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Unformatted text preview: 1 CHAPTER 16 Exercises E16.1 The input power to the dc motor is loss out source in P I V + = = Substituting values and solving for the source current we have 3350 746 50 220 + = A 8 . 184 = Also we have % 76 . 91 3350 746 50 746 50 % 100 = + = = % 35 . 4 % 100 1150 1150 1200 % 100 regulation speed = = = load full no 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 R i N 24 40 ) 3 . ( 2 ) ( ) ( = = + = + Bli f m/s 333 . 3 ) 3 . ( 2 2 = = = Bl u (b) A 667 . 6 ) 3 . ( 2 4 = = = 2 V 667 . 1 ) 667 . 6 ( 05 . 2 = = = A T I R V e m/s 778 . 2 ) 3 . ( 2 667 . 1 = = = Bl u W 11 . 11 ) 778 . 2 ( 4 = = = f p load m W 222 . 2 2 = = i W 33 . 13 ) 667 . 6 ( 2 = = = t % 33 . 83 33 . 13 11 . 11 % 100 = = = (c) A 333 . 3 ) 3 . ( 2 2 = = = pull V 167 . 2 ) 333 . 3 ( 05 . 2 = + = + = m/s 611 . 3 ) 3 . ( 2 167 . 2 = = = W 222 . 7 ) 611 . 3 ( 2 = = = W 667 . 6 ) 333 . 3 ( 2 = = = W 5555 . 2 = = % 31 . 92 222 . 7 667 . 6 % 100 = = = E16.4 Referring to Figure 16.15 we see that 125 E V for 2 = F A and . 1200 = n Then for 1500 = , we have V 156 1200 1500 125 = = E16.5 Referring to Figure 16.15 we see that 145 V for 5 . 2 = A and . 1200 = Then for 1500 = , we have V 3 . 181 1200 1500 145 = = rad/s 1 . 157 60 2 = = Nm 49 . 47 1 . 157 746 10 = = = dev P A 15 . 41 3 . 181 746 10 = = = V 6 . 193 ) 15 . 41 ( 3 . 3 . 181 = + = + = 3 E16.6 20 10 10 10 300 = = = F T adj I R V Because remains constant the value of K 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 E rad/s 2 . 131 228 . 2 4 . 292 = = = m rpm 1253 2 60 = = n Thus the motor speeds up when is increased. E16.7 Following Example 16.4, we have A 6 30 10 240 = + = + = Referring to Figure 16.18 we see that 200 V for 6 = A and . 1200 = Thus we have 592 . 1 ) 60 / 2 ( 1200 200 = = = A 3 . 164 592 . 1 5 . 261 = = = dev V 229.3 ) 3 . 164 ( 065 . 240 = = = rad/s . 144 592 . 1 3 . 229 = = = rpm 1376 2 60 = = kW 36 = = out P kW 87 . 40 ) 3 . 164 6 ( 240 )...
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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 Tech.
 Spring '07
 Haris

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