ch_09_gasdyn_of_nozzles

8/7/2019 Ch_09_Gasdyn_of_nozzles

1/24

bjc 9.1 9/16/09

C

HAPTER

9

G

ASDYNAMICS

OF

N

OZZLE

F

LOW

A nozzle is an extremely efficient device for converting thermal energy to kinetic

energy. Nozzles come up in a vast range of applications. Obvious ones are the

thrust nozzles of rocket and jet engines. Converging-diverging ducts also come up

in aircraft engine inlets, wind tunnels and in all sorts of piping systems designed

to control gas flow. The flows associated with volcanic and geyser eruptions are

influenced by converging-diverging nozzle geometries that arise naturally in geo-

logical formations.

9.1 A

REA

-M

ACH

NUMBER

FUNCTION

In Chapter 8 we developed the area-averaged equations of motion.

(9.1)

Assume the only effect on the flow is streamwise area change so that

. (9.2)

Also assume that streamwise normal stresses and heat fluxes are small

enough to be neglected. With these assumptions the governing equations (9.1)

together with the perfect gas law reduce to

(9.3)

Introduce the Mach number

d U( ) mA

------- UdA

A-------=

d P xx

( ) UdU+ 12---U

2 4C

f

dx

D------

Uxm U( )m

A-----------------------------------

Fx

A---------+=

d ht

xx

-------Q

x

U

--------+

q w htm

ht

xx

-------Q

x

U

--------+

m

UA

------------+=

m Cf

Fx

q w 0= = = = =

xx

, Qx

d UA( ) 0=dP UdU+ 0=

CpdT UdU + 0=

P RT=


2/24

Area-Mach number function

9/16/09 9.2 bjc

(9.4)

Each of the equations in (9.3) can be expressed in fractional differential form

(9.5)

Equation (9.4) can also be written in fractional differential form

(9.6)

Use the equations for mass, momentum and energy to replace the terms in the

equation of state.

(9.7)

Solve for

(9.8)

Equation (9.8) shows the effect of streamwise area change on the speed of theflow. If the Mach number is less than one then increasing area leads to a decrease

in the velocity. But if the Mach number is greater than one then increasing area

leads to an increase in flow speed.

Use (9.8) to replace in each of the relations in (9.5).

U2

RT M2

=

d

------dU

2

2U2----------

dA

A-------+ + 0=

dP

P-------

M2

2-----------

dU2

U2

----------+ 0=

dT

T-------

1( )M2

2-------------------------

dU2

U2

----------+ 0=

dP

P-------

d

------dT

T-------+=

dU2

U2

----------dT

T-------

dM2

M2

-----------+=

M2

2-----------

dU2

U2

----------dU2

2U2

----------dA

A-------

1( )M2

2-------------------------

dU2

U2

----------=

dU2

U2

dU2

U2

----------2

M2

1

----------------- dA

A-------=

dU2

U2


3/24


bjc 9.3 9/16/09

(9.9)

Equations (9.9) describe the effects of area change on the thermodynamic state of

the flow. Now use (9.8) and the temperature equation in (9.6).

(9.10)

Rearrange (9.10). The effect of area change on the Mach number is

(9.11)

Equation (9.11) is different from (9.8) and (9.9) in that it can be integrated from

an initial to a final state. Integrate (9.11) from a Mach number to one

(9.12)

The result is

(9.13)

Evaluate (9.13)at the limits

d

------

M2

M2

1

----------------- dA

A-------=

dP

P-------M

2

M2

1-----------------

dAA-------=

dT

T-------

1( )M2

M2

1

------------------------- dA

A-------=

2M

21

----------------- dAA------- 1( )M2

M2

1-------------------------dAA------- dM

2

M2-----------+=

dA

A-------

M2

1

2 1 1

2------------

M2+

---------------------------------------------dM

2

M2

-----------=

M

M2

1

2 1 1

2------------

M2+

---------------------------------------------dM

2

M2

-----------

M2

1

dAA-------A

A*

=

Ln A*A

------- L n M[ ] Ln 1 1

2------------ M

2+

1+

2 1( )

--------------------

Ln 2

1+

2 1( )--------------------

+ +

M2

1

=


4/24


9/16/09 9.4 bjc

(9.14)

which becomes

(9.15)

Exponentiate both sides of (9.15). The result is the all-important area-Mach num-

ber equation.

. (9.16)

Note that for convenience we referenced the integration process to . The

area is a reference area at some point in the channel where although

such a point need not actually be present in a given problem.

