Derivations for Continuity equation of Fluid & Euler's Equation of Fluid Motion

[Shan K]

Will anyone give me the derivations for continuty equation of fluid and euler's equation of fluid motion .

 

[boneh3ad]

So just in case you aren't familiar with the Reynolds Transport Theorem, I will start before that. Also note I am leaving it all as single integrals rather than introducing double and triple integrals. The domain of integration should make it obvious what it means.

Starting with the continuity equation. We can state, in words, the conservation of mass in an Eulerian frame as:

The time rate of change of mass in a control volume, $\mathscr{V}$ plus the net outward flux of mass through its surface, $A$ is identically equal to zero.

First we will talk about the time rate of change of mass in the volume. We define the following:

Mass in volume $d\mathscr{V}$: $$\rho\;d\mathscr{V}$$
Mass in volume $\mathscr{V}$: $$\int\limits_{\mathscr{V}}\rho \; d\mathscr{V}$$
Time rate of change of mass in volume $\mathscr{V}$: $$\int\limits_{\mathscr{V}}\frac{\partial \rho}{\partial t} \; d \mathscr{V}$$

Next we address the outward mass flux across the surface:

Mass flux per area: $$\rho \bar{V}$$
Mass flux through $dA$: $$\rho \bar{V} \cdot \hat{n}\; dA$$
Mass flux through $A$: $$\oint\limits_{A}\rho \bar{V} \cdot \hat{n}\; dA$$

So combining these terms to form the integral form of the continuity equation gives
$$\int\limits_{\mathscr{V}}\frac{\partial \rho}{\partial t} \; d \mathscr{V} + \oint\limits_{A}\rho \bar{V} \cdot \hat{n} \; dA = 0$$

From here, we can use the Divergence Theorem (Gauss' Theorem) to change the second term to a volume integral.
$$\oint\limits_{A}(\rho \bar{V}) \cdot \hat{n} \; dA = \int\limits_{\mathscr{V}}\nabla \cdot (\rho \bar{V}) d\mathscr{V}$$
This can be substituted into the previous equation to get
$$\int\limits_{\mathscr{V}}\left[\dfrac{\partial \rho}{\partial t} + \nabla \cdot (\rho \bar{V})\right] d\mathscr{V} = 0$$
For this to be satisfied in general, the integrand must be zero, so the differential form of the continuity equation is
$$\boxed{\dfrac{\partial \rho}{\partial t} + \nabla \cdot (\rho \bar{V}) = 0}$$
You can also use the definition of the material derivative to put it in its slightly more popular form
$$\boxed{\dfrac{D\rho}{Dt} + \rho \nabla \cdot \bar{V} = 0}$$

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Moving onto the momentum equation, we can state the conservation law in words as

The time rate of change of momentum in $\mathscr{V}$ plus the net outward momentum flux through $A$ equals the net force on the fluid.

Using the same method as before to define each term, we can say that

Momentum per volume: $$\rho\bar{V}$$
Momentum in volume $d\mathscr{V}$: $$\rho\bar{V}\;d\mathscr{V}$$
Momentum in volume $\mathscr{V}$: $$\int\limits_{\mathscr{V}}\rho\bar{V}\;d\mathscr{V}$$
Time rate of change of momentum in $\mathscr{V}$: $$\int\limits_{\mathscr{V}}\dfrac{\partial}{\partial t}(\rho\bar{V})\;d\mathscr{V}$$

Next, the outward momentum flux term can be addressed as follows:

Momentum flux per area: $$(\rho\bar{V})\bar{V}$$
Momentum flux across $dA$: $$(\rho\bar{V})\bar{V}\cdot\hat{n}\;dA$$
Momentum flux across $A$: $$\oint\limits_{A}(\rho\bar{V})\bar{V}\cdot\hat{n}\;dA$$

In addressing the forces on the fluid, we split the possible forces into two categories: body forces and surface forces. The body forces (such as gravity) are handled as follows:

Body force per unit mass: $$\bar{f}$$
Body force per unit volume: $$\rho \bar{f}$$
Body force on volume $d\mathscr{V}$: $$\rho \bar{f}\;d\mathscr{V}$$
Body force on volume $\mathscr{V}$: $$\int\limits_{\mathscr{V}}\rho \bar{f}\;d\mathscr{V}$$

The surface forces typically are represented by the stress tensor, but since we are assuming inviscid flow here (as per the assumptions of Euler's equation), we can ignore viscosity and deal only with pressure forces.

Force due to pressure on $dA$: $$-p\hat{n}\;dA$$
Force due to pressure on $A$: $$-\oint\limits_{A} p\hat{n}\;dA$$

So pulling all this together according to our stated conservation law, we can write the integral form of the momentum equation as
$$\int\limits_{\mathscr{V}}\frac{\partial}{\partial t}(\rho \bar{V}) \; d \mathscr{V} + \oint\limits_{A}(\rho \bar{V})\bar{V} \cdot \hat{n} \; dA = \int\limits_{\mathscr{V}}\rho \bar{f} \; d \mathscr{V} - \oint\limits_{A} p\hat{n}\;dA$$

