Past exam of the viscous flow course of the University of Oxford / 2023 (B5.3, A12164W1)

Solutions to all three questions of the viscous flow course of the University of Oxford Part B examination, sat on 15 June 2023. Source: original released PDF, bundle pages 228–231; paper pages 1–4. The solutions follow every lettered and Roman-numbered part. In Question 1(b), the requested equation implicitly selects steady flow. In Question 2(c), the printed far-field sign must be corrected to describe the specified lower half-space; Question 3(a) also has a harmless component-name typo. These qualifications are explained at the affected steps.

Let $V(t)$ have a boundary moving with velocity ${\mathbf{v}}_b$, and let $\mathbf{n}$ be its outward unit normal vector. For a smooth scalar field $q$, the Reynolds transport theorem isIn the last expression, extend ${\mathbf{v}}_b$ smoothly into $V(t)$ and use the divergence theorem. For a material volume, its boundary moves with the fluid velocity $\mathbf{u}$, so ${\mathbf{v}}_b=\mathbf{u}$. Equivalently, the Reynolds transport theorem for a material volume readswhere $D/Dt$ is the material derivative.

For an arbitrary moving region, mass conservation includes the relative mass flow rate across its boundary:Substitution of the Reynolds transport theorem and the divergence theorem givesSince the region is arbitrary, a continuous integrand must vanish: if it had a positive or negative value at a point, a sufficiently small surrounding region would have an integral of the same sign. This gives the continuity equation. With constant positive mass density $\rho$, it reduces toThus the flow is incompressible. For a material volume, the same argument starts directly from conservation of its total mass, with zero relative boundary flux.

At an oriented internal surface with unit normal vector $\mathbf{n}$, the traction $\mathbf{t}(\mathbf{x},t;\mathbf{n})$ is the limiting contact force per unit surface area exerted by the material on the positive-$\mathbf{n}$ side on the material on the negative-$\mathbf{n}$ side:For a boundary of a chosen material volume, this convention gives the external traction when $\mathbf{n}$ is outward. The Cauchy stress tensor has componentsso the second index specifies the surface orientation and the first specifies the force component. The vectors ${\mathbf{e}}_i$ form the standard Cartesian coordinate system basis of unit vectors.

Cauchy's stress theorem states that the dependence of traction on the normal vector is linear:Repeated indices are summed according to Einstein notation. In particular, reversing the normal vector reverses the traction, and knowing the tractions on three mutually perpendicular coordinate planes determines the traction on every plane. Symmetry of the Cauchy stress tensor, for an ordinary fluid without body couples, follows separately from angular momentum conservation.

For an arbitrary material volume $V(t)$, Newton's second law equates the rate of change of linear momentum to the resultant contact force, since there is no body force:Use the Reynolds transport theorem on the left. Expanding the product and using the continuity equation leavesOn the right, the divergence theorem applied to each row of the Cauchy stress tensor givesConsequently the volume integral of $\rho D{u}_i/Dt-{\partial }_j{\sigma }_{ij}$ vanishes on every material volume. Localizing as in part (i) gives the Cauchy momentum equation:The left side is mass density times material acceleration; the right side is the divergence of the Cauchy stress tensor.

For an isotropic Newtonian fluid in incompressible flow, the constitutive equation isHere $p$ is pressure, ${\delta }_{ij}$ is the Kronecker delta, $\mu$ is the constant dynamic viscosity, and $E$ is the strain-rate tensor. The more general compressible constitutive equation can include a term proportional to $(\nabla \cdot \mathbf{u}){\delta }_{ij}$; that term vanishes here.

Differentiate the Cauchy stress tensor, retaining the constant dynamic viscosity outside the derivatives:The last equality uses incompressibility. Substitution into the Cauchy momentum equation gives the Navier-Stokes equationThe pressure gradient and the Laplacian of the velocity field therefore give the two contact force densities in this model.

For the unidirectional flow $\mathbf{u}=(0,0,w)$, incompressibility gives ${\partial }_{z}w=0$. Hence $w=w(x,y,t)$, and the convective acceleration $w{\partial }_{z}w$ vanishes. The transverse components of the Navier-Stokes equation give $p_x=p_y=0$, while the axial component givesA steady-flow assumption is implicit in the question. Neither unidirectional flow nor a time-independent pressure gradient alone eliminates ${\partial }_{t}w$: a decaying transient can be superposed on the final pipe profile. The native PDF does not explicitly say steady flow in this part. For the intended steady flow, $w_t=0$, and thereforeTaking the pipe walls to be stationary, the no-slip boundary condition isThe no-penetration boundary condition is already satisfied by the assumed velocity field. This is the Poisson equation defining rectangular duct flow.

