Commit bfc8e12a authored by Olivier's avatar Olivier

Exercises 2: proof-read, exercises now workable

parent 54a07273
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\atstartofexercises
\fluidmechexercisestitle
......@@ -38,8 +38,6 @@ Balance of energy in a considered volume with steady flow:
\end{boiboiboite}
\mecafluboxtmp
%%%%
\subsubsection{Pipe expansion without losses}
\wherefrom{\cczero \oc}
......@@ -69,12 +67,13 @@ Balance of energy in a considered volume with steady flow:
\wherefrom{\cczero \oc}
\label{exo_pipe_with_losses}
Water flows in a long pipe which has constant diameter; a valve is installed in the middle of the pipe length. Water comes in the pipe with a uniform velocity of \SI{1,5}{\metre\per\second} and the pipe diameter is \SI{250}{\milli\metre}. The heat capacity of the water is \SI{1}{\kilo\joule\per\kilogram\per\kelvin}.
Water flows in a long pipe which has constant diameter; a valve is installed in the middle of the pipe length. Water comes in the pipe with a uniform velocity of \SI{1,5}{\metre\per\second} and the pipe diameter is \SI{250}{\milli\metre}.
The pipe itself and the valve, together, induce a pressure loss which can be quantified using the \vocab{loss coefficient} $K_\text{valve}$ (we will study this 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:
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} &=& 3
K_\text{valve} &\equiv& \frac{|\Delta p_\text{valve}|}{\frac{1}{2} \rho V_\text{incoming}^2} &=& \num{2,6}
\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}
\begin{enumerate}
\item What is the outlet velocity of the water?\\
......@@ -94,15 +93,14 @@ Balance of energy in a considered volume with steady flow:
The conditions at inlet are as follows:
\begin{itemize}
\item Air mass flow: \SI{0,2}{\kilogram\per\second};
\item Air properties: \SI{20}{\bar}, \SI{240}{\degreeCelsius}, \SI{12}{\metre\per\second}
\item Fuel mass flow: \SI{1}{\milli\gram\per\second}.
\item Air mass flow: \SI{0,5}{\kilogram\per\second};
\item Air properties: \SI{25}{\bar}, \SI{1050}{\degreeCelsius}, \SI{12}{\metre\per\second}
\item Fuel mass flow: \SI{5}{\gram\per\second}.
\end{itemize}
% Fuel mass flow rough calculation: \dot Q = (\dot m * c_p * \Delta T)_air = 250 kW
% \dot Q / c_comb = 0,005 kg/s
At the outlet the conditions are as follows:
\begin{itemize}
\item Gas properties: \SI{20}{\bar}, \SI{1800}{\degreeCelsius}
\end{itemize}
At the outlet, the hot gases have pressure \SI{24,5}{\bar} and temperature \SI{1550}{\degreeCelsius}.
We consider that the air and gas keep the same thermodynamic properties throughout ($c_\text{p} = \SI{1050}{\joule\per\kilogram}$)
......@@ -118,7 +116,7 @@ Balance of energy in a considered volume with steady flow:
\wherefrom{White \smallcite{white2008} Ex3.9}
\label{exo_water_jet}
A horizontal water jet hits a vertical wall and is split in two symmetrical vertical flows (\cref{fig_water_wall}). The water nozzle has a~\SI{3}{\centi\metre\squared} cross-sectional area, and the water speed at the nozzle outlet is $V_\text{jet} = \SI{20}{\metre\per\second}$.\\
A horizontal water jet hits a vertical wall and is split in two symmetrical vertical flows (\cref{fig_water_wall}). The water nozzle has a~\SI{3}{\centi\metre\squared} cross-sectional area, and the water speed at the nozzle outlet is $V_\text{jet} = \SI{20}{\metre\per\second}$.
\begin{figure}
\begin{center}
\includegraphics[width=7.5cm]{nozzle_plate}
......@@ -132,27 +130,36 @@ Balance of energy in a considered volume with steady flow:
\item What is the net force exerted on the water by the wall?
\end{enumerate}
Now, the wall moves longitudinally in the same direction as the water jet, with a speed $V_\text{wall} = \SI{15}{\metre\per\second}$. This conceptual setup allows us to approach the case where water acts on the blades of a turbine.
Now, the wall moves longitudinally in the same direction as the water jet, with a speed $V_\text{wall} = \SI{15}{\metre\per\second}$.\\
(This may be because the wall is the back of a van traveling away from the jet. This is a crude conceptual setup, but it allows us to approach the case where water acts on the blades of a turbine.)
\begin{enumerate}
\shift{1}
\shift{2}
\item What is the new force exerted by the water on the wall?
