# Fluid Mechanics for MAP/Differential Analysis of Fluid Flow

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## Differential relations for a fluid particle

We are interested in the distribution of field properties at each point in space. Therefore, we analyze an infinitesimal region of a flow by applying the RTT to an infinitesimal control volume, or , to a infinitesimal fluid system.

## Conservation of mass

The conservation of mass according to RTT
$\displaystyle {\frac {\partial }{\partial t}}\int _{CV}\rho \ dV+\int _{CS}\rho \;{\vec {U}}\cdot {\vec {n}}\ dA=0$
or in tensor form
$\displaystyle {\frac {\partial }{\partial t}}\int _{CV}\rho \ dV+\int _{CS}\rho \;U_{i}n_{i}\ dA=0$
The differential volume is selected to be so small that density $\displaystyle (\rho )$ can be accepted to be uniform within this volume. Thus the first integral in 1 is:
$\displaystyle {\frac {\partial }{\partial t}}\int _{CV}\rho \ dV\approx {\frac {\partial \rho }{\partial t}}dx_{1}dx_{2}dx_{3}={\frac {\partial \rho }{\partial t}}dV$
The flux term (second integral term) in the equation of conservation of mass can be analyzed in groups:
$\displaystyle \int _{CS}\rho U_{i}n_{i}\ dA=\int _{CS\,x_{1}}\rho U_{i}n_{i}\;dA+\int _{CS\,x_{2}}\rho U_{i}n_{i}\;dA+\int _{CS\,x_{3}}\rho U_{i}n_{i}\;dA$
Let's look to the surfaces perpendicular to $\displaystyle x_{1}-axis$
$\displaystyle \int _{CS\,x_{1}}\rho U_{i}n_{i}\ dA=-\rho U_{1}dx_{2}dx_{3}+\left[\rho U_{1}+{\frac {\partial \left(\rho U_{1}\right)}{\partial x_{1}}}dx_{1}\right]dx_{2}dx_{3}$
$\displaystyle ={\frac {\partial \left(\rho U_{1}\right)}{\partial x_{1}}}dx_{1}dx_{2}dx_{3}={\frac {\partial \left(\rho U_{1}\right)}{\partial x_{1}}}dV$
Similarly, the flux integrals through surfaces perpendicular to $\displaystyle x_{2}-axis$ and $\displaystyle x_{3}-{\textrm {axis}}$ are
$\displaystyle \int _{CS\,x_{2}}\rho U_{i}n_{i}\ dA={\frac {\partial \left(\rho U_{2}\right)}{\partial x_{2}}}dV,$
$\displaystyle \int _{CS\,x_{3}}\rho U_{i}n_{i}\ dA={\frac {\partial \left(\rho U_{3}\right)}{\partial x_{3}}}dV.$
Hence the flux integral reads;
$\displaystyle \int _{CS}\rho U_{i}n_{i}\ dA={\frac {\partial \left(\rho U_{i}\right)}{\partial x_{i}}}dV$
The conservation of mass equation becomes:
$\displaystyle {\frac {\partial \rho }{\partial t}}dV+{\frac {\partial \left(\rho U_{i}\right)}{\partial x_{i}}}dV=0$
Droping the $\displaystyle dV$, we reach to the final form of the conservation of mass:
$\displaystyle {\frac {\partial \rho }{\partial t}}+{\frac {\partial \left(\rho U_{i}\right)}{\partial x_{i}}}=0$
This equation is also called continuity equation. It can be written in vector form as:
$\displaystyle {\frac {\partial \rho }{\partial t}}+\nabla \cdot \left(\rho {\vec {U}}\right)=0\ \ {\text{where}}\ \ \nabla ={\frac {\partial }{\partial x_{1}}}{\vec {e_{1}}}+{\frac {\partial }{\partial x_{2}}}{\vec {e_{2}}}+{\frac {\partial }{\partial x_{3}}}{\vec {e_{3}}}\ \ {\text{gradient operator}}$
For a steady flow, continuity equation becomes:
$\displaystyle {\frac {\partial \left(\rho U_{i}\right)}{\partial x_{i}}}=0$
For incompresible flow, i.e. $\displaystyle \rho ={\text{constant}}$:
$\displaystyle {\frac {\partial U_{i}}{\partial x_{i}}}=0\ \ {\text{i.e.}}\ \ {\frac {\partial U_{1}}{\partial x_{1}}}+{\frac {\partial U_{2}}{\partial x_{2}}}+{\frac {\partial U_{3}}{\partial x_{3}}}=0$

