The insulation that makes a wire lose more heat

An electrician sizing a cable consults a table that gives, for each conductor size and each way of installing it, the largest current it can carry without overheating its insulation. A curious feature of those tables, for thin conductors, is that a wire with its plastic sheath on can carry more current than the same wire would if the sheath were stripped off and it were strung bare in the same air. The sheath, which is a thermal insulator, helps the wire stay cool. A hot water pipe wrapped in the same plastic would lose less heat, as everyone expects. The difference between the two is a matter of size, and it can be found exactly.

The argument is about where heat goes once it leaves a hot surface, and it rests on the same arithmetic as the ice that grows more slowly the thicker it gets: a layer of material between a hot side and a cold side carries heat in proportion to the temperature difference and inversely to a thermal resistance, and layers in series add their resistances. What makes a cylinder different from a flat wall is that its outer surface grows as the layer thickens.

Two resistances that pull opposite ways #

Heat leaving a hot wire of radius $r_i$ through insulation out to radius $r_o$ meets two resistances in series. The first is conduction through the insulation. For a cylindrical shell it is

$$R_{\text{cond}} = \frac{\ln(r_o/r_i)}{2\pi k}$$

per metre of length, where $k$ is the insulation’s thermal conductivity. It grows as the insulation thickens, but only logarithmically, because each extra millimetre is laid down round a larger circumference and so is a wider, easier path than the millimetre before. The second is the transfer from the insulation’s outer surface into the air, by convection and radiation together, described by a surface coefficient $h$:

$$R_{\text{surf}} = \frac{1}{2\pi h\,r_o}.$$

It falls as the insulation thickens, in proportion to the outer radius, because a fatter cylinder has more surface for the heat to leave from.

The sum of a logarithm that rises and an inverse that falls has a minimum, where their slopes cancel: $1/(2\pi k r_o) = 1/(2\pi h r_o^2)$, which gives

$$r_c = \frac{k}{h}.$$

That is the critical radius. A cylinder smaller than it loses more heat when insulated, because the falling surface resistance outweighs the rising conduction resistance; one larger than it loses less. The radius depends only on the insulation and on how readily the air takes heat from its surface, not on the size of whatever is being insulated.

A wire that loses four times more when wrapped #

The numbers for a thin wire make the effect hard to miss.

PVC has a conductivity of about 0.19 watts per metre per kelvin, and still air with radiation takes heat from a surface at about 10 watts per square metre per kelvin, so the critical radius is 19 millimetres. A wire one millimetre in radius is far inside it. Bare, it loses 2.5 watts per metre at 40 degrees above the air; wrapped to the critical radius, 12 — nearly five times as much. And past the critical radius the loss falls back so slowly, because the conduction resistance grows only as a logarithm, that it does not return to the bare wire’s value until the PVC is 178 kilometres thick. For every thickness of plastic a cable could ever carry, the insulated wire loses more heat than the bare one.

The flat-wall intuition fails because it treats insulation as only a resistance. A flat wall’s surface does not grow when it is thickened, so every layer adds resistance and nothing else, and more insulation always means less heat lost. A cylinder’s insulation is also a fin: it spreads the heat from a small surface over a larger one, as the fins on a motorcycle engine or a computer’s heat sink do deliberately, out of metal. PVC is a poor fin material, but the air is a poorer conductor of heat away from a surface than the PVC is of heat through it, over the first several millimetres, and that is enough.

The current a sheath lets a wire carry #

For a wire carrying current, the sheath’s effect becomes a rating. The wire generates heat at $I^2\rho/\pi r_i^2$ per metre, and its temperature rises above the air’s by that heat times the total resistance.

Every millimetre of sheath lowers the wire’s temperature, because every millimetre is still inside the critical radius. A 20-amp current that runs a bare 2 mm copper wire at 60 °C runs the sheathed one at 44. Turned around, the current that brings the wire to 70 °C — the temperature at which ordinary PVC begins to soften and age, and the usual limit in the rating tables — rises from 22.7 amps bare to 31 amps with a 1 mm sheath. The electrician’s table records exactly this: thin insulated conductors are rated for more current than the same conductors bare, and the effect disappears for thick conductors and thick insulation, where the cable is larger than the critical radius and the sheath is a genuine thermal penalty that the rating has to allow for.

