Reynolds transport theorem: Question bank — Physics deep dive

Reminder of the master statement, so no symbol here is unearned:

True or false — justify

Each line: decide true or false, then give the one-line reason.

The storage term

\frac{d}{d t} \int_{C V} ρ b d V

is zero whenever fluid is flowing through the CV.

False. That term measures unsteadiness in time, not flow; it vanishes only when nothing inside the CV changes with time (steady flow), even while fluid streams through.

For a moving control volume, the flux term uses the fluid's ground velocity

v

False. Crossing a moving surface depends on motion relative to it, so you use

v_{r} = v - v_{C S}

; a surface drifting with the fluid sees zero flux.

A control volume must be a rigid, fixed region of space.

False. A CV may be fixed, translating, deforming, or moving with an object (a moving CV just adds the relative-velocity correction

v_{r}

If flow is steady, then

B_{sy s}

(following the matter) cannot change.

False. Storage is zero, but the surface flux can still be nonzero, so

d B_{sy s} / d t

equals that net efflux — steady flow through a bend still changes momentum.

The integrand of the storage term is

ρ b

, not

b

True. The stored amount is (property per mass)

\times

(mass), and

d m = ρ d V

, so you must weight

b

ρ

b = 1

turns the RTT into the Continuity equation.

True. For mass

B = m

, so

b = d B / d m = 1

, and

d m_{sy s} / d t = 0

gives the mass-conservation statement.

The left side of the RTT,

d B_{sy s} / d t

, is an Eulerian (control-volume) quantity.

False. It follows a fixed lump of matter — that is the Lagrangian (system) side; the RTT's whole job is to rewrite it in Eulerian terms. See Eulerian vs Lagrangian description.

The RTT is a special trick for fluids only.

False. It is a purely kinematic transport identity for any extensive property carried by any continuum; fluids are just its most common home.

For steady, incompressible flow with one inlet and one outlet,

A_{in} v_{in} = A_{o u t} v_{o u t}

True. Continuity with

ρ

constant and zero storage gives

\sum (A v)_{o u t} = \sum (A v)_{in}

, so a narrower pipe means faster flow.

Spot the error

Each line contains a subtly wrong claim. State what's wrong.

"At an inlet,

v \cdot \overset{n}{^} > 0

because fluid is entering."

Wrong sign: at an inlet the flow points into the CV, opposite the outward normal

\overset{n}{^}

, so

v \cdot \overset{n}{^} < 0

; only outlets give a positive value.

"The whole fluid speed

∣ v ∣

determines how much property crosses a patch

d A

Only the normal component

v \cdot \overset{n}{^}

pierces the surface; velocity tangent to the boundary slides along it and carries nothing across.

"Because

\sum F

appears on the left of the momentum RTT, the surface term must be a force."

The surface term is a momentum flux (rate at which momentum is carried across CS by mass flow), not a force; forces are what balance that flux plus storage.

"We derived the RTT by shrinking the CV to a point."

We shrank the time interval

d t \to 0

, not the volume; the CV keeps a finite size. Shrinking to a point instead gives the Material derivative.

"For a jet hitting a wall we set the storage term to zero, so the flow must be incompressible."

The storage term drops because the flow is steady (time-independent), not because it is incompressible; these are separate assumptions.

"Since the same molecules define the system,

B_{sy s} (t) = B_{C V} (t)

holds at all times."

It holds only at the single instant

t

where we chose system and CV to coincide; an instant later the system has drifted and the equality breaks.

\oint_{C S}

means integrate only over the inlet and outlet."

It integrates over the entire closed control surface; solid walls just happen to contribute zero because

v \cdot \overset{n}{^} = 0

there (no flow through a wall).

Why questions

Why do we start the derivation by making the system and CV coincide at time

t

So that

B_{sy s} (t) = B_{C V} (t)

exactly, which lets us swap "system" for "CV" at that instant and turn a matter-tracking rate into a region-based one.

