Optimum expansion — P_e = P_a for maximum thrust: Exercises

This page is a self-test. Read each problem, try it on paper, THEN open the collapsible solution. Every symbol used here is built in the parent topic; if a piece feels unfamiliar, revisit Thrust equation derivation and Isentropic flow relations first.

Level 1 — Recognition

L1·Q1 — Name the regime

A nozzle exits gas at $P_{e} = 0.4 bar$ . The rocket sits at sea level where $P_{a} = 1.0 bar$ . Is this nozzle under-expanded, optimally expanded, or over-expanded?

Recall Solution

Compare the two pressures. Here $P_{e} = 0.4 bar < P_{a} = 1.0 bar$ .

$P_{e} > P_{a}$ → under-expanded
$P_{e} = P_{a}$ → optimal
$P_{e} < P_{a}$ → over-expanded ✔

The gas has already dropped below the outside pressure, so the atmosphere squeezes back on the exhaust. Answer: over-expanded.

L1·Q2 — Which term vanishes?

Write the thrust equation, then state which single term becomes zero when a nozzle is perfectly (optimally) expanded, and what the equation reduces to.

Recall Solution

$F = \overset{m}{˙} v_{e} + (P_{e} - P_{a}) A_{e}$ Optimal means $P_{e} = P_{a}$ , so $(P_{e} - P_{a}) = 0$ . The pressure term $(P_{e} - P_{a}) A_{e}$ vanishes. $F_{opt} = \overset{m}{˙} v_{e}$ All thrust now comes purely from the momentum of the exhaust — the cleanest possible conversion.

Level 2 — Application

L2·Q1 — Compute thrust with a pressure term

A rocket has $\overset{m}{˙} = 250 kg/s$ , $v_{e} = 3000 m/s$ , $A_{e} = 1.5 m^{2}$ , and $P_{e} = 0.5 bar$ . Find the thrust at an altitude where $P_{a} = 0.3 bar$ .

Recall Solution

Convert pressures: $P_{e} = 0.5 \times 1 0^{5} = 50, 000 Pa$ , $P_{a} = 0.3 \times 1 0^{5} = 30, 000 Pa$ . $F = \overset{m}{˙} v_{e} + (P_{e} - P_{a}) A_{e}$ $F = (250) (3000) + (50, 000 - 30, 000) (1.5)$ $F = 750, 000 + (20, 000) (1.5) = 750, 000 + 30, 000$ $F = 780, 000 N = 780 kN$ Here $P_{e} > P_{a}$ : the nozzle is under-expanded, so the pressure term is positive and adds a little thrust.

L2·Q2 — Same engine, sea level

Take the identical engine from L2·Q1 ( $\overset{m}{˙} = 250$ , $v_{e} = 3000$ , $A_{e} = 1.5$ , $P_{e} = 0.5 bar$ ) and now fire it at sea level, $P_{a} = 1.0 bar$ . Find the thrust and say what happened.

Recall Solution

$P_{e} = 50, 000 Pa$ , $P_{a} = 1.0 \times 1 0^{5} = 100, 000 Pa$ . $F = (250) (3000) + (50, 000 - 100, 000) (1.5)$ $F = 750, 000 + (- 50, 000) (1.5) = 750, 000 - 75, 000$ $F = 675, 000 N = 675 kN$ Now $P_{e} < P_{a}$ : the nozzle is over-expanded, the pressure term is negative, and thrust drops by $75 kN$ compared to the momentum-only value. See Altitude compensation for why one fixed nozzle can't win everywhere.

Level 3 — Analysis

L3·Q1 — Percentage thrust loss

An engine designed for vacuum has $\overset{m}{˙} = 400 kg/s$ , $v_{e} = 3200 m/s$ , $A_{e} = 3.0 m^{2}$ , $P_{e} = 0.08 bar$ . Compute its vacuum thrust and its sea-level thrust ( $P_{a} = 1.01 bar$ ), then the percentage thrust lost at sea level.