### Example

For a two dimensional, steady and incompressible flow in $\displaystyle x_{1}x_{2}$ plane given by:
$\displaystyle \displaystyle U_{1}=Ax_{1}$
Find how many possible $\displaystyle U_{2}$ can exist.
For incompressible steady flow:
$\displaystyle \rho {\frac {\partial U_{i}}{\partial x_{i}}}=0$
in two diemnsions
$\displaystyle {\frac {\partial U_{1}}{\partial x_{1}}}+{\frac {\partial U_{2}}{\partial x_{2}}}=0\ \ \rightarrow \ \ {\frac {\partial U_{2}}{\partial x_{2}}}=-{\frac {\partial U_{1}}{\partial x_{1}}}$
Thus,
$\displaystyle {\frac {\partial U_{2}}{\partial x_{2}}}=-A$
This is an expression for the rate of change of $\displaystyle U_{2}$ velocity while keeping
$\displaystyle x_{1}$ constant. Therefore the integral of this equation reads
$\displaystyle \displaystyle U_{2}=-Ax_{2}+f(x_{1})$
Thus, any function $\displaystyle f(x_{1})$ is allowable.

### Example

Consider one-dimensional flow in the piston. The piston suddenly moves with the velocity $\displaystyle V_{p}$ . Assume uniform $\displaystyle \rho (t)$ in the piston and a linear change of velocity $\displaystyle U_{1}$ such that $\displaystyle U_{1}=0$ at the bottom ($\displaystyle x_{1}=0$) and $\displaystyle U_{1}=V_{p}$ on the piston ($\displaystyle x_{1}=L$), i.e.
$\displaystyle U_{1}={\frac {x_{1}}{L}}V_{p}$
Obtain a function for the density as a function of time.
The conservation of mass equation is:
$\displaystyle {\frac {\partial \rho }{\partial t}}+{\frac {\partial \left(\rho U_{i}\right)}{\partial x_{i}}}=0$
For one-dimensional flow and uniform $\displaystyle \rho$, this equaiton simplifies to
$\displaystyle {\frac {\partial \rho }{\partial t}}+\rho {\frac {\partial U_{1}}{\partial x_{1}}}=0$
$\displaystyle {\frac {\partial \rho }{\partial t}}=-\rho {\frac {\partial U_{1}}{\partial x_{1}}}=-\rho {\frac {V_{p}}{L}}$
$\displaystyle \displaystyle L=L_{0}+V_{p}t$
$\displaystyle {\frac {\partial \rho }{\partial t}}={\frac {d\rho }{dt}}=-\rho {\frac {V_{p}}{L_{0}+V_{p}t}}$
$\displaystyle \int _{\rho _{0}}^{\rho }{\frac {d\rho }{\rho }}=\int _{0}^{t}-{\frac {V_{p}}{L_{0}+V_{p}t}}dt$
$\displaystyle ln\left({\frac {\rho }{\rho _{0}}}\right)=ln\left({\frac {L_{0}}{L_{0}+V_{p}t}}\right)$
$\displaystyle \rho (t)=\rho _{0}\left({\frac {L_{0}}{L_{0}+V_{p}t}}\right)$
The same problem can be solved by using the integral approach with a deforming control volume.

