Commit a2dd1a26 authored by Olivier's avatar Olivier
Browse files

Exercises 2: added answers

parent 1f971fc4
......@@ -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:
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}
% 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:
\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) $F_\text{net on water} = \SI{-7,5}{\newton}$;
\item p.~\pageref{exo_pipe_expansion}
\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 $\Delta p_{5\to 6} = \SI{+1,29}{\pascal}$
\item p.~\pageref{exo_pipe_with_losses}
\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!)
\item p.~\pageref{exo_combustor}
\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)
\item p.~\pageref{exo_water_jet}
\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 $\dot W_\text{wall} = \SI{112,5}{\watt}$
\item p.~\pageref{exo_high_speed_gas_flow}
\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 $e_{k4} = \SI{82,01}{\kilo\joule\per\kilogram}$
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