Different forms of fluid energy conservation equations

[happyparticle]

Homework Statement

Derive the different forms of the continuity equation

Relevant Equations

$\frac{De}{Dt} + (\gamma - 1)e \nabla \cdot \vec{u} = - \frac{1}{\rho} (\vec{u} \cdot \nabla)p$

$\frac{Dp}{Dt} + \gamma p \nabla\cdot \vec{u} = 0$

$\frac{D}{Dt} \frac{p}{\rho^{\gamma}} = 0$

$e = \frac{1}{\gamma -1} \frac{p}{\rho}$

I asked this question about one year ago, but at that time I didn't really understand what I was doing.

After spending a lot of time in this problem, I still fail to get the asked answer.

Starting with $\frac{De}{Dt} + (\gamma - 1)e \nabla \cdot \vec{u} = - \frac{1}{\rho} (\vec{u} \cdot \nabla)p$ I have to derive the following expressions.

$\frac{Dp}{Dt} + \gamma p \nabla\cdot \vec{u} = 0$

$\frac{D}{Dt} \frac{p}{\rho^{\gamma}} = 0$

I don't know what I'm doing wrong or what I'm missing, but I couldn't get those expressions.

Here is what I did.
$\frac{De}{Dt} + (\gamma - 1)e \nabla \cdot \vec{u} = - \frac{1}{\rho} (\vec{u} \cdot \nabla)p$ (1)

$\frac{De}{Dt} = \frac{\partial}{\partial t} (\frac{P}{(\gamma - 1) \rho}) + (\vec{u} \cdot \nabla) (\frac{P}{(\gamma -1) \rho})$ (2)

Replacing (2) into (1) and multiplying both side by $(\gamma - 1) \rho$
$\frac{\partial p}{\partial t} - \frac{p \partial \rho}{\rho \partial t} + \vec{u} \cdot \nabla p - \frac{p}{\rho} \vec{u} \cdot \nabla \rho + (\gamma -1)p \nabla \cdot \vec{u} = - (\gamma -1) \vec{u} \cdot \nabla p$

Then using the definition of the material derivative.
$\frac{Dc}{Dt} = \frac{\partial c}{\partial t} + \vec{u} \cdot \nabla c$

$\frac{Dp}{Dt} - \frac{P}{\rho} \frac{D\rho}{Dt} + (\gamma -1)p \nabla \cdot \vec{u} = - (\gamma -1) \vec{u} \cdot \nabla p$

So far there are too much terms on the left hand side. For instance, there are two terms with $\gamma$ and in the final answer there is only one.

 

[pasmith]

The continuity equation is <br /> \frac{\partial \rho}{\partial t} + \nabla \cdot (\rho \vec{u}) = 0. This can be rewritten as <br /> \frac{D\rho}{Dt} + \rho \nabla \cdot \vec{u} = 0. Your first relevant equation for De/Dt is the conservation of internal energy, which is an independent result.

From <br /> (\gamma -1 )e\rho = p you can take the material derivative of both sides and use the product rule on the left to get an expression for Dp/Dt in terms of De/Dt and D\rho/Dt, both of which you already know. Then you can use the quotient rule on <br /> \frac{D}{Dt}\left(\frac{p}{\rho^\gamma}\right).

 

[Chestermiller]

$\frac{Dp}{Dt} + \gamma p \nabla\cdot \vec{u} = 0$

This equation is the expression for adiabatic reversible expansion or compression of an ideal gas.

$\frac{D}{Dt} \frac{p}{\rho^{\gamma}} = 0$

This can be derived from the previous equation using the continuity equation.

 

[happyparticle]

I don't know where to begin to get this equation $\frac{Dp}{Dt} + \gamma p \nabla\cdot \vec{u} = 0$

I thought I had to start from the conservation of internal energy.

 

[Chestermiller]

The last equation is derived by the ideal gas relationship:$$e=C_VT= C_V\frac{pv}{R}$$where v is the molar density $v=1/\rho$ and e is the internal energy per mole. So, $$e=\frac{C_Vp}{\rho R}=\frac{p}{\rho}\frac{1}{\gamma-1}$$

The first equation derives from the adiabatic thermal energy balance equation $$\frac{\partial \rho e}{\partial t}+\nabla\centerdot (\rho \vec{u}e)=-p\nabla \centerdot \vec{u}$$

 

[happyparticle]

I knew how to get those equations. Here I have to get the second equation, but I don't know where to begin.


