Fin des exercices à reporter.
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@ -2814,6 +2814,127 @@ 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 moteur à explosion fonctionne avec un gaz polyatomique sans vibrations (NR = 0,2) qui subit quatre transformations (voir figure \ref{exos:moteurexplosion})~:
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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/MoteurExplosion.ps_tex}
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\end{center}
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\caption{Cycle du moteur à explosion\label{exos:moteurexplosion}}
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\end{figure}
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\begin{description}
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\item[AB] Compression adiabatique.
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\item[BC] Chauffage à volume constant (explosion).
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\item[CD] Détente adiabatique.
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\item[DA] Refroidissement à volume constant (+ échange de gaz).
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\end{description}
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Résolvez les points suivants~:
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\begin{enumerate}
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\item Calculez les températures à la fin de chaque étape.
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\item Déterminez \(\Delta U\), \(A\) et \(Q\) pour chaque étape.
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\item Établir le bilan thermique.
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\item Calculez le rendement du cycle et celui de Carnot.
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\end{enumerate}
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\begin{solos}
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Il s'agit d'un moteur à explosion pour lequel nR = 0,2.
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\begin{enumerate}
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\item Les températures à la fin de chaque étapes sont~:
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\begin{align*}
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T_A&=\frac{p_A\cdot V_A}{nR}=\frac{10^5\cdot 600\cdot 10^{-6}}{0,2}=\SI{300}{\kelvin}\\
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T_B&=\frac{12\cdot 10^5\cdot 100\cdot 10^{-6}}{0,2}=\SI{600}{\kelvin}\\
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T_C&=\frac{40\cdot 10^5\cdot 100\cdot 10^{-6}}{0,2}=\SI{2000}{\kelvin}\\
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T_D&=\frac{4\cdot 10^5\cdot 600\cdot 10^{-6}}{0,2}=\SI{1200}{\kelvin}
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\end{align*}
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\item Pour les transformations, on a les énergies internes,
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\begin{align*}
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\Delta U_{AB}&=\frac{i}{2}\cdot (p_B\cdot V_B-p_A\cdot V_A)\\
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&=\frac{6}{2}\cdot (12\cdot 10^5\cdot 100\cdot 10{-6}\\
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&-10^5\cdot 600\cdot 10^{-6})=\SI{180}{\joule}\\
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\Delta U_{BC}&=\frac{i}{2}\cdot nR\cdot \Delta T\\
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&=\frac{6}{2}\cdot 0,2\cdot (1727-327)=\SI{840}{\joule}\\
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\Delta U_{CD}&=\frac{6}{2}\cdot 0,2\cdot (927-1727)=\SI{-480}{\joule}\\
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\Delta U_{DA}&=\frac{6}{2}\cdot 0,2\cdot (27-927)=\SI{-540}{\joule}\\
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\end{align*}
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les travaux,
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\begin{align*}
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A_{AB}&=-\Delta U_AB=\SI{-180}{\joule}\\
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A_{BC}&=0\\
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A_{CD}&=\SI{480}{\joule}\\
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A_{DA}&=0
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\end{align*}
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et les chaleurs échangées
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\begin{align*}
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Q_{AB}&=0\\
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Q_{BC}&=\Delta U_BC=\SI{840}{\joule}\\
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Q_{CD}&=0\\
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Q_{DA}&=\SI{-540}{\joule}
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\end{align*}
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\item Le bilan du cycle est alors~:
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Chaleur reçue par le moteur de la source chaude~:
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\[\sum Q>0=\SI{840}{\joule}\]
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Chaleur rejetée par le moteur vers la source froide~:
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\[\sum Q<0=\SI{-540}{\joule}\]
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Il faut relever que la température de la source froide est de \SI{300}{\kelvin}, soit \SI{27}{\celsius}, c'est-à-dire la température ambiante.
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Travail fourni par le moteur~:
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\[\sum A=\SI{300}{\joule}\]
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\item Le rendement du cycle est alors~:
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\[\eta=\frac{\text{Fourni}}{\text{Investi}}=\frac{300}{840}=35,7\%\]
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et le rendement de Carnot~:
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\[\eta=1-\frac{T_{froide}}{T_{chaude}}=1-\frac{300}{2000}=85\%\]
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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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Le bilan d'énergie d'un moteur thermique est représenté sur le diagramme de la figure \ref{exos:diagsimple1}.
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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/DiagSimple1.ps_tex}
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\end{center}
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\caption{Diagramme du moteur\label{exos:diagsimple1}}
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\end{figure}
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Déterminez le rendement du moteur et le rendement de Carnot.
