Dr (HDR) Sylvain Serra

Équation de continuité : \(div \vec V = 0 \Rightarrow \frac{\partial V_2}{\partial x_2}=0 \Rightarrow V_2(x_3)\)

Équation de Navier-Stokes : \(\rho \frac{d \vec V}{dt}=-\overrightarrow{grad} P+\underbrace{\rho \vec g}_{\text{\normalsize{négligeable}}} +\mu \Delta \vec V=-\nabla P + \mu \Delta \vec V\)

\(\left\{\begin{array}{l}\left.\begin{array}{l} \frac{\partial P}{\partial x_1}=0\\ \frac{\partial P}{\partial x_3}=0\end{array} \right\}\rightarrow P(x_2)\\\frac{\partial P}{\partial x_2}=\mu \frac{\partial^2 V_2}{\partial x_3^2}=\mu\frac{d^2 V_2}{d x_3^3}\end{array}\right.\)

Par symétrie, \( \frac{\partial^2 V_2}{\partial x_1^2}=0\)

\(V_2\) ne dépend que de \(x_3\), \(P\) ne dépend que de \(x_2\) \(\Rightarrow \frac{d P}{d x_2}=\mu\frac{\partial^2 V_2}{\partial x_3^2}=cte\)

\(cte=\frac{\Delta P}{L}=\frac{P_2-P_1}{L}\) que l'on peut aussi retrouver en disant \(\left.\begin{array}{l}P^*=cte. x_2+A\\x_2=0\rightarrow P=P_1=A\\ x_2=L\rightarrow P=P_2=cte.L.A=cte.L.P_1\end{array}\right\}\Rightarrow cte=\frac{P_2-P_1}{L}\)

\(\frac{\partial ^2V_2}{\partial x_3^2}=\frac 1\mu\left( \frac{\partial P}{\partial x_2}\right)\)

\(\frac{\partial V_2}{\partial x_3}=\frac 1\mu x_3\left( \frac{d P}{d x_2}\right)+A\)

\(V_2(x_3)=\frac{1}{2\mu}\left( \frac{d P}{d x_2}\right)x_3^2+Ax_3+B\)

CL :

\(V_2(-y_0)=0=\frac{1}{2\mu}\left( \frac{d P}{d x_2}\right)y_0^2-Ay_0+B\) (1)

\(V_2(y_0)=w_0=\frac{1}{2\mu}\left( \frac{d P}{d x_2}\right)y_0^2+Ay_0+B\) (2)

(1)+(2)\(\Rightarrow 2B+\frac 1\mu\left(\frac{d P}{d x_2}\right)y_0^2=w_0\)

\(B=\frac 12 w_0-\frac{1}{2\mu}\left(\frac{\partial P}{\partial x_2}\right)y_0^2\)

\(A=\frac{1}{2\mu}\cancel{\left(\frac{\partial P}{\partial x_2}\right)}y_0+\frac 12 \frac{w_0}{y_0} -\frac{1}{2\mu}\cancel{\left(\frac{\partial P}{\partial x_2}\right)}y_0\)

\(A=\frac 12 \frac{w_0}{y_0}\)

\(V_2(x_3)=\frac{1}{2\mu}\left(\frac{d P}{d x_2}\right)x_3^2 + \frac{w_0}{2y_0}x_3+\frac{w_0}{2}-\frac{1}{2\mu}\left(\frac{d P}{d x_2}\right)y_0^2\)

\(V_2(x_3)=\frac{1}{2\mu}\left(\frac{d P}{d x_2}\right)\left(x_3^2-y_0^2\right) + \frac{w_0}{2}\left(1+\frac{x_3}{y_0}\right)\)

\(\overrightarrow{grad} P=0\Rightarrow \frac{dP}{dx_2}=0=\frac{\Delta P}{L}\) d'où \(\frac{\partial^2V_2}{\partial x_3^2}=0\)

\(\frac{\partial V_2}{\partial x_3}=A\) et \(V_2=Ax_3+B\)

\(x_3=-y_0\rightarrow V_2=0=-Ay_0+B\rightarrow B=\frac{w_0}{2}\)