## The differential equation of linear momentum

The integral equation for the momentum conservation is
$\displaystyle \sum F_{i}={\frac {\partial }{\partial t}}\int _{CV}\rho \;U_{i}\;dV+\int _{CS}\rho \;U_{i}\;U_{j}\;n_{j}dA$
For the first integral we assume $\displaystyle \rho$ and $\displaystyle U_{i}$ are uniform within dV, and dV is so small that:
$\displaystyle {\frac {\partial }{\partial t}}\int _{CV}\rho \;U_{i}\;dV\approx \ {\frac {\partial }{\partial t}}(\rho \;U_{i})dx_{1}dx_{2}dx_{3}$
Analyze the flux of the $\displaystyle \rho U_{i}$ momentum terms through the faces perpendicular to the axis:
$\displaystyle \int _{CS}\rho \;U_{i}\;U_{j}\;n_{j}\;dA=\int _{CS\,x_{1}}\rho \;U_{i}\;U_{j}\;n_{j}\;dA\ +\int _{CS\,x_{2}}\rho \;U_{i}\;U_{j}\;n_{j}\;dA\ +\int _{CS\,x_{3}}\rho \;U_{i}\;U_{j}\;n_{j}\;dA$
First consider the flux of $\displaystyle \rho \;U_{i}$ (momentum per unit volume in i-direction) through the surfaces perpendicular to $\displaystyle x_{1}$ axis:
$\displaystyle \int _{CS\,x_{1}}\rho \;U_{i}\;U_{j}\;n_{j}\;dA=-\rho \;U_{i}\;U_{1}\;dx_{2}dx_{3}\;+\left[\rho \ U_{i}\ U_{1}\ dx_{2}\ dx_{3}+{\frac {\partial }{\partial x_{1}}}(\rho \ U_{i}\ U_{1})\ dx_{1}\right]dx_{2}\ dx_{3}$
$\displaystyle ={\frac {\partial }{\partial x_{1}}}(\rho \;U_{i}\;U_{1})dV$
Similarly, the momentum flux through the surfaces in other directions read
$\displaystyle \int _{CS\,x_{2}}\rho \;U_{i}\;U_{j}\;n_{j}\;dA={\frac {\partial }{\partial x_{2}}}(\rho \;U_{i}\;U_{2})dV$
$\displaystyle \int _{CS\,x_{3}}\rho \;U_{i}\;U_{j}\;n_{j}\;dA={\frac {\partial }{\partial x_{3}}}(\rho \;U_{i}\;U_{3})dV$
Rearranging the equation for $\displaystyle \sum F_{i}$ we obtain:
$\displaystyle \sum F_{i}=\left[{\frac {\partial }{\partial t}}(\rho \ U_{i})+{\frac {\partial }{\partial x_{j}}}(\rho \ U_{i}\ U_{j})\right]dV$
We can simplify further:
$\displaystyle \sum F_{i}=\left[U_{i}{\frac {\partial \rho }{\partial t}}+\rho {\frac {\partial U_{i}}{\partial t}}+U_{i}{\frac {\partial }{\partial x_{j}}}(\rho \;U_{j})+\rho \;U_{j}\;{\frac {\partial U_{i}}{\partial x_{j}}}\right]dV$
$\displaystyle \sum F_{i}=U_{i}\underbrace {\left[{\frac {\partial \rho }{\partial t}}\;+{\frac {\partial }{\partial x_{j}}}(\rho \;U_{j})\right]} _{continuity\ equation=0}dV\;+\rho \underbrace {\left[{\frac {\partial U_{i}}{\partial t}}\;+U_{j}{\frac {\partial U_{i}}{\partial x_{j}}}\right]} _{{\frac {DU_{i}}{Dt}};subtantial\ derivative\ of\ U_{i}}dV$
Hence
$\displaystyle \sum F_{i}=\rho {\frac {DU_{i}}{Dt}}dV$
Let's look to the forces on the exposed on the diffrential control volume:
$\displaystyle {\frac {dP_{i}}{dt}}=\sum F_{i}=dF_{body\;i}+dF_{surface\;i}=\rho {\frac {dU_{i}}{dt}}dV$
Here, only gravitational force is considered as a body force. Thus,
$\displaystyle dF_{body\;i}=\rho \;dV\;g_{i}$
Surface forces are the stresses acting on the control surfaces. $\displaystyle F_{s}$ can be resolved into three components. $\displaystyle dF_{n}$ is normal to dA. $\displaystyle dF_{t}$ are tangent to dA:
$\displaystyle \sigma _{n}=\lim _{dA\rightarrow 0}{\frac {dF_{n}}{dA}}$
$\displaystyle \sigma _{t}=\lim _{dA\rightarrow 0}{\frac {dF_{t}}{dA}}$
$\displaystyle \sigma _{n}$ is a normal stress whereas $\displaystyle \sigma _{t}$ is a shear stress. The shear stresses are also designated by $\displaystyle \tau$.
Thus, the surface forces are due to stresses on the surfaces of the control surface.
$\displaystyle \sigma _{ij}=\left[{\begin{matrix}\sigma _{11}&\sigma _{12}&\sigma _{13}\\\sigma _{21}&\sigma _{22}&\sigma _{23}\\\sigma _{31}&\sigma _{32}&\sigma _{33}\end{matrix}}\right]$
