Stand a column of fluid up and every layer has to carry the weight of everything above it. That is the entire content of $P = \rho g h$ — read aloud P equals rho g h. $P$ is the gauge pressure in pascals, $\rho$ is the fluid's density in kg/m³, $g$ is 9.81 m/s², and $h$ is the depth below the free surface, in metres. Note what is absent: the width of the tank, its shape, the volume it holds. A thimble of water 10 m deep presses exactly as hard at the bottom as a reservoir 10 m deep. This is the hydrostatic paradox, and it is the reason a slender standpipe can pressurise a whole building.

Read the same relation the other way and it becomes $h = \dfrac{P}{\rho g}$, the pressure head: a pressure re-expressed as the height of fluid that would produce it. For water the exchange rate is worth memorising — one metre of water is 9.81 kPa, so 10 m of head is very nearly one atmosphere. (In the imperial trade the same fact reads 2.31 feet of head per psi, and it is quoted just as often.)

Then there is the question of where zero is. A gauge pressure measures from the local atmosphere, and it is what almost every instrument in a plant room reports — which is why an open tank reads zero at the surface even though the air is pressing on it. Absolute pressure measures from a perfect vacuum: $P_{abs} = P_{gauge} + P_{atm}$, where $P_{atm}$ is the local barometric pressure, 101.3 kPa at sea level. Gauge for pipework and pumps; absolute for anything involving boiling, cavitation or a gas law. Mixing the two is how a perfectly good NPSH calculation gets thrown away.