\(x_3=y_0\rightarrow V_2=w_0=Ay_0+B\rightarrow A=\frac{w_0}{2y_0}\)

\(\displaystyle Q=\iint_S\vec V \vec n dS\)

\(\displaystyle Q=\int_{-y_0}^{y_0}V_2(x_3)ldx_3=\int_{-y_0}^{y_0}\frac{w_0l}{2}\left( 1+\frac{x_3}{y_0}\right) dx_3\)

\(Q=w_0ly_0+\frac{w_0l}{2}\left[\frac12 \frac{x_3^2}{y_0}\right]_{-y_0}^{y_0}=w_0ly_0\)

Tenseur des contraintes visqueuses :

\(\tau=\left(\begin{array}{l c r} 0&0&0\\0&0&\tau_{23}\\0&\tau_{32}&0\end{array}\right)\)

\(\overrightarrow{\sigma_V}(y_0)=\tau(y_0)\vec n\) avec\(\vec n=\) normale sortante \(=\vec e_3=\left(\begin{array}{l}0\\0\\1 \end{array}\right)\) donc :

\(\overrightarrow{\sigma_V}(y_0)=\left(\begin{array}{lcr}0&0&0\\ 0&0&\tau_{23}(y_0)\\ 0&\tau_{32}(y_0)&0 \end{array}\right)\left(\begin{array}{l}0\\ 0\\ 1 \end{array}\right)=\left(\begin{array}{l}0\\ \tau_{23}(y_0)\vec e_2\\ 0 \end{array}\right)\) contrainte selon \(\vec e_2\) sur une facette perpendiculaire à \(\vec e_3\)

\(|| \overrightarrow{\sigma _V}(y_0) ||=\tau_P=|\tau_{23}(y_0)|=\left|\mu\left(\frac{\partial V_2}{\partial x_3}+\cancel{\frac{\partial V_3}{\partial x_2}}\right)\right|_{y_0}=\mu\frac{w_0}{2y_0}\)

\(\displaystyle C_f=\frac{\tau_P}{\left(\rho \frac{\overline{V}}{2}^2\right)}=\frac{\mu w_02}{\rho \overline{V}^2 2y_0}\)

On a \(Re=\frac{\rho V D_H}{\mu}\) avec \(D_H=4\frac{2y_0l}{2l}=4y_0\) (\(2l\) car il n'y a pas de parois latérales)

\(H=P^*+\frac{V_2^2(x_3)}{2g}\)

\(\overline{H}=P^*+\alpha \frac{\overline{V}^2}{2g}\)

\(\displaystyle \alpha=\frac 1S\iint_S\left( \frac{V_2(x_3)}{\overline{V}}\right)^3dS\) avec \(V_2(x_3)=\frac{w_0}{2}\left(1+\frac{x_3}{y_0}\right)=\overline{V}\left(1+\frac{x_3}{y_0}\right)\)

donc

\(\displaystyle \alpha=\frac{1}{2y l}\int_0^l\int_{-y_0}^{y_0}\left(1+\frac{x_3}{y_0}\right)^3dx_1dx_3\)

\(\displaystyle =\frac{1}{2y l}\left[y_0\frac14\left(1+\frac{x_3}{y_0}\right)^4\right]_{-y_0}^{y_0}=\frac{16}{8}=2\)

\(w_0=0\hspace{1cm}\overrightarrow{grad} P\neq 0\)

\(\Rightarrow \mu\frac{\partial^2V_2}{\partial x_3^2}=\frac{\partial P}{\partial x_2}=cte=\frac{\Delta P}{L}<0\) et \(\Delta P=P_2-P_1\)

\(V_2(x_3)=\frac{1}{2\mu}\left(\frac{\Delta P^*}{L}\right)x_3^2+Ax_3+B\)

\(x_3=-y_0\rightarrow V_2(-y_0)=0=\frac{1}{2\mu}\left( \frac{\Delta P}{L}\right)y_0^2-Ay_0+B\)

\(x_3=y_0\rightarrow V_2(y_0)=0=\frac{1}{2\mu}\left( \frac{\Delta P}{L}\right)y_0^2+Ay_0+B\)

