Hydrostatic Pressure

In one application I’m working on I need to supply hydraulic pressure to the unit what placed at the height approx 100 ft. I decided to calculate the pressure losses created by the height only for the pump at the land HPU.

The height difference creates hydrostatic pressure (what is a pressure drop for the pump) what can be calculated from Pascal’s law:

$\triangle p=\rho\cdot g\cdot(h_2-h_1)=\gamma\cdot(h_2-h_1)$

here:

$\begin{array}{l}\rho\;-\;oil\;density\;(slugs/ft^3)\\g\;-\;acceleration\;due\;to\;gravity\;(32.174\;ft/s^2)\\\gamma\;-\;specific\;weight\;of\;the\;oil\;(lb_s/ft^3)\end{array}$

The specific weight (the weight per unit volume of a material) of the mineral oil on Earth is approx.: $53.57\;lb_s/ft^3 = 0.031\;lb_s/in^3$ (at environment temperature 40°C for oil with ISO Grade 46)

So, for 1 ft (for 12 in) of height we get pressure drop in PSI:

$\triangle p=0.031\;\frac{lb_s}{in^3}\cdot12\;in\;=\;0.372\;psi$

Therefore, for 100 ft the pressure drop is 37.2 psi = 2.57 bar only.

The calculations look very easy, but one problem I found in internet: many sources mix density and specific weight. For example, at The Engineering Toolbox you can find value of density of the water $62.4 \;lb_s/in^3$ . But it is not right, because this value is a specific weigh of the water, not a density! The density of the water is:

$\rho_{water}=\frac{\gamma_{water}}g=\frac{62.4\;lb/ft^3}{32.174\;ft/s^2}\;\;=\;1.94\;slugs/ft^3$

It has to be clear: the density is mass per unit volume, this is constant at any places (on Earth, on Moon, on any height., etc.) and it is different from specific weight.

Leave a Comment Cancel reply