A problem in my thermodynamics text reads as follows...

The barometer of a mountain hiker reads $13.8$ $psia$ at the beginning of a hiking trip and $12.6$ $psia$ at the end. Neglecting the effect of altitude on local gravitational acceleration, determine the vertical distance climbed. Assume an average air density of $.074\cdot\frac{lbm}{ft^3}$ and take $g = 31.8\cdot\frac{ft}{s^2}$.

My Solution:

Our solution begins by solving for the change in air pressure and using this to determine elevation climbed...

$$\Delta P=P_1-P_2$$



$$\Delta P=\rho g h$$

rewrite as...

$$h=\frac{\Delta P}{\rho g}$$

But first we needed to convert $ft$ units to $in$ since that is how we traditionally define pressures in the English system...



We can now solve for $h$...

$$h=\frac{\Delta P}{\rho g}=\frac{1.2\cdot lbf}{1.33\times10^{-6}\frac{slugs}{in^3}\cdot381.6\frac{in}{s^2}}=2364\frac{lbf\cdot in^2\cdot s^2}{slugs}$$

Answer in text:

$$h=2363\cdot ft$$


The scalar values are almost identical but the units just don't fit. Where did I go wrong?

  • $\begingroup$ The most likely conclusion is that there is a storm coming. Methodology aside, air pressure is OK as an altitude measure when all else fails but horrendously weather affected. $\endgroup$ – Russell McMahon Apr 5 '15 at 8:05

I'm glad we got rid of that obsolete system of obscure units decades ago. The metric system is so much easier.

From the definition of the Slug,

${1 Slug = 1 {lb}_f.{s}^2/ft}$.

Substituting that into the units you got, the ${slugs}$, ${lb}_f$ & ${s}^2$ go & ${ft}$ comes in.

The only mistake you made was to exclude ${in}^2$ in your final calculation.

Pressure in US units is pounds per square inch, not pounds as you had it. Correct for this and you end up with units of feet in the final calculation.

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