Newton's laws and conservation principles are naturally stated for a system: a fixed collection of matter followed as it moves. Fluid-flow problems are often easier to analyze with a control volume: a region in space through which matter can enter and leave.

The Reynolds transport theorem connects these two descriptions.

Let $B$ be any extensive property of a system, such as mass, momentum, or energy. Let $b$ be the amount of that property per unit mass, so locally

$$dB=b,dm=b\rho,dV.$$

For a fixed control volume $CV$ with outward-pointing normal $\mathbf n$ on its control surface $CS$, the theorem is

$$\boxed{ \frac{dB_{\rm sys}}{dt}

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

It says that the rate of change of a property for the moving material system equals

1. the rate at which that property is accumulating inside the fixed control volume;
2. plus the net outward transport of that property through the control surface.

Storage term

The term

$$\frac{\partial}{\partial t} \int_{CV}\rho b,dV$$

measures how much of the property is changing inside the control volume itself.

For a steady flow, fields at fixed positions do not change with time, so this storage term is zero.

Flux term

The surface term

$$\oint_{CS}\rho b(\mathbf v\cdot\mathbf n),dA$$

measures net outward transport.

Because $\mathbf n$ points outward:

• at an outlet, $\mathbf v\cdot\mathbf n>0$, giving a positive contribution;
• at an inlet, $\mathbf v\cdot\mathbf n<0$, giving a negative contribution.

The sign therefore handles inflow and outflow automatically.

Mass as a special case

Choose

$$B=m,\qquad b=1.$$

A material system cannot gain or lose mass, so

$$\frac{dm_{\rm sys}}{dt}=0.$$

The transport theorem becomes

$$\boxed{ 0=rac{\partial}{\partial t}\int_{CV}\rho,dV +\oint_{CS}\rho\mathbf v\cdot\mathbf n,dA }.$$

This is the integral conservation-of-mass equation.

For steady flow with uniform conditions at a finite set of inlets and outlets, it reduces to

$$\boxed{\sum \dot m_{\rm in}=\sum \dot m_{\rm out}},$$

where

$$\dot m=\rho Av_n$$

for a uniform port whose normal speed is $v_n$.

Worked bookkeeping example

A fixed tank has one inlet carrying

$$\dot m_{\rm in}=3.0,\mathrm{kg/s}$$

and one outlet carrying

$$\dot m_{\rm out}=2.0,\mathrm{kg/s}.$$

For mass, the transport theorem gives

$$0=\frac{dm_{CV}}{dt}+\dot m_{\rm out}-\dot m_{\rm in}.$$

Thus

$$\frac{dm_{CV}}{dt} =3.0-2.0 =\boxed{1.0,\mathrm{kg/s}}.$$

The control volume is accumulating mass even though mass of the material system is conserved.

Why the theorem matters

The same structure applies to different conserved or balanced quantities simply by changing $B$ and $b$:

• mass: $b=1$;
• linear momentum: $\mathbf b=\mathbf v$;
• angular momentum: $\mathbf b=\mathbf r\times\mathbf v$;
• energy: $b$ is specific total energy.

The physics law belongs to the material system; Reynolds transport theorem converts it into a balance over a region in space that fluid can flow through.