Application of Oblique Shock

One of the practical applications of the oblique shock is the design of an inlet suction for a supersonic flow. It is suggested that a series of weak shocks should replace one normal shock to increase the efficiency (see Figure (13.17))^13.26. Clearly, with a proper design, the flow can be brought to a subsonic flow just below $M=1$ . In such a case, there is less entropy production (less pressure loss). To illustrate the design significance of the oblique shock, the following example is provided.

Figure: Schematic for Example (13.4).

Solution

The first configuration is of a normal shock. For which the results^13.27 are

$\rule[-0.1in]{0.pt}{0.3 in}\mathbf{M_x}$	$\mathbf{M_y}$	$\mathbf{T_y \over T_x}$	$\mathbf{\rho_y \over \rho_x}$	$\mathbf{P_y \over P_x}$	$\mathbf{{P_0}_y \over {P_0}_x }$
2.0000	0.57735	1.6875	2.6667	4.5000	0.72087

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In the oblique shock the first angle shown is

$\rule[-0.1in]{0.pt}{0.3 in}\mathbf{M_x}$	$\mathbf{{M_y}_s}$	$\mathbf{{M_y}_w}$	$\mathbf{\theta_s}$	$\mathbf{\theta_w}$	$\mathbf{\delta }$	$\mathbf{{P_0}_y \over {P_0}_x }$
2.0000	0.58974	1.7498	85.7021	36.2098	7.0000	0.99445

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and the additional information by the minimal info in Potto-GDC

$\rule[-0.1in]{0.pt}{0.3 in}\mathbf{M_x}$	$\mathbf{{M_y}_w}$	$\mathbf{\theta_w}$	$\mathbf{\delta }$	$\mathbf{{P_y} \over {P_x}}$	$\mathbf{{T_y} \over {T_x} }$	$\mathbf{{P_0}_y \over {P_0}_x }$
2.0000	1.7498	36.2098	7.0000	1.2485	1.1931	0.99445

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In the new region new angle is $7^\circ + 7^\circ$ with new upstream Mach number of $M_x = 1.7498$ results in

$\rule[-0.1in]{0.pt}{0.3 in}\mathbf{M_x}$	$\mathbf{{M_y}_s}$	$\mathbf{{M_y}_w}$	$\mathbf{\theta_s}$	$\mathbf{\theta_w}$	$\mathbf{\delta }$	$\mathbf{{P_0}_y \over {P_0}_x }$
1.7498	0.71761	1.2346	76.9831	51.5549	14.0000	0.96524

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And the additional information is

$\rule[-0.1in]{0.pt}{0.3 in}\mathbf{M_x}$	$\mathbf{{M_y}_w}$	$\mathbf{\theta_w}$	$\mathbf{\delta }$	$\mathbf{{P_y} \over {P_x}}$	$\mathbf{{T_y} \over {T_x} }$	$\mathbf{{P_0}_y \over {P_0}_x }$
1.7498	1.5088	41.8770	7.0000	1.2626	1.1853	0.99549

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An oblique shock is not possible and normal shock occurs. In such a case, the results are:

$\rule[-0.1in]{0.pt}{0.3 in}\mathbf{M_x}$	$\mathbf{M_y}$	$\mathbf{T_y \over T_x}$	$\mathbf{\rho_y \over \rho_x}$	$\mathbf{P_y \over P_x}$	$\mathbf{{P_0}_y \over {P_0}_x }$
1.2346	0.82141	1.1497	1.4018	1.6116	0.98903

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With two weak shock waves and a normal shock the total pressure loss is

$${{P_0}_4 \over {P_0}_1 } = {{P_0}_4 \over {P_0}_3 } {{P_0}_3 \o... ...} {{P_0}_2 \over {P_0}_1 } = 0.98903 \times 0.96524 \times 0.99445 = 0.9496$$

The static pressure ratio for the second case is

$${P_4 \over P_1 } = {P_4 \over P_3 } {P_3 \over P_2 } {P_2 \over P_1 } = 1.6116 \times 1.2626 \times 1.285 = 2.6147$$

The loss in this case is much less than in direct normal shock. In fact, the loss in the normal shock is 31% larger for the total pressure.

Figure: Schematic for Example (13.5).

