Fini report exo climatiseur
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@ -2885,14 +2885,14 @@ Les valeurs des c\oe fficients de dilatation thermique sont celles du tableau \r
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\end{exos}
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\begin{exos}
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On comprime un gaz diatomique de sorte que sa pression et son volume se déplace sur une droite de A à B (voir figure \ref{lediagpv}). La température initiale vaut \(T_A=\SI{270}{\kelvin}\).
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\begin{figure}[!t]
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\begin{figure}[t]
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\centering
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\caption{Le diagramme PV.\label{lediagpv}}
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\centering
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\def\svgwidth{0.9\columnwidth}
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\input{Annexe-Exercices/Images/DiagPV.ps_tex}
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\end{figure}
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On comprime un gaz diatomique de sorte que sa pression et son volume se déplace sur une droite de A à B (voir figure \ref{lediagpv}). La température initiale vaut \(T_A=\SI{270}{\kelvin}\).
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\begin{enumerate}
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\item Calculez la chaleur reçue par le gaz pour passer de A à B.
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\item Au cours de cette transformation, on a que~:
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@ -2909,7 +2909,8 @@ Les valeurs des c\oe fficients de dilatation thermique sont celles du tableau \r
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\item Puis, on calcule la chaleur échangée lors de la transformation~:
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\begin{align*}
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Q%=\Delta U+A=\frac{i}{2}\cdot n\cdot R\cdot \Delta T+A\\
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&=\frac{5}{2}\cdot 10\cdot (480-270)+\frac{1}{2}\cdot (9\cdot 10^{-3}-2\cdot 10^{-3})\\
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&=\frac{5}{2}\cdot 10\cdot (480-270)\\
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&+\frac{1}{2}\cdot (9\cdot 10^{-3}-2\cdot 10^{-3})\\
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&\cdot (24\cdot 10^5-3\cdot 10^5)+7\cdot 10^{-3}\cdot 3\cdot 10^5\\
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&=5250+7350+2100=\SI{14700}{\joule}
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\end{align*}
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@ -2938,12 +2939,144 @@ Les valeurs des c\oe fficients de dilatation thermique sont celles du tableau \r
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\end{solos}
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\end{exos}
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%\begin{exos}
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% Un énoncé de test.
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% \begin{solos}
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% Un autre corrigé de test.
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% \end{solos}
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%\end{exos}
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\begin{exos}
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Le climatiseur d'un local fonctionne avec un gaz monoatomique qui subit, dans un circuit fermé, trois transformations au cours de son cycle. On prendra \(nR=10\).
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\begin{description}
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\item[A-B]Compression, hors du local, du gaz de \SI{1}{\bar} à \SI{8}{\bar}. Au cours de cette transformation on a~:
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\[p_A\cdot V_A^{3/2}=p\cdot V^{3/2}=p_B\cdot V_B^{3/2}\]
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et le travail \emph{reçu} par le gaz vaut \SI{5,8}{\kilo\joule}.
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\item[B-C]Détente adiabatique (\(p\cdot V^{5/3}=\text{constante}\) jusqu'à \SI{1}{\bar}.
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\item[C-A]Chauffage à pression constante jusqu'à la température intiale de \SI{290}{\kelvin}. En s'échauffant, le gaz refroidit l'air du local.
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\end{description}
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On demande de réaliser les points suivants.
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\begin{enumerate}
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\item Calculez p, V et T au début de chaque étape.
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\item Déterminez \(\Delta U\), \(A\) et \(Q\) pour chaque étape.
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\item Établir un bilan thermique.
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\item Déterminez l'efficacité de la climatisation.
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\item Si, chaque seconde, le local reçoit \SI{1000}{\joule} de chaleur, calculez le nombre de cycles que doit effectuer la climatisation par minute pour évacuer cette chaleur.
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\item Comment peut-on calculer le travail dans l'étape A-B ?
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\end{enumerate}
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\begin{solos}
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Le diagramme PV du cycle est présenté à la figure \ref{climatiseur}.
