Utilisateur:Romary/formule Tex

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

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

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

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

$K={\sqrt {N}}\,$

$S=2\pi RL\,$

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

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

$\int _{R_{1}}^{R}{\frac {dR}{R}}=-{\frac {2\pi \lambda L}{\Phi }}\int _{T_{1}}^{T}dT\,$

$\ln {\frac {R}{R_{1}}}={\frac {2\pi \lambda L}{\Phi }}(T_{1}-T)\,$

$\ T=T_{1}-{\frac {\Phi }{2\pi \lambda L}}\ln {\frac {R}{R_{1}}}\,$

$\ T_{1}-T_{2}={\frac {\Phi }{2\pi \lambda _{A}L}}\ln {\frac {R_{2}}{R_{1}}}\,$

$\ T_{1}-T_{3}={\frac {\Phi }{2\pi L}}\left({\frac {\ln {\frac {R_{2}}{R_{1}}}}{\lambda _{A}}}+{\frac {\ln {\frac {R_{3}}{R_{2}}}}{\lambda _{B}}}\right)\,$

$\ R_{thA}={\frac {\ln {\frac {R_{2}}{R_{1}}}}{\lambda _{A}}}\,$

$\ R_{thB}={\frac {\ln {\frac {R_{3}}{R_{2}}}}{\lambda _{A}}}\,$

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

$T_{1}-T_{2}={\frac {e_{X}}{\lambda _{X}S}}\Phi \,$

$R_{X}={\frac {e_{X}}{\lambda _{X}}}\,$

$\Phi _{A}={\frac {T_{1}-T_{2}}{R_{A}}}S\,$

$\Phi _{B}={\frac {T_{1}-T_{2}}{R_{B}}}S\,$

$\Phi _{C}={\frac {T_{1}-T_{2}}{R_{C}}}S\,$

$\Phi =(T_{1}-T_{2})S({\frac {1}{R_{A}}}+{\frac {1}{R_{B}}}+{\frac {1}{R_{C}}})(\,$

$R={\frac {1}{R_{A}}}+{\frac {1}{R_{B}}}+{\frac {1}{R_{C}}}\,$

$\Phi =(T_{1}-T_{2})S({\frac {1}{R}})\,$

$\varphi =(T_{1}-T_{2})S({\frac {1}{R}})\,$

$I=I_{1}+I_{2}+I_{3}\,$

$I_{X}=RI_{X}\,$

$I=({\frac {1}{R_{1}}}+{\frac {1}{R_{2}}}+{\frac {1}{R_{3}}})\Delta U\,$

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

$T_{1}-T_{2}={\frac {e_{A}}{\lambda _{A}}}\varphi \,$

$T_{1}-T_{4}=(T_{1}-T_{2})+(T_{2}-T_{3})+(T_{3}+T_{4})\,$

$T_{1}-T_{4}=({\frac {e_{A}}{\lambda _{A}}}+{\frac {e_{B}}{\lambda _{B}}}+{\frac {e_{C}}{\lambda _{C}}})\varphi \,$

$T_{1}-T_{4}=(R_{thA}+R_{thB}+R_{thC})\varphi \,$

$T=T_{1}-{\frac {e_{X}}{\lambda _{X}}}\varphi \,$

$C_{th}={\frac {1}{R_{th}}}={\frac {\lambda _{A}}{e_{A}}}+{\frac {\lambda _{B}}{e_{B}}}+{\frac {\lambda _{C}}{e_{C}}}\,$

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

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

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

$\int _{T}^{T_{1}}\,dT\ =-\int _{x}^{x_{1}}{\frac {\Phi }{\lambda S}}dS\,$

$T-T_{1}=-{\frac {\Phi }{\lambda S}}(x-x_{1})\,$

$\Rightarrow T=T-_{1}--{\frac {\Phi }{\lambda S}}(x-x_{1})\,$

$\int _{T_{2}}^{T_{1}}\,dT\ =-\int _{x_{2}}^{x_{1}}{\frac {\Phi }{\lambda S}}dS\,$

