Fluid mechanics: Euler's equation derivation

[happyparticle]

Homework Statement

Derivation of the Euler's equation using $\frac{d}{dt} (\rho \vec{u} \otimes \vec{u}) = \phi + F_v + F_s$

Relevant Equations

$\frac{d}{dt} (\rho \vec{u} \otimes \vec{u}) = \phi + F_v + F_s$

Euleur's equation:
$\frac{D\vec{u}}{Dt} = \frac{-\nabla P + \vec{g}}{\rho}$

For a volume fixed in space with the positive component of the z axis pointing up, I have the force of gravity pointing down.

Using the following equation I have to derive the Euler's equation.
$\frac{d}{dt} (\rho \vec{u} \otimes \vec{u}) = \phi + F_v + F_s$

Where $\phi$ is the flux, $F_v$ and $F_s$ are the volume and surface forces.

I think that the flux moving outside of the volume is:
$- \oint_S \rho \vec{u} \otimes \vec{u} dS$

Also, since the force of gravity is a volume force
$F_v = \rho V \vec{g}$
and
$F_s = \oint_S (-P \hat{n}) dS$ is the pressure on the surfaces, pointing outside of the volume.

So far I have
$\frac{d}{dt} (\rho \vec{u} \otimes \vec{u}) = - \oint_S \rho \vec{u} \otimes \vec{u} dS + \rho V \vec{g} + \oint_S (-P \hat{n}) dS$

I'm wondering if I'm right. Also, I'm not sure which component of the pressure should I keep.
For instance, for a cylinder where the surface S is the top, where the flux is measured. I think I could consider only the z components of both the pressure and gravity.
In this case, I would have $F_s = \oint_S (-P \hat{n}) \cdot \hat{z} dS$ and $F_v = \rho V g \hat{z}$.

 

[Chestermiller]

Homework Statement: Derivation of the Euler's equation using $\frac{d}{dt} (\rho \vec{u} \otimes \vec{u}) = \phi + F_v + F_s$
Relevant Equations: $\frac{d}{dt} (\rho \vec{u} \otimes \vec{u}) = \phi + F_v + F_s$

Euleur's equation:
$\frac{D\vec{u}}{Dt} = \frac{-\nabla P + \vec{g}}{\rho}$

For a volume fixed in space with the positive component of the z axis pointing up, I have the force of gravity pointing down.

Using the following equation I have to derive the Euler's equation.
$\frac{d}{dt} (\rho \vec{u} \otimes \vec{u}) = \phi + F_v + F_s$

Where $\phi$ is the flux, $F_v$ and $F_s$ are the volume and surface forces.

I think that the flux moving outside of the volume is:
$- \oint_S \rho \vec{u} \otimes \vec{u} dS$

Also, since the force of gravity is a volume force
$F_v = \rho V \vec{g}$
and
$F_s = \oint_S (-P \hat{n}) dS$ is the pressure on the surfaces, pointing outside of the volume.

So far I have
$\frac{d}{dt} (\rho \vec{u} \otimes \vec{u}) = - \oint_S \rho \vec{u} \otimes \vec{u} dS + \rho V \vec{g} + \oint_S (-P \hat{n}) dS$

I'm wondering if I'm right. Also, I'm not sure which component of the pressure should I keep.
For instance, for a cylinder where the surface S is the top, where the flux is measured. I think I could consider only the z components of both the pressure and gravity.
In this case, I would have $F_s = \oint_S (-P \hat{n}) \cdot \hat{z} dS$ and $F_v = \rho V g \hat{z}$.

That is not what I get. Here is my analysis of momentum balance on the control volume: $$\int{\frac{\partial(\rho \mathbf{u})}{\partial t}dV}=-\int{\rho \mathbf{u}(\mathbf{u}\centerdot}\mathbf{n})dS+\int{\rho \mathbf{g}dV}-\int{P\mathbf{n}dS}$$The left hand side is the rate of accumulation of momentum within the control volume. The first term on the right hand side is the net rate of momentum flowing into the control volume. The second term on the right hand side is the gravitational force on the fluid within the control volume. The third term on the right hand side is the surface force acting on the fluid within the control volume. If we apply the divergence theorem and Gauss' theorem to the surface integrals in this equation, we obtain:
$$\int{\frac{\partial (\rho \mathbf{u})}{\partial t}dV}=-\int{\nabla \centerdot (\rho \mathbf{u} \otimes \mathbf{u})}dV+\int{\rho \mathbf{g}dV}-\int{\nabla P dV}$$

 

[happyparticle]

I see my errors now.

