(In)-Compressibility with temperature changes

cymbourne · April 11, 2015, 16:32

Hi guys!

I am having some philosophical issues with the (in)compressible concept when applied to a low Mach flow with high-temperature changes.

As far as I know, a flow can be considered as incompressible when Mach number is below 0.3. However, I'm not certain about what happens when high-temperature gradients take place. Let's image a duct heated by two close constant-temperature walls (at 300K and 2000K). Let's say the air flow enters at 500K and the Mach number is 0.05. Should I consider this as incompressible?

Thanks!!!!

tas38 · April 11, 2015, 16:57

Quote:

Originally Posted by cymbourne

Hi guys!

I am having some philosophical issues with the (in)compressible concept when applied to a low Mach flow with high-temperature changes.

As far as I know, a flow can be considered as incompressible when Mach number is below 0.3. However, I'm not certain about what happens when high-temperature gradients take place. Let's image a duct heated by two close constant-temperature walls (at 300K and 2000K). Let's say the air flow enters at 500K and the Mach number is 0.05. Should I consider this as incompressible?

Thanks!!!!

Incompressible generally refers to negligible density variation with pressure. For the case described, an incompressible solver coupled to the energy equation would suffice. Density and viscosity would be solely functions of temperature.

cymbourne · April 12, 2015, 06:30

Thank you for your prompt response! But I'm still pretty confused

What I don't get is why the density gradient is then neglected in the continuity and momentum equations. For example, let's consider the continuity equation in its compressible form:

$\frac{\partial \rho}{\partial t} + \vec{\nabla} \cdot (\rho \vec{v}) = 0$

when the incompressible hypothesis is applied, the equation reads:

$\vec{\nabla} \cdot (\vec{v}) = 0$

why can we say that $\vec{\nabla} \cdot (\rho) = 0$ ? Isn't $\rho = f(\vec{x})$ ?

Thank you!

Regards,

FMDenaro · April 12, 2015, 06:56

Quote:

Originally Posted by cymbourne

Thank you for your prompt response! But I'm still pretty confused

What I don't get is why the density gradient is then neglected in the continuity and momentum equations. For example, let's consider the continuity equation in its compressible form:

$\frac{\partial \rho}{\partial t} + \vec{\nabla} \cdot (\rho \vec{v}) = 0$

when the incompressible hypothesis is applied, the equation reads:

$\vec{\nabla} \cdot (\vec{v}) = 0$

why can we say that $\vec{\nabla} \cdot (\rho) = 0$ ? Isn't $\rho = f(\vec{x})$ ?

Thank you!

Regards,

I see some confusion in your comment....

If you ASSUME the flow has a homogeneous density, then rho=rho0 does no longer depend on time and space. That leads to Div v =0.

Conversely, if you assume only Div v =0 then

d rho /dt + v.Grad rho = 0

and the density remains a function of time and space. If you assume a steady condition then Grad rho = 0.

Incompressible model means you consider dp/drho = a^2 -> Inf so that M->0. In other words, you assume that wave pressure travels at infinity speed, so that any pressure disturbance is propagated instantaneously.

As rho is also a function of the temperature, you can have a relation rho=rho(T) that can be introduced into the incompressible model, for example using Bousinnesq for buoyancy flows

PS: Div rho does not mathematically exist.....

cymbourne · April 12, 2015, 17:52

Thanks for your reply

It was very helpful but I still do not get it

I think that perhaps I did not explain my question clearly, so I will try to reformulate it in a simpler way:

- Can I use the following simplified continuity equation in the aforementioned duct ?

${\nabla \cdot \mathbf{u}} = 0.$

- If yes, why?

- If no, which form of the continuity equation should I use?

Thanks!!

PS: I consider only stationary flows.

PS2: Sorry about the divergence, I didn't pay enough attention to the divergence/gradient nabla symbols

FMDenaro · April 12, 2015, 18:01

In case of high temperature gradient, Bousinnesq can not be applied, you need to formulate the system in such a way to take into account for the variation of density in terms of temperature.

You can find in literature formulations adopted for reactive flows at low Mach number

April 11, 2015, 16:32	(In)-Compressibility with temperature changes	#1
cymbourne New Member Marion Join Date: Apr 2013 Posts: 15 Rep Power: 13	Hi guys! I am having some philosophical issues with the (in)compressible concept when applied to a low Mach flow with high-temperature changes. As far as I know, a flow can be considered as incompressible when Mach number is below 0.3. However, I'm not certain about what happens when high-temperature gradients take place. Let's image a duct heated by two close constant-temperature walls (at 300K and 2000K). Let's say the air flow enters at 500K and the Mach number is 0.05. Should I consider this as incompressible? Thanks!!!!

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April 12, 2015, 06:30		#3
cymbourne New Member Marion Join Date: Apr 2013 Posts: 15 Rep Power: 13	Thank you for your prompt response! But I'm still pretty confused What I don't get is why the density gradient is then neglected in the continuity and momentum equations. For example, let's consider the continuity equation in its compressible form: $\frac{\partial \rho}{\partial t} + \vec{\nabla} \cdot (\rho \vec{v}) = 0$ when the incompressible hypothesis is applied, the equation reads: $\vec{\nabla} \cdot (\vec{v}) = 0$ why can we say that $\vec{\nabla} \cdot (\rho) = 0$ ? Isn't $\rho = f(\vec{x})$ ? Thank you! Regards,

April 12, 2015, 17:52		#5
cymbourne New Member Marion Join Date: Apr 2013 Posts: 15 Rep Power: 13	Thanks for your reply It was very helpful but I still do not get it I think that perhaps I did not explain my question clearly, so I will try to reformulate it in a simpler way: - Can I use the following simplified continuity equation in the aforementioned duct ? ${\nabla \cdot \mathbf{u}} = 0.$ - If yes, why? - If no, which form of the continuity equation should I use? Thanks!! PS: I consider only stationary flows. PS2: Sorry about the divergence, I didn't pay enough attention to the divergence/gradient nabla symbols

April 12, 2015, 18:01		#6
FMDenaro Senior Member Filippo Maria Denaro Join Date: Jul 2010 Posts: 6,768 Rep Power: 71	In case of high temperature gradient, Bousinnesq can not be applied, you need to formulate the system in such a way to take into account for the variation of density in terms of temperature. You can find in literature formulations adopted for reactive flows at low Mach number