$e=x_{2}-x_{1}\,$

$\Rightarrow \Phi ={\frac {\lambda S}{e}}(T_{1}-T_{2})\,$

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

$\varphi ={\frac {\lambda }{e}}(T_{1}-T_{2})\,$

$(U_{1}-U_{2})=RI\,$

$(T_{1}-T_{2})={\frac {e}{\lambda }}\varphi \,$

$(U_{1}-U_{2})\leftrightarrow (T_{1}-T_{2}),$

$I\leftrightarrow \varphi ,$

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

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

$\lambda =\lambda _{0}(1+a\Theta )\,$

$dQ_{x}=-\lambda _{x}\ {\frac {\delta T}{\delta x}}dS_{x}dt\,$

$d\Phi ={\frac {dQ_{x}}{dt}}=-\lambda _{x}{\frac {\delta T}{\delta x}}dS_{x}\,$

$\varphi ={\frac {d\Phi }{dS_{x}}}\,$

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

$\Phi =KS(T_{A}-T_{B})\,$

$(U_{A}-U_{B})=RI\,$

$(T_{A}-T_{B})={\frac {1}{KS}}\cdot \ \Phi \,$

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

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

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

$\delta \ Q_{A}\ \left({\frac {1}{T_{A}}}-{\frac {1}{T_{B}}}\right)>0\,$

$\delta \ Q\ =m\cdot \ C_{p}\cdot \ dT\,$

$\Rightarrow m_{A}\cdot \ C_{pA}\cdot \ dT_{A}\left({\frac {1}{T_{A}}}-{\frac {1}{T_{B}}}\right)>0\,$

$\delta \ Q_{B}\ \left({\frac {1}{T_{B}}}-{\frac {1}{T_{A}}}\right)>0\,$

$\Rightarrow m_{B}\cdot \ C_{pB}\cdot \ dT_{B}\left({\frac {1}{T_{B}}}-{\frac {1}{T_{A}}}\right)>0\,$

$T_{A}>T_{B}\,$

$\left\{{\begin{matrix}dT_{A}<0\Rightarrow \delta \ Q_{A}<0,\\dT_{B}>0\Rightarrow \delta \ Q_{B}>0,\end{matrix}}\right.\,$

$dS=dS_{A}+dS_{B}\,$

$dS>0\,$

$\Rightarrow dS={\frac {\delta \ Q_{A}}{T_{A}}}+{\frac {\delta \ Q_{B}}{T_{B}}}\,$

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

$\Rightarrow \delta \ Q\ =0=Q_{A}+Q_{B}\,$

$\Rightarrow Q_{A}=-Q_{B}\,$

$\ \delta \ Q_{A}=-\delta \ Q_{B}\,$

$dU=0\,$

$\delta \ L\ =0\,$

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

$H=100-r\,$

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

avec

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

$d={\frac {d_{2}+d_{1}}{2}}$

$J={\bar {v}}-V$

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

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

$U_{max}=100,2V$

$U_{min}=99,7V$

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

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

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

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

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

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

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

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

avec

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

$d={\frac {d_{2}+d_{1}}{2}}$

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

$R_{e}={\frac {F}{S}}\,\!$

$P={\frac {F}{S}}\,\!$

$S={\frac {dI}{dG}}\,\!$

$S={\frac {dI}{dG}}=\operatorname {Arctan} \ \varphi \,\!$

${\begin{matrix}&I&=&{\rm {indication\ donn{\acute {e}}e\ par\ l'essai}}\end{matrix}}$

${\begin{matrix}&G&=&{\rm {quantit{\acute {e}}\ de\ grandeur\ {\grave {a}}\ mesurer}}\end{matrix}}$

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

$S={\frac {10mm}{5cl}}={\frac {2mm}{cl}}\,\!$

${\begin{matrix}&m&=&{\frac {\rm {(1)}}{(96500)}}\cdot {\frac {\rm {(A)}}{(n)}}\cdot &{\rm {I}}\cdot &{\rm {t}}\end{matrix}}$

${\begin{matrix}&K&=&{\frac {\rm {(Surface\ de\ la\ cathode)}}{(Surface\ de\ l'anode)}}\end{matrix}}$

$\Delta E_{c}=m\cdot g\cdot h$

$K=m\cdot g\cdot h-m\cdot g\cdot h'$