Under-expanded nozzle — Prandtl-Meyer expansion, efficiency loss

What Is an Under-Expanded Nozzle?

Contrast with other cases:

Perfectly expanded: $p_e = p_\infty$ → no waves, maximum axial thrust
Over-expanded: $p_e < p_\infty$ → oblique shock waves compress the flow, also losing efficiency (covered in 3.3.14-Over-expanded-nozzle-shock-diamonds)
Under-expanded: $p_e > p_\infty$ → Prandtl-Meyer expansion fans turn flow outward

Why Prandtl-Meyer Expansion Waves Form

First Principles: Pressure Imbalance at the Exit Plane

At the nozzle exit lip, the exhaust at pressure $p_e$ meets ambient air at $p_\infty$ . If $p_e > p_\infty$ , the high-pressure exhaust "sees" a sudden drop in external resistance. To equilibrate, it must expand.

But why expansion waves and not a uniform expansion?

In supersonic flow, information cannot travel upstream (disturbances propagate along Mach waves at angle $\mu = \arcsin(1/M)$ ). So the expansion must happen through a centered expansion fan—a continuous family of Mach waves emanating from the nozzle lip.

Formula Prandtl-Meyer Function

The Prandtl-Meyer function $\nu(M)$ relates Mach number to the cumulative flow turning angle through an isentropic expansion: $\nu(M) = \sqrt{\frac{\gamma+1}{\gamma-1}} \arctan\left(\sqrt{\frac{\gamma-1}{\gamma+1}(M^2-1)}\right) - \arctan\left(\sqrt{M^2-1}\right)$ What it means: If flow at Mach $M_1$ turns through angle $\theta$ , the new Mach number $M_2$ satisfies: $\nu(M_2) = \nu(M_1) + \theta$

Derivation from first principles:

For an isentropic turn, total pressure and total temperature are constant.
The Mach wave angle is $\mu = \arcsin(1/M)$ . Each infinitesimal wave turns the flow by $d\theta$ .
The Mach number change $dM$ and turn angle $d\theta$ are related by geometry and the isentropic relations.
Starting from the area-velocity relation in isentropic flow: $\frac{dA}{A} = \left(\frac{1}{M^2} - 1\right) \frac{dV}{V}$ For a turn (no area change in the fan itself), we use the velocity hodograph and momentum balance.
Integrating the differential relation (derived from oblique shock relations with $p_2/p_1 \to 1$ ): $d\theta = \frac{\sqrt{M^2-1}}{1 + \frac{\gamma-1}{2}M^2}\,\frac{dM}{M}$
Integrate from $M=1$ (where $\theta=0$ , define $\nu(1)=0$ ) to arbitrary $M$ : $\nu(M) = \int_1^M \frac{\sqrt{M^2-1}}{M\left(1 + \frac{\gamma-1}{2}M^2\right)}\,dM$
Using trig substitution $M = \sec\phi$ and simplifying yields the closed form above.

Why this step? The Prandtl-Meyer function encapsulates the maximum turning angle possible in supersonic flow before the flow "straightens out" at infinite Mach. For $\gamma=1.4$ , $\nu_{\max} \approx 130.5°$ .

Physical Process at Nozzle Exit

At the exit lip ( $x=x_e$ , $r=r_e$ ): Flow exits at Mach $M_e$ , pressure $p_e > p_\infty$ .
Expansion fan nucleates: Mach waves spread from the lip, each turning the flow slightly outward.
Flow accelerates and turns: As the exhaust passes through the fan, Mach number increases to $M_\text{plume}$ , pressure drops to $p_\infty$ , and velocity vector rotates by angle $\theta_\text{turn}$ .
Beyond the fan: Flow is uniform again at higher Mach, lower pressure, but now has a radial velocity component.

The total turning angle $\theta_\text{turn}$ satisfies: $\nu(M_\text{plume}) = \nu(M_e) + \theta_\text{turn}$ where $M_\text{plume}$ is found from the isentropic relation (total pressure $p_0$ is conserved): $\frac{p_e}{p_\infty} = \frac{\left(1 + \frac{\gamma-1}{2}M_\text{plume}^2\right)^{\gamma/(\gamma-1)}}{\left(1 + \frac{\gamma-1}{2}M_e^2\right)^{\gamma/(\gamma-1)}}$ i.e. the plume expands until its static pressure matches $p_\infty$ , reaching a higher Mach $M_\text{plume} > M_e$ .

