This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: CHAPTER 16 Solutions for Exercises E16.1 The input power to the dc motor is loss out source source in 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 V 667 . 1 ) 667 . 6 ( 05 . 2 = = = A A T A I R V e 1 m/s 778 . 2 ) 3 . ( 2 667 . 1 = = = Bl e u A W 11 . 11 ) 778 . 2 ( 4 = = = u f p load m W 222 . 2 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 W 5555 ....
View
Full
Document
This note was uploaded on 11/09/2008 for the course EE EE100 taught by Professor Briggs during the Spring '08 term at UCLA.
 Spring '08
 Briggs
 Electrical Engineering

Click to edit the document details