# 0.4 Stress in fluids  (Page 3/4)

 Page 3 / 4
$\begin{array}{c}\underset{v}{\phantom{\rule{0.277778em}{0ex}}\int \int \int \phantom{\rule{0.277778em}{0ex}}}\rho \mathbf{a}dV=\underset{v}{\phantom{\rule{0.277778em}{0ex}}\int \int \int \phantom{\rule{0.277778em}{0ex}}}\rho \mathbf{f}dV+\underset{s}{\phantom{\rule{0.277778em}{0ex}}\int \int \phantom{\rule{0.277778em}{0ex}}}{\mathbf{t}}_{\left(n\right)}dS\hfill \\ \underset{v}{\phantom{\rule{0.277778em}{0ex}}\int \int \int \phantom{\rule{0.277778em}{0ex}}}\rho \left(\mathbf{x}×\mathbf{a}\right)dV=\underset{v}{\phantom{\rule{0.277778em}{0ex}}\int \int \int \phantom{\rule{0.277778em}{0ex}}}\rho \left(\mathbf{x}×\mathbf{f}\right)dV+\underset{s}{\phantom{\rule{0.277778em}{0ex}}\int \int \phantom{\rule{0.277778em}{0ex}}}\left(\mathbf{x}×{\mathbf{t}}_{\left(n\right)}dS\hfill \end{array}$

The surface integral has as its ${i}^{th}$ component

$\begin{array}{ccc}\hfill \underset{s}{\phantom{\rule{0.277778em}{0ex}}\int \int \phantom{\rule{0.277778em}{0ex}}}\left(\mathbf{x}×{\mathbf{t}}_{\left(n\right)}dS& =& {\epsilon }_{ijk}{x}_{j}{T}_{kp}{n}_{p}dS\hfill \\ & =& \underset{v}{\phantom{\rule{0.277778em}{0ex}}\int \int \int \phantom{\rule{0.277778em}{0ex}}}{\epsilon }_{ijk}{\left({x}_{j}{\mathbf{T}}_{pk}\right)}_{,p}dV\hfill \end{array}$

by Green's theorem. However, since ${x}_{j,p}={\delta }_{jp}$ , this last integrand is

$\begin{array}{ccc}\hfill {\epsilon }_{ijk}{\left({x}_{j}{T}_{pk}\right)}_{,p}& =& {\epsilon }_{ijk}{x}_{j}{T}_{pk,p}+{\epsilon }_{ijk}{T}_{jk}\hfill \\ & \text{or}& \\ & =& \mathbf{x}×\left(\nabla •\mathbf{T}\right)+{\mathbf{T}}_{×}\hfill \end{array}$

where $\mathbf{T}$ is the vector ${\epsilon }_{ijk}\phantom{\rule{0.277778em}{0ex}}{\mathbf{T}}_{jk}$ .

Since $\mathbf{v}×\mathbf{v}=0,\phantom{\rule{0.277778em}{0ex}}d\left(\mathbf{x}×\mathbf{v}\right)/dt=\mathbf{x}×\mathbf{a}$ , applying the transport theorem to the angular momentum we have

$\frac{D}{Dt}\underset{v}{\phantom{\rule{0.277778em}{0ex}}\int \int \int \phantom{\rule{0.277778em}{0ex}}}\rho \left(\mathbf{x}×\mathbf{v}\right)dV=\underset{v}{\phantom{\rule{0.277778em}{0ex}}\int \int \int \phantom{\rule{0.277778em}{0ex}}}\rho \left(\mathbf{x}×\mathbf{a}\right)dV$

Substituting back into the equation for the angular momentum and rearranging gives

$\underset{V}{\phantom{\rule{0.277778em}{0ex}}\int \int \int \phantom{\rule{0.277778em}{0ex}}}\mathbf{v}×\left(\rho \mathbf{a}-\rho \mathbf{f}-\nabla •\mathbf{T}\right)dV=\underset{V}{\phantom{\rule{0.277778em}{0ex}}\int \int \int \phantom{\rule{0.277778em}{0ex}}}{\mathbf{T}}_{×}dV$

However, the left-hand side vanishes for an arbitrary volume and so

${T}_{×}\equiv 0.$

The components of ${\mathbf{T}}_{×}$ are $\left({T}_{23}-{T}_{32}\right)$ , $\left({T}_{31}-{T}_{13}\right)$ , and $\left({T}_{12}-{T}_{21}\right)$ and the vanishing of these implies

${T}_{ij}={T}_{ji}$

so that $\mathbf{T}$ is symmetric for nonpolar fluids.

