Commit a2dd1a26 by Olivier

 ... ... @@ -89,7 +89,7 @@ Balance of energy in a considered volume with steady flow: The pipe itself and the valve, together, induce a pressure loss which can be quantified using the dimensionless \vocab{loss coefficient} $K_\text{valve}$ (we later will later encounter it as eq.~\ref{eq_def_loss_coeff} p.\pageref{eq_def_loss_coeff}). With this tool, the pressure loss is related to the average incoming speed $V_\text{incoming}$ as: \begin{IEEEeqnarray}{rCcCl} K_\text{valve} &\equiv& \frac{|\Delta p_\text{valve}|}{\frac{1}{2} \rho V_\text{incoming}^2} &=& \num{2,6} K_\text{valve} &\equiv& \frac{|\Delta p_\text{valve}|}{\frac{1}{2} \rho V_\text{incoming}^2} &=& \num{2,6}\label{eq_k_valve} \end{IEEEeqnarray} % Note: K = 2,6 is guesstimate: 2 for swing check valve (from White p.401) + {f L/D = 0.05 * 3 / 0.25 = 0,6} ... ... @@ -213,13 +213,54 @@ Balance of energy in a considered volume with steady flow: \end{enumerate} \begin{comment} \item [\ref{exo_water_jet}]% \tab 1) $F_\text{net on water} = \SI{-120}{\newton}$; \tab 2) $\vec F_\text{water/wall} = -\vec F_\text{net on water}$; \tab 3) $F_\text{net on water} = \SI{-7,5}{\newton}$; \end{comment} \clearpage \subsubsection*{Answers} \startofanswers \begin{enumerate} \item p.~\pageref{exo_pipe_expansion} \begin{enumerate} \item $\dot m = \SI{7,53}{\kilogram\per\second}$; $\dot \vol = \SI{7,53}{\liter\per\second}$ \itme $V_2 = \SI{0,375}{\metre\per\second}$ \item $\Delta p_{1\to 2} = \SI{+1054}{\pascal}$ \item $\Delta p_{3\to 4} = \SI{+4218}{\pascal}$ \item $\Delta p_{5\to 6} = \SI{+1,29}{\pascal}$ \end{enumerate} \item p.~\pageref{exo_pipe_with_losses} \begin{enumerate} \item $V_2 = V_1 = \SI{1,5}{\metre\per\second}$ by application of the mass balance equation; although a mis-application of the energy equation would suggest otherwise \item With eq.~\ref{eq_k_valve}, $\Delta p_\text{valve} = \SI{-2925}{\pascal}$ \item $\dot W_\text{injection} = \SI{-215,37}{\watt}$ \item With eq.~\ref{eq_sfee}, $\Delta T = \SI{+7}{\milli\kelvin}$ (very small!) \end{enumerate} \item p.~\pageref{exo_combustor} \begin{enumerate} \item $\dot \vol_1 = \SI{0,0759}{\metre\cubed\per\second}$ \& $\dot \vol_2 = \SI{0,1078}{\metre\cubed\per\second}$ (there is no volume balance equation!) \item $\dot Q = \SI{+261}{\kilo\watt}$ (using $V_2 = \SI{50}{\metre\per\second}$) \item $F_\net = \SI{+19,25}{\newton}$ (in flow-wise direction) \end{enumerate} \item p.~\pageref{exo_water_jet} \begin{enumerate} \item $F_\text{net on water} = \SI{-120}{\newton}$ \item $\vec F_\text{water/wall} = -\vec F_\text{net on water}$ \item $F_\text{net on water} = \SI{-7,5}{\newton}$ \item $\dot W_\text{wall} = \SI{112,5}{\watt}$ \end{enumerate} \item p.~\pageref{exo_high_speed_gas_flow} \begin{enumerate} \item With eq.~\ref{eq_sfee}, $V_2 = \SI{417,2}{\metre\per\second}$, and so $\dot m_2 = \dot m = \SI{14,1}{\kilogram\per\second}$ \item $\ma_1 = \num{0}$ (zero velocity) and $\ma = \num{1}$ (a classical feature of compressible flow expansions) \item $F_{\net 1 \to 2} = \SI{+5,882}{\kilo\newton}$ (flow-wise) \item $e_{k2} = \SI{87,03}{\kilo\joule\per\kilogram}$ \item With eq.~\ref{eq_sfee}, $V_3 = \SI{638,71}{\metre\per\second}$ (you may then calculate $\rho_3$ and obtain $A_3 > A_2$ even though $V_3 > V_2$, a classical feature of supersonic flows) \item $\ma_3 = \num{1,795}$ \item $F_{\net 2 \to 3} = \SI{+3,137}{\kilo\newton}$ \item $e_{k3} = \SI{204,61}{\kilo\joule\per\kilogram}$ \item $V_4 = \SI{405}{\metre\per\second}$ \item $\rho_4 = \SI{0,7952}{\kilogram\per\second}$ and so $A_4 = \SI{0,0438}{\metre\squared}$ \item $\ma_4 = \num{0,96}$ \item $F_{\net 3 \to 4} = \SI{-3,309}{\kilo\newton}$ (so, against the flow direction) \item $e_{k4} = \SI{82,01}{\kilo\joule\per\kilogram}$ \end{enumerate} \end{enumerate} \atendofexercises