We define the positive direction for the stress as the positive coordinate direction on the surfaces (e.g. on ABCD) for which the outwards normal is in the positive coordinate direction . If the outward normal represents the negative direction (A'B'C'D'), then the stresses are considered positive if directed in the negative coordinate directions.
The stresses on the surface $\displaystyle (\sigma _{ij})$ are the sum of pressure plus the viscous stresses which arise from motion with velocity gradients:
$\displaystyle \sigma _{ij}=\left[{\begin{matrix}-P&0&0\\0&-P&0\\0&0&-P\end{matrix}}\right]+\left[{\begin{matrix}\tau _{11}&\tau _{12}&\tau _{13}\\\tau _{21}&\tau _{22}&\tau _{23}\\\tau _{31}&\tau _{32}&\tau _{33}\end{matrix}}\right]=\left[{\begin{matrix}-P+\tau _{11}&\tau _{12}&\tau _{13}\\\tau _{21}&-P+\tau _{22}&\tau _{23}\\\tau _{31}&\tau _{32}&-P+\tau _{33}\end{matrix}}\right]$
$\displaystyle p$ has a minus sign since the force due to pressure acts opposite to the surface normal.
Let us look to the differential surface force in the $\displaystyle x_{1}$ direction:
$\displaystyle dF_{surface\ 1}={\frac {\partial \sigma _{11}}{\partial x_{1}}}\;dx_{1}dx_{2}dx_{3}+{\frac {\partial \sigma _{21}}{\partial x_{2}}}\;dx_{1}dx_{2}dx_{3}+{\frac {\partial \sigma _{31}}{\partial x_{3}}}\;dx_{1}dx_{2}dx_{3}$
Noting that $\displaystyle dV=dx_{1}dx_{2}dx_{3}$ and $\displaystyle \sigma _{ij}=-p\delta _{ij}+\tau _{ij}$ (4),
$\displaystyle dF_{surface\ 1}=\left(-{\frac {\partial P}{\partial x_{1}}}+{\frac {\partial \tau _{11}}{\partial x_{1}}}+{\frac {\partial \tau _{21}}{\partial x_{2}}}+{\frac {\partial \tau _{31}}{\partial x_{3}}}\right)dV$
Thus in tensor form the differential surface forces in $\displaystyle i$'th direction can be written as
$\displaystyle dF_{surface\ i}=\left(-{\frac {\partial P}{\partial x_{i}}}+{\frac {\partial \tau _{ji}}{\partial x_{j}}}\right)dV$
Note that $\displaystyle \tau _{ij}$ is a symmetric tensor, i.e.
$\displaystyle \tau _{ji}=\tau _{ij}$
Hence, the diffential surface forces reads:
$\displaystyle dF_{surface\ i}=\left(-{\frac {\partial P}{\partial x_{i}}}+{\frac {\partial \tau _{ij}}{\partial x_{j}}}\right)dV$
Inserting $\displaystyle dF_{body\ i}$ and $\displaystyle dF_{surface\ i}$ into (2),
$\displaystyle \rho {\frac {DU_{i}}{Dt}}dV=\rho \ g_{i}\ dV+\left(-{\frac {\partial P}{\partial x_{i}}}+{\frac {\tau _{ij}}{\partial x_{j}}}\right)\;dV$
and canceling $\displaystyle dV$ we obtain
$\displaystyle \rho {\frac {DU_{i}}{Dt}}=\rho \;g_{i}-{\frac {\partial P}{\partial x_{i}}}+{\frac {\partial \tau _{ij}}{\partial x_{j}}}.$
Expanding the substatial derivative at the left hand side,
$\displaystyle \rho {\frac {\partial U_{i}}{\partial t}}+\rho U_{j}{\frac {\partial U_{i}}{\partial x_{j}}}=\rho g_{i}-{\frac {\partial P}{\partial x_{i}}}+{\frac {\partial \tau _{ij}}{\partial x_{j}}}$
We obtain the the most general form of momentum equation which is valid for any fluid (Newtonian, Non-newtonian, Compressible, etc.). It is non-linear due to the $\displaystyle 2^{nd}$ term at the LHS. Efect of Newtonian and Non-newtonian properties appears in the formulation of the viscous stresses $\displaystyle \tau _{ij}$. $\displaystyle \tau _{ij}$ will introduce also non-linearity when the fluid is non-Newtonian.
It should be noted that these formulations are based on stress conception which was thought to exist in fluids in motion. However it is known that $\displaystyle \tau _{ij}$ can be expressed as momentum transfer per unit area and time. Thus it can be considered as molecular momentum transport term. Derivations based on this concept requires a molecular approach (which is lengthy). The students should be aware that $\displaystyle \tau _{ij}$ causes momentum transport when there is a gradient of velocity.

