Past exam of the theoretical mechanics course of the University of Oxford / 2010 / A / 1

Let $\mathbf{x}(\mathbf{X},t)$ be the particle flow map from the fixed reference volume $V_0$ to the material volume, and let $J=\det (\partial \mathbf{x}/\partial \mathbf{X})$. Change variables and differentiate on the fixed domain:Use the two permitted identities, $DJ/Dt=J\nabla \cdot \mathbf{u}$ and $Df/Dt=f_t+\mathbf{u}\cdot \nabla f$. Changing back givesThis is the Reynolds transport theorem: both the value carried by each particle and the changing volume element contribute.

The traction $\mathbf{t}(\mathbf{n})$ is force per unit area across an oriented surface with unit normal $\mathbf{n}$. The stress component ${\sigma }_{ij}$ is its $i$th component on a surface normal to the $j$th coordinate axis. Cauchy's stress theorem states that this dependence is linear:For the boundary of a material volume, take the outward normal and the traction exerted by the surrounding fluid. Newton's law and the divergence theorem giveConstant density and incompressibility make the transport-theorem integrand on the left $\rho D{u}_i/Dt$. The volume is arbitrary, soSubstitute the Newtonian constitutive law. With constant viscosity,Hence

For a steady unidirectional field $\mathbf{u}=w\mathbf{k}$, incompressibility gives ${\partial }_{z}w=0$. Thus $w=w(x,y)$ and the convective acceleration vanishes. The transverse momentum equations give $p_x=p_y=0$, so $p=p(z)$. The axial equation isIts left side is independent of $z$ and its right side independent of $x,y$, so both equal a constant $G$. No slip against the translating wall gives $w=U$ on the boundary. Thus the asserted Poisson problem follows.

For the circular pipe, the axisymmetric solution isIt satisfies the boundary value and ${\Delta }_{\perp }{r}^2=4$. Integrating over the cross-section givesTherefore the laboratory-frame volume flux vanishes precisely whenFor a pressure decrease in the positive axial direction, this requires upstream wall translation.