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\begin{figure}[t]
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\centering
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\caption{Diagramme PV du climatiseur.\label{climatiseur}}
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\centering
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\def\svgwidth{0.9\columnwidth}
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\input{Annexe-Exercices/Images/Climatiseur.ps_tex}
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\end{figure}
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Voici les réponses au différents points demandés.
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\begin{enumerate}
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\item Tout d'abord les grandeurs associées aux états du gaz.
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\begin{description}
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\item[État A] Les grandeurs caractéristiques sont~:
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\begin{align*}
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p_A&=\SI{1e5}{\pascal}\\
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T_A&=\SI{290}{\kelvin}\\
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V_A&=\frac{nR\cdot T_A}{p_A}=\frac{10\cdot 290}{1\cdot 10^5}=\SI{0.029}{\metre\cubed}
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\end{align*}
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\item[État B] Les grandeurs caractéristiques sont~:
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\begin{align*}
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p_B&=\SI{8}{\bar}\\
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V_B^{3/2}&=\frac{p_A}{p_B}\cdot V_A^{3/2}\\
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&=\frac{10^5}{8\cdot 10^5}\cdot (0,0029)^{3/2}\\
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&=6,17\cdot 10^{-4}\\
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V_B&=\SI{0,00725}{\metre\cubed}\\
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T_B&=\frac{p_B\cdot V_B}{nR}\\
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&=\frac{8\cdot 10^5\cdot 0,00725}{10}=\SI{580}{\kelvin}
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\end{align*}
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\item[État C] Les grandeurs caractéristiques sont~:
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\begin{align*}
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p_C&=\SI{1e5}{\pascal}\\
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V_C^{5/3}&=\frac{p_B}{p_C}\cdot V_B^{5/3}\\
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&=\frac{8\cdot 10^5}{1\cdot 10^5}\cdot (0,00725)^{5/3}\\
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&=2,1726\cdot 10^{-3}\\
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V_C&=\SI{0,02524}{\metre\cubed}\\
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T_C&=\frac{p_C\cdot V_C}{nR}\\
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&=\frac{10^5\cdot 0,02524}{10}=\SI{252,4}{\kelvin}
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\end{align*}
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\end{description}
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\item Ensuite les grandeurs associées aux transformations.
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\begin{description}
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\item[AB] Les grandeurs caractéristiques sont~:
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\begin{align*}
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\Delta U&=\frac{i}{2}\cdot nR\cdot \Delta T\\
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&=\frac{3}{2}\cdot 10\cdot (580-290)=\SI{4350}{\joule}\\
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A&=\SI{-5800}{\joule}\;\text{Donné}\\
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Q&=\Delta U+A=4350-5800=\SI{-1450}{\joule}
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\end{align*}
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\item[BC] Pour une adiabatique, les grandeurs caractéristiques sont~:
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\begin{align*}
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Q&=0\\
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\Delta U&=\frac{i}{2}\cdot nR\cdot \Delta T\\
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&=\frac{3}{2}\cdot 10\cdot (252,4-580)=\SI{-4914}{\joule}\\
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A&=-\Delta U=\SI{4914}{\joule}
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\end{align*}
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\item[CA] Pour une isobare, les grandeurs caractéristiques sont~:
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\begin{align*}
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\Delta U&=\frac{3}{2}\cdot 10\cdot (290-252,4)=\SI{564}{\joule}\\
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A&=p\cdot \Delta V=10^5\cdot (0,029-0,02524)\\
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&=\SI{376}{\joule}\\
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Q&=\Delta U+A=564+376=\SI{940}{\joule}
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\end{align*}
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\end{description}
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\item Le bilan du cycle est alors déduit du tableau des transformations.
