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@ -2627,6 +2627,8 @@ if you have $\tau_0 = 0$ (the Kelvin-Voigt model of a viscoelastic solid) $\delt |
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\pagebreak |
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\pagebreak |
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\chapter{Continuum Mechanics in 3D} |
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\chapter{Continuum Mechanics in 3D} |
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\section{Navier-Stokes Derivation} |
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\begin{description} |
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\begin{description} |
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\item[Material Coordinates] (lagrangian) |
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\item[Material Coordinates] (lagrangian) |
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@ -2674,7 +2676,7 @@ At $t = 0$, $X(A, t= 0) = A \implies F = \mathbb I \implies J = 1$ |
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Since $J(t = 0) = 1$, $J \ne 0$ always implies $J(t) > 0$ |
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Since $J(t = 0) = 1$, $J \ne 0$ always implies $J(t) > 0$ |
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\week{} |
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\week{} |
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\lecture |
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\lecture{} |
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\begin{ex} Uniform dilatation |
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\begin{ex} Uniform dilatation |
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@ -3050,8 +3052,6 @@ Gives $\vec t = T^T \vec n$ where $T$ is the \emph{Cauchy stress tensor} |
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\lecture {} |
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The left hand side of $( * * ) $ can be written as |
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The left hand side of $( * * ) $ can be written as |
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$$\varrho D_t \vec v + \vec v \cdot D_t \varrho + \vec v \varrho(\nabla \cdot |
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$$\varrho D_t \vec v + \vec v \cdot D_t \varrho + \vec v \varrho(\nabla \cdot |
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@ -3119,6 +3119,7 @@ So the full system we have so far is |
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What is missing is the constitutive relation for $T$ |
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What is missing is the constitutive relation for $T$ |
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\lecture{} |
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\subsection{Material frame indifference} |
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\subsection{Material frame indifference} |
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Any constitutive law should not depend on translation or rotation: any rigid body motion. |
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Any constitutive law should not depend on translation or rotation: any rigid body motion. |
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@ -3134,9 +3135,520 @@ smooth functions, and $Q$ is a rotation matrix. $Q Q^T = \mathbb I \forall t > |
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other coordinate frame. You should be able to compute the stress tensor |
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other coordinate frame. You should be able to compute the stress tensor |
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in either and get the same result. |
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in either and get the same result. |
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\item Form invariance: any constitutive law $T = G(T) $ stress tensor in |
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terms of another quantity $R$ should be the same in the transformed |
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coordinate frame. |
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\end{enumerate} |
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These conditions give: |
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$$G(R^*) = T^* = Q G(R) Q^T \qquad\text{For all } R \text{ and } T$$ |
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\begin{ex} |
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$X^* = QX + b(t) \quad |_{\p t}$ |
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$V^* - Q'(t) X + QV + b'(t) $ this is how the velocity field transforms. |
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And a constitutive law such as $T = G(v)$ now in spatial coordinates. (We are |
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using $R = \vec v$ and $R^* = \vec v^*$). |
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MFI in this case is |
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$$Q G(v) Q^T = G(Q'(t) x + Q v + b'(t))$$ |
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For all rotations $Q(t)$ and shifts $b(t)$, but even if $Q = \mathbb I$ |
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$C(v) = G(v + b'(t))$ is only true if $b'(t) = 0$ ie $b$ is constant. |
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\end{ex} |
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\begin{ex} |
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$T = \mu (\nabla v + (\nabla v)^T) $ ie $R = \nabla v + \nabla v ^T$ is |
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materially frame indifferent. |
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\end{ex} |
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\begin{thm}{}{} Rivilin Ericson Theorem |