Solution

The detached shock is a normal shock and the results are

$\rule[-0.1in]{0.pt}{0.3 in}\mathbf{M_x}$	$\mathbf{M_y}$	$\mathbf{T_y \over T_x}$	$\mathbf{\rho_y \over \rho_x}$	$\mathbf{P_y \over P_x}$	$\mathbf{{P_0}_y \over {P_0}_x }$
3.5000	0.45115	3.3151	4.2609	14.1250	0.21295

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Now utilizing the isentropic relationship for $k=1.4$ yields

$\rule[-0.1in]{0.pt}{0.3 in}\mathbf{M}$	$\mathbf{T \over T_0}$	$\mathbf{\rho \over \rho_0}$	$\mathbf{A \over A^{\star}}$	$\mathbf{P \over P_0}$	$\mathbf{A\times P \over A^{*} \times P_0}$
0.45115	0.96089	0.90506	1.4458	0.86966	1.2574

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Thus the area ratio has to be 1.4458. Note that the pressure after the weak shock is irrelevant to area ratio between the normal shock and the ``throat'' according to the standard nozzle analysis.

Figure: $\;$ Schematic of two angles turn with two weak shocks.

Solution

The shock BD is an oblique shock which response to total turn of $6^\circ$ . The condition for this shock are:

$\rule[-0.1in]{0.pt}{0.3 in}\mathbf{M_x}$	$\mathbf{{M_y}_s}$	$\mathbf{{M_y}_w}$	$\mathbf{\theta_s}$	$\mathbf{\theta_w}$	$\mathbf{\delta }$	$\mathbf{{P_0}_y \over {P_0}_x }$
3.0000	0.48013	2.7008	87.8807	23.9356	6.0000	0.99105

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The transition for shock AB is

$\rule[-0.1in]{0.pt}{0.3 in}\mathbf{M_x}$	$\mathbf{{M_y}_s}$	$\mathbf{{M_y}_w}$	$\mathbf{\theta_s}$	$\mathbf{\theta_w}$	$\mathbf{\delta }$	$\mathbf{{P_0}_y \over {P_0}_x }$
3.0000	0.47641	2.8482	88.9476	21.5990	3.0000	0.99879

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For the Shock BC the results are:

$\rule[-0.1in]{0.pt}{0.3 in}\mathbf{M_x}$	$\mathbf{{M_y}_s}$	$\mathbf{{M_y}_w}$	$\mathbf{\theta_s}$	$\mathbf{\theta_w}$	$\mathbf{\delta }$	$\mathbf{{P_0}_y \over {P_0}_x }$
2.8482	0.48610	2.7049	88.8912	22.7080	3.0000	0.99894

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And the isentropic relationship for $M = 2.7049, \; 2.7008$ are

$\rule[-0.1in]{0.pt}{0.3 in}\mathbf{M}$	$\mathbf{T \over T_0}$	$\mathbf{\rho \over \rho_0}$	$\mathbf{A \over A^{\star}}$	$\mathbf{P \over P_0}$	$\mathbf{A\times P \over A^{*} \times P_0}$
2.7049	0.40596	0.10500	3.1978	0.04263	0.13632
2.7008	0.40669	0.10548	3.1854	0.04290	0.13665

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The combined shocks AB and BC provides the base to the calculation of the total pressure ratio at zone 3. The total pressure ratio at zone 2 is

$${{P_0}_2 \over {P_0}_0} = {{P_0}_2 \over {P_0}_1} {{P_0}_1 \over {P_0}_0} = 0.99894 \times 0.99879 = 0.997731283$$

On the other hand, the pressure at 4 has to be

$${P_4 \over {P_0}_1} = {P_4 \over {P_0}_4} {{P_0}_4 \over {P_0}_1} = 0.04290 \times 0.99105 = 0.042516045$$

The static pressure at zone 4 and zone 3 have to match according to the government suggestion; hence, the angle for the BE shock which causes this pressure ratio needs to be found. To that, check whether the pressure at 2 is above or below or above the pressure (ratio) in zone 4.

$${P_2 \over {P_0}_2} = {{P_0}_2 \over {P_0}_0} {P_2 \over {P_0}_2} = 0.997731283 \times 0.04263 = 0.042436789$$

Since ${P_2 \over {P_0}_2} < {P_4 \over {P_0}_1}$ a weak shock must occur to increase the static pressure (see Figure 5.4). The increase has to be

$${P_3/ P_2} = {0.042516045 / 0.042436789 } = 1.001867743$$

To achieve this kind of pressure ratio, the perpendicular component has to be

$\rule[-0.1in]{0.pt}{0.3 in}\mathbf{M_x}$	$\mathbf{M_y}$	$\mathbf{T_y \over T_x}$	$\mathbf{\rho_y \over \rho_x}$	$\mathbf{P_y \over P_x}$	$\mathbf{{P_0}_y \over {P_0}_x }$
1.0008	0.99920	1.0005	1.0013	1.0019	1.00000

