Nozzles, Diffusers, and Wind Tunnels

Section 10.2

Governing equations for quasi-one-dimensional flow

In truly one-dimensional flow the area is constant and p, ρ, T, u vary only with x. In quasi-one-dimensional flow the area also varies, A = A(x), but the flow properties are still assumed uniform over each cross section: p = p(x), ρ = ρ(x), T = T(x), u = u(x). This is an approximation; the real flow in a variable-area duct is three-dimensional. For slowly varying ducts it predicts nozzle performance remarkably well.

Apply the integral conservation laws to a control volume bounded by two cross sections, A₁ and A₂, and the duct wall. The wall is a streamline surface, so there. For steady, inviscid, adiabatic flow:

The term is the pressure force of the duct wall on the gas. It is the only new term compared with the normal-shock equations, and it requires A(x).

Together with p = ρRT and h = c_pT, these five equations determine the five unknowns ρ₂, p₂, u₂, T₂, h₂. A direct solution takes substantial algebra. The differential forms below lead to the physical insight much faster.

Derivation of the integral forms (notes pp. 1–3)

1

Continuity, for steady flow. On face a, ; on face c, ; on face b (the wall) it is zero. With uniform properties over each face:

2

The x-component of momentum, . On the wall, , so the wall contributes :

3

Energy, . Divide by and use h = e + p/ρ:

Applied to an infinitesimal control volume of length dx, the same three laws give the differential forms:

Equation (10.18) is the Euler equation: the momentum equation for steady, inviscid flow. It holds for quasi-one-dimensional flow even though A varies, because the area terms cancel against continuity.

Derivation of the Euler equation (notes p. 4)

1

Momentum on the slab between x and x + dx, with :

Drop products of differentials:

2

Expand continuity and multiply by u:

3

Subtract (10.17) from (10.16): , which is (10.18).

Section 10.2 · Physical characteristics

The area–velocity relation

Divide continuity by ρuA: . For flow without shocks (isentropic flow), , so the Euler equation becomes . Eliminating dρ/ρ:

The same area change does opposite things in subsonic and supersonic flow. Subsonic: a converging duct speeds the flow up and a diverging duct slows it down. Supersonic: a diverging duct speeds the flow up and a converging duct slows it down. Sonic: dA = 0, so M = 1 can occur only at a minimum area, the throat. As M → 0, Au = const, the incompressible continuity equation.

Interactive 1 · Same duct, two Mach numbersEach duct changes its area by the same 5% over its length.

Bar lengths are proportional to the fractional changes du/u, dp/p and dρ/ρ from (10.25), (10.18) and continuity; the outlet arrow shows whether u rises or falls: , , . They grow without bound as M → 1.

Section 10.3 · Nozzle flows

The area–Mach number relation

Apply continuity between the sonic section A* (where M = 1 and u* = a*) and any other section A: . Using , and :

The Mach number at any location in the duct depends only on the ratio of the local area to the sonic area. Every A/A* > 1 has two solutions, one subsonic and one supersonic. Which one the flow takes depends on the pressures at the ends of the duct.

Interactive 2 · Two Mach numbers for every area ratio

Section 10.3 · Figures 10.12–10.16

A converging-diverging nozzle and its back pressure

A reservoir at p₀, T₀ feeds a nozzle with throat area A_t and exit area A_e; the nozzle exhausts into a back pressure p_B. With p₀ fixed, lowering p_B moves the flow through a sequence of regimes. Three back pressures divide them:

p_e1: the throat just reaches M = 1 and the flow returns to subsonic speed in the diverging section, from the subsonic root of (10.32) at A_e/A_t.
p_e5: the supersonic flow ends in a normal shock standing exactly at the exit plane.
p_e6: the design pressure, from the supersonic root of (10.32). The flow is isentropic and supersonic all the way to the exit.

Once the throat is sonic, nothing downstream can change the flow upstream of it: the mass flow is choked at . Lowering p_B further changes only the flow downstream of the throat.

The choked mass flow in terms of reservoir conditions

1

At the sonic throat, , with , , and .

2

Combine:

The mass flow of a choked nozzle is proportional to p₀A_t and inversely proportional to . It does not depend on the back pressure or on the diverging section.

3

Across a normal shock T₀ and are unchanged, so the same expression written at a sonic reference area on each side gives . The total-pressure loss enlarges the sonic area downstream of the shock.

Interactive 3 · Nozzle explorer

Current flowDesign (supersonic isentropic) solutionCritical subsonic solution (p_e1)Normal shock

Reading the pressure plot

Pressure falls wherever the flow accelerates, subsonic or supersonic: in the converging section, and in the diverging section once the flow is supersonic.
Pressure jumps across a normal shock, then rises further as the subsonic flow behind it decelerates in the diverging section.
A subsonic jet leaves at the back pressure, p_e = p_B. A supersonic jet can leave above or below p_B and adjusts outside the nozzle through oblique shocks (p rises) or expansion fans (p falls).
Once the nozzle is choked, the pressure in the converging section, and the mass flow, no longer change with back pressure.
The Mach number, pressure and temperature at each section depend only on A/A* and on whether the flow there is subsonic or supersonic. The shape of the nozzle only decides where along x a given area, and so a given shock strength, occurs. Switch between the contoured and conical diverging sections to see this.

Worked examples

Nozzle flows: Examples 10.1–10.3

The Exact column comes from the relations above; the Notes column is the hand solution from Appendix A. Change any given value and the exact solution updates. The notes column applies to the original values only.

