Fluid Mechanics for Mechanical Engineers/Differential Analysis of Fluid Flow

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Differential vs Integral Approach

Integral approach for a Control Volume (CV) is interested in a finite region and it determines gross flow effects such as force or torque on a body or the total energy exchange. For this purpose, balances of incoming and outgoing flux of mass, momentum and energy are made through this finite region. It gives very fast engineering answers, sometimes crude but useful.

File:Fluid Dynamics 1.png

Flow over an airfoil: Lagrangian vs Eulerian approach and differential vs integral approach.

Differential Approach seek solution at every point $(x_1,x_2,x_3)$, i.e describe the detailed flow pattern at all points. In other words, when we use differential relations, 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.

Lagrangian versus Eulerian Approach: Substantial Derivative

Let $\alpha$ be any flow variable (pressure, velocity, etc.). Eulerian approach deals with the description of $\alpha$ at each location $(x_i)$ and time (t). For example, measurement of pressure at all $x_i$ defines the pressure field: $P(x_1,x_2,x_3,t)$. Other field variables of the flow are:

Template:Center top$U_j(x_i,t),P(x_i,t),\rho (x_i,t),T(x_i),{\tau }_{jk}(x_i,t)\Rightarrow \alpha (x_i,t)$Template:Center bottom

File:Fluid Dynamics 3.png

Path of a fluid element

Lagrangian approach tracks a fluid particle and determines its properties as it moves.

Template:Center top$x_i{|}_p(t+\mathrm{\Delta }t)=x_i{|}_p(t)+\underset{t}{\overset{t+\mathrm{\Delta }t}{\int }}{U}_i{|}_p(t^{\prime })d{t}^{\prime }$Template:Center bottom

Oceanographic measurements made with floating sensors delivering location, pressure and temperature data, is one example of this approach. X-ray opaque dyes, which are used to trace blood flow in arteries, is another example.

Let ${\alpha }_p$ be the variable of the particle (substance) P, this ${\alpha }_p$ is called "substantial variable".

For this variable:

${\alpha }_p(\overrightarrow{x_p},t)$ and $\overrightarrow{x_p}=\overrightarrow{x_p}(t)\to {\alpha }_p(\overrightarrow{x_p},t)={\alpha }_p(t)$

In other words, one observes the change of variable $\alpha$ for a selected amount of mass of fixed identity, such that for the fluid particle, every change is a function of time only.

In a fluid flow, due to excessive number of fluid particles, Lagrangian approach is not widely used.

Thus, for a particle P finding itself at point $x_i$ for a given time, we can write the equality with the field variable:

Template:Center top${\alpha }_p(t)=\alpha [(x_i{)}_p,t]$Template:Center bottom

Along the path of the particle:

Template:Center top${\alpha }_p(t+\mathrm{\Delta }t)=\alpha [(x_i+\mathrm{\Delta }{x}_i{)}_p,t+\mathrm{\Delta }t]$Template:Center bottom

Hence,

Template:Center top$d{\alpha }_p=\alpha ((x_i+\mathrm{\Delta }{x}_i{)}_p,t+\mathrm{\Delta }t)-\alpha ((x_i{)}_p,t)$Template:Center bottom

Using Taylor series approximation for $\alpha ((x_i+\mathrm{\Delta }{x}_i{)}_p,t+\mathrm{\Delta }t)$, the $d{\alpha }_p$ can be written as

Template:Center top$d{\alpha }_p=\frac{\partial \alpha }{\partial t}dt+\frac{\partial \alpha }{\partial x_i}d{x}_{ip}$Template:Center bottom

Template:Center top$\frac{d{\alpha }_p}{dt}=\frac{\partial \alpha }{\partial t}+\left(\frac{\partial \alpha }{\partial x_i}\right){\left(\frac{d{x}_i}{dt}\right)}_p=\underset{Localchangeintime}{\underbrace{\frac{\partial \alpha }{\partial t}}}+\underset{Changeinspace}{\underbrace{\frac{\partial \alpha }{\partial x_i}{U}_i}}$Template:Center bottom

The local change in time is the local time derivative (unsteadiness of the flow) and the change in space is the change along the path of the particle by means of the convective derivative.

Template:Center top$\frac{d{\alpha }_p}{dt}=\frac{D\alpha }{Dt}=(\frac{\partial }{\partial t}+U_i\frac{\partial }{\partial x_i})\alpha$Template:Center bottom

The substantial derivative connects the Lagrangian and Eulerian variables.