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The shock angle, $\theta$ , can be calculated from

$$\theta = \sin^{-1} {1.0008 / 2.7049} = 21.715320879^\circ$$

The deflection angle for such a shock angle with the Mach number is

$\rule[-0.1in]{0.pt}{0.3 in}\mathbf{M_x}$	$\mathbf{{M_y}_s}$	$\mathbf{{M_y}_w}$	$\mathbf{\theta_s}$	$\mathbf{\theta_w}$	$\mathbf{\delta }$	$\mathbf{{P_0}_y \over {P_0}_x }$
2.7049	0.49525	2.7037	0.0	21.72	0.026233	1.00000

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For the last calculation it is clear that the government's proposed schematic of the double wedge is in conflict with the boundary condition. The flow in zone 3 will flow into the wall in about $2.7^{\circ}$ . In reality, the flow of a double wedge produces curved shock surfaces with several zones. Only far away for the double wedge the flow behaves as the only angle that exist of $6^{\circ}$ .

Solution

The angle of Mach angle of $34^\circ$ while below maximum deflection means that it is a weak shock. Yet the upstream Mach number, $M_1$ , has to be determined

$$M_1 = { U_1 \over \sqrt{kRT} } = {1000 \over 1.4 \times 287 \times 300} = 2.88$$

With this Mach number and the Mach deflection angle, either using the table or the figure or Potto-GDC results in

$\rule[-0.1in]{0.pt}{0.3 in}\mathbf{M_x}$	$\mathbf{{M_y}_s}$	$\mathbf{{M_y}_w}$	$\mathbf{\theta_s}$	$\mathbf{\theta_w}$	$\mathbf{\delta }$	$\mathbf{{P_0}_y \over {P_0}_x }$
2.8800	0.48269	2.1280	0.0	34.00	15.78	0.89127

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The relationship for the temperature and pressure can be obtained by using equations (13.15) and (13.13) or simply by converting the $M_1$ to a perpendicular component.

$$M_{1n} = M_1 * \sin \theta = 2.88 \sin (34.0 ) = 1.61$$

From Table (5.1) or the Potto-GDC the following can be obtained:

$\rule[-0.1in]{0.pt}{0.3 in}\mathbf{M_x}$	$\mathbf{M_y}$	$\mathbf{T_y \over T_x}$	$\mathbf{\rho_y \over \rho_x}$	$\mathbf{P_y \over P_x}$	$\mathbf{{P_0}_y \over {P_0}_x }$
1.6100	0.66545	1.3949	2.0485	2.8575	0.89145

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The temperature ratio combined with upstream temperature yield

$$T_2 = 1.3949 \times 300 \sim 418.5 K$$

and it is the same for the pressure

$$P_2 = 2.8575 \times 3 = 8.57 [bar]$$

And for the velocity

$$U_{n2} = {{M_y}_w} \sqrt{ kRT} = 2.128 \sqrt { 1.4 \times 287 \times 418.5} = 872.6 [m/sec]$$

Solution

Utilizing GDC for Mach number 2.5 and angle of $11^\circ$ results in

$\rule[-0.1in]{0.pt}{0.3 in}\mathbf{M_x}$	$\mathbf{{M_y}_s}$	$\mathbf{{M_y}_w}$	$\mathbf{\theta_s}$	$\mathbf{\theta_w}$	$\mathbf{\delta }$	$\mathbf{{P_0}_y \over {P_0}_x }$
2.5000	0.53431	2.0443	85.0995	32.8124	11.0000	0.96873

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Solution

It has to be recognized that without any other boundary condition the shock is weak shock. For weak shock the maximum pressure ratio is obtained when at the deflection point because it is the closest to normal shock. The obtain the maximum point for 2.5 Mach number either use Maximum Deflection Mach number's equation or POTTO-GDC

$\rule[-0.1in]{0.pt}{0.3 in}\mathbf{M_x}$	$\mathbf{{M_y}_{max}}$	$\mathbf{\theta_{max}}$	$\mathbf{\delta }$	$\mathbf{{P_y} \over {P_x}}$	$\mathbf{{T_y} \over {T_x} }$	$\mathbf{{P_0}_y \over {P_0}_x }$
2.5000	0.94021	64.7822	29.7974	4.3573	2.6854	0.60027

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In these calculation Maximum Deflection Mach's equation was used to calculate Normal component of the upstream than Mach angle was calculated using the geometrical relationship of $\theta = \sin ^{-1} {M_{1n} / M_1}$ . With these two quantities utilizing equation (13.12) the deflection angle, $\delta$ is obtained.