Consider the isentropic supersonic flow through a convergent-divergent nozzle with an exit-to-throat area ratio of . The reservoir pressure and temperature are atm and K. Calculate M, p and T at the nozzle exit.

Isentropic flow has no shock, so the throat is the sonic section: A_t = A*, and A_e/A* = A_e/A_t. Take the supersonic root of (10.32).

Example 10.4

Rocket engine thrust

Applying the integral momentum equation to a control volume wrapped around a rocket engine gives its thrust:

where is the mass flow, u_e the exit velocity, p_e the exit pressure, p_∞ the ambient pressure and A_e the exit area.

Liquid hydrogen and oxygen burn in the combustion chamber at atm and K. The throat area is m². The exit is designed so that the exit pressure equals the ambient pressure at a standard altitude of 20 km, p_∞ = Pa. Assume isentropic flow with γ = and R = J/(kg·K). (a) Calculate the thrust. (b) Calculate the exit area.

Try it first. The rocket uses its own γ and R, not the γ in the page header.

Sections 10.4–10.5 · Figures 10.17–10.20

Diffusers and supersonic wind tunnels

A diffuser slows a flow to low speed with as large a static-pressure rise, and as small a total-pressure loss, as possible. The ideal supersonic diffuser is a converging-diverging duct run in reverse: supersonic flow decelerates isentropically to M = 1 at the throat and then subsonically, with p_0,2 = p_0,1. In practice the oblique waves and boundary layer make this impossible. A real diffuser decelerates the flow through a pattern of reflected oblique shocks and a weak normal shock near the throat, so p_0,2 < p_0,1 and s₂ > s₁.

A supersonic wind tunnel must supply a reservoir pressure p₀ high enough to drive the test-section Mach number M_e against the back pressure p_B. The diffuser lowers that requirement. Compare three layouts:

Nozzle exhausting directly to the atmosphere (Figure 10.18): the exit must be at design, p_e = p_B, so p₀ = p_B(p₀/p_e).
Constant-area duct with a normal shock at its exit (Figure 10.19): the shock raises the pressure by p₂/p₁, so p₀ is lower by that factor.
Diffuser with a second throat (Figure 10.20): during start-up a normal shock passes through the test section, so the second throat must swallow the reduced total pressure behind it: . Once started, the shock sits in the diffuser at a lower Mach number, and its loss is smaller.

Interactive 4 · How much reservoir pressure does the tunnel need?

Section 10.5 · Supersonic wind tunnels

Starting a wind tunnel: swallowing the shock

When the tunnel is switched on, p₀/p_B rises from 1. First the whole tunnel is subsonic. Once the nozzle throat chokes, a normal shock forms in the nozzle and moves downstream as p₀/p_B rises. The test section reaches its design Mach number only after this shock has passed through the test section and through the second throat into the diverging part of the diffuser. The shock has then been swallowed and the tunnel is started.

The strongest shock, and the largest total-pressure loss, occurs while the shock stands in the test section at M_e. The second throat must pass the full mass flow at that reduced total pressure, so it must be at least

If the second throat is smaller, it chokes first. The shock then stays in the nozzle, at the Mach number whose total-pressure ratio equals A_t,1/A_t,2, and raising p₀ cannot move it: the tunnel is unstarted. Once started, the shock sits in the diffuser at a lower Mach number, so the tunnel can run at a lower p₀/p_B than it needed to start. Lowering p₀/p_B below the running minimum makes the shock jump back upstream.

Interactive 5 · Start-up and shutdown

Model: inviscid quasi-one-dimensional flow; the diffuser exit area is 1.5 times the test-section area, and the exit static pressure equals p_B. Interactive 4 instead lets the flow behind the shock come to rest, so its running pressures are a little lower. Real tunnels need more, because of the viscous losses described in Section 10.6.

Section 10.6

Viscous flow: shock–boundary-layer interaction inside nozzles

The quasi-one-dimensional model puts a plane normal shock across the whole duct. In a real nozzle a boundary layer grows along the walls, and its inner part is subsonic. The pressure rise of the shock travels upstream through that subsonic layer, so the boundary layer thickens ahead of the shock and, for strong enough shocks, separates.

The thickened or separated layer turns the outer supersonic flow into itself, so an oblique shock, the front leg, forms ahead of the main shock. The front leg and a rear leg meet the nearly normal central shock at a triple point, and the pattern takes the shape of the Greek letter λ: a lambda shock. Behind it the flow near the wall is separated, the effective flow area shrinks, and the pressure recovers more gradually than the one-dimensional jump.

Consequences for practice:

The shock position, and the wall pressure downstream of it, differ from the one-dimensional prediction. The interactives on this page are the inviscid limit.
In strongly overexpanded rocket nozzles, for example a large-expansion engine at sea level, the flow separates from the wall inside the nozzle. Unsteady separation can produce large side loads.
In wind-tunnel diffusers the same interaction replaces the single weak normal shock with a train of oblique and λ shocks, which is why a real diffuser recovers less total pressure than the ideal one.

Schematic of the upper half of a diverging nozzle. Supersonic core flow from the left; the boundary layer separates at S under the oblique front leg of the lambda shock, and reattaches downstream or leaves the nozzle separated.

Self-check

Check your understanding

Pick an answer to see whether it is right and why. Use the nozzle explorer to test your reasoning.

Summary