Elect machine total_2012

Courses In
Electrical
Engineering

Volume IV

EECTRICAL MACHINES

EXAM QUESTIONS WITH SOLUTIONS
(2012 academic year)

By

Jean-Paul NGOUNE

DIPET I (Electrotechnics), DIPET II (Electrotechnics)
M.Sc. (Electrical Engineering)
Teacher in the Electrical Department, GTHS KUMBO, Cameroon.

Exam questions with solutions_2012_Jean-Paul NGOUNE 1

Foreword

This is a compilation of some exam questions that I gave to my students during this
academic year. They are accompanied by solutions proposed by me. I will be delight
if this book can be of any use for you. I will also be very happy to receive any critic or
suggestion from you. I dedicate this book to my students of Class 6, Electrical
Technology, GTHS Kumbo, 2012 batch. They are a bit stubborn, but I like to teach
them. May you be blessed as you are using this book.

NGOUNE Jean-Paul.
17 May 2012.


Acknowledgement

Most of the questions treated in this book are “Probatoire Technique” past questions
proposed by the Cameroon General Certificate of Education Board (GCEB) and the
“Office du Baccalaureat du Cameroun” (OBC).


Courses In
Electrical
Engineering

Volume IV

ELECTRICAL MACHINES

FIRST SEQUENCE EXAM WITH SOLUTION

By

Jean-Paul NGOUNE



REPUBLIC OF CAMEROON FIRST SEQUENCE EXAM
Peace – Work – Fatherland
Class: F36
……………
GTHS KUMBO/ ELECT DPT Option: Electrotechnology
Duration: 2H30
Coefficient: 3
ELECTRICAL MACHINES
No document is allowed except the one given to
the candidates by the examiners

I TECHNOLOGY
1. Define electrical generator.
2. Give two functions of yoke in DC machines.
3. What is the role of the commutator in DC generators?
4. Explain why the armature of electrical machines is made up of substances
having low hysteresis coefficient.
5. How can the strength of the magnetic field of an electromagnet be increased?
6. Cite the three main types of magnetic materials and give one example of each
of them.
7. Why are the armature core and the pole cores of a dc machine made up of
laminated steel?

II ELECTROTECHNOLOGY

Exercise 1: Shunt generator.
An asynchronous three phase motor drives a shunt generator which supplies in full
load a current of 40A under 320V. The useful power of the driving motor is equal to
20.614 kW at full load. Its armature resistance is 1.25Ω and its field resistance is
200Ω. Determine:
1. The useful power of the generator.
2. The current in the field circuit and in the armature.
3. The emf of the generator.

Exercise 2: Long shunt compound DC generator.
A 60kW long shunt compound wound dc generator delivers a rated current of 150A
at its rated voltage. Calculate:
1. The rated voltage.


2. The resistance of the connected load.
If series field resistance, shunt field resistance and armature resistance are 0.075Ω,
220 Ω and 0.15 Ω respectively. Calculate:
3. The shunt and series field currents.
4. The voltage across the armature.
5. The emf generated.

Exercise 3: Self inductance, AC circuits.
The electrical model of an asynchronous motor can be given by the following circuit:

r1 x
A

V(t) Xm Rt r2

C
B

v(t ) 230 2 sin 314t ; Xm = 110Ω; Rt = 900Ω; r1 = 1.5Ω; r2 = 48 Ω; x = 6 Ω.
1. Determine the complex impedance of the branches AB and AC of the circuit.
2. Express the total impedance of the circuit in the form: Z = R + jX.
3. Determine the complex expression of the current I consumed by the circuit.
4. Determine the active and the reactive power consumed by the circuit.
5. Deduce the complex expression of the currents passing through branches AB
and AC.
Proposed by Mr. NGOUNE Jean-Paul,
PLET Electrotechnics, GTHS KUMBO.


PROPOSITION OF SOLUTION

I TECHNOLOGY
1. An electrical generator is an electromagnetic converter that permits to
transform mechanical energy into electrical energy.
2. Two functions of the yoke:
- It provide a mechanical protection to the machine;
- It carries coils and other parts such as mechanical parts;
- It acts as a magnetic circuit where magnetic flux circulates (it canalize the
flux).
3. The commutator permits to convert alternating current from the armature into
direct current in the external load.
4. The armature of electrical machines is made up of substances having low
hysteresis coefficient in order to reduce hysteresis losses in the armature.
5. The strength of the magnetic field of an electromagnet can be increased using
the following methods:
- By increasing the current supplying the electromagnet;
- By inserting an iron core within the electromagnet;
- By adding the number of turns of coil of the electromagnet.
6. Magnetic materials:
- Ferromagnetic materials: iron, nickel;
- Paramagnetic materials: Oxygen, aluminium, platinum;
- Diamagnetic materials: Nitrogen, water, silver, bismuth.
7. Armature core and pole core of DC machines are made up of laminated steel
in order to reduce Eddy current losses.

