Commit 36ce996b authored by Olivier's avatar Olivier

Exercises 3: fix notation of answers

parent 11bebf52
\renewcommand{\lastedityear}{2019}
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\atstartofexercises
......@@ -375,10 +375,10 @@ Angular momentum balance through an arbitrary volume:
\tab 2) $\vec F_\net = \left(\num{-0,5681};\num{-1,2184}\right) \si{\newton}$
\tab 3) The force is quadrupled.
\item [\ref{exo_exhaust_gas_deflector}]%
\tab 1) $F_{\net\ x} = \SI{-9,532}{\kilo\newton}$ \& $F_{\net\ y} = \SI{+1,479}{\kilo\newton}$ : $F_\net = \SI{9,646}{\kilo\newton}$;
\tab 3) $F_\text{net 2} = \SI{8,525}{\kilo\newton}$.
\tab 1) $\vec F_\net = \left(\num{-9,532}; \num{+1,479}\right) \si{\kilo\newton}$ : $||\vec F_\net|| = \SI{9,646}{\kilo\newton}$ (force on ground is opposite: $\vec F_\text{fluid on pipe} = -\vec F_\text{pipe on fluid}$);
\tab 3) $||\vec F_\text{net 2}|| = \SI{8,525}{\kilo\newton}$.
\item [\ref{exo_pelton_turbine}]%
\tab 1) $F_\net = \SI{-883,6}{\newton}$;
\tab 1) $F_\net = \left(\num{-883,6}; 0\right) \si{\newton}$;
\tab 2) $M_{\net\, X} = |F_\net| R = \SI{883,6}{\newton\metre}$;
\tab 3) $\dot W_\text{rotor} = \SI{0}{\watt}$;
\tab 4) $\omega = \SI{143,2}{rpm}$ ($F_\net = \SI{0}{\newton}$);
......@@ -399,7 +399,7 @@ Angular momentum balance through an arbitrary volume:
\tab 1) Work eq.~\ref{eq_rtt_angularmom} down to scalar equation (in $y$-direction), solve for $\theta$: $\theta = \SI{123,1}{\degree}$.
\tab 2) There are multiple solutions which allow both moment and force equations to be solved at the same time. $r_\C$ can be shortened, the flow in C can be split into forward and rearward components, or tilted downwards etc. Reductions in $\dot m_\B$ or $V_\C$ are also possible, but quantifying them requires solving both equations at once.
\item [\ref{exo_pressure_losses_pipe_flow}]%
\tab $V_\text{center} = \num{1,2245} U$; $F_\net = \SI{+393}{\newton}$ (positive!)
\tab $V_\text{center} = \num{1,2245} U$; $F_{\text{net} x} = \SI{+393}{\newton}$ (positive!)
\item [\ref{exo_thrust_reverser}]%
\tab $\dot m_\text{cold} = \SI{297}{\kilogram\per\second}$, $\dot m_\text{hot} = \SI{59,4}{\kilogram\per\second}$;
\tab $F_\text{cold flow, normal, bench \& runway} = \SI{+74,25}{\kilo\newton}$,\\ $F_\text{hot flow, normal \& reverse, bench \& runway} = \SI{+8,316}{\kilo\newton}$;
......
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