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Exercises 6.9 Practice Exercises

Remark 6.9.1 Solutions and feedback.

Full solutions to the exercises in this chapter can be found in the solutions to this chapter A.5 in Appendix A at the back of the book.
If you would like feedback on your work, the tutor will by default concentrate on the exercises highlighted with a yellow background in the online notes (the homework exercises), unless you specifically ask for feedback on a different exercise instead.

Warning 6.9.2.

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1. Reynolds number estimation.

Based on Problem 2.1 in [4].
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);

(b)

blood flow in a large artery (diameter 2 cm) at 0.3 m/s;

(c)

a thick layer of honey draining off a spoon at 1 cm/s;

(d)

a spermatozoon with tail length \(10^{-3}\) m swimming at 100 \(\mu\)m/s in water.

2. Nondimensionalisation.

Based on Problem 2.2 in [4]
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.

(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.

(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?

(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 \(\tau\text{?}\)

(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\(^3\text{.}\)

(g)

Comment on the experimental set up needed to measure the terminal velocity.

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 6.34.
Figure 6.34. Sketch of plate.

(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).

(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*}

(c)

In this part, find the drag force on the plate by using the stress tensor, and check your answer against the previous part.

5. Stress on a surface due to the pressure.

In this problem, you will calculate the resultant force on the cube shown in Figure 6.18 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.
Note that, in this question, as with the derivation of the Navier–Stokes equations (6.5.1), you may use some sleight of hand, so as to avoid carefully integrating the force over the surface. However, as in that case, fortunately the errors cancel and the result is correct.

(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.

(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*}

(c)

Check that you get the same result using the stress tensor of an inviscid fluid (6.4.10).

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 \(w=0\text{.}\)

(c)

Use the continuity equation to show that the \(y\)-component of the velocity is also zero (\(v=0\)).

(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.

(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.

(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 \(x=0\text{.}\)

(g)

Compare the stresses on the two plates. Why is this to be expected?

7. Force on a blob of fluid.

Based on Exercises 6.1 and 6.2 in [4].
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.
You may use the fact that
\begin{equation*} \hat{\bn}\times\left(\nabla\times\bu\right) =\nabla(\hat{\bn}\cdot\bu)-(\hat{\bn}\cdot\nabla)\bu, \end{equation*}
where, in the first bracket, the operator \(\nabla\) acts on \(\bu\) but not \(\bn\text{.}\)

(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*}

(c)

Deduce that, if the blob is small, the net force on it per unit volume of fluid equals \(-\nabla p+\mu\nabla^2\bu\text{.}\)

8. General deformation of a fluid element.

Based on Exercise 6.7 in [4].
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*}

(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{6.9.1} \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 \(\bg\)

(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.

9. Decomposition of a shear flow.

Based on Exercise 6.8 in [4].
In this problem, you will apply the equation from the previous problem Exercise 6.9.8 to a shear flow \(\bu=(\beta y,0,0)\text{.}\)

(a)

Use (6.9.1) 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.

10. Derivation of the Cauchy equation.

In this problem, you will derive the Cauchy equation (6.5.2) for the conservation of linear momentum in an incompressible fluid that is not necessarily Newtonian (the Navier–Stokes equation only applies to Newtonian fluids).

(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 6.18, and apply the Reynolds Transport Theorem (3.1.2) to the fluid in this volume.

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 (6.5.1) 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 \(\bg\text{.}\)

(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?

(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.

(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.

12. Hagen–Poiseuille flow.

In this exercise you will review the derivation of Poiseuille flow and prove the Hagen–Poiseuille formula (6.8.4).

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*}

(b)

Show that the Navier–Stokes equations reduce to
\begin{equation*} \pd{p}{r}=\frac{\rho a^2U^2}{r^3}. \end{equation*}

(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.

(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.

(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*}

(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 6.30.
Examples of the solution for three different Womersley numbers are shown in Figure 6.37, Figure 6.38 and Figure 6.39. 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.
Womersley \(\alpha=1\)
Figure 6.37. Womersley flow profile for \(\alpha=1\text{.}\)
Womersley \(\alpha=10\)
Figure 6.38. Womersley flow profile for \(\alpha=10\text{.}\)
Womersley \(\alpha=100\)
Figure 6.39. Womersley flow profile for \(\alpha=100\text{.}\)

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.

(b)

What is the volumetric flux of fluid leaving the sphere?
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