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Unformatted text preview: CHAPTER 17 Solutions for Exercises E17.1 From Equation 17.5, we have ) 240 cos( ) ( ) 120 cos( ) ( ) cos( ) ( gap  + + = t Ki t Ki t Ki B c b a Using the expressions given in the Exercise statement for the currents, we have ) 240 cos( ) 120 cos( ) 120 cos( ) 240 cos( ) cos( ) cos( gap  + + = t KI t KI t KI B m m m Then using the identity for the products of cosines, we obtain )] 360 cos( ) 120 cos( ) 360 cos( ) 120 cos( ) cos( ) [cos( 2 1 gap  + + + + + + + + + = t t t t t t KI B m However we can write ) 120 cos( ) 120 cos( ) cos( = + + + t t t ) cos( ) 360 cos( + = + t t ) cos( ) 360 cos( + = + t t Thus we have ) cos( 2 3 gap + = t KI B m which can be recognized as flux pattern that rotates clockwise. E17.2 At 60 Hz, synchronous speed for a fourpole machine is: ( 29 rpm 1800 4 60 120 120 = = = P f n s The slip is given by: % 778 . 2 1800 1750 1800 = = = s m s n n n s The frequency of the rotor currents is the slip frequency. From Equation 17.17, we have s = slip . For frequencies in the Hz, this becomes: 1 Hz 667 . 1 60 02778 . slip = = = sf f In the normal range of operation, slip is approximately proportional to output power and torque. Thus at half power, we estimate that % 389 . 1 2 2.778 = = s . This corresponds to a speed of 1775 rpm. E17.3 Following the solution to Example 17.1, we have: rpm 1800 = s n 02 . 1800 1764 1800 = = = s m s n n n s The per phase equivalent circuit is: ( 29 8 . 4 . 29 6 . 50 8 . 4 . 29 6 . 50 2 2 . 1 j j j j j Z s + + + + + + + = 15.51 + 22.75 j = 29...
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This note was uploaded on 11/09/2008 for the course EE EE100 taught by Professor Briggs during the Spring '08 term at UCLA.
 Spring '08
 Briggs
 Electrical Engineering

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