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HIGH-SPEED AERODYNAMICSMACE 31321

Lecture 10 Compressible flow through nozzles

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OBJECTIVES OF THIS LECTURE

• To examine different scenarios of nozzle flows

• To understand how mass flow rate through a nozzle will change with the exit pressure

• To acquire a understanding of the geometry of a supersonic wind tunnel

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NOZZLE FLOW• A pressure difference

between p0 and pe is required to generate the nozzle flow.

• For a purely subsonic flow throughout the nozzle, At ≠A*.

In this case, A* would be the area when the flow were accelerated to sonic velocity.

Throat

At A*

Beware that the entry of A/A* in Table A is not always equal to A/At

When pe is slightly smaller than p0 ,the flow is essentially subsonic the nozzle.

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NOZZLE FLOW

• At a certain pe/p0(= pe6/p0), isentropic expansion of gas to supersonic flow through the nozzle is achieved– The local Mach number is a

function of A/A*.– At the throat, M=1 and

At=A*.– The variations of p/p0 and

T/T0 are given in Table A.

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NOZZLE FLOW• If pe6/p0 < pe/p0 < pe3/p0

– The flow at and upstream of the throat remains as that at pe=pe3.

– A normal shock forms in the divergent section since pe is too high to allow isentropic supersonic flow.

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NOZZLE FLOW

• At a certain pe/p0 (=pe5/p0),– The normal shock stands right

at the nozzle exit.– The flow is isentropic before

the exit.

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SUMMARY• The graphs to the right

illustrate the six different scenarios of nozzle flow as the exit pressure decreases from p0 to pe6.

• Given the nozzle geometry, the theory gives a reasonable prediction of the flow inside the duct.

• To obtain the proper contour for a supersonic nozzle which is shock-free, the method of characteristics should be used (not covered here).

Me4

Me5

Me6

Pe6/p0

Pe4/p0

Pe5/p0

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MASS FLOW RATE• The mass flow rate through the nozzle

• When pe=pe3, sonic velocity is achieved at the throat. Hence

• When pe<pe3, the condition at the throat remains unchanged.

• The nozzle is “chokedsince stays constant.

• Flow upstream is not affectedby what happens downstream.

ttt Aum ρ=&

∗∗∗= Aam ρ&

m&

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EFFECT OF BACK PRESSURE• The pressure in the

surrounding downstream of the nozzle is called the BACK PRESSURE, pB– When Me<1, pe=pB as no

pressure jump is allowed in subsonic flow.

– When Me>1, depending on whether pe is greater or smaller than pB, either oblique shock waves or expansion waves will result.

Overexpanded

Underexpanded10

QUESTION • Consider the flow through a convergent-divergent

nozzle with an exit-to-throat area ratio of 2. The reservoir pressure and temperature are 1atm and 288oK, respectively. Calculate

1. The Mach number at the throat and the exit when the exit pressure is 0.973atm. (Answer: Me=0.2, Mt=0.44.)

2. Sketch the distribution of Mach number throughout the nozzle if the exit pressure is 0.2atm.

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SOLUTION Q1 • At the nozzle exit .

• From Table A, if the flow is sonic at the throat but subsonic elsewhere, for Ae/A*=2,

• Since the exit pressure is higher than that allows M=1 at the throat, the flow must be subsonic throughout the nozzle. From Table A, for po/pe=1.028, Me=0.2.

• Note: At ≠A* in this case. At Me=0.2, Ae/A*=2.964• But

• From Table A, for At/A*=1.482, Mt=0.44.

028.1973.010 ==

epp

482.1964.221

=×== ∗∗ AA

AA

AA e

e

tt

10 064.1 PR

pp

e

==

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SOLUTION Q2• Since , at least a portion of the

flow in the divergent nozzle will be supersonic.• However, for the flow being isentropic supersonic

in the entire divergent nozzle, we need

• Since , a normal shock wave forms somewhere in the divergent nozzle resulting a Mach number distribution as shown below.

10 5

2.01 PR

pp

e

>==

20 69.10 PR

pp

e

==

20 5 PR

pp

e

<=

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• To create a Mach 2.5 uniform flow for testing a model of a supersonic vehicle. This requires p0/pe=17.09, Ae/A* = 2.637. Compare the following three designs and comment on which design is least and more efficient.

• Design 1: Nozzle exhausting directly to the atmosphere

SUPERSONIC WIND TUNNELS

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• Design 2: Nozzle exhausting into a constant-area-duct

SUPERSONIC WIND TUNNELS

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• Design 3: A diffuser section is added to the end of the test section

SUPERSONIC WIND TUNNELS

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• Equating mass flow rate through the two throats,

• It can be proved that

SUPERSONIC WIND TUNNELS

222111 tt AaAa ∗∗∗∗ = ρρ ∗∗ = 21 aaFor adiabatic flow:

12,0

1,0

1,

2, >=pp

AA

t

t due to the losses in total pressure

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REFERENCES

• In “Fundamentals of Aerodynamics” by Andersons, 2nd edition.–§10.3–§10.5