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Archive for March 3rd, 2012

PHY454H1S Continuum Mechanics. Lecture 13: Hydrostatics. Surface normals and tangent vectors. Taught by Prof. K. Das.

Posted by peeterjoot on March 3, 2012

[Click here for a PDF of this post with nicer formatting and figures if the post had any (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]

Disclaimer.

Peeter’s lecture notes from class. May not be entirely coherent.

Midterm discussion.

We had been asked to solve a 1D flow problem with one fixed surface, and one moving surface, as illustrated in figure (\ref{fig:continuumL13:continuumL13Fig1})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL13Fig1}
\caption{Return flow configuration. One fixed surface, one moving surface.}
\end{figure}

we end up showing that our solutions are of the forms

\begin{itemize}
\item
Shear flow (\ref{fig:continuumL13:continuumL13Fig2a}). U \ne 0, G = 0.
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL13Fig2a}
\caption{Shear Flow.}
\end{figure}
\item Channel flow (\ref{fig:continuumL13:continuumL13Fig2b}). U = 0, G \ne 0.
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL13Fig2b}
\caption{Channel flow}
\end{figure}
\item Return flow (\ref{fig:continuumL13:continuumL13Fig2c}). U \ne 0, G \ne 0.
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL13Fig2c}
\caption{Return flow.}
\end{figure}
\end{itemize}

Note that the last sort of solution, that of return flow, was discussed in the context of surface tension.

Hydrostatics

Consider a sample volume of water, not moving with respect to the rest of the surrounding water. If it is not moving the forces must be in balance. What are the forces acting on this bit of fluid, considering a cylinder of the fluid above it as in figure (\ref{fig:continuumL13:continuumL13Fig3a})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL13Fig3a}
\caption{A control volume of fluid in a fluid.}
\end{figure}

In the column of fluid above the control volume (\ref{fig:continuumL13:continuumL13Fig3b}) we have
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL13Fig3b}
\caption{column of fluid above a control volume.}
\end{figure}

\begin{aligned}h A_w \rho g + p_A A_w = p_w A_w\end{aligned} \hspace{\stretch{1}}(3.1)

so

\begin{aligned}p_w = h \rho g + p_A\end{aligned} \hspace{\stretch{1}}(3.2)

If we were to replace this blob of water with something of equal density, it should not change the dynamics (or statics) of the situations and that would not move.

We call this the

\begin{definition}
\emph{(Buoyancy force)}

Buoyancy force =
weight of the equivalent volume of water – weight of the foreign body.
\end{definition}

If the densities are not equal, then we’d have motion of the new bit of mass as depicted in figure (\ref{fig:continuumL13:continuumL13Fig4})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL13Fig4}
\caption{A mass of different density in a fluid.}
\end{figure}

Consider a volume of ice floating on the surface of water, one with solid ice and one with partially frozen ice (with water or air or dirt or an anchor or anything else in it) as in figure (\ref{fig:continuumL13:continuumL13Fig5})

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL13Fig5}
\caption{Various floating ice configurations on water.}
\end{figure}

No matter the situation, the water level will not change if the ice melts, because the total weight of the displaced water must have been matched by the weight of the unmelted ice plus additives.

Now what happens when we have fluid flows? Consider figure (\ref{fig:continuumL13:continuumL13Fig6})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL13Fig6}
\caption{flow through channel with different apertures.}
\end{figure}

Conservation of mass is going to mean that the masses of fluid flowing through any pair of cross sections will have to be equal

\begin{aligned}\rho_1 A_1 v_1 = \rho_2 A_2 v_2,\end{aligned} \hspace{\stretch{1}}(3.3)

With incompressible fluids (\rho = \rho_1 = \rho_2) we have

\begin{aligned}A_1 v_1 = A_2 v_2,\end{aligned} \hspace{\stretch{1}}(3.4)

so that if

\begin{aligned}A_1 > A_2,\end{aligned} \hspace{\stretch{1}}(3.5)

we must have

\begin{aligned}v_1 < v_2,\end{aligned} \hspace{\stretch{1}}(3.6)

to balance this.

