Incompressible flow

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Latest revision as of 18:57, 19 August 2013

A flow is said to be incompressible if the density of a fluid element does not change during its motion. It is a property of the flow and not of the fluid. The rate of change of density of a material fluid element is given by the material derivative

$\frac{D \rho}{D t} = \frac{\partial \rho}{\partial t} + u_j \frac{\partial \rho}{\partial x_j}$

From the continuity equation we have

$\frac{D \rho}{D t} = - \rho \frac{\partial u_j}{\partial x_j}$

Hence the flow is incompressible if the divergence of the velocity field is identically zero. Note that the density field need not be uniform in an incompressible flow. All that is required is that the density of a fluid element should not change in time as it moves through space. For example, flow in the ocean can be considered to be incompressible even though the density of water is not uniform due to stratification.

Compressible flow can with good accuracy be approximated as incompressible for steady flow if the Mach number is below 0.3.

Dimensional analysis

Assume $L$ is the characteristic length, $U$ is the characteristic velocity, the magnitude of velocity gradient is $U/L$ . The physical meaning of the incompressible flow is

$\frac{\partial u_j}{\partial x_j} \approx 0$

or

$|\frac{\partial u_j}{\partial x_j}|=|\rho^{-1}\frac{d\rho}{dt}| << U/L.$

It is know that

$\rho^{-1}\frac{d\rho}{dt}=\rho^{-1}\frac{d\rho}{dp}\frac{dp}{dt}=(\rho c^2)^{-1}\frac{dp}{dt},$

where $c$ is the sound speed. Hence, the incompressible condition is,

$|\frac{1}{\rho c^2}\frac{dp}{dt}|<<U/L.$

In general, for the liquid, $\rho=const$ and $\frac{d\rho}{dt}=0$ . So the question is that under what contion, the gas flow can be taken as incompressible flow?

We have

$\frac{dp}{dt}=\frac{\partial p}{\partial t}+u_i\frac{\partial p}{\partial x_i}=\frac{\partial p}{\partial t}-\rho u_i\left(\frac{\partial u_i}{\partial t}+u_j\frac{\partial u_i}{\partial x_j}\right)+\rho u_i f_i+u_i \frac{\partial \sigma_{i,j}}{\partial x_j},$

where $f_i$ is the body force (taken as gravitational force in the present document), $\sigma_{i,j}$ is the stress tensor defined as,

$\sigma_{i,j}=\eta e_{kk}\delta_{ij}+2\mu e_{ij},$

where $e_{ij}=\frac{1}{2}(\frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i})$ .

Then we can summarize as following,

Steady incompressible flow

The steady flow can be taken as incompressible flow under the following conditions,

$\frac{1}{c^2}|u_i\frac{\partial u_ju_j}{\partial x_i}|<<U/L;\quad \frac{1}{c^2}|u_if_i|<<U/L;\quad \frac{1}{\rho c^2}|u_i \frac{\partial \sigma_{i,j}}{\partial x_j}|<<U/L.$

Then it arrives,

$U^2<<c^2;\quad Lg<<c^2;\quad \frac{\nu U}{L}<<c^2,$

where $\nu=\frac{\mu,\eta}{\rho}$ .

If $L<10m$ , $Lg<<c^2$ holds; $\nu=c\lambda$ , so $U<<\frac{Lc}{\lambda}$ ( $\lambda$ is the molecular free path) holds naturally for the continuous medium.

So for the steady flow, it can be taken as incompressible flow when $U<<c$ , i.e. $Ma<<1$ . In general, the flow will be taken as incompressible flow when $Ma\le 0.3$ .

Unsteady incompressible flow

For unsteady flow, if $|\frac{\partial }{\partial t}|\sim|u_i\frac{\partial }{\partial x_i}|$ , the flow is incompressible when $Ma<<1$ . But if $|\frac{\partial }{\partial t}|>>|u_i\frac{\partial }{\partial x_i}|$ , it needs to figure out the condition. It is known that

$|\frac{\partial u_i}{\partial t}|\sim|\frac{1}{\rho}\frac{\partial p}{\partial x_i}|$

Hence,

$p\sim \rho U L/\tau,$

where $\tau$ is the characteristic time.

