Shock-Expansion Theory

Abstract

This chapter develops the foundation of inviscid nonlinear shock-expansion theory on pointed bodies in supersonic flows. Fundamental equations of shock and expansion waves are derived and discussed. To extend the analysis to three-dimensional bodies, conical shock theory is derived. Examples from diamond and bi-convex airfoils are used to highlight the shock-expansion techniques in developing lift and wave drag on two-dimensional pointed bodies. Drag polar and the aerodynamic performance parameter, $(L/D)_{\text {max}}$ are discussed. Axisymmetric and slender bodies are the basis of 3-D shock-expansion application. Approximate methods based on the method of local cones is introduced to estimate the effect of 3-D expansion, using conical shocks associated with smaller nose angle. The problem of asymmetrical flows is analyzed through the method of local cones. The shock-expansion theory is extended to the upper transonic regime. Here, the principle of stationary local Mach numbers near sonic freestream flow is demonstrated and applied to wave drag of axisymmetric bodies at sonic speed. This chapter contains 9 examples and 42 practice problems at the end of the chapter.

4.1 Introduction

Supersonic flow over bodies involves the creation of waves of compression and expansion to achieve flow turning. Blunt bodies, for example, create a detached, or bow shock that forms ahead of the blunt body and the flow downstream of the shock is a mixed subsonic-supersonic flow. Such mixed flows are not amenable to analytical solutions. However, there is a class of problems in inviscid supersonic flow over pointed bodies that lend itself to exact analytical solution (known as shock-expansion theory). This class of inviscid supersonic flow problems requires the oblique shock (either plane or conical) to be attached to the body in flight at the leading edge/nose. This necessitates first the body to be pointed and second the turning imposed by the body at the leading edge not to exceed the maximum turning angle corresponding to a plane oblique (or conical) shock at the designated flight Mach number. Given these stipulations, shock-expansion theory may be applied to any inviscid flow above sonic, which includes the upper transonic regime.

Our presentation of shock-expansion theory in this chapter is based on inviscid fluid assumption. As the fluid viscosity is ignored, the no-slip boundary condition is lost and thus the boundary layers are not formed. Consequently, the effect of boundary layer separation on aerodynamic forces and moments is clearly beyond the scope of inviscid flow theory, such as shock-expansion theory. In transonic flows over airfoils, wings and bodies the local Mach number exceeds the sonic value, therefore shocks can be formed. These shocks interact with the low momentum, low energy zone next to the body, i.e., the boundary layer, and may cause boundary layer separation. This subject is further elaborated in Chap. 6.

Before we get started with the applications of shock-expansion theory, we briefly review the plane oblique and conical shocks as well as the Prandtl-Meyer expansion wave theory. Waves of infinitesimal strength in supersonic flow are called Mach waves, which are capable of turning the flow only infinitesimally. The angle of the Mach wave with respect to the flow is called Mach angle and is given the symbol, $\mu$. Since the wave strength is established by the normal component of the flow upstream of the wave, the normal component of a supersonic flow to a Mach wave is therefore sonic. This geometric property of Mach wave is shown in a right triangle in Fig. 4.1.

Fig. 4.1

Definition sketch of the Mach wave and the Mach angle, $\mu$, in supersonic flow

From the definition sketch of Fig. 4.1, we deduce:

$$\begin{aligned} \sin \mu&= \frac{1}{M} \end{aligned}$$

(4.1a)

$$\begin{aligned} \tan \mu&= \frac{1}{\sqrt{M^2-1}} \end{aligned}$$

(4.1b)

$$\begin{aligned} \mu&= \sin ^{-1} \left( \frac{1}{M}\right) = \tan ^{-1} \left( \frac{1}{\sqrt{M^2-1}}\right) \end{aligned}$$

(4.1c)

For example in a Mach-2 flow, the Mach angle is $\sin ^{-1}\left( \frac{1}{2}\right)$, which is $30^{\circ }$. In sonic flow at Mach 1, the Mach angle is $90^{\circ }$, i.e., normal to the flow and for Mach 1.1, the Mach angle is reduced to $65.38^{\circ }$ with respect to the local flow direction. However a single Mach wave due to its infinitesimal strength is unable to provide finite turning of the supersonic flow that is needed to make room for a body (of finite thickness) in flight. Consequently, finite compression and expansion waves are created to turn the supersonic flow. The coalescence of compression Mach waves creates shocks and finite expansion waves are created through many expansion Mach waves, known as the Prandtl-Meyer expansion waves. Figure 4.2a shows the coalescence of compression Mach waves on a concave surface that culminates in the shock formation and Fig. 4.2b shows the spreading of the expansion Mach waves on a convex surface that create finite turning in the flow at supersonic speeds.

Fig. 4.2

Waves on concave and convex surfaces in supersonic flow. a Waves generated on a concave surface. b Waves generated on a convex surface

We note that the compression Mach waves on a concave surface coalesce, as shown in Fig. 4.2a, and form an oblique shock wave whereas the expansion Mach waves on a convex surface (Fig. 4.2b) spread/diverge and thus never form a surface of discontinuity of rarefaction. The coalescence of compression Mach waves that form the shock thus creates a region of very steep pressure and temperature gradients in the gas where normal viscous stress and thermal conductivity of the gas play an important role in establishing the equilibrium thickness for the shock wave. It is therefore the dominance of viscous effects within the shock that creates an entropy increase and thus non-isentropic flow conditions across the shock. We also know from kinetic theory that the thickness of the shock is proportional to the mean-free path of the gas at reference sonic temperature condition (e.g., see classical textbooks of Shapiro [6], Liepmann-Roshko [5], among others). The steep gradients in pressure, temperature, density, normal component of velocity, entropy and stagnation pressure across the shock are depicted graphically in Fig. 4.3.

