Chapter 17 - CHAPTER 17 Exercises E17.1 From Equation 17.5,...

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1 CHAPTER 17 Exercises E17.1 From Equation 17.5, we have ) 240 cos( ) ( ) 120 cos( ) ( ) cos( ) ( gap ° ° + + = θ 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 ° ° ° ° + + = ω KI 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 ° ° ° ° + + + + + + + + + = However we can write 0 ) 120 cos( ) 120 cos( ) cos( = + + + ° ° θ ω θ ω θ ω ) cos( ) 360 cos( θ ω θ ω + = + ° ) cos( ) 360 cos( θ ω θ ω + = + ° Thus we have ) cos( 2 3 gap θ ω + = which can be recognized as flux pattern that rotates clockwise. E17.2 At 60 Hz, synchronous speed for a four-pole machine is: () rpm 1800 4 60 120 120 = = = P f n s The slip is given by: % 778 . 2 1800 1750 1800 = = = The frequency of the rotor currents is the slip frequency. From Equation

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2 17.17, we have ω ω s = slip . For frequencies in the Hz, this becomes: 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 = = . This corresponds to a speed of 1775 rpm. E17.3 Following the solution to Example 17.1, we have: rpm 1800 = n 02 . 0 1800 1764 1800 = = = m The per phase equivalent circuit is: () 8 . 0 4 . 29 6 . 0 50 8 . 0 4 . 29 6 . 0 50 2 2 . 1 j Z + + + + + + + = 15.51 + 22.75 = 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 0 440 = = = V I A rms For a delta-connected machine, the magnitude of the line current is rms A 68 . 27 3 98 . 15 3 = = = line I and the input power is kW 43 . 17 cos 3 in = = θ V P
3 Next, we compute r x I V and . () 8 . 0 4 . 29 6 . 0 50 8 . 0 4 . 29 6 . 0 50 j s + + + + + = I V 6 . 15 2 . 406 = o 2 . 2 4 . 406 = V rms 4 . 29 6 . 0 8 . 0 + + = V I o 727 . 3 54 . 13 = A rms The copper losses in the stator and rotor are: 2 3 I R P = ( ) 2 98 . 15 2 . 1 3 = W 3 . 919 = and 2 3 = ( ) 2 54 . 13 6 . 0 3 = W 0 . 330 = Finally, the developed power is: 2 dev 1 3 × = ( ) 2 54 . 13 4 . 29 3 = kW 17 . 16 = kW 27 . 15 rot dev out = = The output torque is: meters newton 82.66 out out = = m T ω The efficiency is: % 61 . 87 % 100 in out = × = η

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4 E17.4 The equivalent circuit is: () 8148 . 0 162 . 1 8 . 0 2 . 1 50 8 . 0 2 . 1 50 eq eq eq j jX R Z + = + + + = + = The impedance seen by the source is: eq 2 2 . 1 s + + = 8148 . 0 162 . 1 2 2 . 1 + + + = o 00 . 50 675 . 3 = Thus, the starting phase current is o o 00 . 50 675 . 3 0 440 starting , = = V I o 00 . 50 7 . 119 starting , = I A rms and for a delta connection, the line current is rms A 3 . 207 3 7 . 119 3 starting , starting , = = = line I The power crossing the air gap is (three times) the power delivered to the right of the dashed line in the equivalent circuit shown earlier. 2 starting , eq ag 3 P = kW 95 . 49 =
5 Finally, the starting torque is found using Equation 17.34. s P T ω ag starting dev, = 2 60 2 49950 π = meters newton 0 . 265 = E17.5 This exercise is similar to part (c) of Example 17.4. Thus, we have 1 3 1 3 sin sin = δ δ 50 200 168 . 4 sin sin 3 = ° δ which yields the new torque angle o 90 . 16 3 = δ . E r remains constant in magnitude, thus we have o 90 . 16 9 . 498 3 = E V rms o o 045 . 1 6 . 103 4 . 1 90 . 16 9 . 498 480 3 3 =

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This note was uploaded on 11/02/2010 for the course ECE 3710 taught by Professor Haris during the Spring '07 term at Georgia Institute of Technology.

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Chapter 17 - CHAPTER 17 Exercises E17.1 From Equation 17.5,...

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