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\begin{solos}
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On a, très simplement~:
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\begin{align*}
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&\eta=\frac{A}{Q_{investi}}=\frac{1,2}{8,25}=0,145=14,5\%\\
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&\eta_{Carnot}=1-\frac{T_f}{T_c}=1-\frac{300}{3000}=90\%
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\end{align*}
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\end{solos}
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\end{exos}
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\begin{exos}
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Une pompe à chaleur présente le bilan d'énergie représenté sur le diagramme de la figure \ref{exos:diagsimple2}.
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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/DiagSimple2.ps_tex}
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\end{center}
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\caption{Diagramme de la pompe\label{exos:diagsimple2}}
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\end{figure}
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Déterminez l'efficacité de la pompe. Quelle serait l'efficacité maximale ?
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\begin{solos}
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Pour une pompe à chaleur, au lieu de rendement, on parle de \emph{c\oe fficient d'amplification frigorifique} ou CAF.
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\begin{align*}
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&CAF=\frac{utile}{investi}=\frac{65,1}{5,1}=12,76=1276\%\\
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&\eta_{max}=\frac{1}{\eta_{Carnot}}=\frac{1}{1-T_f/T_c}\\
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&=\frac{T_c}{T_c-T_f}=\frac{300}{300-290}=3000\%
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\end{align*}
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\end{solos}
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\end{exos}
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\begin{exos}
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Une machine thermique cyclique travaille avec un gaz parfait monoatomique qui subit quatre transformations. Le tableau suivant présente les échange d'énergie au cours du cycle.
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\begin{center}
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@ -3233,6 +3354,100 @@ 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 frigo consomme \SI{20}{\joule} de travail par cycle. Son efficacité est de 3. Déterminez le bilan d'énergie (dessin, calcul de A et des Q).
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\begin{solos}
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Le bilan est présenté sur la figure \ref{exos:diagsimple3}.
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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/DiagSimple3.ps_tex}
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\end{center}
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\caption{Diagramme de la pompe\label{exos:diagsimple3}}
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\end{figure}
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Avec un travail de \SI{20}{\joule}, on a simplement~:
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\[\eta=CAF=3=\frac{\sum Q>0}{20}\;\Rightarrow\;\sum Q>0=\SI{60}{\joule}\]
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\end{solos}
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\end{exos}
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\begin{exos}
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On donne aux figures \ref{exos:machine1}, \ref{exos:machine2} et \ref{exos:machine3} le bilan de trois machines thermiques. Trouvez s'il s'agit d'un moteur, d'un frigo ou d'une pompe à chaleur et calculez soit le rendement, soit les efficacités (pratiques, non idéales).
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\begin{figure}
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\subfigure[Premier diagramme.\label{exos:machine1}]{
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\def\svgwidth{7cm}
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\centering
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\input{Annexe-Exercices/Images/Machine1.ps_tex}
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}\\
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\subfigure[Deuxième diagramme.\label{exos:machine2}]{
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\def\svgwidth{7cm}
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\centering
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\input{Annexe-Exercices/Images/Machine2.ps_tex}
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}\\
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\subfigure[Troisième diagramme.\label{exos:machine3}]{
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\def\svgwidth{7cm}
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\centering
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\input{Annexe-Exercices/Images/Machine3.ps_tex}
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}
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\caption{Les trois machines.}
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\end{figure}
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\begin{solos}
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Le diagramme de la figure \ref{exos:machine1} est celui d'une pompe à chaleur ou un climatiseur. En effet, les températures froide et chaude sont respectivement \SI{26,85}{\celsius} et \SI{126,85}{\celsius}.
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S'il s'agit d'un climatiseur, le \emph{c\oe fficient d'amplification frigorifique} ou CAF vaut~: \[CAF=\frac{3000}{2000}=1,5\]
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S'il s'agit d'une pompe à chaleur, on utilise plutôt le \emph{c\oe fficient de performance} ou COP qui vaut~: \[COP=\frac{5000}{2000}=2,5\]
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Le diagramme de la figure \ref{exos:machine2} est celui d'un réfrigérateur. En effet, les températures froide et chaude sont respectivement \SI{6,85}{\celsius} et \SI{26,85}{\celsius}.
