NEC Changes For #14 Ampacity

[FionaZuppa]

ok, after some chit'n around in some other forums with math/physics/derivatives, got an equation that describes the temp of the wire (less the heatsink factor). right now my measurements match the math, which is good. now working on adding in the heatsink factor. according to my tests of both short and long wire, the heatsink factor seems to be insignificant, but need the math to confirm, etc.

Omega connectors may come tomorrow, but are on UPS schedule for monday :rant:

 

[mbrooke]

ok, after some chit'n around in some other forums with math/physics/derivatives, got an equation that describes the temp of the wire (less the heatsink factor). right now my measurements match the math, which is good. now working on adding in the heatsink factor. according to my tests of both short and long wire, the heatsink factor seems to be insignificant, but need the math to confirm, etc.

Omega connectors may come tomorrow, but are on UPS schedule for monday :rant:

Can you post this equation here? (I like to play with math)

 

[mbrooke]

open air test.
well, meter says 83.2F, was sitting there for past couple of hours. if i normalize it back to 90F the #'s are:

114.7F (9" of same wire @26A)
118.2F (84" of same wire @26A)

there's a 3.5F (3.1%) diff after normalization, i dont see this diff as significant, which leads me to believe the heatsink affect on the clamped ends are not significant.

but now lets extrapolate out the % diff per inch = 3.1/75 = 0.0413%/inch, for all intensive purpose this in %/in between heatsink affecting the temp vs infinite long wire (no heatsink affect).

the wire in foam experiment is 48", my free air test of NM was 24", do the diff is 24"
24x0.0413 = 0.9912%

so thats my crude adjustment factor, 1%, regardless of any other error. if i measure 100F we will tack on 1F,........... not significant.

So to just verify my thinking, the heatsink effect is linear? It stays the same even when temperature goes up?

 

[mbrooke]

One more reason which might be behind 240.4 (D). I brought it up in another thread but it slipped out of my mind. Conduit installations where inductive heating is taking place could further motivate that restriction. Inductive heating can come from crossed circuits or neutral to ground faults out on the circuit which cause high net currents. These high net currents will heat the conduit and in turn impair heat dissipation. A home in Chicago might benefit from 240.4 (D).

But, we are still back at square 1. NM is not inductive...

 

[FionaZuppa]

have a read here
http://www.roymech.co.uk/Related/Thermos/Thermos_insulation.html

but i am working with this equation (will be adding a heatsink term shortly). need to add in a -f(heatsink)
this was developed around cylinder conductor wrapped by tubular insulation 1ft long.

all items are in units of ft hr btu F (so be ready to have your conversion skills abused)
a couple of tricky things to know
Q and k are normalized to 1ft
h is a coefficient to the boundary condition at the insulation to air interface
Tw is wire temp
T0 is ambient
ro is radius of insulation to centerline
ri is radius of wire to centerline

i approximated the k value of insulation to be 0.17W/mK given that the nylon layer is approx 10x thinner than the pvc layer
h is very tricky #, it is sensitive to environmentals. from test data we found h=8btu/hr-ft-F

the equation says Tw=86.7F, measurement says 83F (for the "long wire" #14 @26amps). the diff is acceptable given the use of a metal TC probe.

 

[FionaZuppa]

let me address the other two questions, heatsink is linear? and induction modes?

yes, the heatsink is linear regardless of temp, heatsink mode is governed by the physicals and properties of the material. delta-T only tells us about the rate, which is linear with delta-T.

as for inductive modes. the 60Hz mag and electric fields do couple to surrounding items, the effects are more noticeable in metals that contain iron. but, for conduit, diversity of the amps i think would be a significant factor, and for a NM "bundle" the diversity is usually very low. am i off base here? (hah, maybe go to aluminum conduit for bundles that have more than 4 CCC's).

 

[wwhitney]

Ah, that's the equation for cylindrical heat flow that I never got around to posting about. A few comments:

That last term with the h in the denominator represents how well the outer surface of the cylinder dissipates heat to the environment. So h takes into account convection, etc. If you just want the heat flow across the cylindrical shell, then you can omit the last term, which I will do for the rest of my comments.