Fig. 4.3

Schematic drawing of flow gradients across a shock, with shock thickness, $\delta$ (not-to-scale). a Normal component of velocity. b Static pressure. c Static temperature. d Density. e Entropy. f Total pressure

Fig. 4.4

Variation of flow properties within a normal shock (from [6]). Note that $x$ indicates the conditions ahead of the shock wave, while $y$ indicates the conditions behind the shock wave

The structure of a normal shock in standard sea level pressure and temperature conditions with upstream Mach number of 1.1 is shown in Fig. 4.4 (from [6]). Subscripts $x$ and $y$ denote the flow conditions upstream and downstream of the shock, respectively. The streamwise coordinate $x$ shows the region where the steep gradients exist. Therefore, noting that the $x$-axis has a $10^{-5}$ inch scale, we appreciate the thickness, or rather thinness of the shock in standard air. To put these numbers in perspective, we recall that the mean-free path in air at standard conditions is ${\sim }3.7 \times 10^{-5}$ cm (or ${\sim }1.46 \times 10^{-5}$ in.), which makes the thickness of the shock between 3–5 times the mean-free path in air at standard conditions. The static pressure and temperature across the normal shock are shown in ratios and the normal velocity component in units of ft/s. The density ratio is not shown in the graph, but it rises as well as the gas is compressed across the shock. It is thus the product of density and temperature ratios following the perfect gas law that produces the static pressure ratio across the shock, as shown in Fig. 4.4. To demonstrate the variation of normal shock thickness with Mach number at standard pressure and temperature, we examine Fig. 4.5 (from [6]). The normal shock thickness on a log-log scale indicates an exponential drop in shock thickness as shock strength or Mach number increases. As a result of shock thinning with increasing Mach number (at a given altitude), the gradients increase and thus total pressure losses mount.

Fig. 4.5

Normal shock thickness at standard atmospheric pressure and temperature, as a function of shock strength parameter, $M_x-1$, in transonic flow (from [6])

The normal shock relations are derived through the application of the great conservation laws of nature, namely conservation of mass, linear momentum and the energy across the shock. The conservation of mass applied to a normal shock with stations 1 and 2 identifying up- and downstream of the normal shock, gives birth to the first constant of motion, related to normal shock flows, namely

$$\begin{aligned} \rho _1 u_1 = \rho _2 u_2 = \text {const.} \end{aligned}$$

(4.2)

The application of the Newton’s second law of motion to a control volume shown in Fig. 4.6, gives birth to the second constant of motion for a flow through a normal shock, namely

$$\begin{aligned} p_1 + \rho _1 u_1^2 = p_2 + \rho _2 u_2^2 = \text {const.} \end{aligned}$$

(4.3)

Fig. 4.6

Definition sketch of the control volume used in the analysis of normal shocks in steady flow

Since the flow across a shock is adiabatic, i.e., there is no heat exchange between an external source and the fluid in the control volume, and since the fluid within the control volume undergoes no chemical reaction, the law of conservation of energy for non-chemically reacting, steady flows across a shock produces the third constant of motion, namely

$$\begin{aligned} h_{t_1} = h_{t_2} = \text {const.} \end{aligned}$$

(4.4)

Here, $h_t$ is the stagnation or total enthalpy in the flow, as measured by a stationary observer with respect to the shock. In hypersonic flow where molecular collisions are highly energetic, the law of conservation of energy is no longer represented by Eq. (4.4) as the collisions may produce dissociation and even ionization in air. There are a few other constants of motion that emerge directly from the conservation of energy, namely stagnation speed of sound, $a_t$, which is based on the stagnation temperature of the gas, remains constant as well for calorically-prefect gas where $c_{p} = \text {constant}$, i.e.,

$$\begin{aligned} a_{t_1} = a_{t_2} = a_t = \text {const.} \end{aligned}$$

(4.5)

Since the sonic state may be reached adiabatically in a gas, the total enthalpy of the gas at sonic state that is reached adiabatically and the constant stagnation enthalpy of the flow are equal, namely,

$$\begin{aligned} h^* + \frac{a^{*2}}{2} = c_p T^* + \frac{a^{*2}}{2} = \frac{a^{*2}}{\gamma -1} + \frac{a^{*2}}{2} = \left( \frac{\gamma +1}{\gamma -1}\right) \frac{a^{*2}}{2} = h_t = \text {const.} \end{aligned}$$

(4.6)

This yields the fifth constant of motion, namely

$$\begin{aligned} a_1^* = a_2^* = a^* = \text {const.} \end{aligned}$$

(4.7)

By introducing the concept of maximum velocity that is achieved through expansion of a gas to zero static enthalpy, or inversely, by converting the entire stagnation enthalpy of the gas to kinetic energy, we get the sixth constant of motion, namely

$$\begin{aligned} V_{\text {max}} \equiv \sqrt{2 h_t} = \text {const.} \end{aligned}$$

(4.8)

Any of the constant speeds, $a_t$, $a^*$ or $V_{\text {max}}$ can serve as our reference speed in gas dynamic calculations dealing with normal (or oblique) shocks. For example $a^*$ has served as the reference speed in shock polar and hodograph solutions in gas dynamics and $V_{\text {max}}$ is used in Taylor-Maccoll formulation of conical shocks, as we will briefly review. Also $a^*$ is used in the definition of a characteristic Mach number , $M^*$ using the following definition:

$$\begin{aligned} M^* \equiv \frac{u}{a^*} \end{aligned}$$

(4.9)

The characteristic Mach number , is related to the Mach number through the law of conservation of energy equation, namely,

$$\begin{aligned} M^{*2} = \frac{\left( \gamma +1\right) M^2}{2+\left( \gamma -1\right) M^2} \end{aligned}$$

(4.10a)

Or inversely,

$$\begin{aligned} M^2 = \frac{2}{\left( \gamma +1\right) /M^{*2} - \left( \gamma -1\right) } \end{aligned}$$

(4.10b)

Note that subsonic flows with $M<1$, have $M^*<1$, sonic flow with $M=1$, has $M^*=1$ and supersonic flow with $M>1$ has $M^*>1$. The added advantage of $M^*$ over $M$ is that $M^*$ is bounded whereas $M$ may tend to infinity. This is seen in the following limit:

$$\begin{aligned} M^* \rightarrow \sqrt{\frac{\gamma +1}{\gamma -1}} \quad \quad \text {as} \quad \quad M \rightarrow \infty \end{aligned}$$

(4.10c)

In summary, primary constants of motion for a normal shock are:

$$\begin{aligned} \rho u&= \text {const.} \nonumber \\ p+\rho u^2&= \text {const.} \nonumber \\ h_t&= \text {const.} \nonumber \\ a_t&= \text {const.} \nonumber \\ a^*&= \text {const.} \nonumber \\ V_\text {max}&= \text {const.} \nonumber \end{aligned}$$

The simultaneous solution of the conservation laws yields the celebrated Prandtl’s relation relating the product of gas speeds across a normal shock to the square of the (constant) speed of sound at sonic state, $a^*$, following

$$\begin{aligned} u_1 u_2 = a^{*2} \end{aligned}$$

(4.11)

The immediate result of the Prandtl’s relation is the inverse relationship between the upstream and downstream characteristic Mach numbers, namely

$$\begin{aligned} M_1^* = \frac{1}{M_2^*} \quad or \quad M_2^* = \frac{1}{M_1^*} \end{aligned}$$