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Le CAF est alors~: \[CAF=\frac{3000}{1000}=3\]
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Finalement, le diagramme de la figure \ref{exos:machine3} est celui d'un moteur (en raison du sens des flèches) et son rendement est~: \[\eta=\frac{200}{1200}=16,6\%\]
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\end{solos}
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\end{exos}
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\begin{exos}
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Une pompe à chaleur prélève à chaque cycle une chaleur de \SI{1000}{\joule} dans un lac à \SI{17}{\celsius}. La machine exige \SI{300}{\joule} de travail par cycle. Quelle est la température maximale de la source chaude ?
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\begin{solos}
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La chaleur fournie à l'habitation vaut~:
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\[Q=1000+300=\SI{1300}{\joule}\]
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Le \emph{c\oe fficient de performance} COP vaut alors~:
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\[COP=\frac{\text{fourni}}{\text{investi}}=\frac{1300}{300}=4,33\]
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Mais le rendement maximal pour une pompe à chaleur vaut~:
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\[COP_{max}=\frac{1}{1-T_f/T_c}=\frac{T_c}{T_c-T_f}\]
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Si on fait l'hypothèse que nous sommes au maximum, on peut écrire~:
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\begin{align*}
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&COP=\frac{T_c}{T_c-T_f}\;\Rightarrow\\
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&4,33=\frac{T_c}{T_c-290}\;\Rightarrow\\
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&T_c=\SI{380}{\kelvin}=\SI{107}{\celsius}
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\end{align*}
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\end{solos}
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\end{exos}
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\begin{exos}
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Une pompe à chaleur fournit à chaque cycle une chaleur de \SI{400}{\joule} à l'air ambiant d'une pièce se trouvant à \SI{20}{\celsius}. Quel travail minimum la machine doit-elle recevoir si la chaleur est puisée~:
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\begin{enumerate}
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\item dans un lac à \SI{7}{\celsius} et
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\item dans l'air froid à \SI{-20}{\celsius} ?
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\item Calculez dans chaque cas l'efficacité de la pompe.
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\end{enumerate}
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\begin{solos}
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On suppose une efficacité maximale.
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\begin{enumerate}
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\item \SI{7}{\celsius}=\SI{280,15}{\kelvin}. Ainsi, l'efficacité vaut~:
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\[COP=\frac{1}{1-280,15/293,15}=22,55\]
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Par définition de l'efficacité, on a~:
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\[COP=\frac{400}{A}\;\Rightarrow\;A=\frac{400}{22,55}=\SI{17,738}{\joule}\]
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\item \SI{-20}{\celsius}=\SI{253,15}{\kelvin}. Ainsi, l'efficacité vaut~:
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\[COP=\frac{1}{1-253,15/293,15}=7,328\]
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Par définition de l'efficacité, on a~:
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\[COP=\frac{400}{A}\;\Rightarrow\;A=\frac{400}{7,328}=\SI{54,58}{\joule}\]
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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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% \begin{solos}
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@ -2814,6 +2814,127 @@ 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 moteur à explosion fonctionne avec un gaz polyatomique sans vibrations (NR = 0,2) qui subit quatre transformations (voir figure \ref{exos:moteurexplosion})~:
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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/MoteurExplosion.ps_tex}
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\end{center}
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\caption{Cycle du moteur à explosion\label{exos:moteurexplosion}}
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\end{figure}
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\begin{description}
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\item[AB] Compression adiabatique.
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\item[BC] Chauffage à volume constant (explosion).
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\item[CD] Détente adiabatique.
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\item[DA] Refroidissement à volume constant (+ échange de gaz).
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\end{description}
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Résolvez les points suivants~:
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\begin{enumerate}
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\item Calculez les températures à la fin de chaque étape.
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\item Déterminez \(\Delta U\), \(A\) et \(Q\) pour chaque étape.
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\item Établir le bilan thermique.
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\item Calculez le rendement du cycle et celui de Carnot.
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\end{enumerate}
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\begin{solos}
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Il s'agit d'un moteur à explosion pour lequel nR = 0,2.
