18 Problem 1. 7 d Sketch a cross section of a switched reluctance machine and explain the principle of operation.

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1 Exam Electrical Machine and Drive (ET4117) 9 November 01 from to Thi exam conit of 3 roblem on 3 age. Page 5 can be ued to anwer roblem 4 quetion a. The number before a quetion indicate how many credit you can earn by anwering that quetion. A artly correct anwer may give a art of the credit. Thi examination ha to be made without uing a book, old examination, note, dictionarie or rogrammable calculator; a ocket calculator may be ued. 18 Problem 1 A DC machine with indeendent electrical excitation i connected to a DC voltage ource. 5 a Sketch a DC machine cro ection. Include the interole and the comenating winding. Indicate current direction with dot and croe. 3 b Draw the equivalent circuit. 3 c Give at leat 1 imortant advantage and imortant diadvantage of uing DC machine when comared to induction machine. 7 d Sketch a cro ection of a witched reluctance machine and exlain the rincile of oeration. 1

2 38 Problem The figure deict a linear ermanent-magnet machine. The tator conit of back-iron and ermanent magnet. The moving art (the tranlator) ha three teeth and three coil wound around the teeth. The width of a tooth i b t = 6 mm. The number of turn of one coil i N = 600. The ole itch (the ditance between two magnet) i τ = 1 mm. The tack length of the motor (the length of the magnet and the teeth in the lane erendicular to the lane of the drawing) i l = 50 mm. The flux linkage of the tranlator coil i a linear ueroition of the flux linkage due to the magnet and the flux linkage due to the tranlator current (i a, i b, and i c ): ˆ a ma ia mco( x/ ) Lia ˆ b mb ib mco( x / / 3) Lib ˆ co( x / 4 / 3) Li c mc ic m c The inductance i L = 70 mh. The reitance of the coil i = 1 Ω. Other loe than the coer loe are neglected. In quetion a to d, the current in the tranlator coil are zero. 10 a Write down Amere law and ue it to derive an exreion the flux denity in the air ga below the tooth of hae a. Sketch the contour to which you aly Amere law; you can ue the lat age with the cro ection. Make your aumtion exlicit and exlain the difficultie. In the ret of thi quetion, the amlitude of the flux linkage due to the magnet i ˆ m =180 mwb. b Calculate the amlitude of the magnetic flux going through a tooth. c Calculate the amlitude of the magnetic flux denity going through a tooth. 3 d I thi a realitic value for the flux denity with regard to aturation in the teeth and magnetization of the magnet? Why or why not? What are the tyical value for aturation and magnetization of the magnet? In quetion e to g, the tranlator move with a eed of 1 m/, the amlitude of the three inuoidal hae current i î = A and the hae of the current i o that the electromagnetic force i maximum. 8 e Calculate the amlitude of the no-load voltage (the back emf), the voltage dro over the inductance, the voltage dro over the reitance, and the terminal voltage. 4 f Sketch the haor diagram. 3 g Calculate the electromagnetic force develoed by thi motor. h Until now, iron loe have been neglected. Are they reent in thi motor? Why or why not? 4 i Which kind of iron loe do you know, how do they deend on frequency and flux denity?

3 44 Problem 3 Thi quetion deal with a three-hae tarconnected quirrel-cage induction machine for an electric car. The equivalent circuit i given. 11 a If thi machine i available for teting, how can the arameter of the equivalent circuit be determined in an exerimental way? Mention the tet, the quantitie that have to be meaured and the equation that have to be ued to determine the arameter. You can ue the aumtion that are normally ued. The arameter of the equivalent circuit are given by = 10 mω, L σ = 0.55 mh, L = 11 mh. The tator coer loe are neglected: = 0. The core loe are alo neglected. The number of ole i =4. The nominal line voltage i 400 V. The nominal electrical inut ower i P=30 kw. The machine i driven by a voltage ource inverter, which kee the voltage level roortional to the frequency below the nominal frequency, and contant at the nominal value above the bae frequency. The bae frequency i f=00 Hz. b Why i the tator voltage roortional to the tator frequency below the bae frequency? c Calculate the no-load eed (in rm) at the bae frequency. 3 d Calculate the nominal no-load current (the magnetizing current) at the bae frequency. 4 e Calculate the nominal torque at the bae frequency. 6 f Calculate the nominal li at the bae frequency from the given data. 3 g Calculate the angular rotor frequency at which the torque i maximum. Thi motor i ued to drive an electric vehicle with a ma of 100 kg including the driver. The wheel radiu i 300 mm. Between the motor and the wheel, there i an ideal gearbox (without loe) with a gear ratio of 10 (the wheel rotate lower than the motor). It i aumed that the only friction force acting on the vehicle the i air friction, and the air friction force i given by 1 Faf Cd Aairv Kaf v where Kaf 0.5 N /m. 3 h Calculate the acceleration at tandtill if the motor develo a torque of 150 Nm for a hort time during tarting. i Give a good aroximation of the eed of the car at the bae frequency. 3 j Give a good aroximation of the maximum continuou eed of the car. 5 k Sketch a ower electronic converter (witche and diode) that can connect thi machine to the battery and that enable regenerative braking. 3

4 Examination Electrical Machine and Drive ET4117 Friday, November 9, 01 from to Name: Student number: Anwer to quetion a 4

