WireAmpacity

Safe wire current carrying capacity is normally easily found on National Electric Code (United States) tables that are easy to find. The problem is that the smallest wire on them is normally about 14 gauge or about 15A. These tables are intended for wiring buildings and industrial equipment. But what about a 6A heater on a hotplate? Is 14 gauge wire required for that? It is bigger than the pins on the TO220 triac that controls it. Can I use 24 gauge hookup wire for a 3A load?

Disclaimer:

Even assuming there are no errors in this analysis, this involves heat transfer, which is complex outside the lab. The environmental conditions an object is in are rarely well known. That means there is a large amount of uncertainty in any calculated result in the real world. In short, use at your own risk.

Assumptions:

While usually ignored in circuit design, wires do have a small amount of resistance. That means when wires carry current, the wires are heated by the I^2 R losses. Ignoring heat loss/gain by conduction to the components or terminals at the ends of the wire, that means the heat can only escape radially through the insulation into the surrounding environment.

Heat Transfer Basics:

Conduction:

The rate of heat transfer by conduction through a solid, q, is proportional to the temperature difference across it. As an example, the heat transfer rate by conduction across a flat wall is

q = - k A (T2 - T1) / L

where

q = heat transfer rate

A = area of the wall

T2, and T1 = surface temperatures on each side of the wall

L = wall thickness.

k = thermal conductivity. It is a property of the material.

The negative sign is there to keep it consistent with the convention that heat flows in the direction of decreasing temperature. While this discussion will probably play fast and loose with the sign convention, it is important to remember that heat flows “downhill.”

The equation for conduction out of a hollow tube, like a tube of PVC insulation surrounding a wire is

q = 2 pi L (T2 - T1) / ln (r1/r2)

Convection:

For very thick walls with low thermal conductivity, the surface temperatures are very close to the ambient temperatures. However, touching a window on a cold winter day shows this is not always the case. In those cases the air next to the wall/window is heated or cooled and its temperature changes accordingly. That produces a temperature profile that looks something like what is shown below.

The heat transfer across this region is usually described by q = -h A (Ta - Ts) where h is the heat transfer coefficient and Ta and Ts are the surface and ambient temperatures. So for the winter window example, heat has to travel from the indoor ambient air, through the inside convection region, through the glass, and then through the convection region on the outdoor side, to the outdoor ambient.

But what is the value of h? This depends on the conditions and is almost always determined experimentally. There are whole book chapters, if not whole books written on that. For many common and relatively simple situations, such as flat walls or horizontal cylinders, such as wires, the experiments were done and results reported in those books.

Radiation:

The third form of heat transfer is radiation. Any object whose temperature is above absolute zero, radiates energy by emitting electromagnetic radiation. However, if it is sitting in a room, everything in the room, including the walls, is also radiating energy. That means the object of interest is also absorbing electromagnetic energy from the other objects in the room. In this most basic situation the net energy transferred is q = sig A (Tamb ^4 - Tobj ^4).

Our Problem:

In the case of an insulated wire, the heat generated by I^2 R losses must be conducted from the inside surface of the insulation layer to the outside surface. Naturally the inside surface of the insulation will be hotter than the outside surface, and this is the temperature we are trying to keep less than or equal to our insulation temperature limit. The equation of interest is

q = -2 pi k (Ti - T0) / ln(r0/ri)

Once the heat reaches the outside surface of the wire, it has to continue on its journey by radiation and convection. That also means that qcond = qrad + qconv.

The applicable, good enough formula, for qrad is

q = sig A (Ts^4 - To^4)

where

Ts is the surface temperature (on an absolute scale)

To is the “outside” temperature or ambient (on an absolute scale)

A is the surface area of the wire

sig is a constant (normally represented as Greek letter sigma).

The equation for qconv is

qconv = h A (Ts - To)

and one of the experimentally determined formulas for h mentioned above says

h = 1.3 ((Ts - To)/do) ^ 0.25

where h is in W/m2

and temperatures are in degrees C

and do is the outside wire diameter.

(see Ref 1)

This is a simplified expression for natural convection for horizontal cylinders in air.

Therefore q = sig A (Ts^4 - To^4) + h A (Ts - To)

This is good news. There are three equations and three unknowns, h, q and Ts. This is has a solution, and a very good mathematician might be able to solve it, but mere mortals have to cheat.

Put

A = pi * do * L

Note: we are assuming L = 1m

h = 1.3 ((Ts - To)/do) ^ 0.25

qs = sig A (Ts^4 - To^4) + h A (Ts - To)

qcond = 2 pi k L (Ti - Ts) / ln( ds / di )

Note: the raio of inner vs outer diameters is the same as the radii.

into a spreadsheet, guess at Ts, until the two values of q match. Spare a moment to think of those who did this sort of thing with slide rules and logarithm tables.

That q is equal to the P in P = I^2 R in the wire, which when rearranged gives I = sqrt(P/R) which is another way of saying I = sqrt(q/R).

Results:

Using these values for 14 Gauge wire

do = 0.0027 m measured

di = 0.00163 m Ref 2 pg. 114

k = 0.16 W / (m K) Ref 3

Ti = 90 C

To = 30 C

resitivity = 2.97 Ohm / ft Ref 2 pg. 114

sig = 5.67E-8 W / (m2 K^4)

gives an ampacity of 34A verses 35A per the NEC (Ref 2).

Using the values for 18 Gauge wire

do = 0.0023 m measured

di = 0.00102 m Ref 2 pg. 114

resitivity = 2.97 Ohm / ft Ref 2 pg. 114

gives 19A verses 18A per the per the NEC (Ref 2).

A calculations spreadsheet is available from the link below.

https://docs.google.com/spreadsheets/d/1tqR6rdOTfC18W9NtsCX-T_ADEaI3_ieH/edit?usp=sharing&ouid=114487456932587964222&rtpof=true&sd=true

References:

1. Holman, J. P. “Heat Transfer” McGraw Hill 1963. Pg 168.

The formula is actually presented there in U.S. customary units there.

2. Glover, Thomas J. “Pocket Ref” Sequoia Publishing 2000. Pg 110

3. Lienhard and Lienhard "A Heat Transfer Textbook" Phlogiston Press, 2020 (version 5.10). Pg 718.

As of this writing, a free copy is available at https://ahtt.mit.edu/