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90
Annexe-Exercices/Annexe-Exercices.tex
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@ -2939,6 +2939,94 @@ Les valeurs des c\oe fficients de dilatation thermique sont celles du tableau \r
\end{solos}
\end{exos}
\begin{exos}
Un cycle Diesel se déroule en principe de la façon suivante. Un gaz parfait diatomique est initialement dans l'état \(p_1=\SI{1}{\bar}\), \(V_1=\SI{1000}{\centi\metre\cubed}\) et \(T_1=\SI{333,3}{\kelvin}\) et on lui fait subir quatre transformations successives~:
\begin{description}
\item[Compression adiabatique] jusqu'à l'état \(p_2=\,?\), \(V_2=\SI{70}{\centi\metre\cubed}\) et \(T_2=\SI{965,6}{\kelvin}\).
\item[Échauffement à pression constante] jusqu'à l'état \(p_3=p_2\), \(V_3=\SI{140}{\centi\metre\cubed}\) et \(T_3=\,?\).
\item[Détente adiabatique] jusqu'à l'état \(p_4=\,?\), \(V_3=V_1\) et \(T_3=\SI{879,6}{\kelvin}\).
\item[Refroidissement à volume constant] jusqu'à l'état initial.
\smallskip
On demande de~:
\begin{enumerate}
\item Calculer nR, \(p_2\), \(T_3\) et \(p_4\).
\item Esquisser le diagramme PV du cycle.
\item Calculer pour chaque transformation \(\Delta U\), \(A\) et \(Q\).
\item Établir le bilan du cycle.
\item Calculer le rendement.
\end{enumerate}
\end{description}
\begin{solos}
Voici l'analyse du cycle.
\begin{enumerate}
\item Les grandeurs manquantes.
\begin{align*}
nR&=\frac{p_1\cdot V_1}{T_1}=\frac{10^5\cdot 1000\cdot 10^{-6}}{333,3}=0,3\\
p_2&=\frac{nR\cdot T_2}{V_2}=\frac{0,3\cdot 965,6}{70\cdot 10^{-6}}=\SI{41,4}{\bar}\\
T_3&=\frac{p_3\cdot V_3}{nR}=\frac{41,4\cdot 10^5\cdot 140\dot 10^{-6}}{0,3}\\
&=\SI{1931,2}{\kelvin}\\
p_4&=\frac{nR\cdot T_4}{V_4}=\frac{0,3\cdot 879,6}{1000\cdot 10^{-6}}=\SI{2,6}{\bar}
\end{align*}
\item Le diagramme PV est celui de la figure \ref{exos:diesel}.
\begin{figure}
\def\svgwidth{7cm}
\begin{center}
\input{Annexe-Exercices/Images/Diesel.ps_tex}
\end{center}
\caption{Le diagramme PV\label{exos:diesel}}
\end{figure}
\item Les grandeurs des transformations.
\begin{description}
\item[1-2] Adiabatique
\begin{align*}
\Delta U&=\frac{i}{2}\cdot \Delta (pV)\\
&=\frac{5}{2}\cdot (41,4\cdot 10^5\cdot 70\cdot 10^{-6}\\
&-10^5\cdot 1000\cdot 10^{-6})=\SI{474,5}{\joule}\\
A&=-\Delta U=\SI{-474,5}{\joule}\\
Q&=0
\end{align*}
\item[2-3] Isobare
\begin{align*}
\Delta U&=\frac{i}{2}\cdot nR\cdot \Delta T\\
&=\frac{5}{2}\cdot 0,3\cdot (1931,2-965,6)=\SI{724,2}{\joule}\\
A&=p_2\cdot (V_3-V_2)\\
&=41,4\cdot 10^5\cdot (140\cdot 10^{-6}-70\cdot 10^{-6})\\
&=\SI{289,7}{\joule}\\
Q&=\Delta U+A=724,2+289,7=\SI{1013,9}{\joule}
\end{align*}
\item[3-4] Adiabatique
\begin{align*}
\Delta U&=\frac{i}{2}\cdot \Delta (pV)\\
&=\frac{5}{2}\cdot (p_4\cdot V_4-p_3\cdot V_3)\\
&=\SI{789,3}{\joule}\\
A&=-\Delta U=\SI{-789,3}{\joule}\\
Q&=0
\end{align*}
\item[4-1] Isochore
\begin{align*}
\Delta U&=\frac{i}{2}\cdot nR\cdot \Delta T=\frac{i}{2}\cdot nR (T_1-T_4)\\
&=\frac{5}{2}\cdot 0,3\cdot (333,3-879,6)\\
&=\SI{-409,7}{\joule}\\
A&=0\\
Q&=\Delta U=\SI{-409,7}{\joule}
\end{align*}
\end{description}
\item Le bilan du cycle.