Recall Solution

Momentum term (same everywhere): $\overset{m}{˙} v_{e} = (400) (3200) = 1, 280, 000 N$ .

Vacuum ( $P_{a} = 0$ ), $P_{e} = 0.08 \times 1 0^{5} = 8000 Pa$ : $F_{vac} = 1, 280, 000 + (8000 - 0) (3.0) = 1, 280, 000 + 24, 000 = 1, 304, 000 N$

Sea level, $P_{a} = 1.01 \times 1 0^{5} = 101, 000 Pa$ : $F_{SL} = 1, 280, 000 + (8000 - 101, 000) (3.0) = 1, 280, 000 + (- 93, 000) (3.0)$ $F_{SL} = 1, 280, 000 - 279, 000 = 1, 001, 000 N$

Loss: $\frac{F _{vac} - F _{SL}}{F _{vac}} \times 100% = \frac{303 , 000}{1 , 304 , 000} \times 100% \approx 23.2%$ Nearly a quarter of the thrust vanishes at sea level — the classic penalty of a big-area vacuum nozzle used low in the atmosphere.

L3·Q2 — Find the altitude of optimum operation

The engine of L3·Q1 has $P_{e} = 0.08 bar$ (fixed by geometry). At what ambient pressure $P_{a}$ is this nozzle perfectly expanded, and is that a low or high altitude?

Recall Solution

Optimum needs $P_{e} = P_{a}$ , so $P_{a} = 0.08 bar = 8000 Pa$ . Sea level is $\approx 1.01 bar$ ; ambient pressure falls as you climb, so $0.08 bar$ is reached only at high altitude (roughly $17$ – $18 km$ up, where air is thin). That is exactly why this nozzle is over-expanded at sea level: it is waiting for the thin air it was designed for.

Level 4 — Synthesis

Before starting this level, study Figure 1 below. It plots thrust $F$ (vertical axis) against exit pressure $P_{e}$ (horizontal axis) as we gradually open up the nozzle. The horizontal axis reads exit pressure from left = low $P_{e}$ (large area, over-expanded) to right = high $P_{e}$ (small area, under-expanded). The yellow dashed line marks $P_{e} = P_{a}$ ; the yellow dot on the blue curve is the single peak. Notice the curve rises from the left, tops out at the dashed line, and falls again to the right — that is the whole story of this level.

Figure — Optimum expansion — P_e = P_a for maximum thrust

L4·Q1 — Prove the peak is at $P_{e} = P_{a}$ from the slope

Show, from the thrust equation and Euler's relation for steady flow, that the slope of thrust with respect to exit pressure is $\frac{d F}{d P _{e}} = (P_{e} - P_{a}) \frac{d A _{e}}{d P _{e}}$ , then explain why the only interior maximum is at $P_{e} = P_{a}$ , and locate the under-expanded regime on Figure 1.

Recall Solution

Recap of the cancellation (so it is self-contained). Differentiate $F = \overset{m}{˙} v_{e} + (P_{e} - P_{a}) A_{e}$ with respect to $P_{e}$ (everything varies along the nozzle-opening family): $\frac{d F}{d P _{e}} = \overset{m}{˙} \frac{d v _{e}}{d P _{e}} + A_{e} + (P_{e} - P_{a}) \frac{d A _{e}}{d P _{e}} .$ Euler's momentum relation for steady 1-D flow gives $\overset{m}{˙} d v_{e} = - A_{e} d P_{e}$ , hence $\overset{m}{˙} \frac{d v _{e}}{d P _{e}} = - A_{e}$ . Substituting: $\frac{d F}{d P _{e}} = - A_{e} + A_{e} + (P_{e} - P_{a}) \frac{d A _{e}}{d P _{e}} = (P_{e} - P_{a}) \frac{d A _{e}}{d P _{e}} .$ The two $A_{e}$ terms cancel exactly — that is the heart of the result.

