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Slip in induction motor

  1. Jun 4, 2014 #1
    Hello all
    My question is:
    Induction motor with nominal load rotates with nominal speed.
    How the slip will be changed if we decrease input voltage in 10%?

    I think that if we decrease the voltage, than the current will be decreased also.
    Than, the moment will be decreased and mechanical speed ωm will be decreased also.
    So, the slip will be increased...

    Am I right?
  2. jcsd
  3. Jun 8, 2014 #2
    The power produced by motor it is:
    Pm=m1*Irot^2*Rrot/s=kp*Irot^2/s where : m1=number of stator phases[m1=3] ;Irot=rotor current at slip s.
    Irot=s*Vr/sqrt(Rrot^2+s^2*Xrot) where Xrot is the rotor leakage reactance ; Vr=rotor EMF at start[s=1] ; Vr=krot/stat*Es~=k*Vs ; Vs=stator supply voltage.
    Since s^2*Xrot<<Rrot :
    Irot=~s*Vr/Rrot ; Irot^2=s^2*Vr^2/Rrot^2
    If Pload remains constant then :
    If Pm1=Pload=Pm2 then Vs1^2*s1=Vs2^2*s2 or: s2=Vs1^2/Vs2^2*s1
    If Vs2<Vs1 then s2>s1
    If Vs2=0.9*Vs1 then s2=1/0.9^2*s1=1.235*s1
  4. Jun 11, 2014 #3
    Sorry! The rotor current formula is [of course]:
    Then s^2*Xrot^2<<Rrot^2.
    Usually Xrot/Rrot=sk =~5*sn where sk=the slip corresponding to maximum torque and sn=rated [nominal] slip. sn=~0.05 [usually].
    So (0.05*5)^2*Rrot^2=0.0625*Rrot ^2
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