Warning 7.9.1.
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Exercises 7.9 Practice Exercises
1. Reynolds number estimation.
Give an order-of-magnitude estimate of the Reynolds number in each of the following situations:
(a)
flow past the wing of a jumbo jet at 150 m/s (roughly half the speed of sound);
Solution.
The kinematic viscosity of air at standard temperature and pressure is about \(1.5\times10^{-5}~\textrm{m}^2/\textrm{s}\text{.}\) The wingspan of a jumbo jet is about \(60~\textrm{m}\text{,}\) so the Reynolds number is approximately
\begin{equation*}
\textrm{Re}=\frac{UL}{\nu}
=\frac{150\times60}{1.5\times10^{-5}}
=6\times10^8.
\end{equation*}
(b)
blood flow in a large artery (diameter 2 cm) at 0.3 m/s;
Solution.
The kinematic viscosity of blood is about \(2.4\times10^{-6}~\textrm{m}^2/\textrm{s}\text{.}\) The diameter of the artery is about \(0.02~\textrm{m}\text{,}\) so the Reynolds number is approximately
\begin{equation*}
\textrm{Re}=\frac{UL}{\nu}
=\frac{0.3\times0.02}{2.4\times10^{-6}}
=2500.
\end{equation*}
Although this is above 2000, blood flow in arteries is typically laminar.
(c)
a thick layer of honey draining off a spoon at 1 cm/s;
Solution.
(d)
Solution.
The kinematic viscosity of water at room temperature is about \(10^{-6}~\textrm{m}^2/\textrm{s}\text{.}\) The length of the tail is about \(10^{-3}~\textrm{m}\text{,}\) so the Reynolds number is approximately
\begin{equation*}
\textrm{Re}=\frac{UL}{\nu}
=\frac{100\times10^{-6}\times10^{-3}}{10^{-6}}
=0.1.
\end{equation*}
2. Nondimensionalisation.
In this problem you will work through the nondimensionalisation of a typical fluid flow problem to identify the Reynolds number as a key dimensionless parameter.
Consider the two-dimensional steady flow of a viscous incompressible fluid past a circular cylinder of radius \(a\text{.}\) Far from the cylinder, the fluid velocity is uniform with magnitude \(U\) in the \(x\)-direction. The governing equations are the incompressible Navier–Stokes equations and continuity equation:
\begin{equation*}
\left(\bu\cdot\nabla\right)\bu=-\frac{1}{\rho}\nabla p+\nu\nabla^2\bu,\quad
\nabla\cdot\bu=0,
\end{equation*}
where \(\bu=(u,v)\) is the velocity field, \(p\) is the pressure field, \(\rho\) is the fluid density, and \(\nu\) is the kinematic viscosity of the fluid.
(a)
Write down suitable boundary conditions at the cylinder surface and far from the cylinder.
Solution.
At the cylinder surface, we have the no-slip and no-penetration boundary conditions:
\begin{equation*}
u=v=0.
\end{equation*}
Far from the cylinder, the velocity approaches the uniform flow:
\begin{equation*}
u\rightarrow U,\quad v\rightarrow0\quad\textrm{as }\sqrt{x^2+y^2}\rightarrow\infty.
\end{equation*}
(b)
Introduce nondimensional variables
\begin{equation*}
x^*=\frac{x}{a},\quad y^*=\frac{y}{a},\quad
u^*=\frac{u}{U},\quad v^*=\frac{v}{U},\quad
p^*=\frac{p}{\rho U^2}.
\end{equation*}
Rewrite the governing equations and boundary conditions in terms of these nondimensional variables.
Solution.
We have:
\begin{equation*}
\frac{U^2}{L}\left(\bu^*\cdot\nabla^*\right)\bu^*
=-\frac{U^2}{L}\nabla^*p^*
+\frac{\nu U}{L^2}\nabla^{*2}\bu^*,\quad
\frac{U}{L}\nabla^*\cdot\bu^*=0,
\end{equation*}
which simplify to
\begin{equation*}
\mathrm{Re}\left(\bu^*\cdot\nabla^*\right)\bu^*
=-\mathrm{Re}\nabla^*p^*
+\nabla^{*2}\bu^*,\quad
\nabla^*\cdot\bu^*=0,
\end{equation*}
while the boundary conditions become:
\begin{equation*}
u^*=v^*=0
\end{equation*}
at \(x^{*2}+y^{*2}=1\) and
\begin{equation*}
u^*\rightarrow 1,\quad v^*\rightarrow0\quad\textrm{as }\sqrt{x^{*2}+y^{*2}}\rightarrow\infty.
\end{equation*}
(c)
You should have found that the nondimensional problem depends only on a single parameter: the Reynolds number. Thus flows at equal Reynolds numbers are geometrically similar, and one only needs to know the Reynolds number to understand the flow.
This was a particularly simple problem that has an unambiguous length and velocity scale, and so nondimensionalisation of the problem works particularly well. However, it is frequently helpful to nondimensionalise problems before solving so that key dimensionless groups of parameters (such as the Reynolds number) can be identified.
3. Falling ball viscometer.
For slowly moving spheres, the resistive force is given by Stokes drag, and equals \(3\pi\mu Du\text{,}\) where \(D\) is the diameter of the sphere, \(\mu\) is the dynamic viscosity of the fluid, and \(u\) is the speed of the sphere (this is a famous result).
(a)
Check the dimensions of the Stokes drag. Is the dimension what you expect?
Solution.
The dimensions of the terms are as follows:
\begin{equation*}
[3\pi]=1,\quad
[\mu]=ML^{-1}T^{-1},\quad
[D]=L,\quad
[u]=LT^{-1},
\end{equation*}
meaning that
\begin{equation*}
\left[3\pi\mu Du\right]=1\times ML^{-1}T^{-1}\times L\times LT^{-1}
=MLT^{-2}.
\end{equation*}
Thus the expression \(3\pi\mu Du\) for the Stokes drag has the dimensions of force, as expected.
(b)
A steel ballbearing of mass \(m\text{,}\) diameter \(D\) falls slowly through a viscous fluid of dynamic viscosity \(\mu\text{.}\) In this question you can neglect the buoyancy force. What are the other two forces acting on the ballbearing?
(c)
Write down a differential equation for the time-evolution of the velocity \(u\) of the ballbearing, assuming that the acceleration due to gravity is \(g\text{.}\)
(d)
Given that the ballbearing is released from rest, solve the differential equation to show that
\begin{equation*}
u=u_t\left(1-e^{-t/\tau}\right),
\end{equation*}
where \(u_t\) and \(\tau\) are constant parameters that you should define. What is the physical significance of \(u_t\) and \(\tau\text{?}\)
Solution.
We have
\begin{equation*}
\frac{du}{dt}+\frac{3\pi\mu D}{m}u=g,
\end{equation*}
for which the integrating factor is \(e^{\frac{3\pi\mu D}{m}t}\text{,}\) and
\begin{equation*}
ue^{\frac{3\pi\mu D}{m}t}=\int ge^{\frac{3\pi\mu D}{m}t}dt
=\frac{mg}{3\pi\mu D}e^{\frac{3\pi\mu D}{m}t}+c,
\end{equation*}
where \(c\) is a constant of integration. Hence
\begin{equation*}
u=\frac{mg}{3\pi\mu D}+ce^{-\frac{3\pi\mu D}{m}t}
\end{equation*}
The initial condition is \(u=0\) at \(t=0\text{,}\) and, substituting this into the above equation, \(0=\frac{mg}{3\pi\mu D}+c\text{,}\) giving \(c=-\frac{mg}{3\pi\mu D}\text{.}\) Hence
\begin{equation*}
u=\frac{mg}{3\pi\mu D}\left(1-e^{-\frac{3\pi\mu D}{m}t}\right).
\end{equation*}
We substitute \(u_t=\frac{mg}{3\pi\mu D}\) and \(\tau=\frac{m}{3\pi\mu D}\text{.}\) With these substitutions we get
\begin{equation*}
u=u_t\left(1-e^{-t/\tau}\right).
\end{equation*}
The physical significance of these choices is that \(u_t\) is the terminal velocity of the sphere and \(\tau\) is the timescale over which the velocity of the sphere approaches the terminal velocity. Every \(\tau\) units of time, the velocity of the sphere becomes closer to the terminal velocity \(u_t\) by a factor \(e\approx2.7\text{.}\)
(e)
Sketch the velocity as a function of time.
(f)
If the ballbearing has diameter 2 mm and is falling through glycerol at 20\(^\circ\)C, what is its terminal velocity in mm/s and what is the timescale taken to reach the terminal velocity? You may assume that the dynamic viscosity of glycerol at 20\(^\circ\)C is 1.2 Pa s and the density of steel is 8,000 kg/m\(^3\text{.}\)
Solution.
