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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
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ForewordThis is a compilation of some exam questions that I gave to my students during thisacademic year. They are accompanied by solutions proposed by me. I will be delightif this book can be of any use for you. I will also be very happy to receive any critic orsuggestion from you. I dedicate this book to my students of Class 6, ElectricalTechnology, GTHS Kumbo, 2012 batch. They are a bit stubborn, but I like to teachthem. May you be blessed as you are using this book.NGOUNE Jean-Paul.17 May 2012.Exam questions with solutions_2012_Jean-Paul NGOUNE 2
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AcknowledgementMost of the questions treated in this book are “Probatoire Technique” past questionsproposed by the Cameroon General Certificate of Education Board (GCEB) and the“Office du Baccalaureat du Cameroun” (OBC).Exam questions with solutions_2012_Jean-Paul NGOUNE 3
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Courses In Electrical Engineering Volume IV ELECTRICAL MACHINES FIRST SEQUENCE EXAM WITH SOLUTION 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 4
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REPUBLIC OF CAMEROON FIRST SEQUENCE EXAMPeace – 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. Exam questions with solutions_2012_Jean-Paul NGOUNE 5
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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 Bv(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.Exam questions with solutions_2012_Jean-Paul NGOUNE 6
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PROPOSITION OF SOLUTIONI 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 ELECTROTECHNOLOGYExercise 1: Shunt generator.Data: I = 40A; U = 320V; Pin = 20.614kW Ra = 1.25Ω; Rsh = 200Ω.Exam questions with solutions_2012_Jean-Paul NGOUNE 7
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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 372VExercise 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 150Exam questions with solutions_2012_Jean-Paul NGOUNE 8
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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 0Ua 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 0.15 151.81 411.38 434.151VExercise 3: Self inductance, AC circuits.Let us consider the following circuit: r1 x A V(t) Xm Rt r2 C Bv(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 jXmZ AB jXm Rt Rt jXmExam questions with solutions_2012_Jean-Paul NGOUNE 9
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900 j110 9900 j 9900 j 90 11 j 891000 j 108900Z AB 900 110 j 90 11 j 8100 121 8221Z AB 13.24 108.38 j 109.18 83.03o - Branch AC:Z AC r1 r 2 jxZ 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 89.91oZT Z AB Z AC Z AB Z AC 62.74 114.38 j 130.45 61.25oZT 41.73 28.69o 36.60 20.03 j 3. Total current consumed by the circuit: V 230 0oI 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 sinP RI 2 36.6 5.512 1111.17WQ XI 2 20.03 5.512 608.11VarWe can still proceed as follows:P VI cos 230 5.51cos 28.69 1111.71WQ VI sin 230 5.51cos 28.69 608.39Var 5. Currents in the branches AB and AC: I Ia b Ia c ZAB ZACExam questions with solutions_2012_Jean-Paul NGOUNE 10
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Using current divider theorem, we have: Z AC 49.86 6.91o 5.51 28.69oI AB I 2.106 83.03o A Z AB Z AC 130.45 61.25o Z AB 109.18 83.03o 5.51 28.69oI AC I 4.61 6.91A Z AB Z AC 130.45 61.25o END.Exam questions with solutions_2012_Jean-Paul NGOUNE 11
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Courses In Electrical Engineering Volume IV ELECTRICAL MACHINES THIRD SEQUENCE EXAM WITH SOLUTION 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 12
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REPUBLIC OF CAMEROON THIRD SEQUENCE EXAMPeace – 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 Exam questions with solutions_2012_Jean-Paul NGOUNE 13
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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 motorThe 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 bythe starter. b) Its efficiency. c) Its useful torque. d) The joule losses. e) The totalresistance (armature + field windings). f) The back emf of the motor. g) Determine thedirect starting current. h) What is the value of the resistance to be connected in serieswith the motor to limit the starting current to 240A?Exercise 3: Three Phase asynchronous motorA three phase asynchronous motor, with the stator coupled in delta, is supplied by anetwork supply of 380V, 50Hz. Each phase of the stator has a resistance 0.4Ω. At noload the motor rotates at 1500rpm and absorbs a power of 1150W; the current in aline is 3.2A. A test at a nominal load, under the same voltage of 380V, 50Hz hasgiven 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 rotatesat 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 copperlosses. f) The useful power. g) The efficiency of the motor. h) The useful torque SUBJECT MASTER: NGOUNE Jean-Paul, PLET Electrotechnics, GTHS KUMBO.Exam questions with solutions_2012_Jean-Paul NGOUNE 14