Conservation of mass

File:Conservation of mass.png

The differential control volume dV and the mass flux through its surfaces

The conservation of mass according to RTT

Template:Center top$\frac{d}{dt}\underset{CV}{\int }\rho dV+\underset{CS}{\int }\rho \overrightarrow{U}\cdot \overrightarrow{n}dA=0$Template:Center bottom

or in tensor form

Template:Center top$\frac{d}{dt}\underset{CV}{\int }\rho dV+\underset{CS}{\int }\rho U_i{n}_{i}dA=0$Template:Center bottom

The differential volume is selected to be so small that density $(\rho )$ can be accepted to be uniform within this volume and it can be taken out of the integral. Since the volume is constant, the first integral in the above equation is:

    Template:Center top$\frac{d}{dt}\underset{CV}{\int }\rho dV\approx \frac{d}{dt}\rho \underset{CV}{\int }dV=\frac{d}{dt}(\rho dV)=\rho \underset{=0}{\underbrace{\frac{\partial dV}{\partial t}}}+dV\frac{\partial \rho }{\partial t}=\frac{\partial \rho }{\partial t}dV$Template:Center bottom

where $dV=d{x}_{1}d{x}_{2}d{x}_3$.The mass flow rate term (second integral term) in the equation of conservation of mass can be analyzed in groups:

     Template:Center top$\underset{CS}{\int }\rho U_i{n}_{i}dA=\underset{CS{x}_1}{\int }\rho U_i{n}_{i}dA+\underset{CS{x}_2}{\int }\rho U_i{n}_{i}dA+\underset{CS{x}_3}{\int }\rho U_i{n}_{i}dA$Template:Center bottom

Let's look to the surfaces perpendicular to $x_1-axis$

Template:Center top$\underset{CS{x}_1}{\int }\rho U_i{n}_{i}dA=-\rho U_{1}d{x}_{2}d{x}_3+[\rho U_1+\frac{\partial \left(\rho U_1\right)}{\partial x_1}d{x}_1]d{x}_{2}d{x}_3$Template:Center bottom

Template:Center top$=\frac{\partial \left(\rho U_1\right)}{\partial x_1}d{x}_{1}d{x}_{2}d{x}_3=\frac{\partial \left(\rho U_1\right)}{\partial x_1}dV$Template:Center bottom

Similarly, the flux integrals through surfaces perpendicular to $x_2-axis$ and $x_3-\text{axis}$ are

      Template:Center top$\underset{CS{x}_2}{\int }\rho U_i{n}_{i}dA=\frac{\partial \left(\rho U_2\right)}{\partial x_2}dV,$Template:Center bottom

      Template:Center top$\underset{CS{x}_3}{\int }\rho U_i{n}_{i}dA=\frac{\partial \left(\rho U_3\right)}{\partial x_3}dV.$Template:Center bottom

Hence the integral of the mass flux reads;

        Template:Center top$\underset{CS}{\int }\rho U_i{n}_{i}dA=\frac{\partial \left(\rho U_i\right)}{\partial x_i}dV$Template:Center bottom

The conservation of mass equation becomes:

        Template:Center top$\frac{\partial \rho }{\partial t}dV+\frac{\partial \left(\rho U_i\right)}{\partial x_i}dV=0$Template:Center bottom

Dropping the $dV$, we reach to the final form of the conservation of mass:

Template:Center top$\frac{\partial \rho }{\partial t}+\frac{\partial \left(\rho U_i\right)}{\partial x_i}=0$Template:Center bottom

This equation is also called continuity equation. It can be written in vector form as:

Template:Center top$\frac{\partial \rho }{\partial t}+\mathrm{\nabla }\cdot \left(\rho \overrightarrow{U}\right)=0\text{where}\mathrm{\nabla }=\frac{\partial }{\partial x_1}\overrightarrow{e_1}+\frac{\partial }{\partial x_2}\overrightarrow{e_2}+\frac{\partial }{\partial x_3}\overrightarrow{e_3}\text{gradient operator}$Template:Center bottom

For a steady flow, continuity equation becomes:

      Template:Center top$\frac{\partial \left(\rho U_i\right)}{\partial x_i}=0$Template:Center bottom.

