The angular-momentum law for a material system states that the net external torque about a chosen origin $O$ equals the rate of change of angular momentum about that origin:

$$\boxed{\sum\boldsymbol\tau_{O,\rm ext}=\frac{d\mathbf L_{O,\rm sys}}{dt}}.$$

For a fluid system, the specific angular momentum about $O$ is

$$\mathbf r\times\mathbf v,$$

where $\mathbf r$ is measured from $O$.

Applying Reynolds transport theorem to a fixed control volume gives

$$\boxed{ \sum\boldsymbol\tau_{O,\rm ext}

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

The first term is angular-momentum accumulation inside the control volume. The second is net outward angular-momentum flux.

Steady flow through uniform ports

For steady flow, the storage term vanishes. If velocity and density are approximately uniform across each inlet and outlet,

$$\boxed{ \sum\boldsymbol\tau_{O,\rm ext}

\sum_{\rm out}\dot m,(\mathbf r\times\mathbf v)_{\rm out}

\sum_{\rm in}\dot m,(\mathbf r\times\mathbf v)_{\rm in} }.$$

This is the angular-momentum counterpart of the control-volume linear-momentum equation.

The origin matters. A force whose line of action passes through $O$ contributes no torque about $O$, while the same force can contribute torque about another origin.

Axial component

Many rotating-flow problems have a natural axis, such as the shaft axis of a turbine or pump. In cylindrical coordinates about the $z$-axis, let

• $r$ be distance from the axis;
• $V_\theta$ be the tangential component of the absolute fluid velocity.

The specific angular momentum about the $z$-axis is

$$\boxed{\ell_z=rV_\theta}.$$

Therefore the steady axial torque balance becomes

$$\boxed{ \tau_z

\sum_{\rm out}\dot m,rV_\theta

\sum_{\rm in}\dot m,rV_\theta }.$$

Only the tangential velocity component contributes to angular momentum about the axis. Radial and axial velocity components can carry mass through the device without directly contributing to $L_z$.

Worked example: turning a tangential jet

A steady water jet with mass flow rate

$$\dot m=4.0,\mathrm{kg/s}$$

enters a device at radius

$$r_1=0.30,\mathrm m$$

with tangential velocity

$$V_{\theta1}=12,\mathrm{m/s}.$$

It leaves at the same radius with

$$V_{\theta2}=2.0,\mathrm{m/s}.$$

Taking positive torque in the direction of positive $V_\theta$, the net external torque on the fluid is

$$\tau_z =\dot m(r_2V_{\theta2}-r_1V_{\theta1}).$$

Thus

$$\tau_z =(4.0)(0.30)(2.0-12) =-12.0,\mathrm{N,m}.$$

So

$$\boxed{\tau_{\rm hardware\ on\ fluid}=-12.0,\mathrm{N,m}}.$$

By Newton's third law, the fluid exerts the opposite torque on the hardware:

$$\boxed{\tau_{\rm fluid\ on\ hardware}=+12.0,\mathrm{N,m}}.$$

The fluid loses angular momentum, so the hardware receives torque in the positive direction.

External torques to include

The torque balance must include every external torque acting on the fluid in the control volume, such as

• shaft or blade forces;
• pressure forces whose lines of action do not pass through the chosen origin;
• gravity when it has a nonzero moment about the origin;
• wall shear or other contact forces.

Symmetry often makes some of these contributions vanish, but they should not be omitted without justification.

The control-volume angular-momentum equation is the natural tool whenever a flow exchanges torque with hardware. It underlies sprinklers, rotating jets, turbines, pumps, compressors, propellers, and many other rotating-flow devices.