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\medskip
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\begin{tabular}{|c|c|c|c|}
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\hline
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Étapes/(J) & \(Q\) & \(\Delta U\) & \(A\) \\
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\hline\hline
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AB & -1450 & 4350 & -5800 \\
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\hline
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BC & 0 & -4914 & 4914 \\
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\hline
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CA & 940 & 564 & 376 \\
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\hline\hline
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\(\sum\) & - & 0 & 510 \\
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\hline
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\end{tabular}
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\medskip
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Alors, on peut obtenir les chaleurs entrant et sortant du gaz~:
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\begin{align*}
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&\sum Q>0 : \SI{940}{\joule}\\
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&\sum Q<0 : \SI{-1450}{\joule}
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\end{align*}
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Le diagramme du bilan est alors~:
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\begin{figure}
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\def\svgwidth{7cm}
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\begin{center}
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%\input{Annexe-Exercices/Images/cycle2.eps_tex}
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\input{Annexe-Exercices/Images/cycle3.ps_tex}
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\end{center}
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\caption{Bilan du cycle\label{exos:cycle3}}
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\end{figure}
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\item Le rendement du climatiseur est~:
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\[\eta=\frac{\text{utile}}{\text{investi}}=\frac{940}{510}=184,3\%\]
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En effet, ce qui est utile pour un climatiseur est de retirer \SI{940}{\joule} de la pièce chaude et pour cela, il faut investir en électricité \SI{510}{\joule}.
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\item À chaque cycle, on retire \SI{940}{\joule}. Si on doit retirer \SI{1000}{\joule\per\second}, le nombre de cycle est donné par~:
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\begin{align*}
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&n\cdot 940=1000\;\Rightarrow\;n=\frac{1000}{940}=1,064\,\text{cycles/s}\\
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&\Rightarrow\;1,064\cdot 60=63,83\,\text{cycle/min}
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\end{align*}
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\item Enfin, pour calculer le travail, on passe par sa définition~:
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\begin{align*}
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A&=\int_A^B p\cdot dV=\int_A^B \frac{p_A\cdot V_A^{3/2}}{V^{3/2}}\cdot dV\\
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&=p_A\cdot V_A^{3/2}\cdot\int_A^B\frac{dV}{V^{3/2}}\\
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&=p_A\cdot V_A^{3/2}\cdot\int_A^B V^{-3/2}\cdot dV\\
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&=p_A\cdot V_A^{3/2}\cdot[\frac{1}{-3/2+1}\cdot V^{-3/2+1}]_A^B\\
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&=p_A\cdot V_A^{3/2}\cdot[-2\cdot V^{-1/2}]_A^B\\
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&=-2\cdot p_A\cdot V_A^{3/2}\cdot (V_B^{-1/2}-V_A^{-1/2})\\
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&=-2\cdot 10^5\cdot 0,029^{1,5}\cdot (\frac{1}{\sqrt{0,00725}}-\\
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&\;\frac{1}{\sqrt{0,029}}=\SI{-5800}{\joule}
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\end{align*}
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\end{enumerate}
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\end{solos}
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\end{exos}
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%\begin{exos}
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% Un énoncé de test.
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@ -2885,14 +2885,14 @@ Les valeurs des c\oe fficients de dilatation thermique sont celles du tableau \r
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\end{exos}
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\begin{exos}
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On comprime un gaz diatomique de sorte que sa pression et son volume se déplace sur une droite de A à B (voir figure \ref{lediagpv}). La température initiale vaut \(T_A=\SI{270}{\kelvin}\).
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\begin{figure}[!t]
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\begin{figure}[t]
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\centering
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\caption{Le diagramme PV.\label{lediagpv}}
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\centering
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\def\svgwidth{0.9\columnwidth}
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\input{Annexe-Exercices/Images/DiagPV.ps_tex}
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\end{figure}
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On comprime un gaz diatomique de sorte que sa pression et son volume se déplace sur une droite de A à B (voir figure \ref{lediagpv}). La température initiale vaut \(T_A=\SI{270}{\kelvin}\).
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\begin{enumerate}
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\item Calculez la chaleur reçue par le gaz pour passer de A à B.
