EE 255 - Chapter 8

# EE 255 - Chapter 8 - Chapter 8 DC Machinery Fundamentals...

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204 Chapter 8 : DC Machinery Fundamentals 8-1. The following information is given about the simple rotating loop shown in Figure 8-6: 0.8 T B = 24 V B V = 0.5 m l = 0.4 R = Ω 0.125 m r = 250 rad/s ω = (a) Is this machine operating as a motor or a generator? Explain. (b) What is the current i flowing into or out of the machine? What is the power flowing into or out of the machine? (c) If the speed of the rotor were changed to 275 rad/s, what would happen to the current flow into or out of the machine? (d) If the speed of the rotor were changed to 225 rad/s, what would happen to the current flow into or out of the machine?

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205 (a) If the speed of rotation ω of the shaft is 500 rad/s, then the voltage induced in the rotating loop will be ind 2 e rlB = ( )( )( )( ) ind 2 0.125 m 0.5 m 0.8 T 250 rad/s 25 V e = = Since the external battery voltage is only 24 V, this machine is operating as a generator , charging the battery. (b) The current flowing out of the machine is approximately ind 25 V 24 V 2.5 A 0.4 B e V i R - - = = = Ω ( Note that this value is the current flowing while the loop is under the pole faces. When the loop goes beyond the pole faces, ind e will momentarily fall to 0 V, and the current flow will momentarily reverse. Therefore, the average current flow over a complete cycle will be somewhat less than 2.5 A.) (c) If the speed of the rotor were increased to 275 rad/s, the induced voltage of the loop would increase to ind 2 e rlB = ( )( )( )( ) ind 2 0.125 m 0.5 m 0.8 T 275 rad/s 27.5 V e = = and the current flow out of the machine will increase to ind 27.5 V 24 V 8.75 A 0.4 B e V i R - - = = = Ω (d) If the speed of the rotor were decreased to 450 rad/s, the induced voltage of the loop would fall to ind 2 e rlB = ( )( )( )( ) ind 2 0.125 m 0.5 m 0.8 T 225 rad/s 22.5 V e = = Here, ind e is less than B V , so current flows into the loop and the machine is acting as a motor. The current flow into the machine would be ind 24 V - 22.5 V 3.75 A 0.4 B V e i R - = = = Ω 8-2. Refer to the simple two-pole eight-coil machine shown in Figure P8-1. The following information is given about this machine: B = 10 . T in air gap l = 03 . m (length of coil sides) coils) of (radius m 08 . 0 = r CCW r/min 1700 = n The resistance of each rotor coil is 0.04 Ω . (a) Is the armature winding shown a progressive or retrogressive winding? (b) How many current paths are there through the armature of this machine? (c) What are the magnitude and the polarity of the voltage at the brushes in this machine? (d) What is the armature resistance R A of this machine?
206 (e) If a 10 Ω resistor is connected to the terminals of this machine, how much current flows in the machine? Consider the internal resistance of the machine in determining the current flow. (f)

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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 8 - Chapter 8 DC Machinery Fundamentals...

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