Section 6: Home Work Solution Part 1

1MAE 5420 - Compressible Fluid Flow

Section 6: Home Work Solution

• 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°

! !!

Stephen Whitmore

Stephen Whitmore

Part 1


Compute First Corner

entrance mach wave angle

• M1=4

Corner entrance mach angle

µ1= sin

!1 1

M1

"

#$

%

&' =

1

4! "# $

asin180

%& =14.48°

M1

M215°

Stephen Whitmore



exit mach number

• M1=4

M1

M215°

!(M2) = " +!(M

1)

!(M1) =

" +1

" #1tan

#1 " #1" +1

M1

2 #1( )$%&

'&

()&

*&# tan#1

M1

2 #1

= 1.41710radians

=15!

1801.6789+ =1.6789 radians

• Iterative Solver ---->

M2 = 4.935

Stephen Whitmore

Stephen Whitmore

1.417



exit Total and static pressure

• M1=4.0, M2=4.935

• Because each mach wave is infinitesimal, expansion is isentropic

P02= P0

1= p

11+

! "1

2M

1

2#

$%&

'(

!! "1

=

=2.43 atm1

1.25 1!2

42

( )+" #$ %

1.25

1.25 1!" #$ %

0.01( )

p2= p

1

1+! "12

M1

2

1+! "12

M2

2

#

$

%%%

&

'

(((

!! "1

=

11.25 1!2

42

+

11.25 1!2

4.9352

+" #$ %$ %$ %$ %& ' 1.25

1.25 1!" #& '

0.01( )

=0.002245 atm



exit Total and static temperature

• M1=4.0, M2=4.935

• Because each mach wave is infinitesimal, expansion is isentropic

=651 °K

=160.97 °K

T 02= T 0

1= T

11+

! "1

2M

1

2#

$%&

'(= 1

1.25 1!2

42

+" #$ %

217( )

T2= T

1

1+! "12

M1

2

1+! "12

M2

2

#

$

%%%

&

'

(((

=1

1.25 1!2

42

+

11.25 1!2

4.9352

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

217( )



exit Mach angle (relative to freestream)

• M1=4.0, M2=4.935

µ2= sin

!1 1

M2

"

#$

%

&' !( =

1

4.935! "# $

asin180

%& 15' = -3.309 °

M1

M215°

M1= 4

p1= 0.01atm

T1= 217

oK

µ1= 14.48

o

!

"

##

$

##

%

&

##

'

##

M2= 4.935

p2= 0.0025atm

T2= 160.97

oK

µ2= !3.309o

"

#

$$

%

$$

&

'

$$

(

$$

.002245 atm

Stephen Whitmore


Compute Second Corner

Shock AngleM2= 4.935

p2= 0.0025atm

T2= 160.97

oK

µ2= !3.309o

"

#

$$

%

$$

&

'

$$

(

$$

= 1 (15°) + 2 (15°) = 30°

tan !( ) =2 M

1

2sin

2 "( ) #1{ }tan "( ) 2 + M

1

2 $ + cos 2"( )%& '(%& '(

2=39.409° =

24.409° (to freestream)

M1

M220°

M3

15°

!=30°

!! !

!

.002245 atm

15o

Stephen Whitmore



exit mach numberM

2= 4.935

p2= 0.0025atm

T2= 160.97

oK

µ2= !3.309o

"

#

$$

%

$$

&

'

$$

(

$$

M1

M220°

M3

15°

!=30°

2=39.409°

• Compute Normal Component of M2

Mn2= M

2sin!

2

4.935!180

39.409" #$ %

sin = 3.133

!.002245 atm

15o



exit mach number (cont’d)

M2= 4.935

p2= 0.0025atm

T2= 160.97

oK

µ2= !3.309o

"

#

$$

%

$$

&

'

$$

(

$$

• Change in Normal Component across

Shock wave

=0.428218

=

11.25 1!2

3.1332

( )+

1.25 3.1332

( )1.25 1!2

!" #$ %$ %$ %$ %& ' 0.5

Mn3=

1+! "1( )2

Mn2

2#$%

&'(

!Mn2

2 "! "1( )2

#$%

&'(

.002245 atm



exit mach number (concluded)

