Thrust coefficient C_F = F - (P_c A - ) — derivation: Worked examples

This page is the exhaustive drill-book for the thrust-coefficient derivation. We take the master formula and fire it through every case class: perfectly matched nozzles, over-expanded ones, under-expanded ones, the vacuum limit, the infinite-expansion ceiling, degenerate zero-pressure inputs, a word problem, and an exam twist that hides a trap.

Before any numbers, meet the one formula every example below leans on.

Formula The master formula (from the parent derivation)

$C_{F} = momentum term C_{F, mom} \frac{2 γ ^{2}}{γ - 1} (\frac{2}{γ + 1})^{\frac{γ + 1}{γ - 1}} [1 - (\frac{P _{e}}{P _{c}})^{\frac{γ - 1}{γ}}] + pressure term \frac{P _{e} - P _{a}}{P _{c}} \frac{A _{e}}{A ^{*}}$

$γ$ = ratio of specific heats (a pure number, e.g. $1.2$ ).
$P_{c}$ = chamber (stagnation) pressure, $P_{e}$ = nozzle exit pressure, $P_{a}$ = ambient pressure.
$A^{*}$ = throat area (the narrowest cross-section of the nozzle, in m², where the flow is choked at Mach 1 — see Choked Flow and the Throat).
$A_{e}$ = nozzle exit area (m²); the ratio $A_{e} / A^{*}$ is the expansion ratio (exit area ÷ throat area), see Nozzle Expansion Ratio Ae-over-Astar.
$C_{F}$ multiplies the reference force $P_{c} A^{*}$ to give thrust: $F = C_{F} P_{c} A^{*}$ .
$C_{F, m a x}$ = the largest possible momentum term, reached when the gas expands infinitely ( $P_{e} \to 0$ ); it equals $K$ and depends only on $γ$ . We meet it in Cell E.

Figure — Thrust coefficient C_F = F - (P_c A - ) — derivation

In the figure above, the left panel is the nozzle: the throat $A^{*}$ is the pinch-point (violet), the exit $A_{e}$ is the wide mouth (orange). The right panel stacks the three see-saw states — read the sign of $(P_{e} - P_{a})$ straight off which pan is lower.

The scenario matrix

Here is every distinct case class this topic can throw at you. Each worked example below is tagged with the cell it fills.

#	Case class	Distinguishing condition	Filled by
A	Perfectly matched	$P_{e} = P_{a}$ → pressure term $= 0$	Ex 1
B	Under-expanded	$P_{e} > P_{a}$ → pressure term $> 0$	Ex 2
C	Over-expanded	$P_{e} < P_{a}$ → pressure term $< 0$	Ex 3
D	Vacuum limit	$P_{a} = 0$ → pressure term always $> 0$	Ex 4
E	Infinite-expansion ceiling	$P_{e} \to 0$ → $C_{F} \to C_{F, m a x}$	Ex 5
F	Back-out from data	given $F$ , find $C_{F}$	Ex 6
G	Same engine, two altitudes	sign of pressure term flips	Ex 7
H	Word problem (design)	choose $A_{e} / A^{*}$ for a target	Ex 8
I	Exam twist (hidden trap)	student uses $T_{c}$ — it must vanish	Ex 9

Throughout we use $γ = 1.2$ (typical hot rocket gas) unless stated. Let us first nail the two building-block numbers we will reuse.

Cell A — Perfectly matched ( $P_{e} = P_{a}$ )

Worked example Ex 1 — matched sea-level nozzle

$γ = 1.2$ , $P_{c} = 7$ MPa, $P_{e} = P_{a} = 0.1$ MPa, $A_{e} / A^{*} = 25$ . Find $C_{F}$ and thrust with $A^{*} = 0.01$ m².

Forecast: the pressure term is $(P_{e} - P_{a}) = 0$ , so $C_{F}$ should equal the momentum term alone — guess it lands near $1.5$ – $1.6$ .

Pressure ratio power. $(\frac{P _{e}}{P _{c}})^{n} = (0.1/7)^{0.1667} = (0.01429)^{0.1667} = 0.4933$ . Why this step? This bracket measures how far the gas expanded; smaller $P_{e} / P_{c}$ means more expansion.
Momentum term. $C_{F, mom} = 5.06 (1 - 0.4933) = 5.06 \times 0.5067 = 2.564 = 1.601$ . Why this step? This is the directed-jet contribution set only by $γ$ and $P_{e} / P_{c}$ .
Pressure term. $\frac{P _{e} - P _{a}}{P _{c}} \frac{A _{e}}{A ^{*}} = \frac{0}{7} \times 25 = 0$ . Why this step? Matched nozzle → the exit gas and atmosphere are in balance, nothing extra to add.
Total. $C_{F} = 1.601 + 0 = 1.601$ . Thrust $F = 1.601 \times 7 \times 1 0^{6} \times 0.01 = 112, 070$ N $\approx 112$ kN.

