Utilisateur:Romary/formule Tex

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$h = \frac{w}{K}$

$\begin{matrix}& G & = & {\rm valeurs\ maximum\ -\ valeurs\ minimum\ }\end{matrix}$

$\bar S=\frac{\Delta I}{\Delta G}\,\!$

$K= 1 + \frac{10 \log(N)}{3}$

$K = \sqrt{N}\,$

$S= 2 \pi R L\,$

$\Phi= - \lambda 2 \pi R L \frac{dT}{dR}\,$

$\frac{dR}{R}= - \frac{2 \pi \lambda L dT}{\Phi}\,$

$\int_{R_1}^{R} \frac{dR}{R}= - \frac{2 \pi \lambda L }{\Phi} \int_{T_1}^{T} dT\,$

$\ln \frac{R}{R_1}= \frac{2 \pi \lambda L }{\Phi} (T_1-T)\,$

$\ T= T_1 - \frac{\Phi}{2 \pi \lambda L } \ln \frac{R}{R_1}\,$

$\ T_1-T_2= \frac{\Phi}{2 \pi \lambda_A L } \ln \frac{R_2}{R_1}\,$

$\ T_1-T_3= \frac{\Phi}{2 \pi L } \left ( \frac {\ln \frac {R_2}{R_1} }{\lambda_A} + \frac {\ln \frac {R_3}{R_2} }{\lambda_B}\right )\,$

$\ R_{thA}= \frac {\ln \frac {R_2}{R_1} }{\lambda_A}\,$

$\ R_{thB}= \frac {\ln \frac {R_3}{R_2} }{\lambda_A}\,$

$\ R_{thT}= R_{thA} + R_{thB}\,$

$T_1- T_2= \frac{e_X}{\lambda_X S} \Phi\,$

$R_X= \frac{e_X}{\lambda_X}\,$

$\Phi_A= \frac{T_1- T_2}{R_A} S\,$

$\Phi_B= \frac{T_1- T_2}{R_B} S\,$

$\Phi_C= \frac{T_1- T_2}{R_C} S\,$

$\Phi= (T_1- T_2) S (\frac{1}{R_A}+ \frac{1}{R_B}+ \frac{1}{R_C})(\,$

$R=\frac{1}{R_A}+ \frac{1}{R_B}+ \frac{1}{R_C}\,$

$\Phi= (T_1- T_2) S (\frac{1}{R})\,$

$\varphi= (T_1- T_2) S (\frac{1}{R})\,$

$I= I_1+ I_2+I_3\,$

$I_X= RI_X\,$

$I= (\frac{1}{R_1}+ \frac{1}{R_2}+ \frac{1}{R_3}) \Delta U\,$

$\varphi= (\frac{1}{R_{th1}}+ \frac{1}{R_{th2}}+ \frac{1}{R_{th3}}) \Delta T\,$

$T_1- T_2= \frac{e_A}{\lambda_A} \varphi\,$

$T_1- T_4= (T_1-T_2)+(T_2-T_3)+(T_3+T_4)\,$

$T_1- T_4= (\frac{e_A}{\lambda_A}+ \frac{e_B}{\lambda_B}+ \frac{e_C}{\lambda_C}) \varphi\,$

$T_1- T_4= (R_{thA}+ R_{thB}+ R_{thC}) \varphi\,$

$T= T_1- \frac{e_X}{\lambda_X}\varphi\,$

$C_{th}= \frac{1}{R_{th}}= \frac{\lambda_A}{e_A}+ \frac{\lambda_B}{e_B}+ \frac{\lambda_C}{e_C}\,$