For incompressible flow, i.e. $\rho =\text{constant}$:

      Template:Center top$\frac{\partial U_i}{\partial x_i}=0\text{i.e.}\frac{\partial U_1}{\partial x_1}+\frac{\partial U_2}{\partial x_2}+\frac{\partial U_3}{\partial x_3}=0$Template:Center bottom

Examples

Example 1

For a two dimensional, steady and incompressible flow in $x_1{x}_2$ plane given by:

Template:Center top$U_1=A{x}_1$Template:Center bottom

Find how many possible $U_2$ can exist.

For incompressible steady flow:

Template:Center top$\rho \frac{\partial U_i}{\partial x_i}=0$Template:Center bottom

in two dimensions

Template:Center top$\frac{\partial U_1}{\partial x_1}+\frac{\partial U_2}{\partial x_2}=0\to \frac{\partial U_2}{\partial x_2}=-\frac{\partial U_1}{\partial x_1}$Template:Center bottom

Thus,

Template:Center top$\frac{\partial U_2}{\partial x_2}=-A$Template:Center bottom

This is an expression for the rate of change of $U_2$ velocity while keeping

$x_1$ constant. Therefore the integral of this equation reads

Template:Center top$U_2=-A{x}_2+f(x_1)$Template:Center bottom

Thus, any function $f(x_1)$ is allowable.

Example 2

Consider one-dimensional flow in the piston. The piston suddenly moves with the velocity $V_p$ .

File:Example-5-2.png

Compressible and unsteady flow inside a piston (example 2)

Assume uniform $\rho (t)$ in the piston and a linear change of velocity $U_1$ such that $U_1=0$ at the bottom ($x_1=0$) and $U_1=V_p$ on the piston ($x_1=L$), i.e.

Template:Center top$U_1=\frac{x_1}{L}{V}_p$Template:Center bottom

Obtain a function for the density as a function of time.

The conservation of mass equation is:

Template:Center top$\frac{\partial \rho }{\partial t}+\frac{\partial \left(\rho U_i\right)}{\partial x_i}=0$Template:Center bottom

For one-dimensional flow and uniform $\rho$, this equation simplifies to

Template:Center top$\frac{\partial \rho }{\partial t}+\rho \frac{\partial U_1}{\partial x_1}=0$Template:Center bottom

Template:Center top$\frac{\partial \rho }{\partial t}=-\rho \frac{\partial U_1}{\partial x_1}=-\rho \frac{V_p}{L}$Template:Center bottom

Template:Center top$L=L_0+V_{p}t$Template:Center bottom

Template:Center top$\frac{\partial \rho }{\partial t}=\frac{d\rho }{dt}=-\rho \frac{V_p}{L_0+V_{p}t}$Template:Center bottom

Template:Center top$\underset{{\rho }_0}{\overset{\rho }{\int }}\frac{d\rho }{\rho }=\underset{0}{\overset{t}{\int }}-\frac{V_p}{L_0+V_{p}t}dt$Template:Center bottom

Template:Center top$ln\left(\frac{\rho }{{\rho }_0}\right)=ln\left(\frac{L_0}{L_0+V_{p}t}\right)$Template:Center bottom

Template:Center top$\rho (t)={\rho }_0\left(\frac{L_0}{L_0+V_{p}t}\right)$Template:Center bottom

The same problem can be solved by using the integral approach with a deforming control volume.

The differential equation of linear momentum

File:Linear momentum image.png

The differential control volume dV and the flux of $\rho U_i$ (momentum per unit volume in i-direction) through the surfaces perpendicular to $x_j$ axis

The integral equation for the linear momentum (2nd law of Newton) is:

Template:Center top$\sum F_i=\frac{\partial }{\partial t}\underset{CV}{\int }\rho U_{i}dV+\underset{CS}{\int }\rho U_i{U}_j{n}_{j}dA$Template:Center bottom

For the first integral we assume $\rho$ and $U_i$ are uniform within $dV$, and $dV$ is so small that:

Template:Center top$\frac{\partial }{\partial t}\underset{CV}{\int }\rho U_{i}dV\approx \frac{\partial }{\partial t}(\rho U_i)d{x}_{1}d{x}_{2}d{x}_3$Template:Center bottom

Analyze the flow rate of the $\rho U_i$ momentum terms on the faces perpendicular to each axis:

Template:Center top$\underset{CS}{\int }\rho U_i{U}_j{n}_{j}dA=\underset{CS{x}_1}{\int }\rho U_i{U}_j{n}_{j}dA+\underset{CS{x}_2}{\int }\rho U_i{U}_j{n}_{j}dA+\underset{CS{x}_3}{\int }\rho U_i{U}_j{n}_{j}dA$Template:Center bottom

