Class 15 speed control of DC Motor by ashu468

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									                 Welcome to

            Lectures on D.C.Motor



4/15/2010
                                    1
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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                                          2
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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                                                              3
                                            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

                      No load Speed  Full Load Speed
 % Speed Regulation                                   100%
                              Full Load Speed




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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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                                                            6
                                 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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                                                              7
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.




  4/15/2010
                                                                     8
Solution :

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

                                                  V




                                                      -




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                                                          9
                   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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                                             10
   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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                                          11
(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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                                      12
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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