Past exam of the theoretical mechanics course of the University of Oxford / 2011 / A / 1

For a material volume moving with the fluid, the Reynolds transport theorem gives, componentwise,For a scalar $q$ this is also ${\int }_{V(t)}(Dq/Dt+q\nabla \cdot \mathbf{u})dV$. Apply the component formula to momentum density $q=\rho u_i$. Constant density and incompressibility make the integrand $\rho D{u}_i/Dt$.

The traction $\mathbf{t}(\mathbf{n})$ is the force per unit area exerted by the material outside a surface on the material inside, with outward unit normal $\mathbf{n}$. Reversing the normal exchanges the two sides of the same internal surface. Thus $\mathbf{t}(-\mathbf{n})=-\mathbf{t}(\mathbf{n})$ expresses equal and opposite internal contact forces, the local form of Newton's third law.

With no body force, Newton's second law equates the rate of change of the material volume's total momentum to its surface force. Combining this balance with the transport theorem givesThis is a vector identity, since the component argument applies to each direction.

Define ${\sigma }_{ij}={\mathbf{e}}_i\cdot \mathbf{t}({\mathbf{e}}_j)$: the second index labels the normal direction of a coordinate face, and the first labels the force component. These nine entries form the Cauchy stress tensor.

Take a small tetrahedron with oblique face of area $S$ and outward normal $\mathbf{n}$. First suppose $n_j>0$; its other three faces have normals $-{\mathbf{e}}_j$ and areas $S_j=n_{j}S$, by orthogonal projection of the oblique face. The total surface force isIf its linear size is $\ell$, its inertia is $O({\ell }^3)$ while its surface forces are $O({\ell }^2)$. Divide the momentum balance by $S$ and shrink the tetrahedron to the point. Smoothness of the stress and the odd-normal relation then give $\mathbf{t}(\mathbf{n})={\sum }_j{n}_j\mathbf{t}({\mathbf{e}}_j)$. Choosing coordinate faces with the appropriate reversed normals gives the same identity in the other octants. In components,The face-area relations used are exactly $S_j=|n_j|S$, with each coordinate face oriented opposite the corresponding component of the oblique normal.

Insert the traction formula into the material-volume balance and apply the divergence theorem to each row of the stress tensor:The volume is arbitrary, so continuity of the fields gives the local Cauchy momentum equationFor an incompressible Newtonian fluid, the constitutive law isHere the pressure enforces incompressibility, and the viscous stress is twice the viscosity times the symmetric rate-of-strain tensor. With constant $\mu$,Using $D/Dt={\partial }_t+\mathbf{u}\cdot \nabla$ proves the stated Navier-Stokes equations.

The only velocity component is axial, so incompressibility gives ${\partial }_{z}w=0$. Thus $w=w(x,y)$, and the steady convective acceleration is $w{\partial }_z(w\mathbf{k})=0$. The transverse momentum equations give $p_x=p_y=0$, so $p=p(z)$. The axial momentum equation is thereforeIts left side depends only on $x,y$, while its right side depends only on $z$. Equality throughout the pipe forces both to be one constant, say $G$. Consequently the axial pressure gradient is constant, and $p(z)=Gz+p_{\ast }$. This is absence of dependence on a coordinate, rather than probabilistic independence.

For equilateral triangular pipe flow, the polynomial vanishes on all three walls, so it satisfies no-slip boundary conditions. Direct differentiation givesThusFor $G<0$, $A<0$ and the velocity is positive in the triangular interior.

At the wall $x=a$, the outward normal of the fluid is ${\mathbf{e}}_x$. The axial traction exerted by the wall on the fluid is $\mu w_x$, so the fluid exerts the opposite axial traction on the wall. Since $w_x(a,y)=A(a^2-3y^2)$, the force per unit axial length isIt points with the flow for a negative axial pressure gradient. As a check, the triangular area is $a^2/\sqrt{3}$, so the total pressure driving force per unit length is $-G{a}^2/\sqrt{3}$; each of the three equivalent walls receives one third.