The bare-wire figure here is, if anything, generous to the bare wire. The surface coefficient includes radiation, and bare copper is a poor radiator — a polished metal surface has an emissivity of a few per cent, against about 0.9 for plastic, the difference the glow that says nothing about the surface traced to how a metal reflects. A real bare copper wire loses less heat than the figure allows, and its sheath helps it more.

Why the pipe is insulated and the wire is not #

A domestic hot-water pipe is a copper tube about 22 millimetres across, 11 in radius. Glass wool’s critical radius in still air is 4 millimetres, so the pipe is already well outside it, and every layer of wool lowers its loss. Bare, it loses about 0.69 watts per metre for every degree it is above the air; with a 20 mm jacket of glass wool, the outer radius 31 mm, the two resistances add to 4.6 kelvin metres per watt and the loss falls to 0.22 — a factor of three. That is the familiar case, and it is the only case most people ever meet, because almost everything that gets insulated deliberately — pipes, tanks, walls, bodies — is far larger than the critical radius of the material used on it.

The wire is the exception because it is small and because its insulation is chosen for a different job. PVC is on the wire to keep the current in and fingers out, and it happens to be a moderate conductor of heat by the standards of insulation, ten times worse an insulator than glass wool. Both facts put the wire inside the critical radius. Had the wire been sheathed in glass wool to keep it warm, the sheath would still have had to reach 4 millimetres before it began to help.

The same arithmetic runs in reverse for cold lines. The thin copper tubes that carry refrigerant to an air-conditioner’s indoor unit are a few millimetres across and much colder than the air, so they gain heat and sweat with condensation. Wrapping them in a thin layer of a moderately conductive material increases the heat they gain, by the same argument, and the foam rubber used on them is chosen thick enough, and of low enough conductivity, to be well past the critical radius. An installer who economises on the thickness can make the sweating worse.

The same equation as charge on a conductor #

The geometry has an exact electrical counterpart. Steady heat conduction obeys Laplace’s equation for the temperature, the same equation the electrostatic potential obeys in empty space, with conductivity in place of permittivity and heat flow in place of electric flux. A sphere held at a temperature $\Delta T$ above a vast surrounding medium of conductivity $k$ conducts heat away at

$$Q = 4\pi k\, r\, \Delta T,$$

exactly as a sphere at potential $V$ holds charge $4\pi\varepsilon_0 r V$ — the capacitance of a sphere, which how much charge a shape will hold found growing with the sphere’s radius rather than its area. The thermal conductance of a body into an infinite medium is its capacitance with a change of constant, and the resistance that is a length found the electrical version of the same fact, the spreading resistance into a large conductor set by a length rather than an area.

That analogy says something the flat-wall picture cannot: there is a minimum to how well any finite thickness of insulation can insulate a small body, because even an infinitely thick coat still conducts. An infinitely thick shell of insulation round a sphere conducts $4\pi k r\Delta T$; the bare sphere loses $4\pi h r^2 \Delta T$ from its surface. If $r < k/h$, the first is larger than the second, and no amount of the insulating material wrapped round the ball will ever bring its loss down to what it was bare.

A wall, a pipe and a ball #

The shape decides all of this, and the three simplest shapes show three different behaviours.

The wall loses less with every layer. The cylinder loses more until $k/h$ and less beyond, eventually returning — very slowly — to below its bare value. The sphere has a critical radius of $2k/h$, twice the cylinder’s, because its surface grows as the square of its radius rather than as the radius; its loss peaks higher, and it never returns at all if the ball is smaller than $k/h$, approaching the conduction of an infinite medium from above. A ball smaller than $k/h$ is uninsulatable by that material.

That has a biological reading that should be made carefully. A warm-blooded animal cannot be smaller than a certain size and stay warm, because heat loss scales with surface and heat production, roughly, with volume — the familiar scaling argument that the size at which a body becomes round made for gravity against strength. Fur helps a mouse because fur is mostly still air, with a conductivity of about 0.03 W/m K, giving a critical radius for a sphere in still air of a few millimetres — well below a mouse. For a body of a few millimetres, a coat would add surface faster than resistance and cool it, but no warm-blooded animal is that small: the smallest shrews and hummingbirds are a couple of centimetres long, and they live at the edge of what their metabolism can sustain for other reasons. The critical radius does not set their size; it is one more reason the small end of warm-blooded life is hard.

