section6.2.pdf

1MAE 5420 - Compressible Fluid Flow

Section 6 Lecture 2:Prandtl-Meyer ExpansionWaves

• Anderson, Chapter 4 pp.167-190


What Happens when

•M1 = 3.0, p1=1atm, =1.4, T1=288°K, =0.00001°

• Explicit Solver for

! = M1

2 "1( )2

" 3 1+# "12

M1

2$%&

'()1+

# +1

2M

1

2$%&

'()tan

2 *( ) =8.0

! =

M1

2 "1( )3

" 9 1+# "12

M1

2$%&

'()1+

# "12

M1

2+# +1

4M

1

4$%&

'()tan

2 *( )

+ 3=1.0

!!

!


What Happens when (cont’d)

tan !( ) =

M1

2 "1( ) + 2# cos4$% + cos

"1 &( )3

'()

*+,

3 1+- "12

M1

2./0

123tan 4( )

=19.47° µ =180

!sin

"1 1

M1

#

$%

&

'( = 19.47

o

• “mach line”

•M1 = 3.0, p1=1atm, =1.4, T1=288°K, =0.00001°!!

!


What Happens when (cont’d)

• Compute Normal Component of Free stream mach Number

Mn1= M

1sin! = 1.0000

• p2

p1

= 1+2!

! +1( )Mn

1

2"1( ) = 1.0 (NO COMPRESSION!)

•M1 = 3.0, p1=1atm, =1.4, T1=288°K, =0.00001°!!


Expansion Waves• So if

>0 .. Compression around corner

=0 … no compression across shock

!

"

M1

M2

!

!


Expansion Waves (concluded)

• Then it follows that <0 .. We get an expansion wave (Prandtl-Meyer)

• Flow accelerates around corner

• Continuous flow region …sometimes called “expansion fan”

• Each mach wave is infinitesimally weak

isentropic flow region

• Flow stream lines are curved and smooth through fan

!


Prandtl-Meyer Expansion Fan: MathematicalAnalysis

• Consider flow expansion around an infinitesimal corner

Infinitesimal Expansion Fan Flow Geometry

V

V+dV dθ

µ

Mach Wave

dθµdθ V

• From Law of SinesV

sin!2" µ " d#$

%&'()

=V + dV

sin!2+ µ

$%&

'()


Prandtl-Meyer Expansion Fan: MathematicalAnalysis (cont’d)

• Using the trigonometric identities

• And …

sin!2+ µ

"#$

%&'= sin

!2

"#$

%&'cos µ( ) + sin µ( )cos

!2

"#$

%&'= cos µ( )

sin!2( µ

"#$

%&'= sin

!2

"#$

%&'cos µ( ) ( sin µ( )cos

!2

"#$

%&'= cos µ( )

sin!2" µ " d#$

%&'()= sin

!2" µ

$%&

'()cos d#( ) " cos

!2" µ

$%&

'()sin d#( ) =

cos µ( )cos d#( ) " cos!2

$%&

'()cos µ( ) + sin µ( )sin

!2

$%&

'()

*

+,

-

./sin d#( ) =

cos µ( )cos d#( ) " sin µ( )sin d#( )



• Substitution gives

V

cos µ( )cos d!( ) " sin µ( )sin d!( )=V + dV

cos µ( )#

1+dV

V=

cos µ( )cos µ( )cos d!( ) " sin µ( )sin d!( )

• Since dθ is considered to be infinitesimal

cos d!( ) = 1

sin d!( ) = d!



• and the equation reduces to

1+dV

V=

cos µ( )cos µ( ) ! sin µ( ) d"( )

=1

1! tan µ( ) d"( )

• Exploiting the form of the power series (expanded about x=0)

1

1! x= 1! x( )

|x=0!

1

1! x( )2

|x=0

(!1)"

#$$

%

&''x ! 0( ) + ....O x

2( )(



• Then

• Since dV is infinitesimal … truncate after first order term

1

1!dV

V

= 1+1

1!dV

V

"#$

%&'2

|dV

V=0

dV

V! 0"

#$%&'+O

dV

V

2"

#$%

&'

1

1!dV

V

" 1+dV

V#

1

1!dV

V

"1

1! tan µ( ) d$( )



• Solve for d in terms of dV/V

1

1!dV

V

" 1+dV

V#

1

1!dV

V

"1

1! tan µ( ) d$( )

1! tan µ( ) d$( ) = 1!dV

V# d$ =

1

tan µ( )

dV

V

• Since disturbance is infinitesimal (mach wave)

sin µ( ) =1

M

!



