Ch 13 Unsteady Waves

7/30/2019 Ch 13 Unsteady Waves

1/26

bjc 13.1 9/12/11

C

HAPTER

13

U

NSTEADY

W

AVES

IN

C

OMPRESSIBLE

F

LOW

13.1 G

OVERNING

EQUATIONS

One of the most important applications of compressible flow theory is to the anal-

ysis of the generation and propagation of sound. In Chapter 6 we worked out the

equations for homentropic, unsteady flow that are the starting point for the deri-

vation of the acoustic equations. For , and , the equations of

motion are,

. (13.1)

The role of the energy equation is played by the isentropic relation between pres-

sure and density.

13.2 T

HE

ACOUSTIC

EQUATIONS

Now consider an unsteady flow where disturbances in density and pressure are

extremely small compared to constant background values.

. (13.2)

In addition, assume that the velocity field involves very small fluctuations about

zero. Linearize (13.1) by dropping higher order nonlinear terms.

. (13.3)

! U 0= !s 0=

"#

"t------

"

"xk-------- #U

k( )+ 0=

#"U

i

"t--------- #

"

"xi

-------U

kU

k

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

$ %& ' "P

"xi

-------+ + 0=

P P0

# #0

( )(

=

# #0

#' ; #' #0

1+=

P P0

P' ; P' P0

1+=

"#

"t------ #0

"Uk

"xk

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

#0

"Ui

"t---------

"P

"xi

-------+ 0=


2/26

The acoustic equations

9/12/11 13.2 bjc

Now define the dimensionless density disturbance

. (13.4)

In terms of this variable the equations of motion become.

. (13.5)

Slight rearrangement produces the so-called acoustic equations.

(13.6)

where

. (13.7)

Now differentiate the linearized continuity equation with respect to time and the

linearized momentum equation with respect to space.

. (13.8)

Subtract one from the other in (13.8).

r# #

0

#0---------------=

"r

"t-----

"Uk

"xk

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

#0

"Ui

"t---------

"P

"xi

-------+ 0=

P

P0

------ 1 r+( )(

1 (r+)=

"r

"t-----

"Uk

"xk

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

"Ui

"t--------- a

0

2 "r

"xi

-------+ 0=

a0

2(

P0

#0

------=

"2r

"t2

--------"

2U

k

"xk"t

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

"2U

k

"xk

"t-------------- a

0

2 "2r

"xk

"xk

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


3/26

The acoustic equations

bjc 13.3 9/12/11

. (13.9)

The dimensionless density disturbance satisfies the linear wave equation.

Now repeat the process with the differentiation reversed. The result is

. (13.10)

Now subtract. The velocity fluctuation satisfies the equation

. (13.11)

Recall the vector identity . Using this identity

and the fact that the flow is irrotational, (13.11) can be expressed as the linear

wave equation

(13.12)

Similarly, the pressure satisfies the wave equation. Substitute

. (13.13)

into (13.9).

(13.14)

Finally, the temperature also satisfies the wave equation

"2r

"t2

-------- a0

2 "2r

"xk"xk------------------ 0=

"2r

"t"xi

-------------"

2U

k

"xk

"xi

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

"2U

i

"t2

------------ a0

2 "2r

"xi"

t

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

"2U

i

"t2

------------ a0

2 "2U

k

"xk

"xi

----------------- 0=

! ! U( ) ! ! U( ) !2U=

"2U

i

"t2

------------ a0

2 "2U

i

"xk

"xk

------------------ 0=

r1

(---

P

P0

------ 1$ %& '=

"2P

"t2

--------- a0

2 "2P

"xk

"xk

------------------ 0=


4/26

Propagation of acoustic waves in one space dimension

9/12/11 13.4 bjc

. (13.15)

13.3 PROPAGATIONOFACOUSTICWAVESINONESPACEDIMENSION

In one dimensional flow the acoustic equations become,

. (13.16)

The wave equation,

, (13.17)

can be solved in a very general form,

(13.18)

where and are arbitrary functions. The character of the solution can be

understood by considering first the case . The figure below illustrates the

evolution of an initial distribution of propagating to the right in the plane.