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\begin{enumerate}
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\item Les températures à la fin de chaque étapes sont~:
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\begin{align*}
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T_A&=\frac{p_A\cdot V_A}{nR}=\frac{10^5\cdot 600\cdot 10^{-6}}{0,2}=\SI{300}{\kelvin}\\
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T_B&=\frac{12\cdot 10^5\cdot 100\cdot 10^{-6}}{0,2}=\SI{600}{\kelvin}\\
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T_C&=\frac{40\cdot 10^5\cdot 100\cdot 10^{-6}}{0,2}=\SI{2000}{\kelvin}\\
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T_D&=\frac{4\cdot 10^5\cdot 600\cdot 10^{-6}}{0,2}=\SI{1200}{\kelvin}
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\end{align*}
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\item Pour les transformations, on a les énergies internes,
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\begin{align*}
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\Delta U_{AB}&=\frac{i}{2}\cdot (p_B\cdot V_B-p_A\cdot V_A)\\
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&=\frac{6}{2}\cdot (12\cdot 10^5\cdot 100\cdot 10{-6}\\
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&-10^5\cdot 600\cdot 10^{-6})=\SI{180}{\joule}\\
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\Delta U_{BC}&=\frac{i}{2}\cdot nR\cdot \Delta T\\
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&=\frac{6}{2}\cdot 0,2\cdot (1727-327)=\SI{840}{\joule}\\
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\Delta U_{CD}&=\frac{6}{2}\cdot 0,2\cdot (927-1727)=\SI{-480}{\joule}\\
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\Delta U_{DA}&=\frac{6}{2}\cdot 0,2\cdot (27-927)=\SI{-540}{\joule}\\
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\end{align*}
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les travaux,
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\begin{align*}
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A_{AB}&=-\Delta U_AB=\SI{-180}{\joule}\\
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A_{BC}&=0\\
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A_{CD}&=\SI{480}{\joule}\\
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A_{DA}&=0
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\end{align*}
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et les chaleurs échangées
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\begin{align*}
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Q_{AB}&=0\\
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Q_{BC}&=\Delta U_BC=\SI{840}{\joule}\\
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Q_{CD}&=0\\
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Q_{DA}&=\SI{-540}{\joule}
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\end{align*}
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\item Le bilan du cycle est alors~:
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Chaleur reçue par le moteur de la source chaude~:
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\[\sum Q>0=\SI{840}{\joule}\]
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Chaleur rejetée par le moteur vers la source froide~:
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\[\sum Q<0=\SI{-540}{\joule}\]
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Il faut relever que la température de la source froide est de \SI{300}{\kelvin}, soit \SI{27}{\celsius}, c'est-à-dire la température ambiante.
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Travail fourni par le moteur~:
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\[\sum A=\SI{300}{\joule}\]
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\item Le rendement du cycle est alors~:
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\[\eta=\frac{\text{Fourni}}{\text{Investi}}=\frac{300}{840}=35,7\%\]
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et le rendement de Carnot~:
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\[\eta=1-\frac{T_{froide}}{T_{chaude}}=1-\frac{300}{2000}=85\%\]
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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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Le bilan d'énergie d'un moteur thermique est représenté sur le diagramme de la figure \ref{exos:diagsimple1}.
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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/DiagSimple1.ps_tex}
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\end{center}
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\caption{Diagramme du moteur\label{exos:diagsimple1}}
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\end{figure}
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Déterminez le rendement du moteur et le rendement de Carnot.
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\begin{solos}
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On a, très simplement~:
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\begin{align*}
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&\eta=\frac{A}{Q_{investi}}=\frac{1,2}{8,25}=0,145=14,5\%\\
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&\eta_{Carnot}=1-\frac{T_f}{T_c}=1-\frac{300}{3000}=90\%
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\end{align*}
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\end{solos}
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\end{exos}
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\begin{exos}
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Une pompe à chaleur présente le bilan d'énergie représenté sur le diagramme de la figure \ref{exos:diagsimple2}.
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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/DiagSimple2.ps_tex}
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\end{center}
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\caption{Diagramme de la pompe\label{exos:diagsimple2}}
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\end{figure}
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Déterminez l'efficacité de la pompe. Quelle serait l'efficacité maximale ?
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\begin{solos}
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Pour une pompe à chaleur, au lieu de rendement, on parle de \emph{c\oe fficient d'amplification frigorifique} ou CAF.
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\begin{align*}
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&CAF=\frac{utile}{investi}=\frac{65,1}{5,1}=12,76=1276\%\\
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&\eta_{max}=\frac{1}{\eta_{Carnot}}=\frac{1}{1-T_f/T_c}\\
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&=\frac{T_c}{T_c-T_f}=\frac{300}{300-290}=3000\%
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\end{align*}
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\end{solos}
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\end{exos}
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\begin{exos}
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Une machine thermique cyclique travaille avec un gaz parfait monoatomique qui subit quatre transformations. Le tableau suivant présente les échange d'énergie au cours du cycle.
