**1. Introduction**

Oblique shock waves occur when supersonic flow is turned into itself. In contrast, when a supersonic flow is turned *away *from itself along a convex corner, an expansion fan is formed, as illustrated in Figure 1.

*Figure 1. Expansion fan.*

The expansion fan is a *continuous *expansion region that can be visualized as an infinite number of Mach waves, each making a Mach angle with the local flow direction. The expansion fan is bounded upstream by a Mach wave which makes the angle with respect to the upstream flow such that = *arcsin*(1/*M _{1}*), where

*M*is the normal Mach number of flow upstream of the fan. The expansion fan is bounded downstream by another Mach wave, in this case one that makes an angle with respect to the downstream flow such that =

_{1}*arcsin*(1/

*M*), where

_{2}*M*is the normal Mach number of flow just downstream of the fan. Thus, the expansion can be visualized as a continuous succession of Mach waves; since Mach waves are isentropic, the expansion itself is also isentropic. This is in direct contrast to flow across an oblique shock, which is always accompanied by some entropy increase.

_{2}An expansion fan emanating from a sharp convex corner is called a *Prandtl-Meyer expansion wave*, so named in honor of Ludwig Prandtl (1875 -1953) and Theodor Meyer (1882 – 1972). Prandtl is one of the most celebrated aerodynamicists of all time, known for his seminal contributions to the field in the first third of the 20th century. Meyer was a doctoral student of Prandtl’s at Göttingen University; his doctoral dissertation, published in 1908, laid the foundations of expansion wave theory.

**2. Deriving the Prandtl-Meyer function**

*Figure 2. Expansion wave.*

Let us turn to a mathematical treatment of expansion wave phenomena. Consider first a single Mach wave, expanding and deflecting the supersonic flow through a very small angle of magnitude = *dv*, as shown in Figure 2*. *With no pressure gradient in the tangential direction, again, there is no change of the tangential velocity component across the wave. Equating the expressions for *V _{t}* upstream and downstream of the Mach wave (Figure 2), we may write

Since *dv *is a very small angle, we may employ the approximations *cos dv* 1 and *sin dv* * dv, *so that equation (1) can be restated as

Expanding the product on the right-hand side,

The last term on the RHS contains a product of differentials and can be neglected. Simplifying further leads to

Since *sin ** *= 1/*M*, we may write

Substituting in (4),

As in the mathematical treatment of oblique shocks, our aim is to express the variation of flow properties as a function of turning angle and Mach number only. From the definition of Mach number for a perfect gas with constant specific heats, we may write

Taking the logarithm of each side and differentiating,

However, for this adiabatic flow, there is no change in stagnation temperature. Therefore, denoting stagnation conditions with a “*o*” subscript and using an elementary compressible flow relation,

Taking the logarithm and differentiating,

Equations (8) and (10) can be combined to yield

or

Substituting into (6) yields

Now, let

or, equivalently, *M ^{2}* = 1 +

*x*. It follows that 2

^{2}*MdM*= 2

*xdx*, or

Using the relation above and identity (14), we substitute into (13) to obtain

Next, use partial fractions to divide the right-hand side into two groups of terms,

so that

Equating the two sides yields the equations

Solving this system of equations yields *A *= + 1 and *B *= –1. Thus, the transformed equation can be arranged into two groups and leads to the following two integrals:

At this point, we evoke the tabulated integral

which ultimately gives

We can return to equation (13) with this relationship, integrating and finding the change of Mach number associated with the turning of a supersonic flow through a finite angle = *v _{2}* –

*v*,

_{1}Using (23),

For the purpose of tabulating this result, it is convenient to define a reference state such that *M *= *M _{ref}* at

*v*, giving

_{ref}Letting the reference state be *v *= 0 at *M *= 1, we may write

The expression above, *v(M)*, is known as the Prandtl-Meyer function. It represents the angle through which a stream, initially at Mach 1, must be expanded to reach a supersonic Mach number *M*. Importantly, the Prandtl-Meyer function is a monotonically increasing function of *M*.

Two limiting cases for *v *can be considered, namely *M ** *1 and *M **. *The former is elementary and yields *v ** *0. In the latter case, it can be shown that

The expression above yields the maximum turning angle for Prandtl-Meyer flow and clearly depends only on the ratio of specific heats. For air, = 1.4 and *v _{max}* = 130.45

^{o}. This maximum value is of academic interest only, since long before a supersonic flow is deflected through this angle the continuum assumption would become invalid as the static pressure attains very low values.

