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Basics of Fluid Mechanics, 2014a

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502 CHAPTER 12. COMPRESSIBLE FLOW 2–DIMENSIONAL<br />

12.2.2.4 Given Two Angles, δ and θ<br />

It is sometimes useful to obtain a relationship where the two angles are known. The<br />

first upstream Mach number, M 1 is<br />

Mach Number Angles Relationship<br />

M 1 2 =<br />

2 (cot θ + tan δ)<br />

sin 2θ − (tan δ)(k + cos 2θ)<br />

(12.59)<br />

The reduced pressure difference is<br />

2(P 2 − P 1 )<br />

ρU 2<br />

=<br />

2 sin θ sin δ<br />

cos(θ − δ)<br />

(12.60)<br />

The reduced density is<br />

ρ 2 − ρ 1 sin δ<br />

=<br />

ρ 2 sin θ cos(θ − δ)<br />

(12.61)<br />

For a large upstream Mach number M 1 and a small shock angle (yet not approaching<br />

zero), θ, the deflection angle, δ must also be small as well. Equation (12.51)<br />

can be simplified into<br />

θ ∼ = k +1 δ (12.62)<br />

2<br />

The results are consistent with the initial assumption which shows that it was an appropriate<br />

assumption.<br />

12.2.2.5 Flow in a Semi–2D Shape<br />

Example 12.2:<br />

In Figure 12.9 exhibits wedge in a supersonic flow with unknown Mach number. Examination<br />

<strong>of</strong> the Figure reveals that it is in angle <strong>of</strong> attack. 1) Calculate the Mach number<br />

assuming that the lower and the upper Mach angles are identical and equal to ∼ 30 ◦<br />

each (no angle <strong>of</strong> attack). 2) Calculate the Mach number and angle <strong>of</strong> attack assuming<br />

that the pressure after the shock for the two oblique shocks is equal. 3) What kind are<br />

the shocks exhibits in the image? (strong, weak, unsteady) 4) (Open question) Is there<br />

possibility to estimate the air stagnation temperature from the information provided in<br />

the image. You can assume that specific heats, k is a monotonic increasing function <strong>of</strong><br />

the temperature.<br />

Solution<br />

Part (1)<br />

The Mach angle and deflection angle can be obtained from the Figure 12.9. With this<br />

data and either using equation (12.59) or potto-GDC results in

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