Resistance @ 95 from Resistance @ 35

  1. Please explain the below:
    Resistance@95 =R@35 x (1+α*(95x(I/Io)²+20))/(1+75x α) at the actual current A (I Act).
     
  2. jcsd
  3. berkeman

    Staff: Mentor

    Welcome to the PF.

    Why don't you start by telling us what you know about that equation. :smile:
     
  4. I know the basic equation to find resistance at different temp from the following:

    R@T2 = R@20(1+a@20(T2-20),

    where,
    "T2" is the temperature at which the new resistance to be found,
    "a" (alpha) is the temperature coefficient @ 20deg C (0.00393)

    I am trying to understand the posted equation as it was found in a voltage drop calculation sheet for Busduct.

    I would like to know "(95x(I/Io)²+20))" this part. whether 95 is the "T2" temperature if so, how that "+20" came there.

    Reply would be highly appreciated.
     
  5. Baluncore

    Baluncore 2,963
    Science Advisor

    Is this a question about the resistance of metallic conductors at different temperatures ?

    Please post a link to the “voltage drop calculation sheet for Busduct” you are using.
     
  6. Baluncore

    Baluncore 2,963
    Science Advisor

    We now have the definitions for most of the terms in the equation.
    We do not know what the symbol Io refers to. Maybe it is the design rating current?
    It seems that the 35 and 95 are probably conductor temperatures 35°C and 95°C.

    You were unable to understand the equation given the “Voltage drop sheet”.
    At PF we are only human, like you, we need more information or context from the manufacturers website.
    Where did you get the “Voltage drop sheet” page from ?

    If you cannot give us a link to the manufacturers website, then we can only guess and you will need to direct your question to the equipment supplier.
     
  7. Thank you for your guess. I searched the web for this equation; all i found was the standard equation. Wanted to know whether is there any standard equation like R@T2 = R@20(1+α@20(T2-20).

    As you mentioned "Io" is the rated design current and "I" is the actual load current.
     
  8. Baluncore

    Baluncore 2,963
    Science Advisor

    Attached is another version of the bus duct equation, downloaded from;
    http://furutec.com.my/home/wp-content/uploads/2012/07/standard09.pdf

    The equation boils down to the following;
    Code (Text):

             1 + a * ( 55 * ( I/Io )^2 + 20 )
    R = Ro * --------------------------------
             1 + a * ( 55              + 20 )
     
    The 75 = 55 + 20 is being used to blend two conditions.
    Your original version of the equation seems confused?
     

    Attached Files:

    1 person likes this.
  9. Baluncore

    Baluncore 2,963
    Science Advisor

    Looks like the 95 in your equation was from a typo in the original document. It should be 55.
     
  10. jim hardy

    jim hardy 5,207
    Science Advisor
    Gold Member
    2014 Award

    read up on "Callendar van Dusen" equation
    http://static.elitesecurity.org/uploads/2/6/2694332/CalVan.pdf

    alpha is the slope at some particular temperature

    and your term, (95x(I/Io)²+20))/(1+75x α),

    appears to be modifying alpha for some different starting temperature


    While most explanations of Callendar focus on platinum resistance thermometry, it's a really handy bit of algebra. When you can convert the quadratic coefficients A B and C into [itex]\alpha[/itex] and [itex]\delta[/itex] of eq(6) in that link , you've got it.
    ...............................................................................................

    now - at zero current your equation reduces to
    Resistance@95 =R@35 x (1+α*(95x(I/Io)²+20))/(1+75x α)
    = R@35 x (1 + 20[itex]\alpha[/itex])/(1+75[itex]\alpha[/itex])
    = R@35X(1.0786)/(1.29475)
    = 0.833R@35


    which doesn't make much sense unless there's active cooling of that duct.
    so check my arithmetic....

    I suspect a misplaced paren.


    OOps i see Baluncore fixed it while i was typing !

    oh well as Sophie says, cross posts mean it's an interesting topic.
    I learned something about bus duct - thanks, guys

    old jim
     
    Last edited: Jul 19, 2014
    1 person likes this.
  11. Thank you everyone for the reply. And its getting interesting. Sorry to go further, becoz i would like to have a proper idea behind that equation.

    I am attaching few docs for your reference, which i found in the web, which shall be useful to you all.

    >>> supporting doc for R@t2 = R@20(1+α(t2-20). https://www.dropbox.com/s/41oadn016oea1su/The temperature coefficient of resistance of copper.pdf
    >>> load-temperature correction factors.

    I would like to know whether the value "55" shall change based on ambient temp. As per IEC the max temp rise allowed for busducts is 105degC, if so, for an operating temp of 50degC, the value "55" remains OK; provided we put the resistance value Ro as the R@50 deg C.

    Based on the equation,

    R = Ro * (1 + a * ( 55 * ( I/Io )^2 + 20 ))/ (1 + a * 75)


    can we conclude that the "R" is the resistance at 95 deg rise at load current at an ambient temp of 50deg C, again the value Ro has to be R@50.

    If so, i think my equation also is correct as

    R = R@20 * (1 + a * ( 105-20) * ( I/Io )^2 + 20 ))/ (1 + a * 75)

    as here we put the Ro as R@20 considering the ambient temp to be 20degC.

    Please comment.
     

    Attached Files:

    Last edited: Jul 20, 2014
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