Problems set 7 to hand in

Exercises 7.10 Problems set 7 to hand in

Please hand in the problems in this Section for Set 7 (due in the week commencing 1 Dec).

1. Analysis of flow in a pipe.

It might help to try Exercise 7.9.6 before attempting this problem.

A viscous incompressible fluid flows steadily along an infinitely long pipe of radius \(a\) under a constant pressure gradient \(P=-\de p/\de z\text{.}\) Work in cylindrical polar coordinates \((r,\theta,z)\) with corresponding velocity components \((u_r,u_\theta,u_z)\) with the axis of the pipe aligned with \(r=0\text{.}\)

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(a)

Write down the boundary conditions to be applied.

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Solution.

The no-slip boundary conditions give

\begin{equation*} \left.\bu\right|_{r=a}=0. \end{equation*}

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(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 \(\theta\) and \(z\text{,}\) and also that the \(\theta\)-component of the velocity is zero.

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Solution.

Assuming axisymmetric flow implies there is no dependence on \(\theta\) and that the \(\theta\)-component of the velocity is zero \(u_\theta=0\text{.}\) Since the pipe is infinitely long in the \(z\)-direction, and the pressure gradient is constant, there is no preferred location in this directions, and hence the velocity profile is independent of \(z\text{.}\)

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(c)

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

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In cylindrical polar coordinates, the continuity equation for an incompressible fluid is

\begin{equation*} \frac{1}{r}\pd{}{r}(ru_r)+\frac{1}{r}\pd{u_\theta}{\theta}+\pd{u_z}{z}=0. \end{equation*}

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Solution.

We have no \(\theta\)- or \(z\)-dependence, meaning that the continuity equation reduces to

\begin{equation*} \frac{1}{r}\pd{}{r}(ru_r)=0, \end{equation*}

and hence \(u_r=A/r\text{,}\) where \(A\) is constant. Since the cylinder is a solid boundary, we need \(u_r=0\) at \(r=a\text{.}\) Therefore, \(A=0\text{,}\) and so \(u_r=0\) everywhere.

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(d)

Hence the only non-zero component of the velocity is the \(z\)-component, \(u_z\text{,}\) and this only depends on \(r\text{,}\) \(u_z(r)\text{.}\) Check that the profile

\begin{equation*} u_z=\frac{P}{4\mu}\left(a^2-r^2\right) \end{equation*}

satisfies the boundary conditions.

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Note that we also need to check that the velocity is smooth at \(r=0\text{.}\) This is easier to see if we rewrite the velocity profile in Cartesian coordinates. Converting to Cartesian coordinates, the flow is \(u=(0,0,u_z)\) with

\begin{equation*} u_z=\frac{P}{4\mu}\left(a^2-(x^2+y^2)\right), \end{equation*}

which is smooth at \(r=0\text{.}\) Note, however, that for example the profile

\begin{equation*} u_z=\frac{P}{4\mu}\left(a-r\right) \end{equation*}

would not be smooth at \(r=0\text{.}\)

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Solution.

At \(r=a\text{,}\) we have

\begin{equation*} u_z=\frac{P}{4\mu}\left(a^2-a^2\right)=0, \end{equation*}

which satisfies the no-slip boundary condition.

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(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.

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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.

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(f)

Calculate the three components of the stress vector, the force per unit area that the fluid exerts, on the inside of the pipe. You may assume that the pressure equals \(p_0\) at \(z=0\text{.}\) You may also assume that the rate-of-strain tensor in cylindrical polar coordinates is given by (7.4.7), and that, being a Newtonian fluid, the stress tensor is given by (7.4.2).

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Solution.

The pressure at a general point is given by

\begin{equation*} p=p_0+Pz. \end{equation*}

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The stress vector is given by \(\btau=\bsigma\bn\text{,}\) with \(\bn=-\hat{\br}\) and \(\bsigma=-p\bI+2\mu\be\text{.}\) Simplifying (7.4.7), we have

\begin{equation*} \be=\left(\begin{matrix} 0&0&\frac12\pd{u_z}{r}\\ 0&0&0\\ \frac12\pd{u_z}{r}&0&0 \end{matrix}\right) =\left(\begin{matrix} 0&0&-\frac{Pr}{4\mu}\\ 0&0&0\\ -\frac{Pr}{4\mu}&0&0 \end{matrix}\right). \end{equation*}

Hence,

\begin{equation*} \bsigma=\left(\begin{matrix} -p_0-Pz&0&-Pr/2\\ 0&-p_0-Pz&0\\ -Pr/2&0&-p_0-Pz \end{matrix}\right). \end{equation*}

