Incompressible Fluid With Variation of Tempeture

lucaswsn · April 8, 2022, 07:50

Dear All,

I have a issues about definition of a incompressible fluids, is it possible a variation with temperature? I has been reading this in a book, image below.

Link image

LuckyTran · April 8, 2022, 08:31

Any material where density is at most an arbitrary state function of temperature is still incompressible, the partial derivative of density with respect to pressure is 0.

Incompressible means the substance cannot be compressed by mechanical forces (i.e. mechanical pressure). Its density and volume are allowed to change via other means. Incompressible and constant density do not mean the same thing.

lucaswsn · April 8, 2022, 09:43

Quote:

Originally Posted by LuckyTran

Any material where density is at most an arbitrary state function of temperature is still incompressible, the partial derivative of density with respect to pressure is 0.

Incompressible means the substance cannot be compressed by mechanical forces (i.e. mechanical pressure). Its density and volume are allowed to change via other means. Incompressible and constant density do not mean the same thing.

In continuum mechanics incompressible material mean that don't have variation in volume (Isochoric process)? In consequence div v = 0 and rho = cte?

How include temperature in that definition?

LuckyTran · April 8, 2022, 10:31

Are we talking about fluids or continuum mechanics? In fluids, the Navier-Stokes equations contain only pressure, velocity, and density and don't explicitly contain temperature and other state variables. Hence, there's no need to differentiate between difference sources of compressibility and it is enough that the partial derivative with density is zero.

In continuum mechanics, everything is compressible and you have to keep track of the types of compressibility. If there's no volume change then it's trivial, nothing happened. If you want you can apply this same approach to fluid mechanics and just forget about incompessibility.

lucaswsn · April 8, 2022, 11:03

Quote:

Originally Posted by LuckyTran

Are we talking about fluids or continuum mechanics? In fluids, the Navier-Stokes equations contain only pressure, velocity, and density and don't explicitly contain temperature and other state variables. Hence, there's no need to differentiate between difference sources of compressibility and it is enough that the partial derivative with density is zero.

In continuum mechanics, everything is compressible. If there's no volume change then it's trivial, nothing happened.

Solution a problem that have flow and variation of temperature was necessary solution of Navier-Stokes, energy equation, continuity equation and equation of state, simultaneously. I this problem rho depend of pressure and temperature, my issues is with rho variate with temperature this problems will be classify a incompressibel flow.

LuckyTran · April 8, 2022, 11:12

If rho depends only on temperature (i.e. it doesn't depend on pressure) then the material derivative of rho will always be zero and it will always satisfy the incompressible flow criteria.

lucaswsn · April 8, 2022, 11:27

Quote:

Originally Posted by LuckyTran

If rho depends only on temperature (i.e. it doesn't depend on pressure) then the material derivative of rho will always be zero and it will always satisfy the incompressible flow criteria.

So in this case, don't have coupling velocity-pressure and consequently don't have necessary to use SIMPLE, PRIME,etc. algorithms, corrects? The equation of state will be necessary to describe evolution of rho?

LuckyTran · April 8, 2022, 11:31

You always have a pressure-velocity coupling problem because you don't have an independent equation for pressure.

Rho always has an equation of state unless this a degenerate case with a constant density. The difference is, if rho depends only on temperature then even the energy equation is weakly coupled and I can replace rho(T) with rho(x,y,z) and just prescribe the background density field and you wouldn't even know.

lucaswsn · April 8, 2022, 11:58

Quote:

Originally Posted by LuckyTran

You always have a pressure-velocity coupling problem because you don't have an independent equation for pressure.

Rho always has an equation of state unless this a degenerate case with a constant density. The difference is, if rho depends only on temperature then even the energy equation is weakly coupled and I can replace rho(T) with rho(x,y,z) and just prescribe the background density field and you wouldn't even know.

So rho = cte and rho = rho(T) was the same problems in numerical perspective? This characterize a incompressivel flow?

LuckyTran · April 8, 2022, 12:17

For fluids... yes. To give an example of what this looks like conceptually... a flow where rho is a function of temperature is like a mixture of a bunch of different but otherwise constant density fluids all mixing with one another.

It might be helpful to distinguish between compressibility of the medium and the divergence free velocity constraint (which we call an incompressible flow).

lucaswsn · April 8, 2022, 12:44

Quote:

Originally Posted by LuckyTran

For fluids... yes. To give an example of what this looks like conceptually... a flow where rho is a function of temperature is like a mixture of a bunch of different but otherwise constant density fluids all mixing with one another.