"2T

"t2

--------- a0

2 "2T

"xk"xk------------------ 0=

"r

"t-----

"U

"x-------+ 0=

"U

"t------- a

0

2"r

"x------+ 0=

"2r

"t2

-------- a0

2"2r

"x2

-------- 0=

r x t,( ) F x a0t( ) G x a

0t+( )+=

F G

G 0=

F x t


5/26


bjc 13.5 9/12/11

Figure 13.1 Right propagating wave.

The disturbance moves to the right unchanged in shape. Acoustic waves are non-

dispersive meaning that the wave speed does not depend on the wavelength.

The figure below illustrates the evolution of an initial distribution of propagat-

ing to the left in the plane.

Figure 13.2 Left propagating wave.

In the absence of dispersion and dissipation the waves simply translate while

retaining their form. The pressure disturbance produced by the wave is related to

the density by

2 4 6 8

-1

1

2

3

4

5F

F

x

t1

0

t2

t

G

x t

-2 2 4 6

-1

1

2

3

4

5

x0

t1

t2

G

G

Gt


6/26


9/12/11 13.6 bjc

(13.19)

or, in differential form,

. (13.20)

The velocity induced by the acoustic disturbance can also be written in a very gen-

eral form

(13.21)

Substitute the expressions for and into the acoustic equations

(13.22)

Let

(13.23)

Equation (13.22) becomes

(13.24)

Add and subtract the relations in (13.24)

(13.25)

from which we can conclude that The result (13.25)

gives us the relationship between density and velocity in left and right running

waves. Thus

(13.26)

P P0

P0

---------------- (r (# #0

#0

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

dP

P------- (

d#

#------=

U f x a0t( ) g x a

0t+( )+=

r U

" F x a0t( ) G x a0t+( )+( )"t

--------------------------------------------------------------------- " f x a0t( ) g x a0t+( )+( )

"x--------------------------------------------------------------------+ 0=

" f x a0t( ) g x a

0t+( )+( )

"t-------------------------------------------------------------------- a

0

2" F x a0t( ) G x a0t+( )+( )

"x---------------------------------------------------------------------+ 0=

* x a0t= + x a

0t+=

a0F* a0G+ f* g++ + + 0=

f* g+ a0F* a0G++ + + 0=

a0G+ g++ 0=

a0F* f* 0=

g a0G ; f a

0F= =

U a0F x a

0t( ) a

0G x a

0t+( )=


7/26

Isentropic, finite amplitude waves

bjc 13.7 9/12/11

Now

. (13.27)

In a right running wave

(13.28)

and in a left running wave

. (13.29)

In differential form

. (13.30)

13.4 ISENTROPIC,FINITEAMPLITUDEWAVES

The speed of sound may vary from one point to another and in a general, 1-D,

isentropic flow.

. (13.31)

Locally, the particle disturbance velocity is given by (13.30) with the speed of

sound regarded as a function of space and time,

. (13.32)

Equation (13.32) is one of the most important equations in unsteady gas dynamics

and is often introduced as an intuitively obvious extension of equation (13.30).

U

a0r

--------F G

F G+---------------=

U

a0r

-------- 1=

U

a0r

-------- 1=

dU a0

d#

#------=

dP

P------- (

d#

#------=

a a0

#

#0

------$ %& '

( 1

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

=

dU ad##------=


8/26


9/12/11 13.8 bjc

We can put (13.32) on a firmer foundation by returning to the full nonlinear equa-

tions for one-dimensional, isentropic flow. In one dimension (13.1) becomes

(13.33)

Now assume that the density can be written as a function of the velocity.

(13.34)

The derivatives of the density are

(13.35)

Substitute (13.35) into (13.33).

(13.36)

Rearrange (13.36).