If $\tau>> L/c$ , then $\tau>> LU/c^2$ , and we get

$|\frac{1}{\rho c^2}\frac{\partial p}{\partial t}|<<U/L; \quad |\frac{1}{c^2}\frac{\partial u_i u_i}{\partial t}|<<U/L.$

So for this case, the flow can be taken as incompressible flow when $\tau>>L/c$ .

Low speed atmospheric motion

For the atmospheric motion, one more condition is needed, $|\frac{u_i f_i}{c^2}|<<U/L$ , i.e. $L<<c^2/g\sim 10^4$ . But the characteristic length for the atmospheric motion is more than 10 km, the low speed atmospheric motion is not incompressible flow.

Question for thinking: is the sound wave compresible or incompressible wave? Why?

Governing Equations

The Navier-Stokes equations for incompressible flow are

Continuity equation

$\frac{\partial u_j}{\partial x_j} = 0$

Momentum equation

$\frac{\partial u_i}{\partial t} + u_j \frac{\partial u_i}{\partial x_j} + \frac{1}{\rho} \frac{\partial p}{\partial x_i} = \nu \Delta u_i$

Energy equation

$\frac{\partial E}{\partial t} + u_j \frac{\partial E}{\partial x_j} = \Phi + \frac{1}{\rho} \frac{\partial}{\partial x_j} \left( k \frac{\partial T}{\partial x_j} \right)$

where

$\Delta$ is the Laplacian operator
E is the internal energy per unit mass
$\Phi$ is the rate of dissipation of mechanical energy per unit mass or called more often the viscous dissipation function per unit mass

$\Phi = \frac{\nu}{2} \left( \frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i} \right) \left( \frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i} \right)$

$\nu$ is the kinematic viscosity
k is the coefficient of thermal conductivity
T is the temperature

However, the momentum equation is a composite equation, being the sum of the two equations,

$\frac{\partial u_i}{\partial t} = \pi^S (-u_j \frac{\partial u_i}{\partial x_j} + \nu \Delta u_i) ,$

$\frac{1}{\rho} \frac{\partial p}{\partial x_i} = \pi^I (-u_j \frac{\partial u_i}{\partial x_j} + \nu \Delta u_i),$

which result from the Helmholtz decomposition. $\pi^S$ and $\pi^I$ are the solenoidal (divergence-free) and irrotational projection operators.

The divergence-free governing equation for the velocity is pressure-free, calling into question the interpretation of the pressure as the enforcer of continuity. Rather, this equation should be considered a kinematic equation with continuity as a conservation law. The equation for the pressure as a functional of the velocity can be recognized as a form of the pressure poisson equation.

If expressions for the projection operators are inserted, it is clear that these are integro-differential equations, rather than the differential-algebraic form of the composite equation. This inconvenience can be eliminated by use of the equivalent weak form,

$\left(w_k,\frac{\partial u_i}{\partial t}\right) = -\left(w_k,u_j \frac{\partial u_i}{\partial x_j}\right)-\nu \left(\frac{\partial w_k}{\partial x_j},\frac{\partial u_i}{\partial x_j}\right) ,$

where the weight functions $w$ are divergence-free. The function of the projection operator has been replaced by the orthogonality of solenoidal and irrotational function spaces. Cast in discrete form, this forms the basis for finite element computations.

A question immediately comes to mind: If the governing equation does not involve the pressure, how does one specify boundary conditions for pressure-driven (Poiseuille) flow? A divergence-free field implies the existence of a stream function (or vector potential in 3D) $\Psi$ such that the velocity field is the curl of $\Psi$ . Then specifying $\Psi$ on a part of the boundary determines the flow from the Stokes theorem.