II ELECTROTECHNOLOGY

Exercise 1: Shunt generator.

Data: I = 40A; U = 320V; Pin = 20.614kW
Ra = 1.25Ω; Rsh = 200Ω.


Ish

U
I
RA
RSH
LOAD

E

1. Useful power of the generator:

P U I
P 320 40
P 12800W

2. Current in the field circuit and in the armature:
U 320
I sh 1 .6 A
Rsh 200

Ia I sh I 41.6 A

3. Emf of the generator:
E Ra I a U 1.25 41.6 320 372V

Exercise 2: Long shunt compound dc generator.
Data: P = 60 kW; I = 150A; Rs = 0.075Ω; Rsh = 220 Ω; Ra = 0.15 Ω.
Ish

RS U
I
RA
R SH Ua LO AD

E

1. Rated voltage:
P 60000
U 400V
U 150


2. Resistance of the connected load:
U 400
R 2.66
I 150
3. Shunt an series field currents:
U 400
I sh 1.81A
Rsh 220

IS I sh I 151.81A

4. Voltage across the armature:
Ua RS I S U 0
Ua RS I S U 0.075 151.81 400 411.38V
5. Emf generated:
E Ra I a U a 0
E Ra I a U a 0.15 151.81 411.38 434.151V

Exercise 3: Self inductance, AC circuits.
Let us consider the following circuit:

r1 x
A

V(t) Xm Rt r2

C
B

v(t ) 230 2 sin 314t ; Xm = 110Ω; Rt = 900Ω; r1 = 1.5Ω; r2 = 48 Ω; x = 6 Ω.
1. Determination of impedances:
- Branch AB:
Rt jXm
Z AB jXm Rt
Rt jXm


900 j110 9900 j 9900 j 90 11 j 891000 j 108900
Z AB
900 110 j 90 11 j 8100 121 8221
Z AB 13.24 108.38 j 109.18 83.03o

- Branch AC:
Z AC r1 r 2 jx
Z AC 49.5 6 j 49.86 6.91o

2. Total impedance of the circuit:
Z AB Z AC 109.18 83.03o 49.56 6.91o 5443.71 89.91o
ZT Z AB Z AC
Z AB Z AC 62.74 114.38 j 130.45 61.25o
ZT 41.73 28.69o 36.60 20.03 j

3. Total current consumed by the circuit:
V 230 0o
I 5.51 28.69o A
ZT 41.73 28.69o

4. Active and reactive power consumed by the circuit:
ZT = R + jX
P RI 2 VI cos
Where is the power factor of the circuit.
Q XI 2 VI sin

P RI 2 36.6 5.512 1111.17W
Q XI 2 20.03 5.512 608.11Var
We can still proceed as follows:
P VI cos 230 5.51cos 28.69 1111.71W
Q VI sin 230 5.51cos 28.69 608.39Var
5. Currents in the branches AB and AC:
I

Ia b Ia c
ZAB ZAC


Using current divider theorem, we have:
Z AC 49.86 6.91o 5.51 28.69o
I AB I 2.106 83.03o A
Z AB Z AC 130.45 61.25o

Z AB 109.18 83.03o 5.51 28.69o
I AC I 4.61 6.91A
Z AB Z AC 130.45 61.25o

END.


Courses In
Electrical
Engineering

Volume IV

ELECTRICAL MACHINES

THIRD SEQUENCE EXAM WITH SOLUTION

By

Jean-Paul NGOUNE



REPUBLIC OF CAMEROON THIRD SEQUENCE EXAM
Peace – Work – Fatherland
Class: F36
……………
GTHS KUMBO/ ELECT DPT Option: Electrotechnology
Duration: 03H
Coefficient: 3
Written paper

ELECTRICAL MACHINES

No document is allowed except the one given to
the candidates by the examiners.

