Different forms of fluid energy conservation equations

  • #1
happyparticle
442
20
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.
 
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  • #2
The continuity equation is [tex]
\frac{\partial \rho}{\partial t} + \nabla \cdot (\rho \vec{u}) = 0.[/tex] This can be rewritten as [tex]
\frac{D\rho}{Dt} + \rho \nabla \cdot \vec{u} = 0.[/tex] Your first relevant equation for [itex]De/Dt[/itex] is the conservation of internal energy, which is an independent result.

From [tex]
(\gamma -1 )e\rho = p[/tex] you can take the material derivative of both sides and use the product rule on the left to get an expression for [itex]Dp/Dt[/itex] in terms of [itex]De/Dt[/itex] and [itex]D\rho/Dt[/itex], both of which you already know. Then you can use the quotient rule on [tex]
\frac{D}{Dt}\left(\frac{p}{\rho^\gamma}\right).[/tex]
 
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  • #3
happyparticle said:
##\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.
happyparticle said:
##\frac{D}{Dt} \frac{p}{\rho^{\gamma}} = 0
##
This can be derived from the previous equation using the continuity equation.
 
  • #4
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.
 
  • #5
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 $$\frac{\partial \rho e}{\partial t}+\nabla\centerdot (\rho \vec{u}e)=-p\nabla \centerdot \vec{u}$$
 
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  • #6
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.
 
  • #7
happyparticle said:
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.
 
  • #8
Chestermiller said:
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?
 
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  • #9
happyparticle said:
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?
 
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  • #10
Ah yeah I see. I don't understand why it takes me so long to understand.

Thank you!
 
  • #11
happyparticle said:
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?
 
  • #12
Chestermiller said:
So we don’t need to show the rest, right?
No, I'm fine. Thank you.
 
  • #13
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?
 
  • #14
happyparticle said:
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.
 
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  • #15
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
##
 
  • #16
happyparticle said:
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. $$\rho\frac{D\hat{k}}{Dt}=-\vec{u} \centerdot \nabla p$$
 
  • #17
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##
 
  • #18
happyparticle said:
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.
 
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  • #19
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.
 
  • #20
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.
 
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  • #21
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##
 
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  • #22
Chestermiller said:
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?
 
  • #23
Chestermiller said:
Do you agree with equations P and Q?
Yes, if the energy is ##\vec{K} + \vec{U}##
 
  • #24
happyparticle said:
Yes, if the energy is ##\vec{K} + \vec{U}##
Aside from potential energy, that is what it is.
 
  • #25
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##
 
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  • #26
happyparticle said:
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.
happyparticle said:
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.
 
  • #27
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
## ?
 
  • #28
happyparticle said:
I see. So basically, my energy balance equation was wrong all along

Yes
 
  • #29
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.


##
 
  • #30
happyparticle said:
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.
 
  • #31
Chestermiller said:
This equation is the expression for adiabatic reversible expansion or compression of an ideal gas.

This can be derived from the previous equation using the continuity equation.
I though I had to start with##
\frac{Dp}{Dt} + \gamma p \nabla\cdot \vec{u} = 0

##
The starting point has to be the energy equation and the mass balance equation, also known as continuity equation.
Basically, I have to do what @pasmith says in post #2?
 
  • #32
happyparticle said:
I though I had to start with##
\frac{Dp}{Dt} + \gamma p \nabla\cdot \vec{u} = 0

##
I don’t think so. This equation can be derived.
 
  • #33
Chestermiller said:
I don’t think so. This equation can be derived.
I guess I misread post #3

I guess I finally found both expression. I think I was completely wrong since the beginning. More than I thought.
 
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