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Unformatted text preview: CHAPTER 17 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 expressions given in the Exercise statement for the currents, we have ) 240 cos( ) 120 cos( ) 120 cos( ) 240 cos( ) cos( ) cos( gap ° ° ° ° − − + − − + = θ ω θ ω θ ω 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 ω θ ω θ ω θ ω θ ω θ ω t t t t t t KI B m However we can write ) 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 fourpole machine is: ( ) 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 575 17.17, we have ω ω s = slip . For frequencies in the Hz, this becomes: 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 n = s n n s The per phase equivalent circuit is: ( ) 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 = o 29 . 34 53 . 27 ∠ = ( ) lagging % 62 . 82 34.29 cos factor power = = o o o o 29 . 34 98 . 15 29 . 34 53 . 27 440 − ∠ = ∠ ∠ = = s s s Z V I A rms For a deltaconnected machine, the magnitude of the line current is rms A 68 . 27 3 98 . 15 3 = = = s line I I and the input power is kW 43 . 17 cos 3 in = = θ s s V I P 576 Next, we compute r x I V ′ and . ( ) 8 . 4 . 29 6 . 50 8 . 4 . 29 6 . 50 j j j j s x + + + + + = I V 6 . 15 2 . 406 j − = V rms o 2 . 2 4 . 406 − ∠ = 4 . 29 6 . 8 . + + = ′ j x r V I A rms o 727 . 3 54 . 13 − ∠ = The copper losses in the stator and rotor are: 2 3 s s s I R P = ( )( ) 2 98 . 15 2 . 1 3 = W 3 . 919 = and ( ) 2 3 r r r I R P ′ ′ = ( )( ) 2 54 . 13 6 . 3 = W . 330 = Finally, the developed power is: ( ) 2 dev 1 3 r r I R s s P ′ ′ − × = ( )( 2 54 . 13 4 . 29 3 = ) kW 17 . 16 = kW 27 . 15 rot dev out = − = P P P The output torque is: meters newton 82.66 out out = = m P T ω The efficiency is: % 61 . 87 % 100 in out = × = P P η 577 E17.4 The equivalent circuit is: ( ) 8148 . 162 . 1 8 . 2 . 1 50 8 . 2 . 1 50 eq eq eq j j j j j jX R Z + = + + + = + = The impedance seen by the source is: eq 2 2 . 1 Z j Z s + + = 8148 . 162 . 1 2 2 . 1 j j + + + = = o 00 . 50 675 . 3 ∠ Thus, the starting phase current is...
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This note was uploaded on 10/08/2009 for the course AERO 215 taught by Professor Tyler during the Spring '09 term at Texas A&M.
 Spring '09
 Tyler

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