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\begin{center}
|
||||
@ -3233,6 +3354,93 @@ Les valeurs des c\oe fficients de dilatation thermique sont celles du tableau \r
|
||||
\end{solos}
|
||||
\end{exos}
|
||||
|
||||
\begin{exos}
|
||||
Un frigo consomme \SI{20}{\joule} de travail par cycle. Son efficacité est de 3. Déterminez le bilan d'énergie (dessin, calcul de A et des Q).
|
||||
\begin{solos}
|
||||
Le bilan est présenté sur la figure \ref{exos:diagsimple3}.
|
||||
|
||||
\begin{figure}
|
||||
\def\svgwidth{7cm}
|
||||
\begin{center}
|
||||
\input{Annexe-Exercices/Images/DiagSimple3.ps_tex}
|
||||
\end{center}
|
||||
\caption{Diagramme de la pompe\label{exos:diagsimple3}}
|
||||
\end{figure}
|
||||
|
||||
Avec un travail de \SI{20}{\joule}, on a simplement~:
|
||||
\[\eta=CAF=3=\frac{\sum Q>0}{20}\;\Rightarrow\;\sum Q>0=\SI{60}{\joule}\]
|
||||
\end{solos}
|
||||
\end{exos}
|
||||
|
||||
\begin{exos}
|
||||
On donne aux figures \ref{exos:machine1}, \ref{exos:machine2} et \ref{exos:machine3} le bilan de trois machines thermiques. Trouvez s'il s'agit d'un moteur, d'un frigo ou d'une pompe à chaleur et calculez soit le rendement, soit les efficacités (pratiques, non idéales).
|
||||
|
||||
\begin{figure}
|
||||
\subfigure[Premier diagramme.\label{exos:machine1}]{
|
||||
\def\svgwidth{7cm}
|
||||
\centering
|
||||
\input{Annexe-Exercices/Images/Machine1.ps_tex}
|
||||
}\\
|
||||
\subfigure[Deuxième diagramme.\label{exos:machine2}]{
|
||||
\def\svgwidth{7cm}
|
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\centering
|
||||
\input{Annexe-Exercices/Images/Machine2.ps_tex}
|
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}\\
|
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\subfigure[Troisième diagramme.\label{exos:machine3}]{
|
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\def\svgwidth{7cm}
|
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\centering
|
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\input{Annexe-Exercices/Images/Machine3.ps_tex}
|
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}
|
||||
\caption{Les trois machines.}
|
||||
\end{figure}
|
||||
\begin{solos}
|
||||
Le diagramme de la figure \ref{exos:machine1} est celui d'une pompe à chaleur ou un climatiseur. En effet, les températures froide et chaude sont respectivement \SI{26,85}{\celsius} et \SI{126,85}{\celsius}.
|
||||
|
||||
S'il s'agit d'un climatiseur, le \emph{c\oe fficient d'amplification frigorifique} ou CAF vaut~: \[CAF=\frac{3000}{2000}=1,5\]
|
||||
|
||||
S'il s'agit d'une pompe à chaleur, on utilise plutôt le \emph{c\oe fficient de performance} ou COP qui vaut~: \[COP=\frac{5000}{2000}=2,5\]
|
||||
|
||||
Le diagramme de la figure \ref{exos:machine2} est celui d'un réfrigérateur. En effet, les températures froide et chaude sont respectivement \SI{6,85}{\celsius} et \SI{26,85}{\celsius}.
|
||||
|
||||
Le CAF est alors~: \[CAF=\frac{3000}{1000}=3\]
|
||||
|
||||
Finalement, le diagramme de la figure \ref{exos:machine3} est celui d'un moteur (en raison du sens des flèches) et son rendement est~: \[\eta=\frac{200}{1200}=16,6\%\]
|
||||
\end{solos}
|
||||
\end{exos}
|
||||
|
||||
\begin{exos}
|
||||
Une pompe à chaleur prélève à chaque cycle une chaleur de \SI{1000}{\joule} dans un lac à \SI{17}{\celsius}. La machine exige \SI{300}{\joule} de travail par cycle. Quelle est la température maximale de la source chaude ?