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Assuming that $T$ and $R$ are symmetric and objective and $T = G(R)$ is form |
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invariant, then it can be written as |
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$$T = \alpha_0 \mathbb I + \alpha_1 R + \alpha_2 R^2$$ |
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Where $\alpha_0, \alpha_1,\alpha_2 \in \R$ are functions of |
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$I_R = tr(R) = |
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\lambda_1 + \lambda_2 + \lambda_3$ |
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$II_R = \frac 12(tr(R)^2 - tr(R^2)) = \lambda_1 \lambda_2 + \lambda_1 |
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\lambda_3 + \lambda_2 \lambda_3$ |
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$III_R = \det(R) = \lambda_1 \lambda_2 \lambda_3$ |
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Which are the principle invariants of $R$, ie, they are conserved under |
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rotations and $\lambda_1 , \lambda_2 \lambda_3$ are eigenvalues of $R$. |
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How to show this: |
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Either constrctive by picking a range of potential rotations, or assuming |
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that $G$ has a (taylor) expansion $G(R) = \sum_{n = 0}^\infty \kappa_n R^n$ |
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with constants $\kappa_n$, and using the Cayley-Hamilton theorem: |
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$R \in \R^{3\times 3} $ satisfies its characteristic equation: $\det(R - X |
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\mathbb I) = 0 \implies R^3 I_R R^2 - II_R + III_R \mathbb I$. |
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$... = \alpha_{2,n} R^2 + \alpha_{1,n} R + \alpha_{0,n} \mathbb I$. |
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\end{thm} |
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\week{} |
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\lecture{} |
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\subsection{Newtonian Fluids} |
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The stress depends linearly on the strain rate. (In 1D strain rate was $\p_t |
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\varepsilon$). In 3D strain rate is pressure. |
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\begin{itemize} |
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\item Pressure is modelled by $T = -p \mathbb I$. Acting through an |
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arbitrary plane with unit normal $\vec n$, it is given by $-p \vec n$. |
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Pressure is isotripic. |
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\item Viscous stress. In 1D this was $\varepsilon_t = \p_t (U_A) = V_A$ |
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to model relative velocity differences between neighbouring particles. |
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In 3D spatial coordinates an analog would be $\p_x v, \p_y v, \p_z v$ ie |
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$\nabla \vec v$. |
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Decompose this into its symmetric and anti-symmetric component: $\nabla |
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\vec v = D + W $ where: |
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$D$ is the rate of deformation tensor defined as: |
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$$ |
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\begin{aligned} |
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{D} & \equiv \frac{1}{2}\left(\nabla \vec{v}+\nabla \vec{v}^{T}\right) \\ |
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&=\left(\begin{array}{ccc} |
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\frac{\partial v_{1}}{\partial x} & \frac{1}{2}\left(\frac{\partial v_{1}}{\partial y}+\frac{\partial v_{2}}{\partial x}\right) & \frac{1}{2}\left(\frac{\partial v_{1}}{\partial z}+\frac{\partial v_{3}}{\partial x}\right) \\ |
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\frac{1}{2}\left(\frac{\partial v_{1}}{\partial y}+\frac{\partial v_{2}}{\partial x}\right) & \frac{\partial v_{2}}{\partial y} & \frac{1}{2}\left(\frac{\partial v_{2}}{\partial z}+\frac{\partial v_{3}}{\partial y}\right) \\ |
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\frac{1}{2}\left(\frac{\partial v_{1}}{\partial z}+\frac{\partial v_{3}}{\partial x}\right) & \frac{1}{2}\left(\frac{\partial v_{2}}{\partial z}+\frac{\partial v_{3}}{\partial y}\right) & \frac{\partial v_{3}}{\partial z} |
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\end{array}\right) |
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\end{aligned} |
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$$ |
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And $W$ is the vorticity or spin tensor: |
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$$ |
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\begin{aligned} |
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{W} & \equiv \frac{1}{2}\left(\nabla \vec{v}-\nabla \vec{v}^{T}\right) \\ |
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&=\left(\begin{array}{ccc} |
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0 & \frac{1}{2}\left(\frac{\partial v_{1}}{\partial y}-\frac{\partial v_{2}}{\partial x}\right) & \frac{1}{2}\left(\frac{\partial v_{1}}{\partial z}-\frac{\partial v_{3}}{\partial x}\right) \\ |
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\frac{1}{2}\left(\frac{\partial v_{2}}{\partial x}-\frac{\partial v_{1}}{\partial y}\right) & 0 & \frac{1}{2}\left(\frac{\partial v_{2}}{\partial z}-\frac{\partial v_{3}}{\partial y}\right) \\ |
