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Unit 5 · Lesson 5-2
Nozzle Performance and Convergent–Divergent Flow
1h 50m1300 XP
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Nozzle Performance and Convergent–Divergent Flow

Nozzle pressure ratio, under- and over-expansion, shock diamonds, the gross thrust coefficient, shocks inside the nozzle and divergence loss

Diamonds in the plume
Behind an F-16 or an F-22 on afterburner, behind an SR-71's J58s, behind a missile's rocket motor, the exhaust plume carries a row of bright, evenly spaced diamonds. They form when the supersonic jet leaves its nozzle over-expanded — at a static pressure below the air around it. The surrounding air squeezes the jet through a pattern of oblique shocks; each one compresses and reheats the gas, the jet overshoots, expands again and is squeezed again (shock diamonds). The exhaust is typically over-expanded at low altitude, where the air pressure is highest. The diamonds are a nozzle telling you that it is not quite the right shape for the pressure it is working against — and that some of the thrust it could have given is being spent on shocks.
The nozzle's job is to turn the stagnation pressure left after the turbine into jet velocity, and to deliver the jet at ambient pressure. Its operating condition is the nozzle pressure ratio, NPR=pt/p0\mathrm{NPR} = p_t/p_0.
Below the critical NPR, (γ+12)γ/(γ−1)=1.85\left(\frac{\gamma + 1}{2}\right)^{\gamma/(\gamma - 1)} = 1.85 for hot gas, a convergent nozzle expands the jet fully and stays subsonic. Above it the throat chokes at Mach 1: a convergent nozzle then leaves the jet at a pressure above ambient, while a convergent–divergent (C-D) nozzle can carry on expanding it supersonically. Turbofans in cruise run NPRs of about 2–4; afterburning fighters 10 and more.
Fg=m˙ Ve+(pe−p0)Ae,CFG=Fgm˙ Ve, idealF_g = \dot m\,V_e + (p_e - p_0)A_e, \qquad C_{FG} = \frac{F_g}{\dot m\,V_{e,\,ideal}}

Gross thrust of a nozzle and its gross thrust coefficient: the actual gross thrust over the thrust of the same flow expanded isentropically to ambient pressure.

Derivation
Why full expansion gives the most thrust
Follow a nozzle of growing exit area. Steady, frictionless flow obeys m˙ dV=−A dp\dot m\,dV = -A\,dp (momentum for a slice of the nozzle). The gross thrust at an exit area AeA_e is F=m˙Ve+(pe−p0)AeF = \dot m V_e + (p_e - p_0)A_e, so extending the nozzle by dAdA changes it by
dF=m˙ dV+(pe−p0) dA+A dp=(pe−p0) dA.dF = \dot m\,dV + (p_e - p_0)\,dA + A\,dp = (p_e - p_0)\,dA .
While pe>p0p_e > p_0 (under-expanded) a longer nozzle adds thrust; once pe<p0p_e < p_0 (over-expanded) it subtracts. The thrust is greatest when pe=p0p_e = p_0 — perfect expansion. That is the nozzle the gross thrust coefficient compares against.
💡Four regimes of a C-D nozzle
For a fixed area ratio Ae/A∗A_e/A^*, lowering the ambient pressure (raising NPR) takes the nozzle through:
  • Unchoked: subsonic everywhere, a venturi.
  • Choked with a normal shock inside: supersonic just past the throat, a normal shock in the divergent part, subsonic exit. In real nozzles the flow usually separates from the wall before this.
  • Over-expanded: supersonic exit at a pressure below ambient; oblique shocks and diamonds outside.
  • Perfectly expanded, and then under-expanded: the jet leaves above ambient pressure and expands outside through expansion fans.
📐The cost of the wrong area ratio
Gross thrust coefficient of an isentropic nozzle (hot gas, Lab 2):
  • A convergent nozzle at NPR 15 loses 11 % (CFG=0.893C_{FG} = 0.893); at NPR 4 only 2 %.
  • Area ratio 2 at NPR 8: 0.998. At NPR 2 it is badly over-expanded: 0.606.
  • Area ratio 3 at NPR 2 would push a shock into the nozzle.
  • The right area ratio grows with NPR: 1.25 at 4, 1.78 at 8, 2.60 at 15.
No fixed nozzle is right everywhere — which is why fighters have variable C-D nozzles, and airliners, at low NPR, settle for convergent ones.
📐A shock inside the nozzle
A C-D nozzle with Ae/A∗=2A_e/A^* = 2 (air, Lab 3) is designed for an exit Mach number of 2.197 at pe/pt=0.094p_e/p_t = 0.094. It runs subsonic and choked up to pb/pt=0.937p_b/p_t = 0.937; a normal shock stands exactly at the exit at pb/pt=0.513p_b/p_t = 0.513. In between, the shock sits inside: at pb/pt=0.7p_b/p_t = 0.7 it stands where A/A∗=1.51A/A^* = 1.51. brentq finds the position by matching the subsonic exit pressure behind the shock to the back pressure.
Divergence loss. A real nozzle's exit flow is not all axial: a conical nozzle of half-angle α\alpha throws the jet out over a cone, and only the axial component of its momentum is thrust. Lab 1 tabulates the factors below: a 15° cone loses 1.7 %, and keeping the loss under 0.5 % needs a half-angle under 8°. Long, slender nozzles lose little to divergence but are heavy and have more friction — a design compromise, often met with contoured (bell-shaped) walls.
λconical=1+cos⁡α2,λ2D=sin⁡αα\lambda_{conical} = \frac{1 + \cos\alpha}{2}, \qquad \lambda_{2D} = \frac{\sin\alpha}{\alpha}