Efficiency Loss: Thrust Reduction

Axial vs. Radial Momentum

The thrust coefficient measures how much of the ideal momentum flux is converted to axial thrust. For an under-expanded nozzle:

$C_F = \frac{F}{\dot{m}\, v_e^\text{ideal}} < C_{F,\text{optimum}}$

Why does thrust drop?

Momentum redirection: The expansion fan turns velocity outward by $\theta_\text{turn}$ . Only the axial component $v \cos\theta_\text{turn}$ contributes to thrust.
No nozzle wall to react against: In a perfectly expanded nozzle, the entire expansion happens inside where walls redirect flow. Here, the outside expansion is "free" — no wall pushes back.

Formula Thrust of an Under-Expanded Nozzle

The standard rocket thrust equation, applied at the physical exit plane: $F = \dot{m}\, v_e + A_e\,(p_e - p_\infty)$ Derivation:

Apply the momentum theorem to a control volume bounding the exhaust up to the exit plane. Momentum flux out is $\dot{m} v_e$ ; the net pressure force on the exit plane is $A_e(p_e - p_\infty)$ .
When $p_e > p_\infty$ (under-expanded), the pressure term $A_e(p_e-p_\infty) > 0$ adds to thrust, but $v_e$ is lower than the fully-expanded exit velocity because the gas has not yet expanded to $p_\infty$ inside the nozzle.
The optimum occurs when $p_e = p_\infty$ : the pressure term vanishes but $v_e$ is maximal, and one can show $\partial F/\partial A_e = 0$ there — the perfectly-expanded case gives the largest $F$ .

The efficiency loss relative to the perfectly-expanded ideal is: $\Delta F = F_\text{opt} - F_\text{actual} > 0$ Physically, the external Prandtl-Meyer expansion turns momentum off-axis: the axial momentum in the plume is $\dot m\, v_\text{plume}\cos\theta_\text{turn}$ , and the factor $(1-\cos\theta_\text{turn})$ quantifies the fraction of the additional expansion momentum lost to the radial direction. This term grows with $p_e/p_\infty$ (larger $\theta_\text{turn}$ ), so bigger pressure mismatches cost more.

Practical numbers: If $p_e/p_\infty = 1.5$ and $M_e = 3$ , typical $\theta_\text{turn} \approx 3-5°$ , yielding $\lesssim 1\%$ thrust loss. Larger pressure ratios → larger angles → worse loss.

Worked Examples

Worked example Example 1: Sea-Level Under-Expansion

Given: Nozzle with exit Mach $M_e = 3.5$ , exit pressure $p_e = 1.5\text{ atm}$ , ambient $p_\infty = 1\text{ atm}$ , $\gamma = 1.3$ . Find: Plume Mach $M_\text{plume}$ , turning angle $\theta_\text{turn}$ , and efficiency loss.

Solution:

Check regime: $p_e/p_\infty = 1.5 > 1$ → under-expanded. Good.
Plume Mach from isentropic relation. Total pressure $p_0$ is conserved, so: $\frac{p_e}{p_\infty} = \frac{\left(1+\tfrac{\gamma-1}{2}M_\text{plume}^2\right)^{\gamma/(\gamma-1)}}{\left(1+\tfrac{\gamma-1}{2}M_e^2\right)^{\gamma/(\gamma-1)}}$ With $\gamma=1.3$ : exponent $=\gamma/(\gamma-1)=1.3/0.3=4.333$ , and $\tfrac{\gamma-1}{2}=0.15$ . $1+0.15 M_e^2 = 1+0.15(12.25)=2.8375$ . $1.5 = \left(\frac{1+0.15 M_\text{plume}^2}{2.8375}\right)^{4.333}$ Why this step? The plume must expand further (Mach rises, static pressure falls to $p_\infty$ ), so the numerator carries $M_\text{plume}$ . Solve: $1.5^{1/4.333} = 1.0964$ , so $1+0.15 M_\text{plume}^2 = 1.0964 \times 2.8375 = 3.111$ . $0.15 M_\text{plume}^2 = 2.111 \Rightarrow M_\text{plume}^2 = 14.07 \Rightarrow M_\text{plume} \approx 3.75$
Turning angle from Prandtl-Meyer ( $\gamma=1.3$ , $\sqrt{(\gamma+1)/(\gamma-1)}=\sqrt{2.3/0.3}=2.769$ ): $\nu(3.5) = 2.769\arctan\!\left(\tfrac{1}{2.769}\sqrt{12.25-1}\right) - \arctan\sqrt{11.25}$ $=2.769\arctan(1.2111) - \arctan(3.354) = 2.769(50.46°) - 73.42° = 66.3°$ $\nu(3.75) = 2.769\arctan\!\left(\tfrac{1}{2.769}\sqrt{14.06-1}\right) - \arctan\sqrt{13.06}$ $=2.769\arctan(1.3055) - \arctan(3.614) = 2.769(52.55°) - 74.53° = 71.0°$ $\theta_\text{turn} = 71.0° - 66.3° = 4.7°$
Thrust loss: $\cos(4.7°) \approx 0.9966$ , so the off-axis momentum loss on the extra expansion is $\sim 0.3\%$ . Mild under-expansion → small penalty.