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\frac{1}{2}\left(\frac{\partial v_{3}}{\partial x}-\frac{\partial v_{1}}{\partial z}\right) & \frac{1}{2}\left(\frac{\partial v_{3}}{\partial y}-\frac{\partial v_{2}}{\partial z}\right) & 0 |
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\end{array}\right) |
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\end{aligned} |
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$$ |
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\end{itemize} |
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This suggests a constitutive law of the form: |
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$$T = -p \mathbb I + G(D,w) \qquad(*)$$ |
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However under rigid body motion $X^* = Q(t) X + b(t)$ and $D$ and $w$ transform |
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according to $D^* = QDQ^T$ $w^* = QwQ^T + |
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\underbrace{Q'Q^T}_{\text{an antisimmetric |
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matrix}}$ |
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$(*)$ is MFI: $QG(R)Q^T = G(R^*)$ if |
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$$QG(D,w) Q^T = G(D^*, w^*) = G(Q D Q^T, QwQ^T, + Q'Q^T)\quad \forall Q(t)$$ |
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But if $Q = \mathbb I$ and $Q' = M$ (anti-symmetric), |
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$$G(D,w) = G(D, w + M)$$ |
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Which is only true if $M = \hat 0 \in \R^{3 \times 3}$ is the zero matrix. |
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$\implies T = -p \mathbb I + G(D)$. |
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The rivilin-erickson theorem now tells us that |
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$$G(D) = \alpha_0 \mathbb I + \alpha_q D + \alpha_2 D^2$$ |
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And $\alpha_0 = 0$ if $D = \mathbf 0$, there is no viscous stress if $D = \mathbf 0$. |
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Finally, since linearity characterises newtonian fluids, we simplify the model |
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by assuming linear dependence of $T$ on $D$. |
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$\implies \alpha_2 = 0$ because $D^2$ is not linear. |
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$\alpha_1 = 2\mu$, a constant, since $D$ is linear. |
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$\alpha_0 = \lambda I_D = \lambda tr(D) = \lambda \nabla \cdot \vec v$ since |
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$I_d$ is the only principle invariant that is linear. |
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\begin{itemize} |
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\item [$\mu$] Dynamic shear velocity |
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\item [$p$] pressure |
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\item [$\lambda$] bulk viscosity |
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\end{itemize} |
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$\implies \nabla \cdot T = - \nabla p + \lambda \nabla (\nabla \cdot \vec v ) |
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+ \mu(\Delta \vec v + \nabla (\nabla \cdot \vec v))$ |
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And substitute into the momentum equation to obtain the Navier-Stokes equation. |
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$$\varrho D_t \vec v = -\nabla p (\lamba + \mu) \nabla (\nabla \cdot \vec v) + |
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\mu \Delta \vec v + \varrho \vec f$$ |
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Typically couple to the continuity equation: |
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$$D_t \varrho + \varrho \nabla \cdot \vec v = 0$$ |
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Ideal gas law (with known temperature $T$): |
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$$p = \varrho R T$$ |
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Usually this system is solved numerically, for example using OpenFoam software |
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package. |
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\lecture{} |
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A common simplification is to assume incompressibility: $\p_t \varrho = 0 = |
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\p_{\vec x_i } \varrho \implies \p_t \varrho + \Nabla \cdot (\varrho \vec v) = 0$ |
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$\implies \nabvla \cot \vec v = 0$ |
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The stress in this case is $T = -p \I + 2\mu D$ and we obtain the incompressible |
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navier-stokes equation: |
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$$\varrho D_t \vec v = - \nabla p + \mu \Delta \vec v + \varrho \vec f$$ |
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Coupled to : $\nabvla \cdot \vec v = 0$ |
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And treat $p$ as lagrangian multiplier to enforce $\nabla\cdot \vec v = 0 $ |
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Then for an incompressible newtonian fluid in cartesian coordinates, letting |
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$\vec v = (u,v,w)$ and $\vec f = (f,g,h)$: |
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\begin{gathered} |
