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Next: Optimization of Suction Section Up: Detached Shock Previous: Oblique Shock Examples   Index

## 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 . 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.

Solution

The first configuration is of a normal shock. For which the results13.27 are
 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

 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

 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 with new upstream Mach number of results in

 1.7498 0.71761 1.2346 76.9831 51.5549 14.0000 0.96524
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 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:

 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

The static pressure ratio for the second case is

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.

Solution

The detached shock is a normal shock and the results are
 3.5000 0.45115 3.3151 4.2609 14.1250 0.21295
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Now utilizing the isentropic relationship for yields

 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.

Solution

The shock BD is an oblique shock which response to total turn of . The condition for this shock are:
 3.0000 0.48013 2.7008 87.8807 23.9356 6.0000 0.99105
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The transition for shock AB is

 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:

 2.8482 0.48610 2.7049 88.8912 22.7080 3.0000 0.99894
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And the isentropic relationship for are

 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

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

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.

Since a weak shock must occur to increase the static pressure (see Figure 5.4). The increase has to be

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

 1.0008 0.99920 1.0005 1.0013 1.0019 1.00000
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The shock angle, , can be calculated from

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

 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 . 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 .

Solution

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

With this Mach number and the Mach deflection angle, either using the table or the figure or Potto-GDC results in
 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 to a perpendicular component.

From Table (5.1) or the Potto-GDC the following can be obtained:
 1.6100 0.66545 1.3949 2.0485 2.8575 0.89145
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The temperature ratio combined with upstream temperature yield

and it is the same for the pressure

And for the velocity

Solution

Utilizing GDC for Mach number 2.5 and angle of results in
 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

 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 . With these two quantities utilizing equation (13.12) the deflection angle, is obtained.

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.

 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

 4.0000 2.9290 27.0629 15.0000 1.7985 1.7344 0.80382
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With Mach number of the second deflection angle is also . with these values the following can be obtained.

 2.9290 0.51367 2.2028 84.2808 32.7822 15.0000 0.90041
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 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.

Solution

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

 2.8500 2.3575 29.0000 10.5131 1.4089 1.3582 0.96263
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With the deflection Angle of the so called reflective shock provide the following information

 2.3575 0.54894 1.9419 84.9398 34.0590 10.5100 0.97569
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 2.3575 1.9419 34.0590 10.5100 1.3984 1.3268 0.97569
<>

Solution

For the normal shock the results are
 5.0000 0.41523 5.8000 5.0000 29.0000 0.06172
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While the results for the oblique shock are

 5.0000 0.41523 3.0058 0.0 30.00 20.17 0.49901
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 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

 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.

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.

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 and Mach number =5.

 5.0000 0.43914 3.5040 86.0739 24.3217 15.0000 0.69317
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 5.0000 3.5040 24.3217 15.0000 1.9791 1.9238 0.69317
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For the second direction is for and Mach number =5.

 5.0000 0.43016 3.8006 86.9122 21.2845 12.0000 0.80600
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 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 which results in

 3.5040 0.47413 2.6986 85.6819 27.6668 13.5000 0.88496
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 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 and Mach 3.8006

 3.8006 0.46259 2.9035 85.9316 26.3226 13.5000 0.86179
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 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

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 to a larger number.

Solution

 Waiting for the solution

Next: Optimization of Suction Section Up: Detached Shock Previous: Oblique Shock Examples   Index
Created by:Genick Bar-Meir, Ph.D.
On: 2007-11-21