# 0.5 Equations of motion and energy in cartesian coordinates  (Page 2/12)

 Page 2 / 12

We will consider the case of single-phase flow with conservative body forces (e.g., gravitational) and density a single valued function of pressure. The pressure and potential from the body force can be combined into a single potential.

$\begin{array}{c}\mathbf{f}-\frac{1}{\rho }\nabla p=-\nabla \Omega \hfill \\ where\hfill \\ \Omega ={\int }^{p}\frac{dp}{\rho }-gz\hfill \end{array}$

If the change in density is small enough, the potential can be approximated by potential that has the units of pressure.

$\begin{array}{c}\Omega \approx \frac{P}{\rho },\phantom{\rule{4.pt}{0ex}}\text{small}\phantom{\rule{4.pt}{0ex}}\text{change}\phantom{\rule{4.pt}{0ex}}\text{in}\phantom{\rule{4.pt}{0ex}}\text{density}\phantom{\rule{4.pt}{0ex}}\hfill \\ \text{where}\hfill \\ P=p-\rho gz\hfill \end{array}$

Suppose that the flow is characterized by a certain linear dimension, $L$ , a velocity $U$ , and a density $\rho$ . For example, if we consider the steady flow past an obstacle, $L$ may be it's diameter and $U$ and $\rho$ the velocity and density far from the obstacle. We can make the variables dimensionless with the following

$\begin{array}{ccc}\hfill {\mathbf{v}}^{*}& =& \frac{\mathbf{v}}{U},\phantom{\rule{0.277778em}{0ex}}{\mathbf{x}}^{*}\frac{\mathbf{x}}{L},\phantom{\rule{0.277778em}{0ex}}{t}^{*}=\frac{U}{L}t,\phantom{\rule{0.277778em}{0ex}}{P}^{*}=\frac{P}{\rho {U}^{2}}\hfill \\ \hfill {\nabla }^{*}& =& L\nabla ,\phantom{\rule{0.277778em}{0ex}}{\nabla }^{*2}={L}^{2}{\nabla }^{2}\hfill \end{array}$

The conservative body force, Navier-Stokes equation is made dimensionless with these variables.

$\begin{array}{ccc}\hfill \rho \frac{D\mathbf{v}}{Dt}& =& -\nabla P+\left(\lambda +\mu \right)\nabla \Theta +\mu {\nabla }^{2}\mathbf{v}\hfill \\ \hfill \rho \frac{{U}^{2}}{L}\frac{D{\mathbf{v}}^{*}}{D{t}^{*}}& =& -\rho \frac{{U}^{2}}{L}{\nabla }^{*}{P}^{*}+\frac{\mu \phantom{\rule{0.277778em}{0ex}}U}{{L}^{2}}\left(\lambda /\mu +1\right){\nabla }^{*}{\Theta }^{*}+\frac{\mu \phantom{\rule{0.277778em}{0ex}}U}{{L}^{2}}{\nabla }^{*2}{\mathbf{v}}^{*}\hfill \\ \hfill \frac{\rho UL}{\mu }\left[\frac{D{\mathbf{v}}^{*}}{D{t}^{*}},+,{\nabla }^{*},{P}^{*}\right]& =& \left(\lambda /\mu +1\right){\nabla }^{*}{\Theta }^{*}+{\nabla }^{*2}{\mathbf{v}}^{*}\hfill \\ \hfill {N}_{Re}\left[\frac{D{\mathbf{v}}^{*}}{D{t}^{*}},+,{\nabla }^{*},{P}^{*}\right]& =& \left(\lambda /\mu +1\right){\nabla }^{*}{\Theta }^{*}+{\nabla }^{*2}{\mathbf{v}}^{*}\hfill \\ & \text{where}& \\ \hfill {N}_{Re}& =& \frac{\rho U\phantom{\rule{0.277778em}{0ex}}L}{\mu }=\frac{\rho {U}^{2}}{\mu U/L}\hfill \end{array}$

The Reynolds number partitions the Navier -Stokes equation into two parts. The left side or inertial and potential terms, which dominates for large NRe and the right side or viscous terms, which dominates for small NRe. The potential gradient term could have been on the right side if the dimensionless pressure was defined differently, i.e., normalized with respect to $\left(\mu U\right)/L$ , the shear stress rather than kinetic energy. Note that the left side has only first derivatives of the spatial variables while the right side has second derivatives. We will see later that the left side may dominate for flow far from solid objects but the right side becomes important in the vicinity of solid surfaces.