(4.12)

The emergence of subsonic flow downstream of a normal shock in supersonic flow is directly based on the above inverse relation between $M_2^*$ and $M_1^*$. In terms of the upstream Mach number, $M$, we may express the “jump” conditions across a normal shock as follows:

$$\begin{aligned} \frac{\rho _2}{\rho _1}&= \frac{u_1}{u_2} = \frac{\left( \gamma +1\right) M_1^2}{2+\left( \gamma -1\right) M_1^2} \end{aligned}$$

(4.13)

$$\begin{aligned} \frac{p_2}{p_1}&= 1 + \frac{2\gamma }{\gamma +1}\left( M_1^2 - 1\right) \end{aligned}$$

(4.14)

$$\begin{aligned} \frac{T_2}{T_1}&= \left[ 1 + \frac{2\gamma }{\gamma +1}\left( M_1^2 - 1\right) \right] \left[ \frac{2+\left( \gamma -1\right) M_1^2}{\left( \gamma +1\right) M_1^2} \right] \end{aligned}$$

(4.15)

$$\begin{aligned} \frac{s_2 - s_1}{R}&= \frac{\gamma }{\gamma -1} \ln \left[ 1 + \frac{2\gamma }{\gamma +1}\left( M_1^2 - 1\right) \right] \left[ \frac{2+\left( \gamma -1\right) M_1^2}{\left( \gamma +1\right) M_1^2} \right] \nonumber \\&\quad - \ln \left[ 1 + \frac{2\gamma }{\gamma +1}\left( M_1^2 - 1\right) \right] \end{aligned}$$

(4.16)

$$\begin{aligned} \frac{p_{t2}}{p_{t1}}&= e^{-\frac{\varDelta s}{R}} = \left[ 1 + \frac{2\gamma }{\gamma +1}\left( M_1^2 - 1\right) \right] \left[ \frac{2+\left( \gamma -1\right) \left( \frac{2+\left( \gamma -1\right) M_1^2}{2\gamma M_1^2 - \left( \gamma -1\right) }\right) }{2+\left( \gamma -1\right) M_1^2} \right] ^{\frac{\gamma }{\gamma -1}}\nonumber \\ \end{aligned}$$

(4.17)

In addition, downstream Mach number is related to upstream Mach number through the use of Prandtl’s relation, namely

$$\begin{aligned} M_2^2 = \frac{1+\left[ \left( \gamma -1\right) /2\right] M_1^2}{\gamma M_1^2 - \left( \gamma -1\right) /2} \end{aligned}$$

(4.18)

The graphical depiction of the flow properties across a normal shock is very instructive and the levels and scales of these parameters may be used as an instructional tool. For example in Fig. 4.7 we have plotted the static pressure, temperature and density ratios across a normal shock for a calorically-perfect gas with the ratio of specific heats, $\gamma =1.4$. Note that the upstream Mach number range is intentionally limited to the supersonic range of up to Mach 5.

Fig. 4.7

Normal shock parameters for a calorically-perfect gas, $\gamma =1.4$

We note that the level of pressure jump is higher than both density and temperature ratios, since in fact it is the product of the two jumps (in density and temperature) that create the pressure jump. In the transonic range, we note that the behavior of these jumps in static pressure, temperature and density is nearly linear. Limiting the Mach number range to 1.3 and re-plotting the ratios of pressure, temperature and density in Fig. 4.8 we can show this linear behavior.

Fig. 4.8

The linear behavior of normal shock (jump) properties in transonic range

The Mach number downstream of a normal shock (for $\gamma =1.4$) is shown in Fig. 4.9. We note that the initial drop in downstream Mach number is rapid and it gradually plateaus with higher Mach numbers. In fact, there is a limiting downstream Mach number that can be deduced from expression (4.18) in the limit of upstream Mach number tending to infinity, namely

$$\begin{aligned} M_2 \rightarrow \sqrt{\frac{\gamma -1}{2\gamma }} \quad \text {as} \quad M_1 \rightarrow \infty \end{aligned}$$

(4.19)

The behavior of Mach number downstream of a normal shock in transonic range is also linear, as expected (see Fig. 4.10). The normal shock relations up to a Mach number of 10 are tabulated in Appendix B.

Fig. 4.9

Normal shock parameters for a calorically-perfect gas, $\gamma =1.4$

Fig. 4.10

Downstream Mach number in normal shock flows in transonic range (as compared to a straight line)

Considering a moving source (i.e., a small disturbance) in a compressible fluid, we learned that the source creates spherical acoustic waves that propagate at sonic speed in the medium. The case of a source moving at sonic speed, $a$, and the source moving faster than the sonic speed, $V > a$, i.e., moving supersonically, are shown in Fig. 4.11. The labels for the zones of action and silence signify the space where disturbance is felt and the space corresponding to “forbidden” signal propagation, or zone of silence. The Mach cone downstream of the moving source thus represents the zone of action and the vast zone upstream of the Mach cone represents the zone of silence.

Fig. 4.11

Delineation of zones of action and silence for a moving source (of small disturbance) in a compressible medium. a Source moving at the speed of sound. b Source moving at supersonic speed, $V > a$

The motion of finite disturbances at supersonic speeds, as in bodies with finite thickness moving faster than the speed of sound, involves the creation of oblique shocks. The analysis of plane oblique shocks is based on the application of conservation principles in the normal and tangential directions to the shock. The schematic drawing of an oblique shock in a supersonic flow, shown in Fig. 4.12, identifies the wave angle $\beta$, the flow turning angle, $\theta$, and the normal and tangential components of flow with respect to the oblique shock wave.

