Effective exhaust velocity c = v_e + (P_e − P_a)A_e - ṁ: Worked examples

The scenario matrix

Before touching numbers, let us name what "every scenario" even means. Our formula has five inputs — $v_{e}, P_{e}, P_{a}, A_{e}, \overset{m}{˙}$ — and one output $c$ . The interesting behaviour is driven almost entirely by the sign of the pressure difference

$Δ P \equiv P_{e} - P_{a}$

(the exit gas pressure minus the outside air pressure) and by whether any input hits an extreme value (zero, huge, or a designed-for match). Here is the full grid.

Cell	Scenario class	Trigger	What we expect
A	Optimally expanded	$P_{e} = P_{a}$ ( $Δ P = 0$ )	$c = v_{e}$ exactly
B	Underexpanded	$P_{e} > P_{a}$ ( $Δ P > 0$ )	$c > v_{e}$ (bonus thrust)
C	Overexpanded	$P_{e} < P_{a}$ ( $Δ P < 0$ )	$c < v_{e}$ (thrust penalty)
D	Vacuum limit	$P_{a} \to 0$	largest $c$ for that engine
E	Ascent (variable $P_{a}$ )	$P_{a}$ drops with altitude	$c$ rises during climb
F	Degenerate area	$A_{e} \to 0$	pressure term vanishes, $c \to v_{e}$
G	Solve-backwards	$c$ given, find $v_{e}$ or $A_{e}$	rearrange the definition
H	Real-world word problem	data buried in prose	translate → same formula
I	Exam twist / units trap	kPa vs Pa, gauge vs absolute	consistency of units
J	Degenerate $\overset{m}{˙}$	$\overset{m}{˙} \to 0$ or $\overset{m}{˙} \to \infty$	$c \to \infty$ or $c \to v_{e}$
K	Degenerate $v_{e}$	$v_{e} \to 0$	$c$ from pressure term alone

Every worked example below is tagged with the cell(s) it covers. Together they fill the whole grid.

Reading the scenario map (Figure 1)

To see the grid rather than just read it, look at the quadrant map below. Here is exactly how to read it, region by region:

The horizontal axis is the pressure difference $Δ P = P_{e} - P_{a}$ : negative to the left, zero at the centre line, positive to the right.
The vertical axis is the resulting correction to velocity, $\frac{Δ P A _{e}}{m ˙}$ — i.e. how far $c$ sits above or below the bare gas speed $v_{e}$ .
The pale-yellow horizontal line through the middle is $c = v_{e}$ : the "no pressure help" baseline.
The blue region (right) is where $Δ P > 0$ → correction positive → $c > v_{e}$ : this is underexpanded (cell B).
The pink region (left) is where $Δ P < 0$ → correction negative → $c < v_{e}$ : this is overexpanded (cell C).
The centre dot where the sloped line crosses the baseline is matched (cell A), $Δ P = 0$ , $c = v_{e}$ .
The straight sloped line is $\frac{A _{e}}{m ˙} Δ P$ ; its steepness is set by $A_{e} / \overset{m}{˙}$ — a bigger exit area or smaller mass flow tilts it up, amplifying the correction (that is cells F and J in disguise).

Figure — Effective exhaust velocity c = v_e + (P_e − P_a)A_e - ṁ

Below that, look at the three nozzle mouths: each shows one sign of $Δ P$ with a single arrow for which way the leftover pressure shoves — outward (blue, bonus), none (yellow, matched), or inward (pink, penalty).

Examples covering every cell

Worked example Example 2 — Cell B (underexpanded,

$Δ P > 0$ ) Same engine as Ex. 1 ( $\overset{m}{˙} = 400$ kg/s, $v_{e} = 3000$ m/s, $A_{e} = 1.5 m^{2}$ ) but flown higher where $P_{a} = 20$ kPa while $P_{e} = 45$ kPa. Find $c$ .

