Section 6: Home Work Solution Part 1
Transcript of 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°
! !!
2MAE 5420 - Compressible Fluid Flow
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°
3MAE 5420 - Compressible Fluid Flow
Compute First Corner
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
4MAE 5420 - Compressible Fluid Flow
Compute First Corner
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
5MAE 5420 - Compressible Fluid Flow
Compute First Corner
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( )
6MAE 5420 - Compressible Fluid Flow
Compute First Corner
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
7MAE 5420 - Compressible Fluid Flow
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
8MAE 5420 - Compressible Fluid Flow
Compute Second Corner
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
9MAE 5420 - Compressible Fluid Flow
Compute Second Corner
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
10MAE 5420 - Compressible Fluid Flow
Compute Second Corner
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
11MAE 5420 - Compressible Fluid Flow
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( )
12MAE 5420 - Compressible Fluid Flow
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( )
13MAE 5420 - Compressible Fluid Flow
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
14MAE 5420 - Compressible Fluid Flow
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) =
15MAE 5420 - Compressible Fluid Flow
Compute Mach Number behind Shock
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"# $%
16MAE 5420 - Compressible Fluid Flow
Compute Mach Number behind Shock
using Alternate method (cont’d)
M3M3
Mn3Mt3
!"#
Mt3= Mt
2
T2
T3
= M2cos(!
2)T2
T3
=
4.935!180
39.409" #$ %
cos" #$ % 160.97
350.4" #$ %
0.5
=2.584
17MAE 5420 - Compressible Fluid Flow
Compute Mach Number behind Shock
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!
18MAE 5420 - Compressible Fluid Flow
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