Using the fact that the volume of a sphere is \(\pi D^3/6\text{,}\) where \(D\) is the diameter, we have \(m=\pi\rho_sD^3/6\text{,}\) where \(\rho_s=8,000\) kg/m\(^3\text{.}\) Thus
\begin{align*}
u_t=&\frac{mg}{3\pi\mu D}=\frac{mg}{3\pi\mu D}=\frac{\rho_sgD^2}{18\mu}\\
=&\frac{8000\times9.81\times(4\times10^{-6})}{18\times1.2}
=0.0146\,\textrm{m/s}=14.6\,\textrm{mm/s}.
\end{align*}
The timescale \(\tau\) is given by
\begin{equation*}
\tau=\frac{m}{3\pi\mu D}=\frac{\rho_sD^2}{18\mu}
=\frac{8000\times4\times10^{-6}}{18\times1.2}
=0.00148\,\textrm{s}.
\end{equation*}
(g)
Comment on the experimental set up needed to measure the terminal velocity.
Solution.
The ball would attain its terminal velocity very quickly (in much less than a second), and since \(u_t\) is on order of centimetres per second, a container of about 10 cm deep would be suitable for measurement. In fact this idea is commonly used to measure the viscosity of fluids (since the viscosity can be calculated from the terminal velocity); this is called a falling ball viscometer.
4. Stress on a plate.
A rigid plate of length \(L\) is fixed along the line \(y=kx\) and has width \(W\) in the \(z\)-direction. An incompressible fluid of density \(\rho\) and viscosity \(\mu\) flows steadily in the shaded region \(y>kx\text{,}\) shown in Figure 7.9.3.
(a)
Given that the velocity field is
\begin{equation*}
u=\alpha y+Ax+B,\quad v=-Ay+Dx+C,
\end{equation*}
what are \(A\text{,}\) \(B\text{,}\) \(C\text{,}\) \(D\) in terms of \(\alpha\) and \(k\text{?}\) Simplify the velocity components \(u\) and \(v\text{.}\) (Note that the fact that the coefficients on the terms \(Ax\) in \(u\) and \(-Ay\) in \(v\) are equal and opposite is required to ensure incompressibility of the fluid).
Solution.
We apply no-slip boundary conditions at the plate. Thus if \(y=kx\) then we must have \(u=0\) and \(v=0\text{.}\) Hence
\begin{equation*}
0=\alpha kx+Ax+B,\quad 0=-Akx+Dx+C,
\end{equation*}
for all \(x\text{,}\) which implies that \(\alpha k+A=0\text{,}\) \(B=0\text{,}\) \(-Ak+D=0\) and \(C=0\text{.}\) Solving, we obtain
\begin{equation*}
A=-\alpha k,\quad
B=0,\quad
D=-\alpha k^2,\quad
C=0.
\end{equation*}
Hence
\begin{equation*}
u=\alpha\left(y-kx\right),\quad v=\alpha k\left(y-kx\right).
\end{equation*}
(b)
Find the drag force on the plate.
Hint.
The drag force can be found by changing coordinates from \((x,y)\) to \((\xi,\eta)\text{,}\) which are Cartesian coordinates where the \(\xi\)-axis runs along the plate and the \(\eta\)-axis goes from the plate perpendicularly into the fluid. You may use the fact that
\begin{equation*}
\boldsymbol{i}=\cos\theta\hat{\boldsymbol{\xi}}-\sin\theta\hat{\boldsymbol{\eta}},\qquad
\boldsymbol{j}=\sin\theta\hat{\boldsymbol{\xi}}+\cos\theta\hat{\boldsymbol{\eta}},
\end{equation*}
where \(\boldsymbol{i}\text{,}\) \(\boldsymbol{j}\) are unit vectors in the \(x\)- and \(y\)-directions, respectively, and \(\hat{\boldsymbol{\xi}}\text{,}\) \(\hat{\boldsymbol{\eta}}\) are unit vectors in the \(\xi\)- and \(\eta\)-directions, respectively, and \(\theta\) is the angle between the plate and the \(x\)-axis. You may use the fact that \(\cos\theta=1/(\sqrt{1+k^2})\text{,}\) \(\sin\theta=k/(\sqrt{1+k^2})\) and
\begin{equation*}
x=\frac{\xi}{\sqrt{1+k^2}}-\frac{k\eta}{\sqrt{1+k^2}},\quad
y=\frac{k\xi}{\sqrt{1+k^2}}+\frac{\eta}{\sqrt{1+k^2}}.
\end{equation*}
Solution.
We perform the coordinate transform illustrated in Figure 7.9.4 from the \((x,y)\)-plane to the \((\xi,\eta)\)-plane:
We use
\begin{align*}
\bu=&u\boldsymbol{i}+v\boldsymbol{j}\\
=&u\left(\cos\theta\hat{\boldsymbol{\xi}}-\sin\theta\hat{\boldsymbol{\eta}}\right)
+v\left(\sin\theta\hat{\boldsymbol{\xi}}+\cos\theta\hat{\boldsymbol{\eta}}\right)\\
=&\left(u\cos\theta+v\sin\theta\right)\hat{\boldsymbol{\xi}}
+\left(-u\sin\theta+v\cos\theta\right)\hat{\boldsymbol{\eta}}\\
=&\frac{\alpha(y-kx)+\alpha k^2(y-kx)}{\sqrt{1+k^2}}\hat{\boldsymbol{\xi}}
+\frac{-\alpha k(y-kx)+\alpha k(y-kx)}{\sqrt{1+k^2}}\hat{\boldsymbol{\eta}}\\
=&\alpha\sqrt{1+k^2}(y-kx)\hat{\boldsymbol{\xi}}\\
=&\alpha\sqrt{1+k^2}\sqrt{1+k^2}\eta\hat{\boldsymbol{\xi}}\\
=&\alpha(1+k^2)\eta\hat{\boldsymbol{\xi}}
\end{align*}
and so the velocity components are \(u_\xi=\alpha(1+k^2)\eta\) and \(u_\eta=0\text{.}\)
Thus the shear stress is given by
\begin{equation*}
\mu\pd{u_\xi}{\eta}=\mu\alpha(1+k^2),
\end{equation*}
and hence the drag force equals
\begin{equation*}
\mu\alpha(1+k^2)LW.
\end{equation*}
(c)
In this part, find the drag force on the plate by using the stress tensor, and check your answer against the previous part.
Solution.
We have the normal vector to the plate
\begin{equation*}
\hat{\bn}=\frac1{\sqrt{1+k^2}}(-k,1,0).
\end{equation*}
The rate-of-strain tensor is given by
\begin{equation*}
\be=\left(\begin{matrix}
-\alpha k&\frac12\alpha(1-k^2)&0\\
\frac12\alpha(1-k^2)&\alpha k&0\\
0&0&0
\end{matrix}\right).
\end{equation*}
Hence the stress tensor equals
\begin{equation*}
\bsigma=-p\bI+2\mu\be
=\left(\begin{matrix}
-p-2\mu\alpha k&\mu\alpha(1-k^2)&0\\
\mu\alpha(1-k^2)&-p+2\mu\alpha k&0\\
0&0&-p
\end{matrix}\right).
\end{equation*}
The stress on the surface is given by
\begin{align*}
\tau=&\bsigma\hat{\bn}\\
=&\frac1{\sqrt{1+k^2}}\left(kp+2\mu\alpha k^2+\mu\alpha(1-k^2),-\mu k\alpha(1-k^2)-p+2\mu\alpha k,0\right)\\
=&\frac1{\sqrt{1+k^2}}\left(kp+\mu\alpha(1+k^2),-p+\mu k\alpha(1+k^2),0\right)\\
=&\frac1{\sqrt{1+k^2}}\left(p(k,-1,0)+\mu\alpha(1+k^2)(1,k,0)\right)\\
=&-p\hat{\bn}+\mu\alpha(1+k^2)\hat{\bt},
\end{align*}
where \(\hat{\bt}\) is the unit tangential vector to the plate in the \((x,y)\)-plane.
This is the pressure pushing into the plate (direction \(-\hat{\bn}\)) plus the drag \(\mu\alpha(1+k^2)\) along the plate. Hence, the total drag force is the drag times the area:
\begin{equation*}
\mu\alpha(1+k^2)
\end{equation*}
which agrees with the previous answer.
5. Stress on a surface due to the pressure.
In this problem, you will calculate the resultant force on the cube shown in Figure 7.5.1 due to the pressure in the surrounding fluid acting on the fluid within the cube. We will consider two different methods of calculating this force.
(a)
Assuming that the pressure \(p(x,y,z)\) is uniform over the surface \(x=x_0\) and over the surface \(x=x_0+\delta\text{,}\) calculate the force on the fluid in the cube due to the pressure of the surrounding fluid on these two faces and show that it is approximately
\begin{equation*}
-\delta^3\pd{p}{x}\bi=-V\pd{p}{x}\bi,
\end{equation*}
where \(V=\delta^3\) is the volume of the cube.
Solution.