For a profile depending only on $y$, the Poisson equation becomes $w_1^{\prime \prime }=-G/\mu$. Integrating twice givesThe no-slip boundary conditions $w_1(b)=w_1(-b)=0$ imply $C_1=0$ and $C_2=G{b}^2/(2\mu )$. ThusThis is plane Poiseuille flow between the two horizontal walls. It is positive in the interior for $G>0$, so the velocity points down the pressure gradient. It does not satisfy the side-wall no-slip boundary conditions, which is precisely what the correction in part (iii) repairs.

Subtract the particular plane Poiseuille flow profile from the full rectangular duct flow. The remainder is a harmonic function:The data are even in $x$ and $y$. The boundary value problem is unique: if two solutions differed by $v$, Green's first identity and the zero Dirichlet boundary condition would giveThus $v$ is constant, and its zero boundary values force $v=0$. Reflection of a solution in either coordinate axis therefore gives the same solution. We may seek even separated eigenfunctions.

Use separation of variables withEach summand solves the Laplace equation, vanishes at $y=\pm b$, and has the required evenness. The hyperbolic cosine ratio equals one at either side wall. The functions $\cos (k_{n}y)$ have orthogonality relationThey are the even Dirichlet eigenfunctions on $[-b,b]$, so their Fourier series represents the even boundary data. Projecting that data onto the $n$th eigenfunction yields the Fourier coefficientFor completeness, two integrations by parts give the quadratic arch expansion in half-integer cosine modes: the displayed integral is $4(-1{)}^n/k_n^3$, agreeing with the identity supplied in the paper.

Adding the particular solution givesThe coefficients are absolutely summable and the hyperbolic cosine ratio is at most one throughout the closed rectangle, giving uniform convergence and the prescribed boundary values. On compact subsets of the interior, the hyperbolic cosine ratio supplies exponential decay in $n$, justifying differentiation of the series and the Laplace equation. The side-wall correction slows the plane Poiseuille flow enough to satisfy all four no-slip boundary conditions.

The volumetric flow rate is the integral of axial velocity over the cross-section:The plane Poiseuille flow contribution isFor each correction eigenfunction,Uniform convergence permits termwise integration. Multiplying these two integrals by $A_n$ givesConsequentlyEvery summand in the subtracted quantity is positive. Thus the side walls reduce the flux below the infinite-width plane Poiseuille flow prediction. Its dimensions are volume per unit time, as required for a volumetric flow rate.

For the proposed velocity field, $u_x+w_z=1-1=0$, so incompressibility holds. Its convective acceleration isand both components have zero Laplacian. The Navier-Stokes equation therefore requires $p_x=-x$ and $p_z=-z$. Hencewhere $p_{\ast }$ is an arbitrary constant reference pressure. The same pressure gradient balances the same convective acceleration in the Euler equations for an inviscid fluid. Since the viscous terms vanish identically, this is an exact solution for every positive Reynolds number, as well as for the inviscid limit.

At $z=0$, the normal velocity is zero, and the prescribed dimensionless tangential traction isThis stagnation point flow therefore has zero surface shear stress, although its strain-rate tensor and its viscous normal stresses are nonzero.

The held-flat interface imposes the no-penetration boundary condition, while its specified tangential traction gives a shear stress condition:A differentiable boundary value $w(x,0)\equiv 0$ implies $w_x(x,0)=0$, so the second boundary condition is also $u_z(x,0)=\beta x$. There is no imposed no-slip boundary condition at this interface; tangential velocity is determined by the shear stress and the fluid equations.

The solution in part (a) has zero tangential traction, and so remains valid for the whole interface if and only if $\beta =0$. At the single point $x=0$ the imposed shear stress vanishes for any $\beta$, but this does not satisfy a boundary condition that must hold for every $x$.

In a thin boundary layer, take streamwise velocity and distance to be order one, set $z=\delta \zeta$, and use incompressibility to set $w=\delta W$. The streamwise convective acceleration is order one, whereas its dominant viscous contribution after division by the Reynolds number is ${\mathrm{Re}}^{-1}{\delta }^{-2}{u}_{\zeta \zeta }$. Their dominant balance selectsWith this choice, the exact rescaled Navier-Stokes equations areThus the leading pressure is independent of $\zeta$. Matching it to the outer solution from part (a), evaluated near $z=0$, gives $p=p_{\ast }-x^2/2$, hence $-p_x=x$. The leading boundary-layer equation is thereforeThe exact surface shear stress condition rescales to $u_{\zeta }+{\delta }^2{W}_x=\delta \beta x=\gamma x$. Since $W(x,0)=0$ for all $x$, its tangential derivative also vanishes there. The leading surface boundary conditions areMatching to the outer solution requiresThe second matching condition specifies the leading growth of $W$, allowing a constant displacement correction. Requiring $W+\zeta \to 0$ would impose an extra condition not supplied by the leading outer solution. This is the stress-driven stagnation-point boundary layer.

