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Unformatted text preview: CHAPTER 16 Exercises E16.1 The input power to the dc motor is in loss out source source P P I V P + = = Substituting values and solving for the source current we have 3350 746 50 220 + × = source I A 8 . 184 = source I Also we have % 76 . 91 3350 746 50 746 50 % 100 = + × × = × = in out P P η % 35 . 4 % 100 1150 1150 1200 % 100 regulation speed = × − = × − = − − − load full load full load no n n n E16.2 (a) The synchronous motor has zero starting torque and would not be able to start a highinertia load. (b) The seriesfield dc motor shows the greatest amount of speed variation with load in the normal operating range and thus has the poorest speed regulation. (c) The synchronous motor operates at fixed speed and has zero speed regulation. (d) The ac induction motor has the best combination of high starting torque and low speed regulation. (e) The seriesfield dc motor should not be operated without a load because its speed becomes excessive. E16.3 Repeating the calculations of Example 16.2, we have (a) A 40 05 . 2 ) ( = = = + A T A R V i N 24 40 ) 3 . ( 2 ) ( ) ( = = + = + A Bli f m/s 333 . 3 ) 3 . ( 2 2 = = = Bl V u T (b) A 667 . 6 ) 3 . ( 2 4 = = = Bl f i load A 540 V 667 . 1 ) 667 . 6 ( 05 . 2 = − = − = A A T A I R V e m/s 778 . 2 ) 3 . ( 2 667 . 1 = = = Bl e u A W 11 . 11 ) 778 . 2 ( 4 = = = u f p load m 2 W 222 . 2 = = R i p A R W 33 . 13 ) 667 . 6 ( 2 = = = A T t i V p % 33 . 83 33 . 13 11 . 11 % 100 = = × = t m p p η (c) A 333 . 3 ) 3 . ( 2 2 = = = Bl f i pull A V 167 . 2 ) 333 . 3 ( 05 . 2 = + = + = A A T A I R V e m/s 611 . 3 ) 3 . ( 2 167 . 2 = = = Bl e u A W 222 . 7 ) 611 . 3 ( 2 = = = u f p pull m W 667 . 6 ) 333 . 3 ( 2 = = = A T t i V p 2 W 5555 . = = R i p A R p % 31 . 92 222 . 7 667 . 6 % 100 = = × = m t p η E16.4 Referring to Figure 16.15 we see that 125 ≅ A E V for 2 = F I A and . 1200 = n Then for 1500 = n , we have V 156 1200 1500 125 = × = A E E16.5 Referring to Figure 16.15 we see that 145 ≅ A E V for 5 . 2 = F I A and . 1200 = n Then for 1500 = n , we have V 3 . 181 1200 1500 145 = × = A E rad/s 1 . 157 60 2 = × = π ω n m Nm 49 . 47 1 . 157 746 10 = × = = m dev dev P T ω A 15 . 41 3 . 181 746 10 = × = = A dev A E P I V 6 . 193 ) 15 . 41 ( 3 . 3 . 181 = + = + = A A A T I R E V 541 E16.6 Ω 20 10 10 10 300 = × − = − = F F F T adj I I R V R Because I F remains constant the value of K ϕ is the same value as in Example 16.4, which is 2.228. Furthermore the loss torque also remains constant at 11.54 Nm, and the developed torque remains at 261.5 Nm. Thus the armature current is still 117.4 A. Then we have V 4 . 292 ) 4 . 117 ( 065 . 300 = − = − = A A T A I R V E rad/s 2 . 131 228 . 2 4 . 292 = = = φ ω K E A m rpm 1253 2 60 = = π ω m m n Thus the motor speeds up when V T is increased....
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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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