As I said, I thought I had to start from the conservation of internal energy.

 

[Chestermiller]

I knew how to get those equations. Here I have to get the second equation, but I don't know where to begin.


As I said, I thought I had to start from the conservation of internal energy.

The final equation in my post is the energy conservation equation.

 

[happyparticle]

The final equation in my post is the energy conservation equation.

Exactly, @pasmith wrote it. However I thought I could get $\frac{Dp}{Dt} + \gamma p \nabla\cdot \vec{u} = 0$ from it.

In my first post I started with the energy conservation equation to get the expression above. However, I think they are independent. Basically, I'm trying to understand how to get $\frac{Dp}{Dt} + \gamma p \nabla\cdot \vec{u} = 0$, then as you said, I can derive $\frac{D}{Dt} \frac{p}{\rho^{\gamma}} = 0$ from it.

I think I understand...
The term $\nabla P$ is 0 for the energy conservation equation and then from there I can get what I'm looking for. Am I correct?

 

[Chestermiller]

Exactly, @pasmith wrote it. However I thought I could get $\frac{Dp}{Dt} + \gamma p \nabla\cdot \vec{u} = 0$ from it.

In my first post I started with the energy conservation equation to get the expression above. However, I think they are independent. Basically, I'm trying to understand how to get $\frac{Dp}{Dt} + \gamma p \nabla\cdot \vec{u} = 0$, then as you said, I can derive $\frac{D}{Dt} \frac{p}{\rho^{\gamma}} = 0$ from it.

I think I understand...
The term $\nabla P$ is 0 for the energy conservation equation and then from there I can get what I'm looking for. Am I correct?

No.
So you accept my two equations?
$$e=\frac{C_Vp}{\rho R}=\frac{p}{\rho}\frac{1}{\gamma-1}\tag{1}$$
and$$\frac{\partial \rho e}{\partial t}+\nabla\centerdot (\rho \vec{u}e)=-p\nabla \centerdot \vec{u}\tag{2}$$If I substitute Eqn. 1 into Eqn.2, I get $$\frac{1}{\gamma-1}\left[\frac{\partial p}{\partial t}+\nabla\centerdot (p \vec{u})\right]=-p\nabla \centerdot \vec{u}\tag{3}$$OK so far?

 

[happyparticle]

Ah yeah I see. I don't understand why it takes me so long to understand.

Thank you!

 

[Chestermiller]

Ah yeah I see. I don't understand why it takes me so long to understand.

Thank you!

So we don’t need to show the rest, right?

 

[happyparticle]

So we don’t need to show the rest, right?

No, I'm fine. Thank you.

 

[happyparticle]

I don't know if it's too late, but I realized something.

To get $\frac{De}{Dt} + (\gamma - 1)e \nabla \cdot \vec{u} = - \frac{1}{\rho} (\vec{u} \cdot \nabla)p$

I have to use the adiabatic thermal energy balance equation. $\frac{\partial \rho e}{\partial t}+\nabla\centerdot (\rho \vec{u}e)=-\nabla(p \vec{u})$

However, I have to use what you wrote. $\frac{\partial \rho e}{\partial t}+\nabla\centerdot (\rho \vec{u}e)=-p\nabla \centerdot \vec{u}\tag{2}$ To get $\frac{Dp}{Dt} + \gamma p \nabla\cdot \vec{u} = 0$

Knowing that $\nabla(p \vec{u}) = p \nabla \cdot \vec{u} + \vec{u} \cdot \nabla p$. It looks like the last term is 0. Why?

 

[Chestermiller]

I don't know if it's too late, but I realized something.

To get $\frac{De}{Dt} + (\gamma - 1)e \nabla \cdot \vec{u} = - \frac{1}{\rho} (\vec{u} \cdot \nabla)p$

Your starting equation is not correct if e is the internal energy per unit mass. The right hand side should be $-p\nabla \centerdot \vec{u}$. Your first equation does not include the kinetic energy terms. If you include the KE terms, e needs to be replaced by $e+\frac{1}{2}u^2$, and you need to subtract the velocity vector dotted with the equation of motion.

