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Basics of Fluid Mechanics, 2014a

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12.2. OBLIQUE SHOCK 513<br />

M x M y<br />

T y<br />

T x<br />

ρ y<br />

ρ x<br />

P y<br />

P x<br />

P 0y<br />

P 0x<br />

1.6100 0.66545 1.3949 2.0485 2.8575 0.89145<br />

The temperature ratio combined upstream temperature yield<br />

T 2 =1.3949 × 300 ∼ 418.5K<br />

and the same for the pressure<br />

P 2 =2.8575 × 3=8.57[bar]<br />

And the velocity<br />

√ √<br />

U n2 = M yw kRT =2.128 1.4 × 287 × 418.5 = 872.6[m/sec]<br />

End Solution<br />

Example 12.9:<br />

For Mach number 2.5 and wedge with a total angle <strong>of</strong> 22 ◦ , calculate the ratio <strong>of</strong> the<br />

stagnation pressure.<br />

Solution<br />

Utilizing GDC for Mach number 2.5 and the angle <strong>of</strong> 11 ◦ results in<br />

M x M ys M yw θ s θ w δ<br />

P 0y<br />

P 0x<br />

2.5000 0.53431 2.0443 85.0995 32.8124 11.0000 0.96873<br />

End Solution<br />

Example 12.10:<br />

What is the maximum pressure ratio that can be obtained on wedge when the gas is<br />

flowing in 2.5 Mach without any close boundaries? Would it make any difference if the<br />

wedge was flowing into the air? If so, what is the difference?<br />

Solution<br />

It has to be recognized that without any other boundary condition, the shock is weak<br />

shock. For a weak shock the maximum pressure ratio is obtained at the deflection<br />

point because it is closest to a normal shock. To obtain the maximum point for 2.5<br />

Mach number, either use the Maximum Deflection Mach number’s equation or the<br />

Potto–GDC<br />

M x M ymax θ max δ<br />

P y<br />

P x<br />

T y<br />

T x<br />

P 0y<br />

P 0x<br />

2.5000 0.94021 64.7822 29.7974 4.3573 2.6854 0.60027

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