• Performing some algebraic and trigonometric voodoo

sin µ( ) =1

M! sin

2 µ( ) =1

M2=sin

2 µ( ) + cos2 µ( )M

2!

M2 =

sin2 µ( ) + cos2 µ( )sin

2 µ( )= 1+

1

tan2 µ( )

!1

tan2 µ( )

= M 2"1

1

tan µ( )= M

2"1

• and …. d! =

1

tan µ( )

dV

V= M

2"1

dV

V• Valid forReal and ideal gas



• For a finite deflection the O.D.E is integrated over the complete expansion fan

! = M2 "1

dV

VM1

M2

#

• Write in terms of mach by …

V = M ! c" dV = dM ! c + M ! dc"

dV

V=dM ! c + M ! dc

M ! c=dM

M+dc

c



• Substituting in

! = M2 "1

dV

VM1

M2

# = M2 "1

dM

M+dc

c

$%&

'()

M1

M2

#

• For a calorically perfect adiabatic gas flow

And T0 is constantc0= ! R

gT0"

c0

c

#$%

&'(=

T0

T

#$%

&'(= 1+

! )12

M2*

+,-./



• Solving for c, differentiating, and normalizing by c

dc

c=

!1

2

"#$

%&'

1

1+( !12

M2"

#$%&'

(( !1)M dM( )

1+( !12

M2"

#$%&'

= !

(( !1)2

M dM( )

1+( !12

M2"

#$%&'



• Returning to the integral for

! = M2 "1

dV

VM1

M2

# = M2 "1

dM

M+ "

($ "1)2

M dM( )

1+$ "12

M2%

&'()*

%

&

'''

(

)

***M1

M2

#

!



• Simplification gives

! = M2 "1

dM

M1+ "

(# "1)2

M2

1+# "12

M2$

%&'()

$

%

&&&

'

(

)))M1

M2

* =

M2 "1

dM

M

1+# "12

M2$

%&'()"(# "1)2

M2

1+# "12

M2$

%&'()

$

%

&&&

'

(

)))=

M1

M2

*M

2 "1dM

M

1+# "12

M2$

%&'()

$

%

&&&

'

(

)))

M1

M2

*



• Evaluate integral by performing substitution

M2 !1

dM

M

1+" !12

M2#

$%&'(

) * Ln[M ] + u*dM

M= du,M

2= e

2u,-.

/01

M2 !1

dM

M

1+" !12

M2#

$%&'(

) =e2u !1

1+" !12

e2u#

$%&'(

) du



• Standard Integral Table Form

e2u !1

1+" !12

e2u#

$%&'(

) du*emx !1

1+ bemx( )) du

emx !1

1+ bemx( )" du = !

2

mtan

!1emx !1 !

2 b +1( )m

tan!1 b

b +1emx !1( )

b +1 bm

• From tables (CRC math handbook)



• Substituting m = 2,b =! "12,e

mx= M

2#$%

&'(

M2 !1

dM

M

1+" !12

M2#

$%&'(

) = !2

2tan

!1M

2 !1 +2

2

" !12

+1

" !12

tan!1

" !12

" !12

+1

M2 !1( )

*

+,,

-,,

.

/,,

0,,

=

" +1

" !1tan

!1 " !1" +1

M2 !1( )

*+,

-,

./,

0,! tan!1

M2 !1



• Collected Equations

• Or more simply

! =

" +1

" #1tan

#1 " #1" +1

M2

2 #1( )$%&

'&

()&

*&# tan#1

M2

2 #1+

,--

.

/00#

" +1

" #1tan

#1 " #1" +1

M1

2 #1( )$%&

'&

()&

*&# tan#1

M1

2 #1+

,--

.

/00

! ="(M2) #"(M

1)$"(M ) = % +1

% #1tan

#1 % #1% +1

M2 #1( )

&'(

)(

*+(

,(# tan#1

M2 #1

“Prandtl-Meyer Function”

Implicit function … more Newton!



• And we already know the derivative

d

dM!(M )[ ] =

1

M

M2 "1

1+# "12

M2$

%&'()


Newton Solver Algorithm

! ="(M2) #"(M

1)$"(M

2) = ! +"(M

1)

! +"(M1) ="(M

2) ="(M

2 j( ) ) +#"#M

$%&

'()

j( )M

2* M

2 j( )( ) +O(M 2) + ....

