# Class 15 speed control of DC Motor by ashu468

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```									                 Welcome to

Lectures on D.C.Motor

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Speed Control of a D.C. Motor

Eb
N

 V  I a Ra   
N  k
              

             
So Speed can be varied by varying

1. Flux/Pole – field Control

2. Armature Drop – Armature Control

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Speed Control of D.C. Shunt Motor

1. Field Control                    Ish          I
+
Ia

Eb
N                                              V
      Shunt Field
Field
Rheostat
Eb
N1 
1                                           -

Eb
N2 
2
N 2 1

N1  2

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Ish
2. Armature Control                                       I
+
Ia
Eb
N                                         Rheostat V
          Shunt Field

Eb1
N1 
                                            -

Eb 2
N2 


N2   Eb 2

N1   Eb1

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Speed Regulation

It is the change in speed when the load on the motor is
reduced from rated value to zero

% Speed Regulation                                   100%

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Efficiency of D.C. Motor

Output
Efficiency           100%
Input
Input - Losses
                 100%
Input

Losses in a D.C.Motor

Pi = Iron Loss
Pwf = Windage and Friction Loss
Pcu = Copper Loss (in Armature and field windings mainly)

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Ish              I
+
Ia
Ra
Shunt Field                              V
Armature
Rsh

-

Input - Losses
Efficiency                   100%
Input
VI - Pi  Pwf  Pcu
                      100%
Input

Pcu = Armature copper loss + Field copper loss
Armature Copper loss = Ia2 Ra
Shunt Field copper loss = Ish2 Rsh

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Ex. A D.C. series motor operates at 800 rpm with a line current of
120A from a 250V mains. Determine the motor speed at a
current of 60A at 250V assuming that the flux/pole at 60A is
70% of its value at 120A. The armature and series field
resistances are 0.15Ω and 0.1Ω respectively.

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Solution :

Ia=Ise=I
Rse = 0.1Ω
Series Field
Ra = 0.15Ω   +

V

-

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Eb
N

Case1 : At 250V, 120A, 800rpm
Eb1
800 
1
V  I a1 Ra  Rse 
or, 800 
1
250  1200.15  0.1
or, 800 
1
220
or, 800                            (1)
1
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Case2 : At 250V, 60A

Let speed be N2 rpm

Eb 2
N2 
2
V  I a 2 Ra  Rse 
or, N 2 
0.71
250  600.15  0.1
or, N 2 
0.71
235
or, N 2                         (2)
0.71

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(2) ÷ (1) →   N2     235

800 220  0.7

235  800
or , N 2 
220  0.7

or, N 2  1220.78 rpm

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Ex. A 25kW, 250V D.C.Shunt Machine has armature and
field resistances of 0.06Ω and 100Ω respectively. Determine the
total armature power developed when working
i) as a generator delivering 25kW output
ii) as a motor taking 25kW input

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