With $u=x{f}^{\prime }(\zeta )$ and $W=-f(\zeta )$, incompressibility is automatic because $u_x=f^{\prime }$ and $W_{\zeta }=-f^{\prime }$. In the streamwise boundary-layer equation,Equating coefficients of $x$ givesThis coefficient comparison determines the equation even though the streamwise equation vanishes identically at the stagnation point $x=0$.

The no-penetration boundary condition gives $f(0)=0$, the surface shear stress gives $x{f}^{\prime \prime }(0)=\gamma x$, and matching the tangential velocity to the outer solution gives $f^{\prime }\to 1$. ThusIf the correction $f^{\prime }-1$ is integrable at the far end, then $f=\zeta +O(1)$ there, which is consistent with $W\sim -\zeta$. The small-shear stress solution below has precisely this property.

At zero surface shear stress, the solution connected to part (a) isIndeed, it has $f_0^{\prime }=1$, $f_0^{\prime \prime }=f_0^{\prime \prime \prime }=0$ and $f_0(0)=0$. Insert the regular perturbation $f=\zeta +\gamma f_1+O({\gamma }^2)$ into the nonlinear ordinary differential equation. The terms areThe coefficient of $\gamma$ therefore givesExpanding the surface boundary conditions and the far-field matching condition givesThe printed $+\infty$ limit is a sign error. It is present in the native PDF on paper page 3, not introduced by transcription. The fluid occupies $z<0$, the inner variable is $\zeta =z/\delta <0$, and part (b)(iii) correctly matches at $-\infty$. We retain that physical domain throughout. A change to the positive coordinate $\eta =-\zeta$ would also change the differential equation and surface derivative signs; it cannot be made by changing only the far-field sign.

The inequality ${\delta }^2\ll \gamma$ ensures that effects discarded at order ${\delta }^2$ in the boundary-layer scaling are smaller than the retained order-$\gamma$ perturbation. The inequality $\gamma \ll 1$ justifies this regular perturbation. The two conditions control different approximations.

Set $g=f_1^{\prime }$. The boundary value problem becomesFor $y_1=1+{\zeta }^2$, direct substitution gives $y_1^{\prime \prime }+\zeta y_1^{\prime }-2y_1=2+2{\zeta }^2-2(1+{\zeta }^2)=0$. This growing polynomial solution is useful for reduction of order.

Write $g=y_{1}v$. Substitution into the linear ordinary differential equation cancels the terms proportional to $v$ and givesAn integrating factor for this first-order equation for $v^{\prime }$ yieldsThe far-field boundary condition selects the convergent integral from $-\infty$:Any additional multiple of $1+{\zeta }^2$ would grow at $-\infty$. The chosen expression tends to zero: for $r=-\zeta \ge 1$,Here $(1+t^2{)}^{-2}\le (1+r^2{)}^{-2}$ and ${\int }_r^{\infty }{e}^{-t^2/2}dt\le r^{-1}{\int }_r^{\infty }t{e}^{-t^2/2}dt$. Thus the far-field boundary condition is verified without a cancellation of large terms.

At $\zeta =0$, differentiating gives $g^{\prime }(0)=C$, so $C=1$. Using the antiderivative supplied in the paper gives the explicit answerHere $\mathrm{erf}$ is the error function. Finally,recovers the unique correction satisfying $f_1(0)=0$. The decay bound also shows that $f_1$ approaches a finite constant at $-\infty$, as allowed by the normal velocity matching. This is the weak surface-shear perturbation of stagnation flow.

If the printed $+\infty$ condition is instead imposed literally on a positive half-line, reduction of order givesIt satisfies $g_{+}^{\prime }(0)=1$ but grows as $\zeta \to -\infty$, so it cannot describe the fluid domain specified in this paper. This makes the consequence of the printed sign error explicit.

At the interface, the tangential velocity is $u(x,0)=x{f}^{\prime }(0)$ and the normal velocity remains zero. The preceding error function expression givesHence, within the leading boundary layer model,Allowing the formally discarded finite-Reynolds number terms gives an additional order-${\delta }^2$ error, smaller than the order-$\gamma$ correction under the assumed ordering. The coefficient is $\sqrt{\pi /8}\approx 0.626657$.

Thus positive surface shear increases the outward surface speed linearly: for $x>0$ it increases $u$, and for $x<0$ it makes $u$ more negative. The stagnation point remains at $x=0$. A negative coefficient would result from using the printed $+\infty$ condition literally, and would describe the different half-line problem identified in part (ii).

