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An example where lubrication theory comes short: hydraulic jumps in a flow down an inclined plate

Published online by Cambridge University Press:  29 December 2014

E. S. Benilov*
Affiliation:
Department of Mathematics, University of Limerick, Ireland
V. N. Lapin
Affiliation:
School of Mathematics, University of Leeds, Leeds LS2 9JT, UK
*
†Email address for correspondence: eugene.benilov@ul.ie
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Abstract

We examine two-dimensional flows of a viscous liquid on an inclinedplate. If the upstream depth $h_{-}$ of the liquid is larger than its downstream depth $h_{+}$ , a smooth hydraulic jump (bore) forms and starts propagating down the slope. If the inclination angle of the plate is small, the bore can be described by the so-called lubrication theory. In this work we demonstrate that bores with $h_{+}/h_{-}<(\sqrt{3}-1)/2$ either are unstable or do not exist as steady solutions of the governing equation (physically, these two possibilities are difficult to distinguish). The instability/evolution occurs near a stagnation point and, generally, causes overturning – sometimes on the scale of the whole bore, sometimes on a shorter, local scale. The overturning occurs because the flow advects disturbances towards the stagnation point and, thus, ‘compresses’ them, increasing the slope of the free surface. Interestingly, this effect is not captured by the lubrication theory, which formally yields a well-behaved stable solution for all values of $h_{+}/h_{-}$ .

Information

Type
Papers
Copyright
© 2014 Cambridge University Press 
Figure 0

Figure 1. The setting: a smooth hydraulic jump (bore) on an inclined plate.

Figure 1

Figure 2. Numerical results on the existence of steady bores on the $({\it\eta},{\it\alpha})$ plane (${\it\alpha}$ is the plate’s inclination angle, ${\it\eta}=h_{+}/h_{-}$ is the downstream-to-upstream depth ratio). The regions where micro-overturning and overturning proper occur (see figure 3) are labelled ‘MO’ and ‘Overturning’, respectively. The vertical part of the boundary between the existence/MO regions corresponds to condition (4.1). The four ‘black squares’ correspond to the solutions presented in figure 3.

Figure 2

Figure 3. Examples of steady bores: the depth $h$ of the liquid versus $x$. The corresponding points in the $({\it\alpha},{\it\eta})$ plane are shown in figure 2 by ‘black squares’.

Figure 3

Figure 4. Examples of steady bores: the surface velocity $u_{s}$ (normalised by the plate’s slope $s=\tan {\it\alpha}$) versus $x$. The corresponding graphs of $h(x)$ are shown in figure 3.

Figure 4

Figure 5. Two patterns of overturning for ${\it\alpha}=7^{\circ }$: (a) overturning proper (${\it\eta}=0.15$), (b) micro-overturning (${\it\eta}=0.22$). The stagnation points are shown by black dots, the insets show blow-ups of the regions near the stagnation points (note that the $x$ and $y$ ranges of inset (a) are approximately 16 and 173 times larger than those of inset (b)). The solutions shown as solid and dotted lines are separated in time by (a) $t=488$; (b) $t=326$, with the dotted line corresponding to the later $t$. The corresponding points in the $({\it\eta},{\it\alpha})$ plane are shown in figure 6 by ‘black squares’.

Figure 5

Figure 6. Existence of steady bores on the $({\it\eta},{\it\alpha})$ plane (as in figure 2) and the applicability of the lubrication theory. The thin solid line corresponds to the maximum slope of the free surface (calculated using expression (3.1)) being 0.6. The two ‘black squares’ correspond to the solutions presented in figure 5.

Figure 6

Figure 7. The streamlines for steady bores as described by the lubrication theory (2.18)–(2.19), (2.10). The plate’s inclination angle is ${\it\alpha}=5^{\circ }$, the non-dimensional upstream depth is $h_{-}=1$, the downstream depth is (a) $h_{+}=0.5$ and (b) $h_{+}=0.2$. In (b), observe the recirculation region (its boundary shown as a dotted line) and the SP shown by a black dot. Another SP is located infinitely far to the left, so that the two SPs are connected by the dotted line.

Figure 7

Figure 8. The asymptotic solution (5.25) for ${\it\eta}=0.15$, ${\it\alpha}=3^{\circ }$, and two different initial conditions: (a) $\tilde{h}(x,0)=-x\exp (-x^{2}/2)$ and (b) $\tilde{h}(x,0)=\exp (-x^{2}/2)$. The curves labelled 0–2 correspond to $t=0$, 0.2, 0.4.