Forecast: now $P_{e} > P_{a}$ , so $Δ P > 0$ → $c$ should be above 3000 m/s.

$Δ P = 45000 - 20000 = 25000$ Pa (positive — underexpanded). Why this step? Positive $Δ P$ means the gas mouth is still squeezed harder than outside air; that leftover push helps.
Pressure term $= \frac{25000 \times 1.5}{400} = \frac{37500}{400} = 93.75$ m/s. Why this step? Divide the pressure force $Δ P A_{e}$ by $\overset{m}{˙}$ to convert a force into a velocity boost (that's the definition of $c$ ).
$c = 3000 + 93.75 = 3093.75$ m/s. Why this step? Add the positive boost to the bare gas speed — that is the whole formula assembled.

Verify: $c > v_{e}$ as forecast ✓. Sign of the correction is $+$ , consistent with the "Underexpanded → Up" mnemonic.

Worked example Example 3 — Cell C (overexpanded,

$Δ P < 0$ ) Same engine, now fired at sea level: $P_{a} = 101.3$ kPa, $P_{e} = 45$ kPa.

Forecast: $P_{e} < P_{a}$ , so $Δ P < 0$ → correction is negative, $c < 3000$ m/s.

$Δ P = 45000 - 101300 = - 56300$ Pa (negative — overexpanded). Why this step? The outside air is now squeezing harder than the exhaust, so it pushes back into the nozzle.
Pressure term $= \frac{- 56300 \times 1.5}{400} = \frac{- 84450}{400} = - 211.125$ m/s. Why this step? Same "force ÷ mass-flow → velocity" conversion as before; the minus sign is carried through because the net push now points inward.
$c = 3000 - 211.125 = 2788.875$ m/s. Why this step? Add the (negative) correction to $v_{e}$ — adding a negative number subtracts, giving a $c$ below the bare gas speed.

Verify: $c < v_{e}$ as forecast ✓. This is the sea-level penalty; the same engine (Ex. 2) does better up high. Physically this over-expansion can even cause flow separation — see Nozzle Expansion (Under/Over/Optimal).

Worked example Example 4 — Cell D + E (vacuum limit and ascent trend)

Take the engine of Ex. 1–3 ( $v_{e} = 3000$ , $A_{e} = 1.5$ , $\overset{m}{˙} = 400$ , $P_{e} = 45$ kPa) and tabulate $c$ at three altitudes: sea level ( $P_{a} = 101.3$ kPa), high altitude ( $P_{a} = 20$ kPa), and vacuum ( $P_{a} = 0$ ).

Forecast: as we climb, $P_{a}$ shrinks, so $- P_{a} A_{e} / \overset{m}{˙}$ gets less negative → $c$ should strictly increase, biggest in vacuum.

Sea level ( $P_{a} = 101.3$ kPa): reuse Ex. 3's arithmetic, $c = 2788.875$ m/s. Why this step? Ex. 3 already evaluated this exact engine at sea-level pressure; re-deriving it would just repeat the same $Δ P = - 56300$ Pa computation, so we quote it as the low anchor of the trend.
High altitude ( $P_{a} = 20$ kPa): reuse Ex. 2's arithmetic, $c = 3093.75$ m/s. Why this step? Ex. 2 is the identical engine at $P_{a} = 20$ kPa; quoting it lets us line up three altitudes on one axis without redoing the $Δ P = 25000$ Pa step.
Vacuum ( $P_{a} = 0$ ): $Δ P = 45000 - 0 = 45000$ Pa, $c = 3000 + \frac{45000 \times 1.5}{400} = 3000 + 168.75 = 3168.75 m/s .$ Why this step? Setting $P_{a} = 0$ removes all back-push; the full exit pressure now helps — the theoretical maximum $c$ for this geometry.

Verify: $2788.875 < 3093.75 < 3168.75$ — monotonically rising as forecast ✓. This is exactly why vacuum $I_{s p}$ is quoted higher; see Atmospheric Pressure vs Altitude. The rising staircase is drawn below.

