You ask a simple question to this group and stunning brilliance floods
my inbox. RE-Wrenches, You are truly amazing. Thanks to all that
discussed this for the knowledge I have gained.
Larry
On 12/2/14 3:47 PM, Daniel Young wrote:
It all boils down to how ampacity is determined in the NEC.
Ampacity is really related to temperature as far as the NEC is
concerned. The ampacity of a 1/0 wire at 90c is the constant current
it can carry in free air (30 C air) and not achieve an internal
temperature of more than 90 C. the 75 c ampacity is the same, the
amperage it can carry while not going over 75 C.
Now, if you double the diameter of a circle, the circumference also
doubles, but the cross sectional Area actually goes up much more (4x
more in fact) because cross sectional area is based on the square of
the diameter and circumference is simply based on the diameter ^1 power.
Wire dissipates heat from its surface only, so the dimension critical
for the amount of heat a wire can dissipate is circumference, not
cross sectional area. So even though the wire is larger and has a much
lower resistance, the heat dissipating area does not increase by as
much, so in the end the larger wire has a lower current carrying
capacity per unit cross sectional area, than a smaller wire.
Here is an example that lets me keep the math simple: (the #’s are all
round #’s and not based on real ampacities/resistances, just to keep
the math simple.)
Wire 1:
A diameter of 10 units and can carry 100A through it and stay at 90C.
It has a resistance of 1 ohm/1000ft.
Wire 2:
A diameter of 20 units. It has 4times the cross sectional area, and
double the circumference (which means 2x the outer surface area) to
dissipate heat. It has a resistance of .25 ohm/1000ft (1/4 that of
wire 1 since it has 4x the amount of copper to carry current).
For wire 1 to stay at 90 C, it has to dissipate (P=I^2*R),
P=(100amps^2)*1 omh=10,000 watts per unit of outer surface area.
So if Wire 2 has double the surface area to dissipate heat, it can
dissipate about 2x the energy, or 20,000watts. So if we work it
backwards (P=I^2*R is the same as I=sqrt[P/R])
I=sqrt[20,000/.25]=sqrt[80000]=282amps
So wire 2 can handle (282amps/100amps)=2.8 times the amperage, even
though it has 4times the cross sectional area, all because it only has
2 times the surface are to get rid of heat. There are other factors
with heat transfer that make the larger wire have even lower ampacity,
but this demonstrates the main contributing factor.
[In the NEC table 310.15(B) we see that a 250kcmill copper wire
handles 255A @75C, and a 100kcmill (4x the area) can only take 545A,
or 2.2 times the current, so not too far from my example].
I hope the above helps more than it hurts.
With Regards,
Daniel Young,
NABCEP Certified PV Installation Professional^TM : Cert #031508-90
NABCEP Certified Solar Heating Installer^TM : Cert #SH031409-13
_______________________________________________
List sponsored by Redwood Alliance
List Address: [email protected]
Change listserver email address & settings:
http://lists.re-wrenches.org/options.cgi/re-wrenches-re-wrenches.org
List-Archive:
http://www.mail-archive.com/[email protected]/maillist.html
List rules & etiquette:
www.re-wrenches.org/etiquette.htm
Check out or update participant bios:
www.members.re-wrenches.org
