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Commits (6)
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\label{chap_one}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Concept of a fluid}
......@@ -82,27 +81,34 @@
\begin{IEEEeqnarray}{rCl}
\vec V &\equiv& \timederivative{\vec x}
\end{IEEEeqnarray}
The magnitude~$V$ of velocity~$\vec V$ is called \vocab{speed} and measured in \si{\meter\per\second}. Contrary to speed, in order to express velocity completely, three distinct values (also each in \si{\metre\per\second}) must be expressed. Generally, this is done by using one of the following notations:
\begin{IEEEeqnarray*}{rCl}
The length $V$ of the velocity vector~$\vec V$ is measured in \si{\meter\per\second}:
\begin{IEEEeqnarray}{rCl}
V &\equiv& ||\vec V||
\end{IEEEeqnarray}
In this document, $V$ always is a positive number. Its formal name is \vocab{speed}, but in practice the term \vocab{velocity} is used to designate either the vector or its length, according to context.
In order to express the velocity vector completely, three distinct values (each having positive or negative values in \si{\metre\per\second}) must be expressed. In this document, this is done with different notations. In Cartesian coordinates, we have:
\begin{IEEEeqnarray*}{rCcCl}
\vec V &=& \left(\begin{array}{c}
V_x\\
V_y\\
V_z
\end{array}\right) %
&=& \left(\begin{array}{c}
u\\
v\\
w
\end{array}\right)\\
&=& \left(\begin{array}{c}
u_x\\
u_y\\
u_z
\end{array}\right)\\
&=& \left(u_i\right) \equiv \left(u_1, u_2, u_3\right)\\
&=& \left(\begin{array}{c}
\end{array}\right)
\end{IEEEeqnarray*}
In cylindrical coordinates, we write:
\begin{IEEEeqnarray*}{rCCl}
\vec V &=& \left(\begin{array}{c}
u_r\\
u_\theta\\
u_z
\end{array}\right)\\
\end{IEEEeqnarray*}
The \vocab{acceleration} $\vec a$ of the object is the rate of change in time of its velocity:
\begin{IEEEeqnarray}{rCl}
\vec a &\equiv& \timederivative{\vec V}
......@@ -128,6 +134,7 @@
\end{IEEEeqnarray}
\subsection{Energy}
\label{ch_energy}
\vocab{Energy}, measured in \si{joules} (\si{\joule}), is in most general terms the ability of a body to set other bodies in motion. It can be accumulated or spent by bodies in a large number of different ways. The most relevant forms of energy in fluid mechanics are:
\begin{description}
......@@ -186,17 +193,17 @@
\subsection{Perfect gas model}
Under a specific set of conditions (most particularly at high temperature and low pressure), several properties of a gas can be related easily to one another. Their absolute temperature~$T$ is then modeled as a function of their pressure~$p$ with a single, approximately constant parameter $R \equiv pv/T$:
When a gas has relatively simple molecules, moderate temperature and low pressure, several of its properties can be related easily to one another with the \vocab{perfect gas model}. Their absolute temperature~$T$ is then modeled as a function of their pressure~$p$ with a single, approximately constant parameter $R \equiv p/\rho T$:
\begin{IEEEeqnarray}{rCl}
p v &=& R T\\
\frac{p}{\rho} &=& R T
\end{IEEEeqnarray}
\begin{equationterms}
\item where \tab $R$ \tab depends on the state and nature of the gas (\si{\joule\per\kelvin\per\kilogram});
\item and \tab $p$ \tab is the pressure (\si{\pascal}), defined later on.
\end{equationterms}
Note that $R$ here is a \emph{specific} gas constant (whose value depends on the gas); chemists often instead use a \emph{universal} definition of $R$ in \si{\joule\per\mole\per\kelvin}.
This type of model (relating temperature to pressure and density) is called an \vocab{equation of state}. Where $R$ remains constant, the fluid is said to behave as a \vocab{perfect gas}. The properties of air can satisfactorily be predicted using this model. Other models exist which predict the properties of gases over larger property ranges, at the cost of increased mathematical complexity.
This type of model (relating temperature to pressure and density) is called an \vocab{equation of state}. When $R$ remains approximately constant, the fluid is said to behave as a perfect gas. The properties of air can satisfactorily be predicted using this model. Other models exist which predict the properties of gases over larger property ranges, at the cost of increased mathematical complexity.
Many fluids, especially liquids, do not follow this equation and their temperature must be determined in another way, most often with the help of laboratory measurements.
......@@ -335,13 +342,15 @@
\begin{equationterms}
\item where $\dot \vol$ is the volume flow (\si{\metre\cubed\per\second}).
\end{equationterms}
\item [Mechanical power] is a time rate of energy transfer. Compression and expansion of fluids involves significant power. When the flow is incompressible (see \S \ref{ch_classification_of_fluid_flows} below), the mechanical power $\dot W$ necessary to force a fluid through a fixed volume where pressure losses occur is:
\begin{IEEEeqnarray}{rCcCl}
\dot W &=& \vec F_\text{net, pressure} \cdot \vec V_\text{fluid, average} &=& \dot \vol \ |\Delta p|_\text{loss} \label{eq_power_deltap}
\item [Mechanical power] is a time rate of energy transfer. Compression and expansion of fluids involves significant power. The mechanical power $\dot W$ necessary to force a fluid through a chosen surface at a given pressure is:
\begin{IEEEeqnarray}{rCl}
\dot W &=& \vec F_\text{pressure} \cdot \vec V_\text{fluid} \\
&=& \dot \vol \ p \label{eq_power_deltap}\\
&=& \frac{\dot m}{\rho} \ p
\end{IEEEeqnarray}
\begin{equationterms}
\item where \tab $\dot W$ \tab\tab\tab is the power spent as work (\si{\watt});
\item and \tab $|\Delta p|$ \tab is the pressure loss occurring due to fluid flow (\si{\pascal}).
\item where \tab $\dot W$ \tab is the power spent as work (\si{\watt});
\item and \tab $p$ \tab\tab is the mean pressure at the surface (\si{\pascal}).
\end{equationterms}
\item [Power as heat] is also a time rate of energy transfer. When fluid flows uniformly and steadily through a fixed volume, its temperature $T$ increases according to the net power as heat $\dot Q$:
\begin{IEEEeqnarray}{rCl}
......@@ -442,7 +451,7 @@
\end{description}
%\clearpage%handmade
\clearpage%handmade
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Limits of fluid mechanics}
......