It might be helpful to distinguish between compressibility of the medium and the divergence free velocity constraint (which we call an incompressible flow).

What's it the difference between compressibility of the medium and incompressible flow?

LuckyTran · April 8, 2022, 13:27

Because you can have a very compressible medium, e.g. a gas, that flows in such a way that the divergence free velocity constraint is still satisfied.

FMDenaro · April 8, 2022, 13:52

Just consider the density equation:

d rho/dt + v.Grad rho = - rho div v

in case of a simple divergence-free assumption, you get D rho/Dt=0, that is the particle retains its density constant along the trajectory. But density can have time and space gradients.

On the other hand if you assume rho constant in time and space then the velocity is divergence-free.

The "incompressible flow" assumption is a mathematical model, it assumes that dp/rho=a^2 is such that the sound velocity is >> then the convective velocity. Note that for a velocity field going to zero the density equation says that d rho/dt = 0 but Grad rho could be not zero! That is a case for a stratification due to a temperature field.

lucaswsn · April 8, 2022, 14:35

Quote:

Originally Posted by FMDenaro

Just consider the density equation:

d rho/dt + v.Grad rho = - rho div v

in case of a simple divergence-free assumption, you get D rho/Dt=0, that is the particle retains its density constant along the trajectory. But density can have time and space gradients.

On the other hand if you assume rho constant in time and space then the velocity is divergence-free.

The "incompressible flow" assumption is a mathematical model, it assumes that dp/rho=a^2 is such that the sound velocity is >> then the convective velocity. Note that for a velocity field going to zero the density equation says that d rho/dt = 0 but Grad rho could be not zero! That is a case for a stratification due to a temperature field.

$P = P(\rho , T)$ , invert that equation $\rho = \rho(P , T)$ , for a incompressible flow $\rho = \rho(T)$ or $\rho = cte$ , that imples $dP / d\rho = 0$ , this argument is valid?

FMDenaro · April 8, 2022, 14:49

Quote:

Originally Posted by lucaswsn

$P = P(\rho , T)$ , invert that equation $\rho = \rho(P , T)$

dp=(dp/rho) d rho + (dp/dT) d T

Ok, for example perfect gas model rho RT=p. So what? We are focusing on the assumptions for dp/drho.

Incompressible flow model assumes that a small perturbation of the density (drho) produces an infinite variation of the pressure (dp). Only if you add also the omothermal assumption the density is modelled as constant in time and space.

(Of course dp/drho at constant entropy is the correct sound velocity…)

That is:

- if you assume rho=rho0, the flow is incompressible and omothermal. In this case the pressure gradient in the momentum equation has no physical meaning but it is only a gradient field that ensures the divergence-free constraint.

- if you assume only an incompressible flow model you can admit variation of density due to temperature (as it it assumed in buoyancy-driven flow, in combustion, etc). Pressure field still ensures the divergence-free constraint.

hnemati · April 11, 2022, 07:42

Quote:

Originally Posted by lucaswsn

$P = P(\rho , T)$ , invert that equation $\rho = \rho(P , T)$ , for a incompressible flow $\rho = \rho(T)$ or $\rho = cte$ , that imples $dP / d\rho = 0$ , this argument is valid?

It is called low-Mach number approximation, where the density is only a function of temperature.
Have a look at this paper.
https://www.sciencedirect.com/scienc...17931014011478

lucaswsn · April 11, 2022, 07:56

Quote:

Originally Posted by hnemati

It is called low-Mach number approximation, where the density is only a function of temperature.
Have a look at this paper.
https://www.sciencedirect.com/scienc...17931014011478

I'm concluding that variation of pressure with density classifies a fluid in incompressivel or compressivel and not information about density only;

LuckyTran · April 11, 2022, 08:39

It shouldn't be surprising... The state postulate gives that substance is fully defined by two independent intensive properties.

If you take pressure as a function of density and temperature:

The variation of pressure with respect to temperature is the gas constant R. The variation of pressure with density is related to sound and compressibility

That's it. There are no more derivatives. You are free to use other interpretations of course and consider density as a function of pressure and temperature and what-not but at the end of the day those variants have to be equivalent because they describe the same system.