At \(r=a\text{,}\) we have

\begin{align*} \tau=&\bsigma\bn\\ =&\left(-p\bI+2\mu\be\right)(-\hat{\br})\\ =&\left(p\bI-2\mu\be\right)\hat{\br}\\ =&p\hat{\br}-\mu\pd{u_z}{r}\hat{\bz}. \end{align*}

Hence the normal component of the stress vector is

\begin{equation*} \tau_r=p=p_0+Pz, \end{equation*}

and the axial component of the shear stress vector is

\begin{gather*} \tau_z=-\mu\pd{u_z}{r}\\ =-\mu\pd{}{r}\left[\frac{P}{4\mu}(a^2-r^2)\right]\\ =-\mu\left(-\frac{P}{2\mu}r\right)\\ =\frac{Pa}{2} \end{gather*}

The \(\theta\)-component of the stress vector is zero:

\begin{equation*} \tau_\theta=0. \end{equation*}

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(g)

You should have found that the normal stress is radially outward, while the shear stress points along the pipe in the direction of increasing \(z\text{.}\) Explain why you would expect these directions.

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Solution.

The normal stress is due to the pressure in the fluid, which acts equally in all directions, and hence the normal stress on the inside of the pipe wall points radially outward into the wall. The shear stress points along the pipe in the direction of increasing \(z\) since this is the direction of the flow, and hence the fluid exerts a tangential force on the pipe wall in this direction.

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2. Rotating flow.

Based on Exercise 6.6 in [4].

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Consider an incompressible Newtonian fluid flowing with velocity \(\bu=\bOmega\times\bx\text{,}\) where \(\bOmega\) is a constant vector. Since \(\bOmega\) is constant, for simplicity you may assume that \(\bOmega=\Omega\bk\text{.}\)

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(a)

Calculate the velocity components in Cartesian coordinates in the case \(\bOmega=\Omega\bk\text{.}\)

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Solution.

We have

\begin{equation*} \bu=\bOmega\times\bx =\begin{vmatrix} \bi & \bj & \bk \\ 0 & 0 & \Omega \\ x & y & z \end{vmatrix} =-\Omega y\bi+\Omega x\bj+0\bk. \end{equation*}

Hence the velocity components are

\begin{equation*} u=-\Omega y,\quad v=\Omega x,\quad w=0. \end{equation*}

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(b)

Show that this flow is incompressible.

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Solution.

We have

\begin{equation*} \nabla\cdot\bu=\pd{u}{x}+\pd{v}{y}+\pd{w}{z}=0+0+0=0, \end{equation*}

and hence the flow is incompressible.

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(c)

Find the components of the stress tensor.

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Solution.

We have

\begin{align*} \sigma_{11}=&-p+2\mu\pd{u}{x}=-p+0=-p,\\ \sigma_{12}=&\mu\left(\pd{u}{y}+\pd{v}{x}\right) =\mu(-\Omega+ \Omega)=0,\\ \sigma_{13}=&\mu\left(\pd{u}{z}+\pd{w}{x}\right)=\mu(0+0)=0,\\ \sigma_{21}=&\mu\left(\pd{v}{x}+\pd{u}{y}\right) =\mu(\Omega -\Omega)=0,\\ \sigma_{22}=&-p+2\mu\pd{v}{y}=-p+0=-p,\\ \sigma_{23}=&\mu\left(\pd{v}{z}+\pd{w}{y}\right)=\mu(0+0)=0,\\ \sigma_{31}=&\mu\left(\pd{w}{x}+\pd{u}{z}\right)=\mu(0+0)=0,\\ \sigma_{32}=&\mu\left(\pd{w}{y}+\pd{v}{z}\right)=\mu(0+0)=0,\\ \sigma_{33}=&-p+2\mu\pd{w}{z}=-p+0=-p. \end{align*}

Hence the deviatoric stress tensor is

\begin{equation*} \bsigma =\left(\begin{matrix} -p & 0 & 0 \\ 0 & -p & 0 \\ 0 & 0 & -p \end{matrix}\right), =-p\bI, \end{equation*}

where \(\bI\) is the identity tensor.

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(d)

Without doing a calculation, explain why you would have expected this answer.

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Solution.

The result shows that the deviatoric stress tensor is zero, meaning there are no viscous stresses in this flow. The only stress is due to the pressure in the fluid. This is as expected since the flow represents a solid body rotation with angular velocity \(\bOmega\text{.}\) In a solid body rotation, there is no relative motion between adjacent fluid particles, and hence no viscous stresses.

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3. Flow outside a rotating cylinder.

Based on problem 2.8 in [4].

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An incompressible Newonian fluid fills \(\mathbb{R}^3\text{.}\) An infinite solid cylinder of radius \(a\) is placed in the fluid with its axis on the line \(r=0\) in cylindrical polar coordinates.