The nature of the flow field can also be seen form the definition of the Reynolds number. The second expression is the ratio of the characteristic kinetic energy and the shear stress.

The alternate form of the dimensionless Navier-Stokes equation with the other definition of dimensionless pressure is as follows.

$\begin{array}{c}{N}_{Re}\frac{D{\mathbf{V}}^{*}}{D{t}^{*}}=-{\nabla }^{*}{P}^{**}+\left(\lambda /\mu +1\right){\nabla }^{*}{\Theta }^{*}+{\nabla }^{*2}{\mathbf{v}}^{*}\\ {P}^{**}=\frac{P}{\mu U/L}\hfill \end{array}$

## Dissipation of energy by viscous forces

If there was no dissipation of mechanical energy during fluid motion then kinetic energy and potential energy can be exchanged but the change in the sum of kinetic and potential energy would be equal to the work done to the system. However, viscous effects result in irreversible conversion of mechanical energy to internal energy or heat. This is known as viscous dissipation of energy. We will identify the components of mechanical energy in a flowing system before embarking on a total energy balance.

The rate that work $W$ is done on fluid in a material volume $V$ with a surface $S$ is the integral of the product of velocity and the force at the surface.

$\begin{array}{ccc}\hfill \frac{dW}{dt}& =& \underset{s}{∯}\mathbf{v}•{\mathbf{t}}_{\left(n\right)}\phantom{\rule{0.277778em}{0ex}}dS\hfill \\ & =& \underset{s}{∯}\mathbf{v}•\mathbf{T}•\mathbf{n}\phantom{\rule{0.277778em}{0ex}}dS\hfill \\ & =& \underset{v}{\phantom{\rule{0.277778em}{0ex}}\int \int \int \phantom{\rule{0.277778em}{0ex}}}\nabla •\left(\mathbf{v}•\mathbf{T}\right)\phantom{\rule{0.277778em}{0ex}}dV\hfill \end{array}$

The last integrand is rather complicated and is better treated with index notation.

$\begin{array}{ccc}\hfill {\left({v}_{i}{T}_{ij}\right)}_{,j}& =& {T}_{ij}{v}_{i,j}+{v}_{i}{T}_{ij,j}\hfill \\ & =& {T}_{ij}{v}_{i,j}+{v}_{i}\left[\rho ,\frac{D{v}_{i}}{Dt},-,\rho ,{f}_{i}\right]\hfill \\ & =& {T}_{ij}{v}_{i,j}+\frac{1}{2}\rho \frac{D{v}_{2}}{Dt}-\rho {f}_{i}{v}_{i}\hfill \\ \hfill \nabla •\left(\mathbf{v}•\mathbf{t}\right)& =& \mathbf{T}:\nabla \mathbf{v}+\frac{1}{2}\rho \frac{D{v}^{2}}{Dt}-\rho \mathbf{f}•\mathbf{v}\end{array}$

We made use of Cauchy's equation of motion to substitute for the divergence of the stress tensor. The integrals can be rearranged as follows.

so some one know about replacing silicon atom with phosphorous in semiconductors device?
how to fabricate graphene ink ?
for screen printed electrodes ?
SUYASH
What is lattice structure?
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials​ and their applications of sensors.
what is nano technology
what is system testing?
preparation of nanomaterial
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
what is system testing
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
Hello
Uday
I'm interested in Nanotube
Uday
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
Prasenjit
can nanotechnology change the direction of the face of the world
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
Good
Got questions? Join the online conversation and get instant answers!