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\item Au cours de cette transformation, on a que~:
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@ -2909,7 +2909,8 @@ Les valeurs des c\oe fficients de dilatation thermique sont celles du tableau \r
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\item Puis, on calcule la chaleur échangée lors de la transformation~:
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\begin{align*}
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Q%=\Delta U+A=\frac{i}{2}\cdot n\cdot R\cdot \Delta T+A\\
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&=\frac{5}{2}\cdot 10\cdot (480-270)+\frac{1}{2}\cdot (9\cdot 10^{-3}-2\cdot 10^{-3})\\
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&=\frac{5}{2}\cdot 10\cdot (480-270)\\
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&+\frac{1}{2}\cdot (9\cdot 10^{-3}-2\cdot 10^{-3})\\
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&\cdot (24\cdot 10^5-3\cdot 10^5)+7\cdot 10^{-3}\cdot 3\cdot 10^5\\
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&=5250+7350+2100=\SI{14700}{\joule}
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\end{align*}
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@ -2938,12 +2939,142 @@ Les valeurs des c\oe fficients de dilatation thermique sont celles du tableau \r
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\end{solos}
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\end{exos}
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%\begin{exos}
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% Un énoncé de test.
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% \begin{solos}
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% Un autre corrigé de test.
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% \end{solos}
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%\end{exos}
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\begin{exos}
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Le climatiseur d'un local fonctionne avec un gaz monoatomique qui subit, dans un circuit fermé, trois transformations au cours de son cycle. On prendra \(nR=10\).
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\begin{description}
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\item[A-B]Compression, hors du local, du gaz de \SI{1}{\bar} à \SI{8}{\bar}. Au cours de cette transformation on a~:
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\[p_A\cdot V_A^{3/2}=p\cdot V^{3/2}=p_B\cdot V_B^{3/2}\]
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et le travail \emph{reçu} par le gaz vaut \SI{5,8}{\kilo\joule}.
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\item[B-C]Détente adiabatique (\(p\cdot V^{5/3}=\text{constante}\) jusqu'à \SI{1}{\bar}.
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\item[C-A]Chauffage à pression constante jusqu'à la température intiale de \SI{290}{\kelvin}. En s'échauffant, le gaz refroidit l'air du local.
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\end{description}
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On demande de réaliser les points suivants.
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\begin{enumerate}
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\item Calculez p, V et T au début de chaque étape.
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\item Déterminez \(\Delta U\), \(A\) et \(Q\) pour chaque étape.
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\item Établir un bilan thermique.
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\item Déterminez l'efficacité de la climatisation.
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\item Si, chaque seconde, le local reçoit \SI{1000}{\joule} de chaleur, calculez le nombre de cycles que doit effectuer la climatisation par minute pour évacuer cette chaleur.
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\item Comment peut-on calculer le travail dans l'étape A-B ?
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\end{enumerate}
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\begin{solos}
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Le diagramme PV du cycle est présenté à la figure \ref{climatiseur}.
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\begin{figure}[t]
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\centering
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\caption{Diagramme PV du climatiseur.\label{climatiseur}}
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\centering
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\def\svgwidth{0.9\columnwidth}
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\input{Annexe-Exercices/Images/Climatiseur.ps_tex}
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\end{figure}
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Voici les réponses au différents points demandés.
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\begin{enumerate}
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\item Tout d'abord les grandeurs associées aux états du gaz.
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\begin{description}
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\item[État A] Les grandeurs caractéristiques sont~:
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\begin{align*}
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p_A&=\SI{1e5}{\pascal}\\
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T_A&=\SI{290}{\kelvin}\\
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V_A&=\frac{nR\cdot T_A}{p_A}=\frac{10\cdot 290}{1\cdot 10^5}=\SI{0.029}{\metre\cubed}
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\end{align*}
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\item[État B] Les grandeurs caractéristiques sont~:
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\begin{align*}
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p_B&=\SI{8}{\bar}\\
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V_B^{3/2}&=\frac{p_A}{p_B}\cdot V_A^{3/2}\\
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&=\frac{10^5}{8\cdot 10^5}\cdot (0,0029)^{3/2}\\
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&=6,17\cdot 10^{-4}\\
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V_B&=\SI{0,00725}{\metre\cubed}\\
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T_B&=\frac{p_B\cdot V_B}{nR}\\
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&=\frac{8\cdot 10^5\cdot 0,00725}{10}=\SI{580}{\kelvin}
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\end{align*}
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\item[État C] Les grandeurs caractéristiques sont~:
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\begin{align*}
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p_C&=\SI{1e5}{\pascal}\\
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V_C^{5/3}&=\frac{p_B}{p_C}\cdot V_B^{5/3}\\
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&=\frac{8\cdot 10^5}{1\cdot 10^5}\cdot (0,00725)^{5/3}\\
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&=2,1726\cdot 10^{-3}\\
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V_C&=\SI{0,02524}{\metre\cubed}\\
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T_C&=\frac{p_C\cdot V_C}{nR}\\
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&=\frac{10^5\cdot 0,02524}{10}=\SI{252,4}{\kelvin}
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\end{align*}
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\end{description}
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\item Ensuite les grandeurs associées aux transformations.