Physical characteristics

A consequence of incompressible flow is that there is no equation of state for pressure, unlike in compressible flow. Since there is no separate equation for pressure, it must be obtained from the continuity and momentum equations. The main role of pressure is to satisfy the zero divergence condition of the velocity field. Note that pressure is only determined up to a constant.

If the viscosity is assumed to be constant, then the energy equation is decoupled from the continuity and momentum equations.

@@ Line 38: / Line 38: @@
 </math>
-In general, for the liquid, <math>\rho=const</math> and <math>\frac{\rho}{dt}=0</math>. So the question is that under what contion, the gas flow can be taken as incompressible flow?
+In general, for the liquid, <math>\rho=const</math> and <math>\frac{d\rho}{dt}=0</math>. So the question is that under what contion, the gas flow can be taken as incompressible flow?
 We have
 :<math>
-\frac{dp}{dt}=\frac{\partial p}{\partial t}+u_i\frac{\partial p}{\partial x_i}=\frac{\partial p}{\partial t}-\rho u_i(\frac{\partial u_i}{\partial t}+u_j\frac{\partial u_i}{\partial x_j})+\rho u_i f_i+u_i \frac{\partial \sigma_{i,j}}{\partial x_j},
+\frac{dp}{dt}=\frac{\partial p}{\partial t}+u_i\frac{\partial p}{\partial x_i}=\frac{\partial p}{\partial t}-\rho u_i\left(\frac{\partial u_i}{\partial t}+u_j\frac{\partial u_i}{\partial x_j}\right)+\rho u_i f_i+u_i \frac{\partial \sigma_{i,j}}{\partial x_j},
 </math>
 where <math>f_i</math> is the body force (taken as gravitational force in the present document), <math>\sigma_{i,j}</math> is the stress tensor defined as,
@@ Line 52: / Line 52: @@
 Then we can summarize as following,
 * Steady incompressible flow
-The steady flow can be take as incompressible flow under the following conditions,
+The steady flow can be taken as incompressible flow under the following conditions,
 :<math>
 \frac{1}{c^2}|u_i\frac{\partial u_ju_j}{\partial x_i}|<<U/L;\quad \frac{1}{c^2}|u_if_i|<<U/L;\quad \frac{1}{\rho c^2}|u_i \frac{\partial \sigma_{i,j}}{\partial x_j}|<<U/L.
@@ Line 62: / Line 62: @@
 where <math>\nu=\frac{\mu,\eta}{\rho}</math>.
-If <math>L<10m<.math>, <math>Lg<<c^2</math> holds; <math>\nu=c\lambda</math>, so <math>U<<\frac{Lc}{\lambda}</math> (<math>\lambda</math> is the molecular free path) holds naturally for the continuous medium.
+If <math>L<10m</math>, <math>Lg<<c^2</math> holds; <math>\nu=c\lambda</math>, so <math>U<<\frac{Lc}{\lambda}</math> (<math>\lambda</math> is the molecular free path) holds naturally for the continuous medium.
-So for the steady flow, it can be taken as incompressible flow when <math>U<<c</math>, i.e. <math>Ma<<1</math>. In general, the flow will be take as incompressible flow when <\math>Ma<0.3<\math>.
+So for the steady flow, it can be taken as incompressible flow when <math>U<<c</math>, i.e. <math>Ma<<1</math>. In general, the flow will be taken as incompressible flow when <math>Ma\le 0.3</math>.
 * Unsteady incompressible flow
+For unsteady flow, if <math>|\frac{\partial }{\partial t}|\sim|u_i\frac{\partial }{\partial x_i}|</math>, the flow is incompressible when <math>Ma<<1</math>. But if <math>|\frac{\partial }{\partial t}|>>|u_i\frac{\partial }{\partial x_i}|</math>, it needs to figure out the condition. It is known that
+:<math>
+|\frac{\partial u_i}{\partial t}|\sim|\frac{1}{\rho}\frac{\partial p}{\partial x_i}|
+</math>
+Hence,
+:<math>
+p\sim \rho U L/\tau,
+</math>
+where <math>\tau</math> is the characteristic time.
+If <math>\tau>> L/c</math>, then <math>\tau>> LU/c^2</math>, and we get
+:<math>
+|\frac{1}{\rho c^2}\frac{\partial p}{\partial t}|<<U/L; \quad |\frac{1}{c^2}\frac{\partial u_i u_i}{\partial t}|<<U/L.
+</math>
