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Po vyzkoušení předchozích hledacích metod jsme uzavřeli jehlový ventil na aparatuře a uvedli do provozu heliový hledač netěsností podle provozního postupu v přiložených deskách. |
Následně nastavili rozsah na nejmenší citlivost a začali zkoušet ofukovat aparaturu heliem z balonku. Po delší době jsme objevili netěsnost v oblasti příruby u Piraniho měrky. |
Následně nastavili rozsah na nejmenší citlivost a začali zkoušet ofukovat aparaturu heliem z balonku. Po chvíli jsme objevili netěsnost v oblasti příruby u Piraniho měrky. |
Urychlovací napětí potřebné k předání správné rychlosti jádrům helia, aby byla jejich dráha zakřivena na poloměr 40mm v magnetickém poli 150mT spočítáme podle Lorentzovy síly a dostředivého zrychlení, které se musejí rovnat. |
\begin{displaymath} F_d = \frac{m v^2}{r} = q v B. \end{displaymath} |
\begin{displaymath} q v B = \frac{m v^2}{r} \end{displaymath} |
Po vyjádření $v$ dostáváme. |
Pro energii částice urychlené v elektrickém poli platí |
\begin{displaymath} v = \frac{r q B}{m}. \end{displaymath} |
\begin{displaymath} E = q U = \frac{m v^2}{2} . \end{displaymath} |
Dosadíme do vztahu pro kinetickou energii a máme. |
Do této rovnice dosadíme rychlost z předchozí rovnice a vyjádříme napětí. |
\begin{displaymath} E = \frac{(r q B)^2}{2 m}. \end{displaymath} |
\begin{displaymath} U = \frac{q r^2 B^2}{2 m} \end{displaymath} |
Po vyčíslení získáme energii \begin{displaymath} E = 2,8247 \times 10^{-16} [J] \end{displaymath} což odpovídá \begin{displaymath} E = 1763 [eV]. \end{displaymath}. A potřebné urychlovací napětí tedy je 1763 V. |
Dostaneme napětí potřebné k urychlení iontu helia, aby dopadl do otvoru detektoru v komoře heliového hledače. \begin{displaymath} U = 438 \textrm{V} \end{displaymath}. |
\section{Závěr} |
V praktiku jsme si tak vyzkoušeli několik zajímavých metod pro hledání netěsností v aparatuře. Z nich některé mne překvapily svou jednoduchostí a přitom vysokou účinností, jako například hledání netěsnosti pomocí ethanolu. |
/dokumenty/skolni/VAK/rotacni_vyveva/vyveva.pdf |
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