Reading the ascent staircase (Figure 3)

How to read it: the horizontal axis lists the three altitudes (sea level → high altitude → vacuum), left to right as the rocket climbs. The vertical axis is $c$ in m/s. The blue dots joined by a line are the three values just computed; watch them step upward as we go right — each step is the shrinking of $P_{a}$ . The pink dashed line marks the bare gas speed $v_{e} = 3000$ m/s: notice the dots start below it (overexpanded penalty), cross it, and end above it (vacuum bonus). The vertical gap between any dot and the pink line is precisely the pressure correction $Δ P A_{e} / \overset{m}{˙}$ for that altitude.

Worked example Example 5 — Cell F (degenerate area

$A_{e} \to 0$ ) An idealised "no-nozzle" thruster: same gas speed $v_{e} = 3000$ m/s, $\overset{m}{˙} = 400$ kg/s, but we shrink the exit area toward zero, $A_{e} \to 0$ . What happens to $c$ at sea level ( $P_{a} = 101.3$ kPa, $P_{e} = 45$ kPa)?

Forecast: if $A_{e} = 0$ there is no ring for pressure to push on, so the correction should die and $c \to v_{e} = 3000$ .

Pressure term $= \frac{( 45000 - 101300 ) \cdot A _{e}}{400}$ . Why this step? $A_{e}$ multiplies the entire pressure force; no area means no force.
As $A_{e} \to 0$ this term $\to 0$ , so $c \to 3000$ m/s. Why this step? A shrinking factor drags the whole product to zero, leaving only $v_{e}$ .

Verify: the limit is independent of the sign of $Δ P$ — whether under- or over-expanded, zero area kills the correction ✓. Sanity: a plain gas jet with no expanding bell has pure momentum thrust only.

Worked example Example 6 — Cell J (degenerate

$\overset{m}{˙}$ : throttle to nothing, or open wide) Same underexpanded numbers as Ex. 2 ( $v_{e} = 3000$ , $Δ P = 25000$ Pa, $A_{e} = 1.5 m^{2}$ ). Ask two limiting questions about the mass flow $\overset{m}{˙}$ .

Forecast: $\overset{m}{˙}$ sits in the denominator of the pressure term. Tiny $\overset{m}{˙}$ should blow the term up ( $c \to \infty$ ); huge $\overset{m}{˙}$ should shrink it to nothing ( $c \to v_{e}$ ).

Write the correction as $\frac{Δ P A _{e}}{m ˙} = \frac{25000 \times 1.5}{m ˙} = \frac{37500}{m ˙}$ . Why this step? Isolating $\overset{m}{˙}$ in the denominator makes both limits obvious.
As $\overset{m}{˙} \to 0^{+}$ : $\frac{37500}{m ˙} \to + \infty$ , so $c \to \infty$ . Why this step? Dividing a fixed positive force by an ever-smaller mass flow means each kilogram carries an unbounded share of the pressure push — the "effective speed" per kg diverges. (Physically an idealisation: you can't have pressure thrust with zero propellant leaving.)
As $\overset{m}{˙} \to \infty$ : $\frac{37500}{m ˙} \to 0$ , so $c \to v_{e} = 3000$ m/s. Why this step? A huge flow dilutes the fixed pressure force across so many kilograms that the per-kg boost vanishes — only the raw gas speed survives.

Verify: the two limits bracket the sensible range; any real $\overset{m}{˙}$ gives a $c$ strictly between $v_{e}$ and $\infty$ for underexpanded flow ✓. Symbolically $lim_{\overset{m}{˙} \to \infty} c = v_{e}$ and $lim_{\overset{m}{˙} \to 0^{+}} c = \infty$ .