Chaleur reçue par le moteur~: \[\sum Q>0=\SI{1014}{\joule}\].
Chaleur rejetée par le moteur~: \[\sum q<0=\SI{-409,7}{\joule}\].
Travail fourni par le moteur~: \[\sum A=\SI{604,5}{\joule}\].
\item Le rendement.
\[\eta=\frac{A}{\sum Q>0}=\frac{604,5}{1014}=60\%\]
\end{enumerate}
\end{solos}
\end{exos}
\begin{exos}
\begin{figure}[t]
\centering
@ -2947,7 +3035,7 @@ Les valeurs des c\oe fficients de dilatation thermique sont celles du tableau \r
\def\svgwidth{0.9\columnwidth}
\input{Annexe-Exercices/Images/Lemoteur.ps_tex}
\end{figure}
On donne le diagramme PV d'un moteur thermique qui fonctionne avec un gaz diatomique dans le sens A → B → C.
On donne le diagramme PV (figure \ref{lemoteur}) d'un moteur thermique qui fonctionne avec un gaz diatomique dans le sens A → B → C.
\begin{enumerate}
\item Calculez les caractéristiques manquantes.
\item Déterminez \(\Delta U\), \(A\) et \(Q\) pour chaque étape.

99
Annexe-Exercices/Annexe-Exercices.tex.bak
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@ -2939,6 +2939,94 @@ Les valeurs des c\oe fficients de dilatation thermique sont celles du tableau \r
\end{solos}
\end{exos}
\begin{exos}
Un cycle Diesel se déroule en principe de la façon suivante. Un gaz parfait diatomique est initialement dans l'état \(p_1=\SI{1}{\bar}\), \(V_1=\SI{1000}{\centi\metre\cubed}\) et \(T_1=\SI{333,3}{\kelvin}\) et on lui fait subir quatre transformations successives~:
\begin{description}
\item[Compression adiabatique] jusqu'à l'état \(p_2=\,?\), \(V_2=\SI{70}{\centi\metre\cubed}\) et \(T_2=\SI{965,6}{\kelvin}\).
\item[Échauffement à pression constante] jusqu'à l'état \(p_3=p_2\), \(V_3=\SI{140}{\centi\metre\cubed}\) et \(T_3=\,?\).
\item[Détente adiabatique] jusqu'à l'état \(p_4=\,?\), \(V_3=V_1\) et \(T_3=\SI{879,6}{\kelvin}\).
\item[Refroidissement à volume constant] jusqu'à l'état initial.
\smallskip
On demande de~:
\begin{enumerate}
\item Calculer nR, \(p_2\), \(T_3\) et \(p_4\).
\item Esquisser le diagramme PV du cycle.
\item Calculer pour chaque transformation \(\Delta U\), \(A\) et \(Q\).
\item Établir le bilan du cycle.
\item Calculer le rendement.
\end{enumerate}
\end{description}
\begin{solos}
Voici l'analyse du cycle.
\begin{enumerate}
\item Les grandeurs manquantes.
\begin{align*}
nR&=\frac{p_1\cdot V_1}{T_1}=\frac{10^5\cdot 1000\cdot 10^{-6}}{333,3}=0,3\\
p_2&=\frac{nR\cdot T_2}{V_2}=\frac{0,3\cdot 965,6}{70\cdot 10^{-6}}=\SI{41,4}{\bar}\\
T_3&=\frac{p_3\cdot V_3}{nR}=\frac{41,4\cdot 10^5\cdot 140\dot 10^{-6}}{0,3}\\
&=\SI{1931,2}{\kelvin}\\
p_4&=\frac{nR\cdot T_4}{V_4}=\frac{0,3\cdot 879,6}{1000\cdot 10^{-6}}=\SI{2,6}{\bar}
\end{align*}
\item Le diagramme PV est celui de la figure \ref{exos:diesel}.