Where the peak is. Setting the slope to zero: $(P_{e} - P_{a}) \frac{d A _{e}}{d P _{e}} = 0$ . Opening the nozzle always changes area, so $\frac{d A _{e}}{d P _{e}} \neq = 0$ . The only way the product is zero is $P_{e} - P_{a} = 0$ , i.e. $P_{e} = P_{a}$ . That single condition is the yellow dot (flat top) in Figure 1.

Direction check (map it onto Figure 1). As we open the nozzle, $A_{e}$ grows and $P_{e}$ drops — so opening the nozzle moves us leftward along Figure 1's horizontal axis, and $\frac{d A _{e}}{d P _{e}} < 0$ .

Right of the peak ( $P_{e} > P_{a}$ , under-expanded): $(P_{e} - P_{a}) > 0$ times a negative $\frac{d A _{e}}{d P _{e}}$ gives $\frac{d F}{d P _{e}} < 0$ . Because $P_{e}$ decreases as we open the nozzle, moving left (toward the peak) raises thrust — the curve climbs toward the dashed line. You are still below the peak; opening the nozzle helps.
Left of the peak ( $P_{e} < P_{a}$ , over-expanded): $(P_{e} - P_{a}) < 0$ times a negative $\frac{d A _{e}}{d P _{e}}$ gives $\frac{d F}{d P _{e}} > 0$ ; opening further (lowering $P_{e}$ , going further left) now loses thrust — the curve falls away. You have overshot the peak.

Both slopes point back toward $P_{e} = P_{a}$ : it is a genuine maximum, exactly the yellow dot. See Nozzle expansion ratio for how $A_{e}$ and $P_{e}$ move together.

L4·Q2 — Optimum expansion ratio from isentropic flow

An engine has $P_{0} = 40 bar$ (chamber), $γ = 1.2$ , and must be optimum at $P_{a} = 0.5 bar$ . Using the isentropic exit-Mach relation $\frac{P _{e}}{P _{0}} = (1 + \frac{γ - 1}{2} M_{e}^{2})^{- \frac{γ}{γ - 1}},$ find the exit Mach number $M_{e}$ (defined above: exit speed ÷ local speed of sound). This feeds the Nozzle expansion ratio $ε$ .

Recall Solution

Optimum ⇒ $P_{e} = P_{a} = 0.5 bar$ , so the pressure ratio is $\frac{P _{e}}{P _{0}} = \frac{0.5}{40} = 0.0125.$ With $γ = 1.2$ : exponent $- \frac{γ}{γ - 1} = - \frac{1.2}{0.2} = - 6$ , and $\frac{γ - 1}{2} = 0.1$ . $0.0125 = (1 + 0.1 M_{e}^{2})^{- 6}$ Raise both sides to the power $- \frac{1}{6}$ to undo the $- 6$ power: $0.012 5^{- 1/6} = 1 + 0.1 M_{e}^{2}$ Compute the left side carefully with logarithms: $ln (0.0125) = - 4.382$ , so $- \frac{1}{6} ln (0.0125) = + 0.7304$ , and $e^{0.7304} = 2.076$ . Thus $2.076 = 1 + 0.1 M_{e}^{2} \Rightarrow 0.1 M_{e}^{2} = 1.076 \Rightarrow M_{e}^{2} = 10.76$ $M_{e} \approx 3.28$ A strongly supersonic exit ( $M_{e} \approx 3.3$ ), exactly what a well-expanded rocket nozzle produces.

Level 5 — Mastery

L5·Q1 — Best of two fixed nozzles for a whole climb

A booster spends its burn split between two conditions of equal duration:

Phase A (low altitude): $P_{a} = 0.9 bar$
Phase B (high altitude): $P_{a} = 0.1 bar$

Two candidate nozzles share $\overset{m}{˙} = 500 kg/s$ , $v_{e} = 3000 m/s$ but differ in exit design:

Nozzle X (sea-level design): $A_{e} = 2.0 m^{2}$ , $P_{e} = 0.9 bar$ .
Nozzle Y (vacuum design): $A_{e} = 5.0 m^{2}$ , $P_{e} = 0.1 bar$ .

