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Chapter 17

# Chapter 17 - CHAPTER 17 Solutions for Exercises E17.1 From...

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CHAPTER 17 Solutions for Exercises E17.1 From Equation 17.5, we have ) 240 cos( ) ( ) 120 cos( ) ( ) cos( ) ( gap ° ° - + - + = θ θ θ t Ki t Ki t Ki B c b a Using the expressionsgiven in the Exercise statement for the currents, we have ) 240 cos( ) 120 cos( ) 120 cos( ) 240 cos( ) cos( ) cos( gap ° ° ° ° - - + - - + = θ ϖ θ ϖ θ ϖ 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 ° ° ° ° - + + + - + - + + - - + + + - = θ ϖ θ ϖ θ ϖ θ ϖ θ ϖ θ ϖ t t t t t t KI B m However we can write 0 ) 120 cos( ) 120 cos( ) cos( = + - + - - + - ° ° θ ω θ ω θ ω t t t ) cos( ) 360 cos( θ ω θ ω + = - + ° t t ) cos( ) 360 cos( θ ω θ ω + = - + ° t t Thus we have ) cos( 2 3 gap θ ω + = t KI B m which can be recognized as flux pattern that rotates clockwise. E17.2 At 60 Hz, synchronous speed for a four-pole 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 ω ω s = slip . For frequencies in the Hz, this becomes: 1

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Hz 667 . 1 60 02778 . 0 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 correspondsto a speed of 1775 rpm. E17.3 Following the solution to Example 17.1, we have: rpm 1800 = s n 02 . 0 1800 1764 1800 = - = - = s m s n n n s The per phase equivalent circuit is: ( 29 8 . 0 4 . 29 6 . 0 50 8 . 0 4 . 29 6 . 0 50 2 2 . 1 j j j j j Z s + + + + + + + = 15.51 + 22.75 j = 29 . 34 53 . 27 = ( 29 lagging % 62 . 82 34.29 cos factor power = = 29 . 34 98 . 15 29 . 34 53 . 27 0 440 - = = = s s s Z V I A rms For a delta-connected machine, the magnitude of the line current is rms A 68 . 27 3 98 . 15 3 = = = s line I
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