# When can one assume Incompressible Flow?

1. Aug 27, 2006

### Clausius2

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 $$\nabla\cdot \overline{u}=0$$. In other words, the velocity field is solenoidal. For a fluid material volume, that is a volume travelling with the local fluid velocity $$\overbar{u}$$ and getting deformed as the fluid does flowing, the conservation of mass must hold:

$$\frac{d}{dt}\oint_{V_f(t)}\rho dV=0$$, 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):

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

Usually the formulation in terms of the fluid volume $$V_f$$ 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 $$V_c(t)$$ located randomly (usually one chooses the easiest location) and which boundaries translate with a velocity $$\overline{u_c}$$. Let us accomplish the same time derivation than before using the RTT:

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

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

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

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

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:

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

which is true if:

$$\mathbf{\frac{\partial\rho}{\partial t}+\nabla \cdot (\rho\overline{u})=0}$$ (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:

$$\frac{D}{Dt}=\frac{\partial}{\partial t}+\overline{u}\cdot\nabla$$ (6)

equation (5) can be expressed as:

$$\frac{D\rho}{Dt}+\rho \nabla\cdot \overline{u}=0$$ (7)

and rearranging it in a more suitable form:

$$\mathbf{\nabla\cdot \overline{u}=-\frac{1}{\rho}\frac{D\rho}{Dt}}$$ (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 $$Kn=\frac{\lambda}{L}$$, where the mean free path $$\lambda\sim 10^{-8}m$$ 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 $$\rho$$ can be written as a function of two other thermodynamic variables, let say pressure $$P$$ and entropy $$s$$ per unit mass. Then, at first order:

$$d\rho=\frac{\partial \rho}{\partial P}\big)_s dP+\frac{\partial \rho}{\partial s}\big)_P ds$$ (9)

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

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

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

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

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

And from the definition (6) of Material Derivative:

$$\frac{DP}{Dt}=\frac{\partial P}{\partial t}+\overline{u}\cdot\nabla P$$ (11)

Also, the Differential Entropy Conservation equation yields:

$$\frac{Ds}{Dt}=\frac{\phi_v}{\rho T}+K\frac{\nabla^2 T}{\rho T}$$ (12)

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

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

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

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

Last edited: Aug 27, 2006
2. Aug 27, 2006

### Clausius2

Let us remind the equation (13):
$$\mathbf{\nabla\cdot \overline{u}=-\frac{1}{\rho}\left \{\frac{1}{c^2}\left(\frac{\partial P}{\partial t}+\overline{u}\cdot\nabla P\right)+\frac{T^2\beta}{c_p}\left(\frac{\phi_v}{T }+K\frac{\nabla^2 T}{T}\right)\right\}}$$

Let us imagine a two dimensional channel of length $$L$$, in vertical position under the action of gravity $$g$$, and width $$D$$. Inside the channel the flow has a characteristic velocity $$U$$. The temperature of the left wall is $$T_o$$ whereas the temperature of the right wall is $$T_1$$. In addition to that, the channel is fed by a pump which pulses fluid at a frequency $$\omega$$. The fluid has a density $$\rho$$, dynamic viscosity $$\mu$$, thermal conductivity $$K$$, speed of sound $$c$$ and heat pressure constant $$c_p$$.

Do not pay to much attention to the geometrical details. What matters here is the intensity of the nondimensional parameters. All the vital information has been given in the above statement.

I am going to look at the order of magnitude expected of each term of the right hand side of the equation (13), relative to the order of magnitude of the left hand side:

$$\nabla\cdot\overline{u}\sim O\left(\frac{U}{L}\right)$$ (14)

where $$O$$ means of the order of.

