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1MAE 5420 - Compressible Fluid Flow1
Section 7: Home Work Solution
• Calculate FreestreamMach Number
Shock wave
kPa
• Assume�=1.40
Section 7, Homework … Due Friday November 20, 2020Part1
Solution Hints: Assume supersonic flow at probe and Wedge has a 10 deg. half angle, work backwards from probe...Static pressure Measurement is ahead of shock
γ =1.4
“RayleighPitotEquation”
MAE 5420 - Compressible Fluid Flow 2
Section 7: Homework (cont’d)
• Step 1: Compute Mach Number on Lower aft ramp (4)
• From Normal Shock relationships, trial and error,iterative solver for
“Rayleigh Pitot”, (or table A.2 in Anderson)
• M4 = 1.8060
MAE 5420 - Compressible Fluid Flow 3
Section 7: Homework (cont’d)
• Step 2: Compute Prandtl-Meyer Function for flow on Lower aft ramp (4)
= 20.898°
MAE 5420 - Compressible Fluid Flow 4
Section 7: Homework (cont’d)
• Step 3: Compute Turning Angle for Expansion fan onLower Surface, subtract from n(M4)
MAE 5420 - Compressible Fluid Flow 5
Section 7: Homework (cont’d)• Step 4: Compute Mach Number ahead of expansion fan
• Use iterative solver or table A-5 in Anderson
=0.8984
M2 =1.0760
MAE 5420 - Compressible Fluid Flow 6
Section 7: Homework (cont’d)
• Step 5: Compute “wedge angle” for Lower leading ramp
• Iterate to get correct Shock Angle
MAE 5420 - Compressible Fluid Flow 7
Section 7: Homework (cont’d)
• Assume freestream Mach Number~ 2.5
Use exact solverFor b
b~45.602°M=2..5
M=2..0
M=3.0
MAE 5420 - Compressible Fluid Flow 8
Section 7: Homework (cont’d)
b~45.602°
= 1.7862
• Use Normal Shock Equations
= 1.54754
Already too High(M2=1.0760)
MAE 5420 - Compressible Fluid Flow 9
Section 7: Homework (cont’d)
= 1.7045
• Pick new M1 = 2.0, b=58.46°
• Use Normal Shock Equations
= =0.6394
MAE 5420 - Compressible Fluid Flow 10
Section 7: Homework (cont’d)
= 1.7045
• Pick new M1 = 2.0, b=58.46°
• Use Normal Shock Equations
=1.46156
MAE 5420 - Compressible Fluid Flow 11
Section 7: Homework (cont’d)
• Calculate total mach number behind Shock
=
= 1.0760
Check!
MAE 5420 - Compressible Fluid Flow 12
Section 7: Homework (cont’d)
• M¥ = 2.0
MAE 5420 - Compressible Fluid Flow
Alternate Solution
13
q
M2 =1.0760
M4 = 1.8060
Stagnation PressureConstant acrossshock wave à
p2p4=1+ γ−1
2M4
2
1+ γ−12M2
2
⎛
⎝
⎜⎜⎜⎜⎜⎜⎜⎜⎜
⎞
⎠
⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟
γγ−1
=
p2 = 85.1907 kPa
MAE 5420 - Compressible Fluid Flow
Alternate Solution
14
M2 =1.0760
M4 = 1.8060
p2 = 85.1907 kPa
q
p2p∞=85.1907kPa26.436kPa
= 3.2225=1+ 2 ⋅γγ+1
⋅ M1 ⋅sinβ( )2−1⎡⎣⎢
⎤⎦⎥
→ M1 ⋅sinβ=γ+12 ⋅γ
p2p∞−1
⎛
⎝⎜⎜⎜⎜
⎞
⎠⎟⎟⎟⎟⎟+1=
=
MAE 5420 - Compressible Fluid Flow
Alternate Solution
15
M2 =1.07600
q M2n =0.63941M1n =1.70441
M2 =Mn2
sin β−θ( )→ sin β−θ( )=
Mn2
M2
=0.639411.0760
= 0.59245
o
→ β= sin−1Mn2
M2
⎛
⎝
⎜⎜⎜⎜⎜
⎞
⎠
⎟⎟⎟⎟⎟+θ=
M1=M1 ⋅sinβsinβ
⎛
⎝⎜⎜⎜⎜
⎞
⎠⎟⎟⎟⎟=
1.70441
58.4589 π180⎝ ⎠
⎛ ⎞sin= 2.0
M¥ = 2.0Check!
1MAE 5420 - Compressible Fluid Flow
-2.0
-1.5
-1.0
1.0
-0.5
0.5
0.0
NACA 0012 M∞ = 0.6
0.2 0.4 0.6 0.8 1.0
x/c
CpSection 7, Homework, Part 2
• Consider NACA 0012 Airfoil at 1.5oa.
• Calculate the critical drag rise (subsonic) Mach number for zero wing sweep
• Re-Calculate the Mcr assuming 30o
wing sweep, L• Compare the effective fineness ratios
(t/c), for the unswept and swept wing sections
• What do you conclude?
Uppersurface
Lowersurface
NACA0006at0oa
2MAE 5420 - Compressible Fluid Flow
-2.0
-1.5
-1.0
1.0
-0.5
0.5
0.0
NACA 0012 M∞ = 0.6
0.2 0.4 0.6 0.8 1.0
x/c
Cp
NACA 0012 at 1.5oa
2
Cpmin ~ -94
@zerowingweep
Mcrit = 0.612
3MAE 5420 - Compressible Fluid Flow
NACA 0012 at 1.5oa with 30 deg. Wing sweep
3
à Mcrit = 0.707
4MAE 5420 - Compressible Fluid Flow
Compare Fineness ratios of Swept and Unswept Wings
4
tcequiv
= tc cosΛ
Unsweptà t/c=12%
Sweptà
t/c=10.39%
=
equivWingiseffectively13.4%thinner