+So for this case, the flow can be taken as incompressible flow when <math>\tau>>L/c</math>.
 * Low speed atmospheric motion
+For the atmospheric motion, one more condition is needed, <math>|\frac{u_i f_i}{c^2}|<<U/L</math>, i.e. <math>L<<c^2/g\sim 10^4</math>. But the characteristic length for the atmospheric motion is more than 10 km, the low speed atmospheric motion is not incompressible flow.
+* Question for thinking: is the sound wave compresible or incompressible wave? Why?
 == Governing Equations ==
@@ Line 97: / Line 116: @@
 * <math>\Delta</math> is the [[Laplacian | Laplacian operator]]
 * E is the internal energy per unit mass
-* <math>\Phi</math> is the rate of dissipation of mechanical energy per unit mass
+* <math>\Phi</math> is the rate of dissipation of mechanical energy per unit mass or called more often the viscous dissipation function per unit mass
 :<math>
-\Phi = \nu \left( \frac{\partial u_i}{\partial x_j} + \frac{\partial u_i}{\partial x_j} \right)
+\Phi = \frac{\nu}{2} \left( \frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i} \right)
-\left( \frac{\partial u_i}{\partial x_j} + \frac{\partial u_i}{\partial x_j} \right)
+\left( \frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i} \right)
 </math>
@@ Line 107: / Line 126: @@
 * k is the coefficient of thermal conductivity
 * T is the temperature
+However, the momentum equation is a composite equation, being the sum of the two equations,
+:<math>
+\frac{\partial u_i}{\partial t} = \pi^S (-u_j \frac{\partial u_i}{\partial x_j}  + \nu \Delta u_i) ,
+</math>
+:<math>
+\frac{1}{\rho} \frac{\partial p}{\partial x_i}  = \pi^I (-u_j \frac{\partial u_i}{\partial x_j} + \nu \Delta u_i),
+</math>
+which result from the Helmholtz decomposition. <math>\pi^S</math> and <math>\pi^I</math> are the solenoidal (divergence-free) and irrotational projection operators.
+The divergence-free governing equation for the velocity is pressure-free, calling into question the interpretation of the pressure as the enforcer of continuity. Rather, this equation should be considered a kinematic equation with continuity as a conservation law. The equation for the pressure as a functional of the velocity can be recognized as a form of the pressure poisson equation.
+If expressions for the projection operators are inserted, it is clear that these are integro-differential equations, rather than the differential-algebraic form of the composite equation. This inconvenience can be eliminated by use of the equivalent weak form,
+:<math>
+\left(w_k,\frac{\partial u_i}{\partial t}\right) =  -\left(w_k,u_j \frac{\partial u_i}{\partial x_j}\right)-\nu \left(\frac{\partial w_k}{\partial x_j},\frac{\partial u_i}{\partial x_j}\right) ,
+</math>
+where the weight functions <math>w</math> are divergence-free. The function of the projection operator has been replaced by the orthogonality of solenoidal and irrotational function spaces. Cast in discrete form, this forms the basis for finite element computations.
+A question immediately comes to mind: If the governing equation does not involve the pressure, how does one specify boundary conditions for pressure-driven (Poiseuille) flow? A divergence-free field implies the existence of a stream function (or vector potential in 3D) <math>\Psi</math> such that the velocity field is the curl of <math>\Psi</math>. Then specifying <math>\Psi</math> on a part of the boundary determines the flow from the Stokes theorem.
 == Physical characteristics ==

Incompressible flow

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Latest revision as of 18:57, 19 August 2013

Dimensional analysis

Governing Equations

Physical characteristics

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