I still have two questions.
First of all, I think the outer product shouldn't be 0. However, Using the identity $\nabla \cdot (\vec{A} \times \vec{B}) = \vec{B} \cdot (\nabla \times \vec{A}) - \vec{A} \cdot (\nabla \times \vec{B})$ and by replacing A and B with u, I get 0.

Secondly, I'm not sure if I can take out $\nabla P$ outside the integral. Does the pressure depend of the volume?

 

[Chestermiller]

I see my errors now.

I still have two questions.
First of all, I think the outer product shouldn't be 0. However, Using the identity $\nabla \cdot (\vec{A} \times \vec{B}) = \vec{B} \cdot (\nabla \times \vec{A}) - \vec{A} \cdot (\nabla \times \vec{B})$ and by replacing A and B with u, I get 0.

This is not present in the derivation.

Secondly, I'm not sure if I can take out $\nabla P$ outside the integral. Does the pressure depend of the volume?

If you combine the terms, the coefficient of dV must equal zero.

 

[happyparticle]

Is there a identity for the outer product. I thought I could use what I wrote above.

If you combine the terms, the coefficient of dV must equal zero.

I'm not sure to understand. I thought the gradient of the pressure depend of the volume.

 

[Chestermiller]

Is there a identity for the outer product. I thought I could use what I wrote above.


I'm not sure to understand. I thought the gradient of the pressure depend of the volume.

$$\int{(\frac{\partial (\rho \mathbf{u})}{\partial t}}+\nabla \centerdot (\rho \mathbf{u} \otimes \mathbf{u})-\rho \mathbf{g}+\nabla P )dV=0$$

In order for this to apply to any arbitrary control volume, the integrand must be equal to zero.

 

[happyparticle]

Ah true. It took me some time, but I see it.

For the outer product, I was told to use an identity. However, exempt from the one I wrote previously, I don't see any other identity I could use.

 

[Chestermiller]

Ah true. It took me some time, but I see it.

For the outer product, I was told to use an identity. However, exempt from the one I wrote previously, I don't see any other identity I could use.

See appendices of Transport Phenomena by Bird, Stewart, and Lightfoot

 

[happyparticle]

There are some identities p.818, but none involving 2 tensors. Also, there is no explanation about the cross product of 2 tensors. Only tensors with vectors.

 

[hunt_mat]

Homework Statement: Derivation of the Euler's equation using $\frac{d}{dt} (\rho \vec{u} \otimes \vec{u}) = \phi + F_v + F_s$
Relevant Equations: $\frac{d}{dt} (\rho \vec{u} \otimes \vec{u}) = \phi + F_v + F_s$

Euleur's equation:
$\frac{D\vec{u}}{Dt} = \frac{-\nabla P + \vec{g}}{\rho}$

For a volume fixed in space with the positive component of the z axis pointing up, I have the force of gravity pointing down.

Using the following equation I have to derive the Euler's equation.
$\frac{d}{dt} (\rho \vec{u} \otimes \vec{u}) = \phi + F_v + F_s$

Where $\phi$ is the flux, $F_v$ and $F_s$ are the volume and surface forces.

I think that the flux moving outside of the volume is:
$- \oint_S \rho \vec{u} \otimes \vec{u} dS$

Also, since the force of gravity is a volume force
$F_v = \rho V \vec{g}$
and
$F_s = \oint_S (-P \hat{n}) dS$ is the pressure on the surfaces, pointing outside of the volume.

So far I have
$\frac{d}{dt} (\rho \vec{u} \otimes \vec{u}) = - \oint_S \rho \vec{u} \otimes \vec{u} dS + \rho V \vec{g} + \oint_S (-P \hat{n}) dS$

I'm wondering if I'm right. Also, I'm not sure which component of the pressure should I keep.
For instance, for a cylinder where the surface S is the top, where the flux is measured. I think I could consider only the z components of both the pressure and gravity.
In this case, I would have $F_s = \oint_S (-P \hat{n}) \cdot \hat{z} dS$ and $F_v = \rho V g \hat{z}$.