In class this was illustrated with a pair of computer animations, one showing the deformation of patches of the fluid, and another showing how the velocities vary through the channel. This is crudely depicted in figure (\ref{fig:continuumL13:continuumL13Fig7})

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL13Fig7}
\caption{area and velocity flows in unequal aperture channel configuration.}
\end{figure}

We see the same behavior for channels that return to the original diameter after widening as in figure (\ref{fig:continuumL13:continuumL13Fig8})

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL13Fig8}
\caption{velocity variation in channel with bulge}
\end{figure}

If we consider half of such a channel as in figure (\ref{fig:continuumL13:continuumL13Fig9a})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL13Fig9a}
\caption{vorticity induction due to pressure gradients in unequal aperture channel.}
\end{figure}

considering the flow around a small triangular section we must have a pressure gradient, which induces a vorticity flow. We’d see something similar in a rectangular channel where there is a block in the channel, as depicted in figure (\ref{fig:continuumL13:continuumL13Fig9b})

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL13Fig9b}
\caption{vorticity due to rectangular blockage.}
\end{figure}

Back to hydrostatics. Height matching in odd geometries.

Now let’s consider the hydrostatics case again, with an arbitrarily weird channel as in figure (\ref{fig:continuumL13:continuumL13Fig10})

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL13Fig10}
\caption{height matching in odd geometries.}
\end{figure}

This was also illustrated with a glass blown container in class as in figure (\ref{fig:continuumL13:continuumL13Fig11})

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL13Fig11}
\caption{a physical demonstration with glass blown apparatus.}
\end{figure}

In this real apparatus, we didn’t have exactly the same height (because of bubbles and capillary effects (surface tension induced meniscus curves), but we see first hand what we are talking about.

To account for this, we need to consider the situation in pieces as in figure (\ref{fig:continuumL13:continuumL13Fig12})

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL13Fig12}
\caption{column volume element decomposition for odd geometries}
\end{figure}

Breaking down the total pressure effects into individual bits, any column of fluid contributes to the pressure below it, even if that column of fluid is not directly on top of a continuous column of fluid all the way to the “bottom”.

Surface tension: normals and tangents.

Consider a surface with some variation as in figure (\ref{fig:continuumL13:continuumL13Fig13})

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL13Fig13}
\caption{variable surface geometries}
\end{figure}

We can construct an equation for the surface

\begin{aligned}z = h(x, t),\end{aligned} \hspace{\stretch{1}}(4.7)

or equivalently

\begin{aligned}\phi = z - h(x, t) = 0.\end{aligned} \hspace{\stretch{1}}(4.8)

If d is the average height, with the \eta(x,t) the variation of the height from this average, we can also write

\begin{aligned}h = d + \eta(x, t)\end{aligned} \hspace{\stretch{1}}(4.9)

and for the surface

\begin{aligned}\phi = d - \eta(x, t) = 0\end{aligned} \hspace{\stretch{1}}(4.10)

We can generalize this and define a surface function as one that satisfies

\begin{aligned}\phi = d - \eta(x, t) = \text{constant}.\end{aligned} \hspace{\stretch{1}}(4.11)

Consider a small section of a 2D surface as in figure (\ref{fig:continuumL13:continuumL13Fig14})

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL13Fig14}
\caption{a vector differential element}
\end{figure}

With \phi = \text{constant} on the surface, we have for \phi = \phi(x, y, z)

\begin{aligned}d\phi = 0 \end{aligned} \hspace{\stretch{1}}(4.12)

or, in coordinates

\begin{aligned}d\phi &= \frac{\partial {\phi}}{\partial {x}} dx+\frac{\partial {\phi}}{\partial {y}} dy+\frac{\partial {\phi}}{\partial {z}} dz \\ &= \boldsymbol{\nabla} \phi \cdot d\mathbf{r}\end{aligned}

Pictorially we see that d\mathbf{r} is tangential to the surface, but since we also have

\begin{aligned}\boldsymbol{\nabla} \phi \cdot d\mathbf{r} = 0,\end{aligned} \hspace{\stretch{1}}(4.13)

the implication is that the gradient is normal to the surface

\begin{aligned}d\mathbf{r} \perp \boldsymbol{\nabla} \phi.\end{aligned} \hspace{\stretch{1}}(4.14)

We can therefore construct the unit normal by scaling the gradient

\begin{aligned}\hat{\mathbf{n}} = \frac{\boldsymbol{\nabla} \phi}{{\left\lvert{\boldsymbol{\nabla} \phi}\right\rvert}},\end{aligned} \hspace{\stretch{1}}(4.15)

since the direction of \hat{\mathbf{n}} is \boldsymbol{\nabla} \phi.