## Linear momentum equation for Newtonian Fluid: "Navier-Stokes Equation"

For a Newtonian fluid, the viscous stresses are defined as:
$\displaystyle \tau _{ij}=\mu \left[{\frac {\partial U_{i}}{\partial x_{j}}}+{\frac {\partial U_{j}}{\partial x_{i}}}\right]-{\frac {2}{3}}\delta _{ij}\mu {\frac {\partial U_{k}}{\partial x_{k}}}$
Note that derivation of this relation is beyond the scaope of this course.
Thus, the momentum equation (6) becomes
$\displaystyle \rho {\frac {DU_{i}}{Dt}}=\rho g_{i}-{\frac {\partial P}{\partial x_{i}}}+{\frac {\partial }{\partial x_{j}}}\left[\mu \left({\frac {\partial U_{i}}{\partial x_{j}}}+{\frac {\partial U_{j}}{\partial x_{i}}}\right)-{\frac {2}{3}}\mu \ \delta _{ij}\ {\frac {\partial U_{k}}{\partial x_{k}}}\right]$
For a flow with constant viscosity ($\displaystyle \mu ={\textrm {constant}}$):
$\displaystyle \rho {\frac {DU_{i}}{Dt}}=\rho g_{i}-{\frac {\partial P}{\partial x_{i}}}+\mu {\frac {\partial }{\partial x_{j}}}\left[{\frac {\partial U_{i}}{\partial x_{j}}}+{\frac {\partial U_{j}}{\partial x_{i}}}-{\frac {2}{3}}\delta _{ij}\ {\frac {\partial U_{k}}{\partial x_{k}}}\right]$
since,
$\displaystyle {\frac {\partial ^{2}U_{i}}{\partial x_{j}\partial x_{i}}}={\frac {\partial ^{2}U_{i}}{\partial x_{i}\partial x_{j}}}={\frac {\partial }{\partial x_{j}}}\left({\frac {\partial U_{i}}{\partial x_{i}}}\right)={\frac {\partial }{\partial x_{i}}}\left({\frac {\partial U_{i}}{\partial x_{j}}}\right)$
then,
$\displaystyle \rho {\frac {DU_{i}}{Dt}}=\rho \ g_{i}-{\frac {\partial P}{\partial x_{i}}}+\mu {\frac {\partial ^{2}U_{i}}{\partial x_{j}x_{j}}}+\mu {\frac {\partial }{\partial x_{i}}}\left({\frac {\partial U_{j}}{\partial x_{j}}}\right)-{\frac {2}{3}}\delta _{ij}\mu {\frac {\partial }{\partial x_{j}}}\left({\frac {\partial U_{k}}{\partial x_{k}}}\right)$
For an incompressible flow
$\displaystyle {\frac {\partial U_{k}}{\partial x_{k}}}={\frac {\partial U_{j}}{\partial x_{j}}}=0$
hence assuming that the viscosity is constant, it can be easily shown that the momentum equation reduces to
$\displaystyle \rho {\frac {DU_{i}}{Dt}}=\rho \ g_{i}-{\frac {\partial P}{\partial x_{i}}}+\mu {\frac {\partial ^{2}U_{i}}{\partial x_{j}^{2}}}$