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\begin{description}
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\item[AB] Les grandeurs caractéristiques sont~:
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\begin{align*}
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\Delta U&=\frac{i}{2}\cdot nR\cdot \Delta T\\
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&=\frac{3}{2}\cdot 10\cdot (580-290)=\SI{4350}{\joule}\\
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A&=\SI{-5800}{\joule}\;\text{Donné}\\
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Q&=\Delta U+A=4350-5800=\SI{-1450}{\joule}
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\end{align*}
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\item[BC] Pour une adiabatique, les grandeurs caractéristiques sont~:
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\begin{align*}
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Q&=0\\
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\Delta U&=\frac{i}{2}\cdot nR\cdot \Delta T\\
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&=\frac{3}{2}\cdot 10\cdot (252,4-580)=\SI{-4914}{\joule}\\
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A&=-\Delta U=\SI{4914}{\joule}
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\end{align*}
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\item[CA] Pour une isobare, les grandeurs caractéristiques sont~:
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\begin{align*}
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\Delta U&=\frac{3}{2}\cdot 10\cdot (290-252,4)=\SI{564}{\joule}\\
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A&=p\cdot \Delta V=10^5\cdot (0,029-0,02524)\\
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&=\SI{376}{\joule}\\
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Q&=\Delta U+A=564+376=\SI{940}{\joule}
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\end{align*}
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\end{description}
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\item Le bilan du cycle est alors déduit du tableau des transformations.
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\medskip
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\begin{tabular}{|c|c|c|c|}
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\hline
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Étapes/(J) & \(Q\) & \(\Delta U\) & \(A\) \\
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\hline\hline
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AB & -1450 & 4350 & -5800 \\
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\hline
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BC & 0 & -4914 & 4914 \\
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\hline
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CA & 940 & 564 & 376 \\
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\hline\hline
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\(\sum\) & - & 0 & 510 \\
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\hline
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\end{tabular}
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\medskip
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Alors, on peut obtenir les chaleurs entrant et sortant du gaz~:
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\begin{align*}
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&\sum Q>0 : \SI{940}{\joule}\\
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&\sum Q<0 : \SI{-1450}{\joule}
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\end{align*}
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Le diagramme du bilan est alors~:
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\begin{figure}
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\def\svgwidth{7cm}
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\begin{center}
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%\input{Annexe-Exercices/Images/cycle2.eps_tex}
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\input{Annexe-Exercices/Images/cycle3.ps_tex}
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\end{center}
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\caption{Bilan du cycle\label{exos:cycle3}}
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\end{figure}
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\item Le rendement du climatiseur est~:
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\[\eta=\frac{\text{utile}}{\text{investi}}=\frac{940}{510}=184,3\%\]
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En effet, ce qui est utile pour un climatiseur est de retirer \SI{940}{\joule} de la pièce chaude et pour cela, il faut investir en électricité \SI{510}{\joule}.
