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Hydraulic oil density and Specific Gravity

July 5, 2023 0 Articles , , 2350 Views

Hydraulic oil Density is the ratio of its mass to the volume of space it occupies:

## \rho=\frac{m}{V} ##

The accepted units of measurement for density according to ASTM are kilograms per cubic meter (kg/m3, SI unit) or grams per milliliter (g/mL).

Hydraulic oil Specific Gravity (or “Relative Density”) is the ratio of Hydraulic oil density to water density at the specific temperature:

## SG_{oil}=\frac{\rho_{oil}}{\rho_{water}} ##

Substance with SG <1 will float on water (like hydraulic oil); substance with SG >1 will sink in water (like honey).

Next, both hydraulic oil density and Specific Gravity vary with temperature and pressure.

Reference: [https://www.internetchemistry.com/chemical-data/water-density-table.php]
Reference: [https://webbook.nist.gov/chemistry/fluid/]

This is why the ASTM D1298-12b “Standard Test Method for Density, Relative Density, or API Gravity of Crude Petroleum and Liquid Petroleum Products by Hydrometer Method” states that accurate determination of the API gravity, density or relative density (specific gravity) uses a standard temperature of 60 degrees F (15 degrees C).

Reference: [https://www.astm.org/d1298-12b.html]

You can find in the oil specification table of every hydraulic oil manufacture either Oil Specific Gravity or Oil Density at the specific temperature of 60°F or 15°C.

As a result, to calculate Specific Gravity or Hydraulic oil density at 15°C and, respectively, Hydraulic oil density or Specific Gravity at 15°C. By ASTM D1298-12b water density at 15°C is 0.999103 g/ml, therefore:

## SG_{oil}=\frac{\rho_{oil}}{0.999103} ##

## \rho_{oil}=\frac{SG_{oil}}{0.999103} ##

Hydrostatic Pressure

May 20, 2017 0 Articles , , 329 Views

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:

#\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#)

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.

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