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\fluidmechchaptertitle
\label{chap_two}
\mecafluboxen
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Motivation}
todo
In this chapter, we learn to analyze fluid flows for which a lot of information is already available. We want, when confronted to a simple flow (for example, flow entering and leaving a machine), to be able to answer three questions:
\begin{itemize}
\item What is the mass flow in each inlet and outlet?
\item What is the force required to move the flow through the considered volume?
\item What energy transfer is required for this movement?
\end{itemize}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{One-dimensional flow problems}
todo
The method we develop here is called \vocab{integral analysis}, because it involves calculating the overall (integral) effect of the fluid flowing through a considered volume. In this chapter, we consider one-dimensional flows (at least in a loose definition); we will consider more advanced cases in \chapterthree.
For now, we are interested in flows where four conditions are met:
\begin{enumerate}
\item There is a clearly identified inlet and outlet;
\item At inlet and outlet, the flow fluid properties can be evaluated in bulk, because they are uniform (e.g.\ the inlet has only one velocity, one temperature etc.);
\item There are no significant changes in flow direction;
\item A lot of information is available about the fluid properties at inlet.
\end{enumerate}
Providing that those conditions are met, we can answer the question: what is the \emph{net} effect of the fluid flow through the considered volume?
In order to write useful equations, we need to begin with rigorous definitions, with the help of figure~\ref{fig_cv_simple}:
\begin{itemize}
\item We call \vocab{control volume} a certain volume we are interested in. Fluid flows through the control volume.\\
In this chapter, the control volume does not change with time. The fluid flow does not change in time, either (i.e. the flow is steady). The fluid enters and leaves the control volume at a clearly-identifiable inlet and outlet.
\item At a certain instant, the mass of fluid that is inside the control volume is called the \vocab{system}. The system is traveling. At a later point in time, it has moved and deformed.
\end{itemize}
\begin{figure}
\begin{center}
\includegraphics[width=0.6\textwidth]{concept_control_volume_system_simple.png}
\end{center}
\supercaption{A control volume within a flow. The \vocab{system} is the amount of mass included within the control volume at a given time. Because mass enters and leaves the control volume, the system is being moved and deformed (bottom).}{\wcfile{System control volume integral analysis.svg}{Figure} \cczero \oc}
\label{fig_cv_simple}
\end{figure}
The three equations that we write in this chapter state that basic physical laws apply to the system. They are \vocab{balance equations} (see \S\ref{ch_conservation_equations} p.\pageref{ch_conservation_equations}). Each time, we will express \textbf{what is happening to the system, as a function of the fluid properties at the inlet and outlet of the control volume}. This will allow us to answer three questions:
\begin{itemize}
\item What is the mass flow entering and leaving the control volume?
\item What is the force required to move the flow through the control volume?
\item What energy transfer is required to move the flow through the control volume?
\end{itemize}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Balance of mass}
How much mass flow is coming in and out of the control volume? We answer this question by writing a mass balance equation. It compares the rate of change of the mass of the system (which by definition is zero, see eq.~\ref{eq_massconservation} p.\pageref{eq_massconservation}), to the rate of change of mass in the control volume:
\begin{IEEEeqnarray}{CCCCCCC}
\timederivative{m_\sys} & = & 0 & = & \timederivative{} m_\cv & + & \dot m_\net \label{eq_mass_oned_one}\\\nonumber\\
\begin{array}{c} {\scriptstyle \text{the time change}} \\
......@@ -38,11 +73,11 @@ todo
\end{array} \nonumber
\end{IEEEeqnarray}
Since we are looking at a steady flow, and we have clearly-identified inlets and outlets, this becomes:
Since we are looking at a steady flow $\diff m_\cv / \diff t = 0$. Furthermore, we have clearly-identified inlets and outlets allowing us to re-express the net mass flow $\dot m_\net$. Equation~\ref{eq_mass_oned_one} becomes:
\begin{IEEEeqnarray}{CCCCCCC}
0 & = & \Sigma \dot m_\text{incoming} &+& \Sigma \dot m_\text{outgoing} \label{eq_mass_oned_two}\\\nonumber\\
\begin{array}{c} {\scriptstyle \text{time rate of creation}} \\
{\scriptstyle \text{or desctruction of mass}} \end{array} & = &
{\scriptstyle \text{or destruction of mass}} \end{array} & = &
\begin{array}{c} {\scriptstyle \text{the sum of}} \\
{\scriptstyle \text{incoming mass flows}} \\
{\scriptstyle \text{(negative terms)}} \end{array} & + &
......@@ -50,39 +85,52 @@ todo
{\scriptstyle \text{outgoing mass flows}} \\
{\scriptstyle \text{(positive terms)}} \end{array} \nonumber
\end{IEEEeqnarray}
\begin{equationterms}
\item For steady flow through a considered volume.
\end{equationterms}
Note that the sign convention is counter-intuitive: mass flows are negative inwards and positive outwards. This has no consequence in practice.
For a case where there are two inlets and two outlets, we can write:
\begin{IEEEeqnarray}{rCCCCCCCl}
Note that the sign convention is counter-intuitive: mass flows are negative inwards and positive outwards. For example, in a case where there were two inlets and two outlets, we could write:
\begin{IEEEeqnarray*}{rCCCCCCCl}
0 & = & \dot m_\text{in 1} &+& \dot m_\text{in 2} &+& \dot m_\text{out 1} &+& \dot m_\text{in 1}\\
0 & = & -|\dot m|_\text{in 1} &-&|\dot m|_\text{in 2} &+& |\dot m|_\text{out 1} &+& |\dot m|_\text{in 1}
\end{IEEEeqnarray}
\end{IEEEeqnarray*}
We can subsitute $\dot m = \rho V_\perp A$ (eq.~\ref{eq_basic_mass_flow} p.\pageref{eq_basic_mass_flow}) into these two last equations, obtaining:
\begin{IEEEeqnarray}{rCCCCCCCC}
We can substitute $\dot m = \rho V_\perp A$ (eq.~\ref{eq_basic_mass_flow} p.\pageref{eq_basic_mass_flow}) into the equations above, obtaining:
\begin{mdframed}
\begin{IEEEeqnarray}{rCl}
0 & = & \Sigma \left[\rho V_\perp A \right]_\text{incoming} + \Sigma \left[\rho V_\perp A \right]_\text{outgoing} \label{eq_mass_oned}
\end{IEEEeqnarray}
\begin{equationterms}
\item For steady flow through a considered volume.