Worked example Example 7 — Cell K (degenerate

$v_{e} \to 0$ ) A hypothetical "cold-gas" case where the gas barely moves, $v_{e} \to 0$ , but is still squeezed above ambient: $P_{e} = 60$ kPa, $P_{a} = 20$ kPa, $A_{e} = 0.5 m^{2}$ , $\overset{m}{˙} = 100$ kg/s. What is $c$ ?

Forecast: with $v_{e} = 0$ the momentum half is gone; $c$ should equal the pressure term alone — a purely pressure-driven effective speed, still positive because $Δ P > 0$ .

Set $v_{e} = 0$ in $c = v_{e} + \frac{Δ P A _{e}}{m ˙}$ , leaving $c = \frac{Δ P A _{e}}{m ˙}$ . Why this step? Dropping the momentum term isolates what pressure alone contributes.
$Δ P = 60000 - 20000 = 40000$ Pa. Why this step? Only the difference enters; it is positive so the leftover push is outward.
$c = \frac{40000 \times 0.5}{100} = \frac{20000}{100} = 200$ m/s. Why this step? Same force-÷-mass-flow conversion; with no gas speed to add to, this is the whole $c$ .

Verify: $c = 200$ m/s $> 0$ , coming purely from pressure thrust as forecast ✓. Had $Δ P$ been negative here, $c$ would come out negative — a "reverse" effective velocity, i.e. net drag rather than thrust.

Worked example Example 9 — Cell G (solve backwards for

$A_{e}$ ) A designer wants the nozzle to hit $c = 3200$ m/s at an altitude with $P_{a} = 30$ kPa, using $v_{e} = 3050$ m/s, $P_{e} = 60$ kPa, $\overset{m}{˙} = 250$ kg/s. What exit area $A_{e}$ is required?

Forecast: we need a positive boost of $3200 - 3050 = 150$ m/s, and $Δ P > 0$ , so a sensible positive area should exist.

From $c = v_{e} + \frac{Δ P A _{e}}{m ˙}$ , solve $A_{e} = \frac{( c - v _{e} ) m ˙}{Δ P}$ . Why this step? Only $A_{e}$ is unknown; isolate it algebraically.
$Δ P = 60000 - 30000 = 30000$ Pa; needed boost $c - v_{e} = 150$ m/s. Why this step? Both ingredients of the ratio must be in SI before dividing.
$A_{e} = \frac{150 \times 250}{30000} = \frac{37500}{30000} = 1.25 m^{2}$ . Why this step? Divide required boost times mass-flow by the pressure difference to get the area.

Verify: put it back: $3050 + \frac{30000 \times 1.25}{250} = 3050 + 150 = 3200 = c$ ✓. Units: $\frac{( m/s ) ( kg/s )}{Pa} = \frac{kg\cdotm/s ^{2}}{N/m ^{2}} = \frac{N}{N/m ^{2}} = m^{2}$ ✓.

Worked example Example 10 — Cell H (real-world word problem)

"A launch engine burns 1.8 tonnes of propellant every 6 seconds. Its exhaust leaves at 2.9 km/s. At the moment of liftoff the nozzle's 1.1 m² mouth carries gas at 90 kPa while the sea-level air outside sits at 101.3 kPa. Report the effective exhaust velocity, in m/s."

Forecast: exhaust is at 90 kPa, air at 101.3 kPa → $Δ P < 0$ → overexpanded → $c$ slightly below 2900 m/s.

Convert the prose. Mass flow: $\overset{m}{˙} = \frac{1.8 t}{6 s} = \frac{1800 kg}{6 s} = 300$ kg/s. Exit speed: $v_{e} = 2.9 km/s = 2900$ m/s. Why this step? The formula needs SI base units: kg, s, m, Pa. Tonnes and km/s are traps.
$Δ P = 90000 - 101300 = - 11300$ Pa. Why this step? Fix the sign early; negative means overexpanded, matching the forecast.
Pressure term $= \frac{- 11300 \times 1.1}{300} = \frac{- 12430}{300} = - 41.4 \overline{3}$ m/s. Why this step? Force-÷-mass-flow conversion, minus sign carried.
$c = 2900 - 41.433 \dots = 2858.57$ m/s (to 2 dp). Why this step? Add the small negative correction to the bare gas speed.