Here again, we make use of the divergence theorem to say that
$$\oint\limits_{A} p\hat{n}\;dA = \int\limits_{\mathscr{V}}\nabla p\;d\mathscr{V}$$
This gives
$$\int\limits_{\mathscr{V}}\frac{\partial}{\partial t}(\rho \bar{V}) \; d \mathscr{V} + \oint\limits_{A}(\rho \bar{V})\bar{V} \cdot \hat{n} \; dA = \int\limits_{\mathscr{V}}\rho \bar{f} \; d \mathscr{V} - \int\limits_{\mathscr{V}}\nabla p\;d\mathscr{V}$$
This is a vector equation and so really represents 3 equations. For grins, let's look at the x-momentum equation first.
$$\int\limits_{\mathscr{V}}\frac{\partial}{\partial t}(\rho u) \; d \mathscr{V} + \oint\limits_{A}(\rho u)\bar{V} \cdot \hat{n} \; dA = \int\limits_{\mathscr{V}}\rho f_x \; d \mathscr{V} - \int\limits_{\mathscr{V}}\dfrac{\partial p}{\partial x}\;d\mathscr{V}$$
The Divergence Theorem let's us change the flux term to
$$\oint\limits_{A}(\rho u \bar{V})\cdot\hat{n}\;dA = \int\limits_{\mathscr{V}}\nabla \cdot (\rho u \bar{V})\;d\mathscr{V}$$
So we have
$$\int\limits_{\mathscr{V}}\frac{\partial}{\partial t}(\rho u) \; d \mathscr{V} + \int\limits_{\mathscr{V}}\nabla \cdot (\rho u \bar{V})\;d\mathscr{V} = \int\limits_{\mathscr{V}}\rho f_x \; d \mathscr{V} - \int\limits_{\mathscr{V}}\dfrac{\partial p}{\partial x}\;d\mathscr{V}$$
$$\int\limits_{\mathscr{V}}\left[ \frac{\partial}{\partial t}(\rho u) + \nabla \cdot (\rho u \bar{V}) - \rho f_x + \dfrac{\partial p}{\partial x} \right] \;d\mathscr{V} = 0$$
Setting the integrand to zero to solve the equation in general gives
$$\boxed{\frac{\partial}{\partial t}(\rho u) + \nabla \cdot (\rho u \bar{V}) = \rho f_x - \dfrac{\partial p}{\partial x}}$$
Using an identical process, you can get the y- and z-momentum equations
$$\boxed{\frac{\partial}{\partial t}(\rho v) + \nabla \cdot (\rho v \bar{V}) = \rho f_y - \dfrac{\partial p}{\partial y}}$$
$$\boxed{\frac{\partial}{\partial t}(\rho w) + \nabla \cdot (\rho w \bar{V}) = \rho f_z - \dfrac{\partial p}{\partial z}}$$

Now, continuing to play with the x-momentum equation further, we can expand the divergence term and the density derivative
$$\rho\dfrac{\partial u}{\partial t} + u\dfrac{\partial \rho}{\partial t} + u\nabla \cdot (\rho \bar{V}) + \rho\bar{V} \cdot \nabla u = -\dfrac{\partial p}{\partial x} + \rho f_x$$
If we multiply the continuity equation by $u$ we get
$$u\dfrac{\partial \rho}{\partial t} + u\nabla \cdot (\rho \bar{V}) = 0$$
which shows up in our momentum equation and can be zeroed. This leaves behind
$$\rho\dfrac{\partial u}{\partial t} + \rho\bar{V} \cdot \nabla u = -\dfrac{\partial p}{\partial x} + \rho f_x$$
which, after recognizing the material derivative, becomes one component of the Euler equation:
$$\boxed{\rho\dfrac{Du}{Dt} = -\dfrac{\partial p}{\partial x} + \rho f_x}$$
The y- and z-components are
$$\boxed{\rho\dfrac{Dv}{Dt} = -\dfrac{\partial p}{\partial y} + \rho f_y}$$
$$\boxed{\rho\dfrac{Dw}{Dt} = -\dfrac{\partial p}{\partial z} + \rho f_z}$$
This can be denoted in vector form as
$$\boxed{\rho\dfrac{D\bar{V}}{Dt} = -\nabla p + \rho \bar{f}}$$
which is the standard form or Euler's equation.

 

[Shan K]

I can't understand why the mass flux is p(rho)v . I also can't understand why you had taken a minus sign when you calculated the force due to pressure on dA in the second derivation

 

[Studiot]

I can't understand why the mass flux is p(rho)v

I think bone deserves real thanks for the effort he must have put into his post.

Think of a pipe volume/second = volume flux = area cross section x velocity.

so mass flux = density x volume flux

and mass flux / area (which is what Bone wrote) =ρAv / A = ρv

I also can't understand why you had taken a minus sign when you calculated the force due to pressure on dA in the second derivation

Next, the outward momentum flux term can be addressed as follows:

Bone is discussing a control volume with the outward direction positive and the inward direction negative.

That is the force is due to the inward pressure acting on the control volume.

 

[Shan K]

Many many thanks to both of u . And a special thanks to boneh3ad .

 

[boneh3ad]

Haha, no problem. Don't be too impressed though, I just got done (this morning) writing up some notes on this topic for the class I am teaching, so I had probably 60% of the LaTeX already written right here next to me. :)

 

[Studiot]

Still a pretty impressive answer though.