ia = armature current (A)
Where
U = terminal voltage in (V) I = line current in (A).
T N T Pin 60 2 . Ia E Pc Pin Pm . UI P Pm Po J
b) Efficiencies
Mechanical efficiency
Electrical efficiency
Overall efficiency
Remark
Efficiency is maximal when Joule losses equal constant losses (PJ = PC). Application exercise
A 220 DC generator runs at 1500rpm and supplies a load of 15A. If the input torque is 28N.m, calculate the efficiency.
Solution By definition, in o P P But W N T P W UI P in o 4396 60 1500 14 . 3 2 28 60 2 . 3300 15 220 Hence, 0.75068 75.068% 4396 3300 pu Exercise 1.10
A long shunt compound wound generator running at 16.67rps supplies 11kW at a terminal voltage of 220V. The resistances of the armature, shunt and series field windings are 0.05ΩΩ, 110Ω, and 0.06Ω respectively. The overall efficiency is 81.5%. Determine: in m m P P m o e P P e m m o in m in o P P p P P P
c) The torque exerted by the prime mover; d) The electrical and mechanical efficiencies.
Solution to exercise 1.10 R A E Ia Ia R s R h U Is h I
a) Total copper losses
2 2 sh a J Ra Rs I RshI P But A U P R U I I I sh sh a 2 50 52 220 11000 110 220 Hence, PJ
0.050.06
52211022 737.44Wb) Iron and friction losses
Pin Pc Pj Pm Po m in C P P P But 44 . 11737 44 . 737 11000 93252 . 13496 815 . 0 11000 J o m o in P P P W P P Hence, Pc 13496.9311737.441759.43W
c) Torque exerted by the prime mover
m N T N P T n T P in in . 925 . 128 67 . 16 14 . 3 2 9325 . 13496 2 2
d) Electrical and mechanical efficiencies
Electrical efficiency 815 . 0 06 . 0 110 05 . 0 220 11 67 . 16 Rs Rsh Ra V U kW P rps n
% 71 . 93 9371 . 0 44 . 11737 11000 pu P P m o e Mechanical efficiency % 96 . 86 8696 . 0 9325 . 13496 44 . 11737 pu P P in m m Exercise 1.11: (Probatoire F3 2010)
An asynchronous three phase motor drives a shunt generator which supplies in full load a current of 40A under a voltage of 320V. The useful power of the driving motor is equal to 20.614kW at full load; its armature resistance is 1.25Ω and its field resistance is 200Ω. Determine:
a) The useful power of the generator;
b) The current in the field circuit and in the armature; c) The emf of the generator;
d) The constant losses.
Exercise 1.12: Compound wound generator (Probatoire 2008)
A shunt generator is to be transformed into a short shunt compound generator by addition of series field windings. A test carried out only on the shunt field gave the following results:
A current of 4.5A produces a voltage of 250V at no load;
A current of 5.5A produces the same voltage at full load of 40A.
The shunt windings have 1200 turns.
a) calculate the number of series turns necessary to maintain this voltage ;
b) Armature, shunt field and series field resistances are respectively 0.5Ω, 70Ω, and 0.3Ω. The generator supplies a load of 5kW at a voltage of 250V. Calculate:
i) The emf of the generator;
ii) The full load efficiency, if constant losses are estimated to 120W;
iii) The efficiency at 3/4full load.
1.3.11 DC generator characteristics
Emf vs. speed characteristics.
a) No load characteristics (open circuit characteristics): E f(If)
The connection diagram used to record data for the plot of no load characteristics is the following, no matter the type of generator.
A V
G
Eo Rh RfE
If
I=0
As the value of rheostat is varying, so the field current is varying and its value is recorded from the ammeter. The speed is kept constant and the generated emf on no load (I = 0) is measured by the voltmeter V. The corresponding values of E and If are recorded and the graph E f(If)can be plotted. Mathematically the relationship
between the emf E and the field current If can be shown.
