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

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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.

can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
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