Fluid Mechanics for Mechanical Engineers/Transport Equations

>back to Chapters of Fluid Mechanics for Mechanical Engineers

Diffusion Phenomena

The direction of diffusion of mass, energy and momentum

The fluid motion (convection) takes place with two mechanisms:

• The advective flow, which is the bulk motion of the fluid in a certain direction.
• The molecular motion, which persists to exist even if there is no convection.

The molecular motion is responsible for the diffusion phenomena. In fluid mechanics, we consider mostly the diffusion of mass, momentum and energy.

The diffusion of a quantity takes place always from rich to poor regions. For example, when one puts a drop of ink in water, ink molecules moves (diffuses) from ink rich region to ink poor region. Heat diffuses from high temperature regions to colder regions. Similarly, momentum diffuses from momentum rich regions to momentum poor regions. The diffusive flux in one dimension is formulated by Fick's first law as follows for mass, energy and momentum respectively:

$J_m=-k_m\frac{dC}{dx}[\frac{kg}{m^{2}s}]\text{where}C\text{is the concentration with the unit}[mol/m^3].$

$J_E=-k\frac{dT}{dx}[\frac{J}{m^{2}s}]$

$J_{mom}=-\mu \frac{dU}{dx}[\frac{kgm/s}{m^{2}s}]$

The negative sign is used since the gradient is negative opposite to the direction of diffusion, i.e. with the negative sign the direction of diffusion is correctly assesed. The constants of diffusion are the diffusion constant of one species into another, thermal conductivity and the dynamic viscosity of the fluid, respectively. The diffusion of energy in the form of heat is also called as conduction.

General Form of Transport Equations

The RTT theorem is given only with advection and without any sources. In the more general form of transport equation, the diffusion and the sources of the transported quantities have to be considered. For an extensive quantity $B$, the transport equation is:

$$\frac{dB}{dt}{|}_{sys}=\frac{\partial }{\partial t}\underset{cv}{\int }\rho bdV+\underset{cs}{\int }\rho b\overrightarrow{U}\cdot \overrightarrow{n}dA+\underset{cs}{\int }{\overrightarrow{J}}_B\cdot \overrightarrow{n}dA-\underset{cv}{\int }{\dot{q}}_{B}dV$$

where ${\dot{q}}_b$ is the source of $B$ per unit volume within CV and ${\overrightarrow{J}}_b$ is the diffusive flux of $B$ through the CS of CV.

Conservation of Mass Equation (Continuity Equation)

The integral equation for the conservation of mass law is $$\frac{dM}{dt}{|}_{sys}=\frac{\partial }{\partial t}\underset{cv}{\int }\rho dV+\underset{cs}{\int }\rho \overrightarrow{U}\cdot \overrightarrow{n}dA+\underset{cs}{\int }{\overrightarrow{J}}_m\cdot \overrightarrow{n}dA-\underset{cv}{\int }{\dot{q}}_{m}dV=0$$

According to this equation, when advective and diffusive flux are ignored, it can be seen that the rate of change of mass in the CV is due to the mass sources:

$$\frac{\partial }{\partial t}\underset{cv}{\int }\rho dV=\underset{cv}{\int }{\dot{q}}_{m}dV$$In case of no advective flux and mass sources, $$\frac{\partial }{\partial t}\underset{cv}{\int }\rho dV=-\underset{cs}{\int }{\overrightarrow{J}}_m\cdot \overrightarrow{n}dA$$the mass in CV, can change only through diffusive flux. Note that, incoming diffusive flux will have a negative sign and together with the extra negative sign its contribution will be positive.

For a differential volume $dV=d{x}_{1}d{x}_{2}d{x}_3$, each term in the integral equation can be formulated as follows:

The time rate of change of mass in the CV:$$\frac{\partial }{\partial t}\underset{cv}{\int }\rho dV\approx \frac{\partial \rho }{\partial t}dV$$

The advective flow rate of mass through CS:

The advective flow rate can be decomposed into the surfaces having their surface normal vectors along

$x_1$

,

$x_2$

and

$x_3$

axis:

$$\underset{cs}{\int }\rho \overrightarrow{U}\cdot \overrightarrow{n}dA=\underset{c{s}_{x1}}{\int }\rho \overrightarrow{U}\cdot \overrightarrow{n}dA+\underset{c{s}_{x2}}{\int }\rho \overrightarrow{U}\cdot \overrightarrow{n}dA+\underset{c{s}_{x3}}{\int }\rho \overrightarrow{U}\cdot \overrightarrow{n}dA$$

