Warning 7.11.1.
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Exercises 7.11 Problems set 8 to hand in
Please hand in the problems in this Section for Set 8 (the final set of problems, due in the week commencing 8 Dec).
1.
A tethered sphere of radius \(a\) is surrounded by a flowing incompressible Newtonian fluid of density \(\rho\) and shear viscosity \(\mu\text{.}\) In the far field, the fluid velocity is uniform and equal to \(U\) (right to left).
(a)
Write down an expression for the Reynolds number \(\mathrm{Re}\) of the fluid.
(b)
Describe what you can infer about the Reynolds number in each case (a), (b) and (c) from these pictures. Explain your reasoning.
Solution.
-
In case (a), the flow is smooth and steady around the sphere, and the flow is also reversible (left–right symmetry), indicating a low Reynolds number.
-
In case (b), there is some disturbance in the flow behind the sphere, suggesting a moderate Reynolds number, of order 100.
-
In case (c), there is a turbulent wake behind the sphere, indicating a high Reynolds number, \(\mathrm{Re}\gg 10^3\text{.}\)
2. Channel flow.
A Newtonian viscous fluid flows steadily in the \(+x\)-direction in two dimensions between plates at \(y=\pm d\text{.}\) You may neglect gravity in this problem.
(a)
Write down the boundary conditions at each plate.
(b)
Make the assumption that the flow is fully developed and use the continuity equation to show that \(v=0\) everywhere.
Solution.
Fully-developed flow implies the velocity components do not depend on \(x\text{.}\) Thus, the continuity equation reduces to
\begin{equation*}
\pd{v}{y}=0,
\end{equation*}
implying that \(v\) is independent of \(y\text{.}\) The no-slip boundary conditions at each plate implies \(v=0\) there. Hence, \(v=0\) everywhere.
(c)
Simplify the Navier–Stokes equations using the assumptions and \(v=0\) to show that
\begin{equation*}
\mu\pd{^2u}{y^2}=-G,
\end{equation*}
where \(G\) is a constant that you should specify.
Solution.
The three components of the Navier–Stokes equations simplify to
\begin{align*}
&0=-\pd{p}{x}+\mu\pd{^2u}{y^2},\\
&0=-\pd{p}{y},\\
&0=0.
\end{align*}
The \(y\)-component shows the pressure only depends on \(x\text{,}\) and for the two terms in the \(x\)-component:
\begin{align*}
&-\pd{p}{x}\quad\mbox{depends on only},\\
&\mu\pd{^2u}{y^2}\quad\mbox{depends on only}.
\end{align*}
Hence both these terms are constant. Setting
\begin{equation*}
G=-\pd{p}{x},
\end{equation*}
where \(G\) is constant, we have
\begin{equation*}
\mu\pd{^2u}{y^2}=-G.
\end{equation*}
(d)
Solve this applying the boundary conditions to find the flow profile.
Solution.
We have
\begin{equation*}
u=-\frac{G}{2\mu}y^2+C_1y+C_2,
\end{equation*}
where \(C_1\) and \(C_2\) are constants, and, applying the boundary condition at \(y=d\text{,}\)
\begin{equation*}
0=-\frac{G}{2\mu}d^2+C_1d+C_2,
\end{equation*}
and at \(y=-d\text{,}\)
\begin{equation*}
0=-\frac{G}{2\mu}d^2-C_1d+C_2.
\end{equation*}
Subtracting these equations, we see that \(C_1=0\) and then we find \(C_2=Gd^2/(2\mu)\text{.}\) Hence
\begin{equation*}
u=-\frac{G}{2\mu}y^2+\frac{Gd^2}{2\mu}
=\frac{G}{2\mu}\left(d^2-y^2\right),
\end{equation*}
and \(v=w=0\text{.}\)
(e)
Find the stress exerted by the fluid on both plates. If each plates has (large) area \(A\text{,}\) what is the \(x\)-component of the tethering force required to keep each plate fixed in place?
Solution.