\(\Rightarrow B=-\frac{1}{2\mu}\left(\frac{\Delta P}{L}\right)y_0^2\) et \(A=0\)

\(\frac{\partial V_2}{\partial x_3}=\frac{1}{\mu}\left(\frac{\Delta P}{L}\right) x_3=0\) si \(x_3=0\)

\(\displaystyle Q=\int_{-y_0}^{y_0}V_2(x_3)dS=\int_{-y_0}^{y_0}V_2(x_3)ldx_3\)

\(\displaystyle =\frac{1}{2\mu}\left(\frac{\Delta P}{L}\right) l\left[\frac13x_3^2\right]_{-y_0}^{y_0}-\frac{1}{2\mu}\left(\frac{\Delta P}{L}\right) ly_0^2\left[x_3\right]_{-y_0}^{y_0}\)

\(=\displaystyle \frac{\cancel{2}}{3}y_0^3\frac{1}{\cancel{2}\mu}\left(\frac{\Delta P}{L}\right) l-\frac{1}{\cancel{2}\mu}\left(\frac{\Delta P}{L}\right) l\cancel{2}y_0^3\)

Comme précédemment,

\(\tau_P=||\overrightarrow{\sigma_V}(y_0)||=|\tau_{23}(y_0)|\) et \(\overrightarrow{\sigma_V}=\tau_{23}(y_0)\vec e_2\)

\(\displaystyle \Rightarrow \tau_{23}(y_0)=\mu\left(\frac{\partial V_2}{\partial x_3}\right)_{x_3=y_0}=\mu\frac 1\mu \left(\frac{\Delta P}{L}\right) y_0=\left(\frac{\Delta P}{L}\right)y_0<0\)

\(\displaystyle \Rightarrow \tau_P=|\tau_{23}(y_0)|=-\tau_{23}(y_0)\)

\(C_f=\frac{\tau_P}{\left(\rho\frac{\overline{V}^2}{2}\right)}=-\frac{\left(\frac{\Delta P}{L}\right)y_0}{\left(\rho\frac{\overline{V}^2}{2}\right)}=-\frac{\cancel{\left(\frac{\Delta P}{L}\right)y_0}\ 2}{\rho\overline{V}\left(-\frac{1}{3}\cancel{\left(\frac{\Delta P}{L}\right)}\frac{y_0^{\cancel{2}}}{\mu}\right)}\)

Coefficient de pertes de charge

\(H_1=H_2+\Delta H_L\) avec \(\Delta H_L=\) pertes de charges linéaires \(=f\frac{\overline{V}^2}{2g}\frac{L}{D_H}\) avec \(D_H=4\frac SP=4\frac{2y_0l}{2l}=4y_0\)

\(H=\frac{P^*}{\rho g}+\frac{V^2}{2g}\approx \frac{P}{\rho g}+\frac{V^2}{2g}\)

\(\Delta H_L=H_1-H_2=\frac{P_1-P_2}{\rho g}=-\frac{\Delta P}{\rho g}=\frac{3\overline{V}\mu L}{y_0^2 \rho g}\)

\(\Delta H_L=\frac{3\overline{V}\mu \cancel{L}}{y_0^2 \rho \cancel{g}}=f\frac{\overline{V}^2}{2\cancel{g}}\frac{\cancel{L}}{4y_0}\)

\(\overline{H}=P^*\alpha \frac{\overline{V}^2}{2g}\) et \(H=P^* \frac{V^2}{2g}\)

\(\displaystyle{ \alpha = \frac 1S \iint_S \left( \frac{V_2(x_3)}{\overline{V}}\right)^3dS=\frac{1}{2ly_0}\int_0^l\int_{-y_0}^{y_0}\left( \frac 32\right)^3\left( 1-\frac{x_3^2}{y_0^2}\right)^3dx_1dx_3}\)

\(=\frac{l}{2ly_0}\frac{27}{8}\displaystyle \int_{-y_0}^{y_0}\left( 1-\frac{x_3^2}{y_0^2}\right)^3dx_3\)

rappel \((a-b)^3=a^3-3a^2b+3ab^2-b^3\)