\begin{figure}
\def\svgwidth{7cm}
\begin{center}
\input{Annexe-Exercices/Images/Diesel.ps_tex}
\end{center}
\caption{Le diagramme PV\label{exos:diesel}}
\end{figure}
\item Les grandeurs des transformations.
\begin{description}
\item[1-2] Adiabatique
\begin{align*}
\Delta U&=\frac{i}{2}\cdot \Delta (pV)\\
&=\frac{5}{2}\cdot (41,4\cdot 10^5\cdot 70\cdot 10^{-6}\\
&-10^5\cdot 1000\cdot 10^{-6})=\SI{474,5}{\joule}\\
A&=-\Delta U=\SI{-474,5}{\joule}\\
Q&=0
\end{align*}
\item[2-3] Isobare
\begin{align*}
\Delta U&=\frac{i}{2}\cdot nR\cdot \Delta T\\
&=\frac{5}{2}\cdot 0,3\cdot (1931,2-965,6)=\SI{724,2}{\joule}\\
A&=p_2\cdot (V_3-V_2)\\
&=41,4\cdot 10^5\cdot (140\cdot 10^{-6}-70\cdot 10^{-6})\\
&=\SI{289,7}{\joule}\\
Q&=\Delta U+A=724,2+289,7=\SI{1013,9}{\joule}
\end{align*}
\item[3-4] Adiabatique
\begin{align*}
\Delta U&=\frac{i}{2}\cdot \Delta (pV)\\
&=\frac{5}{2}\cdot (p_4\cdot V_4-p_3\cdot V_3)\\
&=\SI{789,3}{\joule}\\
A&=-\Delta U=\SI{-789,3}{\joule}\\
Q&=0
\end{align*}
\item[4-1] Isochore
\begin{align*}
\Delta U&=\frac{i}{2}\cdot nR\cdot \Delta T=\frac{i}{2}\cdot nR (T_1-T_4)\\
&=\frac{5}{2}\cdot 0,3\cdot (333,3-879,6)\\
&=\SI{-409,7}{\joule}\\
A&=0\\
Q&=\Delta U=\SI{-409,7}{\joule}
\end{align*}
\end{description}
\item Le bilan du cycle.
Chaleur reçue par le moteur~: \[\sum Q>0=\SI{1014}{\joule}\].
Chaleur rejetée par le moteur~: \[\sum q<0=\SI{-409,7}{\joule}\].
Travail fourni par le moteur~: \[\sum A=\SI{604,5}{\joule}\].
\item Le rendement.
\[\eta=\frac{A}{\sum Q>0}=\frac{604,5}{1014}=60\%\]
\end{enumerate}
\end{solos}
\end{exos}
\begin{exos}
\begin{figure}[t]
\centering
@ -2947,7 +3035,7 @@ Les valeurs des c\oe fficients de dilatation thermique sont celles du tableau \r
\def\svgwidth{0.9\columnwidth}
\input{Annexe-Exercices/Images/Lemoteur.ps_tex}
\end{figure}
On donne le diagramme PV d'un moteur thermique qui fonctionne avec un gaz diatomique dans le sens A → B → C.
On donne le diagramme PV (figure \ref{lemoteur}) d'un moteur thermique qui fonctionne avec un gaz diatomique dans le sens A → B → C.
\begin{enumerate}
\item Calculez les caractéristiques manquantes.
\item Déterminez \(\Delta U\), \(A\) et \(Q\) pour chaque étape.
@ -2994,7 +3082,14 @@ Les valeurs des c\oe fficients de dilatation thermique sont celles du tableau \r
&=\SI{-5600}{\joule}
\end{align*}
\end{description}
\item
\item On peut calculer le travail total de deux manières~:
\begin{align*}
A_{tot}&=\sum A=0+3218,9-1600=\SI{1618,9}{\joule}\\
A_{tot}&=\sum Q=4000+3218,9-5600\\
&=\SI{1618,9}{\joule}
\end{align*}
\item La puissance est simplement~:
\[P=\frac{A}{t}=\frac{3000\cdot 1618,9}{60}=\SI{80985}{\watt}\]
\end{enumerate}
\end{solos}
\end{exos}

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\end{Solution OS}
\begin{Solution OS}{38}
Voici l'analyse du cycle.