## Euler's equation: Inviscid flow

When the velocity gradients in the flow is negligible and/or the Reynolds number takes very high values, the viscous stresses can be neglected:
$\displaystyle \displaystyle \tau _{ij}=0$
Since, the viscous stresses are proportional to viscosity:
$\displaystyle \tau _{ij}\propto \mu$
for flows, where $\displaystyle \tau _{ij}$ is neglected, the flow is called frictionless or inviscid, although there is a finite viscosity of the flow. Accordingly, the linear momentum equation reduces to
$\displaystyle \rho {\frac {DU_{i}}{Dt}}=\rho g_{i}-{\frac {\partial P}{\partial x_{i}}}.$

## Euler's equation in streamline coordinates

Euler's equation take a special form along and normal to a streamline with which one can see the dependency between the pressure, velocity and curvature of the streamline.
To obtain Euler's equation in s-direction, apply Newton's second law in s-direction in the absence of viscous forces.
$\displaystyle \rho dV\left[{\frac {\partial U_{s}}{\partial t}}+U_{s}{\frac {\partial U_{s}}{\partial s}}\right]=-{\frac {\partial P}{\partial s}}dV-\rho \ gsin\beta dV$
Omitting $\displaystyle dV$ would deliver
$\displaystyle {\frac {DU_{s}}{Dt}}=-{\frac {1}{\rho }}{\frac {\partial P}{\partial s}}-gsin\beta$
Since
$\displaystyle sin\beta \approx {\frac {dx_{2}}{ds}}={\frac {\partial x_{2}}{\partial s}}$
then the Equler's equation along a streamline reads
$\displaystyle {\frac {DU_{s}}{Dt}}=-{\frac {1}{\rho }}{\frac {\partial P}{\partial s}}-g{\frac {\partial x_{2}}{\partial s}}\ \ \ (8)$
For a steady flow and by neglecting body forces,
$\displaystyle {\frac {1}{\rho }}{\frac {\partial P}{\partial s}}=-U_{s}{\frac {\partial U_{s}}{\partial s}}$
it can be seen that decrease in velocity means increase in pressure as is indicated by the Bernoulli equation.
To obtain Euler's equation in n direction, apply Newton's second law in the absence of viscous forces and for a steady flow.
$\displaystyle \rho dV{\frac {DU_{n}}{Dt}}=\left(P-{\frac {\partial P}{\partial n}}{\frac {dn}{2}}\right)ds\ dx_{3}-\left(P+{\frac {\partial P}{\partial n}}{\frac {dn}{2}}\right)ds\ dx_{3}-\rho gdVcos\beta$
$\displaystyle \rho \ {\frac {DU_{n}}{Dt}}=\left(-{\frac {\partial P}{\partial n}}-\rho \ g\ cos\beta \right)$
Since,
$\displaystyle cos\beta \approx {\frac {dx_{2}}{dn}}={\frac {\partial x_{2}}{\partial n}}$
Then,
$\displaystyle {\frac {DU_{n}}{Dt}}=-{\frac {1}{\rho }}{\frac {\partial P}{\partial n}}-g\ {\frac {\partial x_{2}}{\partial n}}$
For a steady flow, the normal acceleration of the fluid is towards the center of curvature of the streamline:
$\displaystyle {\frac {DU_{n}}{Dt}}=-{\frac {U_{s}^{2}}{R}}$
Hence,
$\displaystyle {\frac {U_{s}^{2}}{R}}={\frac {1}{\rho }}{\frac {\partial P}{\partial n}}+g{\frac {\partial x_{2}}{\partial n}}$
For an unsteady flow,
$\displaystyle {\frac {DU_{n}}{Dt}}=-{\frac {U_{s}^{2}}{R}}+{\frac {\partial U_{n}}{\partial t}}$
For steady flow neglecting body forces, the Euler's equation normal to the streamline is
$\displaystyle {\frac {1}{\rho }}{\frac {\partial P}{\partial n}}={\frac {U_{s}^{2}}{R}}$
which indicates that pressure increases in a direction outwards from the center of the curvature of the streamlines. In other words, pressure drops towards the center of curvature, which, consequently creates a potential difference in terms of pressure and forces the fluid to change its direction. For a straight streamline $\displaystyle R\rightarrow \infty$, there is no pressure variation normal to the streamline.