The force on \(x=x_0\) is pressure times area acting into the cube, which is \(\delta^2p|_{x_0}\bi\text{,}\) while the force on \(x=x_0+\delta\) equals \(-\delta^2p|_{x+\delta}\bi\text{.}\) Hence the contribution to the force from these two faces is
\begin{equation*}
\delta^2p|_{x_0}\bi-\delta^2p|_{x+\delta}\bi
\approx-\delta^3\pd{p}{x}\bi=-V\pd{p}{x}\bi.
\end{equation*}
(b)
Similarly, calculate the force on the fluid in the cube due to the pressure of the surrounding fluid on the faces \(y=y_0\) and \(y=y_0+\delta\text{,}\) and on the faces \(z=z_0\) and \(z=z_0+\delta\text{.}\) Hence show that the total force on the fluid in the cube due to the pressure of the surrounding fluid is given by
\begin{equation*}
-V\pd{p}{x}\bi-V\pd{p}{y}\bj-V\pd{p}{z}\bk=-V\nabla p.
\end{equation*}
Solution.
The force on \(y=y_0\) is \(\delta^2p|_{y_0}\bj\) and that on \(y=y_0+\delta\) is \(-\delta^2p|_{y+\delta}\bj\text{.}\) This gives a contribution
\begin{equation*}
\delta^2p|_{y_0}\bj-\delta^2p|_{y+\delta}\bj
\approx-\delta^3\pd{p}{y}\bj=-V\pd{p}{y}\bj.
\end{equation*}
Similarly, the force on \(z=z_0\) is \(\delta^2p|_{z_0}\bk\) and that on \(z=z_0+\delta\) is \(-\delta^2p|_{z+\delta}\bk\text{.}\) This gives a contribution
\begin{equation*}
\delta^2p|_{z_0}\bk-\delta^2p|_{z+\delta}\bk
\approx-\delta^3\pd{p}{z}\bk=-V\pd{p}{z}\bk.
\end{equation*}
Hence the total force on the fluid in the cube due to the pressure of the surrounding fluid is given by
\begin{equation*}
-V\pd{p}{x}\bi-V\pd{p}{y}\bj-V\pd{p}{z}\bk=-V\nabla p.
\end{equation*}
(c)
Solution.
The unit normal to the face \(x=x_0\) is \(\hat{\bn}=-\bi\text{,}\) and hence the stress at a point in this face is
\begin{equation*}
\tau=\bsigma\hat{\bn}=-p\bI(-\bi)=p\bi.
\end{equation*}
Hence the force on this face equals \(\delta^2p_{x=x_0}\bi\text{.}\) Similarly, the unit normal to the face \(x=x_0+\delta\) is \(\hat{\bn}=\bi\text{,}\) and hence the stress at a point in this face is
\begin{equation*}
\tau=\bsigma\hat{\bn}=-p\bI(\bi)=-p\bi.
\end{equation*}
Hence the force on this face equals \(-\delta^2p_{x=x_0+\delta}\bi\text{.}\) Thus the contribution to the force from these two faces is
\begin{equation*}
\delta^2p|_{x_0}\bi-\delta^2p|_{x+\delta}\bi
\approx-\delta^3\pd{p}{x}\bi=-V\pd{p}{x}\bi.
\end{equation*}
The contributions from the other faces can be calculated similarly, and the total force on the fluid in the cube due to the pressure of the surrounding fluid is given by
\begin{equation*}
-V\pd{p}{x}\bi-V\pd{p}{y}\bj-V\pd{p}{z}\bk=-V\nabla p.
\end{equation*}
6. Analysis of flow in a channel.
A viscous incompressible fluid flows steadily between two stationary rigid boundaries at \(y=\pm h\) under a constant pressure gradient \(P=-\de p/\de x\text{.}\)
(a)
Write down the boundary conditions to apply at each of the plates.
(b)
The velocity is steady and therefore independent of \(t\text{.}\) Explain why it is also reasonable to assume that the velocity profile is independent of \(x\) and \(z\text{,}\) and also that the \(z\)-component of the velocity is zero \(w=0\text{.}\)
Solution.
Since the plates are infinite in the \(x\) and \(z\) directions, and the pressure gradient is constant, there is no preferred location in these directions, and hence the velocity profile is independent of \(x\) and \(z\text{.}\) Furthermore, since there is no pressure gradient in the \(z\) direction, there is no driving force for flow in this direction, and hence the \(z\)-component of the velocity is zero (\(w=0\)).
(c)
Use the continuity equation to show that the \(y\)-component of the velocity is also zero (\(v=0\)).
Solution.
The continuity equation for an incompressible fluid is
\begin{equation*}
\pd{u}{x}+\pd{v}{y}+\pd{w}{z}=0.
\end{equation*}
Since the velocity is independent of \(x\) and \(z\text{,}\) the first and third terms are zero, and hence
\begin{equation*}
\pd{v}{y}=0.
\end{equation*}
Since the plates are rigid boundaries, there is no flow through them, and hence the no-penetration boundary conditions give
\begin{equation*}
v(x,-h,z)=0,\quad v(x,h,z)=0
\end{equation*}
for all \(x\) and \(z\text{.}\) Since \(v\) is independent of \(y\) and zero at both \(y=-h\) and \(y=h\text{,}\) it follows that \(v=0\) for all \(y\text{.}\)
(d)
Hence the only non-zero component of the velocity is the \(x\)-component, \(u\text{,}\) and this only depends on \(y\text{,}\) \(u(y)\text{.}\) Check that the profile
\begin{equation*}
u=\frac{P}{2\mu}\left(h^2-y^2\right)
\end{equation*}
satisfies the boundary conditions.
Solution.
At \(y=-h\text{,}\) we have
\begin{equation*}
u(-h)=\frac{P}{2\mu}\left(h^2-(-h)^2\right)
=\frac{P}{2\mu}(h^2-h^2)=0,
\end{equation*}
and at \(y=h\text{,}\) we have
\begin{equation*}
u(h)=\frac{P}{2\mu}\left(h^2-h^2\right)
=\frac{P}{2\mu}(h^2-h^2)=0.
\end{equation*}
Hence the profile satisfies the boundary conditions.
(e)
The velocity profile increases as \(P\) increases and decreases as \(\mu\) increases. Explain why this is to be expected. Thus, although you cannot yet prove this is the correct velocity profile (you will soon be able to do so), this profile does look reasonable from what you already know.
Solution.
A larger pressure gradient \(P\) means a larger driving force for the flow, and hence a larger velocity. A larger viscosity \(\mu\) means a larger resistance to the flow, and hence a smaller velocity. Therefore, it is to be expected that the velocity profile increases as \(P\) increases and decreases as \(\mu\) increases.
(f)
Calculate the three components of the stress vector, the force per unit area that the fluid exerts, on each of the two plates. You may assume that the pressure equals \(p_0\) at \(x=0\text{.}\)
Solution.
At \(y=-h\text{,}\) the normal stress is given by the pressure, pointing into the plate:
\begin{equation*}
\tau_y=-p=-p_0+Px.
\end{equation*}
(note the negative sign as "into the plate" is in the \(-y\)-direction). The \(x\)-component of the shear stress at \(y=-h\) is given by
\begin{equation*}
\tau_x=\mu\pd{u}{y}=\mu\pd{}{y}\left[\frac{P}{2\mu}(h^2-y^2)\right]
=\mu\left(-\frac{P}{\mu}y\right)=Ph
\end{equation*}
(note that this is positive since the fluid is pushing the plate in the positive \(x\)-direction). The \(z\)-component of the shear stress at \(y=-h\) is zero since \(w=0\text{.}\) Hence
\begin{equation*}
\tau=(Ph,-p_0+Px,0).
\end{equation*}
At \(y=h\text{,}\) the normal stress is given by the pressure, pointing into the plate:
\begin{equation*}
\tau_y=p_0-Px
\end{equation*}
(note the sign change compared to the previous part as "into the plate" is in the \(+y\)-direction). The \(x\)-component of the shear stress at \(y=h\) is given by
\begin{equation*}
\tau_x=-\mu\pd{u}{y}=-\mu\pd{}{y}\left[\frac{P}{2\mu}(h^2-y^2)\right]
=-\mu\left(-\frac{P}{\mu}y\right)=Ph.
\end{equation*}
Note that the formula used has a negative sign as "into the fluid" is in the \(-y\)-direction. The overall sign is positive since the fluid is pushing the plate in the positive \(x\)-direction. (It is good to check signs after this kind of calculation by checking the direction of the force.) The \(z\)-component of the shear stress at \(y=h\) is zero since \(w=0\text{.}\) Hence
\begin{equation*}
\tau=(Ph,p_0-Px,0).
\end{equation*}
(g)
Compare the stresses on the two plates. Why is this to be expected?
Solution.
The normal stresses on the two plates are equal in magnitude but opposite in direction, as expected since the pressure acts equally on both plates, but "into the plates" is a different direction.
The shear stresses on the two plates are equal in magnitude and direction, indicating that the fluid is exerting the same tangential force on both plates in the direction of the flow. This is also expected since the plane \(y=0\) is a plane of symmetry.