For example, in our case where \phi = y - h(x, t), we have

\begin{aligned}\boldsymbol{\nabla} \phi = \hat{\mathbf{x}} \left( -\frac{\partial {h}}{\partial {x}} \right) + \hat{\mathbf{y}}\end{aligned} \hspace{\stretch{1}}(4.16)

This has norm

\begin{aligned}{\left\lvert{\boldsymbol{\nabla} \phi}\right\rvert} = \sqrt{ 1 + \left( \frac{\partial {h}}{\partial {x}} \right)^2 }\end{aligned} \hspace{\stretch{1}}(4.17)

and our unit normal is

\begin{aligned}\hat{\mathbf{n}} = \frac{\hat{\mathbf{x}} \left( -\frac{\partial {h}}{\partial {x}} \right) + \hat{\mathbf{y}}}{\sqrt{ 1 + \left( \frac{\partial {h}}{\partial {x}} \right)^2 }}\end{aligned} \hspace{\stretch{1}}(4.18)

By inspection we can also express the unit tangent, and we have for both

\begin{aligned}\hat{\mathbf{n}} &= \frac{1}{{\sqrt{ 1 + \left( \frac{\partial {h}}{\partial {x}} \right)^2 }}}\left( -\frac{\partial {h}}{\partial {x}}, 1 \right) \\ \hat{\boldsymbol{\tau}} &= \frac{1}{{\sqrt{ 1 + \left( \frac{\partial {h}}{\partial {x}} \right)^2 }}}\left( 1, \frac{\partial {h}}{\partial {x}} \right)\end{aligned} \hspace{\stretch{1}}(4.19)

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Steady state inclined flow down a plane of two layers of incompressible viscous equal density fluids.

Posted by peeterjoot on March 3, 2012

[Click here for a PDF of this post with nicer formatting and figures if the post had any (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]

Motivation.

Here’s one of the problems from section 2 of [1], a slight variation on what we did in class. It seemed to me that it would be a good midterm prep problem to try.

Steady state inclined flow down a plane of two layers of incompressible viscous equal density fluids.

Setup of the equations of motion for this system.

Our problem is illustrated in figure (\ref{fig:twoLayerInclinedFlow:twoLayerInclinedFlowFig1}) with a plane set at angle \alpha, fluid depths of h^{(1)} and h^{(2)} respectively, and viscosities \mu^{(1)} and \mu^{(2)}. We’ll write H = h^{(1)} + h^{(2)}. We have a pair of Navier-Stokes equations to solve

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{twoLayerInclinedFlowFig1}
\caption{Two fluids layers in inclined flow.}
\end{figure}

\begin{aligned}\rho \frac{d{{\mathbf{u}^{(i)}}}}{dt} = \rho \frac{\partial {\mathbf{u}^{(i)}}}{\partial {t}} + \rho (\mathbf{u}^{(i)} \cdot \boldsymbol{\nabla}) \mathbf{u}^{(i)} = - \boldsymbol{\nabla} p^{(i)} + \mu^{(i)} \boldsymbol{\nabla}^2 \mathbf{u}^{(i)} + \mu^{(i)} \boldsymbol{\nabla} (\boldsymbol{\nabla} \cdot \mathbf{u}^{(i)}) + \rho \mathbf{g}\end{aligned} \hspace{\stretch{1}}(2.1)

Our steady state and incompressibility constraints break this into a few independent equations

\begin{aligned}\rho \frac{\partial {\mathbf{u}^{(i)}}}{\partial {t}} &= 0 \\ \boldsymbol{\nabla} \cdot \mathbf{u}^{(i)} &= 0 \\ \rho (\mathbf{u}^{(i)} \cdot \boldsymbol{\nabla}) \mathbf{u}^{(i)} &= - \boldsymbol{\nabla} p^{(i)} + \mu^{(i)} \boldsymbol{\nabla}^2 \mathbf{u}^{(i)} + \rho \mathbf{g}\end{aligned} \hspace{\stretch{1}}(2.2)

Let’s require that we initially have no components of the flows in the y, or z directions. As the equivalent of Newton’s law for fluid flows, conservation of linear momentum requires that for our steady state problem we have u_y = u_z = 0 for the flows in both fluid layers.