The flat plate moves tangentially and the curved plate is stationary. The no-slip boundary condition and no-penetration boundary condition giveAt the curved plate, impermeability can also be written $w=u{h}_x$; together with zero tangential velocity, this gives both components zero. The native PDF writes $v^{\prime }=0$ in the scaled wall conditions although the normal component is defined as $w^{\prime }$. We use $w^{\prime }$ consistently.

The parabolic lubrication gap changes by order $h_0$ when $x^2/R$ is order $h_0$. Its natural horizontal scale is thereforeThe proposed normal velocity scale $h_{0}U/L$ then makes the two terms of the continuity equation comparable. Balancing the horizontal pressure gradient against the transverse viscous term in the Navier-Stokes equation requires $P/L=\mu U/h_0^2$, soHere pressure is measured from a common reference value, which is chosen at the outer fluid in part (b).

For clarity, define the dimensionless material derivativeSubstitution of all the scalings gives the exact dimensionless equationsFor example, the factor in the last left side is obtained by dividing the vertical acceleration scale $\rho h_0{U}^2/L^2$ by the vertical pressure gradient scale $P/h_0$.

Taking $\delta \ll 1$ and ${\delta }^2\mathrm{Re}\ll 1$ leaves the lubrication approximationThe scaled gap and boundary conditions areThe Reynolds number itself need not be small: it is ${\delta }^2\mathrm{Re}$ that measures inertia relative to transverse viscous resistance in this thin geometry.

Write $H(x^{\prime })=h^{\prime }(x^{\prime })$ and $p_x^{\prime }=d{p}^{\prime }/d{x}^{\prime }$ below, so the subscript $x$ in this shorthand denotes differentiation with respect to $x^{\prime }$. The vertical pressure gradient is zero, and integrating $u_{z^{\prime }{z}^{\prime }}^{\prime }=p_x^{\prime }$ twice with the wall boundary conditions gives the local Couette-Poiseuille flowThe first two terms are Couette flow due to the moving plate; the last is plane Poiseuille flow due to the local pressure gradient. Its dimensionless volume flux per unit width isUse the Leibniz integral rule and incompressibility to differentiate this integral:Thus the flux is independent of position, andThe normal velocity can then be recovered from incompressibility:It is zero on the lower wall by construction and zero on the upper wall by the constant-volume flux relation. This supplies the remaining component of the leading velocity field.

The endpoint pressure conditions are matching conditions, not an assertion that the lubrication approximation remains valid at arbitrarily large dimensional $x$. On an outer length scale $R$, a viscous pressure variation is of size $\mu U/R$, whose ratio to the inner scale isEven an outer inertial pressure scale $\rho U^2$ has ratioConsequently the gap pressure is much larger than either outer variation. Matching to fluid with a common ambient pressure at both ends, and taking that value as zero gauge pressure, givesThere is an overlap region $1\ll |x^{\prime }|\ll {\delta }^{-1}$ before the wall slope becomes order one. The infinite endpoints represent matching through that region. An externally imposed pressure difference comparable with $P$ would change these conditions and the resulting volume flux; equal outer pressure is the physical assumption being used here.

Let the constant in the integrated Reynolds equation be $C$:Both endpoint pressures are zero, so the integral of $d{p}^{\prime }/d{x}^{\prime }$ on the real line vanishes. DefineThe substitution $x^{\prime }=\sqrt{2}\tan \theta$ givesThe second value follows by integrating the supplied identity for ${\cos }^4\theta$; the nonconstant cosine terms have zero integral over this interval. Hence $12C{I}_3-6I_2=0$, givingThe pressure gradient is nowDirect differentiation showsThe candidate tends to zero at both ends, so the additive constant is zero andTogether with the local Couette-Poiseuille flow profile, this solves the sliding-plate flow through a parabolic gap.

To restore dimensions, integrate $u=U{u}^{\prime }$ across $dz=h_{0}d{z}^{\prime }$. The signed dimensional volume flux per unit width, positive in the $+x$ direction, isThus the magnitude drawn in the direction of the moving plate is $2U{h}_0/3$. For a plate width $B$ large enough to neglect side-wall effects, the total volumetric flow rate is $-2BU{h}_0/3$; without a specified width the answer is per unit width.

The Couette flow and plane Poiseuille flow contributions to the dimensional volume flux per unit width are respectivelyThey point in the same direction precisely when $Q_P<0$, givingAt the endpoints the plane Poiseuille flow contribution is zero. Outside that interval it opposes the Couette flow contribution, while their sum remains the constant negative volume flux per unit width $-2U{h}_0/3$. Thus pressure assists the moving plate near the narrowest gap and drives a compensating return contribution farther away.