Combination of the Oblique Shock and Isentropic Expansion

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

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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

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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

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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

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The pressure ratio at point 3 is

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

The pressure ratio at point 4

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

$$d_L = {2 \over k P_1 {M_1}^2 }(P_4 - P_3) \cos \alpha =$$

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

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

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.