FIXME: should go back and think about momentum conservation in the context of fluids. I’m now so used to thinking about this as a symmetry issue from a Lagrangian context. With no Lagrangian here, what would be a robust way treat this in fluids? What is the Lagrangian for a fluid mechanics system? I’d imagine that it would be possible to set up a field Lagrangian with velocity fields.

Our problem is now reduced to a problem in four quantities (two velocities and two pressures). With \mathbf{u}^{(i)} = (u^{(i)}, 0, 0) we can restate Navier-Stokes in coordinate form as

\begin{subequations}

\begin{aligned}\partial_x u^{(i)}(x,y,z) = 0\end{aligned} \hspace{\stretch{1}}(2.5a)

\begin{aligned}\rho (u^{(i)} \partial_x u^{(i)} = - \partial_x p^{(i)} + \mu^{(i)} (\partial_{xx} + \partial_{yy} + \partial_{zz}) u^{(i)} + \rho g \sin\alpha \end{aligned} \hspace{\stretch{1}}(2.5b)

\begin{aligned}0 = - \partial_y p^{(i)} - \rho g \cos\alpha \end{aligned} \hspace{\stretch{1}}(2.5c)

\begin{aligned}0 = - \partial_z p^{(i)} \end{aligned} \hspace{\stretch{1}}(2.5d)

\end{subequations}

In order to solve this, we have eight simultaneous non-linear PDEs, four unknown functions, plus boundary conditions to consider!

What are the boundary conditions? One is the “no-slip” condition, the experimental observation that velocities match at the interfaces. So we should have zero velocity for the fluid lying against the plane, and velocity matching between the fluids. The air above the fluid will also be flowing along at the rate of the uppermost portion of the top layer, but we’ll neglect that effect (i.e. considering two layers of equal density and not three, with one having a separate density). We also have matching of the traction vectors at the interfaces.

Writing this, it occurred to me that I didn’t fully understand what motivated that last boundary value condition. Talking to our Prof about this, the matching of the traction vectors at any point can be thought of as an observational issue, but this is also a force balance issue. There is an induced velocity in the direction of the traction vector at any given point. For example, when we have unidirectional flow, we must have no normal component of the traction vector, and only a tangential component, because we have only the tangential flow. It is probably reasonable to think about this roughly as the equivalent of matching both acceleration and velocity at the boundary, but because densities and viscosities vary, we have to match the traction vectors and not the acceleration itself.

Before continuing to solve our Navier-Stokes equations let’s express the condition that the tangential component of the traction vectors match algebraically.

Dropping indexes temporarily, for the normal to the surface \hat{\mathbf{n}} = (n_1, n_2, n_3) = (0, 1, 0) we want to compute

\begin{aligned}T_1 &= \sigma_{1 k} n_k \\ &= \sigma_{1 k} \delta_{2 k} \\ &= \sigma_{1 2} \\ &= -p \not{{\delta_{1 2}}} + 2 \mu e^{1 2} \\ &= \mu \left( \frac{\partial {u_1}}{\partial {y}} + \not{{\frac{\partial {u_2}}{\partial {x}}}} \right) \end{aligned}

So the tangential component of the traction vector is

\begin{aligned}\mathbf{T}^{(i)} = \mu^{(i)} \frac{\partial {u^{(i)}}}{\partial {y}} \hat{\mathbf{x}}.\end{aligned} \hspace{\stretch{1}}(2.6)

As noted above, this is in fact, the only component of the traction vector, since we don’t have any non-horizontal flow.

Our boundary value conditions, what we need in addition to the Navier-Stokes equations of 2.5, to solve our problem, are the matching at any interface of the following conditions

\begin{subequations}

\begin{aligned}u^{(i)} = u^{(j)} \end{aligned} \hspace{\stretch{1}}(2.7a)

\begin{aligned}p^{(i)} = p^{(j)} \end{aligned} \hspace{\stretch{1}}(2.7b)

\begin{aligned}\mu^{(i)} \frac{\partial {u^{(i)}}}{\partial {y}} = \mu^{(i)} \frac{\partial {u^{(j)}}}{\partial {y}}.\end{aligned} \hspace{\stretch{1}}(2.7c)

\end{subequations}

There are actually three interfaces to consider, that of the lower layer liquid with the inclined plane, the interface between the two fluid layers, and the interface between the upper layer fluid and the air above it.