**3. Problems involving expansion waves**

To determine the angle through which a flow would have to be turned to expand from some upstream Mach number *M _{1}* to some downstream Mach number

*M*, with

_{2}*M*

_{1}*1, one needs only subtract the PM function*

*v*at

_{1}*M*from the value of

_{1}*v*at

_{2}*M*. That is, the turning angle is related to the Prandtl-Meyer functions by the simple relationship

_{2}Values of the PM function are tabulated in any gas dynamics textbook and can be found here. To solve a simple expansion wave problem, we may proceed as follows:

- For the given
*M*, obtain_{1}*v(M*from the Prandtl-Meyer table._{1}) - Calculate
*v(M*from equation (29) using the known and the value of_{2})*v(M*obtained in step 1._{1}) - Obtain
*M*from the expansion wave table corresponding to the value of_{2}*v(M*from step 2._{2}) - The expansion wave is isentropic; hence, stagnation pressure
*p*and stagnation temperature_{0}*T*are constant through the wave, that is,_{0}*T*=_{0,2}*T*and_{0,1}*p*=_{0,2}*p*. The temperature ratio can then be written as_{0,1}

If only *T _{2}* interests you, use the isentropic flow relation

As for the pressure ratio,

If only *p _{2}* interests you, use the isentropic flow relation

**Example problem 1**

A uniform supersonic flow of air ( = 1.4) at Mach 2.6, with stagnation pressure of 5 MPa and stagnation temperature of 1000 K, expands around a 20^{o} convex corner. Determine

- the downstream Mach number;
- the stagnation pressure and temperature;
- the static pressure and temperature

*Part 1: *Entering a Mach number of 2.6 into the Prandtl-Meyer table, we read *v(M _{1}) *= 41.4147

^{o}. Using this quantity and = 20

^{o}, we write

Entering this angle into the Prandtl-Meyer table, the closest data points we have are for *v* = 61.2990^{o}, which corresponds to *M *= 3.68, and for *v *= 61.5953^{o}, which corresponds to *M *= 3.70. As a rudimentary approximation, we interpolate linearly to obtain *M _{2}* 3.688.

*Part 2: *Expansion waves are isentropic; it follows that the stagnation pressure remains unchanged at 5 MPa. Likewise, the stagnation temperature remains unchanged at 1000 K.

*Part 3: *To compute the static pressure, we appeal to equation (33),

The static temperature, in turn, can be computed with equation (31),

Example problem 1 is the easiest type of problem involving expansion waves one can encounter: enter the upstream Mach number into the Prandtl-Meyer table, read the upstream PM function, compute the downstream PM function, and finally enter it in the Prandtl-Meyer table to read the downstream Mach number. Equipped with the downstream Mach number, we can determine the downstream properties with the isentropic flow relations we all know and love.

Things get a little more involved when we consider supersonic flows with more complex profiles, such as airfoils. In such surfaces, an incoming supersonic flow may form expansion waves as it turns away from itself, but, in a similar manner, the flow may also form oblique shocks as it turns into itself. To analyze such flows, we must refer not only to Prandtl-Meyer theory, but also to our knowledge of oblique shock waves. We close this post with two such examples, both drawn from John and Keith (2006).

**Example problem 2**

A uniform supersonic flow of a perfect gas with = 1.4, Mach number 3.0 and an upstream static pressure of 100 kPa flows over a geometry as shown below. Determine the downstream static pressure for each case.

To analyze case 1, we divide the flow into three regions, as shown.

In region 1, from the Prandtl-Meyer table at = 1.4 and *M _{1}* = 3.0, we read

*v*= 49.7573

_{1}^{o}. Using an isentropic flow table, we read a pressure ratio

*p*= 0.02722. Region 2 is reached by passing through an expansion fan in which the flow is turned back 10

_{1}/p_{01}^{o}. Accordingly, we enter

*v*=

_{2}*v*+ 10

_{1}^{o}= 59.7573

^{o}into the Prandtl-Meyer table to read

*M*3.58 (the closest entry is for an angle of 59.7838

_{2}^{o}). Then, we enter this Mach number into an isentropic flow table to read a pressure ratio

*p*/

_{2}*p*= 0.0117. Because an expansion fan is isentropic, the stagnation pressure remains unchanged and

_{02}*p*/

_{02}*p*= 1. Now, region 3 is reached by passing through an oblique shock in which the flow is turned back 10

_{01}^{o}. Entering

*M*= 3.58 and = 10

_{2}^{o}into an oblique shock table or chart, we read

*M*= 2.982 and

_{3}*p*/

_{3}*p*= 2.315. We are now in position to compute downstream static pressure

_{2}*p*,

_{3}We now turn to case 2, which we divide into three regions as before.