|
||||
\begin{solos}
|
||||
La chaleur fournie à l'habitation vaut~:
|
||||
\[Q=1000+300=\SI{1300}{\joule}\]
|
||||
Le \emph{c\oe fficient de performance} COP vaut alors~:
|
||||
\[COP=\frac{\text{fourni}}{\text{investi}}=\frac{1300}{300}=4,33\]
|
||||
Mais le rendement maximal pour une pompe à chaleur vaut~:
|
||||
\[COP_{max}=\frac{1}{1-T_f/T_c}=\frac{T_c}{T_c-T_f}\]
|
||||
Si on fait l'hypothèse que nous sommes au maximum, on peut écrire~:
|
||||
\begin{align*}
|
||||
&COP=\frac{T_c}{T_c-T_f}\;\Rightarrow\\
|
||||
&4,33=\frac{T_c}{T_c-290}\;\Rightarrow\\
|
||||
&T_c=\SI{380}{\kelvin}=\SI{107}{\celsius}
|
||||
\end{align*}
|
||||
\end{solos}
|
||||
\end{exos}
|
||||
|
||||
\begin{exos}
|
||||
Une pompe à chaleur fournit à chaque cycle une chaleur de \SI{400}{\joule} à l'air ambiant d'une pièce se trouvant à \SI{20}{\celsius}. Quel travail minimum la machine doit-elle recevoir si la chaleur est puisée~:
|
||||
\begin{enumerate}
|
||||
\item dans un lac à \SI{7}{\celsius} et
|
||||
\item dans l'air froid à \SI{-20}{\celsius} ?
|
||||
\item Calculez dans chaque cas l'efficacité de la pompe.
|
||||
\end{enumerate}
|
||||
\begin{solos}
|
||||
On suppose une efficacité maximale.
|
||||
\begin{enumerate}
|
||||
\item \SI{7}{\celsius}=\SI{280,15}{\kelvin}.
|
||||
\end{enumerate}
|
||||
\end{solos}
|
||||
\end{exos}
|
||||
|
||||
%\begin{exos}
|
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% Un énoncé de test.
|
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% \begin{solos}
|
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|
||||
64
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64
Annexe-Exercices/Images/DiagSImple1.ps_tex
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138
SolutionsOS.tex
138
SolutionsOS.tex
@ -1092,6 +1092,74 @@
|
||||
|
||||
\end{Solution OS}
|
||||
\begin{Solution OS}{36}
|
||||
Il s'agit d'un moteur à explosion pour lequel nR = 0,2.
|
||||
\begin{enumerate}
|
||||
\item Les températures à la fin de chaque étapes sont~:
|
||||
\begin{align*}
|
||||
T_A&=\frac{p_A\cdot V_A}{nR}=\frac{10^5\cdot 600\cdot 10^{-6}}{0,2}=\SI{300}{\kelvin}\\
|
||||
T_B&=\frac{12\cdot 10^5\cdot 100\cdot 10^{-6}}{0,2}=\SI{600}{\kelvin}\\
|
||||
T_C&=\frac{40\cdot 10^5\cdot 100\cdot 10^{-6}}{0,2}=\SI{2000}{\kelvin}\\
|
||||
T_D&=\frac{4\cdot 10^5\cdot 600\cdot 10^{-6}}{0,2}=\SI{1200}{\kelvin}
|
||||
\end{align*}
|
||||
\item Pour les transformations, on a les énergies internes,
|
||||
\begin{align*}
|
||||
\Delta U_{AB}&=\frac{i}{2}\cdot (p_B\cdot V_B-p_A\cdot V_A)\\
|
||||
&=\frac{6}{2}\cdot (12\cdot 10^5\cdot 100\cdot 10{-6}\\
|
||||
&-10^5\cdot 600\cdot 10^{-6})=\SI{180}{\joule}\\
|
||||
\Delta U_{BC}&=\frac{i}{2}\cdot nR\cdot \Delta T\\
|
||||
&=\frac{6}{2}\cdot 0,2\cdot (1727-327)=\SI{840}{\joule}\\
|
||||
\Delta U_{CD}&=\frac{6}{2}\cdot 0,2\cdot (927-1727)=\SI{-480}{\joule}\\
|
||||
\Delta U_{DA}&=\frac{6}{2}\cdot 0,2\cdot (27-927)=\SI{-540}{\joule}\\
|
||||
\end{align*}
|
||||
les travaux,
|
||||
\begin{align*}
|
||||
A_{AB}&=-\Delta U_AB=\SI{-180}{\joule}\\
|
||||
A_{BC}&=0\\
|
||||
A_{CD}&=\SI{480}{\joule}\\
|
||||
A_{DA}&=0
|
||||
\end{align*}
|
||||
et les chaleurs échangées
|
||||
\begin{align*}
|
||||
Q_{AB}&=0\\
|
||||
Q_{BC}&=\Delta U_BC=\SI{840}{\joule}\\
|
||||
Q_{CD}&=0\\
|
||||
Q_{DA}&=\SI{-540}{\joule}
|
||||
\end{align*}
|
||||
\item Le bilan du cycle est alors~:
|
||||
Chaleur reçue par le moteur de la source chaude~:
|
||||
\[\sum Q>0=\SI{840}{\joule}\]
|
||||
Chaleur rejetée par le moteur vers la source froide~:
|
||||
\[\sum Q<0=\SI{-540}{\joule}\]
|
||||
Il faut relever que la température de la source froide est de \SI{300}{\kelvin}, soit \SI{27}{\celsius}, c'est-à-dire la température ambiante.
