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Basics of Fluid Mechanics, 2014a

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412 CHAPTER 11. COMPRESSIBLE FLOW ONE DIMENSIONAL<br />

1<br />

0.9<br />

Shock Wave relationship<br />

M y<br />

and P 0y<br />

/P 0x<br />

as a function <strong>of</strong> M x<br />

0.8<br />

0.7<br />

M y<br />

P 0y<br />

/P 0x<br />

0.6<br />

M y<br />

0.5<br />

0.4<br />

0.3<br />

0.2<br />

0.1<br />

0<br />

1 2 3 4 5 6 7 8 9 10<br />

Fri Jun 18 15:47:34 2004<br />

M x<br />

Fig. -11.12. The exit Mach number and the stagnation pressure ratio as a function <strong>of</strong> upstream<br />

Mach number.<br />

M ∗ = U c ∗ = c U<br />

c ∗ c = c<br />

c ∗ M (11.97)<br />

The jump condition across the shock must satisfy the constant energy.<br />

c 2<br />

k − 1 + U 2<br />

2 = c∗2<br />

k − 1 + c∗2<br />

2 = k +1<br />

2(k − 1) c∗2 (11.98)<br />

Dividing the mass equation by the momentum equation and combining it with the<br />

perfect gas model yields<br />

2<br />

c 1<br />

+ U 1 = c 2 2<br />

+ U 2 (11.99)<br />

kU 1 kU 2<br />

Combining equation (11.98) and (11.99) results in<br />

[<br />

1 k +1<br />

c ∗2 − k − 1 ]<br />

U 1 + U 1 = 1 [ k +1<br />

c ∗2 − k − 1 ]<br />

U 2 + U 2 (11.100)<br />

kU 1 2 2<br />

kU 2 2 2<br />

After rearranging and dividing equation (11.100) the following can be obtained:<br />

U 1 U 2 = c ∗2 (11.101)

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