File:Conservation of mass.png

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

$$\underset{c{s}_{x1}}{\int }\rho \overrightarrow{U}\cdot \overrightarrow{n}dA\approx -\rho U_{1}d{x}_{2}d{x}_3+\rho U_{1}d{x}_{2}d{x}_3+\frac{\partial \rho U_1}{\partial x_1}d{x}_{1}d{x}_{2}d{x}_3=\frac{\partial \rho U_1}{\partial x_1}dV$$

Hence the total advective flux is:

$$\underset{cs}{\int }\rho \overrightarrow{U}\cdot \overrightarrow{n}dA\approx \frac{\partial \rho U_1}{\partial x_1}dV+\frac{\partial \rho U_2}{\partial x_2}dV+\frac{\partial \rho U_3}{\partial x_3}dV=\frac{\partial \rho U_i}{\partial x_i}dV$$ In fact, one can see clearly the Divergence theorem: $$\underset{cs}{\int }\rho \overrightarrow{U}\cdot \overrightarrow{n}dA=\underset{cv}{\int }\overrightarrow{\mathrm{\nabla }}\cdot \rho \overrightarrow{U}dV=\underset{cv}{\int }\frac{\partial \rho U_i}{\partial x_i}dV\approx \frac{\partial \rho U_i}{\partial x_i}dV$$ The diffusive flow rate of mass through CS:

By using the divergence theorem, the diffusive flow rate can be written for a differential volume as follows:

$$\underset{cs}{\int }{\overrightarrow{J}}_m\cdot \overrightarrow{n}dA=\underset{cv}{\int }\overrightarrow{\mathrm{\nabla }}\cdot {\overrightarrow{J}}_{m}dV=\underset{cv}{\int }\frac{\partial J_{m_i}}{\partial x_i}dV\approx \frac{\partial J_{m_i}}{\partial x_i}dV$$For isotropic diffusion, the $i^{th}$ component of the diffusion flux is

$$J_{m_i}=-k_m\frac{\partial C}{\partial x_i}$$Hence the diffusive flow rate can be written as

$$\underset{cs}{\int }{\overrightarrow{J}}_m\cdot \overrightarrow{n}dA\approx -k_m\frac{{\partial }^{2}C}{\partial x_i^2}dV$$

The net source of mass in CV:

$$\underset{cv}{\int }{\dot{q}}_{m}dV\approx {\dot{q}}_{m}dV$$The final differential form of the conservation of mass equation reads

$$\frac{\partial \rho }{\partial t}dV+\frac{\partial \rho U_i}{\partial x_i}dV-k_m\frac{{\partial }^{2}C}{\partial x_i^2}dV-{\dot{q}}_{m}dV=0$$and per unit volume

$$\frac{\partial \rho }{\partial t}+\frac{\partial \rho U_i}{\partial x_i}-k_m\frac{{\partial }^{2}C}{\partial x_i^2}-{\dot{q}}_m=0$$

Conservation of Energy Equation

The conservation of energy equation is very interesting since it shows the dissipation of energy and reversible conversion of energy as the fluid flows. The transport equation of energy reads:$$\frac{dE}{dt}{|}_{sys}=[\dot{Q}+\dot{W}{]}_{onsys}=\frac{\partial }{\partial t}\underset{cv}{\int }\rho edV+\underset{cs}{\int }\rho e\overrightarrow{U}\cdot \overrightarrow{n}dA+\underset{cs}{\int }{\overrightarrow{J}}_E\cdot \overrightarrow{n}dA-\underset{cv}{\int }{\dot{q}}_{E}dV$$where $e=\hat{u}+\frac{U_i{U}_i}{2}+G$ is the sum of internal energy, kinetic energy and potential energy per unit mass.