(13.37)

Comparing the continuity and momentum equation in (13.37) we see that in order

for (13.34) to be a solution of (13.33) we must have

(13.38)

which can be rearranged to read

"#

"t------ #

"U

"x------- U

"#

"x------+ + 0=

"U

"t------- U

"U

"x-------

(P0

#02

---------#

#0

------$ %& '

( 2 "#

"x------+ + 0=

# # U( )=

"#

"t------d#

dU-------"U

"t-------="#

"x------d#

dU-------"U

"x-------=

d#

dU-------

"U

"t------- #

"U

"x------- U

d#

dU-------

"U

"x-------+ + 0=

"U

"t------- U

"U

"x-------

(P0

#0

2---------

#

#0

------$ %& '

( 2 d#

dU-------

"U

"x-------+ + 0=

"U

"t------- U

"U

"x-------

#

d# dU------------------

"U

"x-------+ + 0=

"U

"t------- U

"U

"x-------

(P0

#0

2---------

#

#0------

$ %& '

( 2 d#

dU-------

"U

"x-------+ + 0=

#

d# dU------------------

(P0

#0

2---------

#

#0

------$ %& '

( 2 d#

dU-------=


9/26


bjc 13.9 9/12/11

(13.39)

Take the square root of (13.39)

(13.40)

Now substitute the isentropic relation between the speed of sound and density

(13.41)

into (13.40). Equation (13.40) becomes

(13.42)

which confirms the validity of (13.32).

Using the isentropic assumption, (13.32) can be integrated from some initial state

to a final state.

(13.43)

Thus

(13.44)

The local acoustic speed is

(13.45)

and the wave speed at any point is

d#

dU-------

$ %& '

2 #0

2

a0

2------

#

#0

------$ %& '

3 (=

d#

dU-------

#0

a0

------#

#0

------$ %& '

3 (

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

=

a

a0

-----#

#0

------$ %& '

( 1

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

=

dU ad#

#------=

U ad#

#------

#1

#

, a1#

#1

------$ %& '

( 1

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

d#

#------

#1

#

,= =

U2a

1

( 1------------

#

#1

------$ %& '

( 1

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

1

$ %- .- .& '

2

( 1------------ a a

1( )= =

a a1

( 1( )

2----------------- U=


10/26

Centered expansion wave

9/12/11 13.10 bjc

(13.46)

13.5 CENTEREDEXPANSIONWAVE

Now consider the situation shown below where a piston is withdrawn from a fluid

at rest.

Figure 13.3 Withdrawal of a piston from a compressible gas.

The speed of the fluid in region (3) is equal to the piston speed

. (13.47)

Note that is negative. In effect (13.47) gives us the sound speed in region (3)

in terms of the known piston speed and the known speed of sound in region (4).

. (13.48)

From this we can see that the gas in region (3) is cooler than that in region (4),

consistent with what we would expect for an expansion.

The front or leading characteristic of the wave propagates to the right with wave

speed equal to the speed of sound . the tail of the wave (terminating char-

acteristic) moves to the right with wave speed

(13.49)

c a U a1

( 1( )

2----------------- U U a

1

( 1+( )

2------------------ U= = =

(3) (4)Up U=0

x

Up

2

( 1------------ a

3a

4( )=

Up

a3

a4

( 1

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

$ %& ' U

p+=

c4

a4

=

c3

a3

Up

+ a4

( 1+

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

$ %& ' U

p+= =


11/26

Centered expansion wave

bjc 13.11 9/12/11

and the expansion zone widens at the rate . It is useful to visu-

alize the situation on the plane.

Figure 13.4 Propagation of an expansion wave in the x-t plane.

