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.



Differential Approach seek solution at every point $$ \displaystyle (x_{1},x_{2},x_{3})$$, 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 $$ \displaystyle \alpha$$ be any flow variable (pressure, velocity, etc.). Eulerian approach deals with the description of $$ \displaystyle \alpha$$ at each location $$ \displaystyle (x_{i})$$ and time (t). For example, measurement of pressure at all $$ \displaystyle x_{i}$$ defines the pressure field: $$ \displaystyle P(x_{1},x_{2},x_{3},t)$$. Other field variables of the flow are:

$$ \displaystyle U_{j}(x_{i},t), P(x_{i},t), \rho(x_{i},t), T(x_{i}), \tau_{jk}(x_{i},t) \Rightarrow \alpha(x_{i},t)$$



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

$$ \displaystyle x_{i}|_{p}(t+\Delta t) = x_{i}|_{p}(t) + \int^{t+\Delta t}_{t}U_{i}|_{p}(t')dt'$$

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 $$ \displaystyle \alpha_{p}$$ be the variable of the particle (substance) P, this $$ \displaystyle \alpha_{p}$$ is called "substantial variable".

For this variable:

$$ \displaystyle \alpha_{p}(\vec{x_{p}},t)$$ and $$ \displaystyle \vec{x_{p}} = \vec{x_{p}}(t) \rightarrow \alpha_{p}(\vec{x_{p}},t) = \alpha_{p}(t)$$

In other words, one observes the change of variable $$ \displaystyle \alpha$$ 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 $$ \displaystyle x_{i}$$ for a given time, we can write the equality with the field variable:

$$ \displaystyle \alpha_{p}(t) = \alpha \left[(x_{i})_{p},t\right]$$

Along the path of the particle:

$$ \displaystyle \alpha_{p}(t + \Delta t) = \alpha \left[(x_{i} + \Delta x_{i})_{p}, t + \Delta t\right]$$

Hence,

$$ \displaystyle d\alpha_{p} = \alpha \left((x_{i} + \Delta x_{i})_{p}, t + \Delta t\right) - \alpha ((x_{i})_{p},t)$$

Using Taylor series approximation for $$\alpha \left((x_{i} + \Delta x_{i})_{p}, t + \Delta t\right) $$, the $$ \displaystyle d\alpha_{p} $$ can be written as

$$ \displaystyle d\alpha_{p} = \frac{\partial \alpha}{\partial t}dt + \frac{\partial \alpha}{\partial x_{i}}dx_{ip}$$

$$ \displaystyle \frac{d\alpha_{p}}{dt} = \frac{\partial \alpha}{\partial t} + \left(\frac{\partial \alpha}{\partial x_{i}}\right)\left(\frac{dx_{i}}{dt}\right)_{p} = \underbrace{\frac{\partial \alpha}{\partial t}}_{Local\ change\ in\ time} + \underbrace{\frac{\partial \alpha}{\partial x_{i}}U_{i}}_{Change\ in\ space}$$

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.

$$ \displaystyle \frac{d\alpha_{p}}{dt} = \frac{D\alpha}{Dt} = \left(\frac{\partial }{\partial t} + U_{i}\frac{\partial }{\partial x_{i}}\right)\alpha$$

The substantial derivative connects the Lagrangian and Eulerian variables.

Conservation of mass


The conservation of mass according to RTT

$$ \displaystyle \frac{d}{d t}\int_{CV}\rho\ dV + \int_{CS}\rho\;\vec{U}\cdot\vec{n}\ dA = 0 $$

or in tensor form

$$ \displaystyle \frac{d}{d t}\int_{CV}\rho\ dV + \int_{CS}\rho\;U_i n_i\ dA = 0 $$

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

$$ \displaystyle \frac{d}{d t}\int_{CV}\rho\ dV\approx\frac{d}{d t}\rho\int_{CV}\ dV=\frac{d}{d t}(\rho dV)=\rho\underbrace{\frac{\partial dV}{\partial t}}_{=0}+dV\frac{\partial \rho}{\partial t}= \frac{\partial\rho}{\partial t}dV $$

where $$dV=dx_1dx_2dx_3$$.The mass flow rate 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 integral of the mass flux 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:

$$ \displaystyle \frac{\partial \rho}{\partial t}dV + \frac{\partial\left(\rho U_{i}\right)}{\partial x_{i}}dV = 0$$