Below which radius insulation does not insulate #

The critical radius depends on the material and on the surroundings, and a table of the two answers most practical questions at once.

Glass wool and other fibrous insulation, whose conductivity is close to that of the still air trapped in them — the gas conductivity that the viscosity that does not care how much gas there is found independent of pressure — have critical radii of a few millimetres in still air, so they insulate any pipe thicker than a pencil, which is every pipe that is ever insulated. PVC and rubber, ten times more conductive, have critical radii of a couple of centimetres, larger than most wires. Solid glass, at a watt per metre per kelvin, has a critical radius of ten centimetres, which is one reason glass is used for insulation only as a fibre or a foam: a solid sleeve of glass around a small hot pipe would increase its loss. In a breeze or in flowing water, $h$ is much larger, the critical radius shrinks below a millimetre, and any insulation on anything insulates.

The air round a thin wire is itself an insulator #

There is a deeper version of the same effect hidden in the surface coefficient. For a very thin wire — the tens of micrometres of a hot-wire anemometer or a thermocouple — heat leaves not mainly by the convection of air rising past it but by conduction into the still air immediately round it, which forms a cylinder of slowly moving gas many times the wire’s radius. That cylinder of air is an insulating layer in its own right, with the logarithmic resistance of any cylindrical shell, and it makes the effective surface coefficient of a thin wire grow as the wire gets thinner, roughly as $k_{ ext{air}}/r$. A 25-micrometre wire in still air sheds heat at several hundred watts per square metre per kelvin, tens of times the figure for a pipe.

That is why the thinnest wires respond fastest to changes in the air around them, and why hot-wire anemometers can follow turbulent eddies at tens of kilohertz: a thin wire is in intimate thermal contact with its surroundings, through a layer of air whose resistance barely grows with its size. And it shows that the critical radius is not really a property of the insulation alone. It is a contest between two shells, the one that is put there and the one the air makes, and the metal that feels colder than the wood showed the same kind of contest deciding how a surface feels to a hand: what matters is not the conductivity of either material on its own but how the two resistances in series divide the temperature difference between them.

One surface coefficient for every size of wire #

The surface coefficient is taken as a constant, when it is not. Natural convection from a horizontal cylinder depends on its diameter — a thin wire sheds heat into still air more readily per unit area than a thick one, because the boundary layer around it is relatively thinner — and on the temperature difference itself. Radiation depends on the surface’s emissivity and on the fourth power of its temperature. Both effects shift the critical radius by tens of per cent and make the bare-wire case worse than drawn, as noted above, without changing the shape of any curve.

The insulation is taken as uniform and its conductivity as constant, the wire as a perfect conductor with no temperature difference across it, and the system as steady. A wire switched on heats up over a time set by its heat capacity and the same resistances, and a thick sheath slows that warm-up even when it raises the steady loss. Contact resistance between conductor and sheath, which can be significant for loosely fitted insulation, is left out. And the critical radius is a statement about the surface the insulation adds; insulation that is reflective, like the aluminised films of a space blanket, cuts the radiative part of $h$ and with it the critical radius, which is why such films insulate well even round small objects.

Still open: how small a thing can be kept warm #

The critical radius sets a floor for conventional insulation, and the materials that push the floor down are those with the lowest conductivities: aerogels, at about 0.015 W/m K, and vacuum panels, which remove the gas from the pores entirely and reach a few thousandths. A vacuum flask works because its insulation conducts almost nothing — and because, being evacuated rather than thickened, it adds no outer surface. Below a millimetre, the scale of microelectronics and of thermal sensors, the question becomes different again, because the heat is carried by molecules and phonons whose mean free paths are comparable with the dimensions, and the continuum picture behind $k/h$ starts to fail, as the equation that only runs forwards warned that the diffusion equation must at the scale of its walkers’ steps. How to insulate a structure a micrometre across — a hot nanowire, a single sensing element — is an active engineering problem, and its answers look nothing like a sheath.

What the sheathed wire shows is a general warning about intuitions built on flat walls. Insulation adds a resistance and also a surface; around a curved body smaller than k/h, the surface wins, and adding insulation increases the heat lost — for a cylinder until it is far thicker than any cable, and for a small enough sphere for ever. A pipe is insulated by wrapping it. A wire is cooled by it.