1
$\begingroup$

In scientific literature, we know that the condition for incompressible flow is that the particular derivative of the fluid density $\rho$ is zero: $$ \frac{D\rho}{Dt}=0$$

and for steady flows the condition is that the partial derivative with respect to time is zero $\left(\dfrac{\partial \rho}{\partial t}=0\right)$.

However, the definition of the particular derivative is: $$\frac{D \rho}{Dt} = \frac{\partial \rho}{\partial t} + \left(\vec{v} \cdot \vec\nabla\right) \rho $$

The question is: if a flow is incompressible does that mean that the flow is necessarily steady? Can there be an unsteady and incompressible flow?

$\endgroup$
3
$\begingroup$

Yes, there can be. But, what we know is that in an incompressible flow (as you have defined it), the density field is always steady -- which is obvious because we are defining the density to not change.

You've only listed the conservation of mass equation, but remember there are 2 more -- conservation of momentum and conservation of energy. Incompressibility does nothing to remove the time derivatives from these equations. It does, however, decouple the energy equation from the rest of the system and I encourage you to work through the entire derivation and implications of the complete system.

As a final note, and this is being a bit pedantic, but incompressibility really means that:

$$\frac{\partial \rho}{\partial p} \approx 0 $$

and not that the density is constant. It is entirely possible (and common) to solve the "incompressible" equations where density varies due to multi-species flows (think helium and air, or salt water and fresh water mixing) or due to thermal gradients (hot air is less dense than cold). What you have defined is constant density flow, which is more restrictive than incompressible. The terms are commonly used interchangeably, but it still bothers me enough that I felt I should point it out! Low speed combustion and many atmospheric flows rely on this distinction.

$\endgroup$
  • $\begingroup$ "You've only listed the conservation of mass equation" The equation $\partial_t\rho +\mathbf v\cdot\nabla\rho=0$ is not mass conservation equation, but condition of constant density. Conservation of mass is not mentioned in the question. $\endgroup$ – Ján Lalinský Sep 6 '15 at 9:57
1
$\begingroup$

$\frac{D}{Dt}(\cdot)$ operator gives the change in property of a $\textit{specific}$ fluid particle. In particular, $\frac{D\rho}{Dt}$ evaluated at $(x,y,z,t)$ gives change in density of the fluid particle located at position $(x,y,z)$ at time $t$. Therefore $\frac{D\rho}{Dt}=0$ everywhere means that at any spatial point and at any time, density of fluid particle remains constant. This doesn't mean however that density itself is constant with time or with space.

Let me illustrate. Consider two immiscible fluids of different densities in a container, say oil on top of water in a wide vessel. Now you shake the container for some time and leave it. The two liquids will slosh around a bit, and there shall be waves (or more complicated activity) at the interface. If you now place a density-probe near the interface, sometimes it will record density of water and sometimes that of oil. So $\frac{\partial \rho}{\partial t}\neq0$. Also density is obviously not constant across space. Yet $\frac{D\rho}{Dt}=0$ everywhere in this flow, because each fluid particle conserves its density. Therefore $\frac{D\rho}{Dt}=0$ does not imply that flow is steady.

Also as suggested by @tpg2114 incompressible flow strictly means $\frac{\partial \rho}{\partial p}=0$.

$\endgroup$
1
$\begingroup$

https://en.wikipedia.org/wiki/Incompressible_flow

Yes a flow can be incompressible (rather isochoric) and unsteady.

However, the unsteady term in Conservation of Mass equation is cancelled by advection term regardless of whether the flow is incompressible or compressible


CASE 1: Any fluid material that undergoes an incompressible flow

ASSUME:

-No mixing of multiple fluids, etc.

-Unsteady flow

ANALYSIS:

1) Integral Form of Conservation of mass (entire control volume) (see wiki):

  • D(mass)/Dt = 0 = D/Dt[integral of density over the control volume]

2) Ignore integral over control volume

3) => Differential Form of Conservation of Mass (differential control volume)

  • => D/Dt[density] = 0;

    Now here's where the actual fluid material properties comes into the algebra/calc

4) Apply the total derivative to density assuming it's a function of space and time; not pressure

  • Although if you did assume pressure was a variable then the term generated would be zero due to flow restrictions which are independent of the fluid material property (see case 2) i.e. partial derivative of density wrt to pressure is zero

  • Also, if any terms are independently zero here then so they are also zero in the differential form of Conservation of Mass

5) Convert the total derivative to a material derivative (the differential control volume follows the differential fluid element)

6) Set Conservation of Mass (Differential Form) equal to equation from 5)

7) => divergence of flow velocity is zero

RESULTS:

  • the unsteady flow will have it's unsteady term "balanced" by the advection term (dot product of the gradient of density with the control volume's velocity) to achieve incompressible flow (isochoric flow).

  • Which implied that the density has to be non-uniform over that control volume i.e. density is function of space and time

  • Because flow is incompressible: density should not be a function of pressure

    • despite if fluid material is compressible (e.g. air)
    • else the term generated from total derivative would need to be zeroed (see 4)
  • Reiterating: the unsteady and advection terms actually zero out the in conservation of mass you could also think of it as they balance out (sum to zero) according to Wiki (though I disagree) Despite in/compressibility and despite if the flow is un/steady and non/uniform

    • This says nothing about the D/Dt(density) (although it is zero) so the unsteady and advection terms would still be null; as those terms would cancel with themselves after equating the total derivative (coinciding with the control volume) to the material derivative

        - But that would say nothing about the in/compressibility of the flow 
      
         - nor the in/compressibility of the fluid material
      

CASE 2 A Compressible Fluid that undergoes Compressible Flow

ASSUME

  • No mixing of multiple fluids, etc.

  • Density would be a function of space, time and pressure.

ANALYSIS

  • Same analysis as above except .......

  • Apply the total derivative to the density function you'll see an additional term (not seen in wiki)

  • Again unsteady and non-uniform terms will be null

  • the product of density and divergence of flow velocity will be equal to this generated term due to density's dependence on pressure for compressible flow situation

RESULTS

-do very minor algebra and you have the compressibility (beta) formula rho = density, P = normal pressures

beta = 1/rho *(drho/dP)


it's really cool cuz it connects very well with mechanics of materials

  • Recognize that the divergence of the flow velocity is just the normal strain rates

-If you assume the the flow velocity is not a function of time (steady) and you do some more algebra, calculus, and hook's law you can acquire bulk modulus (K) (inverse of compressibility) from mechanics of materials:

K = -E/(3*(1-2nu) which will be equal to the inverted compressibility side K = 1/beta = rho*dP/drho


$\endgroup$
  • $\begingroup$ edit: Just emphasized that D/Dt(density) does not say anything about un/steady or non/uniform flow. $\endgroup$ – Louis Tkach Jul 18 '16 at 0:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.