7. Force on a blob of fluid.
In this exercise, you will find an expression for force per unit volume acting on an incompressible Newtonian fluid. You will do this by finding the force acting on a small blob of the fluid as an integral over its volume. First you will find a new expression for the stress vector at the blob’s surface, which makes the rest of the calculation easier. with normal \(\hat{\bn}\text{,}\) and use this to find an expression for the force on a blob of fluid.
(a)
Recall that \(\btau=\bsigma\cdot\hat{\bn}\) is the stress vector acting on a surface with normal \(\hat{\bn}\text{,}\) and this is given by
\begin{equation*}
\tau_i=-pn_i+\mu\sum_{j=1}^3n_j\left(\pd{u_i}{x_j}+\pd{u_j}{x_i}\right).
\end{equation*}
Show that this is identical to
\begin{equation*}
\btau=-p\hat{\bn}+\mu\left[2(\hat{\bn}\cdot\nabla)\bu
+\hat{\bn}\times(\nabla\times\bu)\right].
\end{equation*}
Hint.
Solution.
The \(i\)th component of
\begin{equation*}
\hat{\bn}\times\left(\nabla\times\bu\right)
=\nabla(\hat{\bn}\cdot\bu)-(\hat{\bn}\cdot\nabla)\bu,
\end{equation*}
is
\begin{align*}
\left[\hat{\bn}\times(\nabla\times\bu)\right]_i
&=\sum_{j=1}^3n_j\pd{u_j}{x_i}-(\hat{\bn}\cdot\nabla)u_i\\
&=\sum_{j=1}^3n_j\pd{u_j}{x_i}-\sum_{j=1}^3n_j\pd{u_i}{x_j}.
\end{align*}
Hence, the \(i\)th component of the target expression is
\begin{align*}
&\left\{-p\bn+\mu\left[2(\hat{\bn}\cdot\nabla)\bu
+\hat{\bn}\times(\nabla\times\bu)\right]\right\}_i\\
&=-pn_i+\mu\left[2\sum_{j=1}^3n_j\pd{u_i}{x_j}
+\sum_{j=1}^3n_j\pd{u_j}{x_i}-\sum_{j=1}^3n_j\pd{u_i}{x_j}\right]\\
&=-pn_i+\mu\left[\sum_{j=1}^3n_j\pd{u_i}{x_j}
+\sum_{j=1}^3n_j\pd{u_j}{x_i}\right]\\
&=\sum_{j=1}^3\left[-p\delta_{ij}+\mu\left(\pd{u_i}{x_j}
+\pd{u_j}{x_i}\right)\right]n_j\\
&=\sum_{j=1}^3\sigma_{ij}n_j\\
&=\btau_i,
\end{align*}
as required.
(b)
Hence show that the force on a blob of incompressible fluid of volume \(V\) due to the surrounding fluid is given by
\begin{equation*}
\bF=\iint_S\btau\de S
=\iiint_V\left(-\nabla p+\mu\nabla^2\bu\right)\de V,
\end{equation*}
where \(S\) is the surface of the blob of fluid.
Hint.
Use the following vector calculus identities:
\begin{align*}
&\iint_S\phi\bn\de S=\iiint_V-\nabla\phi\de V,\\
&\iint_S\bF\times\hat{\bn}\de S=-\iiint_V\nabla\times\bF\de V,\\
&\iint_S(\hat{\bn}\cdot\nabla)\phi\de S=\iiint_V\nabla^2\phi\de V,\\
&\nabla^2\bF=\nabla(\nabla\cdot\bF)-\nabla\times(\nabla\times\bF).
\end{align*}
Solution.
We have
\begin{align*}
\bf=&\iint_S\btau\de S\\
=&\iint_S\left\{-p\hat{\bn}+\mu\left[2(\hat{\bn}\cdot\nabla)\bu
+\hat{\bn}\times(\nabla\times\bu)\right]\right\}\de S.
\end{align*}
We now make use of the vector calculus identities given in the hint. The first term becoems
\begin{equation*}
\iint_S-p\hat{\bn}\de S=\iiint_V-\nabla p\de V,
\end{equation*}
the second term becomes
\begin{equation*}
\iint_S2\mu(\hat{\bn}\cdot\nabla)\bu\de S
=\iiint_V2\mu\nabla^2\bu\de V,
\end{equation*}
and for the third term,
\begin{align*}
\iint_S\mu\hat{\bn}\times(\nabla\times\bu)\de S
=&-\iint_S\mu(\nabla\times\bu)\times\hat{\bn}\de S\\
=&\iiint_V\mu\nabla\times(\nabla\times\bu)\de V,\\
=&\iiint_V\mu\nabla(\nabla\cdot\bu)-\mu\nabla^2\bu\de V.
\end{align*}
Hence, since the fluid is incompressible and \(\nabla\cdot\bu=0\text{,}\) the third term becomes
\begin{equation*}
\\\int_V-\mu\nabla^2\bu\de V.
\end{equation*}
Combining these three results gives
\begin{align*}
\bF=&\iiint_V\left(-\nabla p+2\mu\nabla^2\bu-\mu\nabla^2\bu\right)\de V\\
=&\iiint_V\left(-\nabla p+\mu\nabla^2\bu\right)\de V,
\end{align*}
as required.
(c)
Deduce that, if the blob is small, the net force on it per unit volume of fluid equals \(-\nabla p+\mu\nabla^2\bu\text{.}\)
Solution.
If the blob is small, then the integrand in the expression for the force varies little over the volume of the blob, and so
\begin{equation*}
\bF\approx\left(-\nabla p+\mu\nabla^2\bu\right)V,
\end{equation*}
where \(V\) is the volume of the blob. Hence the force per unit volume is given by
\begin{equation*}
\frac{\bF}{V}=-\nabla p+\mu\nabla^2\bu,
\end{equation*}
as required.
8. General deformation of a fluid element.
In this exercise, you will consider the general deformation of a fluid element in an incompressible Newtonian fluid. You will show that the deformation can be expressed as a combination of three types of motion: translation, rotation and pure strain.
Note that, if the fluid were also compressible, there would be a fourth type of motion, namely expansion or contraction.
(a)
Let \(P\) be a point in the fluid with position vector \(\bx\text{,}\) and let \(Q\) be a nearby point with position vector \(\bx+\bs\text{.}\) Explain why the velocity at point \(Q\) can be expressed in terms of that at point \(P\) as
\begin{equation*}
\bu_Q=\bu_P+(\bs\cdot\nabla)\bu.
\end{equation*}
Solution.
The velocity at point \(Q\) can be expressed as a Taylor series expansion about point \(P\text{:}\)
\begin{equation*}
\bu_Q=\bu_P+\left(\bs\cdot\nabla\right)\bu
+\frac{1}{2!}\left(\bs\cdot\nabla\right)^2\bu
+\frac{1}{3!}\left(\bs\cdot\nabla\right)^3\bu+\cdots.
\end{equation*}
Since \(Q\) is a nearby point to \(P\text{,}\) the higher order terms in \(\bs\) are negligible, and hence we have
\begin{equation*}
\bu_Q=\bu_P+(\bs\cdot\nabla)\bu,
\end{equation*}
as required.
(b)
Show that
\begin{equation}
\bu_Q=\bu_P+\frac12(\nabla_x\times\bu)\times\bs
+\frac12\nabla_s(\sum_{i=1}^3\sum_{j=1}^3e_{ij}s_is_j),\tag{7.9.1}
\end{equation}
where \(\nabla_x\) denotes differentiation with respect to the components of \(\bx\text{,}\) \(\nabla_s\) denotes differentiation with respect to the components of \(\bs\text{,}\) and
\begin{equation*}
e_{ij}=\frac12\left(\pd{u_i}{x_j}+\pd{u_j}{x_i}\right)
\end{equation*}
You may use the fact that \((\nabla\times\bf)\times\bg=\bg\cdot\nabla\bf-\nabla(\bg\cdot\bf)\text{,}\) where, in the second term on the right-hand side, the operator \(\nabla\) acts on \(\bf\) but not \(\bg\)
Solution.
We have that the \(k\)th component of the final term of the target expression is
\begin{align*}
&\left\{\frac12\nabla_s(\sum_{i=1}^3\sum_{j=1}^3e_{ij}s_is_j)\right\}_k\\
=&\frac12\pd{}{s_k}(\sum_{i=1}^3\sum_{j=1}^3e_{ij}s_is_j)\\
=&\frac12(\sum_{i=1}^3e_{ik}s_i+\sum_{j=1}^3e_{kj}s_j)\\
=&\frac12\sum_{i=1}^3(e_{ik}s_i+e_{ki}s_i)\\
=&\sum_{i=1}^3e_{ik}s_i,
\end{align*}
where the final equality uses the fact that the rate-of-strain tensor is symmetric \(e_{ij}=e_{ji}\text{.}\) Substituting for \(e_{ik}\) gives
\begin{align*}
&\sum_{i=1}^3e_{ik}s_i\\
=&\frac12\sum_{i=1}^3\left(\pd{u_i}{x_k}+\pd{u_k}{x_i}\right)s_i.
