If we now choose an **arbitrary** system of material (for a
Lagrangian outlook on the problem):

we can write a "word equation" for the changes in momentum of this system as:

Writing this mathematically we get:

$\displaystyle{\frac{d}{dt}\int\int\int_{SYS}\rho \mathbf{v}dV = \sum \mathbf{F}}$

where $\displaystyle{\sum \mathbf{F}}$ denotes the sum of the forces acting on the fluid element within our system.

In order to mathematically translate between a control system
and a control volume one uses the *Reynolds Transport
Theorem* that stipulates that:

$\displaystyle{\frac{d}{dt}\int\int\int_{SYS}\rho \mathbf{v}dV = \frac{d}{dt}\int\int\int_{CV}\rho \mathbf{v}dV + \int\int_{CV}\rho \mathbf{v}\left(\mathbf{v}\cdot\mathbf{n}\right ) dA}$

so that our momentum balance equation can be written as:

$\displaystyle{\frac{d}{dt}\int\int\int_{CV}\rho \mathbf{v}dV = - \int\int_{CV}\rho \mathbf{v}\left(\mathbf{v}\cdot\mathbf{n}\right ) dA + \sum \mathbf{F}}$

It was easier to rationale the balance of momentum using a system -- since we are all familiar with the equation $\displaystyle{\mathbf{F} = m\mathbf{a} = m\frac{d\mathbf{v}}{dt}}$. It might have also helped when we did a material balance, as a material balance on a system would simply have been $\displaystyle{\frac{d}{dt}\int\int\int_{SYS}\rho dV = 0 }$

Recalling from earlier in the course, we know that the potential forces that act on fluids are the following:

- Pressure force at inlets and outlets (a surface force): $\displaystyle{\mathbf{F_P} = - \int\int_{inlet}^{outlet} P\mathbf{n}dA}$
- Pressure and shear force exerted by wall surfaces on fluid in CV (surface forces): $\displaystyle{\mathbf{R}\equiv - \int\int_{walls}P\mathbf{n}dA + \int\int_{walls}\mathbf{\tau}dA}$
- Gravitational force exerted on all fluid within CV (a body or volumetric force): $\displaystyle{\mathbf{W} \equiv \int\int\int_{CV}\rho\mathbf{g}dV = \rho V\mathbf{g}}$

Combining these we can write a general form of the momentum balance equation as:

$\displaystyle{\frac{d}{dt}\int\int\int_{CV}\rho \mathbf{v}dV = - \int\int_{CV}\rho \mathbf{v}\left(\mathbf{v}\cdot\mathbf{n}\right ) dA - \int\int_{CV}P\mathbf{n}dA + \int\int_{CV}\mathbf{\tau}dA + \int\int\int_{CV}\rho\mathbf{g}dV}$

Recall that the shear stresses affect only the walls of the CV;
the pressure affects *all surfaces* (including inlets and
outlets); and the gravitational forces affect the entire CV.

This is a *vector* equation so that, in practice, there
will be *three* momentum balances equations to write; one in
each of the coordinate directions!

Derive and use a macroscopic momentum balance.