April 8, 2022, 07:50	Incompressible Fluid With Variation of Tempeture	#1
lucaswsn New Member Lucas Nascimento Join Date: Apr 2022 Posts: 8 Rep Power: 4	Dear All, I have a issues about definition of a incompressible fluids, is it possible a variation with temperature? I has been reading this in a book, image below. Link image

April 8, 2022, 11:12		#6
LuckyTran Senior Member Lucky Join Date: Apr 2011 Location: Orlando, FL USA Posts: 5,675 Rep Power: 66	If rho depends only on temperature (i.e. it doesn't depend on pressure) then the material derivative of rho will always be zero and it will always satisfy the incompressible flow criteria. arjun likes this.

April 8, 2022, 13:27		#12
LuckyTran Senior Member Lucky Join Date: Apr 2011 Location: Orlando, FL USA Posts: 5,675 Rep Power: 66	Because you can have a very compressible medium, e.g. a gas, that flows in such a way that the divergence free velocity constraint is still satisfied. FMDenaro likes this.

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April 8, 2022, 08:31		#2
LuckyTran Senior Member Lucky Join Date: Apr 2011 Location: Orlando, FL USA Posts: 5,675 Rep Power: 66	Any material where density is at most an arbitrary state function of temperature is still incompressible, the partial derivative of density with respect to pressure is 0. Incompressible means the substance cannot be compressed by mechanical forces (i.e. mechanical pressure). Its density and volume are allowed to change via other means. Incompressible and constant density do not mean the same thing.

April 8, 2022, 10:31		#4
LuckyTran Senior Member Lucky Join Date: Apr 2011 Location: Orlando, FL USA Posts: 5,675 Rep Power: 66	Are we talking about fluids or continuum mechanics? In fluids, the Navier-Stokes equations contain only pressure, velocity, and density and don't explicitly contain temperature and other state variables. Hence, there's no need to differentiate between difference sources of compressibility and it is enough that the partial derivative with density is zero. In continuum mechanics, everything is compressible and you have to keep track of the types of compressibility. If there's no volume change then it's trivial, nothing happened. If you want you can apply this same approach to fluid mechanics and just forget about incompessibility.

April 8, 2022, 11:31		#8
LuckyTran Senior Member Lucky Join Date: Apr 2011 Location: Orlando, FL USA Posts: 5,675 Rep Power: 66	You always have a pressure-velocity coupling problem because you don't have an independent equation for pressure. Rho always has an equation of state unless this a degenerate case with a constant density. The difference is, if rho depends only on temperature then even the energy equation is weakly coupled and I can replace rho(T) with rho(x,y,z) and just prescribe the background density field and you wouldn't even know.

April 8, 2022, 12:17		#10
LuckyTran Senior Member Lucky Join Date: Apr 2011 Location: Orlando, FL USA Posts: 5,675 Rep Power: 66	For fluids... yes. To give an example of what this looks like conceptually... a flow where rho is a function of temperature is like a mixture of a bunch of different but otherwise constant density fluids all mixing with one another. It might be helpful to distinguish between compressibility of the medium and the divergence free velocity constraint (which we call an incompressible flow).

April 8, 2022, 13:52		#13
FMDenaro Senior Member Filippo Maria Denaro Join Date: Jul 2010 Posts: 6,769 Rep Power: 71	Just consider the density equation: d rho/dt + v.Grad rho = - rho div v in case of a simple divergence-free assumption, you get D rho/Dt=0, that is the particle retains its density constant along the trajectory. But density can have time and space gradients. On the other hand if you assume rho constant in time and space then the velocity is divergence-free. The "incompressible flow" assumption is a mathematical model, it assumes that dp/rho=a^2 is such that the sound velocity is >> then the convective velocity. Note that for a velocity field going to zero the density equation says that d rho/dt = 0 but Grad rho could be not zero! That is a case for a stratification due to a temperature field.

April 11, 2022, 08:39		#18
LuckyTran Senior Member Lucky Join Date: Apr 2011 Location: Orlando, FL USA Posts: 5,675 Rep Power: 66	It shouldn't be surprising... The state postulate gives that substance is fully defined by two independent intensive properties. If you take pressure as a function of density and temperature: The variation of pressure with respect to temperature is the gas constant R. The variation of pressure with density is related to sound and compressibility That's it. There are no more derivatives. You are free to use other interpretations of course and consider density as a function of pressure and temperature and what-not but at the end of the day those variants have to be equivalent because they describe the same system.