Zn E
But the flux is function of the field current: kIf
Hence, E ZnkIf f
If .The general appearance of the open circuit characteristics for DC generators is as follows.
0 E(v)
If(A) Eo
The difference OA represents the emf generated due to the residual magnetism in the poles.
Exercise 1.13: (Probatoire 2011).
The no load test of a DC generator functioning at separated field has given the following results at the constant speed N = 1500rpm.
I(A) 0 0.05 0.1 0.15 0.2 0.25 0.3 0.4 0.5 0.6 0.7
E(V) 10 40 70 100 135 165 195 220 230 235 237
The field resistance is r = 100Ω; the armature resistance R = 0.5Ω. The machine is used as a shunt generator.
1. Establish the no load characteristics E = f(i).
2. Determine the emf due to the residual magnetism in the poles;
3. Calculate the value of field rheostat in order to have an emf at no load of E = 220V.
4. The field current is and has its value of question 3. The generator supplies a load with a constant current of I = 49.6A. Calculate.
a) The total current supplied by the generator; b) The voltage across the generator;
c) The total copper losses;
d) The useful power of the generator; e) The power absorbed by the generator;
f) The efficiency of the generator knowing that constant losses are equal to 100W.
Solution of exercise 1.13
1. Plot of the no load characteristics
Scale: 1cm0.1A (Abscissa);1cm10V (Ordinate). See the following page.
2. Emf due to the residual magnetism in the poles
It is the emf when the field current is equal to zero. According to the curve below, we have Eo = 10V.
3. Value of the field rheostat in order to have an emf at no load of 220V.
For E = 220V, If = 0.4A Hence, 100450 4 . 0 220 h R 0 100 200 250 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 1.0 10 E(V) I(A)
4. a) Total current supplied by the generator.
A I
I
Ia f 49.60.450
b) Voltage across the generator.
V I
R E
VG a a 2200.550195
c) Total Copper losses
Pin Pc Pj Pm Po W I V EI P P PJ m 0 a G 2205019549.61328
W I
V
Po G 19549.69672
e) Power absorbed by the generator.
W EI
P
Pa m a 2205011000
f) Efficiency of the generator.
% 135 . 87 87135 . 0 100 11000 9672 0 Pu P P P P p C m o in
Exercise 1.14: Shunt generator
The open circuit characteristics at 1000rpm of a 4 pole, 220V shunt generator with 72 slots and 8 conductors per slot with armature conductors lap connected is as follows:
Field current (A) 0.25 0.5 1 2 3 4 5
emf (V) 25 50 100 175 220 245 255
The field circuit resistance is 75Ω.
1. Explain how this test was carried out
2. Plot the curve and determine: a) The emf induced due to residual magnetism; b) The emf generated at the given field resistance when the generator is under normal operation.
3. The useful flux per pole. 4. The residual flux.
Solution of exercise 1.14 Data: N = 1000rpm; 2 2 4 p ; U = 220V; Z = 72x8 = 576; lap connection; Rsh = 75Ω; shunt generator.
1. Test procedure for the plotting of open circuit characteristics.
The connection of the DC generators for the determination of the open circuit characteristics is as follows. A V G If 0 A E + -R h
The field current If is varied rheostatically and its value measured by an ammeter. The speed si kept constant and the generated emf in the load is measured by the voltmeter V. The corresponding values are recorded and the graph of E = f(If) is plotted.
2. Plot of the curve.
The curve is sketched on the page below (the curve normally should be plotted son a graph paper.
Scale: 1A = 1cm (abscissa) 10V = 1cm (ordinate)
260 1 2 3 4 5 If (A) E(V) 0 S T 6 7 R A 150 10
a) To know the value of the emf induced due to the residual magnetism, we just have to project the curve back ward to cut the ordinate axis (point A). We obtain E0 = 10V
b) The emf for a field resistance of 75Ω
To know the maximum emf the generator will generate on normal operation, we should drawthe shunt resistance line.
