CivilSolve

Fluid Mechanics & Hydraulics

Fluid Properties (ρ, γ, SG)

Converts between mass density ρ, unit weight γ = ρ·g, and specific gravity SG = ρ/ρ_water — give any one and get the others. Optionally: the mass and weight of a given volume, and the gage pressure p = γ·h at a given depth. Use for: 'the specific gravity of the oil is 0.85, find its unit weight', 'what does 2 m³ of this fluid weigh', quick p = γh checks. Not for: viscosity or temperature-dependent property lookups.

Direct solving is free and needs no account. Have a word problem instead? Submit it as text.

Inputs

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Worked examples

These are the solver's own reference problems — the answers come from published sources or independent hand computation, never from the solver itself. The solution below is the live solver output for each.

Example 1

oil SG = 0.85: density, unit weight, weight of 2 m³

Given

specific gravity
0.85
volume
2 m^3

Assumptions

  • SG referenced to water at ρ = 1000 kg/m³; g = 9.80665 m/s².

Solution steps

  1. Relate the three properties

    Starting from the given specific gravity: unit weight is density times gravity, and SG compares the density to water.

    γ = ρ·g; SG = ρ/ρ_water

    ρ = 850 kg/m^3; γ = 8.336 kN/m^3; SG = 0.85

  2. Mass and weight of the volume

    Scale the properties by the volume.

    m = ρ·V; W = γ·V

    m = 1700 kg; W = 16.67 kN

Results

Mass density ρ

850kg/m^3

Unit weight γ

8.336kN/m^3

Specific gravity SG

0.85

Mass of the volume

1700kg

Weight of the volume

16.67kN

Where this answer was checked
source
ρ = SG·ρw; γ = ρ·g; W = γ·V — property chain from SG
verified by
hand-recomputed
derivation
ρ = 850 kg/m³; γ = 850·9.80665 = 8335.7 N/m³ = 8.336 kN/m³; W = 8335.7·2 = 16.67 kN.

Example 2

US: γ = 62.4 pcf is (almost exactly) water

Given

unit weight
62.4 pcf

Assumptions

  • SG referenced to water at ρ = 1000 kg/m³; g = 9.80665 m/s².

Solution steps

  1. Relate the three properties

    Starting from the given unit weight: unit weight is density times gravity, and SG compares the density to water.

    γ = ρ·g; SG = ρ/ρ_water

    ρ = 999.6 kg/m^3; γ = 62.4 pcf; SG = 0.9996

Results

Mass density ρ

1.939slug/ft^3

Unit weight γ

62.4pcf

Specific gravity SG

0.9996

Where this answer was checked
source
ρ = γ/g; SG = ρ/1000 — recover SG from unit weight
verified by
hand-recomputed
derivation
62.4 pcf = 9802.3 N/m³; ρ = 9802.3/9.80665 = 999.56 kg/m³; SG = 0.9996.

Example 3

mercury column: 0.76 m of SG 13.6 fluid ≈ one atmosphere

Given

density
13600 kg/m^3
depth
0.76 m

Assumptions

  • SG referenced to water at ρ = 1000 kg/m³; g = 9.80665 m/s².

Solution steps

  1. Relate the three properties

    Starting from the given density: unit weight is density times gravity, and SG compares the density to water.

    γ = ρ·g; SG = ρ/ρ_water

    ρ = 13600 kg/m^3; γ = 133.4 kN/m^3; SG = 13.6

  2. Gage pressure at depth

    Hydrostatic pressure grows linearly with depth.

    p = γ·h

    p = 101.4 kPa

    = 101.4 kPa

Results

Mass density ρ

13600kg/m^3

Unit weight γ

133.4kN/m^3

Specific gravity SG

13.6

Gage pressure at the given depth

101.4kPa

Where this answer was checked
source
p = γ·h (the barometer identity) — pressure at depth in mercury
verified by
hand-recomputed
derivation
ρ = 13600 kg/m³; γ = 133.37 kN/m³; p = 133370·0.76 = 101.36 kPa — the classic 760 mmHg atmosphere.