# Determine abs pressure in a tank where the atm and vac pressures are given

Given:

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

A vacuum gauge connected to a tank reads $5.4\cdot psi$ at a location where the barometric reading $P_{Hg}= 28.5\cdot in$. Determine the absolute pressure in the tank. Take $\rho_{Hg} = 848.4\frac{lbm}{ft^3}$.

My Solution:

The formula for a vacuum gauge is as follows...

$$P_{vac}=P_{atm}-P_{abs}$$

Rewrite as...

$$P_{abs}=P_{atm}-P_{vac}$$

Determine atmospheric pressure but first convert $ft$ units to $in$ since that is how we traditionally define pressures in English units...

$$P_{atm}=\rho g h$$

$$\rho=848.4\cdot\frac{lbm}{ft^3}\times\frac{ft^3}{(12in)^3}=.4910\cdot\frac{lbm}{in^3}$$

$$g=32.174\cdot\frac{ft}{s^2}\times\frac{12in}{ft}=386.0\cdot\frac{in}{s^2}$$

$$P_{atm}=.4910\cdot\frac{lbm}{in^3}\times386.0\cdot\frac{in}{s^2}\times28.5\cdot in=5402\cdot\frac{lbm}{in\cdot s^2}$$

Question:

The atmospheric pressure obtained cannot be right because the scalar value $5402$ is no where near $14.7$ and the units obtained ($\frac{lbm}{in\cdot s^2}$) look completely wrong. Shouldn't they be $\frac{lbf}{in^2}$? Where did I go wrong? After this is solved I know how to obtain the absolute pressure in the tank so I shall stop here.

$\rho = 26.3 \ slugs/ft^3$
$\rho g = \gamma= 847\ lb_f/ft^3$ "specific weight"
$ft^3 = (12 in)^3$
$P = \rho g h = \frac{847 lb_f}{(12 in)^3}\cdot 25.8 in$