# Thermal conductivity listed in the rather complicated Btu/s-in.²-°F/in - am I converting this right?

I'm currently working through some rather old documentation which gives the thermal conductivity of a material as 9.8 x 10⁻⁶ Btu/s-in.²-°F/in. There are many typical ways to give a thermal conductivity in imperial units, but this does not appear to be one of them - in fact, I cannot find any other instance of thermal conductivity being expressed this way, and even the document I'm working with only uses it a handful of times, none in specifying a known value or constant I could check against another unit of thermal conductivity.

I've worked through the units, and it seems most likely to me, though I'm far from confident in this, that this unit is equivalent to 2943641 watts per meter kelvin, making the given value equal to 28.85 W/mK (which seems like a plausible value).

However, I'm far from confident in my calculation here, in particular about the meaning of the dashes and period in the unit.

Is 1 Btu/s-in.²-°F/in = 2943641 W/mK the correct conversion factor? How would one read this unit aloud - "BTU per second-inch squared-degree farenheit per inch"?

• That looks about as good as giving disc access rates in furlongs/fortnight... Apr 23 '21 at 7:03
• @SolarMike Agreed. The document I'm reading is very particular about using imperial units everywhere, even in places they don't really make much sense. For example, the Boltzmann constant is given as 0.3337 x 10⁻¹⁴ Btu/in.²-s-(°R)⁴ . Apr 23 '21 at 7:19

Thermal conductivity is commonly denoted as $$k$$ or $$\lambda$$. The units $$\mathrm{\frac{BTU}{s\cdot in^2 \cdot \frac{^\circ F}{in}}}$$ you state are perfectly OK, although awkward. I would have preferred to—at least—simplify the inches:

$$\mathrm{\frac{BTU}{s\cdot in \cdot ^\circ\mkern-5mu F}}$$

or, even better, write the expression as:

$$\mathrm{\frac{\frac{BTU}{s}}{ in \cdot ^\circ\mkern-5muF} = \frac{1055\; W}{ in \cdot ^\circ\mkern-5muF}}$$

The conversion to SI is as follows:

$$1\mathrm{\frac{BTU}{s\cdot in^2 \cdot \frac{^\circ F}{in}} = 74767.7 \frac{W}{m\cdot K}}$$

So the value you've got of $$9.8 \times 10^{-6}\mathrm{\frac{BTU}{s\cdot in^2 \cdot \frac{^\circ F}{in}} }$$ is equal to approximately $$0.7327\mathrm{\frac{W}{m\cdot K}}$$.

• The way given is a lot more intuitive, though. It is simply a heat flux (BTU/second)/Area divided by the temperature gradient (degF/inch). Note the area must be normal to the gradient. It's fairly common to see mixed units in these concoctions, like (BTU/s) per sqft divided by degC/cm. But try to match units for meaningful derived quantities like flux and temp gradient. Apr 23 '21 at 20:37