Divergence factors: the fraction of the jet's momentum that is axial, for a conical exit of half-angle α and for a two-dimensional exit.

Derivation
The conical divergence factor
Treat the exit flow as leaving radially from the apex of the cone, with the same speed and mass flux everywhere on a spherical cap of radius rr. A ring of the cap at angle θ\theta has area 2πr2sin⁡θ dθ2\pi r^2\sin\theta\,d\theta and contributes axial momentum in proportion to cos⁡θ\cos\theta. The axial fraction is
λ=∫0αcos⁡θsin⁡θ dθ∫0αsin⁡θ dθ=12sin⁡2α1−cos⁡α=1+cos⁡α2.\lambda = \frac{\int_0^\alpha \cos\theta\sin\theta\,d\theta}{\int_0^\alpha \sin\theta\,d\theta} = \frac{\tfrac12\sin^2\alpha}{1 - \cos\alpha} = \frac{1 + \cos\alpha}{2}.
In two dimensions the exit is an arc, every element of the same length, and the average of cos⁡θ\cos\theta over −α…α-\alpha\ldots\alpha is sin⁡α/α\sin\alpha/\alpha.
I came to realize that exaggerated concern about what others are doing can be foolish. It can paralyze effort, and stifle a good idea. One finds that in the history of science almost every problem has been worked out by someone else. This should not discourage anyone from pursuing his own path.
– Theodore von Kármán, The Wind and Beyond (1967) (Wikiquote)
⚠Over-expansion and separation
A badly over-expanded nozzle does not usually hold a neat normal shock inside: the boundary layer separates from the wall first, and the jet leaves the wall early. Separation can be asymmetric and unsteady, putting large side loads on rocket nozzles at lift-off — one reason rocket nozzles are designed with ground-level operation in mind.
💡Where you will use this
Lesson 5-3 turns the nozzle into a steering and stopping device. Lesson 8-1 shows that a choked nozzle, like a choked turbine, fixes an engine's operating line. Every rocket engine is designed around these relations.
📚Further reading
  • George P. Sutton and Oscar Biblarz, Rocket Propulsion Elements (9th ed., Wiley, 2017), chapter 3, on nozzle theory.
  • Jack D. Mattingly and Keith M. Boyer, Elements of Propulsion: Gas Turbines and Rockets (2nd ed., AIAA, 2016), on exhaust nozzles.
  • Shock diamond on Wikipedia.
↓But there is a problem
A nozzle can do more than push. Turn its jet forward and it stops the aircraft; turn it sideways and it steers it. On 26 May 1991, over Thailand, a Boeing 767 climbing out of Bangkok found out what happens when that capability switches itself on at the wrong moment.