 

[happyparticle]

Yet starting from $\frac{\partial \rho e}{\partial t}+\nabla\centerdot (\rho \vec{u}e)=-\nabla(p \vec{u})$

=> $e[\frac{\partial \rho}{\partial t} + \nabla \cdot (\rho \vec{u})] + \rho [ \frac{\partial e}{\partial t} + (\vec{u} \cdot \nabla) e] = -p \nabla \cdot \vec{u} - (\vec{u} \cdot \nabla)p$

The first term on the left hand side is the continuity equation and thus equal to 0.

Thus, using the material derivative I get the expression $\frac{De}{Dt} + (\gamma - 1)e \nabla \cdot \vec{u} = - \frac{1}{\rho} (\vec{u} \cdot \nabla)p$

It looks fine to me. However, I couldn't get $\frac{Dp}{Dt} + \gamma p \nabla\cdot \vec{u} = 0$

 

[Chestermiller]

Yet starting from $\frac{\partial \rho e}{\partial t}+\nabla\centerdot (\rho \vec{u}e)=-\nabla(p \vec{u})$

=> $e[\frac{\partial \rho}{\partial t} + \nabla \cdot (\rho \vec{u})] + \rho [ \frac{\partial e}{\partial t} + (\vec{u} \cdot \nabla) e] = -p \nabla \cdot \vec{u} - (\vec{u} \cdot \nabla)p$

The first term on the left hand side is the continuity equation and thus equal to 0.

Thus, using the material derivative I get the expression $\frac{De}{Dt} + (\gamma - 1)e \nabla \cdot \vec{u} = - \frac{1}{\rho} (\vec{u} \cdot \nabla)p$

It looks fine to me. However, I couldn't get $\frac{Dp}{Dt} + \gamma p \nabla\cdot \vec{u} = 0$

The correct equation is
$$\frac{\partial \rho (e+\hat{k})}{\partial t}+\nabla\centerdot (\rho \vec{u}(e+\hat{k}))=-\nabla(p \vec{u})$$
where $\hat{k}$ is the kinetic energy per unit mass: $$\hat{k}=\frac{1}{2}|u|^2$$
In addition, the dot product of the velocity vector with the equation of motion gives the Bernoulli Eqn. $$\rho\frac{D\hat{k}}{Dt}=-\vec{u} \centerdot \nabla p$$

 

[happyparticle]

In the notes I'm following $\vec{k}$ is not taking into account.

Also, With out without $\vec{k}$, I don't see how to get $\frac{Dp}{Dt} + \gamma p \nabla\cdot \vec{u} = 0$

 

[Chestermiller]

In the notes I'm following $\vec{k}$ is not taking into account.

Also, With out without $\vec{k}$, I don't see how to get $\frac{Dp}{Dt} + \gamma p \nabla\cdot \vec{u} = 0$

See Transport Phenomena by Bird et al, Chapter 11, large 2-psge-centerfold Table.

 

[happyparticle]

There are some equations of energy without $\vec{k}$. Pretty similar that what I'm using. As I said, I can't find a way to get neither of these equations. The whole chapter doesn't mention $\vec{k}$ $\frac{De}{Dt} + (\gamma - 1)e \nabla \cdot \vec{u} = - \frac{1}{\rho} (\vec{u} \cdot \nabla)p$ or $\frac{Dp}{Dt} + \gamma p \nabla\cdot \vec{u} = 0$

For instance, without $\vec{k}$ I can get $\frac{De}{Dt} + (\gamma - 1)e \nabla \cdot \vec{u} = - \frac{1}{\rho} (\vec{u} \cdot \nabla)p$ which is something I had to do.

Perhaps I don't express myself well.

 

[Chestermiller]

k is the kinetic energy per unit mass. See Eqns. P, Q, and R of BSL Table 11.4.1. Eqn. R is the result of subtracting Eqn. Q from Eqn. P (neglecting viscous stresses and gravitational terms). Note that Eqn. R differs from your energy balance equation, but is of the right form to get your desired results.

 

[happyparticle]

I'm so confuse. How do you get $\frac{\partial \rho e}{\partial t}+\nabla\centerdot (\rho \vec{u}e)=-p\nabla \centerdot \vec{u}\tag{2}$

The term $\nabla (-p \vec{u})$ in $\frac{\partial \rho e}{\partial t}+\nabla\centerdot (\rho \vec{u}e)=-\nabla(p \vec{u})$ come from $\frac{\Delta W}{ \Delta t}$ which is the work against the pressure.

