Originally Posted by **josephmatchett** Hi,
Yes, that would work. However, if I take $\displaystyle \frac P {\rho} $, the units drop out to just inches. |

Let's do a little dimensional analysis:

In SI the units of P is N/m^2 (otherwise known as Pascals). Recall that 1 Newton = 1 Kg-m/s^2, so that P has fundamental units of Kg/(m-s^2). Rho (density) has units of Kg/m^3. So:

$\displaystyle \frac P {\rho} = \frac {Kg/(m-s^2)}{Kg/m^3} = (m/s)^2$

In imperial units you need to be careful to note the difference between pound-mass and pound-force. I suspect you made this mistake:

$\displaystyle \frac P {\rho} = \frac {lb/in^2}{lb/in^3} = in$

But pressure is units of pound-force/area, whereas density is in units of pound-mass/volume. Recall that 1 lb-force = 1 lb-mass x 32.2 ft/s^2:

$\displaystyle \frac P {\rho} = \frac {lb_f/in^2}{lb_m/in^3} = \frac {32.2 lb_m ft/(s^2 \ in^2)}{lb_m/in^3}$

which has units of distance^2 per second^2.