EE 255 - Chapter 9a

# EE 255 - Chapter 9a - Chapter 9: DC Motors and Generators...

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214 Chapter 9 : DC Motors and Generators Problems 9-1 to 9-12 refer to the following dc motor: P rated = 15 hp I L ,rated = 55 A V T = 240 V N F = 2700 turns per pole n rated = 1200 r/min N SE = 27 turns per pole R A = 0.40 Ω R F = 100 Ω R S = 0.04 Ω R adj = 100 to 400 Ω Rotational losses = 1800 W at full load. Magnetization curve as shown in Figure P9-1. Note: An electronic version of this magnetization curve can be found in file p91_mag.dat , which can be used with MATLAB programs. Column 1 contains field current in amps, and column 2 contains the internal generated voltage E A in volts. In Problems 9-1 through 9-7, assume that the motor described above can be connected in shunt. The equivalent circuit of the shunt motor is shown in Figure P9-2.

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215 Note: Figure P9-2 shows incorrect values for R A and R F in the first printing of this book. The correct values are given in the text, but shown incorrectly on the figure. This will be corrected at the second printing. 9-1. If the resistor R adj is adjusted to 175 Ω what is the rotational speed of the motor at no-load conditions? S OLUTION At no-load conditions, 240 V A T E V = = . The field current is given by adj 240 V 240 V 0.873 A 175 100 250 T F F V I R R = = = = + Ω + Ω Ω From Figure P9-1, this field current would produce an internal generated voltage Ao E of 271 V at a speed o n of 1200 r/min. Therefore, the speed n with a voltage A E of 240 V would be A Ao o E n E n = ( ) 240 V 1200 r/min 1063 r/min 271 V A o Ao E n n E = = = 9-2. Assuming no armature reaction, what is the speed of the motor at full load? What is the speed regulation of the motor? S OLUTION At full load, the armature current is adj 55 A 0.87 A 54.13 A T A L F L F V I I I I R R = - = - = - = + The internal generated voltage A E is ( )( ) 240 V 54.13 A 0.40 218.3 V A T A A E V I R = - = - Ω = The field current is the same as before, and there is no armature reaction, so Ao E is still 271 V at a speed o n of 1200 r/min. Therefore, ( ) 218.3 V 1200 r/min 967 r/min 271 V A o Ao E n n E = = = The speed regulation is nl fl fl 1063 r/min 967 r/min SR 100% 100% 9.9% 967 r/min n n n - - = × = × =
216 9-3. If the motor is operating at full load and if its variable resistance adj R is increased to 250 Ω , what is the new speed of the motor? Compare the full-load speed of the motor with adj R = 175 Ω to the full-load speed with adj R = 250 Ω . (Assume no armature reaction, as in the previous problem.) S OLUTION If adj R is set to 250 Ω , the field current is now adj 240 V 240 V 0.686 A 250 100 325 T F F V I R R = = = = + Ω + Ω Ω Since the motor is still at full load, A E is still 218.3 V. From the magnetization curve (Figure P9-1), the new field current F I would produce a voltage Ao E of 247 V at a speed o n of 1200 r/min. Therefore, ( ) 218.3 V 1200 r/min 1061 r/min 247 V A o Ao E n n E = = = Note that adj R has increased, and as a result the speed of the motor n increased. 9-4.

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## This note was uploaded on 10/15/2009 for the course EE 255 taught by Professor Shaban during the Spring '05 term at Cal Poly.

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EE 255 - Chapter 9a - Chapter 9: DC Motors and Generators...

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