Look at the first term on the right hand side of (13). The term $$\partial P/\partial t$$ represents the unsteadiness of the pressure field. This term is usually enabled in Turbomachinery. Turbomachines generate pulses of pressure that translate across the fluid generating pulses of density. One should expect that depending on how large are the pulses of pressure the assumption of incompressibility of the velocity field loses accuracy. What matters here is the value of the pressure generated and the frequency of oscillation:

$$\frac{\frac{1}{\rho c^2}\frac{\partial P}{\partial t}}{\nabla\cdot\overline{u}}\sim \frac{O\left(\frac{\omega \Delta P}{\rho c^2}\right)}{O\left(\frac{U}{L}\right)}=St\cdot M^2$$ (15)

where $$St=\omega L/U$$ is the Strouhal Number, and $$M=U^2/c^2$$ is the Mach Number. The first condition for incompressibilty is then:

$$\mathbf{St\cdot M^2<<1}$$ (16)

That is, for the Strouhal Number being small the characteristic time of pulsation $$t_o\sim 1/\omega$$ must be much longer than the characteristic time than a fluid particle resides in the pipe $$t_r\sim L/U$$. In addition to that, low velocities compared with the speed of sound also make the Mach Number small collaborating to the accuracy of the Incompressible flow assumption.

2.2 Compressibility of Large Scale Flows.

Look at the second term of the right hand side of the equation (13). The term $$\overline{u}\cdot \nabla P$$ represents the transport of the pressure by the inertia of the fluid. This transport is important in Ocean and Atmospheric Dynamics, where large scales are involved:

$$\frac{\frac{\overline{u}\cdot \nabla P}{\rho c^2}}{\nabla\cdot\overline{u}}\sim \frac{O\left(\frac{g U}{c^2}\right)}{O(U/L)}\sim \frac{gL}{c^2}=\frac{M^2}{Fr^2}$$ (17)

where $$Fr=U^2/\sqrt{gL}$$ is the Froude Number. Thus, for the accuracy of the incompressibility assumption, $$M^2/Fr^2<<1$$, or what is the same:

$$\mathbf{\frac{gL}{c^2}<<1}$$ (18)

Note that the larger the system, the lesser the accuracy of the incompressibility assumption. The length $$c^2/g$$ is called the scale heigth, and it is the characteristic scale in the atmosphere over which there are appreciable variations of pressure. Thus for scales larger than the scale heigth the gradient of pressure causes hydrostatically large gradients of density. In order of (18) to be of order unity, $$L\sim c^2/g$$. Thus, the incompressibility assumption begins to fail on systems of scale of 100 km in the Ocean and 1 km in the atmosphere. The incompressibility assumption in the Ocean works better than in the Atmosphere.

2.3. Compressibility caused by the Coefficient of Volumetric Expansion.

Let us take the third term of the right hand side of the equation (13). The viscous dissipation function $$\phi_v$$ is proportional to the contracted product of the strain tensor with itself.

$$\frac{\frac{\phi_v T\beta}{c_p\rho}}{\nabla\cdot\overline{u}}\sim \frac{O\left(\frac{\mu U^2}{L^2}\frac{\beta}{c_p\rho}\right)}{O(U/L)}\sim Re^{-1}\frac{\beta U^2}{c_p}$$ (19)

where $$Re=\rho UL/\mu$$ is the Reynolds Number. Therefore, for the accuracy of the incompressibility assumption:

$$\mathbf{Re^{-1}\frac{\beta U^2}{c_p}<<1}$$ (20)

which is accomplished for low volumetric expansion coefficients.

2.4 Compressibility caused by Temperature Gradient.

The last term on the right hand side of equation (13) represents an irreversible source of entropy due to heat transfer.

$$\frac{T\beta K\frac{\nabla^2 T}{c_p T}}{\nabla\cdot\overline{u}}\sim \frac{O\left(\frac{\beta K \Delta T}{c_p \rho L^2}\right)}{O(U/L)}\sim Re^{-1}Pr^{-1}\beta \Delta T$$ (21)

Where $$Pr=\mu c_p/K$$ is the Prandtl Number. For air $$Pr\sim 0.7$$ whereas for water $$\Pr\sim 7$$, which means that air is a better heat conductor than water. For the accuracy of the Incompressibility assumption:

$$\mathbf{Re^{-1}Pr^{-1}\beta \Delta T<<1}$$ (22)

