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Reynolds Transport Theorem

Intuition

Newton's laws apply to a fixed lump of matter — but the lump of fluid you care about this second is somewhere else next second. Engineers would much rather do bookkeeping on a fixed region of space (a control volume: the inside of a jet engine, a pipe section). The Reynolds transport theorem (RTT) is the exchange rate between the two kinds of accounting: change following the matter = change inside the region + flux through its boundary.

Formal statement

For a quantity with density \(f\) per unit volume, carried by a flow \(\mathbf{v}\) through a volume \(\Omega(t)\) with boundary \(\partial\Omega(t)\) and outward normal \(\mathbf{n}\):

\[\boxed{\;\frac{D}{Dt}\int_{\Omega(t)} f\, dV = \int_{\Omega(t)} \frac{\partial f}{\partial t}\, dV + \oint_{\partial\Omega(t)} f\,\mathbf{v}\cdot\mathbf{n}\, dS\;}\]

It is the finite-volume sibling of the material derivative: the same Lagrangian→Eulerian translation, integrated over a region. (The analogy in thermodynamics: system vs control volume.)

What you get by choosing \(f\)

Choice of \(f\) Conservation law it yields
\(\rho\) mass — the continuity equation
\(\rho\mathbf{v}\) momentum — Euler / Navier–Stokes with the stress tensor \(\mathbf{T}\)
\(\rho e\) energy — heat flux and work terms
\(\rho\,\mathbf{r}\times\mathbf{v}\) angular momentum — turbomachinery analysis

For momentum, the theorem reads

\[\frac{D}{Dt}\int_{\Omega} \rho\mathbf{v}\, dV = \int_{\Omega} \rho\mathbf{f}\, dV + \oint_{\partial\Omega} \mathbf{T}\cdot\mathbf{n}\, dS\]

where \(\mathbf{f}\) is body force per unit volume and \(\mathbf{T}\) the stress tensor — the integral form from which the differential equations of motion are extracted.

Derivation sketch (mass)

The mass of a material volume is \(m(t) = \int_{\Omega(t)}\rho\,dV\). Its rate of change has two contributions: the density changing inside (\(\partial\rho/\partial t\)), and the boundary moving with the fluid, sweeping volume at rate \(\mathbf{v}\cdot\mathbf{n}\) per unit area. Summing:

\[\frac{Dm}{Dt} = \int_{\Omega} \frac{\partial\rho}{\partial t}\,dV + \oint_{\partial\Omega}\rho\,\mathbf{v}\cdot\mathbf{n}\,dS\]

Setting \(Dm/Dt = 0\) (mass is conserved) and applying the divergence theorem gives the continuity equation.

Common mistakes

  • Forgetting the flux term when the control volume is fixed but fluid crosses its boundary — that term is usually the whole point.
  • Wrong normal direction: \(\mathbf{n}\) points outward; inflow contributes negatively.
  • Applying RTT to a quantity per unit mass. The theorem as written wants \(f\) per unit volume (use \(\rho\times\) specific quantity).

Knowledge graph position

Prerequisites: Material derivative, Eulerian vs Lagrangian. Leads to: Continuity, Euler, Navier–Stokes.

Quiz

Q1 (conceptual). A rocket engine test stand encloses the engine in a fixed control volume. Steady operation: nothing inside changes in time. Where does the thrust show up in the RTT?

Answer

Entirely in the flux term: momentum leaves through the nozzle exit at rate \(\dot m v_e\) (plus a pressure–area term). The surface integral of momentum flux equals the force on the stand.

Q2 (multiple choice). Setting \(f = \rho\) and \(D/Dt \int \rho\, dV = 0\) in the RTT gives:

  • (a) Bernoulli's equation (b) the continuity equation (c) the Navier–Stokes equation (d) Archimedes' principle
Answer

(b). Mass conservation is the simplest application.

Q3 (conceptual). Why is the RTT called a bridge between Lagrangian and Eulerian descriptions?

Answer

The left side is Lagrangian (follows a material volume); the right side is Eulerian (a fixed-region volume integral plus boundary flux). It converts statements about matter into statements about fields in space.