What you do is write $\mathbf{F}=m\mathbf{a}$, that's your starting position. However, you have to choose a control volume $V$ for your acceleration and forces: You write $ma=dp/dt$ The rate of change of acceleration;

$$\frac{dp}{dt}=\frac{d}{dt}\int_{V}\rho\mathbf{u}dV$$

You will have to use the Reynolds transport theorem here, to take the time derivative inside the integral. For the forces, you'll have to write:
$$\mathbf{F}_{b}=\int_{V}\mathbf{f}$$
Where $\mathbf{f}$ is the force per unit volume. You will need to consider two types of forces, body forces(which act on \emph{all} parts of the body, like gravity for example) and surface forces(which only act on the boundary of the control volume(like pressure for example)). For surfaces you will have to write surface integrals in terms of volume integrals.

 

[Chestermiller]

There are some identities p.818, but none involving 2 tensors. Also, there is no explanation about the cross product of 2 tensors. Only tensors with vectors.

It seems to me, Appendix A has all the identities you need to handle Newtonian- as well as inviscid fluid dynamics.

 

[happyparticle]

I think I'm confuse with the notation. Is $\vec{v} \vec{w}$ an outer product?

I this case I have $\nabla (\vec{u} \vec{u}) = (\vec{u} \cdot \nabla) \vec{u} + (\vec{u} \cdot \nabla) \vec{u}$

However, I'm still stuck. One of the two term of the right hand side must be 0, but they are identical.

Maybe there is an error in my professor's notes.

 

[Chestermiller]

I think I'm confuse with the notation. Is $\vec{v} \vec{w}$ an outer product?

I this case I have $\nabla (\vec{u} \vec{u}) = (\vec{u} \cdot \nabla) \vec{u} + (\vec{u} \cdot \nabla) \vec{u}$

No, this is correct. Work it out in component form.

However, I'm still stuck. One of the two term of the right hand side must be 0, but they are identical.

Maybe there is an error in my professor's notes.

Notationally, $\vec{v}\vec{v}$ is the same as $\vec{v}\otimes \vec{v}$; i.e., two vectors placed in juxtaposition with no operation implied between them.

 

[happyparticle]

In component form I have
$\frac{\partial}{\partial_i} (u_i u_j) = \frac{\partial u_i}{\partial_i} u_j + u_i \frac{\partial u_j}{\partial_i} = \vec{u}(\nabla \cdot \vec{u}) + \vec{u} \cdot \nabla \vec{u}$

Which is not the same as in post #12.

 

[Chestermiller]

In 2D, $$\vec{u}\vec{u}=u_xu_x\hat{x}\hat{x}+u_xu_y(\hat{x}\hat{y}+\hat{y}\hat{x})+u_yu_y\hat{y}\hat{y}$$
$$\nabla\centerdot(\vec{u}\vec{u})=\hat{x}\centerdot \frac{\partial}{\partial x}(u_xu_x\hat{x}\hat{x}+u_xu_y(\hat{x}\hat{y}+\hat{y}\hat{x})+u_yu_y\hat{y}\hat{y})$$$$+\hat{y}\centerdot \frac{\partial}{\partial y}(u_xu_x\hat{x}\hat{x}+u_xu_y(\hat{x}\hat{y}+\hat{y}\hat{x})+u_yu_y\hat{y}\hat{y})$$
What do you get next?

 

[happyparticle]

For the $\hat{x}$ component I get
$[\frac{\partial u_x}{ dx} u_x \hat{x} \hat{x} + u_x \frac{\partial u_x}{ dx} \hat{x} \hat{x} + \frac{\partial u_x}{ dx} u_y (\hat{x} \hat{y} + \hat{y} \hat{x} ) + u_x \frac{\partial u_y}{ dx} (\hat{x} \hat{y} + \hat{y} \hat{x} )+ \frac{\partial u_y}{ dx} u_y \hat{y} \hat{y} + u_y \frac{\partial u_y}{ dx} \hat{y} \hat{y}] \hat{x}$


Since $(\vec{u} \cdot \nabla)\vec{u} = u_x \frac{\partial u_x}{dx} \hat{x} + u_x \frac{\partial u_y}{dx} \hat{y} + u_y \frac{\partial u_x}{dy} \hat{x} + u_y \frac{\partial u_x}{dy} \hat{y}$

I don't see how it could work.

I'm convince that $\nabla (\vec{u} \vec{u}) = (\vec{u} \cdot \nabla) \vec{u} + (\vec{u} \cdot \nabla) \vec{u}$ is not good. If one the right term is 0 then both must be 0 because they are exactly the same.