\item How would the force be modified if the volume flow was kept constant, but the diameter of the nozzle was reduced? (briefly justify your answer, e.g. in 30 words or less)
\item What is the mechanical power transmitted to the wall?
\item How would the power be modified if the volume flow was kept constant, but the diameter of the nozzle was reduced? (briefly justify your answer, e.g. in 30 words or less)
\end{enumerate}
%%%%
\subsubsection{High-speed gas flow}
\wherefrom{\cczero \oc}
\wherefrom{\ccbysa \oc}
\label{exo_high_speed_gas_flow}
Scientists build a very high speed wind tunnel. For this, they build a large compressed air tank. Air escapes from the tank into a pipe which decreasing cross-section. The pipe diameter reaches a minimum (at the tunnel \vocab{throat}), and then it expands again, before discharging into the atmosphere.
Scientists build a very high speed wind tunnel. For this, they build a large compressed air tank. Air escapes from the tank into a pipe which decreasing cross-section, as shown in fig.~\ref{fig_simple_converging_diverging_nozzle}. The pipe diameter reaches a minimum (at the tunnel \vocab{throat}), and then it expands again, before discharging into the atmosphere.
\begin{figure}[ht]
\begin{center}
\includegraphics[width=0.8\textwidth]{simple_converging_diverging_nozzle}
\vspace{-0.5cm}
\end{center}
\supercaption{A converging-diverging nozzle. Air flows from the left tank to the right outlet, with a contraction in the middle.}{\wcfile{Simple converging diverging nozzle.svg}{Figure} \cczero \oc}
\label{fig_simple_converging_diverging_nozzle}
\end{figure}
For simplicity, we assume that heat losses through the tunnel walls are negligible, and that the fluid has uniformly-distributed velocity in cross-sections of the pipe.
In the tank, the air is stationary, with pressure \SI{12}{\bar} and temperature \SI{200}{\degreeCelsius}.
At the throat, the pressure and temperature have dropped to \SI{2,1}{\bar} and \SI{-10}{\degreeCelsius}. The throat cross-section is \SI{0,04}{\metre\squared}.
In the tank (point 1), the air is stationary, with pressure \SI{7,8}{\bar} and temperature \SI{246,6}{\degreeCelsius}.
At the throat (point 2), the pressure and temperature have dropped to \SI{4,2}{\bar} and \SI{160}{\degreeCelsius}. The throat cross-section is \SI{0,01}{\metre\squared}.
\begin{enumerate}
\item What is the mass flow through the tunnel?
\item What is the Mach number in the tank and at the throat?
......@@ -160,23 +167,23 @@ Balance of energy in a considered volume with steady flow:
\item What is the kinetic energy per unit mass of the air at the throat?
\end{enumerate}
Downstream of the throat, the pressure keeps dropping. By the time it reaches a point A where the cross-section is \SI{0,08}{\metre\squared}, the air has seen its pressure and temperature drop to \SI{0,8}{\bar} and \SI{-45}{\degreeCelsius}.
Downstream of the throat, the pressure keeps dropping. By the time it reaches a point 3, the air has seen its pressure and temperature drop to \SI{1,38}{\bar} and \SI{43}{\degreeCelsius}.
\begin{enumerate}
\shift{4}
\item What is the fluid velocity at point A?
\item What is the Mach number at point A?
\item What is the net force exerted on the fluid between the throat and point A?
\item What is the kinetic energy per unit mass of the air at point A?
\item What is the fluid velocity at point~3?\\
(if you need to convince yourself that $A_3>A_1$, you may also calculate the cross-section area)
\item What is the Mach number at point~3?
\item What is the net force exerted on the fluid between the points 2 and~3?
\item What is the kinetic energy per unit mass of the air at point~3?
\end{enumerate}
Once it has passed point A, the air undergoes complex loss-inducing evolutions (including going through a \vocab{shock wave}, where its properties change very suddenly), before it exits to the atmosphere with pressure \SI{1}{\bar}. As it exits to the atmosphere, the tunnel cross-section is \SI{0,09}{\metre\squared} and the tunnel air temperature is \SI{12}{\degreeCelsius}.
Once it has passed point 3, the air undergoes complex loss-inducing evolutions (including going through a \vocab{shock wave}, where its properties change very suddenly), before it discharges into the atmosphere with pressure \SI{1}{\bar} and temperature is~\SI{65}{\degreeCelsius}.
\begin{enumerate}
\shift{8}
\item What is the fluid velocity at outlet?
\item What is the outlet cross-section area?
\item What is the Mach number at outlet?
\item What is the net force exerted on the fluid between section A and the outlet?
\item What is the net force exerted on the fluid between section 3 and the outlet?
\item What is the kinetic energy per unit mass of the air at the outlet?
\end{enumerate}
......
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