If only gravitational field is responsible for the potential energy, the gravitation vector is defined as:$$g_i=-\frac{\partial G}{\partial x_i}$$The potential can be defined as $G=-g_j{x}_j$,

$g_i=-\frac{\partial (-g_j{x}_j)}{\partial x_i}=g_j\underset{{\delta }_{ij}}{\underbrace{\frac{\partial x_j}{\partial x_i}}}+x_j\underset{\approx 0}{\underbrace{\frac{\partial g_j}{\partial x_i}}}$

The gradient of the gravitation vector can be neglected for human scale applications. The gradient of the position vector will be unity for equal indices.

For practical purposes, say the gravitational acceleration is in negative $x_2$ direction and defined as $\overrightarrow{g}=[0,-g,0]$ and $g$ is the magnitude of the vector,

$G=-g_2{x}_2=g{x}_2$

Rearranging the above equation for an instant at which CV and the system collapse on each other:

$$\frac{\partial }{\partial t}\underset{cv}{\int }\rho edV+\underset{cs}{\int }\rho e\overrightarrow{U}\cdot \overrightarrow{n}dA=[\dot{Q}+\dot{W}{]}_{oncv}-\underset{cs}{\int }{\overrightarrow{J}}_E\cdot \overrightarrow{n}dA+\underset{cv}{\int }{\dot{q}}_{E}dV$$Heat $\dot{Q}$ can be added to the CV in the form of radiation or a surface heater. The source of energy ${\dot{q}}_E$ can be the energy released or absorbed during chemical reaction or the energy of the added mass from the source of mass. Therefore, $\dot{Q}$ can be treated as a part of energy source ${\dot{q}}_E$ if it is a volumetric addition and/or as a part of conduction ${\overrightarrow{J}}_E$ term on the surface of the CV if it is a heating surface.

The rate of change of energy in a differential volume is$$\frac{\partial }{\partial t}\underset{cv}{\int }\rho edV\approx \frac{\partial \rho e}{\partial t}dV$$The advection term for a differential volume can be formulated by using the divergence theorem$$\underset{cs}{\int }\rho e\overrightarrow{U}\cdot \overrightarrow{n}dA=\underset{cv}{\int }\overrightarrow{\mathrm{\nabla }}\cdot (\rho e\overrightarrow{U})dV\approx \frac{\partial \rho e{U}_i}{\partial x_i}dV$$The work of surface forces on CV $(\dot{W})$$${\dot{W}}_s=\underset{cs}{\int }\overrightarrow{U}\cdot \overrightarrow{\sigma }dA$$where $\overrightarrow{\sigma }$ is the sum of the stress caused by pressure and viscous stresses

$$\overrightarrow{\sigma }=-P\overrightarrow{n}+\overrightarrow{\tau }$$Hence, $$\dot{W}=-\underset{cs}{\int }P\overrightarrow{U}\cdot \overrightarrow{n}dA+\underset{cs}{\int }\overrightarrow{U}\cdot \overrightarrow{\tau }dA$$Since $\overrightarrow{\tau }=\overline{\overline{\tau }}\cdot \overrightarrow{n}$ is the local stress vector and $\overline{\overline{\tau }}$ is the stress tensor.

$$\dot{W}=-\underset{cs}{\int }P\overrightarrow{U}\cdot \overrightarrow{n}dA+\underset{cs}{\int }(\overrightarrow{U}\cdot \overline{\overline{\tau }})\cdot \overrightarrow{n}dA$$

For a differential volume, $\overrightarrow{\tau }=\overline{\overline{\tau }}\cdot \overrightarrow{n}={\tau }_{ij}{n}_i$ is the stress vector on a face having its normal in $i^{th}$-direction.

$$\dot{W}\approx -\frac{\partial P{U}_i}{\partial x_i}dV+\frac{\partial U_j{\tau }_{ij}}{\partial x_i}dV$$In order to have a better decomposition later, one can switch $i$ and $j$ index since ${\tau }_{ij}={\tau }_{ji}$ and use the equality $\frac{\partial U_j{\tau }_{ij}}{\partial x_i}=\frac{\partial U_i{\tau }_{ji}}{\partial x_j}$

$$\dot{W}\approx -\frac{\partial P{U}_i}{\partial x_i}dV+\frac{\partial U_i{\tau }_{ji}}{\partial x_j}dV$$

