Density and Pressure

Density and Pressure

The density ρ of a substance is defined as the mass per unit volume of a sample of the substance.

If a small mass element Δm occupies a volume ΔV, the density is given by

ρ = Δm/ΔV

In general, the density of an object depends on position, so that

ρ = f(x, y, z)

If the object is homogeneous, its physical parameters do not change with position throughout its volume. Thus, for a homogeneous object of mass M and volume V, the density is defined as

ρ = M/V   (12.11)

The SI units of density are kg m^-3.

Specific Gravity

The specific gravity of a substance is the ratio of its density to that of water at 4^oC, which is 1000 kg/m3. Specific gravity is a dimensionless quantity numerically equal to the density quoted in g/cm3. For example, the specific gravity of mercury is 13.6, and the specific gravity of water at 100oC is 0.998.

Example: 12.4

Find the density and specific gravity of gasoline if 51 g occupies 75 cm³?

Solution

Density = mass/volume =  = 680 kg/m³

Sp.gr = = 0.68

orSp. gravity =

Example: 12.5

The mass of a liter of milk is 1.032 kg. The butterfat that it contains has a density of
865 kg/m3 when pure, and it constitutes 4 percent of the milk by volume. What is the density of the fat-free skimmed milk?

Solution

Volume of fat in 1000 cm³ of milk = 4% × 1000 cm³ = 40 cm³

Mass of 40cm³ fat = vρ = (40 × 10^-6 m3)(865 kg/m³) = 0.0346 kg

Density of skimmed milk = mass/volume  = = 1039 kg/m3

Pressure

As ΔA → 0, the element reduces to a point, and thus, pressure at a point is defined as

p =     (12.12)

When the force is constant over the surface A, the above equation reduces to

p = F/A   (12.13)

The SI unit of pressure Nm-2 and is also called pascal (Pa).

The other common pressure unit are the atmosphere and bar.

1 atm = 1.01325 × 105 Pa

1 bar = 1.00000 × 105 Pa

Example: 12.6

Atmospheric pressure is about 1.01 × 10⁵ Pa. How large a force does the atmosphere exert on a 2 cm2 area on the top of your head?

Solution

Because p = F/A, where F is perpendicular to A, we have F = pA. Assuming that 2 cm² of your head is flat (nearly correct) and that the force due to the atmosphere is perpendicular to the surface (as it is), we have

F = pA = (1.01 × 10⁵ N/m²)(2 × 10^-4m²) ≈ 20 N

Pressure is Isotropic

Since the fluid cannot support a shear stress, the force exerted by a fluid pressure must also be perpendicular to the surface of the container that holds it.