# Fluid dynamics question, related to the derivation of Bernoulli's Theorem for steady flow

I'm finding it very challenging to understand a step in the derivation of the 'moment equation', found in these notes

http://www.maths.ox.ac.uk/system/files/coursematerial/2011/989/66/fluids1.pdf

I am struggling with the transition of (1.33) to (1.34). I don't see how:

$$\frac{d}{dt}\int_V \rho \mathbf{u} dV=\int_V \rho \frac{D\mathbf{u} }{Dt}dV$$

This is a result of the transport theorem (1.22) and it's corollary (1.32), but I don't really see how the corollary is derived or how it can be applied to vector functions. If anyone could explain as you would a child I would be very grateful..

EDIT: So this really comes down just the derivation of the corollary: For $f=\rho h$

$$\frac{d}{dt}\int_V \rho h dV=\int_V \rho \frac{Dh }{Dt}dV$$

So this depends on the conservation of mass $$\frac{\partial \rho}{\partial t}+\nabla \cdot(\rho \mathbf{u})=0$$ Therefore from the transport theorem $$\frac{d}{dt}\int_V \rho h dV=\int_V \frac{\partial \rho h}{\partial t} +\nabla \cdot(\rho h \mathbf{u}) dV=\int_V h\frac{\partial \rho }{\partial t} +\rho\frac{\partial h }{\partial t} + \rho h(\nabla\cdot\mathbf{u})+ (\nabla \rho h)\cdot \mathbf{u} dV$$

So all we need to do is show that: $$h\frac{\partial \rho }{\partial t} +\rho\frac{\partial h }{\partial t} + \rho h(\nabla\cdot\mathbf{u})+ (\nabla \rho h)\cdot \mathbf{u} =\star\big[\frac{\partial \rho}{\partial t}+\nabla \cdot(\rho \mathbf{u})\big] +\rho \big[ \frac{\partial h}{\partial t} + (\mathbf{u} \cdot\nabla)h\big]$$

Can anyone help with this step? Many thanks.

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Your problem stems from the fact that also the volume depends on t. This is similar to the one dimensional case but now you have the following equation to hold (eq.(1.22) in the lectures you cite):

$$\frac{\partial}{\partial t}\int_{V(t)}fdxdydz=\int_{V(t)}\left(\frac{\partial f}{\partial t}+\nabla\cdot(f\bf{u})\right)dxdydz.$$

The proof given in your text use the Jacobian and that is all you need.

In your case you have to prove that

$$\frac{d}{dt}\int_V \rho \mathbf{u} dV=\int_V \rho \frac{D\mathbf{u} }{Dt}dV$$

but the volume is dependent on time so you will write

$$\frac{d}{dt}\int_{V(t)} \rho \mathbf{u} dV=\int_{V(0)} \frac{D}{Dt}(J\rho \mathbf{u}) dV'$$

being $J$ the Jacobian. One has the identity

$$\frac{DJ}{Dt}=J\nabla\cdot\mathbf{u}$$

and so

$$\int_{V(0)} \left(\frac{D}{Dt}(\rho \mathbf{u})+\nabla\cdot\mathbf{u}\rho \mathbf{u}\right)J dV'.$$

Now

$$\int_{V(t)} \left(\frac{D}{Dt}(\rho \mathbf{u})+\nabla\cdot\mathbf{u}\rho \mathbf{u}\right)dV=\int_{V(t)} \left(\frac{D\rho}{Dt}\mathbf{u}+\rho\frac{\partial\mathbf{u}}{\partial t}+\rho\mathbf{u}\cdot\nabla\mathbf{u}+\nabla\cdot\mathbf{u}\rho \mathbf{u}\right)dV.\qquad (\dagger)$$

Using continuity equation

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

you will have

$$\frac{D\rho}{Dt}=\frac{\partial\rho}{\partial t}+\mathbf{u}\cdot\nabla\rho=\mathbf{u}\cdot\nabla\rho-\nabla\cdot(\rho\mathbf{u})=-\rho\nabla\cdot\mathbf{u}.$$

Putting this into eq.($\dagger$) you get your result.

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Thanks for your response Jon. Indeed, so in my edit I tried to apply this theorem to $f=\rho h$, but I can't see how this implies that $$\frac{d}{dt}\int_V \rho h dV=\int_V \rho \frac{Dh }{Dt}dV$$ – Freeman Mar 19 '12 at 16:11
@LHS: I have extended the answer. Let me know if you need more. – Jon Mar 19 '12 at 22:55
Thanks so much for this! Could you explain this step: $$\frac{d}{dt}\int_V \rho \mathbf{u} dV= \int_V \frac{D}{Dt} (\rho \mathbf{u} )dV$$ I have not seen this identity before, do you know how it can be justified? Also i'm having a little trouble rearranging your last step into the required result.. but i'm pretty exhausted, I will try again in the morning. – Freeman Mar 19 '12 at 23:19
@LHS: Further improved the answer. I hope that now is more readable. – Jon Mar 20 '12 at 9:47
This is fantastic! thanks so much, this has definitely cleared up a few loose ends that I didn't understand when learning this course. I very much appreciate your help. – Freeman Mar 20 '12 at 12:51