Reynolds transport theorem: Worked examples — Physics deep dive

Everything below hangs on the one boxed law from the parent:

Before any numbers, look at the sign convention that drives everything.

The outward normal $\overset{n}{^}$ always points out of the box. So:

Where fluid leaves, its velocity roughly agrees with $\overset{n}{^}$ ⟹ $v \cdot \overset{n}{^} > 0$ (positive = outflow).
Where fluid enters, its velocity opposes $\overset{n}{^}$ ⟹ $v \cdot \overset{n}{^} < 0$ (negative = inflow).

That single rule is the source of every plus and minus sign on this page.

The scenario matrix

Every RTT problem is one (or a blend) of these cells. Each worked example below is tagged with the cell it kills.

#	Cell class	What makes it tricky	Example
A	Steady, single in / single out	baseline; storage term $= 0$	Ex 1
B	Steady, multiple ports	many $v \cdot \overset{n}{^}$ signs to sum	Ex 2
C	Unsteady storage, no flow	surface term $= 0$ , only $\frac{d}{d t} \int$ survives	Ex 3
D	Unsteady storage + flow together	both terms nonzero at once	Ex 4
E	Sign of momentum flux (inflow minus)	negative $v \cdot \overset{n}{^}$ flips a sign	Ex 5
F	Moving control surface	must use relative velocity $v_{r}$	Ex 6
G	Degenerate / zero input	$v = 0$ or symmetric flow ⟹ terms vanish	Ex 7
H	Real-world word problem	translate prose → CV → formula	Ex 8
I	Exam twist (unsteady + momentum + gauge pressure)	everything at once	Ex 9

Cell A — steady, one inlet, one outlet

Worked example Ex 1 — pipe contraction (continuity,

$b = 1$ ) Water ( $ρ = 1000 kg/m^{3}$ ) flows steadily through a pipe that narrows from area $A_{1} = 0.02 m^{2}$ (inlet speed $v_{1} = 3 m/s$ ) to $A_{2} = 0.008 m^{2}$ . Find the outlet speed $v_{2}$ .

Forecast: narrower pipe ⟹ faster water. Guess a number before reading on.

Pick $b = 1$ (mass) and write RTT. $\frac{d m _{sy s}}{d t} = 0 = \frac{d}{d t} \int_{C V} ρ d V + \oint_{C S} ρ (v \cdot \overset{n}{^}) d A .$ Why this step? Mass is conserved, so the whole left side is zero — this is the Continuity equation.
Kill the storage term. Flow is steady ⟹ nothing inside the CV changes with time ⟹ $\frac{d}{d t} \int_{C V} ρ d V = 0$ . Why? The storage term measures unsteadiness, not flow (parent mistake #2).
Evaluate the surface integral at the two flat ports. Inlet: $v \cdot \overset{n}{^} = - v_{1}$ (inflow, negative). Outlet: $v \cdot \overset{n}{^} = + v_{2}$ . $0 = ρ (- v_{1}) A_{1} + ρ (+ v_{2}) A_{2} .$ Why the signs? Outward normal points backward against the incoming stream at the inlet.
Solve. $v_{2} = \frac{v _{1} A _{1}}{A _{2}} = \frac{3 \times 0.02}{0.008} = 7.5 m/s .$

Verify: volume flow in $= 3 \times 0.02 = 0.06 m^{3} / s$ ; out $= 7.5 \times 0.008 = 0.06 m^{3} / s$ . Equal ✓. Units: $\frac{( m/s ) ( m ^{2} )}{m ^{2}} = m/s$ ✓. Faster, as forecast ✓.

Cell B — steady, multiple ports

Cell C — unsteady storage, no flow across the surface

Worked example Ex 3 — a sealed tank being heated (storage only)

A rigid, sealed tank of volume $V = 0.5 m^{3}$ holds gas whose density rises uniformly at $\frac{d ρ}{d t} = 0.2 kg/m^{3} per s$ . No valve is open. At what rate is mass accumulating inside?

Forecast: sealed ⟹ no flux. Only the storage term can act.