LnA*

A-------

Ln 1+2

------------

1+2 1( )--------------------

Ln 2

1+2 1( )--------------------

+

=

L n M[ ] Ln 1 12

------------ M2+

1+2 1( )

--------------------

Ln 2

1+2 1( )--------------------

+ +

LnA*

A-------

Ln 1+2

------------

1+2 1( )--------------------

M

1 1

2------------

M2+

1+2 1( )--------------------

------------------------------------------------------------=

f M( ) A*

A------

1+

2-------------

1+

2 1( )--------------------

M

1 12

------------M2+

1+

2 1( )--------------------

-------------------------------------------------------= =

M 1=

A*

M 1=


5/24


bjc 9.5 9/16/09

The area-Mach-number function is plotted below for three values of .

Figure 9.1 Area-Mach number function

Note that for smaller values of it takes an extremely large area ratio to generate

high Mach number flow. A value of would be typical of the very high

temperature mixture of gases in a rocket exhaust. Conversely, if we want to pro-

duce a high Mach number flow in a reasonable size nozzle, say for a wind tunnelstudy, an effective method is to select a monatomic gas such as Helium which has

.

9.1.1 MASSCONSERVATION

This result (9.16) can also be derived simply by equating mass flows at any two

points in the channel and using the mass flow relation

(9.17)

This can be expressed as

(9.18)

Insert

1 2 3 4 5

0.2

0.4

0.6

0.8

1

M

f(M)

=1.2

=1.4=1.66

1.2=

1.66=

m UA=

m UAP

RT------- RT( )

1

2---

MA= =


6/24

A simple convergent nozzle

9/16/09 9.6 bjc

(9.19)

into (9.18) to produce

. (9.20)

If we equate the mass flows at any two points in a channel (9.20) gives

. (9.21)

In the case of an adiabatic ( ), isentropic ( ) flow

in a channel (9.16) provides a direct relation between the local area and Mach

number.

(9.22)

9.2 A SIMPLECONVERGENTNOZZLE

The figure below shows a large adiabatic reservoir containing an ideal gas at pres-

sure . The gas exhausts through a simple convergent nozzle with throat area

to the ambient atmosphere at pressure . Gas is continuously supplied to

the reservoir so that the reservoir pressure is effectively constant. Assume the gas

is calorically perfect, ( , and are constant) and assume that

wall friction is negligible.

Lets make this last statement a little more precise. Note that we do not assume

that the gas is inviscid since we want to accommodate the possibility of shock for-

mation somewhere in the flow. Rather, we make use of the fact that, if the nozzle

Tt

T----- 1

1

2------------M

2+=

Pt

P----- 1

1

2------------M

2+

1------------

=

m UA

1+

2-------------

1+

2 1( )--------------------

-------------------------------------P

tA

RTt

----------------

f M( )= =

m1

m2

=

Pt1A1

Tt1

--------------- f M1

( )Pt2A2

Tt2

--------------- f M2

( )=

Tt

cons t tan= Pt

cons t tan=

A1

f M1

( ) A2

f M2

( )=

Pt

Ae

Pambient

p RT= Cp Cv


7/24


bjc 9.7 9/16/09

is large enough, the boundary layer thickness will be small compared to the diam-

eter of the nozzle enabling most of the flow to be treated as irrotational and

isentropic.

Figure 9.2 Reservoir with a convergent nozzle.

The isentropic assumption works quite well for nozzles that are encountered in

most applications. But if the plenum falls below a few centimeters in size with a

nozzle diameter less than a few millimeters then a fully viscous, nonisentropic

treatment of the flow is required. Accurate nozzle design, regardless of size, vir-

tually always requires that the boundary layer on the wall of the plenum andnozzle is taken into account.

If the ambient pressure equals the reservoir pressure there is, of course, no flow.

If is slightly below then there is a low-speed, subsonic, approxi-

mately isentropic flow from the plenum to the nozzle. If is less than

a certain critical value then the condition that determines the speed of the flow at

the exit is that the exit static pressure is very nearly equal to the ambient pressure.

(9.23)

The reason this condition applies is that large pressure differences cannot occur

over small distances in a subsonic flow. Any such difference that might arise, say

between the nozzle exit and a point slightly outside of and above the exit, will be

immediately smoothed out by a readjustment of the whole flow. Some sort of

shock or expansion is required to maintain a pressure discontinuity and this can

Pt

Pambient

Ae

Pe

Pambient

Pt

Pt

Pambient

Pe

Pambient

=


8/24


9/16/09 9.8 bjc

only occur in supersonic flow. Slight differences in pressure are present due to the

mixing zone that exists outside the nozzle but in subsonic flow these differences

are very small. Since the flow up to the exit is approximately isentropic the stag-

nation pressure is approximately constant from the reservoir to the nozzle exit

and we can write

(9.24)

Using (9.23) and (9.24) we can solve for the Mach number at the nozzle exit in

terms of the applied pressure ratio.