To draw the shunt resistance line, take any value of If (for example, let us take 2A), multiply this value by the shunt resistance Rsh = 75Ω. Mark the corresponding point on the ordinate axis. Let that point be R. 75x2 = 150V, hence, R(2A, 150V).
Draw the line joining the origin O and the point R, it cuts the open circuit characteristics at the point S.
Draw a horizontal line from S to T. OT gives the maximum emf generated with 75Ω as shunt resistance.
From the curve we can read: OT = E =210V (almost). 3. Useful flux per pole.
A p ZN E 2 60 ; Lap winding2p A Hence, mWb ZN E ZN E 875 . 21 1000 576 210 60 60 60 4. Residual flux. mWb V E E 0 10 1.04 .
b) Load (or external) characteristics U = f(I)
i) Separately excited generator.
Figure 1.16: Separately excited generator under load.
operates at no load, the terminal voltage E12 is equal to the induced voltage Eo because the voltage drop in the armature resistance is zero.
However, if we connect a load across the armature, the resulting load current I produces a voltage drop across the armature resistance Ra and causes the terminal voltage to drop. As we increase the load, the terminal voltage also decreases progressively. The load characteristic of a separately excited generator is presented in the following figure.
Figure 1.17: Load characteristics of a separately excited generator
ii) Shunt generator
When a shunt generator is loaded, after voltage build up, the terminal voltage U drops with increase in load current I. The decrease depends on the armature drop
IaRa and it is however small except the machine is overloaded.
RA E I U Rh Ish Ia a aI R E U
The sketch of the load characteristics of a shunt generator is as follows. U(V) E Un I(A) 0
Figure 1.18: load characteristics of a shunt generator.
The terminal voltage of a self-excited shunt generator falls off more sharply with increasing load than that of a separately excited generator. The reason is that the field current in a separately excited machine remains constant, whereas in a self-excited generator, the exciting current falls as the terminal voltage drops (
sh ex
R U
I ). 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%. The voltage regulation is said to be 10 percent and 15 percent respectively.
iii) Series generator.
E RA RS LOAD
I
U
When the switch S is closed with the load resistance R comparatively large, the machine does not excite; but as R is reduced, a value is reached when a slight reduction of R is accompanied by a relatively large increase of terminal voltage. When the machine is on open circuit, the terminal voltage is very small (voltage due
0 U(V)
I(A)
Figure 1.19: Load characteristics of a series generator.
Because of the rising voltage characteristics of DC generators, they are mostly used as boosters on DC power lines to compensate for voltage drop. Series wound generator is quite unsuitable when voltage is to be maintained constant or even approximately constant over a wide range of load current.
iv) Compound generators
The compound generator was developed to prevent the terminal voltage of a DC generator from decreasing with increasing load (as it is the case for shunt generator. Thus, although we can usually tolerate a reasonable drop in terminal voltage as the load increases, this has a serious effect on lighting circuits. For example, the distribution system of a ship supplies power to both DC machinery and incandescent lamps. The current delivered by the generator fluctuates continually, in response to the varying loads. These current variations produce corresponding
changes in the generator terminal voltage, causing the light flicker. Compound
generators eliminate this problem
A 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. The figure bellow is a schematic diagram showing the series and shunt field connections
Figure 1.20: Shunt and series field connections of a compound generator.
When the generator runs at no load, the current in the series coils is zero. The shunt coils, however, carry exciting current Ix which produces the field flux, just as in a standard self excited shunt generator. As the generator is loaded, the terminal voltage tends to drop, but load current Ic now flows through the series field coils. The mmf developed by these coils acts in the same direction as the mmf of the shunt field. Consequently, the field flux under load rises above its original no-load value, which raises the value of Eo. 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 drop.
