{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Chapter 17

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

This preview shows pages 1–5. Sign up to view the full content.

1 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 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: ( ) 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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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 = = 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 . 0 1800 1764 1800 = = = s m s n n n s The per phase equivalent circuit is: ( ) 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 = 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 = = = 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 I and the input power is kW 43 . 17 cos 3 in = = θ s s V I 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 j j j s x + + + + + = I V 6 . 15 2 . 406 j = o 2 . 2 4 . 406 = V rms 4 . 29 6 . 0 8 . 0 + + = j x r V I o 727 . 3 54 . 13 = A rms 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 . 0 3 = W 0 . 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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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 j j j j jX R Z + = + + + = + = The impedance seen by the source is: eq 2 2 . 1 Z j Z s + + = 8148 . 0 162 . 1 2 2 . 1 j j + + + = o 00 . 50 675 . 3 = Thus, the starting phase current is o o 00 . 50 675 . 3 0 440 starting , = = s s s Z V I o 00 . 50 7 . 119 starting , = s I A rms and for a delta connection, the line current is rms A 3 . 207 3 7 . 119 3 starting , starting , = = = s line I 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.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}