First consider the flow rate of $\rho U_i$ (momentum per unit volume in i-direction) through the surfaces perpendicular to $x_1$ axis:

Template:Center top$\underset{CS{x}_1}{\int }\rho U_i{U}_j{n}_{j}dA=-\rho U_i{U}_{1}d{x}_{2}d{x}_3+[\rho U_i{U}_{1}d{x}_{2}d{x}_3+\frac{\partial }{\partial x_1}(\rho U_i{U}_1)d{x}_1]d{x}_{2}d{x}_3$Template:Center bottom

      Template:Center top$=\frac{\partial }{\partial x_1}(\rho U_i{U}_1)dV$Template:Center bottom

Similarly, the momentum flow rate through the surfaces in other directions read

        Template:Center top$\underset{CS{x}_2}{\int }\rho U_i{U}_j{n}_{j}dA=\frac{\partial }{\partial x_2}(\rho U_i{U}_2)dV$Template:Center bottom,

        Template:Center top$\underset{CS{x}_3}{\int }\rho U_i{U}_j{n}_{j}dA=\frac{\partial }{\partial x_3}(\rho U_i{U}_3)dV$Template:Center bottom.
        

Rearranging the equation for $\sum F_i$ we obtain:

      Template:Center top$\sum F_i=[\frac{\partial }{\partial t}(\rho U_i)+\frac{\partial }{\partial x_j}(\rho U_i{U}_j)]dV$Template:Center bottom

We can simplify further:

Template:Center top$\sum F_i=[U_i\frac{\partial \rho }{\partial t}+\rho \frac{\partial U_i}{\partial t}+U_i\frac{\partial }{\partial x_j}(\rho U_j)+\rho U_j\frac{\partial U_i}{\partial x_j}]dV$Template:Center bottom

Template:Center top$\sum F_i=U_i\underset{continuityequation=0}{\underbrace{[\frac{\partial \rho }{\partial t}+\frac{\partial }{\partial x_j}(\rho U_j)]}}dV+\rho \underset{\frac{D{U}_i}{Dt};subtantialderivativeof{U}_i}{\underbrace{[\frac{\partial U_i}{\partial t}+U_j\frac{\partial U_i}{\partial x_j}]}}dV$Template:Center bottom

Hence

       Template:Center top$\sum F_i=\rho \frac{D{U}_i}{Dt}dV$Template:Center bottom    

Let's look to the forces on the differential control volume:

Template:Center top$\frac{d{P}_i}{dt}=\sum F_i=d{F}_{bodyi}+d{F}_{surfacei}=\rho \frac{d{U}_i}{dt}dV$Template:Center bottom

Here, only gravitational force is considered as a body force. Thus,

Template:Center top$d{F}_{bodyi}=\rho dV{g}_i$Template:Center bottom

File:Planopablo.png

Differential surface forces

Surface forces are the stresses acting on the control surfaces. $F_s$ can be resolved into three components. $d{F}_n$ is normal to dA. $d{F}_t$ are tangent to dA:

Template:Center top${\sigma }_n=\underset{dA\to 0}{\lim }\frac{d{F}_n}{dA}$Template:Center bottom

Template:Center top${\sigma }_t=\underset{dA\to 0}{\lim }\frac{d{F}_t}{dA}$Template:Center bottom

${\sigma }_n$ is a normal stress whereas ${\sigma }_t$ is a shear stress. The shear stresses are also designated by $\tau$.

File:Surface stresses.png

Stresses on the surface of differential control volume

Thus, the surface forces are due to stresses on the surfaces of the control surface.

Template:Center top${\sigma }_{ij}=\left[\begin{array}{ccc}{\sigma }_{11}& {\sigma }_{12}& {\sigma }_{13}\\ {\sigma }_{21}& {\sigma }_{22}& {\sigma }_{23}\\ {\sigma }_{31}& {\sigma }_{32}& {\sigma }_{33}\end{array}\right]$Template:Center bottom

File:Surface stresses 2.png

The positive stress directions

A stress component is positive on a face, when it is in the same direction as the outwards normal vector as shown on ABCD or A'B'C'D'. Note that on A'B'C'D', the stresses are considered to be positive, tough the surface normal is in the $-x_1$ direction.