The heat diffusion (conduction) term for a differential volume can be written by using the divergence term as:

$$\underset{cs}{\int }{\overrightarrow{J}}_E\cdot \overrightarrow{n}dA=\underset{cv}{\int }\overrightarrow{\mathrm{\nabla }}\cdot {\overrightarrow{J}}_{E}dV=-\underset{cv}{\int }k\frac{{\partial }^{2}T}{\partial x_i^2}dV\approx -k\frac{{\partial }^{2}T}{\partial x_i^2}dV$$Rewriting the energy equation, using the above forms derived for a differential volume:

$$\frac{\partial \rho e}{\partial t}dV+\frac{\partial \rho e{U}_i}{\partial x_i}dV=-\frac{\partial P{U}_i}{\partial x_i}dV+\frac{\partial U_i{\tau }_{ji}}{\partial x_j}dV+k\frac{{\partial }^{2}T}{\partial x_i^2}dV+{\dot{q}}_{E}dV$$or per unit volume

$$\frac{\partial \rho e}{\partial t}+\frac{\partial \rho e{U}_i}{\partial x_i}=-\frac{\partial P{U}_i}{\partial x_i}+\frac{\partial U_i{\tau }_{ji}}{\partial x_j}+k\frac{{\partial }^{2}T}{\partial x_i^2}+{\dot{q}}_E$$

When left side of the equation is expanded,

$$e(\frac{\partial \rho }{\partial t}+\frac{\partial \rho U_i}{\partial x_i})+\rho (\frac{\partial e}{\partial t}+U_i\frac{\partial e}{\partial x_i})=-\frac{\partial P{U}_i}{\partial x_i}+\frac{\partial U_i{\tau }_{ji}}{\partial x_j}+k\frac{{\partial }^{2}T}{\partial x_i^2}+{\dot{q}}_E$$it can be seen that the first term involves one part of the conservation of mass equation. Neglecting the diffusion and source of mass, the continuity equation can be set to zero and the energy equation can be reduced to:$$\rho (\frac{\partial e}{\partial t}+U_i\frac{\partial e}{\partial x_i})=-\frac{\partial P{U}_i}{\partial x_i}+\frac{\partial U_i{\tau }_{ji}}{\partial x_j}+k\frac{{\partial }^{2}T}{\partial x_i^2}+{\dot{q}}_E$$The left side becomes the substantial derivative of the energy per unit time:$$\frac{De}{Dt}=-\frac{1}{\rho }\frac{\partial P{U}_i}{\partial x_i}+\frac{1}{\rho }\frac{\partial U_i{\tau }_{ji}}{\partial x_j}+\frac{1}{\rho }k\frac{{\partial }^{2}T}{\partial x_i^2}+\frac{1}{\rho }{\dot{q}}_E$$

Decomposition of Energy Equation

Inserting the definition of $e=\hat{u}+U_i{U}_i/2+g{x}_2$ and expanding the work done by pressure and viscous stresses, $$\frac{D\hat{u}}{Dt}+\frac{D{U}_i{U}_i/2}{Dt}+\frac{Dg{x}_2}{Dt}=-\frac{P}{\rho }\frac{\partial U_i}{\partial x_i}-\frac{U_i}{\rho }\frac{\partial P}{\partial x_i}+\frac{U_i}{\rho }\frac{\partial {\tau }_{ji}}{\partial x_j}+\frac{{\tau }_{ji}}{\rho }\frac{\partial U_i}{\partial x_j}+\frac{1}{\rho }k\frac{{\partial }^{2}T}{\partial x_i^2}+\frac{1}{\rho }{\dot{q}}_E$$Since $\frac{D{U}_i{U}_i/2}{Dt}=U_i\frac{D{U}_i}{Dt}$, and $$\frac{Dg{x}_2}{Dt}=\frac{\partial g{x}_2}{\partial t}+U_i\frac{\partial g{x}_2}{\partial x_i}$$and assuming the differential volume does not change its potential in time, i.e. $\frac{\partial g{x}_2}{\partial t}\approx 0$$$\frac{Dg{x}_2}{Dt}=U_i\frac{\partial g{x}_2}{\partial x_i}=U_i(g\frac{\partial x_2}{\partial x_i}+x_2\frac{\partial g}{\partial x_i})=U_{2}g=-U_i{g}_i$$Hence, the energy equation reads$$\frac{D\hat{u}}{Dt}+\underset{\_}{U_i\frac{D{U}_i}{Dt}}=\underset{\_}{U_i{g}_i}-\frac{P}{\rho }\frac{\partial U_i}{\partial x_i}\underset{\_}{-\frac{U_i}{\rho }\frac{\partial P}{\partial x_i}}+\underset{\_}{\frac{U_i}{\rho }\frac{\partial {\tau }_{ji}}{\partial x_j}}+\frac{{\tau }_{ji}}{\rho }\frac{\partial U_i}{\partial x_j}+\frac{1}{\rho }k\frac{{\partial }^{2}T}{\partial x_i^2}+\frac{1}{\rho }{\dot{q}}_E$$where $U_i{g}_i$ is the work done by the gravitational force per unit mass and time.