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\rho\left(\frac{\partial u}{\partial t}+u \frac{\partial u}{\partial x}+v \frac{\partial u}{\partial y}+w \frac{\partial u}{\partial z}\right)=-\frac{\partial p}{\partial x}+\mu\left(\frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}+\frac{\partial^{2} u}{\partial z^{2}}\right)+\rho f \\ |
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\rho\left(\frac{\partial v}{\partial t}+u \frac{\partial v}{\partial x}+v \frac{\partial v}{\partial y}+w \frac{\partial v}{\partial z}\right)=-\frac{\partial p}{\partial y}+\mu\left(\frac{\partial^{2} v}{\partial x^{2}}+\frac{\partial^{2} v}{\partial y^{2}}+\frac{\partial^{2} v}{\partial z^{2}}\right)+\rho g \\ |
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\rho\left(\frac{\partial w}{\partial t}+u \frac{\partial w}{\partial x}+v \frac{\partial w}{\partial y}+w \frac{\partial w}{\partial z}\right)=-\frac{\partial p}{\partial z}+\mu\left(\frac{\partial^{2} w}{\partial x^{2}}+\frac{\partial^{2} w}{\partial y^{2}}+\frac{\partial^{2} w}{\partial z^{2}}\right)+\rho h \\ |
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\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}+\frac{\partial w}{\partial z}=0 |
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\end{gathered} |
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Note: $D_t \vec v = \p_t \vec v + (\vec v \cdot \nabla) \vec v = \p_t |
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\vec v + (\nabla\vec v) \vec v $ |
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$\Delta \vec v = \begin{pmatrix} \Delta u\\ \Delta v \\ \Delta w \end{pmatrix} $ |
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\subsubsection{Remarks} |
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\begin{itemize} |
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\item |
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\item |
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Proving global existance in time for the incompressible navier stokes |
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equation is one of the millenium problems. |
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\item |
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This model is the basis for |
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simulations of fluids on all spatial scales: galaxies, earths mantle, oceans, |
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hydraulics, in biology; blood flow, tissue, cytoplasm, etc. |
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\item |
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$\mu$ is the shear viscosity: |
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\begin{center} |
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\begin{array}{l|l|l} |
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\hline \text { Fluid } & \text { Viscosity (Pa-s) } & \text { Density }\left(\mathrm{kg} / \mathrm{m}^{3}\right) \\ |
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\hline \text { Air } & 1.8 \times 10^{-5} & 1.18 \\ |
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\hline \text { Water } & 8.9 \times 10^{-4} & 0.997 \times 10^{3} \\ |
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\hline \text { Mercury } & 1.5 \times 10^{-3} & 1.3 \times 10^{4} \\ |
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\hline \text { Olive oil } & 0.08 & 0.92 \times 10^{3} \\ |
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\hline \text { Honey } & 37 & 1.4 \times 10^{3} \\ |
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\hline \text { Pitch } & 2.3\times 10^8& \\ |
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\hline |
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\end{array} |
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\end{center} |
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\end{itemize} |
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\section{Solutions to steady flow problems} |
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Steady flow is characterised by $\p_t \vec v = 0$, and $\p_t p = 0$ and $\p_t |
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\varrho = 0$. |
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This implies that $D_t \vec v = \p_t \vec v + (\vec v \cdot \nabla) \vec v$ |
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So we now the Newtonian Navier--Stokes becomes: |
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$$\begin{cases} \varrho(\vec v \cdot \nabla) \vec v = - \nabla p + \mu \Delta |
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\vec v + \varrho \vec f \\ \nabla \vec v = 0\end{cases}$$ |
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Note that $\vec v, p \varrho $ do not change with time but single particles |
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still move, and their trajectories must satisfy in general: |
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$$\begin{cases}\frac {d X} {dt} = V(A,t) = v(X(A,t), t)\\ X(A,t=0) = A |
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\end{cases}$$ |
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So in steady flow: |
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$$\begin{cases} \frac {dX} {dt} = V(A) = v(X(A,t)) \\ X(A,t=0) = A\end{cases}$$ |
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The trajectory lines do not change with time. |
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\subsection{Coette plane flow} |
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Steady flow of fluid between two parallel planes. |
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\incfigw{coette}{0.6} |
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Lower plate is stationary, upper plate has speed in x direction of $u_0$. |
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$dt y = h$. |
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Use the boundary condition: assume there is a no-slip condition between the fluid |
