(adsbygoogle = window.adsbygoogle || []).push({}); Incompressible Flow: Assumptions for its validity

After a recent hot discussion brought to the board, I think it would be good to clear up this question.

Firstly, it does not make sense to talk about an incompressible fluid. There are no incompressible fluids in the Nature, we can only model a incompressible flow under the assumptions I clarify here.

1. THE CONTINUITY EQUATION.

1.1. The Integral Continuity Equation.

An incompressible flow field is a field such that the divergence of the velocity is zero [tex]\nabla\cdot \overline{u}=0[/tex]. In other words, the velocity field is solenoidal. For a fluid material volume, that is a volume travelling with the local fluid velocity [tex]\overbar{u}[/tex] and getting deformed as the fluid does flowing, the conservation of mass must hold:

[tex]\frac{d}{dt}\oint_{V_f(t)}\rho dV=0[/tex], that is, there cannot be any mass variation inside the material volume.

For doing that time derivative one cannot permute the integral sign and the derivative because the domain of integration depends on time. In order to accomplish the integration one must use the Leibnitz rule, or in this context the so-called Reynolds Transport Teorem (RTT):

[tex]\frac{d}{dt}\oint_{V_f(t)}\rho dV=\oint_{V_f(t)}\frac{\partial \rho}{\partial t} dV+\oint_{S_f(t)}\overline{u}\cdot \overline{dS}=0[/tex] (1)

Usually the formulation in terms of the fluid volume [tex]V_f[/tex] is useless, because in real applications is hard to know how an initial fluid volume is distorted. On the other hand, it is more feasible do the formulation in terms of a control volume [tex]V_c(t)[/tex] located randomly (usually one chooses the easiest location) and which boundaries translate with a velocity [tex]\overline{u_c}[/tex]. Let us accomplish the same time derivation than before using the RTT:

[tex]\frac{d}{dt}\oint_{V_c(t)}\rho dV=\oint_{V_c(t)}\frac{\partial \rho}{\partial t} dV+\oint_{S_c(t)}\overline{u_c}\cdot \overline{dS}[/tex] (2)

Assume both control and fluid volume coincide in space at some instant [tex]t[/tex]. Then, substracting (2) from (1), one obtains the Continuty Equation:

[tex]\mathbf{\frac{d}{dt}\oint_{V_c(t)}\rho dV+\oint_{S_c(t)}(\overline{u}-\overline{u_c})\cdot \overline{dS}=0}[/tex] (3)

Note that for constant density and fixed volume control, the equation reduces to the conservation of the volume flux [tex]\oint_{S_c}\overline{u}\cdot \overline{dS}=0[/tex].

1.2 The Differential Continuity Equation

Assume a fixed control volume. Applying the Gauss Theorem to the surface integral of the expression (3), one obtains:

[tex]\oint_{V}\left\{\frac{\partial\rho}{\partial t}+\nabla \cdot (\rho\overline{u})\right\}dV=0[/tex] (4)

which is true if:

[tex]\mathbf{\frac{\partial\rho}{\partial t}+\nabla \cdot (\rho\overline{u})=0}[/tex] (5)

which corresponds to the differential version of the Continuity Equation. In terms of the Material Derivative, that is, the time derivative viewed from a reference frame attached to the fluid volume:

[tex]\frac{D}{Dt}=\frac{\partial}{\partial t}+\overline{u}\cdot\nabla[/tex] (6)

equation (5) can be expressed as:

[tex]\frac{D\rho}{Dt}+\rho \nabla\cdot \overline{u}=0[/tex] (7)

and rearranging it in a more suitable form:

[tex]\mathbf{\nabla\cdot \overline{u}=-\frac{1}{\rho}\frac{D\rho}{Dt}}[/tex] (8)

Equation (8) is a general expression with only one previous assumption made. I am assuming that the flow is continuum, that is, the spatial derivatives are well defined, what is not the case in highly rarified flows. The measure of rarification is the Knudsen Number [tex]Kn=\frac{\lambda}{L}[/tex], where the mean free path [tex]\lambda\sim 10^{-8}m[/tex] in air at standard conditions. Therefore, the Knudsen Number is low enough in practical applications for considering continuum flow, except those of reentry of space vehicles and hypersonic flight. Also, equation (8) is valid in Reactive Flow. The total production of mass in a fluid volume is still zero under chemical reaction.

2. INCOMPRESSIBLE FLOW ASSUMPTIONS

Let us introduce the definition of the diferential of the density. According to the Thermodynamics the differential of [tex]\rho[/tex] can be written as a function of two other thermodynamic variables, let say pressure [tex]P[/tex] and entropy [tex]s[/tex] per unit mass. Then, at first order:

[tex]d\rho=\frac{\partial \rho}{\partial P}\big)_s dP+\frac{\partial \rho}{\partial s}\big)_P ds[/tex] (9)

where the subscript means holding constant that variable. Both coefficients [tex]\frac{\partial \rho}{\partial P}\big)_s[/tex] and [tex]\frac{\partial \rho}{\partial s}\big)_P [/tex] are called Thermodynamic Coefficients. Turns out that they correspond to:

[tex]\frac{\partial \rho}{\partial P}\big)_s =\frac{1}{c^2}[/tex] where [tex]c[/tex] is the speed of sound, and

[tex]\frac{\partial \rho}{\partial s}\big)_P=\frac{T\rho\beta}{c_p}[/tex], where [tex]\beta[/tex] is the coefficient of volumetric expansion.

Taking the material derivative of (9) and using the above expressions for the Thermodynamic Coefficients:

[tex] \frac{D\rho}{Dt}=\frac{1}{c^2}\frac{DP}{Dt}+\frac{T^2\beta}{c_p}\frac{Ds}{Dt}[/tex] (10)

And from the definition (6) of Material Derivative:

[tex]\frac{DP}{Dt}=\frac{\partial P}{\partial t}+\overline{u}\cdot\nabla P[/tex] (11)

Also, the Differential Entropy Conservation equation yields:

[tex]\frac{Ds}{Dt}=\frac{\phi_v}{\rho T}+K\frac{\nabla^2 T}{\rho T}[/tex] (12)

(see references). Equation (12) is basically stating that the change of entropy in a fluid volume is caused only by the viscous dissipation [tex]\phi_v[/tex] and the divergence of the heat flux.

Inserting (11) and (12) in (10), and (10) in (8), one obtains the transcendental equation:

[tex]\mathbf{\nabla\cdot \overline{u}=-\frac{1}{\rho}\left \{\frac{1}{c^2}\left(\frac{\partial P}{\partial t}+\overline{u}\cdot\nabla P\right)+\frac{T\beta}{c_p}\left(\frac{\phi_v}{T}+K\frac{\nabla^2 T}{T}\right)\right\}}[/tex] (13).

In the next post I will put an example for working out under what conditions equation (13) can be approximated as [tex]\nabla\cdot \overline{u}\sim 0[/tex] with an error asymptotically small.

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# When can one assume Incompressible Flow?

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