Solving our equations of motion.

Starting with the simplest, the z-coordinate equation, of Navier-Stokes 2.5d, we can conclude that each of the pressures is not a function of z, so that we have

\begin{aligned}p^{(i)} = p^{(i)}(x, y).\end{aligned} \hspace{\stretch{1}}(2.8)

Using this, we can integrate our y-coordinate Navier-Stokes equation 2.5c, to find

\begin{aligned}p^{(i)} = - \rho g y \cos\alpha + f^{(i)}(x)\end{aligned} \hspace{\stretch{1}}(2.9)

At this point we can introduce the first boundary value constraint, that the pressures must match at the interfaces. In particular, on the upper surface, where we have atmospheric pressure p_A our pressure is

\begin{aligned}p^{(2)}(H) = - \rho g H \cos\alpha + f^{(2)}(x) = p_A,\end{aligned} \hspace{\stretch{1}}(2.10)

so f^{(2)} is constant with value

\begin{aligned}f^{(2)}(x) = p_A + \rho g H \cos\alpha,\end{aligned} \hspace{\stretch{1}}(2.11)

which fully determines the density of the upper surface

\begin{aligned}p^{(2)}(y) = \rho g \cos\alpha (H - y) + p_A.\end{aligned} \hspace{\stretch{1}}(2.12)

Matching the pressure between the two layers of fluids we have

\begin{aligned}p^{(1)}(h_1) &= - \rho g h_1 \cos\alpha + f^{(1)}(x) \\ &= p^{(2)}(h_1) \\ &= \rho g \cos\alpha (H - h_1) + p_A,\end{aligned}

so that our undetermined function f^{(1)}(x)

\begin{aligned}f^{(1)}(x) = \rho g H \cos\alpha + p_A.\end{aligned} \hspace{\stretch{1}}(2.13)

This is an intuitively satisfying result. With the densities equal, it seems sensible that the pressure would have a single functional form throughout both layers, dependent only on the total height y, independent of the velocities and viscosities. That is precisely what we find

\begin{aligned}p(y) = \rho g \cos\alpha (H - y) + p_A.\end{aligned} \hspace{\stretch{1}}(2.14)

Having solved for the pressure, we are now set to return to the remaining Navier-Stokes equations 2.5a, and 2.5b for this system. From 2.5a we see that the non-linear term on the LHS of 2.5b is killed and also see that our velocities can only be functions of y and z

\begin{aligned}u^{(i)} = u^{(i)}(y, z).\end{aligned} \hspace{\stretch{1}}(2.15)

While more general solutions can likely be found, we will limit ourselves to looking only for solutions that are functions of y. From our solution to the pressure part of the problem p^{(i)} = p^{(i)}(y), we also see that the pressure term \partial_x p^{(i)} of 2.5b is killed. We are left with just

\begin{aligned}0 &= \mu^{(i)} (\partial_{xx} + \partial_{yy} + \partial_{zz}) u^{(i)} + \rho g \sin\alpha \\ &= \mu^{(i)} \partial_{yy} u^{(i)} + \rho g \sin\alpha \\ &= \mu^{(i)} \frac{d^2}{dy^2} u^{(i)} + \rho g \sin\alpha \end{aligned}

This is directly integrable, and we find for the velocities and traction vectors respectively

\begin{subequations}

\begin{aligned}u^{(i)} = -\frac{\rho g \sin\alpha }{2 \mu^{(i)}} y^2 + A^{(i)} y + B^{(i)}.\end{aligned} \hspace{\stretch{1}}(2.16a)

\begin{aligned}T_x^{(i)} = \mu^{(i)} \frac{\partial {u^{(i)}}}{\partial {y}} = - \rho g y \sin\alpha + \mu^{(i)} A^{(i)} \end{aligned} \hspace{\stretch{1}}(2.16b)

\end{subequations}

Our boundary conditions left to exploit are

\begin{aligned}u^{(1)}(0) &= 0 \\ u^{(1)}(h_1) &= u^{(2)}(h_1) \\ T_x^{(1)}(h_1) &= T_x^{(2)}(h_1) \\ T_x^{(2)}(H) &= 0,\end{aligned} \hspace{\stretch{1}}(2.17)

The first is the no-slip condition with the plane, the last, an approximation assuming that the liquid isn’t producing a measurable force on the air above it, and two more for the interface between the two fluids.