## Hydrostatic pressure

If the stress system is such that an element of area always experiences a stress normal to itself and this stress is independent of orientation, the stress is called hydrostatic . All fluids at rest exhibit this stress behavior. It implies that $\mathbf{n}•\mathbf{T}$ is always proportional to $\mathbf{n}$ and that the constant of proportionality is independent of $\mathbf{n}$ . Let us write this constant $-p$ , then

$\begin{array}{ccc}\hfill {n}_{i}{T}_{ij}& =& -p{n}_{j},\phantom{\rule{4.pt}{0ex}}\text{hydrostatic}\phantom{\rule{4.pt}{0ex}}\text{stress}\hfill \\ \hfill \mathbf{n}•\mathbf{T}& =& p\mathbf{n}\hfill \end{array}$

However, this equation means that any vector is a characteristic vector or eigenvector of $\mathbf{T}$ . This implies that the hydrostatic stress tensor is spherical or isotropic. Thus

$\begin{array}{ccc}\hfill {T}_{ij}& =& -p{\delta }_{ij},\phantom{\rule{4.pt}{0ex}}\text{hydrostatic}\phantom{\rule{4.pt}{0ex}}\text{stress}\hfill \\ \hfill \mathbf{T}& =& -p\mathbf{I}\hfill \end{array}$

for the state of hydrostatic stress.

For a compressible fluid at rest, $p$ may be identified with the classical thermodynamic pressure. On the assumption that there is local thermodynamic equilibrium even when the fluid is in motion this concept of stress may be retained. For an incompressible fluid the thermodynamic, or more correctly thermostatic, pressure cannot be defined except as the limit of pressure in a sequence of compressible fluids. We shall see later that it has to be taken as an independent dynamical variable.

The stress tensor for a fluid may always be written

$\begin{array}{ccc}\hfill {T}_{ij}& =& p{\delta }_{ij}+{P}_{ij}\hfill \\ \hfill \mathbf{T}& =& p\mathbf{I}+\mathbf{P}\hfill \end{array}$

and ${P}_{ij}$ is called the viscous stress tensor . The viscous stress tensor of a fluid vanishes under hydrostatic conditions.

If the external or body force is conservative (i.e., gradient of a scalar) the hydrostatic pressure is determined up to an arbitrary constant from the potential of the body force.

$\begin{array}{ccc}\hfill \nabla •\mathbf{T}& =& -\rho \mathbf{f},\phantom{\rule{4.pt}{0ex}}\text{static}\phantom{\rule{4.pt}{0ex}}\text{conditions}\hfill \\ \hfill \mathbf{f}=\mathbf{g}& =& g\nabla z,\phantom{\rule{4.pt}{0ex}}\text{uniform}\phantom{\rule{4.pt}{0ex}}\text{gravity}\phantom{\rule{4.pt}{0ex}}\text{field}\hfill \\ \hfill \nabla •\mathbf{T}& =& -\nabla p,\phantom{\rule{4.pt}{0ex}}\text{hydrostatic}\phantom{\rule{4.pt}{0ex}}\text{conditions}\hfill \\ \hfill \nabla p=\rho \nabla \Phi & & ⇒\hfill \\ \hfill \Phi \left(p\right)& =& {\int }^{p}\frac{dp}{\rho }+{C}_{1}=g\phantom{\rule{0.277778em}{0ex}}z+{C}_{2}\hfill \\ \hfill p=\rho \phantom{\rule{0.277778em}{0ex}}g\phantom{\rule{0.277778em}{0ex}}z+{C}_{3},\phantom{\rule{4.pt}{0ex}}\text{if}\phantom{\rule{4.pt}{0ex}}\rho \phantom{\rule{4.pt}{0ex}}\text{is}\phantom{\rule{4.pt}{0ex}}\text{constant}\end{array}$

## Buoyancy (deen, 1998)

A consequence of the fact that pressure increases with depth in a static fluid is that the pressure exerts a net upward force on any submerged object. To calculate this force, consider an object of arbitrary shape submerged in a constant-density fluid as shown in the figure. The net force on this object, ${F}_{p}$ , is given by

${\mathbf{F}}_{p}=-\underset{s}{\phantom{\rule{0.277778em}{0ex}}\int \int \phantom{\rule{0.277778em}{0ex}}}p\mathbf{n}dS$

where the minus sign reflects the fact that positive pressure are compressive (i.e., pressure acts in the $-\mathbf{n}$ direction). The pressure force is evaluated most easily by considering the situation in the figure where the solid has been replaced by an identical volume of fluid. Because the pressure in the fluid depends on depth only, this replacement of the solid does not affect the pressure distribution in the surrounding fluid. In particular, the pressure distribution on the fluid control surface in (b) must be identical to that on the solid surface in (a). Situation (b) has the advantage that we can apply the divergence theorem to the integral in the last equation, because $p$ is a continuous function of position within the volume V; this is not necessarily true for (a), because we have said nothing about the meaning of $p$ within a solid. Applying the divergence theorem and using the hydrostatic pressure field, we find that

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