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and the boundary with the plate: the fluid travels at the speed of the plate. |
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At the upper plate $\vec v (x, y = h, z) = (u_0 , 0 , 0)$ |
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At the lower plate $\vec v(x, y= 0, z) = (0,0,0)$ |
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Assume laminar flow: $\vec v = (u,0,0)$: flow is only in the $x$ direction, |
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solution is homogeneous in $z$-direction: $\p_z u = 0, \p_z p = 0$ |
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Now our model becomes: |
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$$ |
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(* * *) |
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\begin{cases} |
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u_x = 0 \\ |
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\varrho u_x u = -p_x + \mu (u_{x x} + u_{y y } + u_{z z}) \\ |
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0 = -py + 0 \\ |
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0 = 0 + 0 \\ |
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\end{cases} |
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$$ |
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$\p_z u = 0 = \p_x u \implies u = u(y)$ |
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$\p_z p = 0 = \p_y p \implies p = p(x)$ |
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So $(* * * ) $ becomes $0 = -p_x + \mu u_{y y}$, i.e., $p'(x) = \mu u''(y)$. |
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Use method of separation: $p'(x) = const = \mu u''(y) \implies p' = const |
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\implies$ |
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$$p(x) = p_0 + x p_1$$ |
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Modelling intuition: $p_1 \ne 0$ would impl;y that the pressure blows up at $x = |
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\pm \infty \implies $ assume that $p_1 = 0$, with that $p(x) = p_0$. |
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\lecture{} |
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The second equation becomes $\mu u''(y) = 0 \implies u(y) = ay + b$. The |
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boundary conditions are $u(y = 0) = 0, u(h = 0) = u_0$, so $a = \frac {u_0} h |
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\implies u = \gamma y$ where $\gamma = \frac {u_0} h$ |
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$\gamma = \frac {u_0} {h} $ is the \emph{shear rate} which has dimension $\frac |
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1t$. |
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So we have a solution: |
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$\vec v = \begin{pmatrix} \gamma y \\ 0 \\0 \end{pmatrix}$ and $p = p_0$ |
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Remark: The stress tensor is given by |
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$$T = -p \mathbb I + 2\mu D = -p_0 \mathbb I + \mu(\nabla v + {\nabla v}^T) = -p_0 |
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\mathbb I + \mu \begin{pmatrix}0 & \gamma & 0 \\ \gamma & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} $$ |
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So the stress through a horizontal plane, for example $ \{ y = 0\} , |
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\vec n = (0,1,0)$, is given by |
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$$ |
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T \vec n = \left(-p_0 \mathbb I + \mu \begin{pmatrix} 0 & \gamma & 0 \\ |
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\gamma & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} \right) \begin{pmatrix} 0 \\ 1 \\ |
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0\end{pmatrix} = \underbrace{-p_0 \begin{pmatrix} 0\\1\\0 |
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\end{pmatrix}}_{\vec t_s} + |
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\underbrace{\mu |
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\begin{pmatrix} \gamma \\ 0 \\ 0 \end{pmatrix} }_{\vec t_p} |
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$$ |
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$\vec t_s$ --- the shear component. |
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$\vec t_p$ --- the normal component. |
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\subsubsection{Shear Test} |
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A Shear test can be used to investgate whether a fluid is Newtonian (where |
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stress is linear with respect to $\gamma$) and what the shear viscosity is. |
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Put the fluid between two plates and move the upper plate with differnet speeds, |
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$\gamma = \frac {u_0}{h}$. Measure the force (stress $\times$ area), to see if |
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it is linear with respect to the velocity of the top plate. |
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Since shear stress $\vec t_s = \mu \gamma \begin{pmatrix} 1 \\ 0 \\ 0 |
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\end{pmatrix} $, the gradient of your shear stress / shear rate graph is the |
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shear viscosity $\mu$. |
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\subsection{Poiseuille Flow} |
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Steady flow through a cylindrical pipe of length $L$ with radius $R$. |
