Density
When dealing with matter regardless of its state it will have a certain mass. Remember that mass is same regardless of gravity. This means that if you are on the moon a liter of water will have the same mass as a liter of water on earth. To further define the mass of an object it is helpful to know what its density is.
(Eq 1) $ρ=\frac{m}{V}$
$m$ = mass
$V$ = Volume
Changes in pressure do not greatly effect the density of a liquid or a solid, neither will temperature. This is because they are considered incompressible. A gas’s density however is greatly effected by changes in pressure and temperature. This will be studied further in the next article.
For fluid mechanics density normally used instead of mass, since mass is dependent on how much fluid is present. This in turn will allow you to analyze a fluid mechanics problem without know the exact quantity of fluid.
Specific Volume
Sometimes specific volume will be used to related an objects mass to its volume. This, however, is just the reciprocal of density, and actually represents the same thing.
(Eq 2) $ν = \frac{1}{ρ}$
Specific Weight
The last characteristic of a fluid that is dependent on density is the specific weight of a fluid. The specific weight considers how the gravitation constant can effect the weight of a fluid. Recall that weight is determined by multiply the mass of an object by the gravitational constant g. This means that the specific weight is determined by multiply the density by the gravitational constant g.
(Eq 3) $γ=ρg$
Specific Gravity
Finally, the specific gravity is used to compare the density of a fluid to the density of water. This is done by taking the ratio the density of the fluid in relation to water. The resulting ratio will be unit less. However, even though it is unit less, since density isn’t unit less you will need to make sure you are using the same system of units for both fluid before finding the specific gravity.
(Eq 4) $SG = \frac{ρ}{ρ_{H_2O}}$
Calculator
$kg$
$m^3$
$\frac{kg}{m^3}$
Specific Volume: $\frac{m^3}{kg}$
Specific Weight: $\frac{N}{m^3}$
Specific Gravity:
Example
An unknown fluid in a 1 liter beaker has a mass of 3 kg. Determine the density, the specific volume, the specific weight, and the specific gravity of the fluid.
Solution
$1L = .001 m^3$
Determine the fluid density.
$ρ = \frac{3~kg}{.001~m^3} = 3000 \frac{kg}{m^3}$
Determine the fluids specific volume.
$ν = \frac{1}{3000\frac{kg}{m^3}}= 3.33 e – 4 \frac{m^3}{kg}$
Determine the Specific Weight of the fluid.
$γ=3000 \frac{kg}{m^3}(9.81\frac{m}{s^2} = 29430\frac{N}{m^3}$
Determine the specific gravity of the fluid.
$SG = \frac{3000 \frac{kg}{m^3}}{997 \frac{kg}{m^3}} = 3.01$