Irodov 1.314: Water Under Pressure Still Stores Energy

Problem Irodov 1.314

Fresh water sits at a depth $h = 1000$ m below its own free surface. Find the volume density $u$ of the elastic deformation energy stored in the water there, due to its own compression under the local pressure.

A body of fresh water, free surface at the top. A small volume element dV at depth h is compressed uniformly on every side by the local pressure, and it is this isotropic compression that stores the elastic energy density u. — A body of fresh water, free surface at the top. A small volume element $dV$ at depth $h$ is compressed uniformly on every side by the local pressure, and it is this isotropic compression that stores the elastic energy density $u$ .

Given

$h = 1000$ m
fresh water

Find

u

, the volume density of elastic energy

Solution

Step 1The pressure at depth

h

In a fluid of density $\rho$ , the pressure in excess of whatever acts at the free surface grows linearly with depth, supported entirely by the weight of the fluid column above.

p = \rho g h. \tag{1}

The atmosphere pressing on the free surface adds a further $1.0\times10^5$ Pa on top of this, about one percent of the column at $h = 1000$ m; it is neglected throughout, so $p$ from here on is the pressure of the water column alone.

Step 2Elastic energy density stored in compression

Compressibility $\beta$ is defined by the fractional volume decrease produced per unit pressure increase, $d(\Delta V/V) = \beta\,dp'$ for a small additional pressure $dp'$ . Building the pressure up quasi-statically from zero to $p$ , exactly as a spring is stretched by a growing force, the work done per unit volume is the integral of the pressure against the fractional volume it produces.

u = \int_0^p p'\,d\!\left(\frac{\Delta V}{V}\right) = \int_0^p p'\,\beta\,dp' = \beta\left[\frac{p'^2}{2}\right]_0^p.

u = \frac12\beta p^2

(2)

Step 3Substituting the depth pressure and evaluating

The pressure at this depth follows from (1).

p = \rho g h = (1.00\times10^3)(9.8)(1000) = 9.8\times10^6\ \text{Pa}.

Putting that pressure and fresh water’s compressibility into (2) gives the energy density itself.

u = \frac12\beta p^2 = \frac12\times(4.9\times10^{-10})\times(9.8\times10^6)^2 \approx 2.35\times10^4\ \text{J/m}^3.

u = \frac12\beta(\rho g h)^2 \approx 23.5\ \text{kJ/m}^3

(3)

Check

Dimensionally, $[\beta][p]^2 = \text{Pa}^{-1}\cdot\text{Pa}^2 = \text{Pa} = \text{J/m}^3$ , correct for an energy density. And the numbers close on themselves: $\Delta V/V = \beta p = 4.8\times10^{-3}$ , so the water at this depth is squeezed by under half a percent and treating $\beta$ as constant across the whole range really is justified; reading the energy off that compression instead, as one half the pressure times the fractional squeeze, $\frac12(9.8\times10^6)(4.8\times10^{-3}) = 2.35\times10^4$ J/m $^3$ , reproduces the box without integrating anything.

Remark

Fresh water is treated as incompressible almost everywhere in everyday mechanics, and the boxed number shows what that idealization quietly discards: over twenty kilojoules stored in every cubic meter a kilometer down, which per kilogram is as much energy as lifting that kilogram about two and a half meters. The volume change responsible is far too small for the incompressible picture to be worth abandoning, which is exactly the bargain a very stiff spring makes: an imperceptible displacement, bought at a very large force, still stores a serious amount of work.