Past exam of the topics in fluid mechanics course of the University of Oxford / 2010 (C6.4a, DMAC 2A74, DMSU 2A74)

Original solutions to the paper released at the source request. The complete editable transcription is retained in the private examination archive.

Choose $x$ vertically upward. Let $G(x,t)$ be the gas-occupied part of a pipe cross-section, of area $\alpha A$. Define the phase mean mass density, mass-weighted velocity, second moment of velocity, and phase mean pressure byThe void fraction $\alpha$ is a volume fraction, not a mass fraction. The gas momentum in a slice of length $dx$ is $A\alpha {\bar{\rho }}_g\bar{v}dx$; its advected momentum flux is $A\alpha {\bar{\rho }}_g\overline{v^2}$. The two end pressures give the net force $-A{\partial }_x(\alpha {\bar{p}}_g)dx$. Include every other vertical force on the gas, including interfacial forces, in $M_{g}dx$. Newton's second law for this fixed slice, followed by division by $Adx$, therefore givesThis is a cross-section averaged phase momentum balance. In particular, mass weighting makes the momentum and its flux exact averages; one must not replace $\overline{v^2}$ by ${\bar{v}}^2$ without an additional closure. With liquid mean velocity $\bar{u}$, constant liquid mass density ${\rho }_{\ell }$, and liquid second moment $\overline{u^2}$, the same calculation givesWith the common pressure $p={\bar{p}}_g={\bar{p}}_{\ell }$, the specified pressure contributions to $M_g,M_{\ell }$ cancel the phase-fraction derivatives from the product rule. The remaining pressure forces per unit mixture volume are $-\alpha p_x$ and $-(1-\alpha )p_x$.

Assume that the bubbles at a given height have the stated radius $a$. Their number per unit mixture volume is $N=3\alpha /(4\pi a^3)$. The drag force on a rising bubble opposes its relative velocity, so multiplication of the single-bubble law by $N$ gives the signed magnitudeGravity acts downward; the drag force on the liquid is the opposite of that on the gas. With no wall force,A bubble retains its mass in the absence of transfer between phases. Consequently $a^3{\bar{\rho }}_g=a_0^3{\rho }_0$ andFor clarity suppress bars on $p,{\rho }_g,v$ in the steady calculation. On a macroscopic variation length $\ell$, gas inertia per unit volume scales like $\alpha {\rho }_g{v}^2/\ell$, whereas the quadratic drag force scales like $\alpha {\rho }_{\ell }{v}^2/a$. Their ratio is $O[({\rho }_g/{\rho }_{\ell })(a/\ell )]$. Thus ${\rho }_g=O({\rho }_0)\ll {\rho }_{\ell }$, together with the scale separation $a\ll \ell$ inherent in this averaging model, justifies the leading-order neglect of gas inertia. Small reference density alone does not justify it in an arbitrarily short or strongly compressed region. Neglect gas weight at the same order, with the liquid fraction not vanishing. For a stationary liquid, summing the two momentum equations givesThe leading gas balance is $0=-\alpha p_x-D_v$. Substitution of the bubble-radius law yieldsThe gas continuity equation, with constant pipe area and no mass exchange, is ${\partial }_t(\alpha {\rho }_g)+{\partial }_x(\alpha {\rho }_{g}v)=0$. Hence the steady gas mass flux $J=\alpha {\rho }_{g}v$ is constant. For $0<\alpha <1$ the pressure equation selects $v>0$, andEvaluate the latter constant at a reference point ${\rho }_g={\rho }_0$, $\alpha ={\alpha }_0$ and divide. The terminal-drift density law for compressible bubbles isThe exponent $3/5$ comes from combining mass conservation with the density dependence of bubble radius; no gas equation of state was needed.