The density ratio across the wave is given by the isentropic relation

(13.50)

or

(13.51)

The pressure ratio is,

. (13.52)

and the temperature ratio is

( 1+( ) 2( ) Up

x t

x

t

piston

path

x=-|Up|t

x=c3t

x=c4t

(4)

(3)

#3

#4

------

a3

a4

-----$ %- .& '

2

( 1------------

=

#3

#4

------ 1( 1

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

$ %& '

Up

a4

-------+$ %- .& '

2

( 1------------

=

P3

P4

------ 1( 1

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

$ %& '

Up

a4

-------+$ %- .& '

2(

( 1------------

=


12/26

Compression wave

9/12/11 13.12 bjc

(13.53)

In summary, our solution of the small disturbance wave equation from the previ-ous section has allowed us to determine all the properties of a finite expansion

wave. Things worked out this way because the expansion wave is isentropic to a

high degree of accuracy. Note the similarity between the 1-D unsteady expansion

wave and the steady Prandtl-Meyer expansion and the presence of space-time

characteristics in the plane along which the properties of the flow are

constant.

13.6 COMPRESSIONWAVESuppose the piston motion is to the right into the gas. None of the above relations

change but the physics does change and this can be seen from the diagram.

Figure 13.5 Compression wave converging to a shock in the x-t plane.

The wave speed at the surface of the piston is,

. (13.54)

In this case the characteristics tend to cross, at which point the isentropic assump-

tion is no longer valid. The system of compression waves catches up to itself

(nonlinear steepening) in a very short time collapsing to form a shock wave.

T3

T4

------P

3

P4

------$ %- .& '

( 1

(------------

1( 1

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

$ %& '

Up

a4

-------+$ %- .& '

2

= =

x t

x t

x

t

pistonpath

x=|Up|t

x=c3t

x=c4t

(1)

(2)

shock

c2

a2

Up

+ a1

( 1+

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

$ %& ' U

p+= =


13/26

The shock tube

bjc 13.13 9/12/11

13.7 THESHOCKTUBE

A very important device used to study shock waves in the laboratory is the shock

tube.

Figure 13.6 Shock tube flow.

Regions (1) and (2) of the shock tube are initially separated into high and low pres-

sure sections by a diaphragm. At the diaphragm bursts and the highpressure driver gas rushes into the low pressure test gas. The contact surface

between the two surfaces acts like a piston and the piston motion produces a shock

wave moving at the speed . The shock wave arises through a process of nonlin-

ear steepening based on the fact that the speed of a pressure wave increases with

an increase in the temperature of the gas. Pressure waves to the rear of the com-

low pressure (1)high pressure (4)

P1

P4

t < 0

P1

P4

T1

T4

T2

T3

(1)(2)(3)(4) cs

shockcontactsurface

expansion

U1 = 0U4 = 0 U2 = U3 = Up

t > 0

t 0=

cs


14/26

The shock tube

9/12/11 13.14 bjc

pression front move more rapidly than those near the front, quickly catching up

to form a very thin pressure jump as illustrated in Figure 13.5. The important gas-

dynamic problem is to determine the pressure ratio across the shock for

a given choice of gases and diaphragm pressure ratio . In

the following we will combine the results for expansion waves with normal shock

relations to derive the so-called shock tube equation.

The conditions at the contact surface are,

(13.55)

where is the speed of the slug of gas set into motion by the opening of the

diaphragm. This is the effective piston speed of the fluid released by the

diaphragm.

In a frame of reference moving with the shock wave the gas velocities are.

. (13.56)

and the shock jump conditions are

(13.57)

where . Using the first relation in (13.57) we can write

P2

P1

(1

a1

(4

a4

, , ,( ) P4

P1

P2

P3

=

U2

U3

Up

= =

U2

U'1

cs

=

U'2

cs

Up

+=

U'2

U'1

--------

1(

11

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

1

2+

(1

1+

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

$ %& 'M

1

2

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

P2

P1

------

(1M12 (1 1

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

(1

1+

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

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

M1

cs

a1

U'1

a1

= =


15/26

The shock tube

bjc 13.15 9/12/11

. (13.58)

The piston velocity is

. (13.59)

Note that is positive. Using the second relation in (13.57), equation (13.59)can be expressed in terms of the shock pressure ratio as,

. (13.60)

Equation (13.60) is the expression for the piston velocity derived using normalshock theory.

Now lets work out an expression for the piston velocity using isentropic expan-

sion theory. The velocity behind the expansion is,

. (13.61)

Equate (13.60) and (13.61).