If the cylindrical shell is very thin, this equation reduces to the usual heat flow equation for a wall element, as follows: if the thickness is t, then r_o = r_i + t. Then ln (r_o/r_i) = ln (1 + t/r_i), and for very small x, ln (1 + x) is about x. So the temperature rise becomes (Q * t ) / (2 * pi * r_i * k). Now 2 * pi * r_i is just the area of this thin cylindrical shell (for a unit length), so that's (Q * t) / (A * k), which is familiar.

My naive calculation several days ago was way off, because 2" of insulation around a #14 wire is the opposite of a thin cylindrical shell. To reprise, 20 amps through a #14 wire generates 0.172 BTUs/hr/ft. Say the wire is is surrounded by R-10 insulation in the form of 2" of R-5/inch foam. Then r_i = 0.032, r_o = 2.032, and k = 1/60. The cylindrical heat flow equation gives a temperature rise of 0.172 * ln (2.032/.032) * 60 / 2pi = 6.8 degrees F. That is in stark contrast to the 103 degree F temperature rise for a wall element!

The effect is so large because 0.032 << 2. For a larger wire, the effect is much less but still significant. E.g., for a 250 kcmil wire (diameter 0.5 in), the equivalent heat flux through its surface area equivalent to the example above would be 0.172 * (0.5 / 0.064) = 1.34 BTUs/hr/ft. If it is surrounded by 2" of the R-5/in foam, then r_i = 0.25, r_o = 2.25, k = 1/60, and the temperature rise is 28 degrees F.

Cheers, Wayne

 

[FionaZuppa]

that last term is very dominant, i dont think we can leave it out. its about 4x bigger than the middle term (for h=8).

 

[wwhitney]

but i am working with this equation (will be adding a heatsink term shortly). need to add in a -f(heatsink)
this was developed around cylinder conductor wrapped by tubular insulation 1ft long.

Heat sinking is not going to show up as an additional term subtracted at the end. The two 'Q' terms above are added together because the thermal resistances are in series: in the applicable model heat has to first transverse the cylindrical insulation and then be convected to the environment to escape. Heat sinking represent an alternate, parallel path for the heat to escape.

The equation above is of the form (delta T) = Q * R1, where R1 is the sum of the two coefficients shown. With an alternate parallel path, it is like resistors in parallel, so the equation will be of the form (delta T) = Q / (1/R1 + 1/R2). Here R2 is the thermal resistance of the heat sink path.

Cheers, Wayne

 

[wwhitney]

that last term is very dominant, i dont think we can leave it out. its about 4x bigger than the middle term (for h=8).

Right, now that you mention it, that is clearly true, as you are measuring temperature rises of much more than 6.8 degrees F. I left it out of my discussion not because I didn't think it was important, but simply because I don't have anything to say about it right now. It is presumably convection dominated, and convection is not something I've read up on.

Cheers, Wayne

 

[FionaZuppa]

My naive calculation several days ago was way off, because 2" of insulation around a #14 wire is the opposite of a thin cylindrical shell. To reprise, 20 amps through a #14 wire generates 0.172 BTUs/hr/ft.

Cheers, Wayne

.172 btu/hr-ft ??

W=20^2 * 0.00259 = 1.036W = 1.036J/sec = 3730J/hr = 3.73 kJ/hr = 3.535Btu/hr-ft
1 kJ = 0.9478 Btu

and for clarity, the equation being used is modeled after a wire in a tubular insulator where the only path(s) for dissipation is infinite # of degrees 0-360 (radials), the ends of the section in study are considered to be infinite insulator. once we allow heat to flow outside the insulation via the Cu conductive path thats when we add in -f(heatsink) because some heat will go that way, thus reducing radial dissipation and reducing the temp in center of wire. how much so is TBD.

post #549, ah, but recall that k=1/R, we have written the equation using k, not R.