PART ONE : TECHNOLOGY
1. Explain why the series field DC motor rotates at high speed at no load when it
is supplied at his nominal voltage.
2. For each of the following electrical machines, cite a magnetic material which is
used in the manufacturing of the magnetic circuit: a) transformer; b) DC
machine stator, c) permanent magnet DC motor.
3. Give the role of the “isoptherme” in an AC electrical machine.
4. Give the role of the excitation resistance in a DC motor.
5. Give the role of auxiliary commutation poles in a DC motor.
6. Give the differences between an asynchronous squirrel cage motor and an
asynchronous wound winding motor.
7. Name the test which is conducted in order to determine iron and friction losses
in a DC motor.

PART TWO: ELCTROTECHNOLOGY
Exercise 1: 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.

Exercise 2: Series motor
The starter of a Peugeot 504 (a series motor) takes 200A under a voltage of 12 V.
Under these conditions, it runs at 1000rpm and delivers a useful power of 1500W.
The constant losses are estimated at 100W. Calculate: a) The power absorbed by
the starter. b) Its efficiency. c) Its useful torque. d) The joule losses. e) The total
resistance (armature + field windings). f) The back emf of the motor. g) Determine the
direct starting current. h) What is the value of the resistance to be connected in series
with the motor to limit the starting current to 240A?

Exercise 3: Three Phase asynchronous motor
A three phase asynchronous motor, with the stator coupled in delta, is supplied by a
network supply of 380V, 50Hz. Each phase of the stator has a resistance 0.4Ω. At no
load the motor rotates at 1500rpm and absorbs a power of 1150W; the current in a
line is 3.2A. A test at a nominal load, under the same voltage of 380V, 50Hz has
given the following results: Slip g = 4%; Power absorbed: Pa = 18.1kW; Line current:
I = 32A.
1. For the no load test, calculate: a) the stator copper losses when the motor rotates
at no load. b) The stator iron losses knowing that the mechanical losses are 510W.
2. For the nominal load test, determine: a) The power factor. b) The speed of rotation.
c) The frequency of the rotor current. d) The stator copper losses. e) The rotor copper
losses. f) The useful power. g) The efficiency of the motor. h) The useful torque

SUBJECT MASTER: NGOUNE Jean-Paul,
PLET Electrotechnics, GTHS KUMBO.


AKNOWLEDGEMENT

Almost all the exercises solved in this document are past “Probatoire Technique”
examination questions proposed by the Cameroon General Certificate of Education
Board (CGCEB).

PROPOSITION OF SOLUTION

PART ONE TECHNOLOGY:
1. At no load, armature current and consequently the field flux is very is very
small. This leads to the increasing of the speed of DC series motor, since the
speed is inversely proportional to the armature current (to the field flux).
U RaIa
Eb K N U RaIa N , so when the armature current (and
K
consequently the field flux) is small, the speed N increases. This is why it is
forbidden to start DC series motor at no load.
2. Magnetic material used for magnetic circuit:
Machine Magnetic material
Transformer Silicon steel
DC machine stator Cast iron
Permanent magnet DC motor Cast iron, ALNICO (Aluminium, Nickel,
Cobalt)

3. The role of “isotherme” is to regulate the temperature of motor coils to a
constant value.
4. The excitation resistance permits to regulate the field current flowing in the
field coil of the machine, and therefore to regulate the field flux of the machine.
5. The role of the auxiliary commutation poles (interpoles) is to improve on the
commutation by combating the emf of commutation induced when a section is
crossing the neutral line.
6. Difference between squirrel cage motor and asynchronous wound winding
motor:


Wound winding rotor Squirrel cage rotor
Higher starting current Moderate starting current because of
starting rheostat.
We can get access to rotor windings Rotor conductors are not accessible
through slip rings
Permits smooth starting Can be directly started

7. The test that permits to determine iron and friction losses is the no load test.

PART TWO: ELECTROTECHNOLOGY
Exercise 1: Shunt generator
4
Data: N = 1000rpm; p 2 ; U = 220V; Z = 72x8 = 576; lap connection;
2
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.
0A
If
A
E
+
Rh G V
-

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)


E(V)

260

T S

150 R

A 10
If (A)
0 1 2 3 4 5 6 7

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 draw the 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.
ZN 2p
E ; Lap winding 2p A
60 A
ZN 60 E
E
60 ZN
Hence,
60 210
21.875mWb
576 1000
4. Residual flux.
E E0 10V 1.04mWb .

Exercise 2: Series motor
The stator of the Peugeot 504 is a series motor having the following characteristics:
I = 200A; U = 12V; N = 1000rpm; Pu = 1500W; Pc= 100W.
Rs

U

Ra

Eb

a) Power absorbed by the starter.
P UI 12 200 2400W
b) Efficiency.
Pu 1500
62.5%
P 2400
c) Useful torque.
2 N 60 Pu 60 1500
Pu 2 nT T T 14.33 N .m
60 2 N 2 3.14 1000
d) Joule losses.
Let us first draw the power stages chart.