$\frac{\Delta W}{ \Delta t} = \int (-p \hat{n}) \cdot \vec{u} ds$

 

[Chestermiller]

k is the kinetic energy per unit mass. See Eqns. P, Q, and R of BSL Table 11.4.1. Eqn. R is the result of subtracting Eqn. Q from Eqn. P (neglecting viscous stresses and gravitational terms). Note that Eqn. R differs from your energy balance equation, but is of the right form to get your desired results.

Do you agree with equations P and Q?

 

[happyparticle]

Do you agree with equations P and Q?

Yes, if the energy is $\vec{K} + \vec{U}$

 

[Chestermiller]

Yes, if the energy is $\vec{K} + \vec{U}$

Aside from potential energy, that is what it is.

 

[happyparticle]

Why exactly in this case $e$ needs to be replaced by $e+\frac{1}{2}u^2$?
That is what I don't understand.

My initial question was how to get $\frac{Dp}{Dt} + \gamma p \nabla\cdot \vec{u} = 0$

Also, I have a similar issue using the Eqn. R. Since Two terms on the right hand side must be 0 to have $\frac{\partial \rho e}{\partial t}+\nabla\centerdot (\rho \vec{u}e)=-p\nabla \centerdot \vec{u}\tag{2}$. Similar to my equation. However, I'm not sure to understand the notation used in the book. At the end of the day, it seems like $- (\nabla \cdot q) - (\tau; \nabla v) = \vec{u} \cdot \nabla p = 0$

 

[Chestermiller]

Why exactly in this case $e$ needs to be replaced by $e+\frac{1}{2}u^2$?
That is what I don't understand.

The total energy of the fluid particles is the sum of internal energy, kinetic energy.

My initial question was how to get $\frac{Dp}{Dt} + \gamma p \nabla\cdot \vec{u} = 0$

Also, I have a similar issue using the Eqn. R. Since Two terms on the right hand side must be 0 to have $\frac{\partial \rho e}{\partial t}+\nabla\centerdot (\rho \vec{u}e)=-p\nabla \centerdot \vec{u}\tag{2}$. Similar to my equation. However, I'm not sure to understand the notation used in the book. At the end of the day, it seems like $- (\nabla \cdot q) - (\tau; \nabla v) = \vec{u} \cdot \nabla p = 0$

The flow is assumed adiabatic, so q = 0. The flow is also assumed inviscid, so the term involving tau is zero. $\vec{u} \cdot \nabla p$ is not assumed to be zero. That term cancels when equations P and Q are combined.

 

[happyparticle]

I see. So basically, my energy balance equation was wrong all along or it was just correct to get $\frac{De}{Dt} + (\gamma - 1)e \nabla \cdot \vec{u} = - \frac{1}{\rho} (\vec{u} \cdot \nabla)p$ ?

 

[Chestermiller]

I see. So basically, my energy balance equation was wrong all along

Yes

 

[happyparticle]

All right.

One more quick question.

To get $\frac{D}{Dt} \frac{p}{\rho^{\gamma}} = 0$

Is it correct to start from $\frac{1}{\gamma p} \frac{Dp}{Dt} + \nabla\cdot \vec{u} = \frac{1}{\rho}[ \frac{\partial \rho}{\partial t} + \nabla \cdot (\rho \vec{u})]$

Since $\frac{Dp}{Dt} + \gamma p \nabla\cdot \vec{u} = 0$

$\frac{\partial \rho}{\partial t} + \nabla \cdot (\rho \vec{u}) = 0.$

 

[Chestermiller]

All right.

One more quick question.

To get $\frac{D}{Dt} \frac{p}{\rho^{\gamma}} = 0$

Is it correct to start from $\frac{1}{\gamma p} \frac{Dp}{Dt} + \nabla\cdot \vec{u} = \frac{1}{\rho}[ \frac{\partial \rho}{\partial t} + \nabla \cdot (\rho \vec{u})]$

Since $\frac{Dp}{Dt} + \gamma p \nabla\cdot \vec{u} = 0$

$\frac{\partial \rho}{\partial t} + \nabla \cdot (\rho \vec{u}) = 0.$

The starting point has to be the energy equation and the mass balance equation, also known as continuity equation.