$\frac{dT}{dx}= {\rm Constante}\,$

$\Rightarrow d\Phi= - \lambda S \frac{dT}{dx}\,$

$\Rightarrow \Phi= - \lambda S \frac{dT}{dR}\,$

$\int_{T}^{T_1}\, dT\ = - \int_{x}^{x_1} \frac{\Phi}{\lambda S} dS\,$

$T-T_1= - \frac{\Phi}{\lambda S} (x-x_1)\,$

$\Rightarrow T= T-_1 - - \frac{\Phi}{\lambda S} (x-x_1)\,$

$\int_{T_2}^{T_1}\, dT\ = - \int_{x_2}^{x_1} \frac{\Phi}{\lambda S} dS\,$

$e= x_2-x_1\,$

$\Rightarrow \Phi = \frac{\lambda S}{e} (T_1-T_2)\,$

$\varphi= \frac{\Phi}{S}\,$

$\varphi= \frac{\lambda}{e} (T_1-T_2)\,$

$(U_1-U_2)= RI\,$

$(T_1-T_2)= \frac{e}{\lambda} \varphi\,$

$(U_1-U_2) \leftrightarrow (T_1-T_2),$

$I \leftrightarrow \varphi,$

$R \leftrightarrow R_{thc}= \frac{e}{\lambda}\,$

$\lambda= \lambda_0 e^{0,08H}\,$

$\lambda= \lambda_0 (1+ a\Theta)\,$

$dQ_x= - \lambda_x\ \frac{\delta T}{\delta x} dS_x dt\,$

$d\Phi= \frac{dQ_x}{dt}= - \lambda_x \frac{\delta T}{\delta x} dS_x\,$

$\varphi= \frac{d \Phi}{d S_x}\,$

$\varphi= - \lambda \frac{\delta T}{\delta x}\,$

$\Phi= KS(T_A-T_B)\,$

$(U_A-U_B)= RI\,$

$(T_A-T_B)= \frac{1}{KS} \cdot\ \Phi\,$

$\Phi= \frac{dQ}{dt}\,$

$\varphi= \frac{d\Phi}{dS}= \frac{1}{S} \frac{dQ}{dt}\,$

$\varphi= \frac{\Phi}{S}\,$

$\delta\ Q_A\ \left( \frac{1}{T_A}- \frac{1}{T_B}\right)>0\,$

$\delta\ Q\ = m \cdot\ C_p \cdot\ dT\,$

$\Rightarrow m_A \cdot\ C_{pA} \cdot\ dT_A \left( \frac{1}{T_A}- \frac{1}{T_B}\right)>0\,$

$\delta\ Q_B\ \left( \frac{1}{T_B}- \frac{1}{T_A}\right)>0\,$

$\Rightarrow m_B \cdot\ C_{pB} \cdot\ dT_B \left( \frac{1}{T_B}- \frac{1}{T_A}\right)>0\,$

$T_A> T_B\,$

$\left\{\begin{matrix} dT_A< 0 \Rightarrow \delta\ Q_A< 0, \\ dT_B> 0\Rightarrow \delta\ Q_B> 0, \end{matrix}\right.\,$

$dS= dS_A +dS_B\,$

$dS>0\,$

$\Rightarrow dS= \frac{\delta\ Q_A}{T_A}+ \frac{\delta\ Q_B}{T_B}\,$

$dS= \frac{\delta\ Q}{T}\,$

$\Rightarrow \delta\ Q\ =0 =Q_A +Q_B\,$

$\Rightarrow Q_A = -Q_B\,$

$\ \delta\ Q_A = - \delta\ Q_B\,$

$dU=0\,$

$\delta\ L\ =0\,$

$dU=\delta\ Q\ +\delta\ L\,$

$H = 100 - r\,$

$\begin{matrix} H_M & = & 0,102 \cdot \frac{4F}{\pi \cdot d^2} \end{matrix}$

avec

${\rm Constante} = \frac{1}{g}=\frac{1}{9,8066}=0,102$

$d=\frac{d_2+d_1}{2}$

$J= \bar v - V$

$\begin{matrix} & \bar v & = & {\rm moyenne\ arithm\acute etique\ d'un\ grand\ nombre\ de\ mesures\ }\end{matrix}$

$\begin{matrix} & V & = & {\rm valeur\ vrai\ (ou\ conventionnellement\ vrai)\ }\end{matrix}$

U m a x = 100,2 V

U m i n = 99,7 V

$\begin{matrix} & F &= & +/- &\frac {100.2-99,7}{2} = & +/- &\frac{0,5}{2} = & +/- &0.25 V \end{matrix}$

$F_{max}=+\frac{V_{min}+V_{max}}{2}$

$F_{min}=-\frac{V_{min}+V_{max}}{2}$

$\begin{matrix} & \bar d & = & \sqrt{{\rm (\sum \cdot \ ( \frac{Ox_i}{n} )^2)} + {\rm (\sum \cdot \ ( \frac{Oy_i}{n} )^2)} {\rm }} \\ & & = & \sqrt{{\rm ( {\overline {Ox}})^2} + {\rm ( {\overline {Oy}})^2}} \end{matrix}$

$a 2 n$ $2^{(2n-1)} \,\!$ $a (2 n - 1)$

$d=k\sqrt[3]{t}\,\!$

$C=\frac{2\pi r^3}{3t}\,\!$

$\begin{matrix} & H_B & = & {\rm Constante} \cdot \frac{\rm ( Charge\ de\ l'appareil ) }{\rm (Aire\ de\ l'empreinte) } \\ & & = & 0,102 \cdot \frac{2F}{\pi \cdot D(D-\sqrt{D^2-d^2})} \end{matrix}$

avec

${\rm Constante} = \frac{1}{g}=\frac{1}{9,8066}=0,102$

$d=\frac{d_2+d_1}{2}$

$\begin{matrix} & H_V & = & {\rm Constante} \cdot \frac{\rm ( Charge\ de\ l'essai ) }{\rm (Aire\ de\ l'empreinte) } \\ \\ & & = & 0,102 \cdot \frac{2F \cdot \sin(\frac{136^\circ}{2})}{d^2}\\ \\ & & = & 0,189 \cdot \frac{2F}{d^2}\\ \end{matrix}$ \\