Why does the swept-volume of fluid crossing a patch equal

(v \cdot \overset{n}{^}) d t d A

and not

∣ v ∣ d t d A

The fluid sweeps a slanted cylinder; its height perpendicular to $d A$ is only the normal velocity component times

d t

, so tangential motion doesn't add volume through the surface.

Why does a single closed surface integral automatically handle both inflow and outflow?

The outward-normal sign convention makes

v \cdot \overset{n}{^}

negative at inlets and positive at outlets, so one

\oint_{C S}

subtracts inflow and adds outflow without separate bookkeeping.

Why is the momentum-flux term quadratic in velocity while the mass-flux term is linear?

Momentum flux is

ρ v (v \cdot \overset{n}{^})

— one

v

is the momentum-per-mass (

b = v

) and the other

v \cdot \overset{n}{^}

is how fast mass crosses — so both velocities appear, giving a

v^{2}

dependence.

Why can the Divergence theorem convert the RTT into a differential (point) law?

It rewrites the surface flux

\oint_{C S} ρ b v \cdot \overset{n}{^} d A

as a volume integral

\int_{C V} \nabla \cdot (ρ b v) d V

; equating integrands (valid for any CV) gives the local conservation equation.

Why is Bernoulli the energy RTT with restrictions rather than a fresh law?

It is the

b = e

case under steady, incompressible, inviscid, along-a-streamline conditions; strip those away and you're back to the general energy control-volume balance.

Why must we distinguish an extensive property

B

from its intensive partner

b

RTT integrates

ρ b

over the CV, so it needs the per-unit-mass form

b

; feeding in the total

B

would double-count the mass already inside

ρ d V

Edge cases

Every quadrant of behaviour — pin down what happens when inputs go to zero, extremes, or degenerate.

A CV whose surface moves exactly with the local fluid everywhere.

The relative velocity

v_{r} = v - v_{C S} = 0

on the surface, so the flux term is zero and the CV becomes a system (Lagrangian) again — no matter ever crosses it.

Fluid inside the CV is completely still (

v = 0

) but the walls are being heated.

The flux term is zero, yet for energy (

b = e

) the storage term

\frac{d}{d t} \int ρ e d V

is nonzero — property can change purely by unsteady storage with no flow.

A closed rigid tank with sealed walls, no inlets or outlets.

v \cdot \overset{n}{^} = 0

over the whole surface, so flux is zero and

d B_{sy s} / d t

equals the storage term alone — RTT reduces to "the CV is the system."

Flow enters and leaves through parallel patches with equal

\overset{m}{˙}

but opposite normal directions.

For mass (

b = 1

) the fluxes cancel (

\sum ρ b v \cdot \overset{n}{^} = 0

); for momentum (

b = v

) they need not cancel, because

v

can differ in direction even when speeds match — that's how bends generate force.

A patch of surface where flow grazes tangentially (

v ⊥ \overset{n}{^}

v \cdot \overset{n}{^} = 0

there, so that patch contributes zero flux even though the fluid is fast — motion parallel to the boundary carries nothing across it.

The limit

d t \to 0

in the derivation.

The overlap-region terms collapse to

\frac{d}{d t} B_{C V}

and the thin inflow/outflow slabs (regions I and III) become the surface integral — this limit is what turns discrete bookkeeping into the exact theorem.

Density

ρ \to

constant (incompressible) in the continuity RTT.

ρ

pulls out of both integrals; steadiness kills storage, leaving the purely geometric

\sum (A v)_{o u t} = \sum (A v)_{in}

— volume, not just mass, is conserved.

An outlet where

v \cdot \overset{n}{^} = 0

momentarily (flow reversing sign).

That instant contributes zero flux there; as it crosses zero the patch switches between acting as inlet and outlet, and the sign of

v \cdot \overset{n}{^}

flips continuously with it.

Connections

Parent: RTT topic note
Continuity equation — the $b = 1$ traps live here.
Momentum equation (control volume) — the $b = v$ sign traps.
Eulerian vs Lagrangian description — the left-vs-right-side confusion.
Material derivative — the point-limit contrast used in "Spot the error."
Divergence theorem — surface-to-volume conversion in "Why questions."
Bernoulli equation — the restricted energy case.