The stresses on the surface $({\sigma }_{ij})$ are the sum of pressure plus the viscous stresses which arise from motion with velocity gradients:

Template:Center top${\sigma }_{ij}=\left[\begin{array}{ccc}-P& 0& 0\\ 0& -P& 0\\ 0& 0& -P\end{array}\right]+\left[\begin{array}{ccc}{\tau }_{11}& {\tau }_{12}& {\tau }_{13}\\ {\tau }_{21}& {\tau }_{22}& {\tau }_{23}\\ {\tau }_{31}& {\tau }_{32}& {\tau }_{33}\end{array}\right]=\left[\begin{array}{ccc}-P+{\tau }_{11}& {\tau }_{12}& {\tau }_{13}\\ {\tau }_{21}& -P+{\tau }_{22}& {\tau }_{23}\\ {\tau }_{31}& {\tau }_{32}& -P+{\tau }_{33}\end{array}\right]$Template:Center bottom

$p$ has a minus sign since the force due to pressure acts opposite to the surface normal.

File:Surface stresses 3.png

Stresses on the surface of the differential control volume in the x1 direction

Let us look to the differential surface force in the $x_1$ direction:

Template:Center top$d{F}_{surface1}=\frac{\partial {\sigma }_{11}}{\partial x_1}d{x}_{1}d{x}_{2}d{x}_3+\frac{\partial {\sigma }_{21}}{\partial x_2}d{x}_{1}d{x}_{2}d{x}_3+\frac{\partial {\sigma }_{31}}{\partial x_3}d{x}_{1}d{x}_{2}d{x}_3$Template:Center bottom

Noting that $dV=d{x}_{1}d{x}_{2}d{x}_3$ and ${\sigma }_{ij}=-p{\delta }_{ij}+{\tau }_{ij}$ (4),

Template:Center top$d{F}_{surface1}=(-\frac{\partial P}{\partial x_1}+\frac{\partial {\tau }_{11}}{\partial x_1}+\frac{\partial {\tau }_{21}}{\partial x_2}+\frac{\partial {\tau }_{31}}{\partial x_3})dV$Template:Center bottom

Thus in tensor form the differential surface forces in $i$'th direction can be written as

Template:Center top$d{F}_{surfacei}=(-\frac{\partial P}{\partial x_i}+\frac{\partial {\tau }_{ji}}{\partial x_j})dV$Template:Center bottom

Note that ${\tau }_{ij}$ is a symmetric tensor, i.e.

Template:Center top${\tau }_{ji}={\tau }_{ij}$Template:Center bottom

Hence, the diffential surface forces reads:

Template:Center top$d{F}_{surfacei}=(-\frac{\partial P}{\partial x_i}+\frac{\partial {\tau }_{ij}}{\partial x_j})dV$Template:Center bottom

Inserting $d{F}_{bodyi}$ and $d{F}_{surfacei}$ into (2),

Template:Center top$\rho \frac{D{U}_i}{Dt}dV=\rho g_{i}dV+(-\frac{\partial P}{\partial x_i}+\frac{{\tau }_{ij}}{\partial x_j})dV$Template:Center bottom and canceling $dV$ we obtain

Template:Center top$\rho \frac{D{U}_i}{Dt}=\rho g_i-\frac{\partial P}{\partial x_i}+\frac{\partial {\tau }_{ij}}{\partial x_j}.$Template:Center bottom

Expanding the substatial derivative at the left hand side,

Template:Center top$\rho \frac{\partial U_i}{\partial t}+\rho U_j\frac{\partial U_i}{\partial x_j}=\rho g_i-\frac{\partial P}{\partial x_i}+\frac{\partial {\tau }_{ij}}{\partial x_j}$Template:Center bottom

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 $2^{nd}$ term (Convective term) at the LHS. Efect of Newtonian and Non-newtonian properties appears in the formulation of the viscous stresses ${\tau }_{ij}$. ${\tau }_{ij}$ 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 ${\tau }_{ij}$ can be expressed as momentum transfer per unit area and time (diffusive momentum flux). Thus it can be considered as molecular momentum transport term. Derivations based on this concept requires a molecular approach. The students should be aware that ${\tau }_{ij}$ causes momentum transport when there is a gradient of velocity.

Closure Problem

The conservation of mass and momentum equation form a system of equations. At isothermal conditions, there are 11 unknowns involved in this 4 equation system namely $\rho ,P,U_i\text{and}{\tau }_{i,j}$ where ${\tau }_{i,j}={\tau }_{j,i}$ and $i,j=1,2,3$. The equation system is not closed, i.e. it is impossible to solve this equation. Extra equations are necessary for the closure. This was achieved by Navier and Stokes by relating the stress term to the deformation rate of the fluid, i.e. to the velocity gradients and, consequently, introducing the viscous effect into the momentum equation.