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\item À chaque cycle, on retire \SI{940}{\joule}. Si on doit retirer \SI{1000}{\joule\per\second}, le nombre de cycle est donné par~:
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\begin{align*}
|
||||
&n\cdot 940=1000\;\Rightarrow\;n=\frac{1000}{940}=1,064\,\text{cycles/s}\\
|
||||
&\Rightarrow\;1,064\cdot 60=63,83\,\text{cycle/min}
|
||||
\end{align*}
|
||||
\item Enfin, pour calculer le travail, on passe par sa définition~:
|
||||
\begin{align*}
|
||||
A&=\int_A^B p\cdot dV=\int_A^B \frac{p_A\cdot V_A^{3/2}}{V^{3/2}}\cdot dV\\
|
||||
&=p_A\cdot V_A^{3/2}\cdot\int_A^B\frac{dV}{V^{3/2}}\\
|
||||
&=p_A\cdot V_A^{3/2}\cdot\int_A^B V^{-3/2}\cdot dV\\
|
||||
&=p_A\cdot V_A^{3/2}\cdot[\frac{1}{-3/2+1}\cdot V^{-3/2+1}]_A^B\\
|
||||
&=p_A\cdot V_A^{3/2}\cdot[-2\cdot V^{-1/2}]_A^B\\
|
||||
&=-2\cdot p_A\cdot V_A^{3/2}\cdot (V_B
|
||||
\end{align*}
|
||||
\end{enumerate}
|
||||
\end{solos}
|
||||
\end{exos}
|
||||
|
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%\begin{exos}
|
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% Un énoncé de test.
|
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|
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71
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369
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After Width: | Height: | Size: 7.1 KiB |
@ -421,6 +421,6 @@
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\BOOKMARK [2][-]{subsection.M.3.5}{\376\377\000R\000\351\000s\000u\000m\000\351}{section.M.3}% 421
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\BOOKMARK [2][-]{subsection.M.3.6}{\376\377\000E\000x\000e\000m\000p\000l\000e\000s}{section.M.3}% 422
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\BOOKMARK [2][-]{subsection.M.3.7}{\376\377\000R\000e\000p\000r\000\351\000s\000e\000n\000t\000a\000t\000i\000o\000n\000\040\000g\000r\000a\000p\000h\000i\000q\000u\000e}{section.M.3}% 423
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\BOOKMARK [0][-]{section*.303}{\376\377\000N\000o\000t\000e\000s}{}% 424
|
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\BOOKMARK [0][-]{appendix*.305}{\376\377\000B\000i\000b\000l\000i\000o\000g\000r\000a\000p\000h\000i\000e}{}% 425
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\BOOKMARK [0][-]{section*.306}{\376\377\000I\000n\000d\000e\000x}{}% 426
|
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\BOOKMARK [0][-]{section*.305}{\376\377\000N\000o\000t\000e\000s}{}% 424
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\BOOKMARK [0][-]{appendix*.307}{\376\377\000B\000i\000b\000l\000i\000o\000g\000r\000a\000p\000h\000i\000e}{}% 425
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\BOOKMARK [0][-]{section*.308}{\376\377\000I\000n\000d\000e\000x}{}% 426
|
||||
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Binary file not shown.
125
SolutionsOS.tex
125
SolutionsOS.tex
@ -1143,7 +1143,8 @@
|
||||
\item Puis, on calcule la chaleur échangée lors de la transformation~:
|
||||
\begin{align*}
|
||||
Q%=\Delta U+A=\frac{i}{2}\cdot n\cdot R\cdot \Delta T+A\\
|
||||
&=\frac{5}{2}\cdot 10\cdot (480-270)+\frac{1}{2}\cdot (9\cdot 10^{-3}-2\cdot 10^{-3})\\
|
||||
&=\frac{5}{2}\cdot 10\cdot (480-270)\\
|
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&+\frac{1}{2}\cdot (9\cdot 10^{-3}-2\cdot 10^{-3})\\
|
||||
&\cdot (24\cdot 10^5-3\cdot 10^5)+7\cdot 10^{-3}\cdot 3\cdot 10^5\\
|
||||
&=5250+7350+2100=\SI{14700}{\joule}
|
||||
\end{align*}
|
||||
@ -1172,6 +1173,126 @@
|
||||
|
||||
\end{Solution OS}
|
||||
\begin{Solution OS}{38}
|
||||
Le diagramme PV du cycle est présenté à la figure \ref{climatiseur}.
|
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\begin{figure}[t]
|
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\centering
|
||||
\caption{Diagramme PV du climatiseur.\label{climatiseur}}
|
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\centering
|
||||
\def\svgwidth{0.9\columnwidth}
|
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\input{Annexe-Exercices/Images/Climatiseur.ps_tex}
|
||||
\end{figure}
|
||||
Voici les réponses au différents points demandés.
