Compressible Flow credits Logo credits
Potto Home Contact Us

Potto Home

About Potto

Chapters:

  Content
  Introduction
  Sound
  Isentropic
  Shock
  Gravity
  Isothermal
  Fanno
  Rayleigh
  Tank
  Piston
  Oblique
  Prandtl-Meyer
  Hard copy
  Gas Dynamics Tables

Other things:
Other resources
Download Area
calculators

Other Resources

  FAQs
  Compare Other Books
  Articles

Potto Statistics

License

Feedback

next up previous index
Next: Computer Program Up: Prandtl-Meyer Function Previous: Examples For Prandtl-Meyer Function   Index

Combination of the Oblique Shock and Isentropic Expansion


\begin{examl}
Consider two-dimensional flat thin plate at an angle of attack of...
...ge is $k=1.3$, calculate
the drag coefficient and lift coefficient.
\end{examl}
Solution

For $ M=3.3$ the following table can be obtained
$ \rule[-0.1in]{0.pt}{0.3 in}\mathbf{M} $ $ \mathbf{\nu} $ $ \mathbf{P \over P_0} $ $ \mathbf{T \over T_0} $ $ \mathbf{\rho \over \rho_0} $ $ \mathbf{\mu } $
3.3000 62.3113 0.01506 0.37972 0.03965 73.1416
<>


With the angle of attack the region 3 will at $ \nu \sim 62.31+4$ for which the following table can be obtain (Potto-GDC)

$ \rule[-0.1in]{0.pt}{0.3 in}\mathbf{M} $ $ \mathbf{\nu} $ $ \mathbf{P \over P_0} $ $ \mathbf{T \over T_0} $ $ \mathbf{\rho \over \rho_0} $ $ \mathbf{\mu } $
3.4996 66.3100 0.01090 0.35248 0.03093 74.0528
<>


On the other side the oblique shock (assuming weak shock) results in

$ \rule[-0.1in]{0.pt}{0.3 in}\mathbf{M_x} $ $ \mathbf{{M_y}_s} $ $ \mathbf{{M_y}_w} $ $ \mathbf{\theta_s} $ $ \mathbf{\theta_w} $ $ \mathbf{\delta } $ $ \mathbf{{P_0}_y \over {P_0}_x } $
3.3000 0.43534 3.1115 88.9313 20.3467 4.0000 0.99676
<>


And the additional information by clicking on the minimal button provides

$ \rule[-0.1in]{0.pt}{0.3 in}\mathbf{M_x} $ $ \mathbf{{M_y}_w} $ $ \mathbf{\theta_w} $ $ \mathbf{\delta } $ $ \mathbf{{P_y} \over {P_x}} $ $ \mathbf{{T_y} \over {T_x} } $ $ \mathbf{{P_0}_y \over {P_0}_x } $
3.3000 3.1115 20.3467 4.0000 1.1157 1.1066 0.99676
<>


The pressure ratio at point 3 is

$\displaystyle {P_3 \over P_1} = {P_3 \over P_{03} } {P_{03} \over P_{01}}{P_{01}\over
P_{1}}= 0.0109 \times 1 \times {1 \over 0.01506} \sim 0.7238
$

The pressure ratio at point 4

$\displaystyle {P_3 \over P_1} = 1.1157
$

  $\displaystyle d_L = {2 \over k P_1 {M_1}^2 }(P_4 - P_3) \cos \alpha =$    
  $\displaystyle {2 \over k {M_1}^2 } \left( {P_4 \over P_1} - {P_3 \over P_1} \ri...
...ha = {2 \over 1.3 3.3^2 } \left( 1.1157 - 0.7238 \right) \cos 4^\circ \sim .054$    

$\displaystyle d_d = {2 \over k {M_1}^2 } \left( {P_4 \over P_1} - {P_3 \over P_...
...
{ 2 \over 1.3 3.3^2 } \left( 1.1157 - 0.7238 \right) \sin 4^\circ
\sim .0039
$

This shows that on the expense of a small drag, a large lift can be obtained. Discussion on the optimum design is left for the next versions.



next up previous index
Next: Computer Program Up: Prandtl-Meyer Function Previous: Examples For Prandtl-Meyer Function   Index
Created by:Genick Bar-Meir, Ph.D.
On: 2007-11-21