\end{align*}
Using the given vector identity, the \(k\)th component of the second term of the target expression is
\begin{align*}
&\left\{\frac12(\nabla_x\times\bu)\times\bs\right\}_k\\
=&\frac12\left\{\bs\cdot\nabla_x\bu-\nabla_x(\bs\cdot\bu)\right\}_k\\
=&\frac12\left\{\sum_{i=1}^3s_i\pd{u_k}{x_i}
-\sum_{i=1}^3s_i\pd{u_i}{x_k}\right\}\\
=&\frac12\sum_{i=1}^3\left(\pd{u_k}{x_i}-\pd{u_i}{x_k}\right)s_i.
\end{align*}
Hence the \(k\)th component of the sum of the second and the third terms of the target expression is
\begin{align*}
&\frac12(\nabla_x\times\bu)\times\bs
+\frac12\nabla_s(\sum_{i=1}^3\sum_{j=1}^3e_{ij}s_is_j)\\
=&\frac12\sum_{i=1}^3\left(\pd{u_k}{x_i}-\pd{u_i}{x_k}\right)s_i
+\frac12\sum_{i=1}^3\left(\pd{u_i}{x_k}+\pd{u_k}{x_i}\right)s_i\\
=&\sum_{i=1}^3\pd{u_k}{x_i}s_i,
\end{align*}
which is the \(k\)th component of \((\bs\cdot\nabla)\bu\text{,}\) thus proving the result.
(c)
Explain why the three terms on the right-hand side of the expression in the previous part represent translation, rotation and pure strain, respectively.
Solution.
-
The second term \(\frac12(\nabla_x\times\bu)\times\bs\) represents rotation since, setting \(\bOmega=\nabla_x\times\bu/2=\bomega/2\text{,}\) this term is of the form \(\bOmega\times\bs\text{,}\) which is purely rotational motion about point \(P\) (see Exercise 7.10.2). Note also that this again shows a result that we used in Subsection 6.1.1 that the vorticity \(\bomega\) is twice the angular velocity since \(\bOmega=\bomega/2\text{.}\)
-
The third term \(\frac12\nabla_s(\sum_{i=1}^3\sum_{j=1}^3e_{ij}s_is_j)\) represents pure strain since it depends on the rate-of-strain tensor \(e_{ij}\text{,}\) which describes how the fluid element is deformed without rotation.
9. Decomposition of a shear flow.
In this problem, you will apply the equation from the previous problem Exercise 7.9.8 to a shear flow \(\bu=(\beta y,0,0)\text{.}\)
(a)
Use (7.9.1) to decompose \(\bu\) into its local (i) translational, (ii) rotational, and (iii) pure straining components. For (iii), find the eigenvalues and eigenvectors, which give the principle directions of strain. Sketch each of the components (ii) and (iii).
Note that this example is a special case in which the rotational and pure straining components are uniform over the fluid domain. More generally, both will vary.
Solution.
(ii) The vorticity
\begin{equation*}
\bomega=(0,0,-\beta),
\end{equation*}
and therefore the rotational component is
\begin{equation*}
\frac12\bomega\times\bs
=\frac12\beta(s_y,-s_x,0),
\end{equation*}
which is clockwise rotation with angular velocity \(\beta/2\text{,}\) see left sketch in Figure 7.9.5.
(iii) We have the rate-of-strain tensor
\begin{equation*}
\be=\left(\begin{matrix}
0&\beta/2&0\\
\beta/2&0&0\\
0&0&0
\end{matrix}\right).
\end{equation*}
Hence
\begin{equation*}
\frac12\nabla_s\left(\sum_{j=1}^3\sum_{i=1}^3e_{ij}s_is_j\right)
=\frac{\beta}2\nabla_s\left(s_1s_2\right)
=\frac{\beta}2\left(s_y,s_x,0\right)
\end{equation*}
(note that we are interchanging between \(s=(s_1,s_2,s_3)\) and \((s_x,s_y,s_z)\)). The eigenvalues \(\lambda\) of \(\be\) satisfy
\begin{equation*}
-\lambda^3+\frac14\beta^2\lambda=0,
\end{equation*}
which has solutions \(\lambda=0,\pm\beta/2\text{.}\) The associated eigenvectors are
\begin{equation*}
\hat{\mathbf{k}},
\quad\left(\frac1{\sqrt{2}},\frac1{\sqrt{2}},0\right),
\quad\left(-\frac1{\sqrt{2}},\frac1{\sqrt{2}},0\right).
\end{equation*}
This corresponds to expansion along the line \(x=y\) and contraction along \(x=-y\text{,}\) see middle sketch in Figure 7.9.5.
The solution is illustrated in Figure 7.9.5.
10. Derivation of the Cauchy equation.
(a)
Consider a volume \(V(t)\) moving with the fluid that, at the time \(t\) is the small cube of fluid as shown in Figure 7.5.1, and apply the Reynolds Transport Theorem (3.1.2) to the fluid in this volume.
Solution.
We consider a volume \(V\) with surface \(S\) consisting of a small cube of side length \(\delta\text{,}\) as shown in Figure 7.5.1. We apply the Reynolds Transport Theorem to the fluid in the cube:
\begin{equation*}
\dd{}{t} \iiint_{V(t)} f \, \de{x} \, \de{y} \, \de{z} = \iiint_{V(t)} \left[
\pd{f}{t} + \nabla \cdot (f\bu)\right] \, \de{x} \, \de{y} \, \de{z},
\end{equation*}
where \(f\) is any function, a vector or a scalar.
Setting \(\bf=\rho\bu\) equal to the momentum per unit volume in the fluid, the integral equals the momentum of the fluid in the cube, and the time derivative is the rate of change of momentum, which is the total force acting on the fluid in the cube.
The total force equals the sum of the surface forces and body forces. The body force equals the force of gravity: \(\rho V\bg\text{.}\) The surface forces are an integral:
\begin{equation*}
\iint_S\btau\,\de S=\iint_S\bsigma\bn\,\de S;
\end{equation*}
and we sum the contributions over the six faces in pairs:
-
Front and back faces: The contribution on the back face (\(\bn=-\bi\)) is\begin{equation*} -\delta^2\left.\left(\begin{array}{c} \sigma_{11}\\\sigma_{12}\\\sigma_{13} \end{array}\right)\right|_{x=x_0} \end{equation*}and on the front face (\(\bn=\bi\)), it is\begin{equation*} \delta^2\left.\left(\begin{array}{c} \sigma_{11}\\\sigma_{12}\\\sigma_{13} \end{array}\right)\right|_{x=x_0+\delta}. \end{equation*}The sum of these is\begin{equation*} \delta^2\left.\left(\begin{array}{c} \sigma_{11}\\\sigma_{12}\\\sigma_{13} \end{array}\right)\right|_{x=x_0+\delta} -\delta^2\left.\left(\begin{array}{c} \sigma_{11}\\\sigma_{12}\\\sigma_{13} \end{array}\right)\right|_{x=x_0} =\delta^3\pd{}{x}\left(\begin{array}{c} \sigma_{11}\\\sigma_{12}\\\sigma_{13} \end{array}\right). \end{equation*}
-
Left and right faces: In a similar way, the contribution is\begin{equation*} \delta^3\pd{}{y}\left(\begin{array}{c} \sigma_{21}\\\sigma_{22}\\\sigma_{23} \end{array}\right). \end{equation*}
-
Bottom and top faces: The contribution is\begin{equation*} \delta^3\pd{}{z}\left(\begin{array}{c} \sigma_{31}\\\sigma_{32}\\\sigma_{33} \end{array}\right). \end{equation*}
Hence, the surface force is given by
\begin{equation*}
\delta^3\pd{}{x}\left(\begin{array}{c}
\sigma_{11}\\\sigma_{12}\\\sigma_{13}
\end{array}\right)
+\delta^3\pd{}{y}\left(\begin{array}{c}
\sigma_{21}\\\sigma_{22}\\\sigma_{23}
\end{array}\right)
+\delta^3\pd{}{z}\left(\begin{array}{c}
\sigma_{31}\\\sigma_{32}\\\sigma_{33}
\end{array}\right)
=\delta^3\nabla\cdot\bsigma,
\end{equation*}
Hence
\begin{equation*}
\rho V\bg+\delta^3\nabla\cdot\bsigma
=\iiint_{V(t)} \left[\pd{(\rho\bu)}{t} + \nabla \cdot (\rho\bu\bu)\right]\,\de{x}\,\de{y}\,\de{z},
\end{equation*}
Since the cube is small, the right-hand side is approximately the volume \(V\) times the integrand evaluated at a point in \(V\text{.}\) Dividing by \(V=\delta^3\) gives
\begin{equation*}
\rho\bg+\nabla\cdot\bsigma
=\pd{(\rho\bu)}{t} + \nabla \cdot ((\rho\bu)\bu).
