Newton's second law for a material system can be written in momentum form as

$$\sum\mathbf F_{\rm ext}=\frac{d\mathbf P_{\rm sys}}{dt}.$$

For linear momentum, the extensive property is

$$B=\mathbf P,$$

and the property per unit mass is the velocity,

$$\mathbf b=\mathbf v.$$

Applying Reynolds transport theorem to a fixed control volume gives the control-volume momentum equation:

$$\boxed{ \sum\mathbf F_{\rm ext}

\frac{\partial}{\partial t} \int_{CV}\rho\mathbf v,dV + \oint_{CS}\rho\mathbf v(\mathbf v\cdot\mathbf n),dA }.$$

The external-force vector includes every force exerted on the fluid inside the control volume, including pressure forces, wall or support forces, gravity, and other body forces.

Steady flow through uniform ports

For steady flow, the momentum stored inside a fixed control volume does not change with time, so

$$\frac{\partial}{\partial t} \int_{CV}\rho\mathbf v,dV=\mathbf0.$$

If velocity and density are approximately uniform across each inlet and outlet, the surface integral reduces to

$$\boxed{ \sum\mathbf F_{\rm ext}

\sum_{\rm out}\dot m,\mathbf v_{\rm out}

\sum_{\rm in}\dot m,\mathbf v_{\rm in} }.$$

This is a vector equation. A flow can keep the same speed while changing momentum because its direction changes.

Forces on the fluid and forces on the hardware

The momentum equation gives the net external force on the fluid in the chosen control volume.

If the goal is to find the force of the fluid on a pipe, vane, nozzle, or support, Newton's third law reverses the interaction force:

$$\boxed{\mathbf F_{\rm fluid\ on\ hardware} =-\mathbf F_{\rm hardware\ on\ fluid}}.$$

Pressure forces on inlet and outlet surfaces must be included unless the chosen conditions make their gauge-pressure contributions zero or they have already been accounted for in another force term.

Worked example: a jet turned through $90^\circ$

A steady free jet has mass flow rate

$$\dot m=5.0,\mathrm{kg/s}$$

and speed

$$V=10,\mathrm{m/s}.$$

A smooth vane turns the jet from the positive $x$ direction into the positive $y$ direction without appreciably changing its speed.

Thus

$$\mathbf v_{\rm in}=10\hat{\mathbf x},\mathrm{m/s},$$

$$\mathbf v_{\rm out}=10\hat{\mathbf y},\mathrm{m/s}.$$

Take a control volume around the portion of jet interacting with the vane. Because the inlet and outlet are free jets exposed to the same ambient pressure, their gauge-pressure forces can be taken as zero. Neglect the weight of fluid in the small control volume.

The momentum equation gives

$$\sum\mathbf F_{\rm ext} =\dot m(\mathbf v_{\rm out}-\mathbf v_{\rm in}).$$

Therefore

$$\sum\mathbf F_{\rm ext} =(5.0)(10\hat{\mathbf y}-10\hat{\mathbf x})$$

$$=\boxed{-50\hat{\mathbf x}+50\hat{\mathbf y},\mathrm N}.$$

This is the force of the vane on the fluid. The force of the fluid on the vane is opposite:

$$\boxed{\mathbf F_{\rm fluid\ on\ vane} =50\hat{\mathbf x}-50\hat{\mathbf y},\mathrm N}.$$

Although the jet's speed is unchanged, redirecting its velocity requires a substantial force because momentum is a vector.

Pressure-force bookkeeping

For a pipe flow whose inlet or outlet pressure differs from the surrounding pressure, pressure contributes to the external force on the control volume.

On a control surface with outward normal $\mathbf n$, pressure exerts an inward traction

$$-p\mathbf n.$$

The net pressure force is therefore

$$\boxed{\mathbf F_p=-\oint_{CS}p\mathbf n,dA}.$$

In simple uniform-port problems this becomes a sum of terms such as $pA$ in the appropriate directions.

Why this balance is useful

The control-volume momentum equation predicts forces generated by changes in fluid momentum without requiring the detailed stress and velocity field everywhere inside the device. It is the natural tool for jets, bends, nozzles, diffusers, propulsive streams, and many other open-flow systems.

Its logic is the same as ordinary momentum mechanics: external force equals rate of change of momentum. Reynolds transport theorem adds the bookkeeping needed when momentum crosses the boundary of a region in space.