U'2

U'1

U'1

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

1(1 1

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

1

2+

(1 1+

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

$ %& 'M

1

2

---------------------------------- 11 M

1

2

(1 1+

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

$ %& 'M

1

2

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

Up

U'2

U'1

U'1

1 M1

2

(1

1+

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

$ %& 'M

1

U'1

a1

--------$ %- .& '

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

$ %- .- .- .- .- .& '

a1

M1

21

(1 1+

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

$ %& 'M

1

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

$ %- .- .- .& '

= = =

Up

Up

a1

P2

P1

------ 1$ %- .& ' 2 (1

(1

1+( )P

2

P1

------$ %- .& '

(1

1( )+

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

$ %- .- .- .- .- .& ' 1 2

=

U3

Up

2a4

(4

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

P3

P4

------$ %- .& '

(4

1

2(4

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

$ %- .- .- .- .& '

= =


16/26

The shock tube

9/12/11 13.16 bjc

(13.62)

Using the following identity

. (13.63)

and noting that solve (13.62) for . The result is the basic

shock tube equation,

(13.64)

plotted in Figure 13.7. Given , , , and we can determine the

shock strength from (13.64).

a1

P2

P1

------ 1$ %- .& ' 2 (1

(1

1+( )P

2

P1

------$ %- .& '

(1

1( )+

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

$ %- .- .- .- .- .& ' 1

2---

2a4

(4

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

P3

P2

------$ %- .& ' P2

P1

------$ %- .& ' P1

P4

------$ %- .& '

$ %- .& '

( 4 1

2( 4---------------

$ %- .- .- .- .& '

=

P3

P4

------P

3

P2

------$ %- .& ' P2

P1

------$ %- .& ' P1

P4

------$ %- .& '

=

P3

P2

1= P4

P1

P4

P1

------P

2

P1

------ 1

(4 1( )a

1

a4

-----$ %- .& ' P2

P1

------ 1$ %- .& '

2(1 2(1 (1 1+( )P

2

P1

------ 1$ %- .& '

+

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

$ %- .- .- .- .- .- .& '

2(4

(4

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

=

P4

P1

(1

(4

a1

a4

P2 P1


17/26

The shock tube

bjc 13.17 9/12/11

Figure 13.7 Shock tube pressure ratio versus shock strength.

Once the shock strength is determined, the shock Mach number can be determined

from

(13.65)

or

(13.66)

Notice that very large values of are required to generate strong shocks.

One way to improve the performance of the shock tube is to select a driver gas

with a large speed of sound. Helium driving into nitrogen will produce a much

stronger shock wave than say Argon driving into nitrogen at a given . In

essence Helium provides a much lighter piston that can translate much more rap-

idly into the test gas. The figure below provides this comparison.

2 3 4 5 6

20

40

60

80

P4

P1

------

P2

P1

(1

(4

1.4= =

a1 a4=

P2

P1

------2(

1

(1

1+----------------M

s

2 (1 1

(1

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

$ %- .& '

=

Ms

(1

1+

2(1

----------------$ %- .& '

1 2 P2

P1

------(

11

(1

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

$ %- .& ' 1 2

=

P4

P1

P4 P1


18/26

The shock tube

9/12/11 13.18 bjc

Figure 13.8 Shock Mach number for several driver gases.

Hydrogen as the driver gas produces the strongest shock waves however the fire

and explosion hazard involved has limited its use.

13.7.1 EXAMPLE - FLOWINDUCEDBYTHESHOCKINASHOCKTUBE

The flow in a shock tube is shown below

The shock wave induces a gas velocity in the laboratory frame,

where

(1)(2)(3)(4) cs

shockcontactsurface

expansion

Up

Up

U2

U3

= =


19/26

The shock tube

bjc 13.19 9/12/11

(13.67)

Suppose the gas in the driver section and test section is Air initially at .

A shock wave with a Mach number of 2 is produced in the tube.

1) Determine the stagnation temperature of the gas in region (2) in the labo-

ratory frame.