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

For a Newtonian fluid, the viscous stresses are defined as:

Template:Center top${\tau }_{ij}=\mu [\frac{\partial U_i}{\partial x_j}+\frac{\partial U_j}{\partial x_i}]-\frac{2}{3}{\delta }_{ij}\mu \frac{\partial U_k}{\partial x_k}$Template:Center bottom

Note that derivation of this relation is beyond the scaope of this course.

Thus, the momentum equation becomes

Template:Center top$\rho \frac{D{U}_i}{Dt}=\rho g_i-\frac{\partial P}{\partial x_i}+\frac{\partial }{\partial x_j}[\mu (\frac{\partial U_i}{\partial x_j}+\frac{\partial U_j}{\partial x_i})-\frac{2}{3}\mu {\delta }_{ij}\frac{\partial U_k}{\partial x_k}]$Template:Center bottom

or when the LHS(substantial derivative) is expanded: Template:Center top$\rho \frac{\partial U_i}{\partial t}+\rho U_j\frac{\partial U_i}{\partial x_j}=\rho g_i-\frac{\partial P}{\partial x_i}+\frac{\partial }{\partial x_j}[\mu (\frac{\partial U_i}{\partial x_j}+\frac{\partial U_j}{\partial x_i})-\frac{2}{3}\mu {\delta }_{ij}\frac{\partial U_k}{\partial x_k}]$Template:Center bottom

For a flow with constant viscosity ($\mu =\text{constant}$):

Template:Center top$\rho \frac{D{U}_i}{Dt}=\rho g_i-\frac{\partial P}{\partial x_i}+\mu \frac{\partial }{\partial x_j}[\frac{\partial U_i}{\partial x_j}+\frac{\partial U_j}{\partial x_i}-\frac{2}{3}{\delta }_{ij}\frac{\partial U_k}{\partial x_k}]$Template:Center bottom

since,

Template:Center top$\frac{{\partial }^2{U}_i}{\partial x_j\partial x_i}=\frac{{\partial }^2{U}_i}{\partial x_i\partial x_j}=\frac{\partial }{\partial x_j}\left(\frac{\partial U_i}{\partial x_i}\right)=\frac{\partial }{\partial x_i}\left(\frac{\partial U_i}{\partial x_j}\right)$Template:Center bottom

then,

Template:Center top$\rho \frac{D{U}_i}{Dt}=\rho g_i-\frac{\partial P}{\partial x_i}+\mu \frac{{\partial }^2{U}_i}{\partial x_j{x}_j}+\mu \frac{\partial }{\partial x_i}\left(\frac{\partial U_j}{\partial x_j}\right)-\frac{2}{3}{\delta }_{ij}\mu \frac{\partial }{\partial x_j}\left(\frac{\partial U_k}{\partial x_k}\right)$Template:Center bottom

For an incompressible flow

Template:Center top$\frac{\partial U_k}{\partial x_k}=\frac{\partial U_j}{\partial x_j}=0$Template:Center bottom

hence assuming that the viscosity is constant, it can be easily shown that the momentum equation reduces to

Template:Center top$\rho \frac{D{U}_i}{Dt}=\rho g_i-\frac{\partial P}{\partial x_i}+\mu \frac{{\partial }^2{U}_i}{\partial x_j^2}$Template:Center bottomor when the LHS(substantial derivative) is expanded:Template:Center top$\rho \frac{\partial U_i}{\partial t}+\rho U_j\frac{\partial U_i}{\partial x_j}=\rho g_i-\frac{\partial P}{\partial x_i}+\mu \frac{{\partial }^2{U}_i}{\partial x_j^2}$Template:Center bottom

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: Template:Center top${\tau }_{ij}=0$Template:Center bottom

Since, the viscous stresses are proportional to viscosity: Template:Center top${\tau }_{ij}\propto \mu$Template:Center bottom for flows, where ${\tau }_{ij}$ 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

Template:Center top$\rho \frac{D{U}_i}{Dt}=\rho g_i-\frac{\partial P}{\partial x_i}.$Template:Center bottom

Euler's equation in streamline coordinates

File:Eulers Eq.png

Differential control volume along streamline coordinates and the forces on it for a inviscid flow

File:Eulers Eq 2.png

Differential control volume along streamline coordinates and the forces on it for a inviscid flow

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.

