Say I were calculating the pressure under 1 foot of water, in psi.
I'm thinking the pressure is force per unit area, and should be equal to the force exerted by the column of water standing over 1 square inch. So
$$F = ma~~~~~~~~~~~~~~~~~~$$
$$~~~~~~~~~~~~~~~~~= volume\times density\times g$$
$$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~= 1 inch \times 1 inch \times height(in inches)\times density \times g$$
$$=h\rho g~~~~~~~~~~~~$$
The Pressure under 1 foot of water is $12 inch^3 \times 0.036 pounds/in^3\times 386 inches/s^2$ which gives me $166.75~\text{pound-force per square inch}$. However, various internet references tell me that 1 foot of water exerts $0.43 \text{ psi}$. Eg. Wolfram Alpha
They seem to be leaving out acceleration due to gravity entirely. Why?
Best Answer
As John Rennie explained, in the American Engineering System, force is expressed in lb$_f$, mass in lb$_m$, and acceleration in ft/sec$^2$. This system is not coherent. Hence, a conversion factor other than one must be used in the equation for force; that is, $F=\frac{ma}{g_c}$, where $g_c=32.174 \frac{lbm \cdot ft}{sec^2 \cdot lbf}$ is a constant, known as the gravitational conversion constant.
Derivation of $g_c$
The principle of conservation of units is used to derive $g_c$. We wish to convert $F=ma$ from SI to AES.
First, find the dimension of the "hidden constant" in the given equation. In other words, $1=\frac{F}{ma}$ has dimension force divided by the product of mass and acceleration. Next, consider the units of the hidden constant -- the given units are $\frac{N \cdot s^2}{kg \cdot m}$ and the required units are $\frac{lbf \cdot sec^2}{lbm \cdot ft}$.
Now, convert the given units of the hidden constant to the required units. Thus, $$\left(\frac{1 N \cdot s^2}{kg \cdot m}\right) \left(\frac{0.45359 kg}{lbm}\right) \left(\frac{0.3048 m}{ft}\right)\left(\frac{lbf}{4.4482 N}\right)=\frac{1}{32.174}\frac{lbf \cdot sec^2}{lbm \cdot ft}$$
The hydrostatic pressure is calculated as follows:
$$p=D\frac{(\rho g_c)}{144g} \ ft \frac{ft^2 \ lbm \ ft \ sec^2 \ lbf}{in^2 \ ft^3 \ lbm \ ft \ sec^2}$$
where $p$ is hydrostatic pressure in psig, $D$ is depth (or height) of the fluid in ft, $\rho$ is the fluid density (lbm/ft^3). Note that the density of fresh water is 62.4 lbm/ft^3, $g_c$ is the gravitational constant of acceleration (32.2 ft/sec^2), and $g$ is units conversion to lbf (32.2 $\frac{lbm \ ft/sec^2}{lbf}$)
In AES the fluid density is often express in lbm/gal. There is 7.48 gal/ft^3, therefore $p=0.05194D\rho$.
It is often convenient to express the hydrostatic pressure as a fluid pressure gradient or hydrostatic pressure developed per unit height of fluid. $\nabla p =0.052\rho$, where $\nabla p$ is the hydrostatic pressure gradient (psig/ft). For fresh water the pressure gradient is: $$\nabla p=0.052(8.33)=0.433\ \text{psi/ft}$$