\end{equationterms}
\end{mdframed}
Looking again at an example case where there were two inlets and two outlets, this equation~\ref{eq_mass_oned} would become:
\begin{IEEEeqnarray*}{rCCCCCCCC}
0 & = & \rho_\text{in 1} V_{\perp \text{ in 1}} A_\text{in 1} &+& \rho_\text{in 2} V_{\perp \text{ in 2}} A_\text{in 2} &+& \rho_\text{out 1} V_{\perp \text{ out 1}} A_\text{out 1} &+& \rho_\text{out 1} V_{\perp \text{ out 2}} A_\text{out 1}\nonumber\\
0 & = & \left(\rho V_\perp A \right)_\text{in 1} &+& \left(\rho V_\perp A \right)_\text{in 2} &+& \left(\rho V_\perp A \right)_\text{out 1} &+& \left(\rho V_\perp A \right)_\text{out 2}\nonumber\\
0 & = & -\left(\rho |V_\perp| A \right)_\text{in 1} &-& \left(\rho |V_\perp| A \right)_\text{in 2} &+& \left(\rho |V_\perp| A \right)_\text{out 1} &+& \left(\rho |V_\perp| A \right)_\text{out 2}\nonumber\\
\end{IEEEeqnarray}
0 & = & -\left(\rho |V_\perp| A \right)_\text{in 1} &-& \left(\rho |V_\perp| A \right)_\text{in 2} &+& \left(\rho |V_\perp| A \right)_\text{out 1} &+& \left(\rho |V_\perp| A \right)_\text{out 2}
\end{IEEEeqnarray*}
In a simple case where there is only one outlet and one outlet, this last equation can be rewritten as
\begin{IEEEeqnarray}{rCl}
\left(\rho |V_\perp| A \right)_1 &=& \left(\rho |V_\perp| A \right)_2
\left(\rho |V_\perp| A \right)_1 &=& \left(\rho |V_\perp| A \right)_2 \label{eq_rho_vee_a}
\end{IEEEeqnarray}
Remarks: leads to the conclusion that if area decreases, velocity increases
1) Only true for incompressible flows
2) No causal relationship, just direct relationship. Reducing A2 may increase V2, but it may also decrease dot m
So: careful evaluation: is density constant? And is mass flow known?
This equation~\ref{eq_rho_vee_a} is interesting, but also treacherous. For example, it is easy to draw the conclusion that “if $A$ increases, then $V$ must decrease”. Two remarks must be made about this statement:
\begin{enumerate}
\item The statement only holds true if $\rho$ remains constant. This is the case for water flows, since water is incompressible in most practical applications. But in compressible flows (for example when heat transfer is involved, or when compressed air is expanded), $\rho$ may vary too. It is in fact a well-known result that in supersonic flows, increases in $A$ can lead to \emph{increases} in $V$, because of a decrease in $\rho$.
\item There is no causal relationship in equation~\ref{eq_rho_vee_a}. In an incompressible flow, it may well be that reducing $A_2$ leads to an increase in $V_2$, but nothing guarantees that the product of the two remains constant. In other words, reducing $A_2$ may both increase $V_2$ \emph{and} decrease $\dot m$. Increases in velocity are not “for free”: they require force be applied and energy be spent. The mass balance equation cannot account for those phenomena.
\end{enumerate}
In conclusion, the mass balance equation is very useful to quantify mass flows. But if you find yourself using it to predict velocity, you should also ask yourself: what force, and what power, are required to generate this change in velocity? The following sections provide the answers we are looking for.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Balance of momentum}
What force is applied to the fluid for it to travel through the control volume? We answer this question by writing a momentum balance equation. It compares the rate of change of the momentum of the system (which by definition is the net force applying to it, see eq.~\ref{eq_secondlaw} p.\pageref{eq_secondlaw}), to the rate of change of momentum in the control volume:
\begin{IEEEeqnarray}{CCCCCCC}
\timederivative{(m \vec V_\text{sys})} & = & \vec F_\net & = & \timederivative{} \left(m \vec V\right)_\cv & + & \left(\dot m \vec V\right)_\net \nonumber\\\label{eq_linearmom_oned_one}\\
\begin{array}{c} {\scriptstyle \text{the change of momentum}} \\
......@@ -98,8 +146,7 @@ todo
\end{array} \nonumber
\end{IEEEeqnarray}
Since we are looking at a steady flow, and we have clearly-indentified inlets and outlets, this becomes:
Since we are looking at a steady flow, and we have clearly-identified inlets and outlets, this becomes:
\begin{IEEEeqnarray}{CCCCCCC}
\vec F_\net & = & \Sigma \left(\dot m \vec V\right)_\text{incoming} &+& \Sigma \left(\dot m \vec V\right)_\text{outgoing} \nonumber\\\label{eq_linearmom_oned_two}\\
\begin{array}{c} {\scriptstyle \text{the net force}} \\
......@@ -115,24 +162,35 @@ todo
\end{array} \nonumber
\end{IEEEeqnarray}
For a case where there is one inlet and one outlet, we can write:
\begin{IEEEeqnarray}{rCCCCCCCl}
The same convention as above is applied for the sign of the mass flow $\dot m$. For example, in a case where there were one inlet and one outlet, we would write:
\begin{IEEEeqnarray*}{rCCCCCCCl}
\vec F_\net & = & \left(\dot m \vec V\right)_\inn &+& \left(\dot m \vec V\right)_\out\\
\vec F_\net & = & -\left(|\dot m| \vec V\right)_\inn &+& \left(|\dot m| \vec V\right)_\inn
\end{IEEEeqnarray}
As before, we can subsitute $\dot m = \rho V_\perp A$ (eq.~\ref{eq_basic_mass_flow} p.\pageref{eq_basic_mass_flow}) into these two last equations, obtaining:
\begin{IEEEeqnarray}{rCCCCCCCC}
\vec F_\net & = & -\left(|\dot m| \vec V\right)_\inn &+& \left(|\dot m| \vec V\right)_\out
\end{IEEEeqnarray*}
As before, we can substitute $\dot m = \rho V_\perp A$ (eq.~\ref{eq_basic_mass_flow} p.\pageref{eq_basic_mass_flow}) into the equations above, obtaining:
\begin{mdframed}
\begin{IEEEeqnarray}{rCl}
\vec F_\text{net on fluid} & = & \Sigma \left[\rho V_\perp A \vec V\right]_\text{incoming} + \Sigma \left[\rho V_\perp A \vec V\right]_\text{outgoing}\label{eq_linearmom_oned}
\end{IEEEeqnarray}
\begin{equationterms}
\item For steady flow through a considered volume,
\item where $V_\perp$ is negative inwards, positive outwards.