Figure: A figure for the next example

Solution

This kind problem is essentially two wedges placed in a certain geometry. It is clear that the flow must be parallel to the wall. For the first shock, the upstream Mach number is known with deflection angle. Utilizing the table or POTTO-GDC, the following can be obtained.

$\rule[-0.1in]{0.pt}{0.3 in}\mathbf{M_x}$	$\mathbf{{M_y}_s}$	$\mathbf{{M_y}_w}$	$\mathbf{\theta_s}$	$\mathbf{\theta_w}$	$\mathbf{\delta }$	$\mathbf{{P_0}_y \over {P_0}_x }$
4.0000	0.46152	2.9290	85.5851	27.0629	15.0000	0.80382

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And the additional information is by using minimal information ratio button in POTTO-GDC

$\rule[-0.1in]{0.pt}{0.3 in}\mathbf{M_x}$	$\mathbf{{M_y}_w}$	$\mathbf{\theta_w}$	$\mathbf{\delta }$	$\mathbf{{P_y} \over {P_x}}$	$\mathbf{{T_y} \over {T_x} }$	$\mathbf{{P_0}_y \over {P_0}_x }$
4.0000	2.9290	27.0629	15.0000	1.7985	1.7344	0.80382

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With Mach number of $M=2.929$ the second deflection angle is also $15^\circ$ . with these values the following can be obtained.

$\rule[-0.1in]{0.pt}{0.3 in}\mathbf{M_x}$	$\mathbf{{M_y}_s}$	$\mathbf{{M_y}_w}$	$\mathbf{\theta_s}$	$\mathbf{\theta_w}$	$\mathbf{\delta }$	$\mathbf{{P_0}_y \over {P_0}_x }$
2.9290	0.51367	2.2028	84.2808	32.7822	15.0000	0.90041

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and the additional information is

$\rule[-0.1in]{0.pt}{0.3 in}\mathbf{M_x}$	$\mathbf{{M_y}_w}$	$\mathbf{\theta_w}$	$\mathbf{\delta }$	$\mathbf{{P_y} \over {P_x}}$	$\mathbf{{T_y} \over {T_x} }$	$\mathbf{{P_0}_y \over {P_0}_x }$
2.9290	2.2028	32.7822	15.0000	1.6695	1.5764	0.90041

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With the combined tables the ratios can be easily calculated. Note that hand calculations requires endless time looking up graphical representation of the solution. Utilizing the POTTO-GDC provides a solution in just a few clicks.

$${P_1 \over P_3} = {P_1 \over P_2} {P_2 \over P_3} = 1.7985 \times 1.6695 = 3.0026$$

$${T_1 \over T_3} = {T_1 \over T_2} {T_2 \over T_3} = 1.7344 \times 1.5764 = 2.632$$

Solution

Here the Mach number and the Mach angle are given. With these pieces of information utilizing the GDC provides the following:

$\rule[-0.1in]{0.pt}{0.3 in}\mathbf{M_x}$	$\mathbf{{M_y}_s}$	$\mathbf{{M_y}_w}$	$\mathbf{\theta_s}$	$\mathbf{\theta_w}$	$\mathbf{\delta }$	$\mathbf{{P_0}_y \over {P_0}_x }$
2.8500	0.48469	2.3575	0.0	29.00	10.51	0.96263

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and the additional information by utilizing the minimal info button in GDC provides

$\rule[-0.1in]{0.pt}{0.3 in}\mathbf{M_x}$	$\mathbf{{M_y}_w}$	$\mathbf{\theta_w}$	$\mathbf{\delta }$	$\mathbf{{P_y} \over {P_x}}$	$\mathbf{{T_y} \over {T_x} }$	$\mathbf{{P_0}_y \over {P_0}_x }$
2.8500	2.3575	29.0000	10.5131	1.4089	1.3582	0.96263

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With the deflection Angle of $\delta = 10.51$ the so called reflective shock provide the following information