## Bernoulli equation: Integration of Euler's equation along a streamline for a steady flow

For a steady flow, Euler's equation along a streamline reads,
$\displaystyle {\frac {1}{\rho }}{\frac {\partial P}{\partial s}}+g{\frac {\partial x_{2}}{\partial s}}=-U_{s}{\frac {\partial U_{s}}{\partial s}}$
If a fluid particle moves a distance ds, along a streamline, since every variable becomes a function of $\displaystyle s$:
$\displaystyle {\frac {\partial P}{\partial s}}ds=dP{\text{:Change in pressure along s,}}$
$\displaystyle {\frac {\partial x_{2}}{\partial s}}ds=dx_{2}\ {\text{:Change in elevation along s,}}$
$\displaystyle {\frac {\partial U_{s}}{\partial s}}ds=dU_{s}\ {\text{:Change in velocity along s.}}$
Integration of the Euler equation between two locations, 1 and 2, along $\displaystyle s$ reads
$\displaystyle \int _{1}^{2}\left({{\frac {1}{\rho }}{\frac {\partial P}{\partial s}}+g{\frac {\partial x_{2}}{\partial s}}+U_{s}{\frac {\partial U_{s}}{\partial s}}}\right)ds=0$
For incompressible flow $\displaystyle \rho ={\textrm {constant}}$ and after changing the notation as: $\displaystyle U=U_{s}$ and $\displaystyle z=x_{2}$, the integration results in
$\displaystyle {\frac {P_{2}-P_{1}}{\rho }}+g(z_{2}-z_{1})+{\frac {U_{2}^{2}-U_{1}^{2}}{2}}=0{\textrm {\ along\ s}}$
or in its most beloved form:
$\displaystyle {\frac {P_{1}}{\rho }}+gz_{1}+{\frac {U_{1}^{2}}{2}}={\frac {P_{2}}{\rho }}+gz_{2}+{\frac {U_{2}^{2}}{2}}$
In other words along a streamline:
$\displaystyle {\frac {P}{\rho }}+gz+{\frac {U^{2}}{2}}={\textrm {constant}}$
Note that due to the assumptions made during the derivation, the following restrictions applies to this equation: The flow should be steady, incompressible, frictionless and the equation is valid only along a streamline.

### Different forms of Bernoulli equation

The common forms of Bernoulli equation are as follows:
Energy form (per unit mass)
$\displaystyle \underbrace {\frac {U^{2}}{2}} _{\text{kinetic energy}}+\underbrace {\frac {P}{\rho }} _{\text{pressure energy}}+\underbrace {gz} _{\text{potential energy}}=\zeta$
Pressure form
$\displaystyle \underbrace {\rho {\frac {U^{2}}{2}}} _{\text{dynamic pressure}}+\underbrace {P} _{\text{static pressure}}+\underbrace {\rho gz} _{\text{geodesic pressure}}=K$
Head form
$\displaystyle \underbrace {\frac {U^{2}}{2g}} _{\text{velocity head}}+\underbrace {\frac {P}{\rho g}} _{\text{pressure head}}+\underbrace {z} _{\text{geodesic head}}=k$