|
||||
|
||||
Travail fourni par le moteur~:
|
||||
\[\sum A=\SI{300}{\joule}\]
|
||||
\item Le rendement du cycle est alors~:
|
||||
\[\eta=\frac{\text{Fourni}}{\text{Investi}}=\frac{300}{840}=35,7\%\]
|
||||
et le rendement de Carnot~:
|
||||
\[\eta=1-\frac{T_{froide}}{T_{chaude}}=1-\frac{300}{2000}=85\%\]
|
||||
\end{enumerate}
|
||||
|
||||
|
||||
\end{Solution OS}
|
||||
\begin{Solution OS}{37}
|
||||
On a, très simplement~:
|
||||
\begin{align*}
|
||||
&\eta=\frac{A}{Q_{investi}}=\frac{1,2}{8,25}=0,145=14,5\%\\
|
||||
&\eta_{Carnot}=1-\frac{T_f}{T_c}=1-\frac{300}{3000}=90\%
|
||||
\end{align*}
|
||||
|
||||
\end{Solution OS}
|
||||
\begin{Solution OS}{38}
|
||||
Pour une pompe à chaleur, au lieu de rendement, on parle de \emph{c\oe fficient d'amplification frigorifique} ou CAF.
|
||||
\begin{align*}
|
||||
&CAF=\frac{utile}{investi}=\frac{65,1}{5,1}=12,76=1276\%\\
|
||||
&\eta_{max}=\frac{1}{\eta_{Carnot}}=\frac{1}{1-T_f/T_c}\\
|
||||
&=\frac{T_c}{T_c-T_f}=\frac{300}{300-290}=3000\%
|
||||
\end{align*}
|
||||
|
||||
\end{Solution OS}
|
||||
\begin{Solution OS}{39}
|
||||
Le tableau complété est le suivant :
|
||||
\begin{center}
|
||||
\begin{tabular}{|c|c|c|c|}
|
||||
@ -1136,7 +1204,7 @@
|
||||
|
||||
|
||||
\end{Solution OS}
|
||||
\begin{Solution OS}{37}
|
||||
\begin{Solution OS}{40}
|
||||
On commence par calculer le produit nR~:
|
||||
\[nR=\frac{p\cdot V}{T}=\frac{3\cdot 10^5\cdot 9\cdot 10^{-3}}{270}=10\]
|
||||
\begin{enumerate}
|
||||
@ -1172,7 +1240,7 @@
|
||||
\end{enumerate}
|
||||
|
||||
\end{Solution OS}
|
||||
\begin{Solution OS}{38}
|
||||
\begin{Solution OS}{41}
|
||||
Voici l'analyse du cycle.
|
||||
\begin{enumerate}
|
||||
\item Les grandeurs manquantes.
|
||||
@ -1241,7 +1309,7 @@
|
||||
\end{enumerate}
|
||||
|
||||
\end{Solution OS}
|
||||
\begin{Solution OS}{39}
|
||||
\begin{Solution OS}{42}
|
||||
Au préalable, calculons nR.
|
||||
\begin{equation*}
|
||||
nR=\frac{p_B\cdot V_B}{T_B}=\frac{10\cdot 10^5\cdot 2\cdot 10^{-3}}{1500}=1,33
|
||||
@ -1290,7 +1358,7 @@
|
||||
\end{enumerate}
|
||||
|
||||
\end{Solution OS}
|
||||
\begin{Solution OS}{40}
|
||||
\begin{Solution OS}{43}
|
||||
Le diagramme PV du cycle est présenté à la figure \ref{climatiseur}.
|
||||
\begin{figure}[t]
|
||||
\centering
|
||||
@ -1410,7 +1478,65 @@
|
||||
\end{enumerate}
|
||||
|
||||
\end{Solution OS}
|
||||
\begin{Solution OS}{41}
|
||||
\begin{Solution OS}{44}
|
||||
Le bilan est présenté sur la figure \ref{exos:diagsimple3}.
