Question

A 24 slot, 2-pole dc machine has 10 turns per coil. The
average flux density per pole is 11. The effective length of
the machine is 25cm and radius of the armature is 10cm.
The magnetic poles are designed to cover 80% of the
armature periphery. If the angular velocity is 183.2 rad/s,
determine the induced emf in the armature winding?​

Answer 1

Explanation:

Given A 24 slot, 2-pole dc machine has 10 turns per coil. The average flux density per pole is 11. The effective length of  the machine is 25 cm and radius of the armature is 10 cm.  The magnetic poles are designed to cover 80% of the  armature periphery. If the angular velocity is 183.2 rad/s,  determine the induced emf in the armature winding?

• So total number of conductors are 2 x 24 x 10
•                                                          = 480
• Number of parallel paths for a 2 pole machine is 2
• So the actual pole area will be  
•                          Pole area = 2 π r l / P
•                                           = 2 x 3.14 x 0.1 x 0.25 / 2
•                                          = 0.0785 sq m
• So the effective pole area will be
•                       Effective pole area = 0.0785 x 0.8
•                                                       = 0.0628
• Now effective flux per pole φ = B x effective pole area
•                                                 = 1 x 0.0628
•                                                  = 0.0628 Wb
• Now induced emf in the armature winding will be
•                                  E = Z / 2π P / α φ ω
•                                     = 480 / 2π x P / α φ ω
•                                      = 480 / 2π x 2 / 2 x 0.0628 x 183.2
•                                 E = 879.35

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https://brainly.in/question/354892