\end{equationterms}
\end{mdframed}
In the example case where there is one inlet and one outlet, we would write:
\begin{IEEEeqnarray*}{rCCCCCCCC}
\vec F_\net & = & \left(\rho V_\perp A \vec V\right)_\inn &+& \left(\rho V_\perp A \vec V \right)_\out\\
\vec F_\net & = & -\left(\rho |V_\perp| A \vec V \right)_\inn &-& \left(\rho |V_\perp| A \vec V \right)_\out
\end{IEEEeqnarray}
\end{IEEEeqnarray*}
To make clear a few things, let us focus on the simple case where a considered volume is traversed by a steady flow with mass flow $\dot m$, with one inlet (point 1) and one outlet (point 2). The net force $\vec F_\net$ applying on the fluid is
To make clear a few things, let us focus on the simple case where a considered volume is traversed by a steady flow with mass flow $\dot m$, with one inlet (point~1) and one outlet (point~2). The net force $\vec F_\net$ applying on the fluid is
\begin{IEEEeqnarray}{rCl}
\vec F_\net & = & |\dot m| \left(\vec V_2 - \vec V_1 \right)
\vec F_\net & = & |\dot m| \left(\vec V_2 - \vec V_1 \right)\label{eq_fnet_twovectors}
\end{IEEEeqnarray}
Three remarks can be made about this equation.
First, we need to be aware that this is not one, but \emph{three} equations, one for each dimension. In order to express $\vec F_\net$, we need to calculate its three components:
\begin{IEEEeqnarray}{rCl}
\left\{ \begin{array}{rcl}
......@@ -142,24 +200,25 @@ todo
\end{array}\right.
\end{IEEEeqnarray}
Second, we can see that there are \emph{two} reasons why we could calculate a non-zero net force on the fluid in eq. XX.
Second, we can see that there are \emph{two} reasons why we could calculate a non-zero net force on the fluid in equation~\ref{eq_fnet_twovectors}.
\begin{enumerate}
\item Even if $\vec V_2$ is aligned and in the same direction as $\vec V_1$, they can be of different magnitude. A force is required to accelerate or decelerate the fluid (more precisely, an acceleration or deceleration of the fluid is equivalent to a force);
\item Even if $\vec V_2$ has the same magnitude as $\vec V_1$, they can have different directions. A force is required to change the direction in which a flow is flowing (or more precisely, a change of direction is equivalent to a force).
\end{enumerate}
We will explore these phenomena in greater detail in \chapterthree. In this current chapter, we are mostly interested in one-dimensional flows, and it will suffice for us to solve eq.XX in one suitable direction only, for example,
We will explore these phenomena in greater detail in \chapterthree. In this current chapter, we are mostly interested in one-dimensional flows, and it will suffice for us to solve equation~\ref{eq_fnet_twovectors} in one suitable direction only, for example,
\begin{IEEEeqnarray}{rCl}
F_{\net x} & = & |\dot m| \left(V_{2x} - V_{1x} \right)
\end{IEEEeqnarray}
Lastly, no cause/effect relationship, no explanation as to what exact mechanism (pressure, shear, through static wall or through the movement of parts) is exerting the net force on the fluid
The final remark is that the equation does not describe a cause-effect relationship. The net force does not cause the change in velocity any more than the change in velocity causes the net force: they are both equivalent and simultaneous. Similarly, we have no way to know what $\vec F_\net$ is made of. The exact mechanism which adds up to a net force (pressure or shear applied through a static wall or through the movement of parts) is “hidden” in the control volume, and unknown to us. In order to find out what happens in the control volume, we need a different type of analysis, which we will approach in \chaptersix.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Balance of energy}
Specific energy $e$ (in \si{\joule\per\kilogram}) multiplied by its mass (\si{\kilogram}). The time rate change of $m e$ is measured in \si{watts} ($\SI{1}{\watt} \equiv \SI{1}{\joule\per\second}$).
At which rate does the energy of the system vary when it transits through the control volume? We answer this question by writing an energy balance equation. It compares the rate of change of the energy of the system to the rate of change of energy in the control volume.
For this we prefer to express the energy $E$ as the specific energy $e$ (in \si{\joule\per\kilogram}) multiplied by the mass (\si{\kilogram}). The time rate change of $E = m e$ is measured in \si{watts} ($\SI{1}{\watt} \equiv \SI{1}{\joule\per\second}$). The energy balance equation is then:
\begin{IEEEeqnarray}{CCCCCCC}
\timederivative{m e_\sys} &=& \Sigma \left(\dot Q + \dot W\right) &=& \timederivative{} \left(m e\right)_\cv & + & \left(\dot m e\right)_\net \nonumber\\\label{eq_energy_oned_one}\\
\begin{array}{c} {\scriptstyle \text{the change of}} \\
......@@ -190,17 +249,17 @@ todo
\begin{itemize}
\item the net power transferred as heat $\dot Q_\net$ (positive inwards);
\item the net power transferred as work with moving solid surfaces $\dot W_\text{surfaces, net}$ (for example, a moving piston, turbine blade, or rotating shaft, positive inwards);
\item and the net power transferred to and from the fluid \emph{by the fluid itself}, in order to enter and leave the considered volume. This power is called \vocab{injection power}; for each inlet or outlet we have $\dot W_\text{injection} = \dot m (p / \rho)$.
\item and the net power transferred to and from the fluid \emph{by the fluid itself}, in order to enter and leave the considered volume. This power is called \vocab{injection power}; for each inlet or outlet we have $\dot W_\text{injection} = -\dot m (p / \rho)$.