Verify: units — $C_{F}$ is dimensionless; $P_{c} A^{*}$ = Pa·m² = N. $1.601 \times 7 MPa \times 0.01$ m² gives $\sim 1.1 \times 1 0^{5}$ N ✓. Sanity: $C_{F}$ sits inside the typical $1.3$ – $2.0$ band ✓.

Cell B — Under-expanded ( $P_{e} > P_{a}$ )

Worked example Ex 2 — high-altitude nozzle, exit still over-pressured

Same engine: $γ = 1.2$ , $P_{c} = 7$ MPa, $P_{e} = 0.1$ MPa, $A_{e} / A^{*} = 25$ , but now flying where $P_{a} = 0.04$ MPa.

Forecast: $P_{e} > P_{a}$ , so pressure term is positive → $C_{F}$ should beat the matched value $1.601$ .

Momentum term unchanged. $P_{e} / P_{c}$ is the same as Ex 1, so $C_{F, mom} = 1.601$ . Why this step? The momentum term never sees $P_{a}$ — only the exit-to-chamber ratio.
Pressure term. $\frac{0.1 - 0.04}{7} \times 25 = \frac{0.06}{7} \times 25 = 0.2143$ . Why this step? The exit gas is still pushing harder than ambient over area $A_{e}$ → extra forward force.
Total. $C_{F} = 1.601 + 0.214 = 1.815$ .

Verify: the pressure term is positive as predicted, and $1.815 > 1.601$ ✓. Because $P_{a}$ dropped from $0.1$ to $0.04$ MPa, we gained thrust with the identical nozzle — exactly what the Thrust Equation promises for rising altitude.

Cell C — Over-expanded ( $P_{e} < P_{a}$ )

Worked example Ex 3 — same nozzle forced down to sea level, but over-expanded

A big vacuum-style nozzle: $γ = 1.2$ , $P_{c} = 6$ MPa, $P_{e} = 0.05$ MPa, $A_{e} / A^{*} = 30$ , tested at sea level $P_{a} = 0.1$ MPa.

Forecast: $P_{e} < P_{a}$ — atmosphere pushes back — so the pressure term goes negative and drags $C_{F}$ below its momentum term.

Momentum term. $(0.05/6)^{0.1667} = (0.008333)^{0.1667} = 0.4500$ . $C_{F, mom} = 5.06 (1 - 0.4500) = 5.06 \times 0.5500 = 2.783 = 1.668$ . Why this step? Deeper expansion (smaller $P_{e} / P_{c}$ ) than Ex 1, so a slightly bigger momentum term.
Pressure term. $\frac{0.05 - 0.1}{6} \times 30 = \frac{- 0.05}{6} \times 30 = - 0.25$ . Why this step? Negative bracket = ambient pressure wins over the weak exit pressure across the big exit area → a real thrust loss. See Over- and Under-Expanded Nozzles.
Total. $C_{F} = 1.668 - 0.25 = 1.418$ .

Verify: sign of pressure term is negative, and $1.418 < 1.668$ ✓. This is the classic over-expansion penalty — the giant vacuum bell is wasteful at sea level.

Cell D — Vacuum limit ( $P_{a} = 0$ )

Worked example Ex 4 — the same big nozzle now in space

Take the Ex 3 engine into vacuum: $γ = 1.2$ , $P_{c} = 6$ MPa, $P_{e} = 0.05$ MPa, $A_{e} / A^{*} = 30$ , $P_{a} = 0$ .

Forecast: with $P_{a} = 0$ the pressure term is $\frac{P _{e}}{P _{c}} \frac{A _{e}}{A ^{*}}$ , always positive — this should be the best $C_{F}$ the nozzle can reach at this $P_{e}$ .

Momentum term. Recompute (same $P_{e} / P_{c}$ as Ex 3): $(0.05/6)^{0.1667} = 0.4500$ , so $C_{F, mom} = 5.06 (1 - 0.4500) = 1.668$ . Why this step? Momentum term ignores $P_{a}$ ; only $P_{e} / P_{c}$ matters, so it matches Ex 3.
Pressure term. $\frac{0.05 - 0}{6} \times 30 = \frac{0.05}{6} \times 30 = 0.25$ . Why this step? No atmosphere pushes back, so the entire exit pressure over $A_{e}$ becomes pure bonus thrust.
Total. $C_{F, vac} = 1.668 + 0.25 = 1.918$ .