Why this step? Modest pressure ratios (1.5:1) give small turning angles—under-expansion is not catastrophic here.

Worked example Example 2: Extreme Under-Expansion

Given: $p_e/p_\infty = 5$ , $M_e = 2.5$ , $\gamma = 1.4$ . Find: Plume Mach and turning angle.

Solution:

Exponent $\gamma/(\gamma-1)=1.4/0.4=3.5$ , and $\tfrac{\gamma-1}{2}=0.2$ . Then $1+0.2 M_e^2 = 1+0.2(6.25)=2.25$ . $5 = \left(\frac{1+0.2 M_\text{plume}^2}{2.25}\right)^{3.5}$ $5^{1/3.5} = 1.5849$ , so $1+0.2 M_\text{plume}^2 = 1.5849 \times 2.25 = 3.566$ . $0.2 M_\text{plume}^2 = 2.566 \Rightarrow M_\text{plume}^2 = 12.83 \Rightarrow M_\text{plume} \approx 3.58$
Prandtl-Meyer ( $\gamma=1.4$ , $\sqrt{2.4/0.4}=\sqrt6=2.449$ ): $\nu(2.5) = 2.449\arctan\!\left(\tfrac{1}{2.449}\sqrt{5.25}\right) - \arctan\sqrt{5.25}$ $=2.449\arctan(0.9354) - \arctan(2.291) = 2.449(43.09°) - 66.42° = 39.1°$ $\nu(3.58) = 2.449\arctan\!\left(\tfrac{1}{2.449}\sqrt{11.82}\right) - \arctan\sqrt{11.82}$ $=2.449\arctan(1.4038) - \arctan(3.438) = 2.449(54.53°) - 73.78° = 59.8°$ $\theta_\text{turn} = 59.8° - 39.1° = 20.7°$
Momentum loss: $\cos(20.7°) \approx 0.9354$ → $\sim 6.5\%$ off-axis loss on the additional expansion; net thrust penalty (after pressure-thrust partial compensation) is several percent. Why this matters: At $p_e/p_\infty = 5$ , the turning angle is large and losses are significant. Altitude-optimized nozzles suffer badly at sea level.

Common Mistakes

Active Recall Flashcards

#flashcards/physics

What does "under-expanded nozzle" mean?

Exit pressure

p_e

exceeds ambient

p_\infty

; exhaust expands outside the nozzle via Prandtl-Meyer waves, turning flow outward and losing axial thrust.

What is the Prandtl-Meyer function

\nu(M)

A function relating Mach number to cumulative isentropic turning angle in an expansion fan;

\nu(M_2) = \nu(M_1) + \theta

for a turn

\theta

Why do expansion waves form at the nozzle exit in under-expansion?

Pressure imbalance (

p_e > p_\infty

) forces expansion, but supersonic flow cannot adjust uniformly—disturbances propagate as Mach waves, forming a centered expansion fan.

How does under-expansion reduce thrust?

Expansion fan turns velocity outward by angle

\theta

; only

v\cos\theta

contributes to axial thrust, and

v_e

is below the fully-expanded optimum.

Under-expanded vs. over-expanded nozzle?

Under-expanded:

p_e > p_\infty

, expansion waves, flow turns outward. Over-expanded:

p_e < p_\infty

, oblique shocks, flow turns inward. Both lose efficiency; optimum is

p_e = p_\infty

What is the thrust equation used at the exit plane?

F = \dot{m}\,v_e + A_e(p_e - p_\infty)

; maximized when

p_e = p_\infty

What is the key assumption for Prandtl-Meyer expansion?

Isentropic flow (entropy constant, total pressure constant). Expansion is smooth and reversible, unlike shocks.

