# Determine elevation from the difference in air pressure

Given:

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...

$$\Delta P=P_1-P_2$$

$$=(13.8-12.6)\frac{lbf}{in^2}=1.2\frac{lbf}{in^2}$$

and...

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

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...

$$\rho=.074\frac{lbm}{ft^3}\times\frac{ft^3}{(12in^3)}\times.03108\frac{slugs}{lbm}=1.33\times10^{-6}\frac{slugs}{in^3}$$

$$g=31.8\frac{ft}{s^2}\times\frac{12in}{ft}=381.6\frac{in}{s^2}$$

We can now solve for $h$...

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

$$h=2363\cdot ft$$

Question:

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

• 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. Apr 5, 2015 at 8:05

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