Control volume and the momentum theorem

[arestes]

I'm studying fluid and propulsion mechanics by myself.

I stumbled upon this website from MIT: http://web.mit.edu/16.unified/www/S...opulsion2/UnifiedPropulsion2.htm#fallingblock

It states that "Newton’s second law for a control volume of fixed mass" is $$\sum \vec{F}=\int_Vρ\frac{D}{dt}(\vec{u}_ 0+\vec{u})dV$$ but it's said that this is valid for a fixed mass control volume. $$\vec{u}_0$$ is the velocity of a reference frame attached to the control volume and $$\vec{u}$$ is the velocity of fluid relative to this moving frame.

The notes then goes on to derive this formula: $$\sum {F}_x-{F_0}_x=\int_V\frac{\partial}{\partial t}(ρ{u_x} dV+\int_S u_x(ρ \vec{u}\cdot \vec{n} dA$$
where $${F_0}_x$$ is basically $$ma_x$$.

So far so good. However, I still don't understand why this equation is only valid for a control volume with fixed mass. Moreover, we're allowing the control volume to change its mass with by having the boundary term.

This is even stressed in the quizz accompanying these notes: https://ocw.mit.edu/courses/aeronau...all-2005-spring-2006/thermo-propulsion/q6.PDF

where the solution starts by remarking the validity of this equation depending on this assumption.

This seems to contradict books on Fluid Mechanics, where the mass can vary and they reach this similar equation (or maybe it's not the same equation?). For example Frank White's book equation 3.35:
$$\sum \vec{F}=\frac{d}{dt}\int_V (ρ{\vec{v}}) dV+\int_S\vec{v}ρ\vec{v}_r \cdot \vec{n} dA$$I can see that White's equation is not exactly the same but I'm trying to prove they are by expanding $$\vec{v}$$ and $$\vec{v}_r =\vec{v}-\vec{v}_{control volume}$$ (V relative to Earth, inertial frame and $$v_r$$ is a relative velocity with respect to the control volume).

I think I'm missing something here.

So basically, I'm wondering: Is the requirement of fixed mass even right? Considering that there is a boundary term.
If so, what would be the general equation for non-fixed mass control volume?

 

[vanhees71]

The point is the exchange of the time derivatives from inside the integral to the outside of the integral. If the volume $V$ is time dependent you have to take care about the temporal change when taking the time derivative out of the integral.

 

[arestes]

The point is the exchange of the time derivatives from inside the integral to the outside of the integral. If the volume $V$ is time dependent you have to take care about the temporal change when taking the time derivative out of the integral.

Yes, but how does that relate to the requirement mentioned in those notes about the mass being fixed? That requirement seems odd to me since this whole momentum balance, I think, considers loss or gain of mass.

 

[vanhees71]

I think what's meant is that you consider the movement of a "material element" of the fluid, and that's why there's the "material time derivative",
$$\mathrm{D}_t \vec{v}=\partial_t \vec{v} + (\vec{v} \cdot \vec{\nabla}) \vec{v}$$
under the integral, and this is the acceleration of a material fluid element.

Obviously you deal with non-relativistic particles, which implies that also mass is conserved, i.e., also the local conservation equation (continuity equation) for mass holds, i.e.,
$$\partial_t \rho + \vec{\nabla} \cdot (\rho \vec{v})=0,$$
where $\rho \vec{v}=\vec{j}$ is the mass-current density.