2. Mass, momentum and stress

The question

Why does fluid accelerate through a constriction, and what balances the load on an obstacle? Track what crosses a fixed region rather than starting with the name of a differential equation.

Mass accounting

For a fixed volume $\Omega$ with outward unit normal $\boldsymbol n$, mass changes because fluid crosses its boundary:

$$\frac{\mathrm d}{\mathrm dt}\int_\Omega\rho\,\mathrm dV +\int_{\partial\Omega}\rho\boldsymbol u\cdot\boldsymbol n\,\mathrm dA=0.$$

Use the divergence theorem and allow arbitrary control volumes to obtain $\partial_t\rho+\nabla\cdot(\rho\boldsymbol u)=0$. When density is constant in space and time, this reduces to $\nabla\cdot\boldsymbol u=0$. It constrains volume changes, not the speed of every particle.

In a channel with impermeable walls, the inlet and outlet volume fluxes agree. If the velocity is uniform over two sections, $A_1U_1=A_2U_2$. A smaller section then has a larger speed. The uniform-profile assumption is needed for that last shortcut; in general integrate the actual profile.

Momentum accounting

Momentum per unit volume is $\rho\boldsymbol u$. Its advective flux is the tensor $\rho\boldsymbol u\otimes\boldsymbol u$, and the surface force per area is the traction $\boldsymbol\sigma\boldsymbol n$. With a volume force density $\boldsymbol f$,

$$\frac{\mathrm d}{\mathrm dt}\int_\Omega\rho\boldsymbol u\,\mathrm dV +\int_{\partial\Omega}\rho\boldsymbol u(\boldsymbol u\cdot\boldsymbol n)\,\mathrm dA =\int_{\partial\Omega}\boldsymbol\sigma\boldsymbol n\,\mathrm dA +\int_\Omega\boldsymbol f\,\mathrm dV.$$

Each term is a force. In two dimensions the corresponding area and line integrals represent force per unit out-of-plane length, in $\mathrm{N/m}$.

For constant density and divergence-free velocity, the local form is

$$\rho\left(\partial_t\boldsymbol u+ (\boldsymbol u\cdot\nabla)\boldsymbol u\right) =\nabla\cdot\boldsymbol\sigma+\boldsymbol f.$$

The material derivative $\partial_t\boldsymbol u+(\boldsymbol u\cdot\nabla)\boldsymbol u$ is the acceleration following fluid. A steady field can accelerate a moving particle: the second term survives when a particle enters a faster region.

Close the stress law

Balance alone does not determine stress. For an incompressible Newtonian fluid,

$$\boldsymbol\sigma=-p\boldsymbol I+2\mu\boldsymbol D, \qquad \boldsymbol D=\tfrac12(\nabla\boldsymbol u+\nabla\boldsymbol u^{\mathsf T}).$$

$\boldsymbol D$ measures deformation rate. Rigid rotation has an antisymmetric velocity gradient and gives $\boldsymbol D=0$: rotation alone produces no Newtonian viscous stress. Compression by positive pressure produces traction $-p\boldsymbol n$, directed into the region.

For constant $\mu$, $\nabla\cdot(2\mu\boldsymbol D)=\mu\nabla^2\boldsymbol u+ \mu\nabla(\nabla\cdot\boldsymbol u)$. Incompressibility removes the last term, yielding the familiar velocity Laplacian. Keep the full stress when evaluating boundary traction; equality of interior differential operators does not make all proposed boundary conditions equivalent.

Structural mechanics uses the same force balance and traction. Its elastic constitutive law instead depends on displacement strain. Substituting displacement for velocity here would change both physics and dimensions.

Why a pressure drop costs power

Dot momentum with velocity. The stress identity $\boldsymbol u\cdot\nabla\cdot\boldsymbol\sigma= \nabla\cdot(\boldsymbol\sigma\boldsymbol u)- \boldsymbol\sigma:\nabla\boldsymbol u$ separates boundary work from internal work. For incompressible Newtonian motion,

$$\boldsymbol\sigma:\nabla\boldsymbol u=2\mu\boldsymbol D:\boldsymbol D\geq0.$$

Viscosity removes kinetic energy and converts it to internal energy. Pressure does boundary work, while its local contraction is zero because $\nabla\cdot\boldsymbol u=0$. In steady pressure-driven channel flow, pressure power supplies viscous dissipation. Applying a lossless Bernoulli formula to that whole channel would omit precisely the mechanism driving the flow.

Heat transfer explains how internal energy subsequently moves. The present isothermal flow model does not solve a temperature equation.

Exercises

1. Check that $\rho\partial_t\boldsymbol u$, $\nabla p$ and $\mu\nabla^2\boldsymbol u$ all have units $\mathrm{N/m^3}$.
2. Evaluate $\boldsymbol D$ for rigid rotation $\boldsymbol u=(-\omega y,\omega x)$. Then evaluate it for $\boldsymbol u=(ay,0)$.
3. Integrate incompressibility over a channel with one inlet, one outlet and impermeable walls. Keep the outward-normal signs explicit.
4. Explain why a steady velocity field can have nonzero advective acceleration.

Check your reasoning. Rigid rotation gives zero strain rate. Simple shear has $D_{xy}=D_{yx}=a/2$, shear stress $\mu a$, and dissipation density $\mu a^2$. Mass conservation gives $-Q_{\rm in}+Q_{\rm out}=0$.

For a complementary derivation of stress and viscosity, see David Tong’s fluid mechanics course.

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