The study of fluids in motion is called fluid dynamics. A flowing fluid always has a pattern of flow; either steady or unsteady. In *steady flow* pattern the particles passing through a fixed point move in the same flow line or in other words the flow pattern does not change with time. But in unsteady flow the flow pattern changes with time and becomes turbulent also called *turbulent flow*. In steady flow as shown in Figure 1 the fluid particles passing through the point $p_1$ must pass through the points $p_2$ and $p_3$.

The flow line in which a tangent at any point gives the direction of fluid flow at that point is called *streamline*. If the cross section of the flow tube varies, the flow speed along a flow line changes. In Figure 1 all the particles which flow in the flow line shown have velocity $\vec v_1$ at the point $p_1$ and have velocity $\vec v_2$ at the point $p_2$ and so on. In steady flow the streamlines coincide with the flow lines. The flow pattern remains steady until a certain velocity called critical velocity. If the velocity of the fluid becomes greater than the critical velocity, the flow line looses its steady nature and becomes irregular and chaotic.

The internal friction in a fluid is called *viscosity*. The motion of one layer of fluid against another is opposed by the viscosity of the fluid. The more viscous fluid is more sticky such as honey; it easily sticks when you touch it. The viscous fluid sticks to the walls of the flow tube and therefore the fluid in contact with the walls of the flow tube remains at rest while the other layers move against each other.

## Continuity Equation

We consider an incompressible fluid with zero viscosity whose flow is steady in a flow tube of varying cross-section as shown in Figure 2. But note that small viscosity is needed to insure that the flow is steady but we neglect it here.

The cross-sectional area of the left end of the flow tube is $A_1$ and that of the right end is $A_2$. The fluid entering the left end at cross-section $A_1$ has speed $v_1$ and the fluid leaving the right end at cross-section $A_2$ has speed $v_2$. The volume of the fluid which enters the left end in a very small interval of time $dt$ is $d{V_1} = {A_1}{v_1}dt$ while the volume of the fluid which leaves the right end in the same time interval is $d{V_2} = {A_2}{v_2}dt$. If $\rho$ is the density of the fluid, the mass entering the left end in the time interval $dt$ is $\rho {A_1}{v_1}dt$ and the mass leaving the right end in the same time interval is $\rho {A_2}{v_2}dt$. Since the fluid flow is ideal (steady flow without viscosity and compressibility), the mass which enters the left end in the time interval $dt$ must be equal to the mass that leaves the right end in the same time interval:

\[\begin{align*} \rho {A_1}{v_1}dt &= \rho {A_2}{v_2}dt\\ {\rm{or,}}\quad {A_1}{v_1} &= {A_2}{v_2} \tag{1} \label{1} \end{align*}\]

The Eq. \eqref{1} is called continuity equation. The quantity $A_1v_1$ is the volume flow rate at the left end and $A_2v_2$ is the volume flow rate at the right end. The continuity equation shows that the volume flow rate at one cross-section is equal to the volume flow rate at another cross-section of the flow tube. Therefore, the quantity $Av$ is the volume flow rate which is constant for steady flow of incompressible fluid with zero viscosity.

## Bernoulli's Equation

We are ready to derive an useful equation known as Bernoulli's equation. First we consider a flow tube as shown in Figure 3 whose left end at height $y_1$ has cross-sectional area $A_1$ and right end at height $y_2$ has cross-sectional area $A_2$. The fluid moves from the left end towards the right end. As in the continuity equation above we consider that the fluid is incompressible (that is the density is constant) with no viscosity and the flow is steady.

Let the speed of the flow on the left and right ends of the flow tube be $v_1$ and $v_2$ respectively. Let the pressure on the left end be $P_1$ and that on the right end be $p_2$. The Bernoulli's equation connects pressure and flow speed at one cross-section at an elevation with pressure and flow speed at another cross-section at another elevation. In a very small time interval $dt$ the work done on the segment of the fluid due to the pressure $P_1$ on the left end is ${W_1} = {P_1}{A_1}{v_1}dt$. The force due to the pressure $P_2$ on the right end does negative work because it opposes the flow so the work done by this force is ${W_2} = -{P_2}{A_2}{v_2}dt$. So the net work done by the pressure of the surrounding fluid on the fluid segment is

\[dW = {W_1} - {W_2} = {P_1}{A_1}{v_1}dt - {P_2}{A_2}{v_2}dt \tag{2} \label{2}\]

The only conservative force acting on the segment of the fluid is gravity so the work done given by Eq. \eqref{2} is equal to the change in total mechanical energy of the system. Since the fluid is incompressible without viscosity and the flow is steady, the mass of the fluid which enters the left end in the time interval $dt$ is the same as the mass of the fluid which leaves the right end in the same time. And therefore, $dm = \rho {A_1}{v_1}dt = \rho {A_2}{v_2}dt$ which implies our continuity equation ${A_1}{v_1} = {A_2}{v_2}$. In other words the volume of the fluid which enters the left end in the time interval $dt$ is the same as the volume of the fluid which leaves the right end in the same time interval and you can write $dV = {A_1}{v_1}dt = {A_2}{v_2}dt$ and again ${A_1}{v_1} = {A_2}{v_2}$. The Eq. \eqref{2} can be rewritten as,

\[dW = ({P_1} - {P_2})dV \tag{3} \label{3}\]

The kinetic and potential energies of the shaded portion of the fluid in Figure 3 remain the same for steady flow. So the change in kinetic energy is only due to the change in speed of the fluid element of mass $dm$. Therefore, the change in kinetic energy $dK$ of the fluid between the two ends is

\[dK = \frac{1}{2}dm{\kern 1pt} {\kern 1pt} v_2^2 - \frac{1}{2}dm\,v_1^2 = \frac{1}{2}\rho dV(v_2^2 - v_2^2) \tag{4} \label{4}\]

In the time interval $dt$ the change in potential energy of the fluid segment means the fluid element of volume $dV$ rises a height of $y_2-y_1$ and therefore, the change in potential energy $dU$ is

\[dU = dm{\kern 1pt} g({y_2} - {y_1}) = \rho dVg({y_2} - {y_1}) \tag{5} \label{5}\]

And the total change in mechanical energy is $dE = dK + dU$ which is equal to the work $dW$ done by the pressures of the surrounding fluid in Eq. \eqref{3}. Hence,

\[\begin{align*} ({P_1} - {P_2})dV &= \frac{1}{2}\rho dV(v_2^2 - v_1^2) + \rho dV{\kern 1pt} g({y_2} - {y_1})\\ {\rm{or,}}\quad ({P_1} - {P_2}) &= \frac{1}{2}\rho (v_2^2 - v_1^2) + \rho {\kern 1pt} g({y_2} - {y_1}) \tag{6} \label{6} \end{align*}\]

The Eq. \eqref{6} is called Bernoulli's equation. Rearranging the equation you'll find,

\[\begin{align*} {P_1} + \frac{1}{2}\rho v_1^2 + \rho g{y_1} &= {P_2} + \frac{1}{2}\rho v_2^2 + \rho g{y_2}\\ \tag{7} \label{7} {\rm{or,}}\quad P + \frac{1}{2}\rho v_{}^2 + \rho gy &= {\rm{constant}} \end{align*}\]

The Eq. \eqref{7} shows that the quantity $P + \frac{1}{2}\rho v_{}^2 + \rho gy$ remains constant throughout the flow tube. Note that Bernoulli's equation is valid for an incompressible fluid with zero viscosity and flow should be steady.