$\rule[-0.1in]{0.pt}{0.3 in}\mathbf{M_x}$	$\mathbf{{M_y}_s}$	$\mathbf{{M_y}_w}$	$\mathbf{\theta_s}$	$\mathbf{\theta_w}$	$\mathbf{\delta }$	$\mathbf{{P_0}_y \over {P_0}_x }$
2.3575	0.54894	1.9419	84.9398	34.0590	10.5100	0.97569

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and the additional information of

$\rule[-0.1in]{0.pt}{0.3 in}\mathbf{M_x}$	$\mathbf{{M_y}_w}$	$\mathbf{\theta_w}$	$\mathbf{\delta }$	$\mathbf{{P_y} \over {P_x}}$	$\mathbf{{T_y} \over {T_x} }$	$\mathbf{{P_0}_y \over {P_0}_x }$
2.3575	1.9419	34.0590	10.5100	1.3984	1.3268	0.97569

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$${P_1 \over P_3} = {P_1 \over P_2} {P_2 \over P_3} = 1.4089 \times 1.3984 \sim 1.97$$

$${T_1 \over T_3} = {T_1 \over T_2} {T_2 \over T_3} = 1.3582 \times 1.3268 \sim 1.8021$$

Solution

For the normal shock the results are

$\rule[-0.1in]{0.pt}{0.3 in}\mathbf{M_x}$	$\mathbf{M_y}$	$\mathbf{T_y \over T_x}$	$\mathbf{\rho_y \over \rho_x}$	$\mathbf{P_y \over P_x}$	$\mathbf{{P_0}_y \over {P_0}_x }$
5.0000	0.41523	5.8000	5.0000	29.0000	0.06172

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While the results for the oblique shock are

$\rule[-0.1in]{0.pt}{0.3 in}\mathbf{M_x}$	$\mathbf{{M_y}_s}$	$\mathbf{{M_y}_w}$	$\mathbf{\theta_s}$	$\mathbf{\theta_w}$	$\mathbf{\delta }$	$\mathbf{{P_0}_y \over {P_0}_x }$
5.0000	0.41523	3.0058	0.0	30.00	20.17	0.49901

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And the additional information is

$\rule[-0.1in]{0.pt}{0.3 in}\mathbf{M_x}$	$\mathbf{{M_y}_w}$	$\mathbf{\theta_w}$	$\mathbf{\delta }$	$\mathbf{{P_y} \over {P_x}}$	$\mathbf{{T_y} \over {T_x} }$	$\mathbf{{P_0}_y \over {P_0}_x }$
5.0000	3.0058	30.0000	20.1736	2.6375	2.5141	0.49901

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The normal shock that follows this oblique is

$\rule[-0.1in]{0.pt}{0.3 in}\mathbf{M_x}$	$\mathbf{M_y}$	$\mathbf{T_y \over T_x}$	$\mathbf{\rho_y \over \rho_x}$	$\mathbf{P_y \over P_x}$	$\mathbf{{P_0}_y \over {P_0}_x }$
3.0058	0.47485	2.6858	3.8625	10.3740	0.32671

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The pressure ratios of the oblique shock with normal shock is the total shock in the second case.

$${P_1 \over P_3} = {P_1 \over P_2} {P_2 \over P_3} = 2.6375 \times 10.374 \sim 27.36$$

$${T_1 \over T_3} = {T_1 \over T_2} {T_2 \over T_3} = 2.5141 \times 2.6858 \sim 6.75$$

Note the static pressure raised is less than the combination shocks as compared to the normal shock but the total pressure has the opposite result.

Figure: Illustration for example (13.13)

Solution

The first two zones immediately after are computed using the same techniques that were developed and discussed earlier.

For the first direction is for $15^\circ$ and Mach number =5.

$\rule[-0.1in]{0.pt}{0.3 in}\mathbf{M_x}$	$\mathbf{{M_y}_s}$	$\mathbf{{M_y}_w}$	$\mathbf{\theta_s}$	$\mathbf{\theta_w}$	$\mathbf{\delta }$	$\mathbf{{P_0}_y \over {P_0}_x }$
5.0000	0.43914	3.5040	86.0739	24.3217	15.0000	0.69317

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And the addition conditions are

$\rule[-0.1in]{0.pt}{0.3 in}\mathbf{M_x}$	$\mathbf{{M_y}_w}$	$\mathbf{\theta_w}$	$\mathbf{\delta }$	$\mathbf{{P_y} \over {P_x}}$	$\mathbf{{T_y} \over {T_x} }$	$\mathbf{{P_0}_y \over {P_0}_x }$
5.0000	3.5040	24.3217	15.0000	1.9791	1.9238	0.69317

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For the second direction is for $12^\circ$ and Mach number =5.