Template:Center top$\rho dV[\frac{\partial U_s}{\partial t}+U_s\frac{\partial U_s}{\partial s}]=-\frac{\partial P}{\partial s}dV-\rho gsin\beta dV$Template:Center bottom

Omitting $dV$ would deliver

Template:Center top$\frac{D{U}_s}{Dt}=-\frac{1}{\rho }\frac{\partial P}{\partial s}-gsin\beta$Template:Center bottom

Since

Template:Center top$sin\beta \approx \frac{d{x}_2}{ds}=\frac{\partial x_2}{\partial s}$Template:Center bottom

then the Equler's equation along a streamline reads

Template:Center top$\frac{D{U}_s}{Dt}=-\frac{1}{\rho }\frac{\partial P}{\partial s}-g\frac{\partial x_2}{\partial s}(8)$Template:Center bottom

For a steady flow and by neglecting body forces,

Template:Center top$\frac{1}{\rho }\frac{\partial P}{\partial s}=-U_s\frac{\partial U_s}{\partial s}$Template:Center bottom

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.

Template:Center top$\rho dV\frac{D{U}_n}{Dt}=(P-\frac{\partial P}{\partial n}\frac{dn}{2})dsd{x}_3-(P+\frac{\partial P}{\partial n}\frac{dn}{2})dsd{x}_3-\rho gdVcos\beta$Template:Center bottom

Template:Center top$\rho \frac{D{U}_n}{Dt}=(-\frac{\partial P}{\partial n}-\rho gcos\beta )$Template:Center bottom

Since,

Template:Center top$cos\beta \approx \frac{d{x}_2}{dn}=\frac{\partial x_2}{\partial n}$Template:Center bottom

Then,

Template:Center top$\frac{D{U}_n}{Dt}=-\frac{1}{\rho }\frac{\partial P}{\partial n}-g\frac{\partial x_2}{\partial n}$Template:Center bottom

For a steady flow, the normal acceleration of the fluid is towards the center of curvature of the streamline:

Template:Center top$\frac{D{U}_n}{Dt}=-\frac{U_s^2}{R}$Template:Center bottom

Hence,

Template:Center top$\frac{U_s^2}{R}=\frac{1}{\rho }\frac{\partial P}{\partial n}+g\frac{\partial x_2}{\partial n}$Template:Center bottom

For an unsteady flow,

Template:Center top$\frac{D{U}_n}{Dt}=-\frac{U_s^2}{R}+\frac{\partial U_n}{\partial t}$Template:Center bottom

For steady flow neglecting body forces, the Euler's equation normal to the streamline is

Template:Center top$\frac{1}{\rho }\frac{\partial P}{\partial n}=\frac{U_s^2}{R}$Template:Center bottom

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 $R\to \mathrm{\infty }$, 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,

Template:Center top$\frac{1}{\rho }\frac{\partial P}{\partial s}+g\frac{\partial x_2}{\partial s}=-U_s\frac{\partial U_s}{\partial s}$Template:Center bottom

If a fluid particle moves a distance ds, along a streamline, since every variable becomes a function of $s$:

Template:Center top$\frac{\partial P}{\partial s}ds=dP\text{: Differential change in pressure along ds,}$Template:Center bottom

Template:Center top$\frac{\partial x_2}{\partial s}ds=d{x}_2\text{:Differential change in elevation along ds,}$Template:Center bottom

Template:Center top$\frac{\partial U_s}{\partial s}ds=d{U}_s\text{:Differential change in velocity along ds.}$Template:Center bottom

Integration of the Euler equation between two locations, 1 and 2, along $s$ reads Template:Center top$\underset{1}{\overset{2}{\int }}\left(\frac{1}{\rho }\frac{\partial P}{\partial s}+g\frac{\partial x_2}{\partial s}+U_s\frac{\partial U_s}{\partial s}\right)ds=0$Template:Center bottom

For incompressible flow $\rho =\text{constant}$ and after changing the notation as: $U=U_s$ and $z=x_2$, the integration results in Template:Center top$\frac{P_2-P_1}{\rho }+g(z_2-z_1)+\frac{U_2^2-U_1^2}{2}=0\text{along}s$Template:Center bottom or in its most beloved form: Template:Center top$\frac{P_1}{\rho }+g{z}_1+\frac{U_1^2}{2}=\frac{P_2}{\rho }+g{z}_2+\frac{U_2^2}{2}$Template:Center bottom

In other words along a streamline:

Template:Center top$\frac{P}{\rho }+gz+\frac{U^2}{2}=\text{constant}$Template:Center bottom

