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    Question 17

    Found this about question 17:

    Positive Post Rise
    As a battery ages, the positive grids corrode. This is also known as “growth” and it’s well understood in the industry. It is the primary determining factor in when it’s time to have a retirement party for a battery. In addition to this condition occurring as the battery approaches its end of life, the plates will eventually extend down to the bottom of the cell and at that point, there’s nowhere to go but up. Refer to Figure 4.

    Here’s one reason why you’ll see the positive terminals pushing up out of the post seal, exposing an unusual amount of lead above the cover. In many cases, the post seal and surrounding area has limited flexibility to allow the added growth. As a result, the cell cover cracks. Don’t bother breaking out your ohmic testers or load banks.

    Batteries that look like this are on death’s doorstep. What’s not expected is to see this occurring early in battery life. This condition can be accelerated with high operating temperatures, excessive and unnecessary cycling and over charging. (here: http://www.alber.com/Papers/TresslerPaperDONE2011_2.pdf)

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    Quote Originally Posted by ddd675 View Post
    10. In insulating oil, what gas would indicate partial discharge in the oil?
    Hydrogen or Acetylene
    Edit: It's methane.

    Last edited by SecondGen; July 27, 2017 at 10:27 AM.

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    Thanks

    Just wanted to thank everyone for the input in this thread. Has for sure helped in my studying, me and the PE of the company are taking the test tomorrow, and two of our other guys have taken it this month. I was told to look out for lots of questions on fiber optics, especially optical time domain reflectometry, good amount on batteries and calculating transformer load, and a random question on knowing when Bellville washer is bad, which I haven't found too much info on. Here's a good article for transformer through fault testing I found helpful, and about FR3

    http://blog.fauske.com/blog/primary-...r-transformers

    https://static1.squarespace.com/stat...3-insulect.pdf

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

    Please correct me if I'm wrong:

    1.A (Optic time domain reflectometry)

    5.C

    7. You did not specify the xfmr connection but assuming it is Delta/Wye my calculation is 3007.12 A primary

    8. 5th Harmonic

    10. Methane (Duval's Triangle)

    11. 0.41 miliohms

    12. 60A / sqrt(3) = 34.64A
    On a Wye connection your phase current is equal to your line current but in a delta it is divided by 1.732


    13. 1825.8 @ 0.91 PF

    14. Requires no connection to ground (earth)

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

    16.C (most likely since CAT6 length limit is 100 m)

    17.B

    18.A

    19.C

    20. Dielectric withstand @ 75% of Factory acceptance testing value (IEEE C37.20.6)

    22.D (C200 since i came up with a burden of aprox 1.25 ohms which exceed the error limit of a C100)

    24.B (0.89 vs 0.92)

    25 The line reactor compensates for line capacitance, mitigates voltage transients and limits the fault currents. I don't know what the answers were there.

    26. They are used in PLC (Power Line Carrier) to trap the high frequency communication signals typical of 400 Hz

    29.D maybe? (Lead-acid is 5 years and VRLA 12-18 months so we need more info here)

    31. 2.3 micro-farad

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    Quote Originally Posted by SecondGen View Post
    Hydrogen or Acetylene

    It is methane, refer to Duval's triangle.

  10. #26
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    Quote Originally Posted by ddd675 View Post
    5. On a main incoming of a 480V switchgear, what class powermeter should be installed?
    a. II
    b. III
    c. IV
    d. V
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    Cat IV is is for 3 phase at utility connection.

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    Quote Originally Posted by Vernon View Post
    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
    Hi Vernon, can you explain why we multiply by 12.5 and why we don't use three phase voltage in the calculation? Thank you.

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    11. Two 1mH inductors are connected in parallel and in series with two 45uF capacitors connected in parallel. What is the total impedance of the circuit?

    Quote Originally Posted by slts1991 View Post
    11. 0.41 miliohms

    How did you come up with 0.41 milliohms? I don't see how to solve this without frequency to calculate reactance, assuming 60hz this is what i got

    1mH = 2x 377 ohms in parallel= 188.50 ohms
    45uF = 2x 58.9 ohms in parallel = 29.45 ohms
    series 188.50 + 29.45 = 217.95 ohms
    help?

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    IEC61850-8-1 Object Name

    Quote Originally Posted by ddd675 View Post
    23. What is the IEC 61850 naming structure?
    Relay1/XCBR1$ST$Loc$stVal

    Relay1 = Logical Device
    XCBR1 = Logical Node
    ST = Functional Constraint
    Loc = Data
    stVal = Attribute

    http://www.ucaiug.org/Meetings/Austi...0-Tutorial.pdf

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    Just keep in mind, the new structure of the test recognizes which questions continuously get correct answers. The new structure throws out those questions and inputs different ones so the test can continuously change. While it is good to know the questions put up on this forum, do not expect them to be on your test.

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