Transformers

Discussion in 'Homework Help' started by Alexj1402, May 3, 2015.

  1. Alexj1402

    Thread Starter New Member

    Jan 24, 2015
    18
    0
    Hey Guys, I require some assistance.

    I have a 6 : 1 Step down transformer with these value:
    • 9kVA Power Rating
    • Primary Current: 3.33A --> Secondary Current: 20A
    • Primary Resistance: 0.35 Ohms --> Secondary Resistance: 0.265 Ohms
    • Primary Voltage: 2700V --> Secondary Voltage: 450V
    • 75W Primary Loss in Open Circuit
    • 110W Copper loss at full load
    I need to

    - Find the Non Load - Load Currents which are within 5% of each other ( Primary Side)
    - Find Non Load - Load Copper Loss with is less than 1/400 (Primary Side)

    Since i have a 3.33A current for the primary which is done with the circuit connected. Can i safely assume that this is my load current? Following a simulation, this is the value i get this. I have raised the current supply to match the values i had calculated. Would this be the wrong approach? Following this i added a load resistor into the circuit of 1000 Ohms which produced a simulated value of 3.31A but i cannot get close with my calculations. Since it is open circuit, my train of though is we only consider the left hand side (0.35Ohms, 75W, 2700V)

    upload_2015-5-3_12-33-28.png

    Since i have my full copper loss at 110W which was given earlier and this is the full copper loss of the transformer which is the same as it is here. The secondary current i calculate from the Load current of the question above, i can use this to calculate the power across this and then divide it by 400.


    Your help will be appreciated
     
  2. MrAl

    Well-Known Member

    Jun 17, 2014
    2,439
    492
    Hi,

    I can imagine that you can hold the core loss as a resistance across the primary, but only at that voltage.
    You can view the copper losses as a resistance in series.
    That gives you a perfect transformer with two resistors to consider, at least with the full primary input voltage.
    If you can take the core loss to be linear then you can extrapolate that too.
     
  3. Alexj1402

    Thread Starter New Member

    Jan 24, 2015
    18
    0
    Hi MrAl, Thanks for the Reply
    Part A) I have managed to get this part. Iron Loss = (Np x V2)^2 / (Ns x Power Drawn) --> (6 x 450v)^2 / (1 x 75) = 97.2K --> I = V/R --> 2700 / 97.2k = 0.028A

    Part B) As i have my Iron Loss Resistance (97.2k), 75W (No Load Power), 0.35, 0.265 Ohms for Coils. I need to achieve around the 30k value in order to achieve the 1/400 figure to get 75w.

    So this would be what i am looking --> Rc (Iron Loss) + Resistance Primary(Np/Ns)^2 which is on the secondary side.


    Thanks
     
  4. MrAl

    Well-Known Member

    Jun 17, 2014
    2,439
    492
    Hi,

    You'll have to explain what you want a little better. Try wording it different and explain everything you need very clearly.

    You either put wire resistance on the primary or the secondary, but you dont have to have it on both. The resistance transforms as the square of the turns ratio, so if you know it in one winding you know it in the other winding after a little calculation.
     
  5. Alexj1402

    Thread Starter New Member

    Jan 24, 2015
    18
    0
    Hi MrAl,

    I am sorry. I will clarify this.

    I am at the point now where I need to calculate the Copper losses under full load on the primary side of the transformer. I currently have the Open Circuit Copper loss which is the 75w, I have the resistance values for the coils Rp = 0.35 ohms and Rs = 0.265 ohms as well as a calculated iron loss value of 97.2k Ohms.

    How would I go about calculating the copper loss under full load for the primary side?

    Thanks
     
  6. MrAl

    Well-Known Member

    Jun 17, 2014
    2,439
    492
    Hi,

    In your first post you stated that the copper loss was 110 watts under full load.
    What more could you want?
    Are you saying that you want to know the loss that appears ONLY in the primary winding and not in the secondary?
    If so, the loss is Ip^2*Rp just like any other power loss.
    The loss in the secondary is Is^2*Rs.
    Ip is primary coil current, Rp is primary coil resistance.
    Is is secondary coil current, Rs is secondary coil resistance.

    Is that what you needed?
     
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