Now we can decompose this equation into the mechanical and thermal energy equations. The underlined terms are nothing but the scalar product of the velocity vector and the momentum equation $(\overrightarrow{U}\cdot \frac{D\overrightarrow{U}}{Dt})$ and form the mechanical energy equation, the remaining terms form the thermal energy equation.

The mechanical energy equation:$$U_i\frac{D{U}_i}{Dt}=U_i{g}_i-\frac{U_i}{\rho }\frac{\partial P}{\partial x_i}+\frac{U_i}{\rho }\frac{\partial {\tau }_{ji}}{\partial x_j}$$Since $$\frac{U_i}{\rho }\frac{\partial {\tau }_{ji}}{\partial x_j}=\frac{1}{\rho }\frac{\partial U_i{\tau }_{ji}}{\partial x_j}-\frac{{\tau }_{ji}}{\rho }\frac{\partial U_i}{\partial x_j}$$and$$\frac{U_i}{\rho }\frac{\partial P}{\partial x_i}=\frac{1}{\rho }\frac{\partial U_{i}P}{\partial x_i}-\frac{P}{\rho }\frac{\partial U_i}{\partial x_i}$$The mechanical energy equation can be written in a more interpretable manner:

$$U_i\frac{D{U}_i}{Dt}=U_i{g}_i-\frac{1}{\rho }\frac{\partial U_{i}P}{\partial x_i}+\frac{P}{\rho }\frac{\partial U_i}{\partial x_i}+\frac{1}{\rho }\frac{\partial U_i{\tau }_{ji}}{\partial x_j}-\frac{{\tau }_{ji}}{\rho }\frac{\partial U_i}{\partial x_j}$$The thermal energy equation:$$\frac{D\hat{u}}{Dt}=-\frac{P}{\rho }\frac{\partial U_i}{\partial x_i}+\frac{{\tau }_{ji}}{\rho }\frac{\partial U_i}{\partial x_j}+\frac{1}{\rho }k\frac{{\partial }^{2}T}{\partial x_i^2}+\frac{1}{\rho }{\dot{q}}_E$$The meaning of each term is as follows:

Term	Physical meaning
$$-\frac{1}{\rho }\frac{\partial U_{i}P}{\partial x_i}$$	Rate of work done by pressure.
$$\frac{P}{\rho }\frac{\partial U_i}{\partial x_i}$$	Rate of reversible conversion of kinetic energy into internal energy. (-) in the case of compression and (+) in the case of expansion, because $\frac{\partial U_i}{\partial x_i}<0$ for compression and larger than 0 for expansion. It appears with a negative sign in the internal energy equation.
$$\frac{1}{\rho }\frac{\partial U_i{\tau }_{ji}}{\partial x_j}$$	Rate of work done by viscous stresses.
$$-\frac{{\tau }_{ji}}{\rho }\frac{\partial U_i}{\partial x_j}$$	Rate of dissipation of mechanical energy, in other words irreversible conversion of kinetic energy to internal energy. It is always negative. In other words, mechanical energy drops due to this term. In the internal energy equation it appears with a possitive sign.
$$-\frac{P}{\rho }\frac{\partial U_i}{\partial x_i}$$	Rate of reversible conversion of kinetic energy into internal energy. (+) in the case of compresion and (-) in the case of expansion, because $\frac{\partial U_i}{\partial x_i}<0$ for compression and larger than 0 for expansion.
$$\frac{{\tau }_{ji}}{\rho }\frac{\partial U_i}{\partial x_j}$$	Rate of increase of internal energy by irreversible viscous dissipation. It is always positive. In other words, internal energy increases due to this term.
$$\frac{1}{\rho }k\frac{{\partial }^{2}T}{\partial x_i^2}$$	Rate of heat added or extracted via conduction (thermal diffusion). It can be (+) or (-).
$$\frac{1}{\rho }{\dot{q}}_E$$	Rate of head added via combustion or radiation.