Verify: compare with Ex 3 — the same nozzle went from $1.418$ (sea level) to $1.918$ (vacuum), a jump of exactly $0.50$ , which equals $\frac{P _{a}}{P _{c}} \frac{A _{e}}{A ^{*}} = \frac{0.1}{6} \times 30 = 0.50$ ✓. That is why space stages carry huge expansion ratios.

Cell E — Infinite-expansion ceiling ( $P_{e} \to 0$ )

Worked example Ex 5 — the hard maximum set by

$γ$ Push expansion to the theoretical limit: $P_{e} \to 0$ , still $γ = 1.2$ . Find $C_{F, m a x}$ .

Forecast: as $P_{e} \to 0$ , the bracket $[1 - (P_{e} / P_{c})^{n}] \to 1$ , so $C_{F, mom}$ hits its ceiling $K$ . Guess $5.06 \approx 2.25$ .

Bracket limit. $(P_{e} / P_{c})^{n} \to 0^{0.1667} = 0$ , so $[1 - 0] = 1$ . Why this step? Infinite expansion extracts all available enthalpy — nothing left in the pressure ratio.
Ceiling value. $C_{F, m a x} = K = 5.06 = 2.249$ . Why this step? This depends only on $γ$ — a fundamental limit no nozzle can beat, tied to Isentropic Nozzle Flow.
Interpret. Real engines with $P_{a} = 0$ approach but never reach $2.25$ because a finite nozzle leaves $P_{e} > 0$ .

Verify: plug the Ex-4 numbers — its $C_{F, vac} = 1.918$ is safely below $2.249$ ✓. Take a bigger $A_{e} / A^{*}$ (smaller $P_{e}$ ) and $C_{F}$ creeps toward $2.25$ but never crosses it, exactly as the ceiling demands.

The figure below shows how $C_{F}$ climbs with expansion ratio in vacuum, flattening toward this ceiling.

In that plot the horizontal axis is the expansion ratio $A_{e} / A^{*}$ and the vertical axis is $C_{F}$ . The magenta curve is the full vacuum $C_{F}$ , the dashed violet curve is the momentum term alone, and the dotted orange line is the ceiling $K = 2.25$ — notice the magenta curve bends toward it but never touches.

Cell F — Back-out from measured data

Cell G — Same engine, two altitudes (sign flip)

Worked example Ex 7 — watch the pressure term flip sign

Engine: $γ = 1.2$ , $P_{c} = 6$ MPa, $P_{e} = 0.05$ MPa fixed, $A_{e} / A^{*} = 30$ . Compute $C_{F}$ at sea level and in vacuum, and the change.

Forecast: the momentum term is identical at both altitudes, so the whole difference is the pressure term. From sea level ( $P_{a} = 0.1$ MPa) to vacuum ( $P_{a} = 0$ ), $P_{a}$ drops by $0.1$ MPa; the swing should equal $\frac{Δ P _{a}}{P _{c}} \frac{A _{e}}{A ^{*}} = \frac{0.1}{6} \times 30 = 0.50$ . Guess a rise of exactly $0.50$ .

Momentum term. Recompute: $(0.05/6)^{0.1667} = 0.4500$ , so $C_{F, mom} = 5.06 (1 - 0.4500) = 1.668$ . Why this step? Same $P_{e} / P_{c}$ as Ex 3/Ex 4 → same momentum term; it is altitude-blind.
Sea level ( $P_{a} = 0.1$ MPa). Pressure term $= \frac{0.05 - 0.1}{6} \times 30 = - 0.25$ → $C_{F} = 1.668 - 0.25 = 1.418$ . Why this step? Over-expanded on the ground — atmosphere pushes back.
Vacuum ( $P_{a} = 0$ ). Pressure term $= \frac{0.05}{6} \times 30 = + 0.25$ → $C_{F} = 1.668 + 0.25 = 1.918$ . Why this step? No back-pressure — full exit pressure becomes thrust.
Swing. $Δ C_{F} = 1.918 - 1.418 = 0.50$ . Why this step? The momentum term cancels, so the entire difference is $\frac{Δ P _{a}}{P _{c}} \frac{A _{e}}{A ^{*}}$ .

Verify: $\frac{0.1}{6} \times 30 = 0.50$ matches the swing exactly ✓. This is the same engine as Ex 3/Ex 4, so the numbers cross-check.

Cell H — Word problem (design choice)

Worked example Ex 8 — size a vacuum nozzle to a thrust target

Mission spec: deliver $F = 150$ kN in vacuum ( $P_{a} = 0$ ) with $P_{c} = 7$ MPa, $A^{*} = 0.01$ m². Ground rules give $γ = 1.2$ and a matched-at- $P_{e} = 0.1$ MPa design so $C_{F, mom} = 1.601$ (from Ex 1). What expansion ratio $A_{e} / A^{*}$ do we need?

