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Energy Systems I

by: Fredy Okuneva

Energy Systems I ECE 320

Fredy Okuneva
GPA 3.81


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Class Notes
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This 9 page Class Notes was uploaded by Fredy Okuneva on Friday October 23, 2015. The Class Notes belongs to ECE 320 at University of Idaho taught by Staff in Fall. Since its upload, it has received 7 views. For similar materials see /class/227734/ece-320-university-of-idaho in ELECTRICAL AND COMPUTER ENGINEERING at University of Idaho.

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Date Created: 10/23/15
ECE 320 Session 23 Page 19 Energy Systems Fall 2003 ECE 320 Lecture 23 Notes Example from Last Time Vdc 100V R Sohm B 2T W2 1ft Motor Case Apply a force in the opposition to the direction that the far is moving called Fload on the gure m 2 mass l feind X B w 29 le T X Vdc Let Fload 3N Note that Fmax BWY FmaX 1219N Note that is FmaX lt Fload it will never start Then in steadystate we need to have force produced electric circuit offset this Fload Iss Iss 492A 13 Then Ess Vdc RIss Ess 7539V Ess m vel ss vel ss 12368 No load veloclty was 7 WB 7 s Vdc m 16404 B s ECE 320 Energy Systems Generator Case Session 23 Page 29 Fall 2003 m 2 mass l W 2 61nd T X Now suppose Fgen 3N Fgen Iss2 Iss2 492A reversed polarlty BW Then Ess2 Vdc RIss2 Ess2 12461 V Larger than Vdc Ess2 m veliss2 veliss2 20441 B s Vd No load velocity was 0 m 16404 B s Moving faster than noload ECE 320 Session 23 Page 39 Energy Systems Fall 2003 Rotating Machine Case Rotating machines are far more common than linear machines Now a loop of wire is placed in a magnetic eld and it is either rotated to produce a voltage or it has current passed through it to produce tangential force which in turn causes it to rotate on its axis North Pole South Pole The ux generated by the eld winding crosses the air gap perpendicular to the surface pole face It cross the air gap and enters the rotor which is also a ferromagnetic material The ux spreads uniformly across the material If one draws a vertical line where the air gap permits the ux to cross the air gap without entering the rotor you notice that the ux crossing the air gap has opposite polarities on either side of this line surface of the rotor We will call this armature or power winding 1 Field Winding We are embedding a loop of wire in the f l Field Current Case 1 Loop rotating in the presence of the magnetic ux Consider the case where the loop is rotated in a counterclockwise direction The figure below shows an end View of the loop and a top View Lable the corners of the loop as shown in the top View f1gure 39 edc eba d a End View ECE 320 Energy Systems Session 23 Page 49 Fall 2003 Assuming the ux divides equally whenever the longitudinal segments ab and cd of the loop will see equal ux The velocity of these segments ab and cd is de ned as shown and it tangential to the surface of the rotor at the location of the loop From Faraday s Law eind d I vel gtlt Blen dN View this on a segment by segment basis segment ab The ux is perpendicular to the motion of the wire By the right hand rule the polarity of the induced voltage will be as shown eba BlenabVelab segment bc Now the eld ux is parallel to the length vector so the cross produce it 0 ecb 0 segment cd The ux is perpendicular to the motion of the wire By the right hand rule the polarity of the induced voltage will be as shown edc BlendcVeldc segment da Now the eld ux is parallel to the length vector so the cross produce it 0 eda BlenabVelab Assume that the rotor has a uniform size Then Velab Veldc Vel and lenab lendc len Therefore adding the voltages around the loop eind 2BlenVel This is the quotarmature voltagequot ea eind Now express this in terms of the rotational velocity vel radius 03 ea 2radiusnBlen ECE 320 Session 23 Page 59 Energy Systems Fall 2003 Note also that the surface area of the cylindrical rotor is Area A 21tradiuslen The area under each pole is half of this AA p 2 Then we can express induced voltage in terms of area and area under each pole 4 2 ea ABoa ApBn TC 117 Now recall the de nition of ux crossing a pole face I BAp So nally we can express the voltage in terms of the ux Notice that the voltage depends on the following 1 Flux 2 Speed 3 Machine characteristic the 211 term will change somewhat when there are more turns in the almature winding When the loop passes the vertical line dividing the pole faces the voltage will reverse polarity and the sign of the induced voltage will reverse resulting in a voltage waveform as shown below Ea A Time ECE 320 Session 23 Page 69 Energy Systems Fall 2003 Since we are hoping for a dc machine this is not desirable We can x this by adding a slotted ring to the terminals The slots are cut so they align with where the voltage reverses polarity This is called a commutator The sections between the gaps are called commutator segments See the gure below In addition we need a sliding contact since the rotor is turning to make sure that any wires we connect don t just twist These are usually pieces of carbon called quotbrushesquot In a machine these have springs to push them against the ring Commutator 0 Now connect the rings to a xed electrical circuit and measure that voltage Now the circuit connection to the commutator reverses at the same point where the voltage reverses resulting in the voltage waveform below which is a much better approximation of DC Ea A 7 Tim e ECE 320 Session 23 Page 79 Energy Systems Fall 2003 Case 11 Run current through a stationary loop in the magnetic eld Now the loop is stationary but current is run into it in the polarity as shown below Note that the commutator will allow the current polarity to reverse at the same point the voltage polarity would reverse Now the current interacting with the magnetic ux density or ux produces a force on the wire Since the current in segments ab and cd are in opposite directions two forces are produced that will try to rotate the wire 100p F ilen X B Again evaluate segment by segment Segment ab B is perpendicular to length Force direction by the right hand rule Fab iablenabB Segment cd B is perpendicular to length Fed icdlencdB Segments bc and da B is parallel to length vector so Fbc Fda 0 Since its a loop the currents are equal call this the armature current ia and as before the length of the parallel segments are equal Fnet 2ilenB ECE 320 Session 23 Page 89 Energy Systems Fall 2003 These forces will serve to force the rotor to turn in the counterclockwise direction We can view the forces as producing torque 139 Fnet radius Therefore we can express torque as 139 2radiusialenB As with the voltage de ne the area of the rotor aligned with the pole face as Ap nradiuslen Then we can express torque in terms of area and area under each pole 2 E Ap1aB it Now recall the definition of ux crossing a pole face I BAp So nally we can express the voltage in terms of the ux Notice that the torque depends on the following 1 Flux 2 Current 3 Machine characteristic the 2 11 term will change somewhat when there are more turns in the armature winding This will cause the machine to rotate Note that if the current doesn39t reverse every half rotation the torque will reverse when the effective direction of the ux reverse when the rotor passes the center line the torque will reverse and halt the rotation However the commutator will prevent this since it will force the current to effectively reverse direction as it passes from one segment to the next T 1a Brush Commutator ECE 320 Session 23 Page 99 Energy Systems Fall 2003 Once the machine starts to spin an induced voltage will be produced just like the speed voltage seen in the linear machine This will serve to limit the current and if there is no load the rotating loop rotating machine will reach a steadystate noload speed One can relate the dc power in to the armature winding to the torque and the mechanical power by the following the proof is assigned as a homework problem Pa Ea Ia 7 03 Pmech Note this holds when torque is in Nm and n is in radsec


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