|
||||
\begin{enumerate}
|
||||
\item Tout d'abord les grandeurs associées aux états du gaz.
|
||||
\begin{description}
|
||||
\item[État A] Les grandeurs caractéristiques sont~:
|
||||
\begin{align*}
|
||||
p_A&=\SI{1e5}{\pascal}\\
|
||||
T_A&=\SI{290}{\kelvin}\\
|
||||
V_A&=\frac{nR\cdot T_A}{p_A}=\frac{10\cdot 290}{1\cdot 10^5}=\SI{0.029}{\metre\cubed}
|
||||
\end{align*}
|
||||
\item[État B] Les grandeurs caractéristiques sont~:
|
||||
\begin{align*}
|
||||
p_B&=\SI{8}{\bar}\\
|
||||
V_B^{3/2}&=\frac{p_A}{p_B}\cdot V_A^{3/2}\\
|
||||
&=\frac{10^5}{8\cdot 10^5}\cdot (0,0029)^{3/2}\\
|
||||
&=6,17\cdot 10^{-4}\\
|
||||
V_B&=\SI{0,00725}{\metre\cubed}\\
|
||||
T_B&=\frac{p_B\cdot V_B}{nR}\\
|
||||
&=\frac{8\cdot 10^5\cdot 0,00725}{10}=\SI{580}{\kelvin}
|
||||
\end{align*}
|
||||
\item[État C] Les grandeurs caractéristiques sont~:
|
||||
\begin{align*}
|
||||
p_C&=\SI{1e5}{\pascal}\\
|
||||
V_C^{5/3}&=\frac{p_B}{p_C}\cdot V_B^{5/3}\\
|
||||
&=\frac{8\cdot 10^5}{1\cdot 10^5}\cdot (0,00725)^{5/3}\\
|
||||
&=2,1726\cdot 10^{-3}\\
|
||||
V_C&=\SI{0,02524}{\metre\cubed}\\
|
||||
T_C&=\frac{p_C\cdot V_C}{nR}\\
|
||||
&=\frac{10^5\cdot 0,02524}{10}=\SI{252,4}{\kelvin}
|
||||
\end{align*}
|
||||
\end{description}
|
||||
\item Ensuite les grandeurs associées aux transformations.
|
||||
\begin{description}
|
||||
\item[AB] Les grandeurs caractéristiques sont~:
|
||||
\begin{align*}
|
||||
\Delta U&=\frac{i}{2}\cdot nR\cdot \Delta T\\
|
||||
&=\frac{3}{2}\cdot 10\cdot (580-290)=\SI{4350}{\joule}\\
|
||||
A&=\SI{-5800}{\joule}\;\text{Donné}\\
|
||||
Q&=\Delta U+A=4350-5800=\SI{-1450}{\joule}
|
||||
\end{align*}
|
||||
\item[BC] Pour une adiabatique, les grandeurs caractéristiques sont~:
|
||||
\begin{align*}
|
||||
Q&=0\\
|
||||
\Delta U&=\frac{i}{2}\cdot nR\cdot \Delta T\\
|
||||
&=\frac{3}{2}\cdot 10\cdot (252,4-580)=\SI{-4914}{\joule}\\
|
||||
A&=-\Delta U=\SI{4914}{\joule}
|
||||
\end{align*}
|
||||
\item[CA] Pour une isobare, les grandeurs caractéristiques sont~:
|
||||
\begin{align*}
|
||||
\Delta U&=\frac{3}{2}\cdot 10\cdot (290-252,4)=\SI{564}{\joule}\\
|
||||
A&=p\cdot \Delta V=10^5\cdot (0,029-0,02524)\\
|
||||
&=\SI{376}{\joule}\\
|
||||
Q&=\Delta U+A=564+376=\SI{940}{\joule}
|
||||
\end{align*}
|
||||
\end{description}
|
||||
\item Le bilan du cycle est alors déduit du tableau des transformations.