2) Determine the stagnation temperature of the gas in region (3) in the labo-

ratory frame.

Answer

The pressure ratio across a Mach number 2 shock in Air is,

(13.68)

and the temperature ratio is,

. (13.69)

The speed of sound in Air is and the heat capacity

at constant pressure is, . The piston velocity is,

(13.70)

The temperature in region (3) is obtained from (13.53).

(13.71)

Now, in region (2)

U2

a1

P2

P1

( ) 1( )2 (

1

(1

1+( ) P2

P1

( ) (1

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

$ %- .& '

1 2

=

300K

P2

P1

4.5=

T2

T1

1.687=

a1

(RT 347 M/sec= =

Cp

1005 M2

S ec2

K=

U2

a1

P2

P1

( ) 1( )2 (

1

(1

1+( ) P2

P1

( ) (1

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

$ %- .& '

1 2

= =

347 3.5( )2

15.68-------------

$ %& '

1 2433.75 M/sec=

T3

T4

------ 1( 1

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

$ %& '

Up

a4

-------+$ %- .& ' 2

1 0.2433.75

347----------------

$ %& '$ %

& '2

0.5625= = =


20/26

Problems

9/12/11 13.20 bjc

(13.72)

and in region (3)

(13.73)

13.8 PROBLEMS

Problem 1 - Consider the expression ; n=1 corresponds to the mass flux, n=2

corresponds to the momentum flux and n=3 corresponds to the energy flux of a

compressible gas. Use the acoustic relation,

(13.74)

to determine the Mach number (as a function of n) at which is a maximum

in a one-dimensional, unsteady expansion wave. The steady case is considered in

one of the problems at the end of Chapter 9.

Problem 2 - In the shock tube example discussed above, determine the stagnation

pressure of the gas in regions (2) and (3). Determine the stagnation pressure in

both the laboratory frame and in the frame of reference moving with the shock

wave.

Problem 3 - The figure below shows a sphere moving over a flat plate in a

ballistic range (Van Dyke figure 271). The sphere has been fired from a gun

and is translating to the left at a Mach number of 3. The static temperature of

the air upstream of the sphere is 300K and the pressure is one atmosphere.

On the plane of symmetry of the flow (the plane of the photo) the shock inter-

sects the plate at an angle of 25 degrees as indicated.

Tt2

T2

1

2---

Up

2

Cp

-------+ 1.687 300433.75

2

2 1005---------------------+ 506.1 93.6+ 599.7 K= = = =

Tt3

T3

1

2---

Up

2

Cp

-------+ 0.5625 300433.75

2

2 1005---------------------+ 168.75 93.6+ 262.35K= = = =

#Un

dU ad#

#------=

#Un


21/26

Problems

bjc 13.21 9/12/11

i) Determine the flow Mach number, speed and angle behind the shock in

region 2close to the intersection with the plate. Work out your results in a

frame of reference moving with the sphere. In this frame the upstream air is

moving to the right at 3 times the speed of sound.

ii) What are the streamwise and plate normal velocity components of the flow

in region 2 referred to a frame of reference at rest with respect to the upstream

gas.

iii) Determine the stagnation temperature and pressure of the flow in region 2

referred to a frame of reference at rest with respect to the upstream gas.

Problem 4 - We normally think of the shock tube as a device that can be used

to study relatively strong shock waves. But shock tubes have also been used

to study weak shocks relevant to the sonic boom problem. Suppose the shock

tube is used to generate weak shock waves with . Show thatfor small the relationship between and is approximated by

. (13.75)

252

P2 P1

1 /+=/ P

2P

1 P

4P

1

P4

P1

1 A/+=


22/26

Problems

9/12/11 13.22 bjc

How does A depend on the properties of the gases in regions 1 and 4? Use the

exact theory to determine the strength of a shock wave generated in a air-air

shock tube operated at . Compare with the approximate result.

