User:Nanditaahuja91/sandbox

= Introduction =

According to the Bernoulli’s principle, the energy posed by an incompressible fluid is the sum of the pressure head, the velocity head and the elevation head. Assuming water to be a perfectly incompressible fluid, the total energy of water in a channel can be expressed mathematically as,

$$ H=\frac{p}{\gamma}+\frac{v^2}{2g}+z$$

Where, H=Total head(L) p=Pressure(F/L2)

g=Acceleration due to gravitaty(L/T2)

v=velocity of flow (L/T)

z=elevation of water surface level above the datum(L)

The flow of water through the channel results in loss of energy due to friction resulting from channel constrictions, channel roughness etc. However, for the purpose of this analysis any such losses have been neglected.Therefore, by the principle of conservation of energy, the energy posed by water in the channel at any two points (1 and 2) is same

$$H_1=H_2$$ $$\frac{p_1}{\gamma}+\frac{v_1^{2}}{2g}+z_1=\frac{p_2}{\gamma}+\frac{v_2^{2}}{2g}+z_2$$ \label{eq:totalenergy}

Since, the flow is in an open channel and flowing water is exposed to the atomsphere at all points the pressure at every point is equal to the atmospheric pressure and thus can be neglected from the equation for total energy equation. Also, considering the channel bottom as the datum, the equation can be re-written as

$$\begin{aligned} \frac{v_1^{2}}{2g}+y_1&=&\frac{v_2^{2}}{2g}+y_2\end{aligned}$$

Where,


 * y1 and y2 =Depth of flow in channel at points 1 and 2 respectively(L)

Thus, specific energy in a channel is expressed as:

$$ E=\frac{v^{2}}{2g}+y$$\label{eq:specificenergy}

=Parabolic Section=

Theory and Derivation of General Relationship
Assuming the channel to be a perfect parabola with the equation

$$ y=ax^2$$\label{eq:para}

Where,
 * y=Depth of water in the channel(L)
 * 2x=Top width of the channel(L)



\begin{figure} \centering \includegraphics[scale=0.8]{parageometry} \caption{Geometry of Parabolic channel} \label{fig:parageometry} \end{figure} \begin{eqnarray} \label{eq:area2} A&=&\frac{2}{3}*width*height \nonumber \\ &=&\frac{2}{3}*2x*y \end{eqnarray}

Where,\\ \indent y=Depth of water in the channel(\emph{L})\\ From equation ~\ref{eq:para}

\begin{eqnarray} x=\pm\sqrt\frac{y}{a} \end{eqnarray}

Substituting in ~\ref{eq:area2}

\begin{eqnarray}\label{eq:area3} A=\frac{4}{3} \left(\frac{y^\frac{3}{2}}{a^\frac{1}{2}}\right) \end{eqnarray}

Velocity of flow in the channel determined by the \href{https://en.wikipedia.org/wiki/Continuity_equation}{continuity equation} is,

\begin{eqnarray} \label{eq:vel} v=\frac{Q}{A} \end{eqnarray}

Where,\\ \indent A=Area of cross-section(\emph{L\textsuperscript{2}})\\

Substituting ~\ref{eq:area3} in ~\ref{eq:vel}\\

\begin{eqnarray} v=&&\frac{Q} {\left(\frac{4}{3}\frac{4 y^\frac{3}{2}}{3 a^\frac{1}{2}}\right)} \nonumber \end{eqnarray}

\begin{eqnarray}\label{eq:vel1} v=\frac{3}{4}\left(\frac{Q a^\frac{1}{2}}{y^\frac{3}{2}}\right) \end{eqnarray}

Substituting ~\ref{eq:vel1} in the specific energy equation ~\ref{eq:specificenergy}

\begin{eqnarray} E=\left(\frac{3}{4}\frac{Q a^\frac{1}{2}}{y^\frac{3}{2}})^2 \frac{1}{2g}\right) + y \nonumber \\ E=\frac{9}{32}\left(\frac{Q^2 a}{gy^3}\right) + y \label{eq:sp} \end{eqnarray}

\subsubsection{E-y Diagram} \noindent From equation ~\ref{eq:sp} for a given discharge Q, and a fixed channel cross-section, energy is a function of depth and can be plotted graphically as shown in figure ~\ref{fig:EyDiagram}

\begin{figure}[h!] \centering \includegraphics[scale=1]{E-y_diagram.png} \caption{E-y Diagram for Parabolic Channel} \label{fig:EyDiagram} \end{figure}

From the figure we can see that line E=y is an asymptote to the curves and for a given depth, increase in discharge results in an increase in associated specific energy.

