Help with DeltaDelta calculations

Discussion in 'Homework Help' started by Randy111, May 7, 2011.

  1. Randy111

    Thread Starter New Member

    May 7, 2011
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    hello all I am a 3rd apprentice and I am having some trouble with Delta/Delta calculations any help would be appreciated. below is one of the questions I am having trouble with.

    A 4160V three Phase three wire system feeds a Delta/Delta Transformer with a rating of 100KVA is used to supply a 600V Balanced three phase load of 60KW. If the power factor is .8 what is the Primary Phase current?

    according to the test answers the primary current is 6A I just can't figure out how they got it.

    Any help would be great.
     
  2. chgy

    New Member

    May 7, 2011
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    Randy111,

    I'm recently out of Electrical Technology school and believe I have the correct way to figure out your problem if you haven't figured it out already. I'm sure someone who has more experience and/or schooling will correct me if I'm wrong.

    The load is 60kw which means the load per phase 1/3, or 20kw. Since the PF is .8, each phase will draw 41.67 amps because the system sees 25kva as the load. The current is stepped up by a ratio of nearly 7 (6.93 to be exact). So if voltage is stepped down, current is stepped up by the same ratio. Of course the problem is assuming a 100% efficient transformer.

    Take the 41.67 amp current from the secondary and divide by 6.93 to get 6.01 amps per phase. If you want to know the current through each winding on a delta than divide 6.01 by 1.732 (or the square root of 3).

    I believe that to be correct. I look forward to any corrections.

    chgy
     
    Randy111 likes this.
  3. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    The reasoning seems perfectly sound.

    One thing that could throw a spanner in the works would be if the stated voltages were line-to-line values. I'm assuming they must be line-to-neutral values based on the stated answer of 6A as the primary line current - if that's what was meant by the term "Primary Phase Current". My personal "preference" would be that the voltages in a 3-wire system would be stated as line-to-line values. I they were indeed line-to-line values then the current in the primary lines would be 10.41A and 6A in the primary delta windings.
     
  4. chgy

    New Member

    May 7, 2011
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    When you say line to neutral for a delta what do you mean? My understanding is that delta connections do not normally have a neutral. When they do it creates a stinger voltage. Can you explain?

    Like I said, I'm new out of school and just starting a new job with lttile experience. In fact, when I started my job and learned that our plant uses a delta-delta ungrounded system I was quite confused and thought that is something that shouldn't be done for safety reasons. I understand that is rare.

    Thanks
     
  5. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    This was one of the points I was trying to get across regarding the solution to the problem. In a 3-wire delta system one would normally quote line-to-line voltages.

    If the voltages quoted in the problem were line-to-line values then my other contention is that the apparent solution of 6A primary side line (phase) current is incorrect.

    This would be my reasoning.

    The per-phase power is 20kW at 0.8pf - thus 25kVA per phase.

    Ignoring transformer losses, the same per-phase kVA is present in the transformer primary.

    The per phase kVA=(V_line/√3)*I_line

    So with a line-to-line voltage [V_line] of 4160V

    I_line=(√3*kVA)/V_line=√3*25000/4160=10.409A

    The corresponding delta transformer winding current would then be I_line/√3=6.01A

    The ambiguity in the problem statement should (hopefully) now be clearer to you - given the answer [6A] provided by the OP.
     
  6. chgy

    New Member

    May 7, 2011
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    "I'm assuming they must be line-to-neutral values based on the stated answer of 6A as the primary line current. . ."

    I didn't find any ambiguity with the calculations when I first considered the problem. Where I'm unclear is what you mean by line-to-neutral. I didn't see any reference to a neutral in the problem. Other than creating a stinger voltage in a delta, when and where would one find a neutral?

    Thanks
     
  7. Randy111

    Thread Starter New Member

    May 7, 2011
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    I think the answer is incorrect as well because we haven't dealt with 4 wire delta circuits yet and the question refers to it as a 3 wire system. I am going to be checking tomorrow with my teacher, but I also got 10.409 using the formula
    I= P/(Vline * Pf * 1.732). I will let you guys know tomorrow whats up with this.
    Thanks for all the help
    Randy111
     
  8. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    As I tried to indicate I take your point. I agree that the likely intention is that the stated voltages are line-to-line. You can go on stating it as long as you wish. I agree there is no neutral. One might however consider a virtual neutral which might assist some folk in doing the analysis. I'm not sure it's going to make any difference making this point again. In essence, I guess I was thinking the unseen teacher's solution, which gave an answer of 6A, was done on the basis of such a line-to-(virtual) neutral potential.

    As to the ambiguity ..... My problem is that the answer given to the OP is most likely just wrong. If the required solution was the primary delta winding current then 6A is OK. If the line current was what was expected then 6A is wrong.

    If you are taking the values as line-to-line values and calculating the line current then your submitted answer is incorrect.
     
    Last edited: May 9, 2011
  9. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Hi chgy,

    If you perhaps think the concept of a virtual neutral is somewhat specious, then one might consider the delta-delta transformer primary is fed from another notional [upline] transformer with a star (wye) configured secondary. The star point on such a notional upline transformer may well be considered as a virtual neutral - albeit unconnected to anything external.

    I can also imagine a virtual star point on the load side of the delta-delta transformer. There is nothing in the question which precludes the balanced 60kW 3-phase load being a star connected case. The star point of a balanced load so connected might also be considered as an effective virtual neutral.

    It will be interesting to read what Randy111's teacher comes up with.
     
  10. chgy

    New Member

    May 7, 2011
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    Since I'm fresh out of school I'm not clear on the virtual neutral concept. When I did the calculations I assumed 4160v phase to phase on the primary with a 25kva balanced load, which would be 6 amps per phase. I failed to divide by 1.732 to get the 10.4amps. That is all I've been exposed to with regard to delta connections, aside from center tapping one winding and creating the stinger voltage I mentioned.

    I'm not at all trying to argue with anyone. I'm trying to learn and understand.

    If one were to measure voltage on the 4160v primary from phase to ground what would they see? At my work we have an ungrounded delta system and when I measure from phase to ground I get essentially the same measurement as if it were a wye, measuring phase to neutral. So phase to phase is around 480v and phase to ground is about 270v. If you could explain in laymans terms it would be appreciated.

    Thanks
     
  11. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Possibly, somewhere in your power system or in your locality there is a connection from the supply system neutral to earth/ground. Perhaps your installation or another in your locality has a grounding transformer to provide an effective neutral / ground point - if required for reasons of system protection and the like. Google "grounding transformers".

    Other than that I can't comment about your particular work circumstance.
     
  12. Randy111

    Thread Starter New Member

    May 7, 2011
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    Okay guys here is the correct answer to this particular problem as provided by my instructor. we use the previously stated formula of
    I Line primary = Power total / (line Voltage Pri. * Power factor * 1.732)

    I Line pri. =60'000w / (4160v * .8 * 1.732)

    I Line pri. = 10.409
    and since we are looking for the phase current all we have to do is take
    10.409/1.732 = 6.0098

    thanks again for all the help
    Randy111
     
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