In some cases we have to compensate not only for the armature voltage drop, but also for the IR drop in the 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 increases as the load current rises. Such machines are
called overcompound generators. If the compounding is too strong, a low resistance
can be placed in series with the series field (The name of this resistance is diverter
resistance). This reduces the current in the series field and has the same effect as reducing the number of turns. For example, if the value of the diverter resistance is equal to that of series field, the current in the latter is reduced by half.
In a 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
in DC arc welders, because they tended to limit the short circuit current and to stabilize the arc during the welding process.
Figure 1.21: Load characteristics of compound generators
c) Saturation curve of a DC generator.
The saturation curve is the plot of the field flux under poles against the exciting current Ix. 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. Because the permeability of the air is constant, the flux increases in direct proportion to the exciting current, as shown by the linear portion Oa of the saturation curve.
Figure 1.22: Saturation curve of a DC generator.
However, as we continue to raise the exciting current, the iron in the poles and the armature begins to saturate. A large increase in the mmf is now required to
produce a small increase in flux, as shown by the portion bc of the curve. The machine is now said to be saturated. Saturation of the iron begins to be important when we reach the so-called “knee” ab of the saturation curve.
Figure 1.23: Saturation curve of a DC generator.
Since the emf is directly proportional to the flux under the poles, the saturation curve can also be seen as the plot of the emf Eo at no load against the exciting current Ix.
Figure 1.24: Saturation curve of a DC generator.
d) Controlling the voltage of a shunt generator.
It is easy to control the induced voltage of a shunt excited generator. We simply vary the exciting current Ix by means of a rheostat connected in series with the shunt field. The following circuit presents such a connection.
G
R n m Ix p a b x y EoFigure 1.25: Voltage control of a shunt generator.
To understand how the output voltage varies, suppose that Eo is 120V when the movable contact p is in the centre of the rheostat. If we move the contact towards
the extremity m, the resistance Rt between p and b decreases, which causes the
exciting current to increase. This increases the flux and, consequently, the induced voltage Eo. On the other hand, if we move the contact towards the extremity n, Rt increases, the exiting current decreases, the flux decreases, and so Eo will fall.
We can determine the no load value of Eo if we know the saturation curve of
the generator and the total resistance Rt of the shunt field circuit between p and b. We draw a straight line corresponding to the slope of Rt 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.
For example, if the shunt field has a resistance of 50Ω and the rheostat is set at the extremity m, then Rt = 50Ω. The line corresponding to Rt must pass through the coordinate point E = 50V, I = 1A. This line intersects the saturation curve where the voltage is 150V (see figure of the following page). That is the maximum voltage the shunt generator can produce.
By changing the setting of the rheostat, the total resistance of the field circuit
increases, causing Eo to decreases progressively. For example, if Rt is increased to
Figure 1.26: The no load voltage depends upon the resistance of
the shunt field circuit.
If we continue to raise Rt, 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 drops to zero and will remain so for any Rt greater than this critical value. The critical resistance corresponds to 200Ω.
Exercise 1.15: (Probatoire 2009)
A DC generator has the following magnetisation characteristic at 1200rpm
Field aimed 1 2 4 6 8 10
Emf (V) 192 312 468 566 626 660
1. Draw the magnetisation curves: a) at 1200rpm; at 1000rpm. Take scale 20mm = 100V; 10mm = 1A.
2. If the generator is shunt excited and driven at 1000rpm, determine: a) The voltage at which it will build up on open circuit.
b) The value of the critical resistance of the shunt field circuit at 1000rpm.
c) The terminal potential difference and the load current for a load resistance of 30Ω.
Note: Field resistance = 60Ω and armature resistance = 0.5Ω
Solution of exercise 1.15:
1. Plot of the magnetisation curves.
For the same value of the field current, the variation of the emf is proportional to that of the speed. Hence, emfs at 1000rpm can be deduced from those at 1200rpm using the following reasoning.
