SUPERSONIC AIRFOILS

When the free-stream Mach number exceeds unity, the flow around an airfoil will appear as shown in Figure 5.21a or 5.21b. If the nose of the airfoil is blunt, a detached bow shock will occur, causing a small region of subsonic flow over the nose of the airfoil. After the flow is deflected subsonically around the nose, it expands again through Mach waves fanning out from the convex surfaces to supersonic conditions. As it leaves the trailing edge, the flows along the upper and lower surfaces are deflected by oblique shock waves and become parallel to each other and to the free stream.

In the case of a sharp leading edge, which is the case for an airfoil designed to operate supersonically, the flow is deflected at the leading edge by oblique shock waves attached to the leading edge.

The diamond-shaped supersonic airfoil illustrated in Figure 5.21b is rela­tively easy to analyze, given the oblique shock and Prandtl-Meyer flow relationships. To begin, since pressure distributions cannot be propagated ahead, the flow will be uniform until it is deflected by the oblique shock waves above and below the leading edge. The streamlines, after passing through the oblique shocks, will remain parallel and straight until they are turned through the expansion fan, after which they are again straight and parallel until they are deflected to approximately the free-stream direction by the oblique shock waves from the trailing edge. This flow is illustrated in detail in Figure 5.22.

The flow from the trailing edge does not necessarily have to satisfy the Kutta conditions, as in the subsonic case. Instead, the final deflection, and hence the strength of the trailing oblique shock waves, is fixed by stipulating that the pressure and flow directions be the same for the flows from the upper and lower surfaces as they meet behind the trailing edge.

As an example, consider a supersonic airfoil having a symmetrical wedge configuration. as shown in Figure 5.22. We are given

M„ = 2.0

a =2°

For this case, the angle e in Figure 5.22 becomes 11.3°. Thus the required deflection angles are as follows.

region 1 to region 2 8 = 3.65°

region 2 to region 3 8 = 11.3°

region 3 to region 4 5 = 7.65° + у

region 1 to region 5 8 = 7.65°

region 5 to region 6 5 = 11.3°

region 6 to region 7 8 = 3.65° – у

У is the unknown angle that must satisfy 8^ = 8^. Figures 5.14 and 5.17 cannot be read with sufficient accuracy, so it is necessary to work with the oblique shock and Prandtl-Meyer relationships on which these figures are based. These relationships are easily programmed on a programmable cal­culator. In this way, the following numbers were readily obtained. For M, = 2.000 and 8^2 = 3.65°,

в = 33.08°

1.224

Pi

= 1.155 Pi

M2= 1.911

Pm = ggg Pm

For M2 = 1.911 and S2_3 = 11.3°, the Prandtl-Меуег relationships are as fol­lows.

— = 0.1467 Po

& = 0.2539 Po

M3 = 2.337

& = 0.07548

Po

& = 0.1579

Po

& = 0.2973 Pi

fe = 0.6219 Pi

For M, = 2.000 and 8 M = 7.65°,

в = 36.86°

— = 1.513 Pi

— = 1.341 Pi

M5= 1.808

For the preceding with S5-6 = 11.3°, the Prandtl-Meyer relationships applied to the lower surface are as follows.

— = 0.1719

Po

— = 0.2843

Po

Mb = 2.220

— = 0.09057 Po

— = 0.1799

Po

— = 0.3186 Pi [2]

With this information, we can determine that

— = 0.3639 —

Pi Рг

Also,

— = 0.4820 —

Pi Рб

Thus, if p4 and p? are to be equal, it must follow that

^ = 1.325 —

Pi Pe

The problem now is finding the value of у that results in oblique shocks trailing from the upper and lower surfaces having compression ratios that satisfy the above.

Through the process of trial and error this was accomplished with the following results.

у = 0.3° 03-4 = 31.89	ви = 29.69і
— = 1.612	22 = 1.219
Pi	Pf,
1.402	— = 1.151
Pi	Ре
M4 = 2.148	M7 = 2.121

The ratio of the actual velocities can be found from

V4 _ M4 a4 _ M5 p4 p7 V7 M7 a7 M7 p7 p4 = 0.982

Thus a shear layer, or vorticity, is generated downstream of the airfoil, since the loss in total temperature is slightly different between the upper and lower surfaces.

The lift coefficient can be expressed as

c, = f'(&–&) dx (5.53)

yMJJo p« pJ

where x is the dimensionless distance along the chord and the subscripts l and и refer to lower and upper surfaces. In this example, the pressures are

constant, but with different values, over each half of the chord. Thus

Unlike the two-dimensional, inviscid, subsonic flow, a drag known as wave drag exists for the supersonic case. This drag can be obtained by resolving the integral of the normal pressure forces over the body in the drag direction.

For the symmetrical wedge pictured in Figure 5.22, the wave drag is; therefore

D = [(p2-Ре) sin 3.65° + (ps ~ Рз) sin 7.65°] |

In dimensionless form, this becomes

The moment coefficient about the leading edge will be given by

From Cm and Ct, the center of pressure for this symmetrical airfoil is seen to be 0.395 chord lengths behind the leading edge.