Region 2 is reached by passing through an oblique shock in which the flow is turned through 10^{o}. Referring to an oblique shock table or chart with *M _{1}* = 3.0 and = 10

^{o}, we read

*M*= 2.505 and

_{2}*p*/

_{2}*p*= 2.054. Also,

_{1}*p*/

_{02}*p*= 0.9631. After the oblique shock, the flow evolves isentropically through region 2 at a Mach number

_{01}*M*. Entering

_{2}*M*= 2.505 into an isentropic flow table, we read

_{2}*p*/

_{2}*p*= 0.0585 (the closest entry I found was for

_{02}*M*= 2.50). After reaching the top of the upwards-sloped section, the flow forms an expansion fan before spreading to region 3. Entering

_{2}*M*= 2.505 into the Prandtl-Meyer table, we read

_{2}*v*= 39.1236

_{2}^{o}(again, the closest entry in the table is for

*M*= 2.50). As the flow is turned 10

_{2}^{o}when expanding to region 3, we may write

*v*=

_{3}*v*+ 10

_{2}^{o}= 49.1236

^{o}. Entering this angle into the Prandtl-Meyer table, we read a Mach number

*M*= 2.96 (the closest entry is for an angle of 48.9796

_{3}^{o}). Referring to an isentropic table one last time, we enter

*M*= 2.96 to read

_{3}*p*/

_{3}*p*= 0.0289. We can glean our results and compute downstream static pressure

_{03}*p*

_{3,}In short, the arrangement in case 1 ultimately leads to a decrease in pressure of about 0.5% relatively to initial pressure, while case 2 culminates with an *increase* close to 1.5%.

**Example problem 3**

A two-dimensional supersonic wing has the profile illustrated below. At zero angle of attack, determine the drag force on the wing per unit length of span at Mach 2. Repeat for lift force. Take the maximum thickness of the airfoil to be 0.2 m.

Let the regions involved in the calculations be numbered as follows. An oblique shock occurs as the flow transitions from 1 to 2; a Prandtl-Meyer expansion fan occurs as the flow transitions from 2 to 3; lastly, a second oblique shock occurs as the flow transitions from 3 to 4.

Flow reaches the airfoil at Mach number *M _{1}* = 2.0 and deflection angle such that

At this Mach number and deflection angle, refer to an oblique shock table or chart to read a wave angle = 38.78^{o}, a downstream Mach number *M _{2}* = 1.659, and a pressure ratio

*p*/

_{2}*p*= 1.664. Entering

_{1}*M*into an isentropic flow table, we get the static-to-stagnation pressure ratio

_{2}*p*/

_{2}*p*= 0.2151 (the closest entry is for

_{02}*M*= 1.66). As the flow transitions from 2 to 3, an expansion fan is formed. Entering

_{2}*M*= 1.66 into the Prandtl-Meyer table, we read

_{2}*v*= 16.6328

_{2}^{o}(again, the closest entry is for

*M*= 1.66). Now, as the flow moves from region 2 to 3 it turns 2 = 2 9.462

_{2}^{o}= 18.924

^{o}, so the Prandtl-Meyer function at 3 becomes

Entering this value into a Prandtl-Meyer table yields *M _{3}* = 2.36 (the closest entry is for

*v*= 35.7715

^{o}). Taking this Mach number into an isentropic flow table one last time gives a static-to-stagnation pressure ratio

*p*/

_{3}*p*= 0.0728. The flow goes on to form a second oblique shock at the trailing edge of the airfoil, but our results so far suffice to compute pressures

_{03}*p*,

_{1}*p*, and

_{2}*p*, which are all we need to compute the lift and drag forces on the foil. Thus, we ignore calculations for the transition from 3 to 4. Gleaning results, we write

_{3}At region 1, flow pressure equals the free-stream pressure, that is,

At region 2, flow pressure is

Lastly, at region 3

Now, the drag force acting on the foil consists of two contributions, one from each half of the upper camber; mathematically,

The lift force, in turn, includes not only the contributions from the two halves of the upper camber, but also one from the flat lower camber; in mathematical terms,

The negative sign indicates that the lift is actually pointing downward.

**Go further**

Those looking for more solved problems on shock theory might want to check out our free quiz on normal and oblique shock waves, which you can find here.

**References**

• ANDERSON, J.D., JR. (2011). *Fundamentals of Aerodynamics*. 5th edition. New York: McGraw-Hill.

• BABU, V. (2021). *Fundamentals of Gas Dynamics. *2nd edition. Berlin/Heidelberg: Springer.

• JOHN, J.E. and KEITH, T.G. (2006). *Gas Dynamics*. 3rd edition. Upper Saddle River: Pearson Prentice Hall.