\end{itemize}
We can thus write:
\begin{IEEEeqnarray}{rCl}
\Sigma \left(\dot Q + \dot W\right) &=& \dot Q_{\net} + \dot W_\text{shaft, net} + \dot W_\text{injection}\\
&=& \dot Q_{\net} + \dot W_\text{shaft, net} + \left(\dot m \ \frac{p}{\rho} \right)_\net
&=& \dot Q_{\net} + \dot W_\text{shaft, net} - \left(\dot m \ \frac{p}{\rho} \right)_\net
\end{IEEEeqnarray}
Turning now to the specific energy $e$, we break it down in terms of three components:
Turning now to the specific energy $e$, we break it down in terms of three components (see also \S\ref{ch_energy} p.\pageref{ch_energy}):
\begin{itemize}
\item the specific internal energy $i$, which represents the energy contained as stored heat within the fluid itself (in thermodynamics, this is often noted $u$). For perfect gas, $i$ is simply proportional to absolute temperature ($i = c_v T$), but for other fluids such as water, it cannot be easily measured, and precomputed tables relating $i$ to other properties must be used;
\item the specific internal energy $i$, which represents the energy contained as stored heat within the fluid itself (in thermodynamics, this is often noted $u$, but in fluid mechanics we reserve this symbol to note the $x$-component of velocity). For perfect gas, $i$ is simply proportional to absolute temperature ($i = c_v T$), but for other fluids such as water, it cannot be easily measured, and precomputed tables relating $i$ to other properties must be used;
\item the specific kinetic energy $e_k$,
\begin{IEEEeqnarray}{rCl}
e_k &\equiv& \frac{1}{2} V^2
......@@ -215,15 +274,26 @@ todo
e &\equiv& i + e_k + e_p
\end{IEEEeqnarray}
Now, we focus on steady flows (for which energy in the control volume does not change with time), and we can come back to eq.~\ref{eq_energy_oned_one} and re-write it as:
Now, we focus on steady flows (for which energy in the control volume does not change with time), and we can come back to eq.~\ref{eq_energy_oned_one} and rewrite it as:
\begin{IEEEeqnarray}{rCl}
\dot Q_{\net} + \dot W_\text{shaft, net} + \dot W_\text{injection} &=& \left(\dot m e\right)_\net \nonumber\\
\dot Q_{\net} + \dot W_\text{shaft, net} &=& \left(\dot m e\right)_\net - \left(\dot m \frac{p}{\rho}\right)\nonumber\\
\dot Q_{\net} + \dot W_\text{shaft, net} &=& \left[\dot m \left(i + \frac{p}{\rho} + \frac{1}{2} V^2 + g z \right) \right]_\net \label{eq_sfee}
\end{IEEEeqnarray}
\dot Q_{\net} + \dot W_\text{shaft, net} &=& \left(\dot m e\right)_\net + \left(\dot m \ \frac{p}{\rho}\right)\nonumber\\
\dot Q_{\net} + \dot W_\text{shaft, net} &=& \left[\dot m \left(i + \frac{p}{\rho} + \frac{1}{2} V^2 + g z \right) \right]_\net \label{eq_sfee_tmp}
\end{IEEEeqnarray}
This equation~\ref{eq_sfee} is known in thermodynamics as the \textit{steady flow energy equation}, and usually expressed there with the help of the concept of \vocab{enthalpy} $h \equiv i + p/\rho$.
Rewriting this into one general, usable form, we obtain:
\begin{mdframed}
\begin{IEEEeqnarray}{rCcl}
\dot Q_{\net} + \dot W_\text{shaft, net} &=& & \Sigma \left[\dot m \left(i + \frac{p}{\rho} + \frac{1}{2} V^2 + g z \right) \right]_\inn \nonumber\\
&& +& \Sigma \left[\dot m \left(i + \frac{p}{\rho} + \frac{1}{2} V^2 + g z \right) \right]_\out \label{eq_sfee}
\end{IEEEeqnarray}
\begin{equationterms}
\item For steady flow through a considered volume,
\item where $\dot m = \rho V_\perp A$ is negative inwards, positive outwards.
\end{equationterms}
\end{mdframed}
This equation~\ref{eq_sfee} is known in thermodynamics as the \textit{steady flow energy equation} (in thermodynamics, it is usually expressed with the help of the concept of \vocab{enthalpy} $h \equiv i + p/\rho$, which we do not use here).
As usual, let us focus on a case where there is only one inlet and one outlet. We obtain:
\begin{IEEEeqnarray}{rCl}
......@@ -231,76 +301,54 @@ todo
\dot Q_{\net} + \dot W_\text{shaft, net} &=& |\dot m| \left[ \Delta i + \Delta \frac{p}{\rho} + \Delta \left(\frac{1}{2} V^2\right) + \Delta (g z) \right]
\end{IEEEeqnarray}
useful in principle, not so much in practice
1) lot of terms. Not possible to predict whether power goes to delta p or delta i or delta v for example
2) disproportionate terms. very large heat capacity makes it not useful to compare i and ke
This equation is very useful in principle, but not so much in practice, for two reasons:
\begin{enumerate}
\item It contains a lot of terms. There are five fluid properties at inlet or outlet which affect energy, and it is difficult to predict which one will be affected by a heat or work transfer. For example, consider a simple water pump with known powers $\dot Q_\inn$ and $\dot W_\text{shaft, in}$. An efficient pump will generate large increases in $p$ (or $V$ and $z$), while an inefficient pump will generate large increases in $i$ and $1/\rho$. The energy balance equation, in this form, tells us nothing about how energy input to the control volume is redistributed.
\item The terms have disproportionate values in practice. The heat capacity of ordinary fluids is very large, and so $i$ is usually hundreds of times larger than the four terms in the brackets of equation~\ref{eq_sfee}. In water for example, an increase of temperature of \SI{0,1}{\degreeCelsius} (with the term $\Delta i$) requires the same energy increasing its velocity from \SI{30}{\kilo\metre\per\hour} to \SI{110}{\kilo\metre\per\hour} (with the term $\Delta e_k$). This is not an issue in thermodynamics, where heat, work and temperature are the most important parameters. But in fluid mechanics, velocity is of great interest, and the energy balance is not always useful to predict its changes.
