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    #11
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    Dissolved gas

    Quote Originally Posted by ddd675 View Post
    10. In insulating oil, what gas would indicate partial discharge in the oil?
    "PARTIAL DISCHARGE is a fault of low level energy which usually occurs in gas-filled voids surrounded by oil impregnated material.
    The main cause of decomposition in partial discharges is ionic bombardment of the oil molecules.

    The major gas produced is Hydrogen. The minor gas produced is Methane."

    source - http://www.satcs.co.za/Transformer_Oil_Analysis.pdf

  2. #12
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    Quote Originally Posted by tallyson71 View Post
    4. When testing insulation resistance on a MV Motors, at what voltage does it become ambiguous?

    I couldn't find any information on this. Could the question have been at what IR value (Megohms) does P.I. become ambiguous? (5,000MOhms I think)
    I agree with your assesment.

    If the IR value (at 40 °C) is greater than 5000 MΩ, the PI may be ambiguous and can be disregarded. http://electrominst.com/test-technol...on-resistance/

  4. #13
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    Quote Originally Posted by tallyson71 View Post
    Found this on one manufacturers documentation:

    "Conduct load test two years[emphasis added] after installation and then every five years. When the
    system’s capacity falls below 90% load test annually."

    http://www.sbsbattery.com/PDFs/Batte...-IEEE-NERC.pdf
    Good chart for reference, thank you for posting.

  6. #14
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    Quote Originally Posted by tallyson71 View Post
    "PARTIAL DISCHARGE is a fault of low level energy which usually occurs in gas-filled voids surrounded by oil impregnated material.
    The main cause of decomposition in partial discharges is ionic bombardment of the oil molecules.

    The major gas produced is Hydrogen. The minor gas produced is Methane."

    source - http://www.satcs.co.za/Transformer_Oil_Analysis.pdf
    Looking at different references, it would appear that both Acetylene or Hydrogen are indicators of partial discharge.

    source -http://www.netaworld.org/sites/default/files/public/neta-journals/NWwtr09_Hamrick.pdf

  8. #15
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    Quote Originally Posted by slts1991 View Post
    Thanks for the input. Here are some answers

    Original Question: A primary injection is to be performed on a transformer 138kV/25kV, 2500kva, 8%Z using a 480V generator connected to the high side with the low side grounded. The expected current in the high side is?


    7. You did not specify the xfmr connection but assuming it is Delta/Wye my calculation is 3007.12 A primary
    Didn't read all the answers but this one jumped out and just slapped me as very high. I think that if no three phase connection is given one assumes a single phase transformer. If I tell you I am standing here with a 480V to 120V transformer in my hand you would assume one transformer with a turns ratio of 4 to 1 and a primary voltage of 480 volts.

    Ok - so the full load current at 138,000V and 2,500,000 VA is 18.116 A. At 8%Z with a shorted secondary you would multiply that by 12.5 so at rated voltage you have 226.45 amps Isc. But you are only applying 480V so you would get, proportionately, 0.787 AMPS. Generally, if I specify that a transformer is (for example) 10%Z that means that with the secondary shorted it will take 10% of the rated primary voltage to get full load current. www.cttestset.com

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    Correction

    [QUOTE=tallyson71;8580]
    Quote Originally Posted by slts1991 View Post

    15.B (not required, NEC 700.6(D))

    The chapter you are citing refers to emergency systems.

    I would think the appropriate chapter is:
    "230.95 Ground-Fault Protection of Equipment. Ground-fault protection of equipment shall be provided for solidly grounded wye electric services of more than 150 volts to ground but not exceeding 1000 volts phase-to-phase for each service disconnect rated 1000 amperes or more. The grounded conductor for the solidly grounded wye system shall be connected directly to ground through a grounding electrode system, as specified in 250.50, without inserting any resistor or impedance device. The rating of the service disconnect shall be considered to be the rating of the largest fuse that can be installed or the highest continuous current trip setting for which the actual over current device installed in a circuit breaker is rated or can be adjusted."

    Thanks
    From a subsequent post it looks like this question may have referred to emergency systems, which should not trip.

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    [QUOTE=tallyson71;8580]
    Quote Originally Posted by slts1991 View Post

    15.B (not required, NEC 700.6(D))

    The chapter you are citing refers to emergency systems.

    I would think the appropriate chapter is:
    "230.95 Ground-Fault Protection of Equipment. Ground-fault protection of equipment shall be provided for solidly grounded wye electric services of more than 150 volts to ground but not exceeding 1000 volts phase-to-phase for each service disconnect rated 1000 amperes or more. The grounded conductor for the solidly grounded wye system shall be connected directly to ground through a grounding electrode system, as specified in 250.50, without inserting any resistor or impedance device. The rating of the service disconnect shall be considered to be the rating of the largest fuse that can be installed or the highest continuous current trip setting for which the actual over current device installed in a circuit breaker is rated or can be adjusted."

    Thanks
    I think I got confused because I had both questions on my tests. Last time I went and passed the test I had the one with emergency systems. You need to read the question properly and not mistake the two of them during your test. On emergency is not required except for audio and visual.

  14. #18
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    Question 17

    Found this on a NETA Handbook series II - Arc Flash vol. 2:
    • Available bolted fault current – The punch behind the arc fault
    magnitude. Recall that the magnitude of a low-voltage arcing
    fault is approximately 43-57% of the bolted fault value. This implies
    that systems with significant bolted fault currents will have
    elevated arcing current levels.
    Considering the options avilable I would say the best is 50%

  16. #19
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    INFO

    Question 18:
    Two parallel feeder cables with individual 400:5 CT’s per parallel run are connected to one overcurrent relay per phase. The proper connection would be:

    The polarity of a transformer is Subtractive when current entering in H1 means current exiting X1. The right answer will be Subtractive polarity and in parallel with relay set at 800 Amps.
    Click image for larger version. 

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    Attached Thumbnails Attached Thumbnails 2a.png  


  18. #20
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    Quote Originally Posted by joaogemal View Post
    Question 18:
    Two parallel feeder cables with individual 400:5 CT’s per parallel run are connected to one overcurrent relay per phase. The proper connection would be:

    The polarity of a transformer is Subtractive when current entering in H1 means current exiting X1. The right answer will be Subtractive polarity and in parallel with relay set at 800 Amps.
    Click image for larger version. 

Name:	1a.png 
Views:	436 
Size:	16.4 KB 
ID:	189
    Usually they are all subtractive polarity as shown in your drawing. My take was that they would be connected so the currents were additive and if you had 1 A out of CT#1 and 1 A from CT#2 you get 2A in the relay. Since they didn't use the term "polarity" I went with the currents adding and not canceling. Connect the polarity marks together so that the secondary currents are in phase (I think that is what I posted). The same polarity might be a correct answer but, yes, both subtractive is how it would usually be. I have never seen an additive polarity CT - it is current in on H1 is current out in X1 as shown in your attachment.

    The relay is looking at the total current in the two feeders and 400 total amps always gives you 5 A secondary. If just feeder one is loaded to 400A you get 5 relay amps. If just feeder 2 is loaded to 400A - you get 5 A in the relay. if you have a balance and 200 A on each feeder then each secondary is sourcing 2.5 amps for a total of 5A. So the CT value in the relay should be set to 400 to 5 because 400 total amps primary always gives you 5 amps secondary just like it would if you just have one feeder. If you had 800 amps you would get 10 amps regardless of the distribution of load between the cables - but the ratio is supposed to be to 5, not 10. If you had one feeder with a 400:5 you would put 400:5 in the relay for the CT ratio and this is the same thing. 400 primary amps gives you 5A. www.cttestset.com

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