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Report d'un exo moteur

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67
Annexe-Exercices/Annexe-Exercices.tex
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@ -2939,6 +2939,73 @@ Les valeurs des c\oe fficients de dilatation thermique sont celles du tableau \r
\end{solos}
\end{exos}
\begin{exos}
\begin{figure}[t]
\centering
\caption{Diagramme PV du moteur.\label{lemoteur}}
\centering
\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.
\begin{enumerate}
\item Calculez les caractéristiques manquantes.
\item Déterminez \(\Delta U\), \(A\) et \(Q\) pour chaque étape.
\item Calculez le travail que ce moteur fournit à chaque cycle.
\item Calculez sa puissance s'il effectue 3000 cycles par minute.
Indication. Pour une transformation isotherme, le travail se calcule par~:
\[A=nR\cdot T\cdot \ln(V_2/V_1)\]
\end{enumerate}
\begin{solos}
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
\end{equation*}
\begin{enumerate}
\item Les grandeurs manquantes.
\begin{align*}
T_1&=\frac{p_1\cdot V_1}{nR}=\frac{2\cdot 10^5\cdot 2\cdot 10^{-3}}{1,33}=\SI{300}{\kelvin}\\
V_C&=\frac{nR\cdot T_C}{p_C}=\frac{1,33\cdot 1500}{2\cdot 10^5}=\SI{10}{\deci\metre\cubed}
\end{align*}
\item Les grandeurs des transformations.
\begin{description}
\item[AB] Isochore
\begin{align*}
\Delta U&=\frac{i}{2}\cdot nR\cdot \Delta T\\
&=\frac{5}{2}\cdot 1,33\cdot (1500-300)=\SI{4000}{\joule}\\
A&=0 \\
Q&= \Delta U+A=\Delta U=\SI{4000}{\joule}
\end{align*}
\item[BC] Isotherme
\begin{align*}
\Delta U&=0 \\
A&=nR\cdot T\cdot \ln(V_C/V_B)\\
&=1,33\cdot 1500\cdot \ln(10/2)=\SI{3218,9}{\joule}\\
Q&= \Delta U+A=A=\SI{3218,9}{\joule}
\end{align*}
\item[CA] Isobare
\begin{align*}
\Delta U&=\frac{i}{2}\cdot nR\cdot \Delta T\\
&=\frac{5}{2}\cdot 1,33\cdot (300-1500)=\SI{-4000}{\joule}\\
A&= p\cdot \Delta V\\
&=2\cdot 10^5\cdot (2\cdot 10^{-3}-10\cdot 10^{-3})\\
&=\SI{-1600}{\joule}\\
Q&= \Delta U+A=-4000-1600\\
&=\SI{-5600}{\joule}
\end{align*}
\end{description}
\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}
\begin{exos}
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 \(nR=10\).
\begin{description}