|
||||
|
||||
\medskip
|
||||
|
||||
\begin{tabular}{|c|c|c|c|}
|
||||
\hline
|
||||
Étapes/(J) & \(Q\) & \(\Delta U\) & \(A\) \\
|
||||
\hline\hline
|
||||
AB & -1450 & 4350 & -5800 \\
|
||||
\hline
|
||||
BC & 0 & -4914 & 4914 \\
|
||||
\hline
|
||||
CA & 940 & 564 & 376 \\
|
||||
\hline\hline
|
||||
\(\sum\) & - & 0 & 510 \\
|
||||
\hline
|
||||
\end{tabular}
|
||||
|
||||
\medskip
|
||||
Alors, on peut obtenir les chaleurs entrant et sortant du gaz~:
|
||||
\begin{align*}
|
||||
&\sum Q>0 : \SI{940}{\joule}\\
|
||||
&\sum Q<0 : \SI{-1450}{\joule}
|
||||
\end{align*}
|
||||
Le diagramme du bilan est alors~:
|
||||
|
||||
\begin{figure}
|
||||
\def\svgwidth{7cm}
|
||||
\begin{center}
|
||||
%\input{Annexe-Exercices/Images/cycle2.eps_tex}
|
||||
\input{Annexe-Exercices/Images/cycle3.ps_tex}
|
||||
\end{center}
|
||||
\caption{Bilan du cycle\label{exos:cycle3}}
|
||||
\end{figure}
|
||||
\item Le rendement du climatiseur est~:
|
||||
\[\eta=\frac{\text{utile}}{\text{investi}}=\frac{940}{510}=184,3\%\]
|
||||
En effet, ce qui est utile pour un climatiseur est de retirer \SI{940}{\joule} de la pièce chaude et pour cela, il faut investir en électricité \SI{510}{\joule}.
|
||||
\item À chaque cycle, on retire \SI{940}{\joule}. Si on doit retirer \SI{1000}{\joule\per\second}, le nombre de cycle est donné par~:
|
||||
\begin{align*}
|
||||
&n\cdot 940=1000\;\Rightarrow\;n=\frac{1000}{940}=1,064\,\text{cycles/s}\\
|
||||
&\Rightarrow\;1,064\cdot 60=63,83\,\text{cycle/min}
|
||||
\end{align*}
|
||||
\item Enfin, pour calculer le travail, on passe par sa définition~:
|
||||
\begin{align*}
|
||||
A&=\int_A^B p\cdot dV=\int_A^B \frac{p_A\cdot V_A^{3/2}}{V^{3/2}}\cdot dV\\
|
||||
&=p_A\cdot V_A^{3/2}\cdot\int_A^B\frac{dV}{V^{3/2}}\\
|
||||
&=p_A\cdot V_A^{3/2}\cdot\int_A^B V^{-3/2}\cdot dV\\
|
||||
&=p_A\cdot V_A^{3/2}\cdot[\frac{1}{-3/2+1}\cdot V^{-3/2+1}]_A^B\\
|
||||
&=p_A\cdot V_A^{3/2}\cdot[-2\cdot V^{-1/2}]_A^B\\
|
||||
&=-2\cdot p_A\cdot V_A^{3/2}\cdot (V_B^{-1/2}-V_A^{-1/2})\\
|
||||
&=-2\cdot 10^5\cdot 0,029^{1,5}\cdot (\frac{1}{\sqrt{0,00725}}-\\
|
||||
&\;\frac{1}{\sqrt{0,029}}=\SI{-5800}{\joule}
|
||||
\end{align*}
|
||||
\end{enumerate}
|
||||
|
||||
\end{Solution OS}
|
||||
\begin{Solution OS}{39}
|
||||
Procédons simplement au calcul de l'incertitude absolue.
|
||||
|
||||
\smallskip
|
||||
@ -1219,7 +1340,7 @@
|
||||
Cette expression est légèrement différente de la précédente. Mais, le second terme est négligeable en raison de la présence de l'incertitude sur le temps au carré. On voit ainsi qu'il est nécessaire de faire attention aux ordres de grandeurs.
|
||||
|
||||
\end{Solution OS}
|
||||
\begin{Solution OS}{39}
|
||||
\begin{Solution OS}{40}
|
||||
\dots
|
||||
|
||||
\end{Solution OS}
|
||||
|
||||
Loading…
Reference in New Issue
Block a user