\end{equation*}
Since \(\rho\) is constant,
\begin{equation*}
\rho\bg+\nabla\cdot\bsigma
=\rho\pd{\bu}{t} + \rho\bu\cdot\nabla\bu+\rho(\nabla\cdot\bu)\bu.
\end{equation*}
Since the fluid is incompressible, we have \(\nabla\cdot\bu=0\text{,}\) and, rearranging, we obtain the Cauchy equation:
\begin{equation*}
\rho\left(\pd{\bu}{t} + \bu\cdot\nabla\bu\right)
=\nabla\cdot\bsigma+\rho\bg
\end{equation*}
11. Limits of low and high Reynolds number.
An incompressible Newtonian fluid with density \(\rho\) and shear viscosity \(\mu\) flows in a space with a typical lengthscale \(L\) and velocity scale \(U\text{.}\)
(a)
Perform a nondimensionalisation of the Navier–Stokes (7.5.1) and continuity (3.4.2) equations, using a scale \(P\) for the pressure and \(T\) for time, and writing the gravitational acceleration as \(\bg=g\hat{\bg}\text{,}\) where \(\hat{\bg}\) is a unit vector in the direction of \(\bg\text{.}\)
Solution.
We have
\begin{equation*}
\rho\pd{\bu}{t}+\rho\bu\cdot\nabla\bu=-\nabla p+\mu\nabla^2\bu+\rho{\bg},\quad
\nabla\cdot\bu=0,
\end{equation*}
which, after nondimensionalising, become
\begin{equation*}
\frac{\rho U}{T}\pd{\bu^*}{t^*}+\frac{\rho U^2}{L}\bu^*\cdot\nabla^*\bu^*
=-\frac{P}{L}\nabla^* p^*+\frac{\mu U}{L^2}\nabla^{*2}\bu^*+\rho g\hat{\bg},\quad
\frac{U}{L}\nabla^*\cdot\bu^*=0.
\end{equation*}
(b)
Choose a suitable timescale \(T\) so that the inertial terms balance and multiply the Navier–Stokes equation by a suitable factor so that the nondimensional viscous term has coefficient 1. What do you notice about the coefficient of the inertial terms?
Solution.
We set the coefficients of the two inertial terms equal:
\begin{equation*}
\frac{\rho U}{T}=\frac{\rho U^2}{L}
\quad\Rightarrow\quad
T=\frac{L}{U}.
\end{equation*}
Multiplying the Navier–Stokes equation by \(L^2/(\mu U)\text{,}\) we get
\begin{equation*}
\mathrm{Re}\DD{\bu^*}{t^*}
=-\frac{LP}{\mu U}\nabla^* p^*+\nabla^{*2}\bu^*+\frac{\rho L^2g}{\mu U}\hat{\bg}.
\end{equation*}
The coefficient multiplying the inertial terms is the Reynolds number \(\mathrm{Re}\text{.}\)
(c)
Hence, in the case of a low-Reynolds-number flow, \(\mathrm{Re}\ll1\text{,}\) show that, to leading order, the equations simplify to one of the following three cases:
\begin{equation*}
0=-\nabla p+\rho{\bg},
\end{equation*}
or
\begin{equation*}
0=-\nabla p+\mu\nabla^2\bu+\rho{\bg},\quad
\nabla\cdot\bu=0,
\end{equation*}
or
\begin{equation*}
0=-\nabla p+\mu\nabla^2\bu,\quad
\nabla\cdot\bu=0.
\end{equation*}
Specify the conditions under which each of these equations is obtained. In the latter two cases, we term the equations governing the fluid velocity the Stokes equations.
Solution.
In this case, the inertial terms are small compared to the viscous terms, so we can neglect them. The gravitational acceleration term has order of magnitude \(\rho L^2g/(\mu U)\text{.}\) We need to consider the cases in which this is large, order one or small:
-
If \(\rho L^2g/(\mu U)\gg1\text{,}\) the pressre have a predominantly hydrostatic pressure profile, with the fluid dynamical problem being a small perturbation to the background static problem. Thus the pressure gradient balances this term:\begin{equation*} \frac{LP}{\mu U}=\frac{\rho L^2g}{\mu U} \quad\Rightarrow\quad P=\rho Lg, \end{equation*}and, to leading order, the Navier–Stokes equation becomes\begin{equation*} 0=-\nabla^* p^*+\hat{\bg}. \end{equation*}Once we have solved this problem, we could in principle solve a secondary problem to find the fluid flow. Thus the dimensional problem is\begin{equation*} 0=-\nabla p+\rho\bg. \end{equation*}
-
If \(\rho L^2g/(\mu U)\) is of order 1, the gravitational acceleration term balances the leading-order terms in the equation, and we set\begin{equation*} \frac{LP}{\mu U}=1\quad\Rightarrow\quad P=\frac{\mu U}{L}. \end{equation*}We obtain the leading-order system\begin{equation*} 0=-\nabla^* p^*+\nabla^{*2}\bu^*+\frac{\rho L^2g}{\mu U}\hat{\bg}, \end{equation*}and redimensionalising, we get\begin{equation*} 0=-\nabla p+\mu\nabla^2\bu, \end{equation*}and the continuity equation is retained in full:\begin{equation*} \nabla\cdot\bu=0. \end{equation*}
-
If \(\rho L^2g/(\mu U)\ll1\text{,}\) the gravitational acceleration term is small compared to the leading-order terms in the equation, and we set\begin{equation*} \frac{LP}{\mu U}=1\quad\Rightarrow\quad P=\frac{\mu U}{L}. \end{equation*}We obtain the leading-order system\begin{equation*} 0=-\nabla^* p^*+\nabla^{*2}\bu^*, \end{equation*}and redimensionalising, we get\begin{equation*} 0=-\nabla p+\mu\nabla^2\bu, \end{equation*}and the continuity equation is retained in full:\begin{equation*} \nabla\cdot\bu=0. \end{equation*}
(d)
In the case of a high-Reynolds-number flow, \(\mathrm{Re}\gg1\text{,}\) show that, to leading order, the equations simplify to one of the following three cases:
\begin{equation*}
0=-\nabla p+\rho{\bg},
\end{equation*}
or
\begin{equation*}
\rho\DD{u}{t}=-\nabla p+\rho{\bg},\quad
\nabla\cdot\bu=0,
\end{equation*}
or
\begin{equation*}
\rho\DD{u}{t}=-\nabla p,\quad
\nabla\cdot\bu=0.
\end{equation*}
Specify the conditions under which each of these equations is obtained. In the latter two cases, we term the equations governing the fluid velocity the Euler equations.
Solution.
In this case, the viscous terms are small compared to the inertial terms, so we can neglect them. Dividing the Navier–Stokes equation by \(\mathrm{Re}\text{,}\)
\begin{equation*}
\DD{\bu^*}{t^*}
=-\frac{P}{\rho U^2}\nabla^* p^*+\frac{Lg}{U^2}\hat{\bg}.
\end{equation*}
The gravitational acceleration term now has order of magnitude \(Lg/U^2\text{.}\) We need to consider the cases in which this is large, order one or small:
-
If \(Lg/U^2\gg1\text{,}\) the pressre have a predominantly hydrostatic pressure profile, with the fluid dynamical problem being a small perturbation to the background static problem. Thus the pressure gradient balances this term:\begin{equation*} \frac{P}{\rho U^2}=\frac{Lg}{U^2} \quad\Rightarrow\quad P=\rho Lg, \end{equation*}and, to leading order, the Navier–Stokes equation becomes\begin{equation*} 0=-\nabla^* p^*+\hat{\bg}. \end{equation*}As in the low-Reynolds-number cases, once we have solved this problem, we could in principle solve a secondary problem to find the fluid flow. Thus the dimensional problem is\begin{equation*} 0=-\nabla p+\rho\bg. \end{equation*}
-
If \(Lg/U^2\) is of order 1, the gravitational acceleration term balances the leading-order terms in the equation, and we set\begin{equation*} \frac{P}{\rho U^2}=1\quad\Rightarrow\quad P=\rho U^2. \end{equation*}We obtain the leading-order system\begin{equation*} \DD{u^*}{t^*}=-\nabla^* p^*+\frac{Lg}{U^2}\hat{\bg}, \end{equation*}and redimensionalising, we get\begin{equation*} \rho\DD{u}{t}=-\nabla p+\rho\bg, \end{equation*}and the continuity equation is retained in full:\begin{equation*} \nabla\cdot\bu=0. \end{equation*}
-
If \(Lg/U^2\ll1\text{,}\) the gravitational acceleration term is small compared to the leading-order terms in the equation, and we set\begin{equation*} \frac{P}{\rho U^2}=1\quad\Rightarrow\quad P=\rho U^2. \end{equation*}We obtain the leading-order system\begin{equation*} \DD{u^*}{t^*}=-\nabla^* p^*, \end{equation*}and redimensionalising, we get\begin{equation*} \rho\DD{u}{t}=-\nabla p, \end{equation*}and the continuity equation is retained in full:\begin{equation*} \nabla\cdot\bu=0. \end{equation*}
12. Hagen–Poiseuille flow.
(a)
Review the derivation Example 7.8.4, making your own notes. Use the flow profile to prove the formula (7.8.4).