 

[FionaZuppa]

post #549, ah, but recall that k=1/R, we have written the equation using k, not R.

and to my own fault, yes, the equation is actually using 1/k. but noted, the terms reduce to degrees F, not R, thus if we add another term for the Q that goes left instead of radial we would need to subtract out some degrees, or, the other terms are modified in terms of R.

 

[mbrooke]

let me address the other two questions, heatsink is linear? and induction modes?

yes, the heatsink is linear regardless of temp, heatsink mode is governed by the physicals and properties of the material. delta-T only tells us about the rate, which is linear with delta-T.

Got it now

as for inductive modes. the 60Hz mag and electric fields do couple to surrounding items, the effects are more noticeable in metals that contain iron. but, for conduit, diversity of the amps i think would be a significant factor, and for a NM "bundle" the diversity is usually very low. am i off base here? (hah, maybe go to aluminum conduit for bundles that have more than 4 CCC's).

Off base. Google "zero sequence currents" and "current cancellation" In a properly wired systems these values are very low having negligible effects on surrounding ferromagnetic items, but during a wiring error create high net currents increasing inductive heating substantially.

(You might have another experiment on your hands :lol:)

But for NM this can be ignored.

 

[mbrooke]

have a read here
http://www.roymech.co.uk/Related/Thermos/Thermos_insulation.html

but i am working with this equation (will be adding a heatsink term shortly). need to add in a -f(heatsink)
this was developed around cylinder conductor wrapped by tubular insulation 1ft long.

all items are in units of ft hr btu F (so be ready to have your conversion skills abused)
a couple of tricky things to know
Q and k are normalized to 1ft
h is a coefficient to the boundary condition at the insulation to air interface
Tw is wire temp
T0 is ambient
ro is radius of insulation to centerline
ri is radius of wire to centerline

i approximated the k value of insulation to be 0.17W/mK given that the nylon layer is approx 10x thinner than the pvc layer
h is very tricky #, it is sensitive to environmentals. from test data we found h=8btu/hr-ft-F

the equation says Tw=86.7F, measurement says 83F (for the "long wire" #14 @26amps). the diff is acceptable given the use of a metal TC probe.

Ah, that's the equation for cylindrical heat flow that I never got around to posting about. A few comments:

That last term with the h in the denominator represents how well the outer surface of the cylinder dissipates heat to the environment. So h takes into account convection, etc. If you just want the heat flow across the cylindrical shell, then you can omit the last term, which I will do for the rest of my comments.

If the cylindrical shell is very thin, this equation reduces to the usual heat flow equation for a wall element, as follows: if the thickness is t, then r_o = r_i + t. Then ln (r_o/r_i) = ln (1 + t/r_i), and for very small x, ln (1 + x) is about x. So the temperature rise becomes (Q * t ) / (2 * pi * r_i * k). Now 2 * pi * r_i is just the area of this thin cylindrical shell (for a unit length), so that's (Q * t) / (A * k), which is familiar.

My naive calculation several days ago was way off, because 2" of insulation around a #14 wire is the opposite of a thin cylindrical shell. To reprise, 20 amps through a #14 wire generates 0.172 BTUs/hr/ft. Say the wire is is surrounded by R-10 insulation in the form of 2" of R-5/inch foam. Then r_i = 0.032, r_o = 2.032, and k = 1/60. The cylindrical heat flow equation gives a temperature rise of 0.172 * ln (2.032/.032) * 60 / 2pi = 6.8 degrees F. That is in stark contrast to the 103 degree F temperature rise for a wall element!

The effect is so large because 0.032 << 2. For a larger wire, the effect is much less but still significant. E.g., for a 250 kcmil wire (diameter 0.5 in), the equivalent heat flux through its surface area equivalent to the example above would be 0.172 * (0.5 / 0.064) = 1.34 BTUs/hr/ft. If it is surrounded by 2" of the R-5/in foam, then r_i = 0.25, r_o = 2.25, k = 1/60, and the temperature rise is 28 degrees F.

