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Pressure in compressible fluids

In working out the pressure at a given depth in water we assume that the water is incompressible that is its density is constant with depth. This is very nearly accurate. However this would not be true of a fluid such as air. In the atmosphere the pressure increases with depth but the density of the air also varies from 1.2 kgm-3 at sea level to virtually zero at the edge of space.

Let the change in pressure due to a small change of height Δh be Δp

Take the density at a height h to be ρ and the density at the height h+Δh to be ρ-Δρ.

If the temperature (T) of the air is constant ρ is directly proportional to the pressure (p) and so:

p/po = ρ/ρo where po and ρo are the pressure and density of air at sea level.

Therefore Δp/Δh = - rg = -gρo[p/po]

When this is integrated we have:

Pressure (p) at a height h = poe-g[ρo/po]h

Density (ρ) at a height h = ρoe-g[ρo/po]h

Substituting for the accepted values of g, po and ro we have:

Pressure (p) at a height h = poe-0.116h

(N.B the height h should be given in kilometres here)

Example problem
1. Calculate the pressure of the air at a height of 10 000 m (10 km) (the height at which many intercontinental jets fly). Take the pressure at sea level to be 105 Pa.

Pressure (p) = 105e-0.116h
Pressure = 105xe-1.16 = 3.13x104 Pa or about a third of that at sea level.

2. Calculate the distance below the surface of the Earth at which a piece of iron (density 7870kgm-3) will float in air if the density of air at sea level is 1.3 kgm-3.

7870 = 1.3e-0.116h
Therefore 6053 = e-0.116h giving h = 75 km!

The temperature of the atmosphere also decreases with height.
The standard variation being a drop of 6 oC for every kilometre above the ground up to 11 km. After that it remains constant at around 55 oC.

© Keith Gibbs 2020