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About Omer

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  1. This post is to discuss about the differential protection on 3 phases transformer. specifically, star delta connection. I think it is a rich topic in the concepts and might well appear on the PE How to connect CTs, 30 degrees phase shift, instantaneous currents, zero sequence currents, triple harmonic currents, etc.. you are all welcome to give your thoughts for the benefit of all. the below figure is for your convenience
  2. PPI Practice exam problem #10

    see the below post for the same problem. I think they got it wrong for this one.
  3. Transformer efficiency

    Yes, I meant you got it right, it is the output, not the input. Nice scenario and indeed interesting..
  4. Transformer efficiency

    I did some research and generally in transformer efficiency problems they give you the LOAD at which you are required to find the efficiency at. Thus, it is the output not the input as said by @rg1. One more thing to consider, sometimes they specify the VA rating of the load and the PF. in this case VA rating is used to calculate currents and losses , however, efficiency calculations is done on the Watts ratings only in this case. I thought it is worth mentioning.
  5. Transformer efficiency

    yes, I think very difficult to notice the 25% after using it to calculate the losses and you are ready to calculate for the efficiency. sometimes it is better to slow down and not rushing fast into solving problems familiar to you. unfortunately, I do. Ok, for the next part, I need you to confirm it for me. Calculating for efficiency, it is: output/input . for this problem and generally the transformer rating, is it input or output. in your calculation and also mine, you take it as output. However, I think the right one should be input. calculation should be : (25000-400)/25000 instead of 25000/25400. for this problem results are close but generally speaking it makes difference. I like the formula : 1 - losses/input
  6. Transformer efficiency

    interestingly, you did the same two mistakes I did. can you figure them out? I think you will..
  7. Transformer efficiency

    @rg1 yes, calculation is straight, however sometimes I miss it in the math part. can you try it, I just want to see your solution before I tell my confusion.
  8. I am posting this problem since I usually get it wrong first time. this post will help me remember not to make the same mistake always. Try it, A single phase 100 kVA, 4.6/0.480 kV transformer no-load loss is measured to be 300 W. The total loss at 75% loading is measured at 1200 W. Transformer efficiency at 25% load is approximately:
  9. Base Voltage Throught Transformer

    Look at this topic, it does have some good graphs also http://peguru.com/2011/06/per-unit-system-practice-problem-solved-for-easy-understanding/
  10. Base Voltage Throught Transformer

    just to mention, in the per unit system, voltage across the transformer is 1 Pu, both in primary and secondary, however actual voltages are different , that mean bases are different, right? this problem is all about getting the new base voltage in the motor side while given base voltage on the generator side. the new V base will be 22 (220/25) (13/sqrt3*130) give you 11.1775 as the Vbase on the motor side. plug the value on the equation and you get .09 Pu.
  11. Base Voltage Throught Transformer

    interesting, I think you should get both answers same. I will concentrate on your first approach since, I think, it is the one required by the question. Base MVA is defined as 200, no problem with this one and it is fixed across the whole system. base voltage is defined as 22 Kv on the generator section, for the base voltage it will change across different sections when you pass across a transformer with the same transformer ratio. (I think this one you missed on your first approach). you should get the .09 Pu. your second approach is interesting and I like it.
  12. Base Voltage Throught Transformer

    As I said, just look at it intuitively and don't concentrate too much on the (ratio or the reverse). moving from LV side to the HV side then your multiplier (ratio) is bigger than 1, in this case 220/25 for T1. moving from HV side to the LV side then your multiplier (ratio) is less than 1, in this case 13/130sqrt(3) for T2. hope it make sense for you.
  13. Base Voltage Throught Transformer

    I, usually think of it intuitively. suppose the base voltage of T1 on the primary side is 25 Kv same as its primary rating, then the base on the secondary is 220 Kv same as its secondary rating. thus, to transfer 25 Kv to secondary you multiply by 220/25 (ratio of transformer). in your case the base is 22 Kv, you do the same multiply by 220/25 (ratio of transformer) to get base on secondary. second transformer the same approach. hope it is clear.
  14. capacitive reactance

    Thanks @rg1 and @TNPE for elaborating on the subject. As you mentioned rg1, the formulation of the question could have been better.
  15. capacitive reactance

    Ok, that mean capacitance required is different depending on the connection. will assume for this specific question star as it is usually the case. this will result in capacitance of 9 micro farad per phase. what confused me from the beginning was the solution provided. I think they are mistaken somehow.