Fig. 4.12

An oblique shock with its normal and tangential flow components. a The components of flow Mach number across an oblique shock. b The structure of flow velocities across an oblique shock

The conservation of tangential momentum along the plane oblique or conical shock (i.e., a straight shock) reveals that the tangential velocity remains constant across the (straight) shock, namely

$$\begin{aligned} w_2 = w_1 \end{aligned}$$

(4.20)

However, as the static temperature and thus the speed of sound are both raised across the shock, the tangential Mach numbers up- and downstream of the oblique shock are unequal. Examining part (a) and part (b) of Fig. 4.12 shows the constancy of $w$, but a drop in $M_t$ across the oblique shock, while maintaining the same shock wave angle, $\beta$ and the deflection angle, $\theta$ in the two graphs. Therefore oblique shocks enjoy an additional constant of motion in comparison to normal shocks and that is the tangential velocity. In summary the primary constants of motion for an oblique shock are as follows:

$$\begin{aligned} w&= \text {const.}\\ \rho u&= \text {const.} \\ p + \rho u^2&= \text {const.} \\ h_t&= \text {const.} \\ a_t&= \text {const.}\\ a^*&= \text {const.} \\ V_\text {max}&=\text {const.} \end{aligned}$$

Since the normal component of conservation equations are the same for the oblique and normal shock, all the jump conditions across an oblique shock are thus solely determined from the normal component of upstream Mach number, namely, $M_1 \sin \beta$. We will thus replace the upstream Mach number, $M_1$, in normal shock equations (4.13)–(4.17) by $M_1 \sin \beta$ to establish the jump conditions across an oblique shock of angle $\beta$. These are summarized as follows:

$$\begin{aligned} \frac{\rho _2}{\rho _1}&= \frac{u_1}{u_2} = \frac{\left( \gamma +1\right) M_1^2 \sin ^2 \beta }{2+\left( \gamma -1\right) M_1^2 \sin ^2 \beta } \end{aligned}$$

(4.21)

$$\begin{aligned} \frac{p_2}{p_1}&= 1 + \frac{2\gamma }{\gamma +1}\left( M_1^2 \sin ^2 \beta - 1\right) \end{aligned}$$

(4.22)

$$\begin{aligned} \frac{T_2}{T_1}&= \left[ 1 + \frac{2\gamma }{\gamma +1}\left( M_1^2 \sin ^2 \beta - 1\right) \right] \left[ \frac{2+\left( \gamma -1\right) M_1^2 \sin ^2 \beta }{\left( \gamma +1\right) M_1^2 \sin ^2 \beta } \right] \end{aligned}$$

(4.23)

$$\begin{aligned} \frac{s_2 - s_1}{R}&= \frac{\gamma }{\gamma -1} \ln \left[ 1 + \frac{2\gamma }{\gamma +1}\left( M_1^2 \sin ^2 \beta - 1\right) \right] \left[ \frac{2+\left( \gamma -1\right) M_1^2 \sin ^2 \beta }{\left( \gamma +1\right) M_1^2 \sin ^2 \beta } \right] \\&\quad - \ln \left[ 1 + \frac{2\gamma }{\gamma +1}\left( M_1^2 \sin ^2 \beta - 1\right) \right] \nonumber \end{aligned}$$

(4.24)

$$\begin{aligned} \frac{p_{t2}}{p_{t1}}&= e^{-\frac{\varDelta s}{R}} = \left[ 1 + \frac{2\gamma }{\gamma +1}\left( M_1^2 \sin ^2 \beta - 1\right) \right] \nonumber \\&\quad \left[ \frac{2+\left( \gamma -1\right) \left( \frac{2+\left( \gamma -1\right) M_1^2 \sin ^2 \beta }{2\gamma M_1^2 \sin ^2 \beta - \left( \gamma -1\right) }\right) }{2+\left( \gamma -1\right) M_1^2 \sin ^2 \beta } \right] ^{\frac{\gamma }{\gamma -1}} \end{aligned}$$

(4.25)

From the velocity triangles up- and downstream of the oblique shock, we get the following two equations:

$$\begin{aligned} \tan \beta = \frac{u_1}{w_1} \end{aligned}$$

(4.26a)

$$\begin{aligned} \nonumber \\ \tan \left( \beta - \theta \right) = \frac{u_2}{w_2} \end{aligned}$$

(4.26b)

The ratio of these two equations eliminates the constant tangential velocity and replaces the normal velocity ratio by inverse density ratio, to produce:

$$\begin{aligned} \frac{\tan \beta }{\tan \left( \beta - \theta \right) } = \frac{\left( \gamma +1\right) M_1^2 \sin ^2 \beta }{2+\left( \gamma -1\right) M_1^2 \sin ^2 \beta } \end{aligned}$$

(4.27)

The range of the wave angle $\beta$ lies between the Mach angle, $\mu$ and $90^{\circ }$ that correspond to a normal shock. The flow turning angle, $\theta$, thus starts from zero when the oblique shock is an infinitesimal-strength Mach wave to a maximum turning angle, $\theta _{\text {max}}$, which is shown in Fig. 4.13 (from [2]).

Example 4.1

The flow upstream of a normal shock is at Mach $M_1=2.0$, in air with $\gamma =1.4$. The stagnation speed of sound is $a_t=300\,\text {m/s}$, calculate:

• (a) Downstream Mach number, $M_2$

• (b) The ratio of speeds of sound, $a_2/a_1$ across the normal shock

• (c) The change in gas speed, $\varDelta u$, across the shock in m/s

Solution:

We may use $M_2^2 = \frac{1+\left[ \left( \gamma -1\right) /2\right] M_1^2}{\gamma M_1^2 - \left( \gamma -1\right) /2}$ to calculate the downstream Mach number, $M_2$. We get: $M_2=0.5774$ The ratio of speeds of sound is equal to square root of the static temperature ratio, i.e., $a_2/a_1 = \sqrt{T_2/T_1} = \sqrt{1.6875} \approx 1.30$

From stagnation speed of sound, we get the speed of sound upstream and downstream of the shock using $a^2 = a_t^2/\left[ 1+\left( \gamma -1\right) M^2/2\right]$, from which we get $a_1 \approx 224$ m/s and $a_2 \approx 290$ m/s, therefore $\varDelta u = u_1 - u_2 = M_1 a_1 - M_2 a_2 \approx 280$ m/s

Fig. 4.13

Oblique shock charts for air are from Ref. [2]

There are two reminders in using the oblique shock charts shown in Fig. 4.13. Namely, for a given turning angle, $\theta$, in a supersonic flow, there are two wave angles that satisfy the conservation equations. Therefore, $\beta$ is a double-valued function of $\theta$ (instead of a single-valued function). The smaller wave angle, known as the weak solution, is the angle to be used in oblique shock calculations. The larger wave angle, known as the strong solution, does not naturally occur for attached shocks, but may be created/induced in a supersonic wind tunnel or in ducts, by raising the back pressure. The reason for the dashed lines in the strong solution portion of the oblique shock graphs (e.g., Fig. 4.13) is to alert the user to select the weak wave solution (shown as solid curves). The second reminder concerns $\theta _{\text {max}}$. The flow turning angle beyond $\theta _{\text {max}}$ may not be supported by an attached oblique shock; rather a detached (bow) shock is needed to support the flow. These are graphically summarized in Fig. 4.14.