For the plate at \(y=k\text{,}\) the \(x\)-component of the stress is
\begin{equation*}
\tau=-\mu\pd{u}{y}
\end{equation*}
(the minus sign occurs since the vector into the fluid points in the \(-y\)-direction) and the \(z\)-component of the stress is zero since the flow is two-dimensional. The \(x\)-component equals
\begin{equation*}
\tau=-\mu \left(-\frac{Gd}{\mu}\right)=Gd.
\end{equation*}
For the plate at \(y=-k\text{,}\) the \(x\)-component of the stress is
\begin{equation*}
\tau=\mu\pd{u}{y}
\end{equation*}
(the plus sign occurs since the vector into the fluid points in the \(+y\)-direction) and the \(z\)-component of the stress is zero since the flow is two-dimensional. The \(x\)-component equals
\begin{equation*}
\tau=\mu \left(-\frac{G(-d)}{\mu}\right)=Gd.
\end{equation*}
Thus the stress at both plates is \(Gd\) in the \(+x\)-direction, and the force on the whole plate is \(GdA\) in the \(+x\)-direction. Hence the tethering force to keep each plate fixed is \(GdA\) in the \(-x\)-direction.
Note that it is very easy to make a sign error on this type of question. I would therefore recommend always reflecting on your answer to see if the direction you get is what you’d expect from the physics. A viscous fluid tries to drag the surface along with the flow, and so the force due to the shear stress is sometimes called the drag force. Thus, the tethering force must oppose this drag force, which is what we found.
3. Wavy-walled pipe.
A long cylindrical pipe with slowly varying radius \(R(z)\) is aligned with its axis pointing in the \(z\)-direction. A much narrower cylinder of constant radius \(a\) is placed coaxially within the larger cylinder. The narrow cylinder has porous walls and emits fluid at a constant velocity \(U\) perpendicular to its surface. A fluid flows steadily in the gap between the two cylinders.
In this question, you will work in cylindrical polar coodinates \((r,\theta,z)\text{,}\) centred on the axis of the pipe.
Despite the fact that there is flow out from the inner cylinder in the radial direction, you may assume that the flow in the annular gap between the cylinders is unidirectional along the pipe in the \(+z\)-direction. (In reality there will be a thin layer next to the inner cylinder in which this is not true.) Furthermore, you may assume that the flow is uniform across the cross-section, that is \(\bu=(u(z),0,0)\text{.}\)
(a)
Consider a narrow slice of the cylinder between \(z\) and \(z+\de z\text{.}\) Find an expression for the volumetric flux of fluid through the surfaces \(z\) and \(z+\de z\text{.}\)
Solution.
The volumetric flux into the left-hand surface of this region equals area times velocity:
\begin{equation*}
\pi (R^2-a^2)u
\end{equation*}
and that out of the right-hand surface equals
\begin{equation*}
\pi \left((R+\de R)^2-a^2\right)(u+\de u)
=\pi \left(\left(R+R'\de z\right)^2-a^2\right)\left(u+u'\de z\right),
\end{equation*}
where we used the chain rule to rewrite the increments in terms of \(\de z\text{.}\) Note that throughout this equation, we neglect \(\de z^2\) and higher powers, as they are small.
(b)
Now find an expression for the volumetric flux of fluid through the section of the inner cylinder between \(z\) and \(z+\de z\text{.}\)
(c)
Now use the principle of mass conservation to show that
\begin{equation*}
\dd{u}{z}=\frac{2(aU-RR'u)}{R^2-a^2}.