|
||||
|
||||
\begin{figure}
|
||||
\def\svgwidth{7cm}
|
||||
\begin{center}
|
||||
\input{Annexe-Exercices/Images/DiagSimple3.ps_tex}
|
||||
\end{center}
|
||||
\caption{Diagramme de la pompe\label{exos:diagsimple3}}
|
||||
\end{figure}
|
||||
|
||||
Avec un travail de \SI{20}{\joule}, on a simplement~:
|
||||
\[\eta=CAF=3=\frac{\sum Q>0}{20}\;\Rightarrow\;\sum Q>0=\SI{60}{\joule}\]
|
||||
|
||||
\end{Solution OS}
|
||||
\begin{Solution OS}{45}
|
||||
Le diagramme de la figure \ref{exos:machine1} est celui d'une pompe à chaleur ou un climatiseur. En effet, les températures froide et chaude sont respectivement \SI{26,85}{\celsius} et \SI{126,85}{\celsius}.
|
||||
|
||||
S'il s'agit d'un climatiseur, le \emph{c\oe fficient d'amplification frigorifique} ou CAF vaut~: \[CAF=\frac{3000}{2000}=1,5\]
|
||||
|
||||
S'il s'agit d'une pompe à chaleur, on utilise plutôt le \emph{c\oe fficient de performance} ou COP qui vaut~: \[COP=\frac{5000}{2000}=2,5\]
|
||||
|
||||
Le diagramme de la figure \ref{exos:machine2} est celui d'un réfrigérateur. En effet, les températures froide et chaude sont respectivement \SI{6,85}{\celsius} et \SI{26,85}{\celsius}.
|
||||
|
||||
Le CAF est alors~: \[CAF=\frac{3000}{1000}=3\]
|
||||
|
||||
Finalement, le diagramme de la figure \ref{exos:machine3} est celui d'un moteur (en raison du sens des flèches) et son rendement est~: \[\eta=\frac{200}{1200}=16,6\%\]
|
||||
|
||||
\end{Solution OS}
|
||||
\begin{Solution OS}{46}
|
||||
La chaleur fournie à l'habitation vaut~:
|
||||
\[Q=1000+300=\SI{1300}{\joule}\]
|
||||
Le \emph{c\oe fficient de performance} COP vaut alors~:
|
||||
\[COP=\frac{\text{fourni}}{\text{investi}}=\frac{1300}{300}=4,33\]
|
||||
Mais le rendement maximal pour une pompe à chaleur vaut~:
|
||||
\[COP_{max}=\frac{1}{1-T_f/T_c}=\frac{T_c}{T_c-T_f}\]
|
||||
Si on fait l'hypothèse que nous sommes au maximum, on peut écrire~:
|
||||
\begin{align*}
|
||||
&COP=\frac{T_c}{T_c-T_f}\;\Rightarrow\\
|
||||
&4,33=\frac{T_c}{T_c-290}\;\Rightarrow\\
|
||||
&T_c=\SI{380}{\kelvin}=\SI{107}{\celsius}
|
||||
\end{align*}
|
||||
|
||||
\end{Solution OS}
|
||||
\begin{Solution OS}{47}
|
||||
On suppose une efficacité maximale.
|
||||
\begin{enumerate}
|
||||
\item \SI{7}{\celsius}=\SI{280,15}{\kelvin}. Ainsi, l'efficacité vaut~:
|
||||
\[COP=\frac{1}{1-280,15/293,15}=22,55\]
|
||||
Par définition de l'efficacité, on a~:
|
||||
\[COP=\frac{400}{A}\;\Rightarrow\;A=\frac{400}{22,55}=\SI{17,738}{\joule}\]
|
||||
\item \SI{-20}{\celsius}=\SI{253,15}{\kelvin}. Ainsi, l'efficacité vaut~:
|
||||
\[COP=\frac{1}{1-253,15/293,15}=7,328\]
|
||||
Par définition de l'efficacité, on a~:
|
||||
\[COP=\frac{400}{A}\;\Rightarrow\;A=\frac{400}{7,328}=\SI{54,58}{\joule}\]
|
||||
\end{enumerate}
|
||||
|
||||
\end{Solution OS}
|
||||
\begin{Solution OS}{48}
|
||||
Procédons simplement au calcul de l'incertitude absolue.
|
||||
|
||||
\smallskip
|
||||
@ -1458,7 +1584,7 @@
|
||||
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}{42}
|
||||
\begin{Solution OS}{49}
|
||||
\dots
|
||||
|
||||
\end{Solution OS}
|
||||
|
||||
Loading…
Reference in New Issue
Block a user