\subsubsection{Critical Flow Relationship and Froude number}

Substituting the value of v and A from equation ~\ref{eq:vel1} and ~\ref{eq:area3} respectively in ~\ref{eq:fr}

\begin{eqnarray} \label{eq:frpara} F_r&&=\frac{3}{4}\frac{Q a^\frac{1}{2}}{y^\frac{2}{3}} \frac{1}{\sqrt{\frac{2}{3}gy}} \nonumber \\ &&=\frac{3^\frac{3}{2}}{2^\frac{5}{2}}\left(\frac{Qa^\frac{1}{2}}{g^\frac{1}{2}y^2}\right) \end{eqnarray}

\textbf{Critical Depth}

Solving the above equation for critical conditions, i.e. F\textsubscript{r}=1

\begin{eqnarray} \label{eq:yc} &&y_c=\sqrt{\frac{3^\frac{3}{2}}{2^\frac{5}{2}}\frac{Qa^\frac{1}{2}}{g\frac{1}{2}}} \nonumber \\ or, &&y_c=\frac{3^\frac{3}{4}}{2^\frac{5}{4}}\left(\frac{Q^\frac{1}{2}a^\frac{1}{4}}{g^\frac{1}{4}}\right) \end{eqnarray}

Where,\\ \indent y\textsubscript{c}=Critical Depth(\emph{L})\\

Alternatively, critical flow condition can also be defined as the condition for minimum energy of flow for a given discharge and can be mathematically derived as:

Taking a derivative of the specific energy for a parabolic channel as expressed in c

\begin{eqnarray} \ \frac{d}{d y}E=\frac{d}{d y} \big( \frac{9}{32}\frac{Q^2 a}{gy^3} + y\big) \nonumber \\ \frac{dE}{d y}= \frac{-27}{32}\left(\frac{Q^2 a}{gy^4}\right) + 1\nonumber \\ \end{eqnarray}

At minimum energy \begin{eqnarray} \frac{dE}{d x}&&=0 \nonumber \\ &&=-\frac{27}{32}\left(\frac{Q^2 a}{gy^4}\right) + 1 \nonumber \\ \nonumber \\ y_c^4&&=\frac{27}{32}\left(\frac{Q^2 a}{g}\right) \nonumber \\ y_c&&=\frac{3^\frac{3}{4}}{2^\frac{5}{4}}\left(\frac{Q^\frac{1}{2} a^\frac{1}{4}}{g^\frac{1}{4}}\right) \label{eq:yc2} \end{eqnarray}

The mathematical expression obtained from the equation ~\ref{eq:yc2} is the same as that obtained from equation ~\ref{eq:yc}.\\

\textbf{Critical Energy}

Substituting the value of critical depth obtained from the equation ~\ref{eq:yc2} in the equation for specific energy (equation ~\ref{eq:sp})

\begin{eqnarray} E_c= \frac{2^\frac{3}{4}}{3^\frac{1}{4}}\left(\frac{Q^\frac{1}{2} a^\frac{1}{4}}{g^\frac{1}{4}}\right) \end{eqnarray}

Where,\\ \indent E\textsubscript{c}=Critical Energy(\emph{L})\\

Also, \begin{eqnarray} E_c= \frac{4}{3}y_c \end{eqnarray}

\begin{figure} \centering \includegraphics[scale=1]{ConjugateDepths.png} \caption{Flow regimes in a parabolic channel flow} \label{fig:flowregimes} \end{figure}

\textbf{Subcritical Depth}

At subcritical flow conditions the F\textsubscript{r} $<$1. i.e. \begin{eqnarray} F_r&&=\frac{3^\frac{3}{2}}{2^\frac{5}{2}}\left(\frac{Qa^\frac{1}{2}}{g^\frac{1}{2}y^2}\right) <1 \nonumber \\ y_{sub}&&>\frac{3^\frac{3}{4}}{2^\frac{5}{4}}\left(\frac{Qa^\frac{1}{4}}{g^\frac{1}{4}}\right) \nonumber \\ or, y_{sub}&&>y_c\nonumber \\ \end{eqnarray}

Where,\\ \indent y\textsubscript{sub}= Subcritical Energy(\emph{L})\\

i.e., subcritical depth is always greater than critical depth.\\

\textbf{ Supercritical Depth}

At supercritical flow conditions the F\textsubscript{r} $>$1. i.e.

\begin{eqnarray} F_r&&=\frac{3^\frac{3}{2}}{2^\frac{5}{2}}\left(\frac{Qa^\frac{1}{2}}{g^\frac{1}{2}y^2}\right) >1 \label{eq:fr1} \\ y_{super}&&<\frac{3^\frac{3}{4}}{2^\frac{5}{4}}\left(\frac{Qa^\frac{1}{4}}{g^\frac{1}{4}}\right) \nonumber \\ or, y_{super}&&<y_c \nonumber \\ \end{eqnarray}

Where,\\ \indent y\textsubscript{super}= Supercritical Energy(\emph{L})\\

i.e., subcritical depth is always greater than critical depth.