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AKNOWLEDGEMENTAlmost all the exercises solved in this document are past “Probatoire Technique”examination questions proposed by the Cameroon General Certificate of EducationBoard (CGCEB). PROPOSITION OF SOLUTIONPART 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 materialTransformer Silicon steelDC machine stator Cast ironPermanent 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:Exam questions with solutions_2012_Jean-Paul NGOUNE 15
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Wound winding rotor Squirrel cage rotorHigher starting current Moderate starting current because of starting rheostat.We can get access to rotor windings Rotor conductors are not accessiblethrough slip ringsPermits smooth starting Can be directly started 7. The test that permits to determine iron and friction losses is the no load test.PART TWO: ELECTROTECHNOLOGYExercise 1: Shunt generator 4Data: N = 1000rpm; p 2 ; U = 220V; Z = 72x8 = 576; lap connection; 2Rsh = 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 circuitcharacteristics 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 thevoltmeter V. The corresponding values are recorded and the graph of E = f(If) isplotted. 2. Plot of the curve.The curve is sketched on the page below (the curve normally should be plotted son agraph paper.Scale: 1A = 1cm (abscissa) 10V = 1cm (ordinate)Exam questions with solutions_2012_Jean-Paul NGOUNE 16
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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 75ΩTo know the maximum emf the generator will generate on normal operation, weshould 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 pointon 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 circuitcharacteristics at the point S.Draw a horizontal line from S to T. OT gives the maximum emf generated with 75Ωas shunt resistance.From the curve we can read: OT = E =210V (almost)Exam questions with solutions_2012_Jean-Paul NGOUNE 17
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3. Useful flux per pole. ZN 2pE ; Lap winding 2p A 60 A ZN 60 E E 60 ZNHence, 60 210 21.875mWb 576 1000 4. Residual flux.E E0 10V 1.04mWb .Exercise 2: Series motorThe 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 1500Pu 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 PcExam questions with solutions_2012_Jean-Paul NGOUNE 18
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From the chart, we have:Pj Pin Pm Pin Pc Po 2400 1500 100 800W e) Total resistance (armature + field) Pj 800Pj ( 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 12I 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 EbWith Eb = 0 at starting, we have: U 0 U 12I st Rst Ra Rs 0.02 0.03 . Ra Rs Rst I st 240Exercise 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 IPjs 3RI p 3R 0 RI 02 0.4 3.22 4.096W 3Exam questions with solutions_2012_Jean-Paul NGOUNE 19
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b) Stator iron losses.The power consumed by the motor during the no load test is equivalent to the sum ofiron losses, mechanical losses (windage + friction) and stator copper losses; hencewe have:Po Pi Pmec Pjs Pi Po Pmec Pjs 1150 510 4.096 635.904W 2. For nominal load test: a) Power factor. Pa 18100Pa UI 3 cos cos 0.8593 UI 3 380 32 2 b) Speed of rotation. Ns Nrg 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 IPjs 3R RI 2 0.4 322 409.6W 3The following formula can also be used. 2 2 I 32Pjs Pjso 4.096 409.6W Io 3 .2 e) Rotor copper lossesThe power stage chart for the asynchronous motor can be drawn as follows: Pi Pinr Pm Pin Po Pjs Pjr PmecPjr 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 15862.316WExam questions with solutions_2012_Jean-Paul NGOUNE 20
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g) Efficiency of the motor. Po 15862.316 0.8763 87.63% Pa 18100 h) Useful torque 2 N 60 Po 60 15862.316Po 2 NT T T 105.24 N .m 60 2 N 6.28 1440 ENDExam questions with solutions_2012_Jean-Paul NGOUNE 21
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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/jngouneExam questions with solutions_2012_Jean-Paul NGOUNE 22
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