Dropping 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 incompressible 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 1
For a two dimensional, steady and incompressible flow in $$ \displaystyle x_1 x_2 $$ plane given by:

$$ \displaystyle \displaystyle U_1=A x_1$$

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 dimensions

$$ \displaystyle \frac{\partial U_{1}}{\partial x_{1}} + \frac{\partial U_{2}}{\partial x_{2}} =0 \ \ \rightarrow \ \ \frac{\partial U_{2}}{\partial x_{2}}=-\frac{\partial U_{1}}{\partial x_{1}} $$

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 2
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 equation simplifies to

$$ \displaystyle \frac{\partial \rho}{\partial t} + \rho\frac{\partial U_{1}}{\partial x_{1}} = 0$$

$$ \displaystyle \frac{\partial \rho}{\partial t}=-\rho\frac{\partial U_{1}}{\partial x_{1}} = -\rho \frac{V_p}{L}$$

$$ \displaystyle \displaystyle L=L_0+V_p t$$

$$ \displaystyle \frac{\partial \rho}{\partial t}=\frac{d\rho}{dt} = -\rho \frac{V_p}{L_0+V_p t}$$

$$ \displaystyle \int^{\rho}_{\rho_0}\frac{d\rho}{\rho} =\int^{t}_{0}-\frac{V_p}{L_0+V_pt}dt $$

$$ \displaystyle ln\left(\frac{\rho}{\rho_0}\right) =ln\left(\frac{L_0}{L_0+V_pt}\right) $$

$$ \displaystyle \rho(t)=\rho_0 \left(\frac{L_0}{L_0+V_pt}\right) $$

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 linear momentum (2nd law of Newton) 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 $$ \displaystyle dV $$, and $$ \displaystyle dV $$ 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 flow rate of the $$\displaystyle \rho U_i$$ momentum terms on the faces perpendicular to each 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 flow rate 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 flow rate 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:

$$ \displaystyle \sum F_{i} = \left[U_{i}\frac{\partial \rho}{\partial t} + \rho\frac{\partial U_{i}}{\partial t} + U_{i} \frac{\partial}{\partial x_{j}}(\rho\;U_{j}) + \rho\;U_{j}\;\frac{\partial U_{i}}{\partial x_{j}}\right]dV$$

$$ \displaystyle \sum F_{i}= U_{i}\underbrace{\left[\frac{\partial \rho}{\partial t}\;+ \frac{\partial}{\partial x_{j}}(\rho\;U_{j})\right]}_{continuity\ equation=0}dV\;+ \rho\underbrace{\left[\frac{\partial U_{i}}{\partial t}\;+ U_{j}\frac{\partial U_{i}}{\partial x_{j}}\right]}_{\frac{D U_{i}}{D t};subtantial\ derivative\ of\ U_i}dV$$

Hence

$$ \displaystyle \sum F_{i} = \rho\frac{D U_{i}}{D t} dV$$

Let's look to the forces on the differential 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:

$$ \displaystyle \sigma_{n} = \lim_{d A\rightarrow 0} \frac{d F_{n}}{d A}   $$

$$ \displaystyle \sigma_{t} = \lim_{d A\rightarrow 0} \frac{d F_{t}}{d A} $$

$$ \displaystyle \sigma_{n}$$ 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.

$$ \displaystyle \sigma_{ij}=\left[ \begin{matrix} \sigma_{11} & \sigma_{12} & \sigma_{13} \\ \sigma_{21} & \sigma_{22} & \sigma_{23} \\ \sigma_{31} & \sigma_{32} & \sigma_{33} \end{matrix} \right] $$



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 $$ \displaystyle (\sigma_{ij})$$ 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 (Convective 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 (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 $$ \displaystyle \tau_{ij}$$ 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:

$$ \displaystyle \tau_{ij} = \mu\left[\frac{\partial U_{i}}{\partial x_{j}} + \frac{\partial U_{j}}{\partial x_{i}}\right] - \frac{2}{3}\delta_{ij}\mu\frac{\partial U_{k}}{\partial x_{k}}$$