\end{enumerate}
\begin{comment}
This equation~\ref{eq_rtt_energy_simple} is particularly attractive, but it necessitates the input of a large amount of experimental data to provide useful results. It is indeed very difficult to predict how the terms $i$ and $p /\rho$ will change for a given flow process. For example, a pump with given power $\dot Q_{\net \ \inn}$ and $\dot W_\text{shaft, net in}$ will generate large increases of terms $p$, $V$ and $z$ if it is efficient, or a large increase of terms $i$ and $1/\rho$ if it is inefficient. This equation~\ref{eq_rtt_energy_simple}, sadly, does not allow us to quantify the net effect of shear and the extent of irreversibilities in a fluid flow.
\end{comment}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{The Bernoulli equation}
\label{ch_bernoulli}
The Bernoulli equation has very little practical use for us; nevertheless it is so widely used that we have to dedicate a brief section to examining it. We will start from equation~\ref{eq_rtt_energy_simple} and add five constraints:
\begin{enumerate}
\item Steady flow.\\
Thus $\timederivative{} \iiint_\cv \rho\ e \diff \vol = 0$.\\
In addition, $\dot m$ has the same value at inlet and outlet;
\item Incompressible flow.\\
Thus, $\rho$ stays constant;
\item No heat or work transfer.\\
Thus, both $\dot Q_{\net\ \inn}$ and $\dot W_\text{shaft, net in}$ are zero;
\item No friction.\\
Thus, the fluid energy $i$ cannot increase due to an input from the control volume;
\item One-dimensional flow.\\
Thus, our control volume has only one known entry and one known exit, all fluid particles move together with the same transit time, and the overall trajectory is already known.
\end{enumerate}
\subsection{Theory}
With these five restrictions, equation~\ref{eq_rtt_energy_simple} simply becomes:
\begin{IEEEeqnarray}{rCl}
0 & = & \sum_\out \left\{ \dot m (i + \frac{p}{\rho} + e_k + e_p)\right\} - \sum_\inn \left\{ \dot m (i + \frac{p}{\rho} + e_k + e_p)\right\} \nonumber\\
& = & \dot m \left[(i + \frac{p_2}{\rho} + \frac{1}{2} V_2^2 + g z_2) - (i + \frac{p_1}{\rho} + \frac{1}{2} V_1^2 + g z_1)\right] \nonumber
\end{IEEEeqnarray}
and we here obtain the \vocab{Bernoulli equation}:
\begin{IEEEeqnarray}{rCl}
\frac{p_1}{\rho} + \frac{1}{2} V_1^2 + g z_1 & = & \frac{p_2}{\rho} + \frac{1}{2} V_2^2 + g z_2 \label{eq_bernoulli}
\end{IEEEeqnarray}
This equation describes the properties of a fluid particle in a steady, incompressible, friction-less flow with no energy transfer.
The Bernoulli equation is the energy equation applied to specific cases.
The Bernoulli equation can also be obtained starting from the linear momentum equation (eq.~\ref{eq_rtt_linearmom} p.\pageref{eq_rtt_linearmom}):
\begin{IEEEeqnarray*}{rCl}
\vec F_\net & = & \timederivative{} \iiint_\cv \rho \vec V \diff \vol + \iint_\cs \rho \vec V \ (\vec V_\rel \cdot \vec n) \diff A
\end{IEEEeqnarray*}
When considering a fixed, infinitely short control volume along a known streamline $s$ of the flow, this equation becomes:
\begin{IEEEeqnarray*}{rCl}
\diff \vec F_\text{pressure} + \diff \vec F_\text{shear} + \diff \vec F_\text{gravity} & = & \timederivative{} \iiint_\cv \rho \vec V \diff \vol + \rho V A \diff \vec V
\end{IEEEeqnarray*}
\begin{equationterms}
\item along a streamline, where the velocity $\vec V$ is aligned (by definition) with the streamline.
\end{equationterms}
To derive the Bernoulli equation, we will start from equation~\ref{eq_sfee} and add five constraints:
\begin{enumerate}
\item Steady flow.\\
(We had already implemented this restriction, when we set $\timederivative{(m e)_\cv}$ from eq.~\ref{eq_energy_oned_one} to zero in order to obtain eq.~\ref{eq_sfee})
\item Incompressible flow.\\
Thus, $\rho$ stays constant;
\item No heat or work transfer.\\
Thus, both $\dot Q_{\net}$ and $\dot W_\text{shaft, net}$ are zero;
\item No friction.\\
Thus, the fluid internal energy $i$ cannot increase;
\item One-dimensional flow.\\
Thus, our considered volume has only one inlet (labeled 1) and one outlet (labeled 2): all fluid particles move together with the same transit time, and the overall trajectory is already known.
\end{enumerate}
Now, adding the restrictions of steady flow ($\diff/\diff{t}=0$) and no friction ($\diff \vec F_\text{shear} = \vec 0$), we already obtain:
\begin{IEEEeqnarray*}{rCl}
\diff \vec F_\text{pressure} + \diff \vec F_\text{gravity} & = & \rho V A \diff \vec V
\end{IEEEeqnarray*}
The projection of the net force due to gravity $\diff \vec F_\text{gravity}$ on the streamline segment $\diff s$ has norm $\diff \vec F_\text{gravity} \cdot \diff \vec{s} = -g \rho A \diff z$, while the net force due to pressure is aligned with the streamline and has norm $\diff F_{\text{pressure}, s} = - A \diff p$. Along this streamline, we thus have the following scalar equation, which we integrate from points~1 to~2:
\begin{IEEEeqnarray*}{rCl}
- A \diff p - \rho g A \diff z & = & \rho V A \diff V \nonumber\\
-\frac{1}{\rho} \diff p - g \diff z & = & V \diff V \nonumber\\
-\int_1^2 \frac{1}{\rho} \diff p - \int_1^2 g \diff z & = & \int_1^2 V \diff V
\end{IEEEeqnarray*}
The last obstacle is removed when we consider flows without heat or work transfer, where, therefore, the density $\rho$ is constant. In this way, we arrive to equation.~\ref{eq_bernoulli} again:
\begin{IEEEeqnarray*}{rCl}
\frac{p_1}{\rho} + \frac{1}{2} V_1^2 + g z_1 & = & \frac{p_2}{\rho} + \frac{1}{2} V_2^2 + g z_2
\end{IEEEeqnarray*}
With these five restrictions, equation~\ref{eq_sfee} simply becomes:
\begin{IEEEeqnarray*}{rCcl}
0 + 0 &=& & \left[\dot m \left(i_\cst + \frac{p}{\rho_\cst} + \frac{1}{2} V^2 + g z \right) \right]_1 \\
&& +& \left[\dot m \left(i\cst + \frac{p}{\rho_\cst} + \frac{1}{2} V^2 + g z \right) \right]_2
\end{IEEEeqnarray*}
Dividing by $|\dot m|$ and canceling $i_\cst$, as follows,
\begin{IEEEeqnarray*}{rCl}
0 &=& -\left(i_\cst + \frac{p}{\rho_\cst} + \frac{1}{2} V^2 + g z \right)_1 + \left(i\cst + \frac{p}{\rho_\cst} + \frac{1}{2} V^2 + g z \right)_2\nonumber\\
0 &=& -\left(\frac{p}{\rho_\cst} + \frac{1}{2} V^2 + g z \right)_1 + \left(\frac{p}{\rho_\cst} + \frac{1}{2} V^2 + g z \right)_2
\end{IEEEeqnarray*}
and multiplying by the (constant and uniform) density $\rho$, we obtain the \vocab{Bernoulli equation}, with all terms having dimensions of pressure:
\begin{IEEEeqnarray}{rCl}
\left(p + \frac{1}{2} \rho V^2 + \rho g z \right)_1 &=& \left(p + \frac{1}{2} \rho V^2 + \rho g z \right)_2 \label{eq_bernoulli}
\end{IEEEeqnarray}
This equation describes the properties of a fluid particle in a steady, incompressible, friction-less flow with no energy transfer.
