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Basics of Fluid Mechanics, 2014a

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

In these calculations, Maximum Deflection Mach’s equation was used to calculate the<br />

normal component <strong>of</strong> the upstream, then the Mach angle was calculated using the<br />

geometrical relationship <strong>of</strong> θ = sin −1 M 1n /M 1 . With these two quantities, utilizing<br />

equation (12.12) the deflection angle, δ, is obtained.<br />

End Solution<br />

Example 12.11:<br />

Consider the schematic shown in the following<br />

figure. Assume that the upstream<br />

Mach number is 4 and the deflection angle<br />

is δ = 15 ◦ . Compute the pressure<br />

ratio and the temperature ratio after the<br />

second shock (sometimes referred to as<br />

the reflective shock while the first shock<br />

is called the incidental shock).<br />

stream line<br />

M 1 = 4<br />

3<br />

2<br />

1<br />

θ<br />

δ<br />

Fig. -12.17. Schematic for Example (12.11).<br />

Solution<br />

This kind <strong>of</strong> problem is essentially two wedges placed in a certain geometry. It is clear<br />

that the flow must be parallel to the wall. For the first shock, the upstream Mach<br />

number is known together with deflection angle. Utilizing the table or the Potto–GDC,<br />

the following can be obtained:<br />

M x M ys M yw θ s θ w δ<br />

P 0y<br />

P 0x<br />

4.0000 0.46152 2.9290 85.5851 27.0629 15.0000 0.80382<br />

And the additional information by using minimal information ratio button in<br />

Potto–GDC is<br />

M x M yw θ w δ<br />

P y<br />

P x<br />

T y<br />

T x<br />

P 0y<br />

P 0x<br />

4.0000 2.9290 27.0629 15.0000 1.7985 1.7344 0.80382<br />

With a Mach number <strong>of</strong> M =2.929, the second deflection angle is also 15 ◦ . With<br />

these values the following can be obtained:<br />

M x M ys M yw θ s θ w δ<br />

P 0y<br />

P 0x<br />

2.9290 0.51367 2.2028 84.2808 32.7822 15.0000 0.90041<br />

and the additional information is

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