From u^{(1)}(0) = 0 we see immediately that we have B^{(1)} = 0. From the traction vector equality in the atmosphere, we have

\begin{aligned}0 = - \rho g H \sin\alpha + \mu^{(2)} A^{(2)},\end{aligned} \hspace{\stretch{1}}(2.21)

or

\begin{aligned}A^{(2)} = \frac{\rho g H \sin\alpha }{ \mu^{(2)} }.\end{aligned} \hspace{\stretch{1}}(2.22)

These reduce the problem to solving for two last integration constants, where our velocities are

\begin{aligned}u^{(1)} &= -\frac{\rho g \sin\alpha }{2 \mu^{(1)}} y^2 + A^{(1)} y \\ u^{(2)} &= \frac{\rho g \sin\alpha }{2 \mu^{(2)}} \left( 2 H y -y^2 \right) + B^{(2)}.\end{aligned} \hspace{\stretch{1}}(2.23)

and our traction vectors are

\begin{aligned}T_x^{(1)} &= - \rho g y \sin\alpha + \mu^{(1)} A^{(1)} \\ T_x^{(2)} &= \rho g \sin\alpha \left( H - y \right).\end{aligned} \hspace{\stretch{1}}(2.25)

Matching both at the interface (y = h_1) gives us

\begin{aligned}-\frac{\rho g \sin\alpha }{2 \mu^{(1)}} h_1^2 + A^{(1)} h_1 &= \frac{\rho g \sin\alpha }{2 \mu^{(2)}} h_1 \left( 2 h_2 + h_1 \right) + B^{(2)} \\ - \rho g h_1 \sin\alpha + \mu^{(1)} A^{(1)} &= \rho g h_2 \sin\alpha \end{aligned} \hspace{\stretch{1}}(2.27)

We find

\begin{aligned}A^{(1)} = \frac{\rho g H \sin\alpha }{ \mu^{(1)} },\end{aligned} \hspace{\stretch{1}}(2.29)

and

\begin{aligned}B^{(2)} &=-\frac{\rho g \sin\alpha }{2 \mu^{(1)}} h_1^2 + \frac{\rho g H \sin\alpha }{ \mu^{(1)} }h_1 - \frac{\rho g \sin\alpha }{2 \mu^{(2)}} h_1 \left( 2 h_2 + h_1 \right) \\ &=\frac{\rho g h_1 \sin\alpha }{2}\left(- \frac{h_1}{ \mu^{(1)}}+ \frac{2 H}{ \mu^{(1)}}- \frac{2 h_2 + h_1}{ \mu^{(2)}}\right) \\ &=\frac{\rho g h_1 \sin\alpha }{2} (2 h_2 + h_1) \left(\frac{1}{{ \mu^{(1)}}}-\frac{1}{{ \mu^{(2)}}}\right)\end{aligned}

So, finally, we have

\begin{aligned}u^{(1)}(y) &= \frac{\rho g \sin\alpha }{2 \mu^{(1)}} (2 H y -y^2) \\ u^{(2)}(y) &= \frac{\rho g \sin\alpha }{2 \mu^{(2)}} \left( 2 H y -y^2 + h_1 (2 h_2 + h_1) \left( \frac{\mu^{(2)}}{\mu^{(1)}} - 1 \right)\right) \\ p(y) &= \rho g \cos\alpha (H - y) + p_A.\end{aligned} \hspace{\stretch{1}}(2.30)

The final result looks reasonable. If the viscosities are equal then we have the same velocity profile in both layers. That makes sense given the equal densities, since there would really be nothing that would then distinguish the two layers.

As a follow-on to this calculation, I’d like to re-do it allowing for different densities. Specifically, I’m curious how much the air in the neighborhood of some flowing water gets dragged by that flow. It never occurred to me that this would occur, and I’d like to plug in some numbers and see what the results are. This should be something that can be modeled with two layers like this, one of fluid, one of air of a specific thickness (allowing pressure to vary due to the velocity gradient), and one final layer of air at a fixed pressure (atmospheric). I wouldn’t expect that problem to be much harder than this one, although it may end up being worthwhile to let a computer algebra system do some of the grunt work to solve all the resulting equations.

References

[1] D.J. Acheson. Elementary fluid dynamics. Oxford University Press, USA, 1990.

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