Verify: $c < v_{e}$ as forecast, and by only a small margin because $Δ P$ is small ✓. Rounded answer $\approx 2858.6$ m/s. Sanity of units: kg, s, m, Pa all SI, so the output is genuinely m/s.

Worked example Example 11 — Cell I (exam twist: gauge vs absolute pressure trap)

"An engine gives $v_{e} = 3200$ m/s, $A_{e} = 0.5 m^{2}$ , $\overset{m}{˙} = 120$ kg/s. A gauge reads the exit pressure as $+ 15$ kPa gauge while the local absolute air pressure is $85$ kPa. Find $c$ ."

Forecast: the "+15 kPa gauge" is the trap — gauge pressure already means "above the surrounding air." So the exit is $15$ kPa above the $85$ kPa outside: $Δ P = + 15$ kPa directly. $c$ should be a little above 3200.

Interpret the gauge reading. Gauge pressure = absolute minus ambient, so a reading of $+ 15$ kPa gauge means $P_{e} - P_{a} = 15$ kPa is exactly $Δ P$ already. No conversion to absolute is needed — only the difference enters the formula. Why this step? The formula uses $P_{e} - P_{a}$ ; if a gauge already reports that difference, use it as-is. (If instead the problem gave two absolute values, we would subtract them.)
$Δ P = 15 kPa = 15000$ Pa. Why this step? Convert kPa to Pa for SI consistency before dividing — a second common exam trap (mixing kPa into a Pa formula).
Pressure term $= \frac{15000 \times 0.5}{120} = \frac{7500}{120} = 62.5$ m/s. Why this step? Standard force-÷-mass-flow conversion.
$c = 3200 + 62.5 = 3262.5$ m/s. Why this step? Add the positive boost to the bare gas speed.

Verify: $c > v_{e}$ as forecast ✓. Guard against the trap: had we wrongly added the $85$ kPa ambient on top of the gauge reading, we would have used $Δ P = 100$ kPa and blown the answer up to $\approx 3617$ m/s — wrong. Only the difference $Δ P$ ever enters; see Nozzle Expansion (Under/Over/Optimal).

Recall

Recall Which cell does each situation hit?

Sea-level firing with $P_{e} < P_{a}$ ::: Cell C — overexpanded, $c < v_{e}$ . Same engine climbing to space ::: Cells D and E — $P_{a} \to 0$ , $c$ rises to its max. A gauge already reading $P_{e} - P_{a}$ ::: Cell I — use it directly as $Δ P$ , no absolute conversion. Shrinking $A_{e}$ toward zero ::: Cell F — pressure term dies, $c \to v_{e}$ regardless of $Δ P$ sign. Throttling $\overset{m}{˙}$ toward zero ::: Cell J — pressure term blows up, $c \to \infty$ (idealisation). Opening $\overset{m}{˙}$ very wide ::: Cell J — pressure term dilutes to zero, $c \to v_{e}$ . Gas barely moving, $v_{e} \to 0$ ::: Cell K — $c$ equals the pressure term alone. Given $c$ , hunting the raw gas speed ::: Cell G — $v_{e} = c - Δ P A_{e} / \overset{m}{˙}$ .

Connections

Effective exhaust velocity $c$ (parent) — the formula these examples exercise.
Specific Impulse — divide any $c$ here by $g_{0}$ to get $I_{s p}$ .
Tsiolkovsky Rocket Equation — feed these $c$ values into $Δ v = c ln (m_{0} / m_{f})$ .
Nozzle Expansion (Under/Over/Optimal) — names cells A/B/C.
Atmospheric Pressure vs Altitude — drives cells D and E.
Conservation of Momentum — the momentum-thrust half behind $v_{e}$ .