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Homework Help: Archived DC motor in series, finding the speed

  1. May 23, 2012 #1
    1. The problem statement, all variables and given/known data
    when running at its rated load, a 40kW series motor takes 74A and 550V. The rated speed is 750 rev/min. the armature and field resistances are 0.35 Ω and 0.15 Ω respectivly. when the load torque is doubled the rated value, the current is 110A. dertermin the moror speed and the power output at 100% torque overload.

    Answers given, worked solution not
    ans = 538 Rev/min and 57.4kW

    2. Relevant equations

    Work= Torque x (2pi x speed over 60)
    torque = K(t) I(a) ψ (k(t) =constant and I(a)= armature current)
    EMF =KNψ (k= constant)
    EMF= v-IR

    3. The attempt at a solution
    Work= Torque x (2pi x speed over 60)
    40000= T x ((2pi x 750) / 60)

    EMF =KNψ (condition 1)
    EMF =KNψ (condition 2)

    divide both and cancel
    k is constant as its the same motor
    flux is constant (as flux is not given a value, i assume that its the same)

    EMF(1) over EMF(2) = N(1) over N(2)
    put both to the power -1 so as to find N(2) easier
    EMF(2) over EMF(1) = N(2) over N(1)

    ((550 - (74 x 0.35+0.15)) over (550 - (74 x 0.35+0.15))) x 750 = N(2)
    N(2) = 723.6 rev/min which is not the value of the answer

    seeing as i can't get the speed right i can't go onto finding the power developed, however using the speed provided in the answer i have done it and correctly

    Work= 2Torque x (2pi x speed over 60)
    Work= 2Torque x (2pi x 538 over 60)
  2. jcsd
  3. Mar 9, 2016 #2


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    Under rated load:
    Back emf Eb=550-I(Rf+Ra)
    Mechanical power developed P1=EbI=513×74 W
    Under 100% overload condition:
    Eb=495V..(same formula with I=110A).
    Mechanical power developed P2=495×110=54.45kW.
    Taking ratio P1/P2, we get
    513x74/495x110=N/2N1...(since T1=2T).
    ∴N1=537.87 rpm≈538 rpm.
    Last edited: Mar 9, 2016
  4. Mar 9, 2016 #3


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    Series motor is not a constant flux motor. Field flux varies with load.
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