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)

Template:Center top$\underset{\text{kinetic energy}}{\underbrace{\frac{U^2}{2}}}+\underset{\text{pressure energy}}{\underbrace{\frac{P}{\rho }}}+\underset{\text{potential energy}}{\underbrace{gz}}=\zeta$Template:Center bottom

Pressure form

Template:Center top$\underset{\text{dynamic pressure}}{\underbrace{\rho \frac{U^2}{2}}}+\underset{\text{static pressure}}{\underbrace{P}}+\underset{\text{geodesic pressure}}{\underbrace{\rho gz}}=K$Template:Center bottom

Head form

Template:Center top$\underset{\text{velocity head}}{\underbrace{\frac{U^2}{2g}}}+\underset{\text{pressure head}}{\underbrace{\frac{P}{\rho g}}}+\underset{\text{geodesic head}}{\underbrace{z}}=k$Template:Center bottom

Static, stagnation and dynamic pressures

How do we measure pressure? When the streamlines are parallel to the wall, we can use pressure taps.

Pitot tube used in racing car

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:

Template:Center top$\frac{P}{\rho }+\frac{U^2}{2}=\text{constant}$Template:Center bottom

Template:Center top$\frac{P_0}{\rho }+\frac{U_0^2}{2}=\frac{P_1}{\rho }+\frac{U_1^2}{2}$Template:Center bottom

when $U_0^2=0$

Template:Center top$P_0=P_1+\rho \frac{U_1^2}{2},$Template:Center bottom

where the last term is the dynamic pressure.

File:Pitot tube principle.svg

Figure: a)Jet is impinging to the wall and stagnating at the point of impingment b)Schematics of a of a pressure tap in a channel c)Static pressure probe

If we know the pressure difference $P_0-P_1$, we can calculate the $U_1$ velocity.

Template:Center top$U_1=\sqrt{\frac{2(P_0-P_1)}{\rho }}$Template:Center bottom

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:

Template:Center top$P_0=P_A+\rho \frac{U_A}{2}$Template:Center bottom

Template:Center top$P_0-P_A=(P_A+\rho \frac{U_A^2}{2})-(P_A)=\rho \frac{U_A^2}{2}$Template:Center bottom

Thus, measuring $P_0-P_A$ one can determine $U_A$.

File:Pitot tube mdfd 02.svg

Pitot-static tube for velocity measurement

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 $P_B=P_C$.

Hence,

$U_B=\sqrt{\frac{2(P_{0_B}-P_B)}{\rho }}.$

Unsteady Bernoulli equation

The Euler's equation along a streamline is:

Template:Center top$-\frac{1}{\rho }\frac{\partial P}{\partial s}-g\frac{\partial z}{\partial s}=\frac{D{U}_s}{Dt}=U_s\frac{\partial U_s}{\partial s}+\frac{\partial U_s}{\partial t}$Template:Center bottom

along ds,

Template:Center top$-\frac{1}{\rho }\frac{\partial P}{\partial s}ds-g\frac{\partial z}{\partial s}ds=U_s\frac{\partial U_s}{\partial s}ds+\frac{\partial U_s}{\partial t}ds$Template:Center bottom

hence,

Template:Center top$-\frac{1}{\rho }dP-gdz=U_{s}d{U}_s+\frac{\partial U_s}{\partial t}ds.$Template:Center bottom

Integration between two points along a streamline is:

Template:Center top$-\underset{1}{\overset{2}{\int }}\frac{dP}{\rho }-\underset{1}{\overset{2}{\int }}gdz=\underset{1}{\overset{2}{\int }}{U}_{s}d{U}_s+\underset{1}{\overset{2}{\int }}\frac{\partial U_s}{\partial t}ds$Template:Center bottom

For incompressible flow, $\rho =\text{constant}$, thus the integral reads

Template:Center top$\frac{P_1-P_2}{\rho }+g(z_1-z_2)=\frac{U_2^2-U_1^2}{2}+\underset{1}{\overset{2}{\int }}\frac{\partial U_s}{\partial t}ds$Template:Center bottom

The unsteady Bernoulli equation involves the integration of the time gradient of the velocity between two points.:

Template:Center top$\frac{P_1}{\rho }+g{z}_1+\frac{U_1^2}{2}=\frac{P_2}{\rho }+g{z}_2+\frac{U_2^2}{2}+\underset{1}{\overset{2}{\int }}\frac{\partial U_s}{\partial t}ds$Template:Center bottom