Comment on the dissipation term

The dissipation term $\frac{{\tau }_{ji}}{\rho }\frac{\partial U_i}{\partial x_j}$ is always positive and extracts energy from the mechanical energy equation and increases the internal energy. This can be shown for an incompressible Newtonian fluid as follows:

$\frac{{\tau }_{ji}}{\rho }\frac{\partial U_i}{\partial x_j}=\nu (\frac{\partial U_i}{\partial x_j}+\frac{\partial U_j}{\partial x_i})\frac{\partial U_i}{\partial x_j}$

The velocity gradient term can be expanded as:

$\frac{\partial U_i}{\partial x_j}\frac{\partial U_i}{\partial x_j}+\frac{\partial U_j}{\partial x_i}\frac{\partial U_i}{\partial x_j}={\left(\frac{\partial U_1}{\partial x_1}\right)}^2+{\left(\frac{\partial U_1}{\partial x_2}\right)}^2+{\left(\frac{\partial U_1}{\partial x_3}\right)}^2+{\left(\frac{\partial U_2}{\partial x_1}\right)}^2+{\left(\frac{\partial U_2}{\partial x_2}\right)}^2+{\left(\frac{\partial U_2}{\partial x_3}\right)}^2+{\left(\frac{\partial U_3}{\partial x_1}\right)}^2+{\left(\frac{\partial U_3}{\partial x_2}\right)}^2+{\left(\frac{\partial U_3}{\partial x_3}\right)}^2+$

$\frac{\partial U_1}{\partial x_1}\frac{\partial U_1}{\partial x_1}+\frac{\partial U_2}{\partial x_1}\frac{\partial U_1}{\partial x_2}+\frac{\partial U_3}{\partial x_1}\frac{\partial U_1}{\partial x_3}+\frac{\partial U_1}{\partial x_2}\frac{\partial U_2}{\partial x_1}+\frac{\partial U_2}{\partial x_2}\frac{\partial U_2}{\partial x_2}+\frac{\partial U_3}{\partial x_2}\frac{\partial U_2}{\partial x_3}+$

$\frac{\partial U_1}{\partial x_3}\frac{\partial U_3}{\partial x_1}+\frac{\partial U_2}{\partial x_3}\frac{\partial U_3}{\partial x_2}+\frac{\partial U_3}{\partial x_3}\frac{\partial U_3}{\partial x_3}$

$={(\frac{\partial U_1}{\partial x_2}+\frac{\partial U_2}{\partial x_1})}^2+{(\frac{\partial U_1}{\partial x_3}+\frac{\partial U_3}{\partial x_1})}^2+{(\frac{\partial U_2}{\partial x_3}+\frac{\partial U_3}{\partial x_2})}^2+2{\left(\frac{\partial U_1}{\partial x_1}\right)}^2+2{\left(\frac{\partial U_2}{\partial x_2}\right)}^2+2{\left(\frac{\partial U_3}{\partial x_3}\right)}^2$

Hence the energy dissipation rate per unit mass reads:

$\frac{{\tau }_{ji}}{\rho }\frac{\partial U_i}{\partial x_j}=\nu [{(\frac{\partial U_1}{\partial x_2}+\frac{\partial U_2}{\partial x_1})}^2+{(\frac{\partial U_1}{\partial x_3}+\frac{\partial U_3}{\partial x_1})}^2+{(\frac{\partial U_2}{\partial x_3}+\frac{\partial U_3}{\partial x_2})}^2+2{\left(\frac{\partial U_1}{\partial x_1}\right)}^2+2{\left(\frac{\partial U_2}{\partial x_2}\right)}^2+2{\left(\frac{\partial U_3}{\partial x_3}\right)}^2]$

As can be seen, all the terms are positive. It should be noted that there would be no dissipation without the viscous stresses or velocity gradients. In fact, viscous stresses do occur only when there is a velocity gradient. Hence, in order to avoid dissipation one should avoid or reduce unnecessary velocity gradients.