RTT with $b = 1$ : $\frac{d m _{sy s}}{d t} = \frac{d}{d t} \int_{C V} ρ d V + \oint_{C S} ρ (v \cdot \overset{n}{^}) d A .$
Kill the surface term. Walls are solid: $v \cdot \overset{n}{^} = 0$ everywhere on the closed surface. Why? Nothing crosses a sealed wall — the parent's tangential-flow argument in the extreme.
Evaluate storage. Volume fixed, $ρ$ uniform: $\frac{d}{d t} \int_{C V} ρ d V = \frac{d ρ}{d t} V = 0.2 \times 0.5 = 0.1 kg/s .$

Verify: for the sealed system $\frac{d m _{sy s}}{d t} = 0.1 kg/s$ and there is no inflow — the mass "appears" only because $ρ$ climbs; the system mass truly grows only if we let the sealed system be exactly this tank (closed) — here the number is the accumulation rate inside, $0.1 kg/s$ ✓. Units $(kg/m^{3} / s) (m^{3}) = kg/s$ ✓.

Cell D — unsteady AND flowing at once

Worked example Ex 4 — filling tank (both terms alive)

An open tank of cross-section $A_{t} = 1 m^{2}$ is filled by an inlet pipe of area $A_{i} = 0.01 m^{2}$ carrying water at $v_{i} = 5 m/s$ . No outlet. How fast does the water level $h$ rise?

Forecast: water piles up ⟹ the surface term feeds the storage term. Both nonzero.

Choose the CV = the water region up to level $h (t)$ . $b = 1$ , $ρ$ constant. $0 = \frac{d}{d t} \int_{C V} ρ d V + \oint_{C S} ρ (v \cdot \overset{n}{^}) d A .$
Storage term: the CV volume is $A_{t} h$ , so $\frac{d}{d t} \int_{C V} ρ d V = ρ A_{t} \frac{d h}{d t} .$ Why? The stored mass changes because the top face rises — a genuine time change.
Surface term: only the inlet pierces the CS, $v \cdot \overset{n}{^} = - v_{i}$ : $\oint_{C S} ρ (v \cdot \overset{n}{^}) d A = ρ (- v_{i}) A_{i} .$
Combine and solve: $0 = ρ A_{t} \frac{d h}{d t} - ρ v_{i} A_{i} \Rightarrow \frac{d h}{d t} = \frac{v _{i} A _{i}}{A _{t}} = \frac{5 \times 0.01}{1} = 0.05 m/s .$

Verify: volume filling rate $= A_{t} \frac{d h}{d t} = 1 \times 0.05 = 0.05 m^{3} / s$ , which equals inflow $v_{i} A_{i} = 0.05$ ✓. Units m/s ✓. Both terms were needed — this is why "always drop storage" is wrong.

Cell E — momentum, the inflow minus sign

Worked example Ex 5 — jet on a flat plate (momentum,

$b = v$ ) A horizontal water jet ( $ρ = 1000$ , $A = 2 \times 1 0^{- 3} m^{2}$ , $v = 10 m/s$ ) strikes a wall head-on and spreads sideways, losing all its $x$ -momentum. Steady flow. Find the $x$ -force on the jet from the wall, and the reaction on the wall.

Forecast: the wall must stop $200$ N of momentum-per-second. Predict the sign.

RTT with $b = v_{x}$ , steady ⟹ storage $= 0$ : $\sum F_{x} = \oint_{C S} ρ v_{x} (v \cdot \overset{n}{^}) d A .$
Mass flow: $\overset{m}{˙} = ρ A v = 1000 (2 \times 1 0^{- 3}) (10) = 20 kg/s .$
Outflow $x$ -momentum: water leaves purely sideways, so $v_{x} = 0$ ⟹ out-flux $= 0$ .
Inflow $x$ -momentum: at the inlet $v_{x} = v = 10$ and $v \cdot \overset{n}{^} = - v$ (inflow, negative): $in-flux = ρ v_{x} (v \cdot \overset{n}{^}) A = ρ v (- v) A = - \overset{m}{˙} v = - 200.$ Why the minus? Outward normal opposes the incoming stream — this is exactly the parent's sign trap.
Sum: $\sum F_{x} = 0 - (- 200) = + 200 N$ on the jet (pointing back, opposing the jet). By Newton's third law the jet pushes the wall with $200 N$ forward.

Verify: $\overset{m}{˙} v = 20 \times 10 = 200 N$ ✓. Units $(kg/s) (m/s) = kg\cdotm/s^{2} = N$ ✓. Matches parent's plug-in ✓. See Momentum equation (control volume).

Cell F — moving control surface (relative velocity)

Worked example Ex 6 — jet hitting a moving cart

The same jet ( $ρ = 1000$ , $A = 2 \times 1 0^{- 3}$ , $v = 10 m/s$ ) now hits a plate mounted on a cart moving away at $v_{C S} = 4 m/s$ in the same direction. Water still spreads sideways. Find the $x$ -force on the plate.