\end{equation*}
Note that you can actually solve this equation by instead considering the volume between \(z=0\) and \(z=z_0\text{.}\) This volume has four surfaces: the inner cylinder surface \(r=a\text{,}\) the outer cylinder surface \(r=R(z)\text{,}\) and the two cross-sectional surfaces at \(z=0\) and \(z=z_0\text{.}\) The inner cylinder surface gives a volume flux in of \(2\pi aUz_0\text{,}\) the outer cylinder surface gives no volume flux, the surface \(z=0\) gives a volume flux in of \(\pi(R(0)^2-a^2)u(0)\text{,}\) and the surface \(z=z_0\) gives a volume flux out of \(\pi(R(z_0)^2-a^2)u(z_0)\text{.}\) Equating total volume flux in and volume flux out gives
\begin{equation*}
2\pi aUz_0+\pi(R(0)^2-a^2)u(0)=\pi(R(z_0)^2-a^2)u(z_0),
\end{equation*}
and hence
\begin{equation*}
u(z_0)=\frac{2aUz_0+(R(0)^2-a^2)u(0)}{R(z_0)^2-a^2}.
\end{equation*}
This is true for all \(z_0\text{,}\) hence (setting \(z=z_0\))
\begin{equation*}
u=\frac{2aUz+(R(0)^2-a^2)u(0)}{R(z)^2-a^2}.
\end{equation*}
This was not asked for in this question.
Solution.
Since the volume of the pipe between \(z\) and \(z+\de z\) is fixed, the total volumetric flux into this region equals the total volumetric flux out.
Thus
\begin{align*}
&\pi (R^2-a^2)u + 2\pi aU\de z = \pi \left(\left(R+R'\de z\right)^2-a^2\right)\left(u+u'\de z\right),\\
\Rightarrow\quad&
2\pi aU\de z = \pi \left((R^2-a^2)u'+2RR'u\right)\de z,\\
\Rightarrow\quad&
2aU = (R^2-a^2)u'+2RR'u,\\
\Rightarrow\quad&
u' = \frac{2(aU-RR'u)}{R^2-a^2},
\end{align*}
thus proving the result.
(d)
Write down an expression for the Reynolds number of this flow, explaining your reasoning. You may assume that the pipe has length \(L\) and that \(R_0\) is a typical value of \(R(z)\text{.}\) Your answer should be in terms of the problem parameters \(\rho\text{,}\) \(\mu\text{,}\) \(a\text{,}\) \(U\text{,}\) \(L\) and \(R_0\text{.}\)
Solution.
A suitable lengthscale is \(R_0\text{.}\) Simplifying the differential equation for \(u\text{,}\) by neglecting variations in pipe radius \(R'\) and the radius of the inner cylinder \(a\text{,}\) gives
\begin{equation*}
\dd{u}{z} \approx \frac{2aU}{R_0^2},
\end{equation*}
suggesting a suitable velocity scale is this times \(L\) (the velocity that would be attained at the end of the pipe):
\begin{equation*}
\frac{2aLU}{R_0^2}.
\end{equation*}
Hence the Reynolds number is
\begin{equation*}
\mathrm{Re}=\frac{\rho R_0}{\mu}\frac{2aLU}{R_0^2}
=\frac{2aLU\rho}{\mu R_0}.
\end{equation*}
Many students did not find an expression for the velocity scale, and just left it as \(u(z)\) or similar. This would mean that we do not have an expression for the Reynolds number in terms of the problem parameters. Also, many students did not approximate \(R-a\) as \(R_0\text{,}\) which simplifies the expression (some did get \(R_0-a\text{,}\) but you can go further than this).
I have added a final sentence to the question to clarify that the answer should be in terms of the problem parameters as very few students attempted to find an order of magnitude for \(u(z)\text{.}\)
(e)
State in a sentence the conditions under which it might be appropriate to assume that the velocity is uniform across the cross section. You do not need to do a calculation.
Solution.
The Reynolds number would need to be large (but not so large that the flow becomes turbulent).
In the limit of large Reynolds number, we would recover the Euler equation from the Navier–Stokes equation (see Exercise 7.9.11). In this limit we only need a no-penetration condition (\(\bu\cdot\hat{\bn}=0\)) and not the full no-slip conditions (\(\bu=\bzero\)), and so in this case uniform flow is possible.
(f)
In reality, in a sentence, what would you observe in the flow near the outer wall?
