Some questions about incompressible flow,thx!

aladdincham · December 28, 2010, 20:52

Hi guys,
i got some words on a book, but i cannot understood them very well, it said "In the low speed fluid and gas,they are treated as imcompressible flow.There was no change in the density,so there is no relationship between momentum equations and mass conservation equations".
Well ,my question is ,why?
THX!

ignat · December 29, 2010, 02:59

Quote:

Originally Posted by aladdincham

Hi guys,
i got some words on a book, but i cannot understood them very well, it said "In the low speed fluid and gas,they are treated as imcompressible flow.There was no change in the density,so there is no relationship between momentum equations and mass conservation equations".
Well ,my question is ,why?
THX!

It is true. It's possible to explain it so.
It is known from compressible gasdynamic theory that
density depends on Mach number M as
$\rho = \rho_0 (1+\frac{\gamma -1}{2}M^2)^{-\frac{1}{\gamma-1}}$
where $\gamma$ is the specific heat ratio,
$\rho_0=const$ is the stagnation density,
$M = \frac{u}{c}$
where u is the velocity and c is the speed of sound.

Thus, when the maximum M (and maximum u) tends to zero, density becomes independent of u and M and may be treated as constant. In practice when maximum M < 0.1 flow may be considered as incompressible.

aladdincham · December 29, 2010, 21:21

Quote:

Originally Posted by ignat

It is true. It's possible to explain it so.
It is known from compressible gasdynamic theory that
density depends on Mach number M as
$\rho = \rho_0 (1+\frac{\gamma -1}{2}M^2)^{-\frac{1}{\gamma-1}}$
where $\gamma$ is the specific heat ratio,
$\rho_0=const$ is the stagnation density,
$M = \frac{u}{c}$
where u is the velocity and c is the speed of sound.

Thus, when the maximum M (and maximum u) tends to zero, density becomes independent of u and M and may be treated as constant. In practice when maximum M < 0.1 flow may be considered as incompressible.

Thanks for your reply, i think i make some mistake,the words are“In the low speed fluid and gas,they are treated as imcompressible flow.There was no change in the density,so momentum equations & mass conservation equations are no relationship with energy equations”
WHY?

otd · December 30, 2010, 06:01

With density constant, momentum and conservation depend only on velocities and pressure. In 3-d flow, that's (for example) u, v, w, and p. With 3 momentum eqs. and continuity, that's 4 equations and 4 unknowns. They can (and are) solved without reference to Temperature, internal energy, etc.

Knowing u, v, w, p, etc, the energy equation is one equation in one unknown (temperature, internal energy, enthalpy, etc depending on the form you select for the equation).

It might be better to say "the momentum equations are independent of the temperature" but temperature depends on the momentum and pressure.

Some other folks might say the coupling between momentum-continuity and energy is one-way only.

That thing about 'no relationship' is sort of confusing.

Hope this helps,

OTD

ignat · December 30, 2010, 06:46

Quote:

Originally Posted by aladdincham

There was no change in the density,so momentum equations & mass conservation equations are no relationship with energy equations”
WHY?

It follows from non-dimensional form of energy equation.
Let non-dimensional pressure be:
$p=\frac{p-p_0}{\rho_r u_r^2}$
where

is the constant background pressure, $\rho_r$ is the reference density and

is the reference velocity.

then the energy equation in non-dimensional form can be written as

$M_r^2 \{ \frac{\partial }{\partial t} [p+(\gamma-1)\rho u^i u^i/2 ] + [u^k (\gamma p+(\gamma-1)\rho u^i u^i/2) ]_{,k}\}+u_{,k}^k=0$

Here
$M_r=\frac{u_r}{\sqrt{\gamma p_0 / \rho_r}}$ is the reference Mach number.
If

tends to zero then the energy equation coincides with continuity equation div(u)=0!

Thus energy equation in itself "disappears" for incompressible flows (M->0)!

Therefore "momentum equations & mass conservation equations are no relationship with energy equations"

Details see for example in Wesseling's papers
1. van der Heul DR, Vuik C, Wesseling P. A conservative pressure correction method for compressible flow at all speeds.// Int. J. Numer. Meth. Fluids, 2002, v. 40, pp. 521-529.

2. I. Wenneker, A. Segal and P. Wesseling. A Mach-uniform unstructured grid method. // Int. J. Numer. Meth. Fluids, 2002, v. 40, pp. 1209-1235.

aladdincham · December 30, 2010, 22:34

thanks for your reply, i got the quetion`s answer, your answer is very clear.

December 28, 2010, 20:52	Some questions about incompressible flow,thx!	#1
aladdincham New Member Aladdin Join Date: Jan 2010 Posts: 26 Rep Power: 16	Hi guys, i got some words on a book, but i cannot understood them very well, it said "In the low speed fluid and gas,they are treated as imcompressible flow.There was no change in the density,so there is no relationship between momentum equations and mass conservation equations". Well ,my question is ,why? THX!

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December 30, 2010, 06:01		#4
otd Member private Join Date: Mar 2009 Posts: 74 Rep Power: 17	With density constant, momentum and conservation depend only on velocities and pressure. In 3-d flow, that's (for example) u, v, w, and p. With 3 momentum eqs. and continuity, that's 4 equations and 4 unknowns. They can (and are) solved without reference to Temperature, internal energy, etc. Knowing u, v, w, p, etc, the energy equation is one equation in one unknown (temperature, internal energy, enthalpy, etc depending on the form you select for the equation). It might be better to say "the momentum equations are independent of the temperature" but temperature depends on the momentum and pressure. Some other folks might say the coupling between momentum-continuity and energy is one-way only. That thing about 'no relationship' is sort of confusing. Hope this helps, OTD

December 30, 2010, 22:34		#6
aladdincham New Member Aladdin Join Date: Jan 2010 Posts: 26 Rep Power: 16	thanks for your reply, i got the quetion`s answer, your answer is very clear.