Fig. 4.14

Oblique and detached shock formation on pointed bodies of different nose angle in supersonic flow. a Two theoretically possible oblique shock angles. b Attached oblique shock on a 2-D compression ramp. c The relative positions of oblique shock and Mach wave in supersonic flow. d The formation of a bow shock ahead of a body in supersonic flow with $\theta _{\text {nose}} > \theta _{\text {max}}$

Fig. 4.15

Formation of the conical shock on a cone at zero-angle-of-attack in supersonic flow

In part (a) of Fig. 4.14, we show the possibility of two oblique wave angles, one corresponding to a strong solution and one corresponding to a weak solution. The strong solution is not used in the calculations; therefore it is shown as dashed line. In part (b) of Fig. 4.14 we are showing an attached oblique shock (the weak solution) to a 2-D wedge or ramp where the nose angle is less than the maximum turning angle at $M_1$. We also show an abrupt turning of the supersonic flow across the plane 2-D oblique shock in Fig. 4.14b. In part (c) of Fig. 4.14 we note the relative position of the oblique shock and the Mach wave at $M_1$. It shows that the wave angle for the oblique shock, $\beta$, is greater than the Mach angle, $\mu$, as expected. Finally, in part (d) of Fig. 4.14, we show a detached shock formation upstream of a blunt body with a nose angle greater than the maximum turning angle, $\theta _{\text {max}}$, allowed for an attached oblique shock at freestream Mach number, $M_1$. We may think of bow shock formation as the wave that is needed to create higher entropy rise across the shock in response to higher aerodynamic drag of a blunt body.

The conical shocks, which are created on cones in supersonic flow at zero angle of attack, create a new class of flows known as conical flowfields. Since the flowfield downstream of an attached conical shock lacks length scale, the flow variables become a pure function of the (cone) angle $\theta$ in spherical coordinate system, as shown in Fig. 4.15. The flow variables, e.g., $p$, $T$, $\rho$, $V$, $M$, are thus constant on constant $\theta$-cones in between the conical shock and the cone. These surfaces are indeed compression Mach cones that provide isentropic compression and continuous turning of the flow downstream of a conical shock to accommodate the changing flow area in the downstream direction. The changing flow area downstream of a conical shock is a new feature compared to the constant flow area downstream of a plane oblique shock (see Fig. 4.14b). It is exactly this feature that leads to 3-D relieving effect, as we shall see. A graphical depiction of the continuous bending of a streamline is also shown in Fig. 4.15.

Here again, we emphasize that the angle of the conical shock is greater than the Mach angle corresponding to $M_1$, as expected. To emphasize this point and also show the zones of action and silence on a cone, we graph Fig. 4.16.

Fig. 4.16

Relative position of the conical shock and the Mach cone

The theoretical formulation of conical flowfields downstream of conical shocks is due to Taylor and Maccoll [7]. The flowfield is best suited for spherical coordinates ($r$, $\theta$, $\varPhi$). The velocity components are $V_r$ and $V_{\theta }$ only as the $V_{\varPhi }$ component identically vanishes for axisymmetric flows corresponding to a cone at zero angle-of-attack. Utilizing Crocco’s theorem, we deduce that the flowfield downstream of the straight conical shock is irrotational and thus the two velocity components $V_r$ and $V_{\theta }$ are related to each other according to the irrotationality condition:

$$\begin{aligned} \nabla \times {\varvec{V}}= 0 \quad \rightarrow \quad V_{\theta } = \frac{\text {d}V_r}{\text {d}\theta } \end{aligned}$$

(4.28)

The continuity equation for a steady, axisymmetric flow in spherical coordinates reduces to:

$$\begin{aligned} 2 \rho V_r + \rho V_{\theta } \cot \theta + \rho \frac{\partial V_{\theta }}{\partial \theta } + V_{\theta } \frac{\partial \rho }{\partial \theta } = 0 \end{aligned}$$

(4.29)

The density is further eliminated from the above equation by the application of Euler momentum equation and the local speed of sound in the flow downstream of the conical shock. The result is a second-order non-linear ordinary differential equation in $V_r$ that is known as the Taylor-Maccoll equation:

$$\begin{aligned}&\frac{\gamma -1}{2} \left[ V_{\text {max}}^2 - V_r^2 - \left( \frac{\text {d}V_r}{\text {d}\theta } \right) ^2 \right] \left[ 2 V_r + \frac{\text {d}V_r}{\text {d}\theta } \cot \theta + \frac{\text {d}^2 V_r}{\text {d}\theta ^2} \right] \nonumber \\&- \frac{\text {d}V_r}{\text {d}\theta } \left[ V_r \frac{\text {d}V_r}{\text {d}\theta } + \frac{\text {d}V_r}{\text {d}\theta } \left( \frac{\text {d}^2 V_r}{\text {d}\theta ^2}\right) \right] = 0 \end{aligned}$$

(4.30)

Note that the only independent variable is $\theta$ and the only dependent variable is $V_r$. The $V_{\text {max}}$ term is the reference speed that we introduced earlier in this chapter, as a constant of motion, and is used in conical flowfields, following Taylor-Maccoll’s theoretical formulations. The Eq. (4.30) is still dimensional and may be non-dimensionalized by dividing both sides of the equation by $V_{\text {max}}^3$. The resulting non-dimensional equation is:

$$\begin{aligned}&\frac{\gamma -1}{2} \left[ 1 - V_r^{'2} - \left( \frac{\text {d}V_r^{'}}{\text {d}\theta } \right) ^2 \right] \left[ 2 V_r^{'} + \frac{\text {d}V_r^{'}}{\text {d}\theta } \cot \theta + \frac{\text {d}^2 V_r^{'}}{\text {d}\theta ^2} \right] \nonumber \\&- \frac{\text {d}V_r^{'}}{\text {d}\theta } \left[ V_r^{'} \frac{\text {d}V_r^{'}}{\text {d}\theta } + \frac{\text {d}V_r^{'}}{\text {d}\theta } \left( \frac{\text {d}^2 V_r^{'}}{\text {d}\theta ^2}\right) \right] = 0 \end{aligned}$$

(4.31a)

where

$$\begin{aligned} V_r^{'} \equiv \frac{V_r}{V_{\text {max}}} \end{aligned}$$

(4.31b)

From energy equation, $V^{'}$ is related to the local Mach number according to:

$$\begin{aligned} V^{'} = \left[ \frac{2}{\left( \gamma -1 \right) M^2} + 1 \right] ^{-1/2} \end{aligned}$$

(4.31c)