\subsubsection{Alternate Depths} From the E-y Diagram for the Parabolic Channel as shown in Figure ~\ref{fig:altdepths} we can see that for a given discharge and energy, there are two possible flow depths except in the case of critical depth/critical energy, namely y\textsubscript{1} and y\textsubscript{2} and are called alternate depths. Where, y\textsubscript{1} corresponds to depth of flow at subcritical conditions and y\textsubscript{2} corresponds to depth at subrcritical conditions. \begin{figure} \centering \includegraphics[scale=1]{FlowRegimes.png} \caption{Alternate Depths in a Parabolic Channel} \label{fig:altdepths} \end{figure}

By, the basic definition of alternate depths the energy E\textsubscript{1} and E\textsubscript{2} corresponding to depths y\textsubscript1 and y\textsubscript2, are equal. Therefore, for a given depth the conjugate depth can be mathematically derived as follows,

\begin{eqnarray} E_1&=&E_2 \nonumber \\ \frac{9}{32}\left(\frac{Q^2 a}{gy_1^3}\right) + y_1&=&\frac{9}{32}\left(\frac{Q^2 a}{gy_2^3}\right) + y_2 \label{eq:altdepth} \end{eqnarray}

If the depth of flow(y\textsubscript{1} or y\textsubscript{2}) is known, the above equation can be used to solve for the other alternate depth. The equation can be solved using a trial and error method or using computer tools like Microsoft Excel goal seek tool). \subsection{Depth-Discharge Relationship}

Figure ~\ref{fig:Q-y} illustrates the relationship between discharge and depth in a parabolic channel for a given energy of flow and channel geometry. It can be seen from the figure, that with an increase in the depth of flow, the discharge in the channel increases until the depth of flow reaches the critical depth. An increase in flow depth beyond the critical depth results in a decrease of discharge through the channel. Therefore, for a given energy and channel geometry, discharge is maximum at critical conditions i.e., when the flow depth is equal to the critical depth.

\begin{figure} \centering \includegraphics[scale=0.7]{Q-y.png} \caption{Discharge-Depth Relationship in a Parabolic Channel} \label{fig:Q-y} \end{figure}

\subsection{Example Problem} The concept of subcritical and supercritical flow regimes in a parabolic channel can be demonstrated by the following sluice gate problem.

Given a flow of 1.50 m\textsuperscript{3}/s and a parabolic channel cross-section of y=3x\textsuperscript{2} and a depth of 1.30 m. The flow encounters a sluice gate lowered into the channel causing change in flow regime from subcritical to  supercritical conditions and there is no loss of energy due to friction. We will determine the alternate depth, specific energy and the Froude number at the upstream and downstream location of the sluice gate.

y\textsubscript{1}=1.30 m, Q=1.50 m\textsuperscript{3}/s, a=3 and g=9.81 m/s\textsuperscript{2}

\textbf{Specific Energy upstream of sluice gate} Using equation ~\ref{eq:sp} and substituting value of y, Q, g and a

\begin{eqnarray} E_1=\frac{9}{32}\left(\frac{1.50^2 \times3}{9.81\times1.3_1^3}\right) + 1.3 = 1.46 m\\ \end{eqnarray}

\textbf{Specific Energy downstream of sluice gate} By the principle of conservation of energy,

\begin{eqnarray} E_1=E_2= 1.46 m\\ \end{eqnarray}

\textbf{Alternate Depth} Using equation ~\ref{eq:altdepth} and substituting value of\\

\begin{eqnarray} \frac{9}{32}\times\left(\frac{1.50^2 \times{3}}{9.81\times1.3^3}\right) + 1.3 &=&\frac{9}{32}\times\left(\frac{1.50^2 \times3}{9.81\times y_2^3}\right) + y_2 \\ \end{eqnarray}

Using goal seek tool in Microsoft Excel or by a trial and error method to solve the above equation,

\begin{eqnarray} y_2=0.81 m \\ \end{eqnarray}

\textbf{The Froude number upstream of the sluice gate}

Using equation ~\ref{eq:fr1}

\begin{eqnarray} F_r&&=\frac{3^\frac{3}{2}}{2^\frac{5}{2}}\frac{1.50\times 3^\frac{1}{2}}{9.81^\frac{1}{2}\times 1.3^2} \\ Fr&&=0.60 \end{eqnarray}

\textbf{The Froude number downstream of the sluice gate}

Using equation ~\ref{eq:fr1}

\begin{eqnarray} F_r&&=\frac{3^\frac{3}{2}}{2^\frac{5}{2}}\frac{0.81\times 3^\frac{1}{2}}{9.81^\frac{1}{2}\times 0.81^2} \\ Fr&&=1.55 \end{eqnarray}