M2= 4.935

p2= 0.0025atm

T2= 160.97

oK

µ2= !3.309o

"

#

$$

%

$$

&

'

$$

(

$$

• from geometry

M3=

Mn3

sin !2"#

2c( )=

0.4282

sin$180

9.409%&'

()*

= 2.6193

M3M3

Mn3Mt3

!"#

.002245 atm

Stephen Whitmore


Compute Static Pressure Behind Oblique Shock

M2= 4.935

p2= 0.0025atm

T2= 160.97

oK

µ2= !3.309o

"

#

$$

%

$$

&

'

$$

(

$$

p3= p

21+

2!! +1( )

Mn2

2 "1( )#

$%

&

'(

=

0.02423 atm

M3= 2.6193

Mn3= 0.4282

!2= 39.409

o

Mn2= 3.133

"

#$$

%$$

&

'$$

($$

.002245 atm

12 1.25!1.25 1+( )

3.13321"( )+# $

% &0.002245( )


Compute Total Pressure Behind Oblique Shock

M2= 4.935

p2= 0.0025atm

T2= 160.97

oK

µ2= !3.309o

"

#

$$

%

$$

&

'

$$

(

$$

• Flow Behind Shock is Isentropic M3= 2.6193

Mn3= 0.4282

!2= 39.409

o

Mn2= 3.133

"

#$$

%$$

&

'$$

($$

P03= p

31+

! "1

2M

3

2#

$%&

'(

!! "1

=

=0.5359 atm

.002245 atm

11.25 1!2

2.61932

( )+" #$ %

1.25

1.25 1!( )0.02423( )


Compute Temperature Behind Oblique Shock Wave

M2= 4.935

p2= 0.0025atm

T2= 160.97

oK

µ2= !3.309o

"

#

$$

%

$$

&

'

$$

(

$$

•M3= 2.6193

Mn3= 0.4282

!2= 39.409

o

Mn2= 3.133

"

#$$

%$$

&

'$$

($$

T3

T2

= 1+2!! +1( )

M2sin"2( )

2

#1( )$

%&

'

()2 + ! #1( ) M 2

sin"2( )2( )

! +1( ) M 2sin"

2( )2

$

%

&&

'

(

))

= 350.4 °K.002245 atm


Compute Mach Number behind Shock

using Alternate method

• Compute Tangential Component

M3M3

Mn3Mt3

!"#

4.935!180

39.409" #$ %

cos =3.813

Mt2= M

2cos(!

2) =



using Alternate method (cont’d)

• Across Oblique Shock

Tangential Velocity Component

is preserved M3M3

Mn3Mt3

!"#

Vt2= Vt

3! Mt

2c2= Mt

3c3!

Mt3=Mt

2c2

c3

= Mt2

T2

T3

!

M3= Mt

3

2+ Mn

3

2"# $%



using Alternate method (cont’d)

M3M3

Mn3Mt3

!"#

Mt3= Mt

2

T2

T3

= M2cos(!

2)T2

T3

=

4.935!180

39.409" #$ %

cos" #$ % 160.97

350.4" #$ %

0.5

=2.584



using Alternate method (concluded)

M3M3

Mn3Mt3

!"#

=2.619

• From Earlier

=0.428218Mn3

2

M3= Mt

3

2+ Mn

3

2!" #$ =

2.5842

( ) 0.42822

+( )0.5

Check!


Summary

M1

M220°

15°

!=30°

M3

M1= 4

p1= 0.01atm

T1= 217

oK

µ1= 14.48

o

!

"

##

$

##

%

&

##

'

##

M2= 4.935

p2= 0.0025atm

T2= 160.97

oK

µ2= !3.309o

"

#

$$

%

$$

&

'

$$

(

$$

M3= 2.6193

Mn3= 0.4282

!2= 39.409

o

p3= 0.0025atm

P03= 0.5364atm

T3= 350.4

oK

"

#

$$$$

%

$$$$

&

'

$$$$

(

$$$$

.002245 atm

0.5359 atm

0.02423 atm

15o

Stephen Whitmore

Section 6: Home Work Solution Part 1

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Transcript of Section 6: Home Work Solution Part 1