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

Thus, the momentum equation becomes

$$ \displaystyle \rho\frac{DU_{i}}{Dt} = \rho g_{i} - \frac{\partial P}{\partial x_{i}} + \frac{\partial}{\partial x_{j}}\left[\mu \left(\frac{\partial U_{i}}{\partial x_{j}} + \frac{\partial U_{j}}{\partial x_{i}}\right) - \frac{2}{3}\mu\ \delta_{ij}\ \frac{\partial U_{k}}{\partial x_{k}}\right]$$

or when the LHS(substantial derivative) is expanded: $$ \displaystyle \rho\frac{\partial U_{i}}{\partial t} + \rho U_{j}\frac{\partial U_{i}}{\partial x_{j}} = \rho g_{i} - \frac{\partial P}{\partial x_{i}} + \frac{\partial}{\partial x_{j}}\left[\mu \left(\frac{\partial U_{i}}{\partial x_{j}} + \frac{\partial U_{j}}{\partial x_{i}}\right) - \frac{2}{3}\mu\ \delta_{ij}\ \frac{\partial U_{k}}{\partial x_{k}}\right]$$

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

$$ \displaystyle \rho\frac{DU_{i}}{Dt} = \rho\ g_{i} -\frac{\partial P}{\partial x_{i}} + \mu\frac{\partial^{2}U_{i}}{\partial x_{j}^{2}} $$or when the LHS(substantial derivative) is expanded:$$ \displaystyle \rho\frac{\partial U_{i}}{\partial t} + \rho U_{j}\frac{\partial U_{i}}{\partial x_{j}} = \rho\ g_{i} -\frac{\partial P}{\partial x_{i}} + \mu\frac{\partial^{2}U_{i}}{\partial x_{j}^{2}} $$

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

$$ \displaystyle sin\beta \approx \frac{d x_{2}}{d s}=\frac{\partial x_{2}}{\partial s}$$

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.

$$ \displaystyle \rho dV\frac{DU_{n}}{Dt}= \left(P-\frac{\partial P}{\partial n}\frac{dn}{2}\right)ds\ dx_{3} - \left(P + \frac{\partial P}{\partial n}\frac{dn}{2}\right)ds\ dx_{3} - \rho g dV cos\beta $$

$$ \displaystyle \rho\ \frac{DU_{n}}{Dt}=\left(-\frac{\partial P}{\partial n} - \rho\ g\ cos\beta\right)$$

Since,

$$ \displaystyle cos\beta \approx \frac{d x_{2}}{d n}=\frac{\partial x_{2}}{\partial n}$$

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 $$ \displaystyle s$$:

$$ \displaystyle \frac{\partial P}{\partial s}ds = d P \text{: Differential change in pressure along ds,} $$

$$ \displaystyle \frac{\partial x_{2}}{\partial s}ds = dx_{2}\ \text{:Differential change in elevation along ds,} $$

$$ \displaystyle \frac{\partial U_{s}}{\partial s}ds = dU_{s}\ \text{:Differential change in velocity along ds.} $$

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: $$ \displaystyle \frac{P_1}{\rho} + g z_1 + \frac{U_1^{2}}{2} = \frac{P_2}{\rho} + gz_2 + \frac{U_2^{2}}{2} $$

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:

$$ \displaystyle \frac{P}{\rho} + \frac{U^{2}}{2} =\;\textrm{constant}$$

$$ \displaystyle \frac{P_{0}}{\rho} + \frac{U^{2}_{0}}{2} = \frac{P_{1}}{\rho} + \frac{U^{2}_{1}}{2}$$

when $$ \displaystyle U^{2}_{0} = 0$$

$$ \displaystyle P_{0} = P_{1} + \rho\frac{U^{2}_{1}}{2},$$

where the last term is the dynamic pressure.



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:

$$ \displaystyle P_{0} = P_{A} + \rho\frac{U_{A}}{2}$$

$$ \displaystyle P_{0} - P_{A} = \left(P_{A} + \rho\frac{U^{2}_{A}}{2}\right) - (P_{A}) = \rho\frac{U^{2}_{A}}{2}$$

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:

$$ \displaystyle -\int^{2}_{1}{\frac{dP}{\rho}} - \int^{2}_{1}{g}dz = \int^{2}_{1}{U_{s}dU_{s}} + \int^{2}_{1}{\frac{\partial U_{s}}{\partial t}}ds$$

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

$$ \displaystyle \frac{P_{1}-P_{2}}{\rho} + g(z_{1}- z_{2}) = \frac{U^{2}_{2}-U^{2}_{1}}{2} + \int^{2}_{1}{\frac{\partial U_{s}}{\partial t}ds}$$

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

$$ \displaystyle \frac{P_{1}}{\rho} + gz_{1} + \frac{U^{2}_{1}}{2} = \frac{P_{2}}{\rho} + gz_{2} + \frac{U^{2}_{2}}{2} + \int^{2}_{1}{\frac{\partial U_{s}}{\partial t}ds} $$