Solution.
We have that the volumetric flux along the pipe equals (\(A\) is a circular cross-section \(z=\mathrm{constant}\) of the pipe)
\begin{align*}
Q=&\iint_Aw\,\de A\\
=&\int_0^{2\pi}\int_0^a\frac{G}{4\mu}\left(a^2-r^2\right)r\,\de r\,\de \theta\\
=&2\pi\frac{G}{4\mu}\int_0^a\left(a^2r-r^3\right)\,\de r\\
=&\frac{\pi G}{2\mu}\left[\frac{a^2r^2}{2}-\frac{r^4}{4}\right]_0^a\\
=&\frac{\pi G}{2\mu}\left(\frac{a^4}{2}-\frac{a^4}{4}\right)\\
=&\frac{\pi Ga^4}{8\mu},
\end{align*}
where \(G=\Delta p/L\) is the pressure gradient. Rearranging,
\begin{equation*}
\Delta p=\frac{8\mu LQ}{\pi a^4},
\end{equation*}
as expected.
13. Flow outside a cylinder.
A Newtonian incompressible fluid of density \(\rho\) and shear viscosity \(\mu\) surrounds a circular cylinder of radius \(a\text{.}\) Working in cylindrical polar coordinates, the cylinder surface is at \(r=a\text{.}\)
The Navier–Stokes and continuity equations in cylindrical polar coordinates read:
\begin{align*}
&\rho\left(\pd{u_r}{t}+u_r\pd{u_r}{r}+\frac{u_{\theta}}{r}\pd{u_r}{\theta}+u_z\pd{u_r}{z}-\frac{u_{\theta}^2}{r}\right)\\
=&-\pd{p}{r}
+\mu\left(\nabla^2u_r-\frac{u_r}{r^2}-\frac{2}{r^2}\pd{u_\theta}{\theta}\right),\\
&\rho\left(\pd{u_{\theta}}{t}+u_r\pd{u_{\theta}}{r}+\frac{u_{\theta}}{r}\pd{u_{\theta}}{\theta}+u_z\pd{u_{\theta}}{z}+\frac{u_ru_{\theta}}{r}\right)\\
=&-\frac{1}{r}\pd{p}{\theta}
+\mu\left(\nabla^2u_\theta+\frac{2}{r^2}\pd{u_r}{\theta}-\frac{u_{\theta}}{r^2}\right),\\
&\rho\left(\pd{u_z}{t}+u_r\pd{u_z}{r}+\frac{u_{\theta}}{r}\pd{u_z}{\theta}+u_z\pd{u_z}{z}\right)\\
=&-\pd{p}{z}
+\mu\nabla^2u_z,\\
&\frac1r\pd{}{r}\left(ru_r\right)+\frac1r\pd{u_\theta}{\theta}+\pd{u_z}{z}=0,
\end{align*}
where the Laplacian is defined by its action on a function \(A(r,\theta,z)\text{:}\)
\begin{equation*}
\nabla^2A=\frac1r\pd{}{r}\left(r\pd{A}{r}\right)+\frac1{r^2}\pd{^2A}{\theta^2}+\pd{^2A}{z^2},
\end{equation*}
(a)
In the case that the surface of the cylinder is porous and there is a uniform suction inside it, which induces a steady flow into the cylinder with inflow velocity \(U\) at the surface, you may assume that the flow is steady, two-dimensional and axisymmetric. With these assumptions, use the continuity equation to show that
\begin{equation*}
u_r=-\frac{Ua}{r}.
\end{equation*}
Solution.
Axisymmetry implies \(u_\theta=0\) and two-dimensional flow implies \(u_z=0\text{.}\) With these assumptions, the continuity equation becomes
\begin{equation*}
\frac1r\pd{}{r}\left(ru_r\right)=0.
\end{equation*}
Multiplying by \(r\) and integrating,
\begin{equation*}
ru_r=A\quad\Rightarrow\quad u_r=\frac{A}{r},
\end{equation*}
where \(A\) is constant. The boundary condition on \(r=a\) reads \(u_r=-U\text{,}\) and hence
\begin{equation*}
-U=\frac{A}{a}\quad\Rightarrow\quad A=-aU.
\end{equation*}
Substituting this in,
\begin{equation*}
u_r=\frac{-aU}{r}.
\end{equation*}
(b)
Show that the Navier–Stokes equations reduce to
\begin{equation*}
\pd{p}{r}=\frac{\rho a^2U^2}{r^3}.
\end{equation*}
Solution.
We substitute the assumptions into the Navier–Stokes equations, which state that all \(t\text{,}\) \(\theta\) and \(z\) derivatives are zero and \(u_\theta=u_z=0\text{.}\) The equations become:
\begin{align*}
\rho u_r\pd{u_r}{r}
=&-\pd{p}{r}
+\mu\left(\nabla^2u_r-\frac{u_r}{r^2}\right),\\
0=&0,\\
0=&0,
\end{align*}
and
\begin{equation*}
\nabla^2u_r
=\frac1r\pd{}{r}\left(r\pd{u_r}{r}\right)
=\pd{^2u_r}{r^2}+\frac1r\pd{u_r}{r}.
\end{equation*}
Hence, the only non-trivial equation is the \(r\)-component, which reduces to
\begin{equation*}
\rho u_r\pd{u_r}{r}
=-\pd{p}{r}
+\mu\left(\pd{^2u_r}{r^2}+\frac1r\pd{u_r}{r}-\frac{u_r}{r^2}\right).
\end{equation*}
Substituting the expression for \(u_r\text{,}\)
\begin{equation*}
\rho \left(-\frac{aU}{r}\right)\left(\frac{aU}{r^2}\right)
=-\pd{p}{r}
+\mu\left(-\frac{2aU}{r^3}+\frac{aU}{r^3}+\frac{aU}{r^3}\right),
\end{equation*}
which simplifies to
\begin{equation*}
\pd{p}{r}=\frac{\rho a^2U^2}{r^3},
\end{equation*}
as required.
(c)
Solve this to find the pressure field, given that the far-field pressure is \(p_0\text{.}\)
14. Oscillating flow in a pipe.
In this exercise, you will derive the equations for flow in a circular pipe, driven by an oscillating pressure gradient
\begin{equation*}
\pd{p}{z}=-G_0\cos\omega t.
\end{equation*}
(a)
In cylindrical polar coordinates \((r,\theta,z)\) the continuity equation is
\begin{equation*}
\frac1r\pd{(ru_r)}{r}+\frac1r\pd{u_\theta}{\theta}+\pd{u_z}{z}=0,
\end{equation*}
and the Navier–Stokes equations are
\begin{align*}
&\rho\left(\pd{u_r}{t}+u_r\pd{u_r}{r}+\frac{u_{\theta}}{r}\pd{u_r}{\theta}+u_z\pd{u_r}{z}-\frac{u_{\theta}^2}{r}\right)\\
=&-\pd{p}{r}
+\mu\left(\nabla^2u_r-\frac{u_r}{r^2}-\frac{2}{r^2}\pd{u_\theta}{\theta}\right)+\rho f_r,\\
&\rho\left(\pd{u_{\theta}}{t}+u_r\pd{u_{\theta}}{r}+\frac{u_{\theta}}{r}\pd{u_{\theta}}{\theta}+u_z\pd{u_{\theta}}{z}+\frac{u_ru_{\theta}}{r}\right)\\
=&-\frac{1}{r}\pd{p}{\theta}
+\mu\left(\nabla^2u_\theta+\frac{2}{r^2}\pd{u_r}{\theta}-\frac{u_{\theta}}{r^2}\right)+\rho f_{\theta},\\
&\rho\left(\pd{u_z}{t}+u_r\pd{u_z}{r}+\frac{u_{\theta}}{r}\pd{u_z}{\theta}+u_z\pd{u_z}{z}\right)\\
=&-\pd{p}{z}
+\mu\nabla^2u_z+\rho f_z,
\end{align*}
where the Laplacian is defined by its action on a function \(A(r,\theta,z)\text{:}\)
\begin{equation*}
\nabla^2A=\frac1r\pd{}{r}\left(r\pd{A}{r}\right)+\frac1{r^2}\pd{^2A}{\theta^2}+\pd{^2A}{z^2},
\end{equation*}
and \(\bf=(f_r,f_\theta,f_z)\) represents the body forces per unit mass acting on the fluid.
Stating your assumptions, use the continuity equation to show that \(u_r\) is everywhere zero.
Solution.
