neutral current

Discussion in 'General Electronics Chat' started by barth99666, May 31, 2011.

  1. barth99666

    Thread Starter New Member

    May 31, 2011
    7
    0
    hi


    i have red seal Q and A (elect)


    The following load is connected to a 208 Volt three-phase four-wire system:
    • A-N has a coil with an impedance of 8 ohms and a power factor of .9 lag.
    • B-N has a single-phase motor with an impedance of 20 ohms and a power factor of .8 lag.
    • C-N has a heater load of 10 ohms
    Calculate the neutral current

    ans:-2.45 amps is this correct and if yes how?
     
  2. GetDeviceInfo

    Senior Member

    Jun 7, 2009
    1,571
    230
    tell us how you calculated it and we'll show you where your having troubles is any.
     
  3. barth99666

    Thread Starter New Member

    May 31, 2011
    7
    0
    [SIZE=+2]Part 2: Using Vectors to Approximate the Neutral Current in a Three Phase Power System[/SIZE]

    [SIZE=+1]for three phase 60 hertz systems with unity power factor and resistive loads[/SIZE]
    by Gerald Newton
    November 5, 1999
    [SIZE=+1]The geometric addition of vectors gives a fast approximation of neutral current in a three phase system. A more precise calculation can be performed using algebra. This can be a long mathematical problem, but by restricting our calculation to A phase at 0 degrees, B phase at 120 degrees, and C phase at 240 degrees the calculation is simplified.[/SIZE]
    [SIZE=+1]The figure below demonstrates how to break a vector down into its horizontal (X) and vertical (Y) components. The magnitude of the vector is its length in whatever units we chose to represent. In our case that would be amperes.[/SIZE]
    [​IMG]


    The following figures demonstrates the algebraic addition of two vectors.
    [​IMG]

    [SIZE=+1]The Pythagorean is used to find the algebraic value of a vector once its X and Y components are known.[/SIZE]

    [​IMG]
    [SIZE=+1]The following demonstrates the use of the Pythagorean theorem to find the magnitude of the sum of two vectors.[/SIZE]

    [​IMG]
    [​IMG]

    [SIZE=+1]For three vectors the magnitude calculation is similar to that for two vectors. A third component is added for the third vector. The process of multiplying the polynomials can be reduced by using the known angles for the A phase, B phase, and C phase vectors. The sines and cosines are known for the three angles, 0 degrees, 120 degrees, and 240 degrees.[/SIZE]

    [​IMG]




    Calculator by electrician2.com

    Last updated: 6/1/2011

    Use the mouse button to lay down up to ten vectors. Click on the Action button to calculate the sum.
    Return to electrician2.com
    Last updated: 6/1/2011
     
  4. barth99666

    Thread Starter New Member

    May 31, 2011
    7
    0
    I change some of vectors ei:-0.9, 0.8 lag
     
  5. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    You appear to have saved a web page to your computer, and then tried to copy/paste it into a post. As a result, all of the image links are trying to point to a file that only exists on your hard drive that nobody else can see.

    Example:
    file://C:/Users/cullens/Documents/canada/canada%20file%20electricians%20info/Using%20Vectors%20to%20Approximate%20the%20Neutral%20Current%20in%20a%20Three%20Phase%20Power%20System.mht!x-usc:http://www.electrician2.com/electa1/vector_1.gif
    Where you actually meant to post this image:
    [​IMG]
     
  6. barth99666

    Thread Starter New Member

    May 31, 2011
    7
    0
    hi


    i have red seal Q and A (elect)


    The following load is connected to a 208 Volt three-phase four-wire system:
    • A-N has a coil with an impedance of 8 ohms and a power factor of .9 lag.
    • B-N has a single-phase motor with an impedance of 20 ohms and a power factor of .8 lag.
    • C-N has a heater load of 10 ohms
    Calculate the neutral current

    ans:-2.45 amps is this correct and if yes how. sorry fo not explianing my problem very well but this is red seal question i got of exam bank and can only get 8 amps, i used phasor and calculations my calc was this sqro(15*1+6*-0.5+12*-0.5)sq+(15*0+6*0.866+12*-0.866)sq if this wrong why. thankyou in advance
     
  7. RGPATTER

    New Member

    Jun 2, 2011
    16
    1
    Assuming a positive phase sequence (ABC), Ia=26 amps at an angle of -25.842 degrees, Ib= 10.4 amps at an angle of 203.130 degrees, and Ic=20.8 amps at a angle of 120 degrees. Converting from polar to rectangular coordinates gives Ia=23.400-j11.333, Ib=-9.564-j4.085, and Ic=-10.4+j18.013. The neutral current is the sum of Ia, Ib, and Ic and is equal to 3.436+j2.595 in rectangular coordinates. Converting to polar coordinates gives the neutral current as 4.306 amps at an angle of 37.060 degrees.

    With a negative phase sequence (ACB), the neutral current would be 23.763 amps at an angle of -53.172 degrees.
     
  8. RGPATTER

    New Member

    Jun 2, 2011
    16
    1
    I may not have understood your comment in your last post. I think you are saying that the answer given is 2.45 amps and you keep coming up with 8 amps. I don't agree with the 2.45 amp answer. The aide you reference in your second post is for a three phase system "with unity power factor and resistive loads". You don't have three unity power factor loads. You are summing three vectors 15 angle 0, 12 angle 240, and 6 angle 120 and coming up with 8 amps. With those angles, they are all resistive. Instead you should be summing 208/8 angle -25.8, 208/20 angle 203.1, and 208/10 angle 120. Your answer of 8 amps is correct for three resistive loads of 15, 12, and 6 amperes
     
  9. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    Perhaps 208V is the line-to-line voltage for a 120V line-to-neutral system.

    If this is the case, then the neutral current magnitude would be 2.485A.

    RGPATTER would divide their answer by √3 ...

    Query to barth99666:

    Was there a typo error with the given answer of 2.45A?
     
    barth99666 likes this.
  10. RGPATTER

    New Member

    Jun 2, 2011
    16
    1
    tnk is correct. My first reply should have read:

    Assuming a positive phase sequence (ABC), Ia=15 amps at an angle of -25.842 degrees, Ib= 6 amps at an angle of 203.130 degrees, and Ic=12 amps at a angle of 120 degrees. Converting from polar to rectangular coordinates gives Ia=13.510-j6.543, Ib=-5.522-j2.359, and Ic=-6.004+j10.400. The neutral current is the sum of Ia, Ib, and Ic and is equal to 1.984+j1.498 in rectangular coordinates. Converting to polar coordinates gives the neutral current as 2.485 amps at an angle of 37.060 degrees.

    With a negative phase sequence (ACB), the neutral current would be 13.720 amps at an angle of -53.172 degrees.

    Sorry about the confusion and thanks to tnk for catching my error.
     
    barth99666 likes this.
  11. barth99666

    Thread Starter New Member

    May 31, 2011
    7
    0
    Using phasor diagrams and your advise I've solved my problem. Thank you RGPatter and T N K you have both been a great help. If i have any future problems can I come back to you both for some advise? Thank you again. barth99666
     
  12. barth99666

    Thread Starter New Member

    May 31, 2011
    7
    0
    Using phasor diagrams and your advise I've solved my problem. Thank you RGPatter and T N K you have both been a great help. If i have any future problems can I come back to you both for some...
     
  13. RGPATTER

    New Member

    Jun 2, 2011
    16
    1
    Sure. If it's on a subject that I can be of help with and provided the guardian angels like tnk are watching over me to keep me from putting my foot in my mouth.
     
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