Physical Properties of Fluids

Density

The density of a fluid, $\rho$, is defined by mass per unit volume:

$$\rho =\frac{m}{\forall }\equiv \frac{M}{L^3}\left[\frac{\text{kg}}{{\text{m}}^3}\right]$$

In standard conditions:

• ${\rho }_{\text{water}}=1000\text{ kg}/{\text{m}}^3$
• ${\rho }_{\text{air}}=1.230\text{ kg}/{\text{m}}^3$

Specific volume

The specific volume is the volume per unit mass and is defined as:

$$v=\frac{1}{\rho }=\frac{\forall }{m}$$

such that $\rho v=1$.

Specific weight

The specific weight of a fluid is defined as the weight per unit volume:

$$\gamma =\frac{mg}{\forall }=\rho g[\text{N}/{\text{m}}^3]$$

where $g$ is the local acceleration by gravity, $g=9.81\frac{\text{m}}{{\text{s}}^2}$.

Specific gravity

The specific gravity of a fluid, $SG$, is defined as the ratio of the density of the fluid to the density of water at some specified temperature, usually $4°\text{C}$. At this point, ${\rho }_{{\text{H}}_2\text{O}@4°\text{C}}=1000\text{ kg}/{\text{m}}^3$.

$$SG=\frac{\rho }{{\rho }_{{\text{H}}_2\text{O}@4°\text{C}}}$$

Since this is a ratio of quantities, it is dimensionless.

We can use specific gravity to calculate the density of a substance. For example, given $S{G}_{\text{Hg}}=13.55$, we can calculate

$${\rho }_{\text{Hg}}=13.55\cdot {\rho }_{{\text{H}}_2\text{O}@4°\text{C}}=1.355\times 10^4\text{ kg}/{\text{m}}^3$$