\(\displaystyle \alpha =\frac{27}{16y_0} \int_{-y_0}^{y_0}\left( 1-3\frac{x_3^2}{y_0^2} +3\frac{x_3^4}{y_0^4}-\frac{x_3^6}{y_0^6} \right)^3dx_3\)

\(=\frac{27}{16y_0} \left[\left( x_3-\frac{x_3^3}{y_0^2} +\frac 35\frac{x_3^5}{y_0^4}-\frac 17 \frac{x_3^7}{y_0^6} \right)^3\right]_{-y_0}^{y_0}\)

\(=\frac{27}{16y_0} \left(\cancel{y_0} -\cancel{y_0}+\frac 35 y_0-\frac 17 y_0+\cancel{y_0}-\cancel{y_0}+\frac 35 y_0-\frac 17 y_0\right)\)

Régime permanent, fluide incompressible

\(\vec V=V_\theta \vec e_\theta\) ou \(V_\theta\) de \(\vec V \left| \begin{array}{l} 0=V_r \\ V_\theta(r,\theta) \\ 0=V_3\end{array}\right.\)

\(\omega =\) vitesse angulaire \(=V_\theta r=Vr\)

\(\nabla \vec V = 0 = \frac 1r \frac{\partial}{\partial r}(rV_r)+\frac 1r \frac{\partial V_\theta}{d\theta}+\frac{\partial V_3}{\partial x_3}=0\)

\(\Rightarrow \frac{\partial V_\theta}{\partial \theta}=0\) donc \(V_\theta=f(r)\)

\(\vec V \left| \begin{array}{l} 0=V_r \\ V_\theta(r) \\ 0=V_3\end{array}\right.\)

Navier-Stokes

(1) \(-\rho \frac{V_\theta^2}{r}=-\frac{\partial P^*}{\partial r}\)

(2) \(0=-\frac 1r \require{cancel}\cancel{\underbrace{\frac{\partial P^*}{\partial \theta}}_{\text{par symétrie}}}+\mu \left[\frac 1r \frac{\partial}{\partial r}\left(r\frac{dV_\theta}{dr} \right)-\frac{V_\theta}{r^2}\right]\)

\(\frac 1r \frac{d}{d r}\left(r\frac{dV_\theta}{dr} \right)=\frac{V_\theta}{r^2}\)

\(\frac{d^2V_\theta}{dr^2}+\frac 1r\frac{dV_\theta}{d r}-\frac{V_\theta}{r^2}=0\)

\(\frac{d^2V_\theta}{dr^2}+\frac{d\frac{V_\theta}{r}}{dr}=0\)

(3) \(0=-\frac{\partial P^*}{\partial x_3}\)

Condition aux limites :

en \(r=R_1\), \(V_\theta =R_1\omega_1\)

en \(r=R_2\), \(V_\theta=R_2\omega_2=0\)

Résolution de l'équation différentielle linéaire du second ordre à coefficients non constants

\(\frac{d^2V_\theta}{dr^2}+\frac{d\frac{V_\theta}{r}}{dr}=0\)

\(\frac{d^2V_\theta}{dr^2}+\frac 1r\frac{dV_\theta}{d r}-\frac{V_\theta}{r^2}=0\)

Solution évidente, \(V_\theta=r\)

Variation de la constante :

\(V_\theta=k(r)r\)

\(V_\theta'=k'r+k\)

\(V_\theta''=k''r+2k'\)

\(\frac{d^2 V_\theta}{dr^2}+\frac 1r \frac{dV_\theta}{dr}-\frac{V_\theta}{r^2}=k''r+2k'+k'+\cancel{\frac kr}-\cancel{\frac kr}=0\)

\(k''r+3k'=0\)

\(\frac{k''}{k'}=-\frac 3r \rightarrow lnk'=-3lnr+C\)

\(k'=c_1\frac{1}{r^3}\)

\(k=-\frac 12 \frac{c_1}{r^2}+c_2\)

\(V_\theta (r)=-\frac 12\frac{c_1}{r}+c_2r=\frac{k_1}{r}+k_2r\)