1 1 2 2 1 2 1 2 .E N N E N N E E Where: E2 = emf at 1000rpm, N2 = 1000rpm, N1 = 1200rpm, E1 = emf at 1200rpm. We obtain the following table of values
Iex(A) 1 2 4 6 8 10 E(V) at 1200rpm 192 312 468 566 626 660 E(V) at 1000rpm 160 260 390 471.67 521.67 550
0 1 2 3 4 5 6 7 8 9 10 100V 200V 300V 400V 500V 600V 700V E(V) If(A) Eo = 60If Eo N = 1200rpm N = 1000rpm 2.
a) The open circuit build up voltage is situated at the intersection of the curves
) tan ( . 60 ) ( celine resis I R I E I f E f f f f
So, if we draw a line passing between the origin of the axis and the point having the coordinates E = 60V, I = 1A; that line cut the magnetisation curve at a point whose ordinate is the open loop build up voltage. We find
Eo = 550V
b)
The critical resistance is the resistance for which the resistance line has the same slope with the linear part of the saturation curve. For Rf = 160Ω, the resistance line
c) Determination of load current and load voltage. These quantities are such that:
RA E I U Rh Ish Ia
f
f f I I R E U RI U I R U
f
f I I U I U I U 5 . 0 550 30 60And we find after calculations the following results
U = 536.58V; I = 17.89A.
f) Emf vs speed characteristics.
The emf generated is directly proportional to the speed (provided the flux is maintained constant). Any increase in speed is accompanied by an increase in generated emf. A P ZN E 2 60
Since P and A are constant quantities, we deduce that:
n n N E N E N E N E ... 2 2 1 1
So, the emf vs speed characteristics can be sketched as follows.
0 E(V)
N(rpm) Nc
Below a certain speed (Known as the critical speed Nc), no emf is generated. So the characteristic is not supposed to start at the origin of the axis.
1.4 DC motors.
1.4.1 Principle of DC motor
A DC motor is a machine which converts electrical energy into mechanical energy. Its action is based on the principle that when a current carrying conductor is placed in a magnetic field, it experiences a mechanical force (called Laplace’s force).
F
B
I
Current carrying conductor.
Figure 1.28: Current carrying conductor in a magnetic field
The magnitude of the force F is given by the following formula
Where
B =Flux density of the magnetic field; I = Current flowing in the conductor; L = Length of the conductor.
There is no basic difference in construction between DC generator and DC motor. In fact, the same machine can theoretically be used interchangeably as a
BIL F
when the machine is motorizing, the generated emf is less than the terminal voltage meanwhile in a generator, the generated emf is greater than the terminal voltage.
1.4.2 Back emf
When the motor’s armature rotates, the conductors wound on it cut the magnetic flux under the poles. An emf is therefore induced in them. The direction of
this induced emf, known as back or counter emf is such that it opposes the applied
voltage. Since the back emf is induced due to the generator action, the magnitude of it is therefore given by the same expression as for DC generators.
Where:
Eb = Back emf in volts;
Z = Number of armature conductors;
N = Rotational speed of the armature in rpm; Ф = Flux per pole in Webers;
P = Number of pairs of poles;
A = Number of parallel paths in the armature (A = 2 for wave winding; A = 2P for lap
winding).
The armature of a DC motor is hence equivalent to a source of emf Eb in series with a resistance Ra. The supply voltage across the armature should therefore be large enough to balance both the voltage drop in the armature and the back emf all the time. Ra Eb U Ia A P ZN Eb 2 60 a a b R I E U
1.4.3 Power relationship in a DC motor
The voltage equation of a DC motor is:
a a b R I E U
Multiplying each term of the voltage equation by Ia, we get: 2 a a a b a E I R I UI
The equation above is known as the power equation of the DC motor. The term UIa represents the power supplied to the motor armature, RaIa2 represents the
power lost in the armature, EbIa represents the mechanical power developed in the armature causing the rotation of the armature. The power developed EbIa is not all available at the shaft since some of it is used to overcome the mechanical power losses of the motor.