Fig. 4.17

Conical shock charts for $\gamma =1.4$ from Shapiro [6]. a Shock angle versus approach Mach number, with cone angle as parameter. b Ratio of surface pressure to free stream stagnation pressure versus free stream Mach number. c Surface Mach number versus free stream Mach number, with cone angle as parameter. d Ratio of surface stagnation pressure to free stream stagnation pressure, with cone angle as parameter. e Regions of shock attachment and detachment for cone and wedge. f Pressure drag coefficient based on projected frontal area

There are many good references on the subject of conical flowfields and conical shocks and the reader may refer to them to fill in the blanks in the above derivation. An excellent contemporary reference is by Anderson [1] that is recommended for further reading. The Taylor-Maccoll equation (4.31a) is solved numerically using the standard numerical approach of Runge-Kutta fourth order technique starting from the shock at $\sigma$ with the known Mach number and flow angle immediately downstream of the shock. These are used to get $V_r^{'}$ and $V_{\theta }^{'}$ that serve as the initial conditions for the Runge-Kutta solution of the Taylor-Maccoll equation. Starting from the immediate downstream of the conical shock, we march in the negative theta direction by taking small incremental steps of $\varDelta \theta$. Per step in the numerical procedure, we calculate the new $V_r^{'}$ and $V_{\theta }^{'}$ until we capture the cone at $V_{\theta }^{'} = 0$. In this approach, we are finding the cone that supports the conical shock at $\sigma$. The numerical solution of Taylor-Maccoll equation has been used to develop conical shock charts shown in Fig. 4.17. In part Fig: 3.17a, only the weak solution is plotted and we clearly note the maximum cone angles that support attached conical shocks. In part 4.17b, the ratio of cone surface static pressure to freestream stagnation pressure is plotted. Part 4.17c is a graph of the surface Mach number on the cone for different cones in supersonic flow. In part 4.17d, we have the ratio of stagnation pressures across a conical shock for different cone angles and freestream Mach number. Part 4.17e highlights the three-dimensional relieving effect by comparing a cone to a wedge and their corresponding wave angles. The wave angle for the wedge is greater than the wave angle for the cone and thus the surface of the wedge experiences higher (static) pressure than the (static) pressure on the corresponding cone. This is the essence of three-dimensional relieving effect. The cone wave drag coefficient, referenced to base area, is also shown in part 4.17f.

Example 4.2

Consider a $20^{\circ }$ (2-D) ramp in a Mach 2 flow of air with $\gamma =1.4$. Also, there is a cone of semi-vertex angle of $20^{\circ }$ in the same supersonic flow condition. Calculate and compare:

• (a) the wave angles on the 2-D ramp and the cone

• (b) the surface pressure on the ramp and the cone.

Solution:

From the plane oblique shock charts (Fig. 4.13) we read the weak shock wave angle corresponding to Mach 2 freestream and a turning angle of twenty degrees as $\beta \approx 53.4^{\circ }$. From the conical shock charts (Fig. 4.17a) we read the corresponding conical shock angle of $\theta _s \approx 38^{\circ }$. Comparison of these two angles reveals the weaker 3-D shock as compared to the 2-D shock.

The surface pressure on the ramp is calculated based on the normal component of the flow to the oblique shock, i.e., $M_1 \sin \beta = 2 \sin (53.4^{\circ }) \approx 1.6$, which produces a shock (static) pressure ratio of ${\sim }2.82$ downstream of the 2-D shock. Using Fig. 4.17b, we read the ratio of the cone surface static pressure to the freestream stagnation pressure of ${\sim }0.25$. The stagnation and static pressures are related by (the isentropic relation):

$$\frac{p_t}{p} = \left( 1 + \frac{\gamma -1}{2} M^2\right) ^{\frac{\gamma }{\gamma -1}} = 7.824 \quad \text {for} \quad M=2$$

Therefore the ratio of cone surface pressure to freestream static pressure is: $p_c/p_1 \sim 0.25 (7.824)=1.956$ The counterpart of this ratio for the 2-D ramp was calculated to be 2.82. Here again we reveal the 3-D relieving effect as evidenced by this example, i.e., lower cone surface pressure.

Example 4.3

Compare the maximum turning angles of Mach-2 flow of air in 2-D and 3-D through plane oblique and conical shocks, respectively.

Solution:

Referring to the plane oblique shock charts, we note that $\theta _{\text {max}} \approx 23^{\circ }$ in 2-D. The implication of this angle is that in two-dimensional flow a plane oblique shock in Mach 2 can at most cause ${\sim }23^{\circ }$ turning.

For the case of 3-D flows, we refer to conical shock charts. From Fig. 4.17e, we note that maximum turning at Mach 2 is nearly $40^{\circ }$. The implication is that conical shocks can stay attached to cones of up to $40^{\circ }$ semi-vertex angle in Mach 2 flow, whereas in 2-D the ramp angle above ${\sim }23^{\circ }$ causes the shock to be detached. This too is due to 3-D relieving effect.

Fig. 4.18

Definition sketch of compression and expansion corners and the corresponding waves that are generated. a Compression corner, oblique shock, LRW. b Compression corner, oblique shock, RRW. c Expansion corner, Prandtl-Meyer expansion fan, LRW. d Expansion corner, Prandtl-Meyer expansion fan, RRW

Fig. 4.19

An exaggerated view of infinitesimal turning across an expansion Mach wave

We now direct our attention to flow expansion. The counterpart of compression waves is the expansion waves, which cause flow acceleration and turning in supersonic flow. The direction of flow turning (with respect to local upstream flow) determines whether an expansion or a compression wave is needed to cause the necessary flow deflection. One description that says “turning the flow into itself” is caused by compression waves and “turning the flow out of itself” is accomplished by expansion waves is useful when we consider right- and left-running waves. We identify right- and left-running waves in supersonic flow through the eyes of an observer who stands on the wave and is facing downstream. The right-running wave (RRW) appears on the right side of the observer and the left-running wave (LRW) appears on the left side of the observer. The expansion waves that are centered at a corner are referred to as expansion fans and are known as Prandtl-Meyer expansion waves. Let us examine some compression and expansion corners in supersonic flow and the waves that cause the flow deflection in Fig. 4.18. The oblique shocks are marked as O.S. and the Prandtl-Meyer expansion fans are abbreviated to P-M E.F. in Fig. 4.18. We also have defined the flow angle, $\theta$, with respect to horizontal, or x-axis and have assigned positive values to $\theta$ when the flow direction is above horizontal and have assigned a negative value to $\theta$ when the flow points in the downward (i.e., $-$y) direction. We can also discern the “flow turning into itself” and “flow turning out of itself” from compression and expansion corners that are schematically shown in Fig. 4.18.