Conditions aux limites :

\(\left\{ \begin{array}{l}R_1\omega_1=\frac{k_1}{R_1}+k_2R_1\\ \underbrace{R_2\omega_2}_{=0}=\frac{k_1}{R_2}+k_2R_2\end{array}\right.\)

\(\left\{ \begin{array}{l}k_1=\frac{R_2^2R_1^2(\omega_2-\omega_1)}{R_1^2-R_2^2}\rightarrow k_1=\frac{-R_2^2R_1^2\omega_1}{R1^2-R2^2} \\ k_2=\frac{R_1^2\omega_1-R_2^2\omega_2}{R_1^2-R_2^2}\rightarrow k_2=\frac{R_1^2\omega_1}{R_1^2-R_2^2} \end{array}\right.\)

Le rhéomètre calcule la viscosité à partir du mouvement du couple appliqué au cylindre tournant.

Soit h la hauteur du cylindre

\(\overrightarrow{\sigma}(R_1)=\tau(R_1)\vec u\) avec \(\vec u =\) nommale sortante du fluide \(=(-\vec e_r)\) et \(\tau(R_1)=\) tenseur des contraintes visqueuses à la paroi du cylindre mobile.

\(\tau(R_1)=\left(\begin{array}{l c r}\tau_{rr}(R_1) & \tau_{r\theta}(R_1)&\tau_{r3}(R_1) \\ \tau_{\theta r}(R_1) & \tau_{\theta\theta}(R_1) & \tau_{\theta 3}(R_1) \\ \tau_{3r}(R_1)& \tau_{3\theta}(R_1) & \tau_{33}(R_1)\end{array}\right)=\left(\begin{array}{l c r}0 & \tau_{r\theta}(R_1)&0 \\ \tau_{\theta r}(R_1) & 0 & 0 \\ 0&0&0\end{array}\right)\)

\(\tau(R_1)(\vec e_r)=\left(\begin{array}{l c r}0 & \tau_{r\theta}(R_1)&0 \\ \tau_{\theta r}(R_1) & 0 & 0 \\ 0&0&0\end{array}\right)\left(\begin{array}{c}-1\\0\\0\end{array}\right)=\left(\begin{array}{c}0\\-\tau_{\theta r}(R_1)\\0\end{array}\right)=-\tau_{\theta r}(R_1)\vec e_\theta\)

\(\tau_{\theta r}=\mu \left(\frac{\partial V_\theta}{\partial r}-\frac{V_\theta}{r}+\cancel{\frac 1r \frac{\partial V_r}{\partial \theta}}\right)=\frac{\mu R_1^2\omega_1}{R_2^2-R_1^2}\left(\cancel{-1}-\frac{R_2^2}{r^2}\cancel{+1}-\frac{R_2^2}{r^2}\right)\)

\(\Rightarrow \tau_{\theta r}(R_1)=-\frac{2\mu\omega_1R_2^2}{R_2^2-R_1^2}\)

\(\displaystyle \vec C = \iint_S \overrightarrow{OM}\land \overrightarrow{\sigma}(R_1)dS\) avec \(\overrightarrow{OM}=R_1\vec e_r\)

\(\left(\begin{array}{c}1\\0\\0\end{array}\right)\land \left(\begin{array}{c}0\\1\\0\end{array}\right)=\left(\begin{array}{c}0\\0\\1\end{array}\right)\)

\(\displaystyle \iint_S ||\overrightarrow{OM}\land\overrightarrow{\sigma}||\ [\vec e_r\land \vec e_\theta]\ dS=\iint_S ||\overrightarrow{OM}|| \ ||\overrightarrow{\sigma}|| sin\left(\overrightarrow{OM},\overrightarrow{dF}\right) \vec e_3dS\)

\(\displaystyle =\vec e_3 \iint_S R_1\frac{2\mu \omega_1R_2^2}{R_2^2-R_1^2}R_1d\theta dx_3=\frac{2\mu \omega_1R_2^2 R_1^2}{R_2^2-R_1^2}\int_0^{2\pi} \int_0^h d_\theta dx_3 \vec e_3\)