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Thread: Voltage Drop

  1. #11
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    In the interest of being complete.....should we mention that power factor of the load (temperature too) has a bearing?

  2. #12
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    Quote Originally Posted by Russs57 View Post
    In the interest of being complete.....should we mention that power factor of the load (temperature too) has a bearing?
    good points
    in this case the i is given so it is safe to assume it includes real and reactive components
    that is why I like using values for Z (pf 0.85) from the tables
    imo a more conservative approach
    either way though the difference is moot in most cases
    one reason v is usually higher than rated
    208 is 212 range
    480 is 490 or so
    a bit higher often
    and motors are rated 460 for example

    temperature is another matter
    if not given I guess safe to assume std conditions of amb 40 C and a rise of 30-35???
    and more to do with insulation rating than change in resistivity



  3. #13
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    Quote Originally Posted by Ingenieur View Post
    ...
    temperature is another matter
    if not given I guess safe to assume std conditions of amb 40 C and a rise of 30-35???
    and more to do with insulation rating than change in resistivity
    The values given in the Chapter 9 Table 9 are at 75°C. We seldom 'drive' a conductor to 75°C. Using the case provided in this thread, 27A through a #8 copper conductor in 30°C ambient will not raise the temperature to anywhere near 75°C in typical operating conditions.
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  4. #14
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    Power lost in conductor = 27 x 6.2 = 168 W over 300 ft ~ 0.56 W/LF
    the surface area of 1 LF #8 ~ 0.0107 sq ft
    or 0.56/0.0107 = 52 w/sq ft
    that will get warm, no?
    think of a large 25 W bulb with ~72 sq in (1/2 sq ft) of area
    maybe not hot but warm

    if amb is say 70F wouldn't the cond get at least that warm and equalize?
    or warmer to radiate to the amb air since heat is continously generated in the conductor as long as current flows?

    but I agree on his scale the effects are moot



  5. #15
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    Quote Originally Posted by Ingenieur View Post
    Power lost in conductor = 27 x 6.2 = 168 W over 300 ft ~ 0.56 W/LF
    the surface area of 1 LF #8 ~ 0.0107 sq ft
    or 0.56/0.0107 = 52 w/sq ft
    that will get warm, no?
    think of a large 25 W bulb with ~72 sq in (1/2 sq ft) of area
    maybe not hot but warm

    if amb is say 70F wouldn't the cond get at least that warm and equalize?
    or warmer to radiate to the amb air since heat is continously generated in the conductor as long as current flows?

    but I agree on his scale the effects are moot
    52W sounds like it would get very warm... but it's just a play on numbers.
    52W*300ft/168W = 93ft of conductor dissipating that 52W.
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  6. #16
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    Quote Originally Posted by Smart $ View Post
    52W sounds like it would get very warm... but it's just a play on numbers.
    52W*300ft/168W = 93ft of conductor dissipating that 52W.
    300 LF = 3.2 sq ft of surface area
    168/3.2 = 52 W/sq ft
    same thing
    I would think as i approaches rated ampacity temp approaches insul rating?
    for a given amb temp and install method, ie, conduit, buried, free air, etc

    I know there is a calculation for temp rise, too lazy to find it lol



  7. #17
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    Quote Originally Posted by Ingenieur View Post
    300 LF = 3.2 sq ft of surface area
    168/3.2 = 52 W/sq ft
    same thing
    I would think as i approaches rated ampacity temp approaches insul rating?
    for a given amb temp and install method, ie, conduit, buried, free air, etc

    I know there is a calculation for temp rise, too lazy to find it lol
    Well, as it approaches the ampacity given in the 75°C column of Table 310.15(B)(16) it will approach 75°C in an enclosure. At least that's the theory behind terminal temperature limitation coordination. Somewhat the same as going by the ampacity adjusted for the conditions of use (installation method). But going by the 90°C ampacity won't do us any good if we can't go over 75°C in general.

    Been down that road trying to compensate for temp rise before. It amounts to being too complex for something that isn't required. Perhaps it would be if all you installed were fire pumps.
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  8. #18
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    Here's another one
    there was an error in my calcs (surface area = dia x Pi x L)
    I forgot the Pi
    so W/sq ft ~ 1/Pi x 52 calculated ~ 15 or so

    30 A at 208/1
    10 awg
    100 LF one way
    3%, 6.2 drop, 186 W
    surface area for 200 LF = 5.3
    W/sq ft = 35 W/sq ft

    curious what temp rise would be in amb of 30 C



  9. #19
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    Quote Originally Posted by Ingenieur View Post
    Here's another one
    there was an error in my calcs (surface area = dia x Pi x L)
    I forgot the Pi
    so W/sq ft ~ 1/Pi x 52 calculated ~ 15 or so

    30 A at 208/1
    10 awg
    100 LF one way
    3%, 6.2 drop, 186 W
    surface area for 200 LF = 5.3
    W/sq ft = 35 W/sq ft

    curious what temp rise would be in amb of 30 C
    I'm not sure how to calculate temp rise accurately. Neher-McGrath?
    As to estimating, can we assume delta t is proportional to delta I².
    Table 310.15(B)(16) gives 35A for #10 at 75°C.
    30²/35² = 73.5%
    ... × (75°C - 30°C) = 33°C
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