Forecast: we need $C_{F} = F / (P_{c} A^{*})$ ; then subtract the momentum term to isolate what the pressure term (hence area ratio) must supply.

Required $C_{F}$ . $C_{F} = \frac{150 , 000}{7 \times 1 0 ^{6} \times 0.01} = \frac{150 , 000}{70 , 000} = 2.143$ . Why this step? The definition sets the total coefficient we must hit.
Required pressure term. $C_{F} - C_{F, mom} = 2.143 - 1.601 = 0.542$ . Why this step? Whatever the momentum term can't provide, the vacuum pressure term must.
Solve for area ratio. Vacuum: $\frac{P _{e}}{P _{c}} \frac{A _{e}}{A ^{*}} = 0.542$ , so $\frac{A _{e}}{A ^{*}} = \frac{0.542}{0.1/7} = 0.542 \times 70 = 37.9$ . Why this step? $P_{e} / P_{c} = 0.1/7$ is fixed by the matched design, so only $A_{e} / A^{*}$ is free.

Verify: rebuild $C_{F} = 1.601 + \frac{0.1}{7} \times 37.9 = 1.601 + 0.542 = 2.143$ ✓, and $F = 2.143 \times 70, 000 = 150, 000$ N ✓. An expansion ratio near $38$ is realistic for an upper stage — see Specific Impulse Isp for why bigger ratios also raise $I_{s p}$ in vacuum.

Cell I — Exam twist (the $T_{c}$ trap)

Worked example Ex 9 — the deliberately useless temperature

Exam gives: $γ = 1.2$ , $P_{c} = 7$ MPa, $P_{e} = P_{a} = 0.1$ MPa, $A_{e} / A^{*} = 25$ , and $T_{c} = 3200$ K, $R = 350$ J/(kg·K). Here $T_{c}$ is the chamber temperature and $R$ is the specific gas constant — the gas constant per kilogram of the exhaust, with units J/(kg·K). Find $C_{F}$ . (Same physical case as Ex 1, but stuffed with extra numbers.)

Forecast: the trap is that $T_{c}$ and $R$ tempt you to compute exit velocity. But $C_{F}$ is independent of $T_{c}$ and $R$ — they cancel in the derivation. Answer should equal Ex 1's $1.601$ .

Spot the cancellation. In the parent's Step 4, $\overset{m}{˙} \propto \frac{1}{R T _{c}}$ and $u_{e} \propto R T_{c}$ , so $\overset{m}{˙} u_{e} / (P_{c} A^{*})$ has $R T_{c}$ cancel. Why this step? $C_{F}$ measures nozzle geometry and ratios only — never absolute temperature or the specific gas constant.
Ignore $T_{c}, R$ ; use the master formula. Identical to Ex 1: $(0.1/7)^{0.1667} = 0.4933$ , so $C_{F, mom} = 5.06 (1 - 0.4933) = 1.601$ , and the pressure term $= 0$ (matched). Why this step? Same $γ$ , $P_{e} / P_{c}$ , $A_{e} / A^{*}$ , and matched exit → same coefficient.
Total. $C_{F} = 1.601 + 0 = 1.601$ . Why this step? The extra data $T_{c}$ and $R$ never entered — proving they are decoration for a $C_{F}$ question.

Verify: unchanged from Ex 1 despite the added $T_{c} = 3200$ K and $R = 350$ ✓. If you had wrongly "used" them you would compute an exit velocity $u_{e} \approx \frac{2 γ R T _{c}}{γ - 1} [1 - (P_{e} / P_{c})^{n}] = \frac{2 ( 1.2 ) ( 350 ) ( 3200 )}{0.2} (0.5067) \approx 2860$ m/s — a real, meaningful number, but it belongs to $c^{*}$ and $I_{s p}$ , not $C_{F}$ .

Recall Quick self-test

Matched nozzle pressure term equals what? ::: Zero (since $P_{e} = P_{a}$ ). Over-expanded means which inequality, and what sign of pressure term? ::: $P_{e} < P_{a}$ , negative term (thrust penalty). Going from sea level to vacuum, the $C_{F}$ gain equals? ::: $\frac{Δ P _{a}}{P _{c}} \frac{A _{e}}{A ^{*}}$ . The infinite-expansion ceiling $C_{F, m a x}$ depends only on? ::: $γ$ (for $γ = 1.2$ it is $\approx 2.25$ ). Does $C_{F}$ depend on chamber temperature $T_{c}$ or specific gas constant $R$ ? ::: No — both cancel; they live in $\overset{m}{˙}$ and $c^{*}$ .