Cheers, Wayne

IMO the cylindrical model is only good for rough worse case estimates, especially in fiberglass, but your math eloquently spelled out.

In the end, doesn't the actual surface area on the other side also play a role (if any)?

 

[FionaZuppa]

IMO the cylindrical model is only good for rough worse case estimates, especially in fiberglass, but your math eloquently spelled out.

In the end, doesn't the actual surface area on the other side also play a role (if any)?

it should be good for any annular insulating study regardless of the insualting material.

when it comes to a flat sandwich which almost all wire in insualtion will look like, the annular equation will be a tad lenient on temp because there is infinite # of radials of varying length in a flat sandwich, this if you use the shortest ro. we could do some analysis to see if choosing a ro that is somewhere between r0(min) and ro(max), perhaps ro(max)-ro(min) ?? this yield a ro value that is more than ro(min) but less than ro(max). this is a simple triangle problem, etc. a flat sandwich doesnt fit the annular equation so well, but i think we can use an adjusted value that helps us use the equation to get approximate answers.

 

[FionaZuppa]

so, the diff or avg of min & max radii may not be the good choice, rather just find the r of a circle that yields same area as the rectangle.

 

[wwhitney]

.172 btu/hr-ft ??

[. . .] 3.535Btu/hr-ft

Mea culpa! In post 531, I used I * R for the power instead of I^2 * R. That was probably the simplest part of the calculation, too. Thanks for catching that.

So post 531 is off by a factor of 20, i.e. .172 * 20 = 3.44 BTU/hr-ft is the correct answer there. I used a slightly different value of resistance of copper there. The fact we are not correcting the copper resistance for temperature is a source of error in our calculations; your value is slightly higher than mine and so probably closer to reality.

Basically all my heat dissipations since post 531 are off by a factor of 20, and so all my temperature rises are also off by a factor of 20. I didn't catch that in post 531, because the cylindrical correction is 103/6.8 = a factor of 15, mostly cancelling my initial error.

The upshot is that in post 547, the answer should be 136 degrees F expected heat rise for a #14 conductor carrying 20 amps surrounded by 2" of R-5/inch insulation.

post #549, ah, but recall that k=1/R, we have written the equation using k, not R.

We can do that, it just moves the complexity in the formula around. The temperature rise equation becomes:

(delta T) = Q/(k1 + k2)

Where k1 is from post 545, and k2 is for the heat sink. The formula for k1 from post 545 is:

1/k1 = ln(r_o/r_i) / (2*pi*k) + 1 /(2*pi*r_o*h)

Anyway, the point is that heat sinking does not show up as an extra additive term in the equation from post 545.

Cheers, Wayne

 

[wwhitney]

I think if you want to accurately model the flat sandwich you just have to solve the differential equations for heat. Or find some who has done it for you already. For an order of magnitude estimate (somewhat underestimate) of heat rise, the cylindrical equation seems good enough to me. (Barring any factor of 20 errors.

It would be interesting to see if the temperature rise is different for the sandwich being flat (horizontal) versus on edge (vertical). I have the impression that convection is better at transferring heat vertically than horizontally, in which case the temperature rise would be greater when the sandwich is on edge.

Cheers, Wayne

 

[wwhitney]

that last term is very dominant, i dont think we can leave it out. its about 4x bigger than the middle term (for h=8).

Now that I corrected my factor of 20 error, here's what I think:

For the bare wire open air test, the term with 'h' is the only term, so you can use the bare wire tests to estimate 'h', once the heat sinking issue is resolved.

For an insulated wire open air test, I expect that the term with 'h' will dominate, as the electrical insulation on the wire is thin and its term won't contribute much.

For the tests of wire in foam, the foam thermal insulation will dominate.

Cheers, Wayne

 

[FionaZuppa]

For an insulated wire open air test, I expect that the term with 'h' will dominate, as the electrical insulation on the wire is thin and its term won't contribute much.

Cheers, Wayne

analysis is good, but it all depends on the k values of the insulation(s) involved
but in this land, pvc is dominant.