# Question about oblique shock and expansion fans

Hi I am an engineering student.

I have a question about oblique shock and expansion fans. In my understanding, oblique shock and expansion fans could happen on supersonic airplane foil. But how often are they happening?

Lets say if there is an airplane moving at constant speed at 3 times sound speed, how do we know how much energy is lost due to expansion fans and oblique shocks if this airplane travels for 3 km? How many shocks should we consider?

• "How many shocks should we consider" - that will depend on the detail of the design. Commented Nov 24, 2022 at 6:28

Not an easy question to answer. The main methods circa 1950s were linear shock theory and shock expansion theory. Here's a link to a decent 1952 NACA report that has a lot of dense graphs of the relevant nondimensional terms for a bunch of airfoil thicknesses, speeds, and airfoil structural constraints. You will need to track down some of the references to dig up all the formula derivations though, this is basically a report on results.

First of all You can use the simulator to study the flow past a wedge.There is more complete shock simulation program that is avaliable on web. The various programs solves for flow past a wedge and for flow past a cone, including the detached normal shock conditions. Another simulation, called ShockModeler, describes the intersection and reflection of multiple shock waves.

Use simulation softwares to study different flows it will definitely help. Further explanation I have mentioned below;

Oblique shocks are generated by the nose and by the leading edge of the wing and tail of a supersonic aircraft. Oblique shocks are also generated at the trailing edges of the aircraft as the flow is brought back to free stream conditions. Oblique shocks also occur downstream of a nozzle if the expanded pressure is different from free stream conditions. In high speed inlets, oblique shocks are used to compress the air going into the engine. The air pressure is increased without using any rotating machinery.

For compressible flows with little or small flow turning, the flow process is reversible and the entropy is constant. The change in flow properties are then given by the isentropic relations (isentropic means "constant entropy"). But when an object moves faster than the speed of sound, and there is an abrupt decrease in the flow area, shock waves are generated in the flow. Shock waves are very small regions in the gas where the gas properties change by a large amount. Across a shock wave, the static pressure, temperature, and gas density increases almost instantaneously. The changes in the flow properties are irreversible and the entropy of the entire system increases. Because a shock wave does no work, and there is no heat addition, the total enthalpy and the total temperature are constant. But because the flow is non-isentropic, the total pressure downstream of the shock is always less than the total pressure upstream of the shock. There is a loss of total pressure associated with a shock wave as shown on the slide. Because total pressure changes across the shock, we can not use the usual (incompressible) form of Bernoulli's equation across the shock. The Mach number and speed of the flow also decrease across a shock wave.

On the slide, a supersonic flow at Mach number M approaches a shock wave which is inclined at angle s. The flow is deflected through the shock by an amount specified as the deflection angle - a. The deflection angle is determined by resolving the incoming flow velocity into components parallel and perpendicular to the shock wave. The component parallel to the shock is assumed to remain constant across the shock, the component perpendicular is assumed to decrease by the normal shock relations. Combining the components downstream of the shock determines the delflection angle. Then:

cot(a) = tan(s) * [{((gam+1) * M^2)/(2 * M^2 * sin^2(s) - 1)} - 1]

where tan is the trigonometric tangent function, cot is the co-tangent function:

cot(a) = tan(90 degrees - a)

and sin^2 is the square of the sine. Gam is the ratio of specific heats. Across the shock wave the Mach number decreases to a value specified as M1:

M1^2 * sin^2(s -a) = [(gam-1)M^2 sin^2(s) + 2] / [2 * gam * M^2 * sin^2(s) - (gam - 1)]

The total temperature across the shock is constant, but the static temperature T increases in zone 1 to become:

T1 / T0 = [2 * gam * M^2 * sin^2(s) - (gam - 1)] * [(gam -1) * M^2 * sin^2(s) + 2] / [(gam + 1)^2 * M^2 * sin^2(s)]

The total pressure pt decreases according to:

pt1 / pt0 = {[(gam + 1) * M^2 * sin^2(s)]/[(gam-1)*M^2 * sin^2(s) + 2]}^[gam/((gam-1)] * {(gam+1)/[2 * gam * M^2 * sin^2(s)-(gam-1)]}^[1/(gam-1)]

The static pressure p increases to:

p1 / p0 = [2 * gam * M^2 * sin^2(s)-(gam -1)] / (gam + 1)

And the density r changes by:

r1 / r0 = [(gam + 1) * M^2 * sin^2(s)] / [(gam -1) * M^2 * sin^2(s) + 2]

The right hand side of all these equations depend only on the free stream Mach number and the shock angle. The shock angle depends in a complex way on the free stream Mach number and the wedge angle. So knowing the Mach number and the wedge angle, we can determine all the conditions associated with the oblique shock.

[The equations describing oblique shocks were published in NACA report (NACA-1135) in 1951.]

• All that and you avoided Prandtl-Meyer expansion and tables. Commented Nov 24, 2022 at 17:47