Mnemonic

Feynman Recap

Recall Explain to a 12-year-old

Imagine you're blowing up a balloon and you let go—the air rushes out, right? Now imagine the balloon neck is a rocket nozzle. If the air inside is squeezed tight (high pressure) and the air outside is loose (low pressure), the air wants to spread out more after it leaves the nozzle.

But here's the thing: in a rocket, we want all the air to shoot straight backward to push the rocket forward. If the air is still "too tight" when it exits, it keeps expanding outside the nozzle, and it spreads out sideways like a fan. That sideways motion doesn't help push the rocket—it's wasted.

Scientists call this "under-expanded" because the nozzle didn't let the gas expand enough inside where we can control it. The gas makes these smooth wave patterns (Prandtl-Meyer waves) as it spreads out. The more it spreads sideways, the less push we get. That's why rocket engineers try to match the exit pressure to the outside air—so all the expansion happens inside the nozzle, pushing straight back!

Connections

3.3.13-Optimal-expansion-ratio — Why $p_e = p_\infty$ maximizes thrust
3.3.14-Over-expanded-nozzle-shock-diamonds — The opposite case: $p_e < p_\infty$
3.2.7-Isentropic-flow-area-Mach-relation — Foundation for expansion math
4.1.3-Oblique-shock-theory — Contrast: expansion waves vs. shocks
3.3.16-Altitude-compensation-nozzles — Designs (aerospike, dual-bell) that mitigate under/over-expansion
2.5.12-Thrust-coefficient-definition — How $C_F$ captures efficiency

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, jab rocket ka nozzle "under-expanded" hota hai, iska matlab hai ki exhaust gas jo nozzle se bahar nikal raha hai uska pressure ( $p_e$ ) surrounding ambient pressure ( $p_\infty$ ) se zyada hota hai. Ab yeh gas abhi bhi "squeezed" state mein hai aur aur expand hona chahta hai, lekin nozzle to already khatam ho gaya. To yeh gas nozzle ke bahar jaake expand karta hai, ek phool ke khilne jaise bahar ki taraf turn karta hua. Yeh sab hota hai Prandtl-Meyer expansion waves ke through, aur main intuition yahi hai ki jab gas bahar ki taraf (sideways) fail jata hai, to hum uska thrust waste kar rahe hote hain kyunki hume pure axial momentum chahiye tha, na ki sideways push.

Ab yeh kyun important hai? Kyunki tumne shayad nozzle ko high altitude ke liye design kiya tha jahan $p_\infty$ kam hota hai, lekin agar tum sea level par fly kar rahe ho ya combustion chamber ka pressure gir gaya, to nozzle current conditions ke liye chhota reh jata hai. Ideal case wahi hai jab $p_e = p_\infty$ ho — isko "perfectly expanded" kehte hain, jahan koi waves nahi bante aur maximum axial thrust milta hai. Under-expanded mein expansion fans banna efficiency loss cause karta hai, isliye engineers ko flight conditions ke hisaab se nozzle design carefully karna padta hai.

Ek key concept jo yaad rakhna hai woh hai ki supersonic flow mein information upstream travel nahi kar sakti — disturbances Mach angle $\mu = \arcsin(1/M)$ par propagate karte hain. Isi wajah se expansion smoothly ek continuous fan ke through hota hai, na ki achanak. Prandtl-Meyer function $\nu(M)$ basically batata hai ki agar flow $M_1$ se $\theta$ angle turn karta hai, to naya Mach number kya hoga: $\nu(M_2) = \nu(M_1) + \theta$ . Yeh function isentropic expansion (jahan total pressure aur temperature constant rehte hain) ko capture karta hai, aur isse tum predict kar sakte ho ki plume kitna expand karega aur kitni efficiency loss hogi. Bas yeh samajhna important hai ki yeh saari physics pressure imbalance ki wajah se hoti hai aur real rocket performance ko directly affect karti hai.

Under-expanded nozzle — Prandtl-Meyer expansion, efficiency loss

What Is an Under-Expanded Nozzle?

Why Prandtl-Meyer Expansion Waves Form

First Principles: Pressure Imbalance at the Exit Plane

Physical Process at Nozzle Exit

Efficiency Loss: Thrust Reduction

Axial vs. Radial Momentum

Worked Examples

Common Mistakes

Active Recall Flashcards

Mnemonic

Feynman Recap

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Rocket Propulsion

Connections