That is, the flow must be fast, the substance should not transfer heat very well for having a high Prandtl Number, and the temperature differences must not be very large. The critical temperature difference should be

$$\Delta T\sim Re\cdot Pr /\beta$$ (23)

3. ESTIMATIONS.

Assume for the problem stated above that the substance is water, with $$\rho=1000kg/m^3$$, $$\mu\sim 10^{-3}Kg/ms$$, and
$$\beta\sim 10^{-8} K^{-1}$$, $$c_p\sim 1000J/KgK$$, $$c\sim 10^3 m/s$$ and an usual flow velocity of $$U\sim 10 m/s$$. And leave the length of the channel, the difference of temperature and the frequency of pulsation free. The criterions for incompressible flow are (23), (20), (18) and (16):

I) $$St \cdot M^2=\frac{\omega L}{10}\frac{10^2}{1000^2}=\omega L 10^3<<1$$

II) $$\frac{gL}{c^2}=\frac{10L}{1000^2}=\frac{L}{100000}<<1$$

This gives a critical length $$L=100km$$ where the incompressibility assumption begins to fail. Thus, the critical frequency for that length is $$\omega\sim 10^{-8}Hz$$. Frequencies larger would cause I) to be of order 1 and the incompressibility assumption fails. For ordinary lengths like $$L\sim 1m$$, the critical frequency is $$\omega \sim10KHz$$. At that frequency the time required for a pressure wave to travel from one end to the other is of the same order than the time of residence of a fluid particle inside the channel. Let us go on:

III) $$Re^{-1}\frac{\beta U^2}{c_p}\sim 10^{-11}<<1$$ for ordinary lengths (1m).

IV) $$Re^{-1}Pr^{-1}\beta \Delta T\sim \Delta T \cdot 10^{-14}$$ for ordinary lengths. For being of order unity one needs a $$\Delta T$$ enormous, what is physically inviable in water.

4. FINAL RESULT: BOUSSINESQ APPROXIMATION.

According to the above analysis, the right hand side of the equation (13) becomes vanishingly small compared with the left hand side under the analyzed conditions.

Let's call
$$\Lambda=St \cdot M^2$$

$$\Pi=\frac{gL}{c^2}$$

$$\Gamma=Re^{-1}\frac{\beta U^2}{c_p}$$

and $$\Theta=Re^{-1}Pr^{-1}\beta \Delta T$$

being $$\Lambda, \Gamma, \Pi, \Theta<<1$$. Let us make an asymptotic expansion of the velocity field in terms of these small parameters:

$$\frac{\overline{u}}{U}=\overline{u}_o+(\Lambda+\Gamma+\Pi+\Theta)\overline{u}_1+ O(\Lambda^2,\Gamma^2,\Pi^2,\Theta^2)$$

and substitute this expansion into (13) letting all these four nondimensional parameters tend to zero, then at first order (i.e. with errors of those non dimensional parameters), the Continuity Equation reads:

$$\mathbf{\nabla \cdot \overline{u}_o=0+ O(\Lambda+\Gamma+\Pi+\Theta)}$$

being the second term on the right vanishingly small under the assumptions made above.

This is what is called the Boussinesq Approximation.

Finally, to end with this tutorial, you have to read my signature:

However, in Science, understanding is achieved through simplification. Liñan A. and Williams F.A. in "Fundamentals Aspects of Combustion".

Last edited: Aug 27, 2006
3. Aug 27, 2006

### Astronuc

Staff Emeritus
Nice job

4. Aug 27, 2006

### Hawknc

Wow...I just assumed Ma < 0.3 usually, but I suppose undergrads can get away with it. Very impressive.

5. Aug 30, 2006

### Norman

I have never taken a fluid dynamics class, but this tutorial is magnificent. Clear, concise and with order of magnitude arguments even someone without a fluids background was able to understand your arguments.

Hats off to you good sir!

6. Sep 5, 2006

### machinest

wow-much efforts-
good work