\subsection{Reality}
Let us insist on the incredibly frustrating restrictions brought by the five conditions above:
\begin{enumerate}
\item Steady flow.\\
......@@ -314,7 +362,18 @@ todo
\item One-dimensional flow.\\
This equation is only valid if we know precisely the trajectory of the fluid whose properties are being calculated.
\end{enumerate}
It becomes possible to correct for most of those shortcomings, by adding one extra (negative) term called “$\Delta p_\text{loss}$” to eq.~\ref{eq_bernoulli}, which we will lump together all of the effects unaccounted for. In this way, we obtain the \vocab{Bernoulli equation with losses}:
\begin{IEEEeqnarray}{rCl}
\left(p + \frac{1}{2} \rho V^2 + \rho g z \right)_1 &=& \left(p + \frac{1}{2} \rho V^2 + \rho g z \right)_2 + \Delta p_\text{loss} \label{eq_bernoulli_losses}
\end{IEEEeqnarray}
There are indeed cases where the pressure loss due to the imperfection of the flow are well-understood, and can be easily quantified. This is true of flow in pipes, for example (we study those in \chaptersevenshort). In those cases, eq.~\ref{eq_bernoulli_losses} is extremely useful.
Nevertheless, this approach is also easily misused! In a fluid flow where several of the restrictions above do not hold —and many such flows can be found in everyday life as well as engineering applications— equation~\ref{eq_bernoulli_losses} will betray its users. Convince yourself that \emph{any wrong equation} can be made correct by adding an unknown “bucket” term at the end: for example $2 + 3 = -18 + \Delta p_\text{loss}$.
In case you are not sure whether the Bernoulli equation applies, \textbf{start from an energy balance equation}. Removing the terms that do not apply will force you to question their relevance (e.g.\ is heat transfer really negligible? etc.). If you do not come to a conclusive end, do not remove terms that are inconvenient. The unfortunate reality is that in fluid mechanics, the energy balance equation contains many terms, with disproportionate values, and using it alone is not enough to solve most practical problems.
Among these, the last is the most severe (and the most often forgotten):\\ \textbf{the Bernoulli equation does not allow us to predict the trajectory of fluid particles}. Just like all of the other equations in this chapter, it requires a control volume with a known inlet and a known outlet.
We finish this chapter with another word of warning. Among the five restrictions listed, the last is the most severe, and the most often forgotten:\\ \textbf{the Bernoulli equation does not allow us to predict the trajectory of fluid particles}. Just like all of the other equations in this chapter, it requires a control volume with a known inlet and a known outlet. If you find yourself drawing out flow streamlines and interpreting the result with the Bernoulli equation, you are running astray. The tools you need to do this correctly are waiting for us in \chaptersix.
\atendofchapternotes
\renewcommand{\lastedityear}{2019}
\renewcommand{\lasteditmonth}{03}
\renewcommand{\lasteditday}{26}
\renewcommand{\lasteditday}{29}
\atstartofexercises
\fluidmechexercisestitle
......@@ -10,10 +10,33 @@
\mecafluexboxen
\begin{boiboiboite}
todo
\end{boiboiboite}
\mecafluboxtmp
Balance of mass in a considered volume with steady flow:
\begin{IEEEeqnarray}{rCl}
0 & = & \Sigma \left[\rho V_\perp A \right]_\text{incoming} + \Sigma \left[\rho V_\perp A \right]_\text{outgoing} \ztag{\ref{eq_mass_oned}}
\end{IEEEeqnarray}
\begin{equationterms}
\item where $V_\perp$ is negative inwards, positive outwards.
\end{equationterms}
Balance of momentum in a considered volume with steady flow:
\begin{IEEEeqnarray}{rCl}
\vec F_\text{net on fluid} & = & \Sigma \left[\rho V_\perp A \vec V\right]_\text{incoming} + \Sigma \left[\rho V_\perp A \vec V\right]_\text{outgoing}\ztag{\ref{eq_linearmom_oned}}
\end{IEEEeqnarray}
\begin{equationterms}
\item where $V_\perp$ is negative inwards, positive outwards.
\end{equationterms}
Balance of energy in a considered volume with steady flow:
\begin{IEEEeqnarray}{rCcl}
\dot Q_{\net} + \dot W_\text{shaft, net} &=& & \Sigma \left[\dot m \left(i + \frac{p}{\rho} + \frac{1}{2} V^2 + g z \right) \right]_\inn \nonumber\\
&& +& \Sigma \left[\dot m \left(i + \frac{p}{\rho} + \frac{1}{2} V^2 + g z \right) \right]_\out \label{eq_sfee}
\end{IEEEeqnarray}
\begin{equationterms}
\item where $\dot m$ is negative inwards, positive outwards.
