Commit a14e6d3e authored by Olivier's avatar Olivier

Exercises 8: added answers

parent 6df20790
......@@ -38,6 +38,7 @@ In homogeneous, isotropic, fully-developed turbulence, the following relationshi
\clearpage
\subsubsection{Hypothetical flow}
\label{exo_hypothetical_flow}
\wherefrom{Non-examinable. From De Nevers \cite{denevers2004} Ex 18.1}
We imagine a turbulent flow described at some point with the equations (in \si{\metre\per\second})
......@@ -54,6 +55,7 @@ In homogeneous, isotropic, fully-developed turbulence, the following relationshi
\subsubsection{Turbulent channel flow}
\label{exo_turbulent_channel_flow}
\wherefrom{Non-examinable. From De Nevers \cite{denevers2004} Ex 18.2}
A wind tunnel carries air through a channel which is \SI{1}{\metre} wide and \SI{0.24}{\metre} high. The average velocity is \SI{0.82}{\metre\per\second}. The pressure drop caused by both friction on the walls and turbulent dissipation is measured at \SI{-0,0286}{\pascal\per\metre}.
......@@ -88,6 +90,7 @@ In homogeneous, isotropic, fully-developed turbulence, the following relationshi
\end{comment}
\subsubsection{Cumulus cloud}
\label{exo_cumulus_cloud}
\wherefrom{Non-examinable. From Tennekes \& Lumley \cite{tennekesetal1972} P1.1}
A cumulus cloud (one of those “fluffy” summer clouds, figure~\ref{fig_fluffy_cloud}) has roughly the size of a sphere of diameter $D = \SI{50}{\meter}$. To a good approximation, it features isotropic, homogeneous, fully-developed turbulence. The largest-scale air currents in the cloud reach a maximum velocity $V = \SI{3}{\metre\per\second}$.
......@@ -105,6 +108,7 @@ In homogeneous, isotropic, fully-developed turbulence, the following relationshi
\end{enumerate}
\subsubsection{Reactor tank}
\label{exo_reactor_tank}
\wherefrom{Non-examinable. \cczero \oc}
A tank used to store chemical reactants has roughly the size of a cube of side length $L = \SI{2}{\metre}$ (figure~\ref{fig_tank_agitator}). The tank is filled with a water-like liquid and vigorously stirred with a large agitator propeller for a prolonged amount of time. The propeller induces a maximum fluid velocity of \SI{1,5}{\metre\per\second}.
......@@ -126,7 +130,39 @@ In homogeneous, isotropic, fully-developed turbulence, the following relationshi
\item What do you expect the new computation time to be?
\end{enumerate}
\clearpage
\subsubsection*{Answers}
\startofanswers
\begin{enumerate}
\item p.~\pageref{exo_hypothetical_flow}
\begin{enumerate}
\item $\overline u = \SI{10}{\metre\per\second}$
\item $u' = \sin (t)$
\item $\overline u' = \SI{0}{\metre\per\second}$ (as always)
\item $I_x = \SI{7,07}{\percent}$
\item $I = \SI{4,08}{\percent}$
\item $k = \SI{0,25}{\joule\per\kilogram}$
\end{enumerate}
\item p.~\pageref{exo_turbulent_channel_flow}
\begin{enumerate}
\item $\dot e_{m \text{main}} = \SI{0,336}{\joule\per\kilogram}$
\item $\epsilon = \SI{0,0191}{\watt\per\kilogram}$
\item $\dot T = \SI{0,02}{\milli\kelvin\per\second}$
\item $k = \SI{5,76}{\milli\joule\per\kilogram}$
\end{enumerate}
\item p.~\pageref{exo_cumulus_cloud}
\begin{enumerate}
\item Since $\re_\Lambda \equiv \num{6,1e3}$, $\eta \approx \SI{0,2}{\milli\metre}$
\item $\epsilon \approx \SI{1,147}{\watt\per\kilogram}$ \& $\dot W_\epsilon \approx \SI{92}{\kilo\watt}$
\item $\eta_2 = \num{0,6} \eta_1$, $\epsilon_2 = 8 \epsilon_1$, $\dot W_{\epsilon 2} = \num{64} \dot W_{\epsilon 1}$
\end{enumerate}
\item p.\pageref{exo_reactor_tank}
\begin{enumerate}
\item $\eta \approx \SI{0,023}{\milli\metre}$ (width of human hair)
\item $\epsilon \approx \SI{3,4}{\watt\per\kilogram}$
\item Increase resolution in all three directions according to $\eta$, and the time resolution according to $t_\eta$: the computation time increases by a factor \num{6,7}.
\end{enumerate}
\end{enumerate}
\atendofexercises
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