# Tag Archive: pressure drop

## Orifice flow calculation

Calculation flow trough orifice

Here is an online calculator of the flow through the orifice. Using this chart you can determine both flow through existing orifice or estimate the orifice diameter for required flow.

## Conduit calculations (pressure losses, velocity)

Pressure losses in the conduit

These calculations can help you to estimate the pressure losses, and flow velocity in the conduit (hose, pipe or tube) and check/correct conduit ID.

Calculations Notes:

• The recommended flow velocity in conduits you can find at the article Recommended flow velocity.
• The assumption: pressure losses on elbows, fittings at calculated conduit is zero.
• Height difference between IN and OUT points needs to calculate hydrostatic pressure what will be added to pressure losses, if IN point is below OUT (use positive value), or subtracted from pressure losses, if IN point is upper than OUT (use negative value). Use value “0” if height difference can be neglected.
• To find Darcy friction factor there are different formulas used:

## 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.