\end{equationterms}
\end{boiboiboite}
%%%%
\subsubsection{Pipe expansion without losses}
......@@ -44,12 +67,13 @@
\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?\\
......@@ -69,15 +93,14 @@
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 ($c_\text{p} = \SI{1050}{\joule\per\kilogram}$)
......@@ -93,7 +116,7 @@
\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}
......@@ -107,27 +130,36 @@
\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?
......@@ -135,23 +167,23 @@
\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}
......
......@@ -2,7 +2,7 @@
\renewcommand{\lasteditmonth}{09}
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\fluidmechchaptertitle
\label{chap_three}
......@@ -32,7 +32,7 @@
\begin{center}
\includegraphics[width=0.6\textwidth]{concept_control_volume_system.png}
\end{center}
\supercaption{A control volume within a flow. The \vocab{system} is the amount of mass included within the control volume at a given time. At a later time, it may have left the control volume, and its shape and properties may have changed. The control volume may also change shape with time, although this is not represented here.}{\wcfile{System control volume intregral analysis.svg}{Figure} \cczero \oc}
\supercaption{A control volume within a flow. The \vocab{system} is the amount of mass included within the control volume at a given time. At a later time, it may have left the control volume, and its shape and properties may have changed. The control volume may also change shape with time, although this is not represented here.}{\wcfile{System control volume integral analysis.svg}{Figure} \cczero \oc}
\label{fig_cv}
\end{figure}
......
......@@ -2,7 +2,7 @@
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\fluidmechchaptertitle
\label{chap_nine}
......
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\section{Notation}
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\clearpage
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\section{Field operators}
\label{appendix_field_operators}
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\end{IEEEeqnarray}
\clearpage
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\section{Derivations of the Bernoulli equation}
\label{appendix_bernoulli}
\subsection{The Bernoulli equation from the energy equation}
\label{appendix_benoulli_energy}
This is covered in section~\ref{ch_bernoulli} p.\pageref{ch_bernoulli}.
\subsection{The Bernoulli equation from the integral momentum equation}
We begin with the integral linear momentum equation (eq.~\ref{eq_rtt_linearmom} p.\pageref{eq_rtt_linearmom}):
\begin{IEEEeqnarray*}{rCl}
\vec F_\net & = & \timederivative{} \iiint_\cv \rho \vec V \diff \vol + \iint_\cs \rho \vec V \ (\vec V_\rel \cdot \vec n) \diff A
\end{IEEEeqnarray*}
When considering a fixed, infinitely short control volume along a known streamline $s$ of the flow, this equation becomes:
\begin{IEEEeqnarray*}{rCl}
\diff \vec F_\text{pressure} + \diff \vec F_\text{shear} + \diff \vec F_\text{gravity} & = & \timederivative{} \iiint_\cv \rho \vec V \diff \vol + \rho V A \diff \vec V
\end{IEEEeqnarray*}
\begin{equationterms}
\item along a streamline, where the velocity $\vec V$ is aligned (by definition) with the streamline.
\end{equationterms}
Now, adding the restrictions of steady flow ($\diff/\diff{t}=0$) and no friction ($\diff \vec F_\text{shear} = \vec 0$), we already obtain:
\begin{IEEEeqnarray*}{rCl}
\diff \vec F_\text{pressure} + \diff \vec F_\text{gravity} & = & \rho V A \diff \vec V
\end{IEEEeqnarray*}
The projection of the net force due to gravity $\diff \vec F_\text{gravity}$ on the streamline segment $\diff s$ has norm $\diff \vec F_\text{gravity} \cdot \diff \vec{s} = -g \rho A \diff z$, while the net force due to pressure is aligned with the streamline and has norm $\diff F_{\text{pressure}, s} = - A \diff p$. Along this streamline, we thus have the following scalar equation, which we integrate from points~1 to~2:
\begin{IEEEeqnarray*}{rCl}
- A \diff p - \rho g A \diff z & = & \rho V A \diff V \nonumber\\
-\frac{1}{\rho} \diff p - g \diff z & = & V \diff V \nonumber\\
-\int_1^2 \frac{1}{\rho} \diff p - \int_1^2 g \diff z & = & \int_1^2 V \diff V
\end{IEEEeqnarray*}
The last obstacle is removed when we consider flows without heat or work transfer, where, therefore, the density $\rho$ is constant. In this way, we arrive to equation.~\ref{eq_bernoulli} p.\pageref{eq_bernoulli} again:
\begin{IEEEeqnarray*}{rCl}
\frac{p_1}{\rho} + \frac{1}{2} V_1^2 + g z_1 & = & \frac{p_2}{\rho} + \frac{1}{2} V_2^2 + g z_2\\
\left(p + \frac{1}{2} \rho V^2 + \rho g z \right)_1 &=& \left(p + \frac{1}{2} \rho V^2 + \rho g z \right)_2
\end{IEEEeqnarray*}
\clearpage
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\section{List of references}
\mecafluboxen
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{\scriptsize These lecture notes are based on textbooks by White \smallcite{white2008}, Çengel \& al.\smallcite{cengelcimbala2010}, and Munson \& al.\smallcite{munsonetal2013}.\par}
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{\scriptsize These lecture notes are based on textbooks by White \smallcite{white2008},\\
Çengel \& al.\smallcite{cengelcimbala2010}, and Munson \& al.\smallcite{munsonetal2013}.\par}
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{\footnotesize Except otherwise indicated, we assume that:\\
Fluids are Newtonian\\
The atmosphere has $p_\atm = \SI{1}{\bar}$; $\rho_\atm = \SI{1,225}{\kilogram\per\metre\cubed}$; $T_\atm = \SI{11,3}{\degreeCelsius}$; $\mu_\atm = \SI{1,5e-5}{\pascal\second}$\\
Air behaves as a perfect gas: $R_\air =\SI{287}{\joule\per\kilogram\per\kelvin}$; $\gamma_\air = \num{1,4}$; $c_{p \text{ air}} = \SI{1005}{\joule\per\kilogram\per\kelvin}$; $c_{v \text{ air}} = \SI{718}{\joule\per\kilogram\per\kelvin}$\\
Liquid water is incompressible: $\rho_{\text{water}} = \SI{1000}{\kilogram\per\metre\cubed}$, $c_{p \text{ water}} = \SI{4180}{\joule\per\kilogram\per\kelvin}$%
}
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