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Basics of Fluid Mechanics, 2014a

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12.9. COMBINATION OF THE OBLIQUE SHOCK AND ISENTROPIC EXPANSION531<br />

The pressure ratio at point 3 is<br />

The pressure ratio at point 4 is<br />

P 3<br />

= P 3 P 03 P 01<br />

1<br />

=0.0109 × 1 ×<br />

P 1 P 03 P 01 P 1 0.01506 ∼ 0.7238<br />

d L =<br />

P 3<br />

P 1<br />

=1.1157<br />

2<br />

2<br />

kP 1 M (P 4 − P 3 ) cos α = 2<br />

2<br />

1 kM 1<br />

d L =<br />

d d = 2 (<br />

P4<br />

2<br />

− P )<br />

3<br />

sin α =<br />

kM 1 P 1 P 1<br />

(<br />

P4<br />

− P )<br />

3<br />

cos α<br />

P 1 P 1<br />

2<br />

1.33.3 2 (1.1157 − 0.7238) cos 4◦ ∼ .054<br />

2<br />

1.33.3 2 (1.1157 − 0.7238) sin 4◦ ∼ .0039<br />

This shows that on the expense <strong>of</strong> a small drag, a large lift can be obtained. Discussion<br />

on the optimum design is left for the next versions.<br />

End Solution<br />

Example 12.19:<br />

To understand the flow after a nozzle consider a flow in a nozzle shown in Figure<br />

12.19. The flow is choked and additionally the flow pressure reaches the nozzle exit<br />

above the surrounding pressure. Assume that there is an isentropic expansion (Prandtl–<br />

Meyer expansion) after the nozzle with slip lines in which there is a theoretical angle<br />

<strong>of</strong> expansion to match the surroundings pressure with the exit. The ratio <strong>of</strong> exit area<br />

to throat area ratio is 1:3. The stagnation pressure is 1000 [kPa]. The surroundings<br />

pressure is 100[kPa]. Assume that the specific heat, k =1.3. Estimate the Mach<br />

number after the expansion.<br />

Solution<br />

The Mach number a the nozzle exit can be calculated using Potto-GDC which provides<br />

M<br />

T<br />

T 0<br />

ρ<br />

ρ 0<br />

A<br />

A ⋆<br />

P<br />

P 0<br />

A×P<br />

A ∗ ×P 0<br />

F<br />

F ∗<br />

1.7632 0.61661 0.29855 1.4000 0.18409 0.25773 0.57478<br />

Thus the exit Mach number is 1.7632 and the pressure at the exit is<br />

P − exit<br />

P exit = P 0 = 1000 × 0.18409 = 184.09[kPa]<br />

P − 0<br />

This pressure is higher than the surroundings pressure and additional expansion must<br />

occur. This pressure ratio is associated with a expansion angle that Potto-GDC provide<br />

as

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