(9.25)

Note that the nozzle area does not appear in this relationship.

9.2.1 THEPHENOMENONOFCHOKING

The exit Mach number reaches one when

(9.26)

For Air with this critical pressure ratio is and

the condition (9.23) holds for .

At the area-Mach number function is at its maxi-

mum value of one. At this condition the mass flow through the nozzle is as large

as it can be for the given reservoir stagnation pressure and temperature. The noz-

zle is said to be choked.

Pt

Pt

Pe

------ 1 1

2------------

Me

2+

1------------

=

Me

2

1------------

1

2--- Pt

Pambient

----------------------

1

------------

1

1

2

---

=

Pt

Pambient

---------------------- 1+

2-------------

1------------

=

1.4= Pt

Pambient

1.893=

1 Pt

Pambient 1.893

Pt

Pambient

1.893= f M( )


9/24

Converging-diverging nozzle

bjc 9.9 9/16/09

If is increased above the critical value the flow from the reservoir

to the nozzle throat will be unaffected; the Mach number will remain

and . However condition (9.23) will no longer hold because now

. The flow exiting the nozzle will tend to expand supersonically

eventually adjusting to the ambient pressure through a system of expansions and

shocks.

If the ambient pressure is much lower than the nozzle exit pressure then the flow

leaving the nozzle will tend to turn through a very large angle as it expands super-

sonically to match the ambient as shown schematically below.

Figure 9.3 Plenum exhausting to very low pressure.

9.3 CONVERGING-DIVERGINGNOZZLE

Now lets generalize these ideas to the situation where the nozzle consists of a

converging section upstream of the throat and a diverging section downstream.

Consider the nozzle geometry shown below.

Figure 9.4 Converging-diverging geometry

Pt

Pambient

Me

1=

Pt

Pe

1.893=

Pe Pambient>

Pambient Pe


10/24


9/16/09 9.10 bjc

The goal is to completely determine the flow in the nozzle given the pressure ratio

and the area ratio . Before analyzing the flow we

should first work out the critical exit Mach numbers and pressures for the selected

area ratio. Solving

. (9.27)

gives two critical Mach numbers and for isentropic flow in the

nozzle with at the throat. The corresponding critical exit pressures are

determined from

(9.28)

There are several possible cases to consider.

9.3.1 CASE 1 - ISENTROPICSUBSONICFLOWINTHENOZZLE

If is not too large then the flow throughout the nozzle will be sub-

sonic and isentropic and the pressure at the exit will match the ambient pressure.

In this instance the exit Mach number is determined using (9.25).

If the pressure ratio is increased there is a critical value that leads to choking at

the throat. This flow condition is sketched below.

Figure 9.5 Onset of choking

Pt

Pambient

Ae

Athroat

Athroat

Ae

------------------ 1+

2-------------

1+

2 1( )--------------------

Me

1 1

2------------M

e

2+

1+

2 1( )--------------------

---------------------------------------------------------=

Mea

1< Meb

1>

M 1=

Pt

Pea

--------- 1 1

2------------

Mea

2+

1------------

=

Pt

Peb

--------- 1 1

2------------

Meb

2+

1------------

=

Pt

Pambient

Pt

Athroat Ae

PambientPeM


11/24


12/24


9/16/09 9.12 bjc

(9.33)

The items on the left-hand-side of (9.33) are known quantities and so one solves

(9.33) implicitly for . With the exit Mach number known, (9.31) is used

to determine the stagnation pressure ratio across the nozzle.

(9.34)

Since the only mechanism for stagnation pressure loss is the normal shock, the

value of determined from (9.34) can be used to infer the shock Mach

number from

(9.35)

Thus all of the important properties of the flow in the nozzle are known given the

plenum to ambient pressure ratio and the nozzle area ratio.As the nozzle pressure ratio is increased, the shock moves more and more down-

stream until it is situated at the nozzle exit. This flow condition is shown below

Figure 9.7 Shock at the nozzle exit.

In this case the Mach number just ahead of the shock is the supersonic critical

value and the Mach number just behind is the corresponding subsonic value

derived from normal shock relations.

Pt

Pambient

---------------------- Athroat

Ae

------------------ 1+

2-------------

1+

2 1( )--------------------

Me

1 1

2------------

Me

2+

1

2---

=

Me

1

ch_09_gasdyn_of_nozzles

Documents