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Write $\Delta T=T_0-T_B$, $\Delta C=C_0-C_B$ and assume these scales are nonzero. The two reference liquidus relations imply $\Delta T=-m\Delta C$. For the similarity variable $\eta =z/(2\sqrt{\kappa t})$ and front $h=2\lambda \sqrt{\kappa t}$,Thus a temperature profile $T=T_{\ast }+\Delta T\theta (\eta )$ has $T_t=-\Delta T\eta {\theta }^{\prime }/(2t)$ and $\kappa T_{zz}=\Delta T{\theta }^{\prime \prime }/(4t)$. Each heat equation becomes ${\theta }^{\prime \prime }+2\eta {\theta }^{\prime }=0$. The solute diffusion equation becomes ${\mathcal{C}}^{l\prime \prime }+2Le\eta {\mathcal{C}}^{l\prime }=0$, withThis Lewis number compares thermal and solutal diffusivities. Inside the mushy layer, $T^s-T_B=-m(C^s-C_B)$ gives $\boxed{{\mathcal{C}}^s={\theta }^s}$. The wall and far-field conditions become ${\theta }^s(0)=0$ and ${\theta }^l,{\mathcal{C}}^l\to 0$ as $\eta \to \infty$.

At the moving phase boundary, equality of temperatures gives ${\theta }^s={\theta }^l+(T_{\infty }-T_B)/\Delta T$. Equality of concentrations gives ${\mathcal{C}}^s={\mathcal{C}}^l+1$. Therefore the dimensionless parameters with the convention used here areNeither concentration nor temperature offsets should be dropped when enforcing continuity. In particular ${\mathcal{R}}_{\theta }$ can be negative. If $k=\rho c_p\kappa$, then $S=L/[c_p\Delta T]$: this is the inverse Stefan number, although the paper calls $S$ its Stefan number.

To check every factor of two in the flux conditions, divide the thermal Stefan condition by $k\Delta T/(2\sqrt{\kappa t})$, and the solute balance by $D\Delta C/(2\sqrt{\kappa t})$. This givesFinally $T_z^l=-m{C}_z^l$ and $-m\Delta C=\Delta T$ imply $\boxed{{\theta }^{l\prime }={\mathcal{C}}^{l\prime }}$. This is marginal equilibrium at a mush-liquid boundary. Together these calculations produce all the rescaled differential equations and all three interface balances.

Integrating $Y^{\prime \prime }+2a\eta Y^{\prime }=0$ once gives $Y^{\prime }=K{e}^{-a{\eta }^2}$. A second integration expresses the solution using the error function when the value at zero is prescribed, and the complementary error function when decay at infinity is prescribed. Write ${\theta }_{\lambda }={\theta }^s(\lambda )$ and take $\lambda >0$, $Le>0$. The wall condition and the interface value giveFor the exterior liquid, continuity at the interface fixes ${\theta }^l(\lambda )={\theta }_{\lambda }-1+{\mathcal{R}}_{\theta }$ and ${\mathcal{C}}^l(\lambda )={\theta }_{\lambda }-1$. Thus the two decaying similarity solutions areThe factor $\sqrt{Le}$ is required because solute and heat diffuse on different length scales. Both denominators are positive for a finite positive $\lambda$. These formulas use the differential equations, wall/far-field data and continuity; the remaining three flux conditions determine their unknown constants.

Introduce three positive functions of the unknown front parameter:Differentiating the error function profiles gives ${\theta }^{s\prime }=2P{\theta }_{\lambda }$, ${\theta }^{l\prime }=-2Q({\theta }_{\lambda }-1+{\mathcal{R}}_{\theta })$ and ${\mathcal{C}}^{l\prime }=-2R({\theta }_{\lambda }-1)$ at the interface. The three flux conditions are therefore exactlyThese undivided equations are the safest description of the constant-fraction mushy-layer similarity solution, including degenerate parameter choices. In the nondegenerate case the third equation first givesThe second then gives the solid fractionUse the third equation to replace the $Q$ term in the first, and eliminate $\varphi$. The front parameter is a positive root of the single transcendental equationThen substitute back for ${\theta }_{\lambda },\varphi$ and $h(t)=2\lambda \sqrt{\kappa t}$. Since the physical parameters have not been assigned numbers, an implicit scalar equation is the appropriate determination of $\lambda$.

The divided formulas require $R\ne Q$ and ${\theta }_{\lambda }-1+{\mathcal{R}}_C\ne 0$. If either vanishes, return to the boxed three-equation system. For example $Le=1$ gives $R=Q$, so the last equation demands ${\mathcal{R}}_{\theta }=0$; otherwise this proposed similarity solution cannot exist. A physical branch must also have $\lambda >0$ and $0\le \varphi \le 1$. The algebra alone does not establish existence or uniqueness of such a branch for arbitrary parameters.