We assume that the flow is axisymmetric and fully developed. The continuity equation reduces to
\begin{equation*}
\frac1r\pd{}{r}\left(ru_r\right)=0,
\end{equation*}
which implies that
\begin{equation*}
u_r=\frac{A}{r}
\end{equation*}
where \(A\) is independent of \(r\text{.}\) Regularity at \(r=0\) implies that \(A=0\text{,}\) and therefore \(u_r=0\) everywhere.
(b)
Simplify the Navier–Stokes equations, and use them to show that
\begin{equation*}
u_z=U(r)e^{\im\omega t}+{\mathrm{c.c.}},
\end{equation*}
where
\begin{equation*}
\dd{^2U}{r^2}+\frac1r\dd{U}{r}-\frac{\im\rho\omega}{\mu} U=-\frac{G_0}{2\mu}.
\end{equation*}
You may neglect gravity.
Solution.
With the assumptions, the Navier–Stokes equations reduce to
\begin{align*}
0=&-\pd{p}{r},\\
0=&0,\\
\rho\im\omega Ue^{\im\omega t}+\mathrm{c.c.}
=G_0\cos\omega t+\frac{\mu}{r}\dd{}{r}\left(rU'\right)e^{\im\omega t}+\mathrm{c.c.}.
\end{align*}
There is thus one non-trivial component, and, taking the part proportional to \(e^{\im\omega t}\text{,}\) we have
\begin{equation*}
\rho\im\omega U=\frac12G_0+\frac{\mu}{r}\dd{}{r}\left(rU'\right),
\end{equation*}
which, dividing by \(\mu\text{,}\) is equivalent to
\begin{equation*}
\frac{\im\rho\omega}{\mu} U=\frac{G_0}{2\mu}+U''+\frac1rU'.
\end{equation*}
Rearranging gives the result.
(c)
Make the substitution \(s=e^{-\im\pi/4}\alpha r\text{,}\) where \(\alpha=\sqrt{\rho\omega/\mu}\text{,}\) to obtain the canonical form
\begin{equation*}
\dd{^2U}{s^2}+\frac1s\dd{U}{s}+U=-\frac{\im G_0}{2\rho\omega}
\end{equation*}
Solution.
With this substitution, we have
\begin{equation*}
\dd{}{r}=e^{-\im\pi/4}\alpha\dd{}{s},
\end{equation*}
and hence
\begin{equation*}
-\im\alpha^2\dd{^2U}{s^2}-\im\alpha^2\frac1s\dd{U}{s}
-\im\alpha^2U=-\frac{G_0}{2\mu},
\end{equation*}
giving
\begin{equation*}
\dd{^2U}{s^2}+\frac1s\dd{U}{s}+U=-\frac{\im G_0}{2\rho\omega}.
\end{equation*}
(d)
Find a particular integral to this equation, i.e. any solution that satisfies the inhomogeneous equation
(e)
The complementary function can be written in terms of a special function of which you are probably not aware called a Bessel function. Regularity implies that we pick the Bessel functions of the the first kind, which are shown in Figure 7.8.22.
Examples of the solution for three different Womersley numbers are shown in Figure 7.9.6, Figure 7.9.7 and Figure 7.9.8. The solution for small Womersley numbers resembles the Poiseuille flow profile, while that for large Womersley numbers is plug (uniform) flow across the core of the pipe with boundary layers exhibiting rapid spatial oscillations.
15. Solving the Navier–Stokes equations in spherical polar coordinates.
A Newtonian incompressible fluid of density \(\rho\) and shear viscosity \(\mu\) flows in three dimensions. In spherical polar coordinates \((r,\theta,\phi)\text{,}\) the continuity equation is
\begin{equation*}
\frac1{r^2}\pd{(r^2u_r)}{r}+\frac1{r\sin\theta}\pd{(u_\theta\sin\theta)}{\theta}+\frac1{r\sin\theta}\pd{u_\phi}{\phi}=0,
\end{equation*}
and the Navier–Stokes equations are
\begin{gather*}
\rho\left(\pd{u_r}{t}+u_r \pd{u_r}{r}+\frac{u_{\theta}}{r}\pd{u_r}{\theta}+\frac{u_{\phi}}{r \sin\theta}\pd{u_r}{\phi}-\frac{u_{\phi}^2+u_{\theta}^2}{r}\right)
=-\pd{p}{r}
+\mu\left(\nabla^2u_r-\frac{2u_r}{r^2}-\frac2{r^2\sin\theta}\pd{}{\theta}\left(u_\theta\sin\theta\right)
+\frac{2}{r^2\sin\theta}\pd{u_{\phi}}{\phi}
\right)
+\rho f_r,\\
\rho\left(\pd{u_\theta}{t}+u_r\pd{u_{\theta}}{r}+\frac{u_{\theta}}{r}\pd{u_{\theta}}{\theta}+\frac{u_{\phi}}{r\sin\theta}\pd{u_{\theta}}{\phi}+\frac{u_ru_{\theta}-u_{\phi}^2\cot\theta}{r}\right)
=-\frac1r\pd{p}{\theta}
+\mu\left(\nabla^2u_\theta-\frac{u_\theta}{r^2\sin^2\theta}+\frac2{r^2}\pd{u_r}{\theta}-\frac{2\cos\theta}{r^2\sin^2\theta}\pd{u_\phi}{\phi}\right)
+\rho f_\theta,\\
\rho\left(\pd{u_{\phi}}{t}+u_r\pd{u_{\phi}}{r}+\frac{u_{\theta}}{r}\pd{u_{\phi}}{\theta}+\frac{u_{\phi}}{r\sin\theta}\pd{u_{\phi}}{\phi}+\frac{u_ru_{\phi}+u_{\phi}u_{\theta}\cot\theta}{r}\right)
=-\frac1{r\sin\theta}\pd{p}{\phi}
+\mu\left(\nabla^2u_\phi-\frac{u_\phi}{r^2\sin^2\theta}+\frac{2}{r^2\sin\theta}\pd{u_r}{\phi}+\frac{2\cos\theta}{r^2\sin^2\theta}\pd{u_\theta}{\phi}\right)
+\rho f_\phi,
\end{gather*}
where the Laplacian is defined by its action on a function \(A(r,\theta,\phi)\text{:}\)
\begin{equation*}
\nabla^2A=\frac1{r^2}\pd{}{r}\left(r^2\pd{A}{r}\right)+\frac1{r^2\sin\theta}\pd{}{\theta}\left(\sin\theta\pd{A}{\theta}\right)+\frac1{r^2\sin^2\theta}\pd{^2A}{\phi^2},
\end{equation*}
and \(\bf=(f_r,f_\theta,f_\phi)\) represents the body forces per unit mass acting on the fluid.
(a)
A sphere of radius \(a\) sits in an infinite vat of fluid of density \(\rho\) and shear viscosity \(\mu\text{.}\) The pressure is \(p_0\) in the far field and the surface of the sphere is at a constant pressure \(p_1\text{.}\) Find the fluid pressure and velocity fields. You may neglect gravity.
Solution.
We assume the flow and pressure fields are steady and spherically symmetric. Thus we neglect all \(t\text{,}\) \(\theta\) and \(\phi\) derivatives, and set \(u_\theta=u_\phi=0\text{.}\) The continuity equation reduces to
\begin{equation*}
\frac1{r^2}\pd{(r^2u_r)}{r}=0,
\end{equation*}
which implies that
\begin{equation*}
u_r=\frac{A}{r^2}
\end{equation*}
where \(A\) is independent of \(r\text{.}\) The Navier–Stokes equations reduce to
\begin{align*}
\rho u_r\pd{u_r}{r}=&-\pd{p}{r}+\mu\left(\frac1{r^2}\pd{}{r}\left(r^2\pd{u_r}{r}\right)-\frac{2u_r}{r^2}\right),\\
0=&0,\\
0=&0,
\end{align*}
Substituting the expression for \(u_r\) into the \(r\)-component gives
\begin{equation*}
-\frac{\rho A^2}{r^5}=-\pd{p}{r}+\mu\left(-\frac{2A}{r^4}+\frac{2A}{r^4}\right),
\end{equation*}
which simplifies to
\begin{equation*}
\pd{p}{r}=\frac{\rho A^2}{r^5}.
\end{equation*}
Integrating and applying the boundary condition as \(r\rightarrow\infty\) gives
\begin{equation*}
p=p_0-\frac{\rho A^2}{4r^4}.
\end{equation*}
Applying the boundary condition at \(r=a\) gives
\begin{equation*}
p_1=p_0-\frac{\rho A^2}{4a^4}\quad\Rightarrow\quad A=a^2\sqrt{\frac{4(p_0-p_1)}{\rho}},
\end{equation*}
and hence the velocity and pressure fields are
\begin{align*}
u_r=&\frac{a^2}{r^2}\sqrt{\frac{4(p_0-p_1)}{\rho}},\\
p=&p_0-\frac{\rho a^4(p_0-p_1)}{r^4}.
\end{align*}
(b)
What is the volumetric flux of fluid leaving the sphere?
