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DC Generator Problem 1

  1. Oct 13, 2013 #1
    1. Find the number of poles for a 1200 kW separately excited generator, if the average voltage between commutator segments is 15 V and the armature ampere-turns per pole on full load is 10000. The generator has single turn coil and is lap connected. Ignore all losses.


    2. Relevant equations



    3.Power rating of an dc electrical machine is given by,
    Power Rating=Voltage at output terminal delivering full load Current (VL) x Full Load Current (IFL)

    Since all the losses are to be considered zero,
    Voltage generated at DC generator commutator brush (Eg) = Voltage available at Output terminal (VL)

    Let P be the number of poles of the generator.

    Since the subject dc generator is Lap wounded,
    Number of parallel path in armature winding (A) = P.

    Let Ip be the current in each parallel path of armature at full load, then
    Ip=Number of parallel path x IFL
    IFL = P x Ip

    In a parallel dc circuit the total power generated is equal to the sum of power generated in each branch, hence
    Total power generated at full load (Pt) = P x Power generated in one parallel path at full load (Pp)
    Note : As each parallel path will have equal number of armature conductors resulting in same voltage generated across the parallel paths i.e Eg AND the current through each parallel path is same i.e Ip.


    As the coils are in series in a given parallel path,
    Voltage generated in one of the parallel path = Voltage generated in one coil x Number of Coils in one parallel path


    Power generate in one of the parallel path at full load(Pp)= Voltage generated in one coil x Number of turns in one parallel path x Current through parallel path at full load (Ip)

    Since the commutator pitch for simplex lap winding is (YG) = ±1, we can say the (assuming simplex lap winding),
    Voltage generated in one coil = Voltage generated between two consecutive commutator segments


    Hence, we can write,
    (Pp)= Voltage generated between two consecutive commutator segments x Number of turns in one parallel path x (Ip)
    (Pp)= Voltage generated between two consecutive commutator segments x Amp-Turns in one parallel path

    In case of Lap winding it can be shown that number of turns under one pole is equal to number of turns in one parallel path as shown below,

    Let Z be number of conductor in a P pole lap wound dc generator, then
    No. of conductors per pole = Z/P
    No. of turns per pole = Z/(2 x P)

    Also,
    Number of conductor in parallel path = Z/Number of parallel path (A)
    Number of turns in a parallel path = Z/(2 x A) = Z/(2 x P), (as No. of Parallel path = No. of Poles)

    So we can re-write the equation as,
    (Pp)= Voltage generated between two consecutive commutator segments x Amp-Turns per pole

    Putting the data given in problem, we get
    (Pp) = 15 V x 10000 AT = 150 kW

    As mentioned above,
    Total power generated at full load (Pt) = P x (Pp)
    1200 kW = P x 150 kW
    P=8
    --------------------------------------------------------------------------

    The underlined concepts used for solution is not discussed in my textbook. I request, to confirm whether it is correct or not.

    Also I would like to state, above problem is from the old question paper, so I don't have solution to it, and my text book doesn't have a similar type of problem. So, please confirm the solution provided above is correct or not
     
    Last edited: Oct 13, 2013
  2. jcsd
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