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(a)

The cylinder rotates at constant angular velocity \(\Omega\) about its axis. In the resulting flow, explain why it is reasonable to assume that \(p=p(r)\) and \(\bu=(u_r(r),u_\theta(r),0)\text{.}\)

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Solution.

The flow and pressure are steady since the cylinder rotates at constant angular velocity. The flow and pressure have rotational symmetry and therefore no dependence on \(\theta\text{.}\) The flow and pressure are independent of \(z\) since the cylinder is infinite in this direction and there is no pressure gradient in this direction. The \(z\)-component of velocity is zero since there is no motion of the cylinder or pressure gradient in this direction.

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(b)

Hence perform a calculation to show that \(u_r=0\text{.}\)

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You may use the continuity equation in cylindrical polar coordinates:

\begin{equation*} \frac{1}{r}\pd{}{r}(ru_r)+\frac{1}{r}\pd{u_\theta}{\theta}+\pd{u_z}{z}=0. \end{equation*}

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Solution.

Since there is no \(\theta\)- or \(z\)-dependence, the continuity equation reduces to

\begin{equation*} \frac{1}{r}\pd{}{r}(ru_r)=0, \end{equation*}

and hence \(u_r=A/r\text{,}\) where \(A\) is constant. Since the cylinder is a solid boundary, we need \(u_r=0\) at \(r=a\text{.}\) Therefore, \(A=0\text{,}\) and so \(u_r=0\) everywhere.

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(c)

The Navier–Stokes equations in cylindrical polar coordinates 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, which you may neglect in this question.

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Show that the flow

\begin{equation*} \bu=\left(0,\frac{a^2\Omega}{r},0\right) \end{equation*}

satisfies the Navier–Stokes equations and boundary conditions.

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Solution.

Since \(u_r=u_z=0\) and there is no \(\theta\)- or \(z\)-dependence, the Navier–Stokes equations reduce to

\begin{align*} &-\rho\frac{u_{\theta}^2}{r}=-\pd{p}{r},\\ &0=\mu\left(\frac1r\pd{}{r}\left(r\pd{u_\theta}{r}\right)-\frac{u_{\theta}}{r^2}\right),\\ &0=-\pd{p}{z}. \end{align*}

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The \(z\)-component shows that \(p=p(r)\text{.}\) Now consider the \(\theta\)-component. Substituting \(u_\theta=a^2\Omega/r\text{:}\)

\begin{align*} \frac1r\pd{}{r}\left(r\pd{u_\theta}{r}\right)-\frac{u_{\theta}}{r^2}\\ =&\frac1r\pd{}{r}\left(r\pd{}{r}\left(\frac{a^2\Omega}{r}\right)\right)-\frac{a^2\Omega}{r^3}\\ =&-a^2\Omega\frac1r\pd{}{r}\left(\frac{1}{r}\right)-\frac{a^2\Omega}{r^3}\\ =&a^2\Omega\frac1r\left(\frac{1}{r^2}\right)-\frac{a^2\Omega}{r^3}\\ =&0. \end{align*}

Hence this flow satisfies the \(\theta\)-component of the Navier–Stokes equation.

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The \(r\)-component of the Navier–Stokes equation is

\begin{align*} \pd{p}{r}=&\rho\frac{u_{\theta}^2}{r}\\ =&\rho\frac{(a^2\Omega/r)^2}{r}\\ =&\rho\frac{a^4\Omega^2}{r^3}. \end{align*}

Hence

\begin{equation*} p(r)=p_0+\frac{\rho a^4\Omega^2}{2}\left(\frac1{a^2}-\frac1{r^2}\right), \end{equation*}

for some constant \(p_0\text{,}\) and therefore (since we have been able to find a suitable pressure), this component of the Navier–Stokes equation is also satisfied.

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Hence the flow \(\bu=(0,a^2\Omega/r,0)\) satisfies the Navier–Stokes equations. Finally, we check the boundary condition at \(r=a\text{:}\)

\begin{equation*} \left.\bu\right|_{r=a}=\left(0,\frac{a^2\Omega}{a},0\right)=(0,a\Omega,0), \end{equation*}

which matches the velocity of the cylinder surface.

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(d)

If the fluid and cylinder are initially at rest, and the cylinder suddenly starts rotating with angular velocity \(\Omega\) at time \(t=0\text{,}\) explain qualitatively how you would expect the flow to evolve with time.

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Solution.

Initially, there will be a thin boundary layer of fluid close to the cylinder surface that starts to rotate with the cylinder due to the no-slip boundary condition. As time progresses, viscous diffusion will cause the rotation to spread outwards from the cylinder surface into the fluid. Eventually, the flow will reach a steady state, which is the flow found in the previous part.

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