Compute the average thrust $\frac{1}{2} (F_{A} + F_{B})$ for each nozzle and decide which booster is better over the full climb.

Recall Solution

Momentum term for both: $\overset{m}{˙} v_{e} = (500) (3000) = 1, 500, 000 N$ . Pressures in Pa: $P_{a}^{A} = 90, 000$ , $P_{a}^{B} = 10, 000$ ; $P_{e}^{X} = 90, 000$ , $P_{e}^{Y} = 10, 000$ .

Nozzle X ( $A_{e} = 2.0$ ):

Phase A: $F = 1, 500, 000 + (90, 000 - 90, 000) (2.0) = 1, 500, 000 N$ (optimum here!)
Phase B: $F = 1, 500, 000 + (90, 000 - 10, 000) (2.0) = 1, 500, 000 + 160, 000 = 1, 660, 000 N$
Average: $\frac{1}{2} (1, 500, 000 + 1, 660, 000) = 1, 580, 000 N$

Nozzle Y ( $A_{e} = 5.0$ ):

Phase A: $F = 1, 500, 000 + (10, 000 - 90, 000) (5.0) = 1, 500, 000 - 400, 000 = 1, 100, 000 N$ (badly over-expanded)
Phase B: $F = 1, 500, 000 + (10, 000 - 10, 000) (5.0) = 1, 500, 000 N$ (optimum here!)
Average: $\frac{1}{2} (1, 100, 000 + 1, 500, 000) = 1, 300, 000 N$

Nozzle X wins, $1, 580 kN$ vs $1, 300 kN$ average. The big vacuum nozzle Y suffers a huge $400 kN$ over-expansion penalty low down (large $A_{e}$ multiplies the negative pressure gap). This is precisely the trade-off that motivates Altitude compensation and Rocket staging: use a compact nozzle low, a large nozzle high.

L5·Q2 — Where flow separation kills the vacuum nozzle

For Nozzle Y in Phase A ( $P_{e} = 0.1 bar$ , $P_{a} = 0.9 bar$ ), the pressure ratio $P_{e} / P_{a} = 0.111$ . A common rule of thumb (Summerfield criterion) says the exhaust separates from the wall when $P_{e} / P_{a} ≲ 0.35$ – $0.4$ . Does Nozzle Y separate at low altitude, and what does that mean physically? (Link: Supersonic flow separation.)

Recall Solution

Compute the ratio: $\frac{P _{e}}{P _{a}} = \frac{0.1}{0.9} = 0.111$ . Since $0.111 < 0.35$ , the flow separates from the nozzle wall well before the exit.

Physically: the ambient pressure is so much higher than the exit pressure that it pushes back up into the diverging section, peeling the supersonic jet off the wall. The gas then effectively exits at a smaller area with a higher pressure than the ideal $0.1 bar$ , and the flow becomes unsteady (side-loads, buffeting). The clean thrust number computed in L5·Q1 is actually an optimistic bound — real over-expanded losses can be worse and can even damage the nozzle. This is why huge vacuum nozzles physically cannot be fired at sea level.

Recall Quick self-quiz (cover the answers)

Optimum thrust condition ::: $P_{e} = P_{a}$ Nozzle with $P_{e} > P_{a}$ is called ::: under-expanded Nozzle with $P_{e} < P_{a}$ is called ::: over-expanded Term that vanishes at optimum ::: the pressure term $(P_{e} - P_{a}) A_{e}$ Meaning of exit Mach $M_{e}$ ::: exit speed divided by the local speed of sound Why a vacuum nozzle is bad at sea level ::: $P_{e} ≪ P_{a}$ ⇒ large negative pressure thrust and flow separation