### Static, stagnation and dynamic pressures

How do we measure pressure? When the streamlines are parallel to the wall we can use pressure taps.
| If the measured location is far from the wall, static pressure measurements can be made by a static pressure probe. The stagnation pressure is the value obtained when a flowing fluid is decelerated to zero velocity by a frictionless flow process. The Stagnation pressure can be calculated as follows: $\displaystyle {\frac {P}{\rho }}+{\frac {U^{2}}{2}}=\;{\textrm {constant}}$ $\displaystyle {\frac {P_{0}}{\rho }}+{\frac {U_{0}^{2}}{2}}={\frac {P_{1}}{\rho }}+{\frac {U_{1}^{2}}{2}}$ when $\displaystyle U_{0}^{2}=0$ $\displaystyle P_{0}=P_{1}+\rho {\frac {U_{1}^{2}}{2}},$ where the last term is the dynamic pressure. | Pitot tube used in racing car |
If we know the pressure difference $\displaystyle P_{0}-P_{1}$, we can calculate the $\displaystyle U_{1}$ velocity.
$\displaystyle U_{1}={\sqrt {\frac {2\left(P_{0}-P_{1}\right)}{\rho }}}$
The stagnation pressure is measured in the laboratory using a probe that faces directly upstream flow.
Such a probe is called a stagnation pressure probe or Pitot tube . Thus, using a pressure tap and a Pitot tube one can measure local velocity:
$\displaystyle P_{0}=P_{A}+\rho {\frac {U_{A}}{2}}$
$\displaystyle P_{0}-P_{A}=\left(P_{A}+\rho {\frac {U_{A}^{2}}{2}}\right)-(P_{A})=\rho {\frac {U_{A}^{2}}{2}}$
Thus, measuring $\displaystyle P_{0}-P_{A}$ one can determine $\displaystyle U_{A}$.
* However, in the absence of a wall with well defined location, the velocity can be measured by a Pitot-static tube. The pressure is measured at B and C; assuming $\displaystyle P_{B}=P_{C}$.
* Hence,
* $\displaystyle U_{B}={\sqrt {\frac {2(P_{0_{B}}-P_{B})}{\rho }}}.$

### Unsteady Bernoulli equation

The Euler's equation along a streamline is:
$\displaystyle -{\frac {1}{\rho }}{\frac {\partial P}{\partial s}}-g{\frac {\partial z}{\partial s}}={\frac {DU_{s}}{Dt}}=U_{s}{\frac {\partial U_{s}}{\partial s}}+{\frac {\partial U_{s}}{\partial t}}$
along ds,
$\displaystyle -{\frac {1}{\rho }}{\frac {\partial P}{\partial s}}ds-g{\frac {\partial z}{\partial s}}ds=U_{s}{\frac {\partial U_{s}}{\partial s}}ds+{\frac {\partial U_{s}}{\partial t}}ds$
hence,
$\displaystyle -{\frac {1}{\rho }}dP-gdz=U_{s}dU_{s}+{\frac {\partial U_{s}}{\partial t}}ds.$
Integration between two points along a streamline is:
$\displaystyle -\int _{1}^{2}{\frac {dP}{\rho }}-\int _{1}^{2}{g}dz=\int _{1}^{2}{U_{s}dU_{s}}+\int _{1}^{2}{\frac {\partial U_{s}}{\partial t}}ds$
For incompressible flow, $\displaystyle \rho ={\textrm {constant}}$, thus the integral reads
$\displaystyle {\frac {P_{1}-P_{2}}{\rho }}+g(z_{1}-z_{2})={\frac {U_{2}^{2}-U_{1}^{2}}{2}}+\int _{1}^{2}{{\frac {\partial U_{s}}{\partial t}}ds}$
The unsetady Bernoulli equation involves the integration of the time gradient of the velocity between two points.:
$\displaystyle {\frac {P_{1}}{\rho }}+gz_{1}+{\frac {U_{1}^{2}}{2}}={\frac {P_{2}}{\rho }}+gz_{2}+{\frac {U_{2}^{2}}{2}}+\int _{1}^{2}{{\frac {\partial U_{s}}{\partial t}}ds}$