Courses In Electrical Engineering Volume IV ELECTRICAL MACHINES FIRST SEQUENCE EXAM WITH SOLUTION
By Jean-Paul NGOUNE DIPET I (Electrotechnics), DIPET II (Electrotechnics) DEA (Electrical Engineering) Teacher in the Electrical Department, GTHS KUMBO, Cameroon.
GTHS KUMBO_Electrical Department_First sequence exam…………………………………oct.2011
1
REPUBLIC OF CAMEROON Peace Peace – Work – Fath Fatherl erland and …………… GTHS KUMBO/ ELECT DPT
FIRST SEQUENCE EXAM
Class: F36 Option: Electrotechnology Duration: 2H30 Coefficient: 3
ELECTRICAL MACHINES
No document is allowed except the one given to the cand candida idates tes b the exa examin miners ers
I
TECHNOLOGY
1. Define Define electr electrica icall gener generato ator. r. 2. Give two functions functions of yoke yoke in DC machines. machines. 3. What is is the role of the commutator commutator in DC generators generators? ? 4. Explain Explain why the the armature armature of electric electrical al machines machines is made made up of substa substances nces having low hysteresis coefficient. 5. How can can the strength strength of the magneti magnetic c field of an electro electromagne magnett be increased increased? ? 6. Cite the three three main main types of of magnetic magnetic materials materials and and give one examp example le of each of them. 7. Why are are the armature armature core core and the pole pole cores cores of a dc machin machine e made up of 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 use useful ful powe powerr of the the genera generator tor.. 2. The curr current ent in in the fiel field d circui circuitt and in the the armatu armature. re. 3. The The emf emf of the the gene genera rato tor. r.
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: GTHS KUMBO_Electrical Department_First sequence exam…………………………………oct.2011
2
1. The The rate rated d volt voltag age. e. 2. The resi resista stance nce of of the conne connecte cted d load. load. If series field resistance, shunt field resistance and armature resistance are 0.075Ω, 220 Ω and 0.15 Ω respect respectively ively.. Calculate Calculate:: 3. The shu shunt nt and and series series field field curre currents nts.. 4. The voltag voltage e acros across s the the armatu armature. re. 5. The The emf emf gen gener erat ated ed..
Exercise 3: Self inductance, AC circuits.
The electrical model of an asynchronous motor can be given by the following circuit:
r1
A
V(t)
Xm
x
Rt
r2
C
B v (t )
230
2 sin 314t ; Xm = 110 ; Rt = 900 ; r1 = 1.5 ; r2 = 48
;x=6
.
1. Determine Determine the comple complex x impedance impedance of of the branches branches AB and AC AC of the circuit. circuit. 2. Express Express the total total impeda impedance nce of the the circuit circuit in the form: form: Z = R + jX. 3. Determine Determine the the complex complex expression expression of the current current I consumed by the circuit. 4. Determine Determine the active active and the the reactive reactive power power consumed consumed by the circuit circuit.. 5. Deduce Deduce the complex complex express expression ion of the currents currents passing passing through through branches branches AB and AC. Proposed by Mr. NGOUNE Jean-Paul, PLET Electrotec Electrotechnics hnics,, GTHS KUMBO. KUMBO.
GTHS KUMBO_Electrical Department_First sequence exam…………………………………oct.2011
3
PROPOSITION OF SOLUTION
I
TECHNOLOGY
1. An electrical electrical generat generator or is an electromag electromagnetic netic conver converter ter that permits permits to transform mechanical energy into electrical energy. 2. Two Two funct function ions s of of the the yoke yoke:: -
It prov provide ide a mech mechani anica call prote protecti ction on to the the machin machine; e;
-
It carr carries ies coils coils and and othe otherr parts parts such such as as mecha mechanic nical al part parts; s;
-
It acts acts as a magneti magnetic c circui circuitt where where magne magnetic tic flux circulates circulates (it canaliz canalize e the the flux).
3. The commutat commutator or permits permits to convert convert alternating alternating current current from the armatur armature e into direct current in the external load. 4. The armature armature of of electrical electrical machines machines is is made up of of substances substances havin having g low hysteresis coefficient in order to reduce hysteresis losses in the armature. 5. The strength strength of the the magnetic magnetic field field of an electromagn electromagnet et can be increas increased ed using the following methods: -
By incr increas easing ing the curr current ent supp supplyi lying ng the the elect electrom romagn agnet; et;
-
By insert inserting ing an iron iron core core with within in the the electr electroma omagn gnet; et;
-
By addi adding ng the the numb number er of of turns turns of of coil coil of the the elect electrom romagn agnet. et.
6. Magn Magnet etic ic mate materi rial als: s: -
Ferr Ferrom omag agne netic tic mate materi rial als: s: iro iron, n, nic nicke kel; l;
-
Parama Paramagn gnetic etic materi materials als:: Oxyge Oxygen, n, alum alumini inium, um, platin platinum; um;
-
Diamag Diamagnet netic ic materi materials als:: Nitrog Nitrogen, en, water water,, silv silver, er, bismut bismuth. h.