Expansion waves, as noted earlier are Mach waves, which cause an infinitesimal turning of the supersonic flow. The analysis of an expansion Mach wave can be carried out using the definition sketch in Fig. 4.19. Note that we had to exaggerate the infinitesimal turning in order to “see” the geometrical relations involving flow angles and velocities across the Mach wave in Fig. 4.19.

By applying the law of sines to the triangle with sides $V$ and $V+ \text {d}V$, we may relate the infinitesimal turning angle, $\text {d}\theta$, and incremental speed change, $\text {d}V$, in the following equation:

$$\begin{aligned} \frac{V + \text {d}V}{\sin \left( \frac{\pi }{2} + \mu \right) } = \frac{V}{\sin \left( \frac{\pi }{2} - \mu - \text {d}\theta \right) } \end{aligned}$$

(4.32)

The Eq. (4.32) may be simplified by expanding the sine terms to get:

$$\begin{aligned}&1 + \frac{\text {d}V}{V} = \frac{\cos \mu }{\cos \left( \mu + \text {d}\theta \right) } \cong \frac{\cos \mu }{\cos \mu - \left( \text {d}\theta \right) \sin \mu } = \frac{1}{1 - \left( \text {d}\theta \right) \tan \mu }\\&\approx 1 + \left( \text {d}\theta \right) \tan \mu \nonumber \end{aligned}$$

(4.33)

By taking isentropic flow conditions across the Mach wave; we may relate incremental turning angle, $\text {d}\theta$, to the incremental Mach number change, $\text {d}M$, according to:

$$\begin{aligned} \text {d}\theta = \frac{\sqrt{M^2 - 1}}{1 + \frac{\gamma -1}{2}M^2} \frac{\text {d}M}{M} \end{aligned}$$

(4.34)

Equation (4.34) may be integrated from zero turning, corresponding to Mach one (sonic flow), to a general turning angle $\nu \left( M\right)$ corresponding to supersonic Mach number, $M$. The angle $\nu \left( M\right)$ is called the Prandtl-Meyer angle and has a closed form solution, i.e.,

$$\begin{aligned} \nu \left( M\right) = \sqrt{\frac{\gamma +1}{\gamma -1}} \tan ^{-1} \sqrt{\frac{\gamma -1}{\gamma +1}\left( M^2-1\right) } - \tan ^{-1} \sqrt{M^2-1} \end{aligned}$$

(4.35)

In addition to a table for the Prandtl-Meyer function and Mach angle, we graph these two functions in terms of Mach number in Fig. 4.20. In Appendix C the relation between $M$, $\mu$ and $\nu$ are tabulated up to a Mach number of 18.

Fig. 4.20

Prandtl-Meyer and Mach angles, in degrees, in supersonic flow with $\gamma =1.4$

The Prandtl-Meyer function asymptotically approaches a limiting angle, which can be established by taking the limit of Eq. (4.35) as freestream Mach number approaches infinity. The implication of the maximum turning angle in $\nu \left( M\right)$ is that the flow will expand to infinite Mach number and thus pure vacuum is achieved. Neither of these two conditions is physically realizable; thus $\nu _{\text {max}}$ represents an ideal limit in high-speed gas dynamics.

$$\begin{aligned} \nu _{\text {max}}&= \nu \left( M_2 \rightarrow \infty \right) = \left( \sqrt{\frac{\gamma +1}{\gamma -1}} - 1 \right) \frac{\pi }{2} ~\text {[rad]} \\ \nu _{\text {max}}&\rightarrow 130.5^{\circ } \quad \text {for} \quad \gamma =1.4 \nonumber \end{aligned}$$

(4.36)

Finally, since the expansion Mach waves never coalesce to create a rarefaction shock, the flow across such waves is isentropic. Consequently, the stagnation pressure remains constant across Prandtl-Meyer expansion waves. The stepwise calculation involving expanding supersonic flows is dictated by the wall geometry/turning, which establishes the Mach number and the pressure. Now we solve a supersonic flow expansion problem using Prandtl-Meyer function, $\nu \left( M\right)$.

Example 4.4

A supersonic flow approaches an expansion corner, as shown. Assuming the medium is air with $\gamma =1.4$, calculate:

• (a) The Mach number downstream of the sharp corner

• (b) The wave envelope, i.e., the angle of the head and tail waves

• (c) The static pressure on the expansion ramp if $p_1=1$ atm.

 

Solution:

The P-M function in region 1 is $\nu _1 \approx 26.5^{\circ }$. The P-M function on the expansion ramp is thus: $\nu _2 = \nu _1 + \varDelta \theta \approx 26.5^{\circ } + 22^{\circ } = 46.5^{\circ }$ Therefore the corresponding Mach number is $M_2 = 2.836$. The leading Mach wave makes an angle of $\mu _1 = \sin ^{-1} \left( 1/2 \right) = 30^{\circ }$ and the tail Mach wave makes an angle of $\mu _2 = \sin ^{-1} \left( 1/2.836 \right) \approx 20.6^{\circ }$ Finally, the ratio of static pressures on the two ramps is related to their respective Mach numbers following the isentropic flow condition up- and downstream of the corner, namely

$$\frac{p_2}{p_1} = \left( \frac{1+\left( \gamma -1\right) M_1^2 / 2}{1+\left( \gamma -1\right) M_2^2 / 2} \right) ^{\frac{\gamma }{\gamma -1}}$$

Upon substitution for the up- and downstream Mach numbers (along with $\gamma =1.4$), we get: $p_2/p_1 \sim 0.2729$. Therefore, flow expansion around a $22^{\circ }$ corner in Mach 2 upstream flow caused a flow acceleration to Mach 2.836 on the downstream ramp and the static pressure was reduced to ${\sim }27.3\,\%$ of upstream value, i.e., $p_2=0.2729$ atm., stagnation pressure, $p_t$, remained constant.

4.2 Lift and Wave Drag

By direct application of the shock waves and Prandtl-Meyer expansion theory to pointed bodies, we are able to calculate local static pressure on a body in supersonic flow. Once the static pressure distribution is obtained; the integration of pressure around the body then establishes the components of resultant force in the lift and drag directions. Since the inviscid drag in supersonic flow is entirely due to pressure, it is called pressure or wave drag. Unlike its subsonic counterpart, this drag is inherently a creature of the supersonic flow where a two-dimensional body in inviscid fluid still produces drag. As the reader recalls D’Alembert paradox where two-dimensional bodies in inviscid fluid and in low speed (i.e., incompressible) flow, even in purely subsonic flow, experienced zero drag!