 

[Chestermiller]

For the $\hat{x}$ component I get
$[\frac{\partial u_x}{ dx} u_x \hat{x} \hat{x} + u_x \frac{\partial u_x}{ dx} \hat{x} \hat{x} + \frac{\partial u_x}{ dx} u_y (\hat{x} \hat{y} + \hat{y} \hat{x} ) + u_x \frac{\partial u_y}{ dx} (\hat{x} \hat{y} + \hat{y} \hat{x} )+ \frac{\partial u_y}{ dx} u_y \hat{y} \hat{y} + u_y \frac{\partial u_y}{ dx} \hat{y} \hat{y}] \hat{x}$


Since $(\vec{u} \cdot \nabla)\vec{u} = u_x \frac{\partial u_x}{dx} \hat{x} + u_x \frac{\partial u_y}{dx} \hat{y} + u_y \frac{\partial u_x}{dy} \hat{x} + u_y \frac{\partial u_x}{dy} \hat{y}$

I don't see how it could work.

I'm convince that $\nabla (\vec{u} \vec{u}) = (\vec{u} \cdot \nabla) \vec{u} + (\vec{u} \cdot \nabla) \vec{u}$ is not good. If one the right term is 0 then both must be 0 because they are exactly the same.

I get $$\nabla\centerdot(\vec{u}\vec{u})=\frac{\partial u_x^2}{\partial x }\hat{x}+\frac{\partial (u_xu_y)}{\partial x }\hat{y}+\frac{\partial (u_xu_y)}{\partial y}\hat{x}+\frac{\partial u_y^2}{\partial y }\hat{y}$$

and $$\vec{u}\centerdot \nabla\vec{u}=u_x\frac{\partial u_x}{\partial x}\hat{x}+u_y\frac{\partial u_x }{\partial y}\hat{x}$$$$+u_x\frac{\partial u_y}{\partial x}\hat{y}+u_y\frac{\partial u_y}{\partial y}\hat{y}$$

But, the identity we are trying to prove is Eqn. A.4-24 of BSL, correct?

 

[happyparticle]

I think I was confuse with the outer product of the unit vectors. Just to be sure $(\hat{y} \hat{x}) \hat{x} = (0,0)^T$

 

[happyparticle]

Also, Eqn. A.4-24 of BSL is not the same as $\nabla (\vec{u} \vec{u}) = (\vec{u} \cdot \nabla) \vec{u} + (\vec{u} \cdot \nabla) \vec{u}$ isn't it?

 

[Chestermiller]

Also, Eqn. A.4-24 of BSL is not the same as $\nabla (\vec{u} \vec{u}) = (\vec{u} \cdot \nabla) \vec{u} + (\vec{u} \cdot \nabla) \vec{u}$ isn't it?

I thought we were trying to prove A.4-24. What is the equation number for the identity you are trying to prove?

 

[happyparticle]

A.4-24 is good. However, in post #12 I wrote $\nabla (\vec{u} \vec{u}) = (\vec{u} \cdot \nabla) \vec{u} + (\vec{u} \cdot \nabla) \vec{u}$ which is not the same.

Also, is my post #18 good?

 

[Chestermiller]

A.4-24 is good. However, in post #12 I wrote $\nabla (\vec{u} \vec{u}) = (\vec{u} \cdot \nabla) \vec{u} + (\vec{u} \cdot \nabla) \vec{u}$ which is not the same.

We showed that this equation is not correct. Plus, the left hand side should be the divergence, not the gradient.

 

[happyparticle]

You are right. Thank you.
One more question. In post #2 where you wrote $\int{P\mathbf{n}dS}$ is it the same as $(P \cdot \hat{n}) ds$, since P is a vector field?

 

[Chestermiller]

You are right. Thank you.
One more question. In post #2 where you wrote $\int{P\mathbf{n}dS}$ is it the same as $(P \cdot \hat{n}) ds$, since P is a vector field?

Pn is a vector normal to the surface. P is a scalar field.

 

[happyparticle]

All right. One more thing.
Still in post #2, you said that you used the Gauss' theorem and the divergence theorem. I thought it was the same thing. What is the difference exactly?

 

[Chestermiller]

All right. One more thing.
Still in post #2, you said that you used the Gauss' theorem and the divergence theorem. I thought it was the same thing. What is the difference exactly?

I'm not sure of what they are called.

 

[happyparticle]

All right, thank you for everything.