(adsbygoogle = window.adsbygoogle || []).push({}); Fluid Mechanics Question::Derivation of Momentum EQ

I have been doing some serious review of my FM. I am working my text from cover to cover in an attempt to solidify what I 'learned' in my 6-week summer FM class. What a load of crap that was.

I am attempting to follow the text's derivation of the Differential form of the Conservation of linear momentum. I will start here from the beginning for those who would like to help, but might not quite remember the derivation off the top of their head

The figure below shows a differential volume element. It was the only one available, so it shows the mass flux instead of the momentum flux on the faces of the element.

In words, we have the momentum equation:

[sum of forces] = [rate of change of momentum in CV] + [momentum flux out of CV] - [momentum flux into CV]

Let (u,v,w) be the scalar (x, y, z) components of velocity of the fluid passing through the differential Control Volume (CV).

[tex]

\sum \mathbf{F} =

\frac{\partial}{\partial{t}}(\rho\mathbf{V})dx dy dz +

\sum(\dot{m}\mathbf{V})_{out} -

\sum(\dot{m}\mathbf{V})_{in}

[/tex]

OR

[tex]

\sum \mathbf{F} =

[\frac{\partial}{\partial{t}}(\rho\mathbf{V}) +

\frac{\partial}{\partial{x}}(\rho u \mathbf{V}) +

\frac{\partial}{\partial{y}}(\rho v \mathbf{V}) +

\frac{\partial}{\partial{z}}(\rho w \mathbf{V})]

dx dy dz

\,\,\,\,\,\,(1)[/tex]

looking at only the term in bracketsfrom (1) we have ... (this is where I lose them)

[tex]

\frac{\partial}{\partial{t}}(\rho\mathbf{V}) +

\frac{\partial}{\partial{x}}(\rho u \mathbf{V}) +

\frac{\partial}{\partial{y}}(\rho v \mathbf{V}) +

\frac{\partial}{\partial{z}}(\rho w \mathbf{V})

=

[\frac{\partial{\rho}}{\partial{t}} + \nabla\cdot(\rho\mathbf{V}] +

\rho(\frac{\partial{\mathbf{V}}}{\partial{t}} +

u\frac{\partial{\mathbf{V}}}{\partial{x}} +

v\frac{\partial{\mathbf{V}}}{\partial{y}} +

w\frac{\partial{\mathbf{V}}}{\partial{z}} )

\,\,\,\,\,\,\,(2)[/tex]

I have no idea how they got to the last line.

Can we just look at one term here at a time? Let's look at just the very first term on the left hand side (LHS) of (2) [itex]\partial({\rho\mathbf{V}})/\partial{t}[/itex]

It appears that this is the last term in the RHS of (2) that is:

[tex]

\frac{\partial}{\partial{t}}(\rho\mathbf{V})=

u\frac{\partial{\mathbf{V}}}{\partial{x}} +

v\frac{\partial{\mathbf{V}}}{\partial{y}} +

w\frac{\partial{\mathbf{V}}}{\partial{z}}

[/tex]

I don't understand. It appears that they have assumed that [itex]\rho[/tex] is constant across the differential Control Volume. I don't like this. Here is why:

a) They did not state that this was an assumption

b) If we are assuming that [itex]\rho[/tex] is constant across the CV, then why don't we

just assume thatVis also constant across the CV and call it a day?

Am I missing something else? Is there some other way that they could have arrived at

[tex]

\frac{\partial}{\partial{t}}(\rho\mathbf{V})=

\rho(u\frac{\partial{\mathbf{V}}}{\partial{x}} +

v\frac{\partial{\mathbf{V}}}{\partial{y}} +

w\frac{\partial{\mathbf{V}}}{\partial{z}})

[/tex]

without assuming constant density?

Let's talk about this.

Thanks,

Casey

EDIT:

I see that I have made an error. The 1st term on the LHS of (2) is thepartialwith respect to time and therefore does not

equal the last term on the RHS of (2) which is clearly thetotalderivative with respect to time.

BUTit still looks like they arefactoring outthe density which doesn't make sense to me. Shouldn't we have to use the

product rule on the term

[tex]\frac{\partial}{\partial{t}}(\rho\mathbf{V})[/tex]

since both rho and V are functions of (x, y, z, t) ?

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# Fluid Mechanics Question: erivation of Momentum EQ

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