Problem 5 - The sketch below shows a one meter diameter tube filled with

Air and divided into two volumes by a heavy piston of weight . The piston

is held in place by a mechanical stop and the pressure and temperature are

uniform throughout the tube at one atmosphere and 300K. Body force effects

in the gas may be neglected.

P4

P1

1.2=

W

g

cs

(1)

(2)

(3)

(4)

(1)

(4)

Up


23/26

Problems

bjc 13.23 9/12/11

The piston is released and accelerates downward due to gravity. After a short

transient the piston reaches a constant velocity . The piston drives a shock

wave ahead of it at a wave speed equal to twice the sound speed in region

1. What is the weight of the piston?

Problem 6 - Each time Stanford makes a touchdown an eight inch diameter, open

ended shock tube is used to celebrate the score. Suppose the shock wave devel-

oped in the tube is required to have a pressure ratio of 2. What pressure is needed

in the driver section? Assume the driver gas is Air. What is the shock Mach num-

ber? Suppose the shock tube is mounted vertically on a cart as shown in the figure

below. Estimate the force that the cart must withstand when the tube fires.

Problem 7 - A moveable piston sits in the middle of a long tube filled with

Air at one atmosphere and 300K. At time zero the piston is moved impul-

sively to the right at .

Up

cs

Up

200m sec=

cs

(1)(2)(3)(4)

(1)(4)

Up


24/26

Problems

9/12/11 13.24 bjc

1) What is the pressure on the right face of the piston (region 2) in

atmospheres?

2) What is the pressure on the left face of the piston (region 3) in atmospheres?

Problem 8 - One of the most versatile instruments used in the study offluid

flow is the shock tube. It can even be used as a wind tunnel as long as one is

satisfied with short test times. The figure below illustrates the idea. An airfoil

is placed in the shock tube and after the passage of the shock it is subject to

flow of the test gas at constant velocity and temperature . In a real

experiment the contact surface is quite turbulent and so the practical useful-

ness of the flow is restricted to the time after the arrival of the shock and before

the arrival of the contact surface, typically a millisecond or so.

Proponents of this idea point out that if the shock Mach number is very largethe flow over the body can be supersonic as suggested in the sketch above.

1) Show that this is the case.

2) For very large shock Mach number, which test gas would produce a higher

Mach number over the body, helium or air. Estimate the Mach number over

the body for each gas.

Problem 9 - The figure below shows a shock wave reflecting from the endwall

of a shock tube. The reflected shock moves to the left at a constant speedinto the gas that was compressed by the incident shock. The gas behind the

reflected shock, labeled region (5), is at rest and at a substantially higher tem-

perature and pressure than it was in state (1) before the arrival of the incident

shock.

Up

T2

(1)(2)

(3)(4) cs

shockcontactsurface

expansion

Up

Up

U

x

crs


25/26

Problems

bjc 13.25 9/12/11

1)Suppose the gas in the driver and test sections is Helium at an initial tem-

perature of 300K prior to opening the diaphragm. The Mach number of the

incident shock wave is 3. Determine the Mach number of the reflected shock.

2)Determine .

Problem 10 - One of the key variables in the design of a shock tube is the

length needed for a shock to develop from the initial compression process.

Suppose a piston is used to compress a gas initially at rest in a tube. During

the startup transient the piston speed increases linearly with time as

shown on the x-t diagram below.

(1)(2)

(3)(4) cs

shockcontactsurface

expansion

Up

Up

U

x

(5)

(2)(3)(4) crs

shockcontact

surface

expansion

Up

Up

U

x

U 0=

U 0=

T5

T1

0 t t1

<


26/26

Problems

In a shock tube the startup time is generally taken to be the time required

for the diaphragm to open. Let the gas be Air at . Use

and to estimate the distance needed

for the shock wave to form.

x

t

piston speed

(1)

(2)

shock

Up = Att1

0 < t < t1

(2) (1)Up

U=0

x

t

1T

1300K=

t1

5 103

sec= A 4 104 M/sec2

=

Ch 13 Unsteady Waves

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Transcript of Ch 13 Unsteady Waves