Forecast: the cart runs from the jet, so it feels less force than the stationary case. Guess.

Attach the CV to the cart. The surface moves, so the flux uses the relative velocity $v_{r} = v - v_{C S} = 10 - 4 = 6 m/s .$ Why this step? Only motion relative to the surface crosses it (parent mistake #1). If the cart matched the jet ( $v_{C S} = 10$ ), zero water would land.
Mass flow that actually strikes the plate: $\overset{m}{˙}_{r} = ρ A v_{r} = 1000 (2 \times 1 0^{- 3}) (6) = 12 kg/s .$
Momentum RTT, steady in the cart frame, using $v_{r}$ for the $x$ -momentum carried across: $F_{x} = 0 - (- \overset{m}{˙}_{r} v_{r}) = \overset{m}{˙}_{r} v_{r} = 12 \times 6 = 72 N .$ Why $v_{r}$ twice? Once for how much mass crosses per second, once for the $x$ -momentum each kilogram brings relative to the moving frame.

Verify: $ρ A v_{r}^{2} = 1000 (2 \times 1 0^{- 3}) (6^{2}) = 72 N$ ✓. Less than the $200$ N stationary case ✓ (forecast right). If $v_{C S} = 10$ : $v_{r} = 0 \Rightarrow F = 0$ ✓ — degenerate check.

Cell G — degenerate / zero input

Cell H — real-world word problem

Cell I — exam twist: unsteady + momentum + gauge pressure

Worked example Ex 9 — thrust on an emptying nozzle

A nozzle discharges water to the atmosphere. Steady flow: inlet area $A_{1} = 0.02 m^{2}$ , gauge pressure $p_{1} = 1.5 \times 1 0^{5} Pa$ , speed $v_{1} = 3 m/s$ ; outlet (to air, gauge $p_{2} = 0$ ) area $A_{2} = 0.008 m^{2}$ . $ρ = 1000$ . Find the axial force the water exerts on the nozzle.

Forecast: pressure pushes forward, the jet speeds up — expect a sizeable force. Predict its sign.

Continuity for $v_{2}$ (Ex 1 reused): $v_{2} = \frac{v _{1} A _{1}}{A _{2}} = \frac{3 ( 0.02 )}{0.008} = 7.5 m/s .$
Mass flow: $\overset{m}{˙} = ρ v_{1} A_{1} = 1000 (3) (0.02) = 60 kg/s .$
Momentum RTT, steady (storage $= 0$ ), including pressure forces on the CS: $p_{1} A_{1} - p_{2} A_{2} + R_{x} = \overset{m}{˙} v_{2} - \overset{m}{˙} v_{1},$ where $R_{x}$ is the axial force of the nozzle on the water. Why pressure terms? Gauge pressure pushes inward on each open face; the wall reaction $R_{x}$ balances the rest.
Insert numbers ( $p_{2} = 0$ ): $1.5 \times 1 0^{5} (0.02) + R_{x} = 60 (7.5) - 60 (3)$ $3000 + R_{x} = 450 - 180 = 270 \Rightarrow R_{x} = 270 - 3000 = - 2730 N .$
Reaction on nozzle: water pushes the nozzle with $- R_{x} = + 2730 N$ , i.e. forward (thrust), $\approx 2.73 kN$ .

Verify: momentum change $\overset{m}{˙} (v_{2} - v_{1}) = 60 (4.5) = 270 N$ ✓. Pressure force $3000 N$ ✓. $R_{x} = 270 - 3000 = - 2730 N$ ✓. Units: $Pa \cdot m^{2} = N$ , $kg/s \cdot m/s = N$ ✓. The nozzle pulls the water back, so the water thrusts the nozzle forward — physically sensible.

Recall Which cell is which?

Steady one-in-one-out is which example? ::: Ex 1 (cell A). Where does only the storage term survive? ::: Ex 3 (cell C, sealed tank). Where are both storage and flux nonzero? ::: Ex 4 (cell D, filling tank). Which example needs relative velocity? ::: Ex 6 (cell F, moving cart). Which example combines pressure, unsteady-check and momentum? ::: Ex 9 (cell I, nozzle thrust).

Connections

2.2.13 Reynolds transport theorem — the parent theorem these examples apply.
Continuity equation — Ex 1, 2, 4, 8 ( $b = 1$ ).
Momentum equation (control volume) — Ex 5, 6, 9 ( $b = v$ ).
Eulerian vs Lagrangian description — why we watch a CV at all.
Bernoulli equation — the energy check hiding behind the nozzle speeds.