1. Junior Member
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Nov 2014
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Miami, Florida, USA
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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. Originally Posted by Russs57
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. Originally Posted by Ingenieur
...
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.

4. 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. Originally Posted by Ingenieur
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.

6. Originally Posted by Smart \$
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. Originally Posted by Ingenieur
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.

8. 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. Originally Posted by Ingenieur
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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