$\rule[-0.1in]{0.pt}{0.3 in}\mathbf{M_x}$	$\mathbf{{M_y}_s}$	$\mathbf{{M_y}_w}$	$\mathbf{\theta_s}$	$\mathbf{\theta_w}$	$\mathbf{\delta }$	$\mathbf{{P_0}_y \over {P_0}_x }$
5.0000	0.43016	3.8006	86.9122	21.2845	12.0000	0.80600

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And the additional conditions are

$\rule[-0.1in]{0.pt}{0.3 in}\mathbf{M_x}$	$\mathbf{{M_y}_w}$	$\mathbf{\theta_w}$	$\mathbf{\delta }$	$\mathbf{{P_y} \over {P_x}}$	$\mathbf{{T_y} \over {T_x} }$	$\mathbf{{P_0}_y \over {P_0}_x }$
5.0000	3.8006	21.2845	12.0000	1.6963	1.6625	0.80600

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The conditions in zone 4 and zone 3 have to have two things that are equal, and they are the pressure and the velocity direction. It has to be noticed that the velocity magnitudes in zone 3 and 4 do not have to be equal. This non continuous velocity profile can occurs in our model because it is assumed that fluid is non-viscous.

If the two sides were equal because symmetry the slip angle was zero. It is to say, for the analysis, that only one deflection angle exist. For the two different deflection angles, the slip angle has two extreme cases. The first case is where match lower deflection angle and second to match the higher deflection angle. In this case, it is assumed that the slip angle moves half of the angle to satisfy both of the deflection angles (first approximation). Under this assumption the continuous in zone 3 are solved by looking at the deflection angle of $12^\circ + 1.5^\circ = 13.5^\circ$ which results in

$\rule[-0.1in]{0.pt}{0.3 in}\mathbf{M_x}$	$\mathbf{{M_y}_s}$	$\mathbf{{M_y}_w}$	$\mathbf{\theta_s}$	$\mathbf{\theta_w}$	$\mathbf{\delta }$	$\mathbf{{P_0}_y \over {P_0}_x }$
3.5040	0.47413	2.6986	85.6819	27.6668	13.5000	0.88496

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with the additional information

$\rule[-0.1in]{0.pt}{0.3 in}\mathbf{M_x}$	$\mathbf{{M_y}_w}$	$\mathbf{\theta_w}$	$\mathbf{\delta }$	$\mathbf{{P_y} \over {P_x}}$	$\mathbf{{T_y} \over {T_x} }$	$\mathbf{{P_0}_y \over {P_0}_x }$
3.5040	2.6986	27.6668	13.5000	1.6247	1.5656	0.88496

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And in zone 4 the conditions are due to deflection angle of $13.5^\circ$ and Mach 3.8006

$\rule[-0.1in]{0.pt}{0.3 in}\mathbf{M_x}$	$\mathbf{{M_y}_s}$	$\mathbf{{M_y}_w}$	$\mathbf{\theta_s}$	$\mathbf{\theta_w}$	$\mathbf{\delta }$	$\mathbf{{P_0}_y \over {P_0}_x }$
3.8006	0.46259	2.9035	85.9316	26.3226	13.5000	0.86179

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with the additional information

$\rule[-0.1in]{0.pt}{0.3 in}\mathbf{M_x}$	$\mathbf{{M_y}_w}$	$\mathbf{\theta_w}$	$\mathbf{\delta }$	$\mathbf{{P_y} \over {P_x}}$	$\mathbf{{T_y} \over {T_x} }$	$\mathbf{{P_0}_y \over {P_0}_x }$
3.8006	2.9035	26.3226	13.5000	1.6577	1.6038	0.86179

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From these tables the pressure ratio at zone 3 and 4 can be calculated

$${P_3 \over P_4} = {P_3 \over P_2} {P_2 \over P_0} {P_0 \over P_1}... ...r P_4} = 1.6247 \times 1.9791 {1 \over 1.6963} { 1 \over 1.6038} \sim 1.18192$$

To reduce the pressure ratio the deflection angle has to be reduced (remember that at weak weak shock there is almost no pressure change). Thus, the pressure at zone 3 has to be reduced. To reduce the pressure the angle of slip plane has to increase from $1.5\circ$ to a larger number.

Solution

Waiting for the solution