Over - under expanded nozzle flows

WHY do we even have this problem?

A converging–diverging (de Laval) nozzle accelerates gas to supersonic speeds. But here is the catch:

• A supersonic exit is "deaf" to downstream conditions. Once the flow is supersonic in the diverging section, pressure information (which travels at the speed of sound) cannot travel upstream against the flow. So the nozzle exit pressure $p_e$ is fixed by the area ratio $A_e/A^*$, not by the back pressure $p_b$.
• The atmosphere outside, however, sits at $p_b$. If $p_e \ne p_b$, nature must reconcile them — and it does so with oblique shocks (if exit pressure is too low) or expansion fans (if exit pressure is too high), outside the nozzle.

WHAT are the three regimes?

HOW does each regime resolve the pressure mismatch?

Under-expanded ($p_e > p_b$)

The jet is at higher pressure than ambient, so it expands further the moment it exits. This happens via Prandtl–Meyer expansion fans anchored at the nozzle lip. The jet bulges outward (diamond/shock-cell pattern forms downstream as fans reflect off the free jet boundary).

Over-expanded ($p_e < p_b$)

Ambient pressure is higher, so it pushes the jet inward. Compression occurs through oblique shocks from the lip.

• Mildly over-expanded: oblique shock pattern (still attached at lip).
• Strongly over-expanded: the required pressure rise is so large that a normal shock or Mach disk forms; if back pressure is high enough the shock moves inside the diverging section (flow separation off the walls).

DERIVATION — Where does $p_e$ come from? (first principles)

We want to show $p_e$ depends only on geometry. Start with three conservation ideas for isentropic, steady, 1-D, adiabatic flow of a perfect gas.

Step 1 — Stagnation relations. Why? Energy conservation ($h_0 = h + \tfrac12 V^2$ constant) plus the isentropic link $p/\rho^\gamma=$const gives, for a perfect gas: $\frac{p_0}{p} = \left(1 + \frac{\gamma-1}{2}M^2\right)^{\frac{\gamma}{\gamma-1}}$ Why this step? This ties local pressure to local Mach number $M$ — the master relation.

Step 2 — Area–Mach relation. Why? Mass conservation $\rho A V =$ const combined with the same isentropic relations gives: $\frac{A}{A^*} = \frac{1}{M}\left[\frac{2}{\gamma+1}\left(1+\frac{\gamma-1}{2}M^2\right)\right]^{\frac{\gamma+1}{2(\gamma-1)}}$ Why this step? $A^*$ is the throat area (where $M=1$, choked). Given the exit-to-throat area ratio $A_e/A^*$, this equation has a unique supersonic root $M_e$.

Step 3 — Combine. Plug $M_e$ (from geometry, Step 2) into Step 1: $\boxed{\,p_e = \frac{p_0}{\left(1 + \frac{\gamma-1}{2}M_e^2\right)^{\gamma/(\gamma-1)}}\,}$

Thrust: the practical WHY this matters

Consequences:

• Perfectly expanded ($p_e=p_b$): pressure term vanishes → maximum thrust for given $V_e$. This is the optimal design point.
• Under-expanded ($p_e>p_b$): positive pressure term adds thrust, but you "left velocity on the table" (could have expanded more) — net efficiency below optimum.
• Over-expanded ($p_e<p_b$): pressure term is negative → drag-like loss; severe over-expansion can cause separation and instability.

Flashcards

Connections

• Isentropic Flow Relations
• Area-Mach Number Relation
• Choked Flow & the Throat (M=1)
• Oblique Shocks
• Prandtl-Meyer Expansion Fans
• Normal Shock & Mach Disk
• Rocket Nozzle Design & Thrust Optimization
• Flow Separation in Nozzles

Concept Map

Hinglish (regional understanding)