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Use the incompressible Navier-Stokes equation written in cylindrical |
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coordinates. |
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Write the problem in polar coordinatnes about the centre of the pipe, with $z$ |
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running along its length: |
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$\vec v = (v_R, v_\theta, v_z) = v_r \vec e_r + v_\theta \vec e_\theta |
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+ v_z + \vec e_z$, and |
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$\vec f = f_r \vec e_r + f_\theta \vec e_\theta + f_z \vec e_z$, |
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$x = r\cos \theta$ and $y = r \sin \theta$ and $z = z$. |
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The incompressible Navier--Stokes system for newtonian fluid in |
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cyndrical coordinates is: |
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\begin{aligned} |
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&\rho\left(\frac{\partial v_{r}}{\partial t}+v_{r} \frac{\partial v_{r}}{\partial r}+\frac{v_{\theta}}{r} \frac{\partial v_{r}}{\partial \theta}-\frac{v_{\theta}^{2}}{r}+v_{z} \frac{\partial v_{r}}{\partial z}\right)\\ |
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&\quad=-\frac{\partial p}{\partial r}+\mu\left[\frac{\partial}{\partial r}\left(\frac{1}{r} \frac{\partial}{\partial r}\left(r v_{r}\right)\right)+\frac{1}{r^{2}} \frac{\partial^{2} v_{r}}{\partial \theta^{2}}-\frac{2}{r^{2}} \frac{\partial v_{\theta}}{\partial \theta}+\frac{\partial^{2} v_{r}}{\partial z^{2}}\right]+\rho f_{r}\\\\ |
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&\rho\left(\frac{\partial v_{\theta}}{\partial t}+v_{r} \frac{\partial v_{\theta}}{\partial r}+\frac{v_{\theta}}{r} \frac{\partial v_{\theta}}{\partial \theta}+\frac{v_{r} v_{\theta}}{r}+v_{z} \frac{\partial v_{\theta}}{\partial z}\right)\\ |
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&\quad=-\frac{1}{r} \frac{\partial p}{\partial |
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\theta}+\mu\left[\frac{\partial}{\partial r}\left(\frac{1}{r} |
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\frac{\partial}{\partial r}\left(r v_{\theta}\right)\right)+\frac{1}{r^{2}} |
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\frac{\partial^{2} v_{\theta}}{\partial \theta^{2}}+\frac{2}{r^{2}} |
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\frac{\partial v_{r}}{\partial \theta}+\frac{\partial^{2} v_{\theta}}{\partial |
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z^{2}}\right]+\rho f_{\theta}\\ \\ |
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&\rho\left(\frac{\partial v_{z}}{\partial t}+v_{r} \frac{\partial v_{z}}{\partial r}+\frac{v_{\theta}}{r} \frac{\partial v_{z}}{\partial \theta}+v_{z} \frac{\partial v_{z}}{\partial z}\right)\\ |
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&\quad =-\frac{\partial p}{\partial z}+\mu\left[\frac{1}{r} |
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\frac{\partial}{\partial r}\left(r \frac{\partial v_{z}}{\partial |
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r}\right)+\frac{1}{r^{2}} \frac{\partial^{2} v_{z}}{\partial |
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\theta^{2}}+\frac{\partial^{2} v_{z}}{\partial z^{2}}\right]+\rho f_{z}\\ \\ |
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&\frac{1}{r} \frac{\partial\left(r v_{r}\right)}{\partial r}+\frac{1}{r} \frac{\partial v_{\theta}}{\partial \theta}+\frac{\partial v_{z}}{\partial z}=0 |
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\end{aligned} |
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Unlike before there is pressure difference across the length ($z$) of the pipe |
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and the flow is coming from the pressure difference. |
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The boundary conditions are $\vec v = \begin{pmatrix} 0\\0\\0 \end{pmatrix} $ at |
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$r = R$ (no-slip boundary condition). $p(z = 0) = p_0$ and $p(z = L) = p_1 < |
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p_0$. |
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$\vec v = \begin{pmatrix} v_r \\ v_\theta \\ v_z \end{pmatrix} $ |
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Again assume laminar flow (trajectory not depending on $\theta$ or $r$). |
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The velocity only has a $z$ component: |
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$v_r(z = 0) = 0$ and $v_\theta(z = 0) = 0$ |
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$v_r(z = L) = 0$ and $v_\theta (z = L) = 0$ |
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$v_\tehta = v_r = 0 \implies \vec v = \begin{pmatrix} 0 \\ 0 \\ v_z |
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\end{pmatrix} $ |
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And assume the solution satisfies rotational symmetry |
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$v_z = v_z(r,z)$ and $p = p(r,z)$. |
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And assume body forces are $0$. |
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Now we can get rid of most of the Navier Stokes terms because of these |
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assumptions which leaves the governing equations |
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$$ |
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\begin{cases} |
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0 = \p_r p \\ |
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\varrho v_z \frac {\p v_z} {\p z} = -p_z + \mu (\frac 1r \p_r( r \p_r |