64
Annexe-Exercices/Annexe-Exercices.tex.bak
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@ -2939,6 +2939,66 @@ Les valeurs des c\oe fficients de dilatation thermique sont celles du tableau \r
\end{solos}
\end{exos}
\begin{exos}
\begin{figure}[t]
\centering
\caption{Diagramme PV du moteur.\label{lemoteur}}
\centering
\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.
\begin{enumerate}
\item Calculez les caractéristiques manquantes.
\item Déterminez \(\Delta U\), \(A\) et \(Q\) pour chaque étape.
\item Calculez le travail que ce moteur fournit à chaque cycle.
\item Calculez sa puissance s'il effectue 3000 cycles par minute.
Indication. Pour une transformation isotherme, le travail se calcule par~:
\[A=nR\cdot T\cdot \ln(V_2/V_1)\]
\end{enumerate}
\begin{solos}
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
\end{equation*}
\begin{enumerate}
\item Les grandeurs manquantes.
\begin{align*}
T_1&=\frac{p_1\cdot V_1}{nR}=\frac{2\cdot 10^5\cdot 2\cdot 10^{-3}}{1,33}=\SI{300}{\kelvin}\\
V_C&=\frac{nR\cdot T_C}{p_C}=\frac{1,33\cdot 1500}{2\cdot 10^5}=\SI{10}{\deci\metre\cubed}
\end{align*}
\item Les grandeurs des transformations.
\begin{description}
\item[AB] Isochore
\begin{align*}
\Delta U&=\frac{i}{2}\cdot nR\cdot \Delta T\\
&=\frac{5}{2}\cdot 1,33\cdot (1500-300)=\SI{4000}{\joule}\\
A&=0 \\
Q&= \Delta U+A=\Delta U=\SI{4000}{\joule}
\end{align*}
\item[BC] Isotherme
\begin{align*}
\Delta U&=0 \\
A&=nR\cdot T\cdot \ln(V_C/V_B)\\
&=1,33\cdot 1500\cdot \ln(10/2)=\SI{3218,9}{\joule}\\
Q&= \Delta U+A=A=\SI{3218,9}{\joule}
\end{align*}
\item[CA] Isobare
\begin{align*}
\Delta U&=\frac{i}{2}\cdot nR\cdot \Delta T\\
&=\frac{5}{2}\cdot 1,33\cdot (300-1500)=\SI{-4000}{\joule}\\
A&= p\cdot \Delta V\\
&=2\cdot 10^5\cdot (2\cdot 10^{-3}-10\cdot 10^{-3})\\
&=\SI{-1600}{\joule}\\
Q&= \Delta U+A=-4000-1600\\
&=\SI{-5600}{\joule}
\end{align*}
\end{description}
\item
\end{enumerate}
\end{solos}
\end{exos}
\begin{exos}
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 \(nR=10\).
\begin{description}
@ -3070,7 +3130,9 @@ Les valeurs des c\oe fficients de dilatation thermique sont celles du tableau \r
&=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
&=-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{solos}

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Annexe-Exercices/Images/Climatiseur.ps_tex.bak
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Annexe-Exercices/Images/Lemoteur.ps_tex Normal file
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\end{Solution OS}
\begin{Solution OS}{38}
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
\end{equation*}
\begin{enumerate}
\item Les grandeurs manquantes.
\begin{align*}
T_1&=\frac{p_1\cdot V_1}{nR}=\frac{2\cdot 10^5\cdot 2\cdot 10^{-3}}{1,33}=\SI{300}{\kelvin}\\
V_C&=\frac{nR\cdot T_C}{p_C}=\frac{1,33\cdot 1500}{2\cdot 10^5}=\SI{10}{\deci\metre\cubed}
\end{align*}
\item Les grandeurs des transformations.
\begin{description}
\item[AB] Isochore
\begin{align*}
\Delta U&=\frac{i}{2}\cdot nR\cdot \Delta T\\
&=\frac{5}{2}\cdot 1,33\cdot (1500-300)=\SI{4000}{\joule}\\
A&=0 \\
Q&= \Delta U+A=\Delta U=\SI{4000}{\joule}
\end{align*}
\item[BC] Isotherme
\begin{align*}
\Delta U&=0 \\
A&=nR\cdot T\cdot \ln(V_C/V_B)\\
&=1,33\cdot 1500\cdot \ln(10/2)=\SI{3218,9}{\joule}\\
Q&= \Delta U+A=A=\SI{3218,9}{\joule}
\end{align*}
\item[CA] Isobare
\begin{align*}
\Delta U&=\frac{i}{2}\cdot nR\cdot \Delta T\\
&=\frac{5}{2}\cdot 1,33\cdot (300-1500)=\SI{-4000}{\joule}\\
A&= p\cdot \Delta V\\
&=2\cdot 10^5\cdot (2\cdot 10^{-3}-10\cdot 10^{-3})\\
&=\SI{-1600}{\joule}\\
Q&= \Delta U+A=-4000-1600\\
&=\SI{-5600}{\joule}
\end{align*}
\end{description}
\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{Solution OS}
\begin{Solution OS}{39}
Le diagramme PV du cycle est présenté à la figure \ref{climatiseur}.
\begin{figure}[t]
\centering
@ -1292,7 +1341,7 @@
\end{enumerate}
\end{Solution OS}
\begin{Solution OS}{39}
\begin{Solution OS}{40}
Procédons simplement au calcul de l'incertitude absolue.
\smallskip
@ -1340,7 +1389,7 @@
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}{40}
\begin{Solution OS}{41}
\dots
\end{Solution OS}