5 Anwer to the exam Electrical Machine and Drive ET Problem 1 5 a 3 b The direction of the current in the comenating winding and the current around the interole i ooite to the direction of the current in the armature. Therefore, in the right half of the machine, there hould be dot in the comenating winding and around the interole, and in the left half of the machine there hould be tar. 3 c Advantage: - eay to control with rectifier or choer Diadvantage: - regular maintenance due to bruhe, - runaway when excitation field i removed, - heavier than induction machine, - more exenive than induction machine, - commutation roblem may occur (arking). 7 d The rincile of oeration i that of magnetic attraction between magnetized teel art. The tator hae are excited one after the other and then attract the cloet rotor teeth. 5

6 38 Problem 10 a Amere law i given by H d JndA Cm Sm In the figure, a ueful contour ha been ketched. We could aume the flux denity to cro the air ga erendicular only on thee lace where there i a magnet below a tooth. We aume the flux denity to be zero if there i no magnet below a tooth. Alication of Amere law give H l H l H l H l 0 ga ga ma ma gb gb mb mb There i a difficulty becaue the magnetic field intenity in the magnet and the ga below the tooth of hae a may be different from the magnetic field intenity in the magnet and the ga below the teeth of hae b and c. However, thi may be olved by auming that they are equal becaue the magnet cover are only half of the urface area of the teeth of hae b and c, o that the um of the fluxe in tooth b and c i equal to the flux in tooth a. In thi cae: Hl g ghl m m 0 The BH curve of the magnet i given by Bm 0 rmh m Brm The BH curve of air i given by Bg 0H g Subtituting thee BH relation for magnet and air in the reult of Amere law give: Bglg ( Bm Brm) lm The equation for flux continuity: B nda 0 Bg Ag Bm Am Bg B Subtitution give Bglg ( Bg Brm) lm lm Bg Brm l l g m m A ˆ b ˆ m m 0. 3mWb N ˆ m c B 1 T btl 3 d Ye, tyical remanent flux denitie of magnet are u to 1.5 T, o that 1 T in a tooth i realitic, and tyical aturation flux denitie in teeth are in the order of 1.5 T to T, o 1 T in a tooth i realitic. 8 e The voltage equation of hae a i given by: da dia ˆ d x ua ia ia L m in( x). dt dt dt The current i in hae with the no-load voltage in order to create maximum force, o 6

7 ia iˆin( x) Subtituting thi in the voltage equation give ˆ ˆ dx ˆ dx ua iin( x) Li co( x) m in( x) dt dt Therefore, the no-load voltage amlitude i ˆ d x eˆ a m 47.1 V dt The voltage dro over the reitance i uˆ ˆ i 4V The voltage dro over the inductance i ˆ d x uˆ L Li 36.65V d t The terminal voltage i uˆ ( ˆ ˆ a ea u) ul 80.01V 4 f 3 g The force develoed by thi motor i Pem 3 ˆ ˆ Fem im N v h Ye, the flux denity in the tranlator varie when the tranlator move. Thi reult in iron loe in the tranlator iron. 4 i Hyterei loe, roortional to frequency and flux denity quared and eddy current loe, roortional to frequency quared and flux denity quared. 7

8 44 Problem 3 11 a eitance meaurement: ut a DC voltage on the terminal of two hae and meaure the DC current while the machine i at tandtill: U I No-load tet: The machine run ynchronouly ( = 0) in no-load at the rated AC voltage. The line voltage U line and the hae current I are meaured. During thi tet, = 0, and therefore, P nl 3I L 1 U 3 line I nl Short-circuit tet: The rotor i blocked ( = 1) and the machine i connected to a reduced AC voltage in uch a way that the current i about the nominal current. The line voltage U line, the hae current I and the three-hae ower P are meaured. Often, the current through the ynchronou inductance i neglected. Therefore, P 3I 1 Uline L ( ) 3I b Below the bae frequency, the voltage i roortional to the eed to avoid aturation of the magnetic circuit. 10 f c n 6000 rm 3 d The magnetizing current through L i given by U hae Uline I m 16.71A noml 3 noml 4 e The ower balance er hae ay: P PCu Pmech PCu mt The air ga ower can be written a: I 1 P I I From thee equation: 1 mt I The mechanical eed can be written a m (1 ) Therefore, 1 I (1 ) T I T P Therefore, the torque of a three-hae machine can be written a 8

9 Pnom Tnom Nm 6 f In the rated oerating oint, we know that 3I Uline Vt Pnom L L Therefore, Pnom L Uline Pnom 0 Therefore, nom U U 4 L P line line nom LP nom 3 g 18.rad/ f 34.7Hz. r L r 3 h The gear increae the torque on the wheel to 1500 Nm. The force on the wheel aly to the car i T F 5000N r The acceleration i F a m/ m 3 i The li i rather low, in the order of 1%. Therefore, neglecting the li, the eed i mrw erw 4 frw vwrw 18.85m/ g g g 3 j The air friction i given by F K v, o the ower to overcome thi friction i 5 k f 3 P Fv K f v At eed above the nominal eed, the machine converter ue field weakening with contant voltage and contant ower. If we neglect all other loe and we alo neglect the loe in the machine, the maximum eed can be aroximated by Pnom v / K f u 3 u 4 9