1.4.4 Types of DC motors
Different types of Dc motors are the following:
Permanent magnet DC motors;
Separately excited DC motor;
Series wound DC motors;
Shunt wound DC motors;
Compound wound DC motors.
a) Permanent magnet DC motors
It consists of an armature and one or several permanent magnet encircling the armature. Field coils are not usually required. However, some of these motors do have coils wound of the poles. If they exist, these coils are intended only for recharging the magnets in the event that they loose their strength. Permanent magnet DC motors work like separately excited motors.
b) Separately excited DC motors
For these motors, field coils and armature conductors are supplied by different sources
Ra Eb U Ia E Rh Field
Figure 1.29: Separately excited DC motor.
a a b a a b R I E U R I E U
a a
a b a a a a J m P P UI R I U R I I E I P 2 c) Series wound DC motors
In series motor, field and armature circuits are connected in series, as shown in the figure 1.30 below; so Ia = If.
Ra Eb
Rs I= Ia =If
U
Figure 1.30: DC series motor.
The field coils consist of a few turns of thick wires. The cross sectional area for the wire of the coils has to be fairly to carry the armature, but owing to the large current, the number of turns of wire in them need not to be large.
R R
IE
U b S a
Power drawn from the main:
UI P
Mechanical power developed:
loss
m P P
R R
I
U
R R
E I IUI
Pm 2 a S a S b.
The speed of DC series motor is very high at no load and very low at heavy
load. The torque is very high at low speed. Therefore DC series motor is suitable for
duties requiring a large starting torque and also frequent starting. Applications for DC series motor are: railway traction, hoist, crane…
d) shunt wound DC motor
The armature circuit and the shunt field circuit are connected across a DC source of fixed voltage U. An external field rheostat is sometimes used in the field circuit to control the speed of the motor.
Ra
Eb
U
Rsh
Ish
I
Ia
Figure 1.31: DC shunt motor.
a a b sh a sh sh I R E U I I I R U I
Power drawn from the main:
UI P
Mechanical power developed:
sh
a a a a a a
a a
a b a a sh m loss m E I I R U I I R UI I R I I U I R UI UI P P P P 2 2 2An important characteristic of shunt motor is that it has a fairly constant speed
for a fairly wide range of loads. Shunt motor applications are found in driving shafts, machine tools, blowers…
e) Compound wound DC motor.
Compound wound DC motor has both series winding and shunt field winding. It is usually connected long shunt.
Ra Eb U Rsh Ish Ia Rs
Figure 1.32: Compound wound DC motor.
a s
a b sh a sh sh I R R E U I I I R U I Power drawn from the main:
UI P
Mechanical power developed by the motor
b a m a s a a a S a a a S a sh a S a sh m loss m E I P I R R U I I R R UI I R R I I U I R R UI UI P P P P 2 2 2Compound motor combines the characteristics of shunt and series motor, and finds applications in elevators, hoists, frequent starting duties such as refrigerators and air compressors.
Exercise 1.16
A 200V series DC motor with armature and field resistance of 0.5Ω and 0.3Ω respectively draws a current of 45A. Calculate the emf generated.
Solution of exercise 1.16 Ra Eb Rs I= Ia =If U
V E I R R U E b S a b 164 164 45 3 . 0 5 . 0 200 Exercise 1.17An 11.19 kW shunt motor draws 51A from the supply. The armature and field resistances are 0.1Ω and 240Ω respectively. The motor efficiency is 91.4%. Calculate:
a) The terminal voltage; b) The emf generated.
Solution of exercise 1.17
Ra
Eb
U
Rsh
Ish
I
Ia
a) Terminal voltage: I P U UI P in in A I R R V U S a 45 3 . 0 5 . 0 200 % 4 . 91 ; 240 ; 1 . 0 ; 51 ; 19 . 11 sh a o R A R A I kW PHence, V I P U o 056 . 240 914 . 0 51 11190 b) Emf generated. a a b a a b R I E U R I E U But Ia I Ish 50A 240 240 51 Hence, Eb 2400.150235V Exercise 1.18
A compound motor running from a 400V DC supply draws a current of 102A. The armature, series and shunt field resistances are 0.08Ω, 0.04Ω and 200Ω respectively. Calculate the generated emf.