• Expand in Taylor’s series

• Truncate after first order terms and solve for M2(j)

M2 j+1( ) =

! +"(M1) #"(M

2 j( ) )$% &'

("(M

)*+

,-.

j( )

+ M2 j( )

• Note: use radians!


M2 versus M1,

M1= 5

M1= 3

M1= 1

!


Pressure and Temperature ChangeAcross Expansion Fan

• Because each mach wave is infinitesimal, expansion is isentropic

- P02 = P01- T02 = T01

p2

p1

=P0

1

p1

!p2

P02

=

1+" #12

M1

2

1+" #12

M2

2

$

%

&&&

'

(

)))

"" #1

T2

T1

=T 0

1

T1

!T2

T 02

=

1+" #12

M1

2

1+" #12

M2

2

$

%

&&&

'

(

)))


Numerical Example

• M1=1.5, p1=81.4 kPa, T1=255.6°K, =1.4, =20°

• Compute M2, p2, P02, T2, T02

M1

M220°

!!


Numerical Example (cont’d)

• M1=1.5

1

1.5! "# $

asin180

%& = 41.81°

!(M1) =

" +1

" #1tan

#1 " #1" +1

M1

2 #1( )$%&

'&

()&

*&# tan#1

M1

2 #1

1.4 1+

1.4 1!" #$ %

0.51.4 1!1.4 1+

1.521!( )" #

$ %0.5

" #& '$ %

atan 1.521!( )0.5

( )atan!

=0.207785 radians

Corner entrance mach angle

Prandtl-Meyer Function

µ1= sin

!1 1

M1

"

#$

%

&' =



• Compute (M2)

!(M2) = " +!(M

1) =

20!

1800.207785+ =0.5569

• Use Iterative Solver to Compute M2

M2 j+1( ) =

! +"(M1) #"(M

2 j( ) )$% &'

("(M

)*+

,-.

j( )

+ M2 j( ) M2=2.2067

!



• Pressure• Because each mach wave is infinitesimal, expansion is isentropic

- P02 = P01- T02 = T01

p2= p

1

1+! "12

M1

2

1+! "12

M2

2

#

$

%%%

&

'

(((

!! "1

=

11.4 1!2

1.52

+

11.4 1!2

2.20672

+" #$ %$ %$ %$ %& ' 1.4

1.4 1!" #& '

81.4 =27.655 kPa



• Temperature• Because each mach wave is infinitesimal, expansion is isentropic

- P02 = P01- T02 = T01

=187.76°K

T2= T

1

1+! "12

M1

2

1+! "12

M2

2

#

$

%%%

&

'

(((

=

11.4 1!2

1.52

+

11.4 1!2

2.20672

+" #$ %$ %$ %$ %& '

255.6



• Exit Mach Angle

µ2= sin

!1 1

M2

"

#$

%

&' !( =

1

2.2067! "# $

asin180

%& 20' = 6.947°


Numerical Example (concluded)

M1

M220°

M1= 1.5

p1= 81.4kPa

T1= 255.56

oK

µ1= 41.81

o

!

"

##

$

##

%

&

##

'

##

M2= 2.2067

p2= 27.655kPa

T2= 187.76

oK

µ2= 6.947

o

!

"

##

$

##

%

&

##

'

##


Maximum Turning Angle

p2! 0" M

2!#

$(#) = % +1

% &1tan

&1 % &1% +1

#2 &1( )'()

*)

+,)

-)& tan&1 #2 &1 =

% +1

% &1&1

.

/01

2342

5max

=% +1

% &1&1

.

/01

2342&

% +1

% &1tan

&1 % &1% +1

M1

2 &1( )'()

*)

+,)

-)& tan&1

M1

2 &1

p2

p1

=P0

1

p1

!p2

P02

=

1+" #12

M1

2

1+" #12

M2

2

$

%

&&&

'

(

)))

"" #1


Maximum Turning Angle (cont’d)• Plotting as a max function of Mach number

• {T2, p2} = 0

!


Maximum Turning Angle (concluded)

Separated Flow Region

• In Reality Viscous Effects Dominate and Flow Separates around steep corner before pressure expands to vacuum


Section 6: Home Work• M1=4, p1=0.01 atm, T1=217°K, =1.25, 1=15°, 2=15°

• Compute After each corner

- Entry and exit Mach wave angles or shock angles- Mach number, - static & total pressure- temperature

M1

M2M315°

15°

! ! !

section6.2.pdf

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