7. Armature Armature core and and pole core core of DC machines machines are are made up of of laminated laminated steel steel in order to reduce Eddy current losses.
II
ELECTROTECHNOLOGY
Exercise 1: Shunt generator.
Data: I = 40A; U = 320V; Pin = 20.614kW
Ra = 1.2 1.25Ω 5Ω ; Rsh Rsh = 20 200Ω .
GTHS KUMBO_Electrical Department_First sequence exam…………………………………oct.2011
4
Ish U I RA
RSH
LOAD E
1. Useful Useful pow power er of of the gener generato ator: r:
U I P 320 40 P 12800W P
2. Current Current in the field field circui circuitt and in the the armature armature:: U
I a
I sh I 41.6 A
Rsh
320
I sh
200
1.6 A
3. Emf Emf of of the the gen gener erat ator or:: E Ra I a
U 1.25 41.6 320 372V
Exercise 2: Long shunt compound dc generator.
Data: P = 60 kW; I = 150 150A; A; Rs = 0.0 0.075Ω ; Rsh = 22 220 0
; Ra = 0.1 0.15 5
.
Is h RS
U I
RA RSH
Ua
LO A D
E
1. Rate Rated d volt voltag age: e: U
P U
60000 150
400V
GTHS KUMBO_Electrical Department_First sequence exam…………………………………oct.2011
5
2. Resist Resistanc ance e of the the conn connecte ected d load: load: R
U I
400 150
2.66
3. Shunt Shunt an an serie series s field field curr current ents: s: I sh
U Rsh
400
1.81 A
220
I S I sh I 151.81 A
4. Voltag Voltage e acros across s the the armatu armature: re: U a RS I S U 0 U a
RS I S U 0.075 151.81 400 411.38V
5. Emf Emf gen gener erat ated ed:: 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
A
V(t)
Xm
x
Rt
B v (t )
230
2 sin 314t ; Xm = 110 ; Rt = 900 ; r1 = 1.5 ; r2 = 48
r2
C
;x=6
.
1. Determ Determina ination tion of impe impedan dances ces:: Z AB
Branch AB:
jXm Rt
Rt jXm Rt jXm
GTHS KUMBO_Electrical Department_First sequence exam…………………………………oct.2011
6
900 j110
Z AB
13.24 108.38 j
900 110 j
-
9900 j
Z AB
90 11 j
9900 j 90 11 j
8100 121
109.18
83.03o
891000 j 108900 8221
Branch AC:
Z AC r 1 r 2
jx
Z AC 49.5 6 j
49.86
o
6.91
2. Total Total imped impedanc ance e of the the circ circuit uit::
Z AC Z AB Z AC
109.18
Z Z 41.73 28.69 36.60 20.03 j Z T
Z AB
Z AB
AC
o
83.03
49.56
o
6.91
62.74 114.38 j
5443.71 89.91 130.45 61.25 o
o
o
T
3. Total Total curren currentt consu consumed med by by the circu circuit: it: I
V Z T
230 0
5.51 28.69
o
41.73
o
A 28.69o A
4. Active Active and reactiv reactive e power power consumed consumed by by the circuit: circuit:
Z T T = R + jX
P RI 2 VI cos Where Q XI 2 VI sin
is the the powe powerr facto factorr of the the circ circui uit. t.
RI 2 36.6 5.512 1111.17W Q XI 2 20.03 5.512 608.11Var P
We can can still still procee proceed d as follo follows ws::
VI cos 230 5.51cos28.69 1111.71W Q VI sin 230 5.51cos28.69 608.39Var P
5. Curren Currents ts in the the branc branche hes s AB and and AC: I Ia b
Ia c Z AB
Z AC
GTHS KUMBO_Electrical Department_First sequence exam…………………………………oct.2011
7
Using current divider theorem, we have: I AB
I AC
Z
Z AC
AB Z AC
Z
49.86 I
Z AB
AB Z AC
109.18 I
6.91o
5.51
130.45
28.69o o
61.25
o
83.03
5.51
130.45
2.106
28.69o
o
61.25
4.61
A 83.03o A
6.91 A
END.
ABOUT THE AUTHOR
NGOUNE Jean-Paul is a teacher in the electrical department department in GTHS KUMBO. KUMBO. He is teaching teaching this this year in the following subjects: Power electronics, Electrical Machines, Digital Circuits, Electricity-Electronics, and Automation. Any suggestion or critic is welcome… welcome…
NGOUNE Jean-Paul, PLET, DEA.(Electrical Engineering). P.O. Box: 102 NSO, Kumbo, Cameroon. Phone: (+237) 7506 2458. Emai Emaill : jngoune@yaho
[email protected] o.fr Web Web sit site e : www.scribd.com/jngoune
GTHS KUMBO_Electrical Department_First sequence exam…………………………………oct.2011
8