Pm
Pin Po = Pu

Pj Pc


From the chart, we have:
Pj Pin Pm Pin Pc Po 2400 1500 100 800W
e) Total resistance (armature + field)
Pj 800
Pj ( Ra Rs ) I 2 Rs Ra 0.02
I2 200 2
f) Bock emf.
U Eb Rs Ra I Eb U Ra Rs I 12 0.02 200 8V
g) Direct starting current.
At starting, Eb = 0, hence,
U 12
I st 600 A
Ra Rs 0.02
h) Value of the resistance to be connected in series with the motor to limit the
starting current to 240A.
Rst Rs

U
Ra

Eb

With Eb = 0 at starting, we have:
U 0 U 12
I st Rst Ra Rs 0.02 0.03 .
Ra Rs Rst I st 240

Exercise 3: Three phase asynchronous motor.

Data: Stator Delta connected; U = 380V; R = 0.4Ω; Ns = 1500rpm; Po = 1150W;
Io = 3.2A; Pmec = 510W (friction +windage); g = 4%; Pa = 18.1kW; I = 32A.
1. For the no load test:
a) Stator copper losses
2
2 I
Pjs 3RI p 3R 0 RI 02 0.4 3.22 4.096W
3


b) Stator iron losses.
The power consumed by the motor during the no load test is equivalent to the sum of
iron losses, mechanical losses (windage + friction) and stator copper losses; hence
we have:
Po Pi Pmec Pjs Pi Po Pmec Pjs 1150 510 4.096 635.904W
2. For nominal load test:
a) Power factor.
Pa 18100
Pa UI 3 cos cos 0.8593
UI 3 380 32 2
b) Speed of rotation.
Ns Nr
g Nr Ns 1 g 1500 1 0.04 1440rpm
Ns
c) Frequency of rotor current.
fr g. f 0.04 50 2 Hz
d) Stator copper losses
2
I
Pjs 3R RI 2 0.4 322 409.6W
3
The following formula can also be used.
2 2
I 32
Pjs Pjso 4.096 409.6W
Io 3 .2
e) Rotor copper losses
The power stage chart for the asynchronous motor can be drawn as follows:
Pi

Pinr Pm
Pin Po

Pjs Pjr Pmec

Pjr g.Pinr g ( Pa Pi Pjs ) 0.04 18100 409.6 635.904 682.18W
f) Useful power.
Po Pinr Pjr Pmec Pinr (1 g ) Pmec 17054.496(1 0.04) 510 15862.316W


g) Efficiency of the motor.
Po 15862.316
0.8763 87.63%
Pa 18100
h) Useful torque
2 N 60 Po 60 15862.316
Po 2 NT T T 105.24 N .m
60 2 N 6.28 1440

END


ABOUT THE AUTHOR

NGOUNE Jean-Paul was born in Foreké-Dschang,
Republic of Cameroon in 1984. He is a holder of a
Master Degree in electrical engineering, obtained in
2010 in the Doctorate School of the University of Douala
(UFD-PSI). He is also a holder of a DIPET II and a
DIPET I respectively obtained in 2009 and 2007 in the
Advanced Teacher Training College for Technical
Education (ENSET de Douala).
He is currently a permanent teacher of Electrical
Engineering at the Government Technical High School
of Kumbo, North-West region, Cameroon. His domain of
research concerns the improvement of energy
conversion techniques for an efficient generation of
electrical energy from renewable sources (especially
wind and solar energy, small and medium scale
hydropower) and digital designing using FPDs.
The author is looking for a Ph.D program in his domain
of research (he has not yet found it). Any suggestion for
this issue will be warmly welcome.

NGOUNE Jean-Paul, M.Sc., PLET.
P.O. Box: 102 NSO, Kumbo, Cameroon.
Phone: (+237) 7506 2458.
Email : jngoune@yahoo.fr
Web site: www.scribd.com/jngoune


Elect machine total_2012

Empfohlen

Empfohlen

Weitere ähnliche Inhalte

Was ist angesagt?

Was ist angesagt? (9)

Ähnlich wie Elect machine total_2012

Ähnlich wie Elect machine total_2012 (20)

Kürzlich hochgeladen

Kürzlich hochgeladen (20)

Elect machine total_2012