\begin{enumerate}
\item Les grandeurs manquantes.
\begin{align*}
nR&=\frac{p_1\cdot V_1}{T_1}=\frac{10^5\cdot 1000\cdot 10^{-6}}{333,3}=0,3\\
p_2&=\frac{nR\cdot T_2}{V_2}=\frac{0,3\cdot 965,6}{70\cdot 10^{-6}}=\SI{41,4}{\bar}\\
T_3&=\frac{p_3\cdot V_3}{nR}=\frac{41,4\cdot 10^5\cdot 140\dot 10^{-6}}{0,3}\\
&=\SI{1931,2}{\kelvin}\\
p_4&=\frac{nR\cdot T_4}{V_4}=\frac{0,3\cdot 879,6}{1000\cdot 10^{-6}}=\SI{2,6}{\bar}
\end{align*}
\item Le diagramme PV est celui de la figure \ref{exos:diesel}.
\begin{figure}
\def\svgwidth{7cm}
\begin{center}
\input{Annexe-Exercices/Images/Diesel.ps_tex}
\end{center}
\caption{Le diagramme PV\label{exos:diesel}}
\end{figure}
\item Les grandeurs des transformations.
\begin{description}
\item[1-2] Adiabatique
\begin{align*}
\Delta U&=\frac{i}{2}\cdot \Delta (pV)\\
&=\frac{5}{2}\cdot (41,4\cdot 10^5\cdot 70\cdot 10^{-6}\\
&-10^5\cdot 1000\cdot 10^{-6})=\SI{474,5}{\joule}\\
A&=-\Delta U=\SI{-474,5}{\joule}\\
Q&=0
\end{align*}
\item[2-3] Isobare
\begin{align*}
\Delta U&=\frac{i}{2}\cdot nR\cdot \Delta T\\
&=\frac{5}{2}\cdot 0,3\cdot (1931,2-965,6)=\SI{724,2}{\joule}\\
A&=p_2\cdot (V_3-V_2)\\
&=41,4\cdot 10^5\cdot (140\cdot 10^{-6}-70\cdot 10^{-6})\\
&=\SI{289,7}{\joule}\\
Q&=\Delta U+A=724,2+289,7=\SI{1013,9}{\joule}
\end{align*}
\item[3-4] Adiabatique
\begin{align*}
\Delta U&=\frac{i}{2}\cdot \Delta (pV)\\
&=\frac{5}{2}\cdot (p_4\cdot V_4-p_3\cdot V_3)\\
&=\SI{789,3}{\joule}\\
A&=-\Delta U=\SI{-789,3}{\joule}\\
Q&=0
\end{align*}
\item[4-1] Isochore
\begin{align*}
\Delta U&=\frac{i}{2}\cdot nR\cdot \Delta T=\frac{i}{2}\cdot nR (T_1-T_4)\\
&=\frac{5}{2}\cdot 0,3\cdot (333,3-879,6)\\
&=\SI{-409,7}{\joule}\\
A&=0\\
Q&=\Delta U=\SI{-409,7}{\joule}
\end{align*}
\end{description}
\item Le bilan du cycle.
Chaleur reçue par le moteur~: \[\sum Q>0=\SI{1014}{\joule}\].
Chaleur rejetée par le moteur~: \[\sum q<0=\SI{-409,7}{\joule}\].
Travail fourni par le moteur~: \[\sum A=\SI{604,5}{\joule}\].
\item Le rendement.
\[\eta=\frac{A}{\sum Q>0}=\frac{604,5}{1014}=60\%\]
\end{enumerate}
\end{Solution OS}
\begin{Solution OS}{39}
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
@ -1221,7 +1290,7 @@
\end{enumerate}
\end{Solution OS}
\begin{Solution OS}{39}
\begin{Solution OS}{40}
Le diagramme PV du cycle est présenté à la figure \ref{climatiseur}.
\begin{figure}[t]
\centering
@ -1341,7 +1410,7 @@
\end{enumerate}
\end{Solution OS}
\begin{Solution OS}{40}
\begin{Solution OS}{41}
Procédons simplement au calcul de l'incertitude absolue.
\smallskip
@ -1389,7 +1458,7 @@
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}{41}
\begin{Solution OS}{42}
\dots
\end{Solution OS}