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v_z ) + \p _{z z} v_z)\\ |
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0 = \p_z v_z |
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\end{cases} |
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$$ |
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$\implies p = p(z) $ and $v_z = (r)$ and $p'(z) = \mu(v_z'' (r) + \frac 1r v_z' |
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(r)$ |
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Use method of separation: $p'(z) = constant = \mu (v_z '' + \frac 1r v_z')$ |
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Use $p (z = 0) = p_0$ and $p(z = L) = p_1$ $\implies$ |
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$p = p_0 + \alpha z$ and $l(L) = p_1 = p_0 + \alpha l \implies \alpha = \frac |
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{p_1 - p_0 } L$ |
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$\implies p(z) = p_0 + \frac {p_1 - p_0}L z$ |
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$\frac {p_1 - p_0} L = \mu (v_z'' + \frac 1r v_z ' ) $ |
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$$\frac r \mu \frac {p_1 - p_0} L = r v_z '' + vz' = (rvz') ' $$ |
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Integrate |
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$$\frac {r^ 2} 2 \frac {p_1 - p_0} {\mu L} = r v_z ' - C$$ |
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Integrate |
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$$v_z ' = \frac r2 \frac {p_1 - p_0} {\mu L} + \frac 1r C$$ |
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Integrate using $v_z(R) = 0$ |
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$$v_z = (\frac {r^2}2 - \frac {R^2} 2) \frac {p_1 - p_0} {2\mu L} + C \log r$$ |
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If $C \ne 0$ the velocity would blow up towards the centre of the pipe, so $C = |
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0$. |
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$$\implies v_z = |
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\begin{pmatrix} 0 \\ 0 \\ R^2 - r^2 \end{pmatrix} |
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\frac {p_0 - p_1} {4 \mu L} $$ |
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So you get a parabolic velocity profile. |
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The stress tensor in cylindrical coordinates is given by: |
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$$T = \begin{pmatrix} -p & 0 & \mu v_z' \\ 0 & -p & 0 \\ |
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\mu v_z ' & 0 & -p \end{pmatrix} $$ |
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Eg the stress at the surface of a cylinder with radius $r < R$, with normal |
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to the cylinder of $\vec n = \begin{pmatrix} 1\\0\\0 \end{pmatrix} |
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\implies$ $T \vec n = \begin{pmatrix} -p \\ 0 \\ 0 \end{pmatrix} + \begin{pmatrix} |
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0 \\ 0 \\ \mu v_z '\end{pmatrix} = \vec t_p + \vec t_s$, shear stress and |
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normal stress. |
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\section{Closing remarks} |
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\begin{enumerate} |
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\item Vorticity: the curl of the velocity at a specific point |
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$$\vec \omega = \nabla \times \vec v = \begin{pmatrix} \frac {\p |
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w} {\p y} - \frac {\p v} {\p t} \\ \frac {\p v} {\p t} - \frac {\p w } |
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{\p x} \\ \frac {\p v} {\p x} - \frac {\p u} {\p y} \end{pmatrix} $$ |
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It is like `how much the particle rotates'. |
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For example in plane Couette flow: $\vec v = \begin{pmatrix} \gamma y \\ 0 |
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\\ 0 \end{pmatrix} $, the vorticity is $\vec \omega = \nabla \times \vec v |
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= \begin{pmatrix} 0 \\ 0 \\ - \gamma\end{pmatrix} $ |
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\item |
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Helmholtz Representation Theorem |
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Let $\vec q (\vex x) : \R^3 \to \R^3$ diffble, then |
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$\vec q (\vec x) = \nabla \phi + \nabla\times \vec g $ where |
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$\phi:\R^3 \to |
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R$ and $\vec g: \R^3 \to \R^3, \nabla \cdot \vec g = 0$. |
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It can be split into the scalar potential with divergence and no rotational |
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component (curl) and |
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the vector potential that is pure rotation. |
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\end{enumerate} |
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\end{enumerate} |
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\week{} |
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\lecture{} |
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\pagebreak |
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\pagebreak |
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\appendix |
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\appendix |
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