Solution of exercise 1.18 Ra Eb U Rsh Ish Ia Rs
A R U I I I I I R R U E I R R E U sh sh a a S a b a S a b 100 200 400 102 Hence,
V Eb 400 0080.04 100400123881.4.5 Condition for maximum power.
The mechanical power developed by the motor is:
a a a m UI I R P 2
Differentiating both sides with respect to Iawe have:
0 2 a a a m R I U dI dP 200 04 . 0 ; 08 . 0 ; 102 ; 400 sh S a R R R A I V U
Then, 2 U R Ia a As U EbIaRa and 2 U R Ia a Then 2 2 U E U E U b b
Thus mechanical power developed by a motor is maximum when back emf is
equal to half the applied voltage. This condition is, however, not realised in practice, because in that case current will be much beyond the normal current of the motor. Moreover, half the input would be wasted in the form of heat and taking other losses into consideration (mechanical and magnetic), the motor efficiency will be well below 50 percent.
1.4.6 Torque.
By the term torque is meant the turning or twisting moment of a force about an axis. It is measured by the product of the force and the radius at which this force acts. Consider a pulley of radius r meters acted upon by a circumferential force F Newton which causes it to rotate at n rps.
F
r
Then, torque T Fr
Work done by this force in one revolution = Force x distance =F*2r
Power developed in one revolution:
t r F t W P 2 , with r T F and n ond t 1sec Then, n T N T T r r T P 2 2 2
We the finally deduce that:
The torque is expressed in Newton metre (N.m)
1.4.7 Power flow and efficiency.
DC motors, like DC generators, experience three main types of losses which influence the power flow. These losses are:
Mechanical losses (friction and windage losses);
Copper losses;
Iron losses (Eddy current losses)
The power flow chart of a DC motor can be drawn as follows:
Pin
Pj
Pm
Po
Pc
Figure 1.33: Power flow chart of a DC motor. a) Powers
Where:
U = Terminal voltage (V); I = line current (A);
Pin = Input power (W).
Where: Eb = Back emf (V); N P T 2 60 UI Pin m a b m E I n T P 2
Ia = armature current (A); n = motor speed in rps
Tm = armature torque, also called total torque or gross torque (N.m) Pm = Mechanical power developed in the armature (W).
Also:
With Pj = Total copper losses.
Where:
Pc = Iron and mechanical losses;
BHP = Brake horse power; 1BHP = 746W (British)
Po = Motor output power (W); T = Shaft torque.
The torque which is available for doing the useful work is known as shaft torque T. It is so called because it is available at the shaft. The horsepower obtained by using the shaft torque is called Brake Horse Power (BHP) because it is the horse power available at the brake.
The difference TmT is known as lost torque. The lost torque is absorbed by the
mechanical losses:
Where TL is the loss torque in (N.m). j in m P P P 5 . 735 2 n T P P BHP Po m c n P H B T T n P H B 2 . . 5 . 735 5 . 735 2 . . n P T T n P c L L c 2 2
b) Efficiencies
Mechanical efficiency:
Electrical efficiency:
Overall efficiency:
Remark: Calculation of armature torque
A p Z I T Finally I A p Zn nT I A p ZN nT I E nT P a m a m a m a b m m , 2 2 2 60 2 2 Exercise 1.19:
A 4 pole DC shunt motor has a lap connected armature with 60 slots, each slot containing 20 conductors. The useful flux per pole is 23 mWb and the armature current is 50A. Calculate the torque developed in the armature.
Solution of exercise 1.19: A I mWb Z p A Lap p p a 50 ; 23 1200 20 60 4 2 2 4 2 m o m P P in m e P P e m in o P P
