Commit 49249d44 by Olivier

With thanks to the students who worked out the problem with me per videocall today.
 \renewcommand{\lastedityear}{2020} \renewcommand{\lasteditmonth}{05} \renewcommand{\lasteditday}{08} \renewcommand{\lasteditday}{15} \renewcommand{\numberofthischapter}{3} \renewcommand{\titleofthischapter}{\namechapterthreetitle} \atstartofexercises ... ... @@ -395,10 +395,11 @@ Angular momentum balance through an arbitrary volume: \tab 3) $\dot W_\text{rotor} = \SI{0}{\watt}$; \tab 4) $\omega = \SI{143,2}{rpm}$ ($F_\net = \SI{0}{\newton}$); \tab 5) $\dot W_\text{rotor} = \SI{0}{\watt}$ again; \tab\tab 6) $\dot W_\text{rotor,\ max} = \SI{1,963}{\kilo\watt}$ @ $V_\text{blade, optimal} = \frac{1}{3} V_\text{water jet}$. \tab 6) $\dot W_\text{rotor,\ max} = \SI{1,963}{\kilo\watt}$ @ $V_\text{blade, optimal} = \frac{1}{3} V_\text{water jet}$. \item [\ref{exo_snow_plow}]% \tab 1) $\dot m = \SI{2500}{\kilogram\per\second}$; $F_{\net\ x} = \SI{+10,07}{\kilo\newton}$, $F_{\net\ z} = \SI{+12,63}{\kilo\newton}$ (force on blade is opposite); \tab 1) $\dot m = \SI{2500}{\kilogram\per\second}$; $F_{\net\ x} = \SI{+10,07}{\kilo\newton}$, $F_{\net\ z} = \SI{-12,63}{\kilo\newton}$ (force on blade is opposite); \tab 2) $\dot W = \vec F_\net \cdot \vec V_\text{plow} = F_{\net x} |V_1| = \SI{69,94}{\kilo\watt}$ \tab 3) $\dot W_2 = \num{1.1}^3 \dot W$ (\SI{+33}{\percent}) \item [\ref{exo_pressure_losses_pipe_flow}]% \tab $V_\text{center} = \num{1,2245} U$%; $F_{\text{net} x} = \SI{+393}{\newton}$ (positive!) \item [\ref{exo_drag_cylindrical_profile}]% ... ...