Fig. 4.21

A pointed 2-D body in supersonic flight with the resultant force (per unit span), $R^{'}$ resolved in normal-axial ($N^{'}$ and $A^{'}$) directions and lift-drag ($L^{'}$ and $D^{'}$) directions

First, we consider a two-dimensional body in Cartesian coordinates, as shown in Fig. 4.21. The x-axis is chosen to be parallel to (or coincide with) the chord and thus the freestream is inclined with respect to the x-axis by the angle of attack, $\alpha$. This approach readily produces the normal and axial force components (per unit span), $N^{'}$ and $A^{'}$ respectively, which are then related to the lift and drag using angle of attack, $\alpha$, according to:

$$\begin{aligned} L^{'}&= N^{'} \cos \alpha - A^{'} \sin \alpha \nonumber \\ D^{'}&= N^{'} \sin \alpha + A^{'} \cos \alpha \end{aligned}$$

(4.37)

In Fig. 4.21, $R^{'}$ is the resultant aerodynamic force, which is then resolved in different orthogonal directions, as shown. The normal force is simply the integral of pressure difference across the airfoil along the chord, namely:

$$\begin{aligned} N^{'} = \int _0^c \left( p_l - p_\text {u} \right) \text {d}x \end{aligned}$$

(4.38)

And the axial force (per unit span) is related to the integral of pressure times the body slopes (along the chord) according to:

$$\begin{aligned} A^{'} = \int _0^c \left[ p_\text {u} \left( \frac{\text {d}y_\text {u}}{\text {d}x} \right) - p_\text {l} \left( \frac{\text {d}y_l}{\text {d}x} \right) \right] \text {d}x \end{aligned}$$

(4.39)

In terms of normal and axial force coefficients, $c_n$ and $c_a$, we may express Eqs. (4.38) and (4.39) in terms of the non-dimensional pressure coefficients on the upper and lower surfaces, namely

$$\begin{aligned} c_n&= \frac{1}{c} \int _0^c \left( C_{p_{l}} - C_{p_\text {u}} \right) \text {d}x \end{aligned}$$

(4.40)

$$\begin{aligned} c_a&= \frac{1}{c} \int _0^c \left[ C_{p_\text {u}} \left( \frac{\text {d}y_\text {u}}{\text {d}x} \right) - C_{p_\text {l}} \left( \frac{\text {d}y_\text {l}}{\text {d}x} \right) \right] \text {d}x \end{aligned}$$

(4.41)

Now, we are ready to apply these principles to a typical supersonic profile, namely, a diamond airfoil.

Example 4.5

A symmetrical diamond airfoil is shown at $5^{\circ }$ angle of attack (not to scale). Use shock-expansion theory to calculate:

• (a) pressure coefficient on all four panels

• (b) normal force coefficient, $c_n$

• (c) axial force coefficient, $c_a$

• (d) lift and wave drag coefficients, $c_l$ and $c_d$

• (e) lift-to-drag ratio, $L^{'}/D^{'}$

 

Solution:

Freestream has to turn an additional $10^{\circ }$ up to reach panel #1 (i.e., from $5^{\circ }$ to $15^{\circ }$). This is the case of flow turning into itself. Therefore, the turning has to occur via an oblique shock wave. From $\theta -\beta -M$ chart for a plane oblique shock, we get:

$$\begin{aligned}&M_{\infty } = 2.4 \text {~and~} \theta = 10^{\circ } \quad \xrightarrow {OS} \quad \beta \approx 33^{\circ } \\&M_{\infty } \sin \beta \approx 1.306 \quad \quad \quad \xrightarrow {NS} \quad M_{n_{1}} \approx 0.786 \text {~and~} p_1/p_{\infty } \approx 1.805 \end{aligned}$$

We use the pressure coefficient definition

$$C_p \equiv \frac{2}{\gamma M_{\infty }^2} \left[ \frac{p}{p_{\infty }} - 1 \right] $$

to get:

$$C_{p_{1}} \approx 0.1996$$

Also, the Mach number on panel 1 is:

$$M_1 = \frac{M_{n_{1}}}{\sin \left( \beta - \theta \right) } \approx \frac{0.786}{\sin \left( 33^{\circ } - 10^{\circ } \right) } \approx 2.01$$

The freestream has to turn a total of $20^{\circ }$ (from $+5^{\circ }$ to $-15^{\circ }$), in order to reach ramp number 2. Reading the plane oblique shock chart, we estimate a wave angle (i.e., the weak solution):

$$M_{\infty } = 2.4 \text {~and~} \theta = 20^{\circ } \quad \xrightarrow {OS} \quad \beta \approx 44.2^{\circ }$$

Again the normal component of flow establishes the shock jumps, namely

$$ M_{\infty } \sin \beta \approx 1.67 \quad \xrightarrow {NS} \quad M_{n_{2}} \approx 0.6458 \text {~and~} p_2/p_{\infty } \approx 3.126 $$

The Mach number on panel 2 is:

$$M_2 = \frac{M_{n_{2}}}{\sin \left( \beta - \theta \right) } \approx \frac{0.6458}{\sin \left( 44.2^{\circ } - 20^{\circ } \right) } \approx 1.71$$

We use the pressure coefficient definition

$$C_p = \frac{2}{\gamma M_{\infty }^2} \left[ \frac{p}{p_{\infty }} - 1 \right] $$

to get:

$$C_{p_{2}} \approx 0.5273$$

To march to ramps 3 and 4, we recognize that the net turning angle is twice the $15^{\circ }$ (i.e., $30^{\circ }$) and since the flow is turning out of itself, we expect an expansion fan to cause the turn. We start with $\nu _1$ which corresponds to $M_1$ and from Prandtl-Meyer table, we get: $\nu _1 \approx 26.38^{\circ }$ and thus

$$ \nu _3 = \nu _1 + 30^{\circ } = 56.38^{\circ } \quad \xrightarrow {P-M} \quad M_3 \approx 3.35 $$

The isentropic conditions between panels 1 and 3 can be used to establish static pressure, $p_3$,

$$ \frac{p_3}{p_1} = \left( \frac{1+\left( \gamma -1 \right) M_1^2 /2}{1+\left( \gamma -1 \right) M_3^2 /2} \right) ^{\frac{\gamma }{\gamma -1}} \approx 0.1272 $$

We use chain rule to write

$$p_3/p_{\infty } = \left( p_3/p_1 \right) \left( p_1/p_{\infty } \right) =0.1272 \cdot 1.805=0.2296$$

Therefore, the pressure coefficient follows as

$$C_p = \frac{2}{\gamma M_{\infty }^2} \left[ \frac{p}{p_{\infty }} - 1 \right] $$

or

$$C_{p_{3}} \approx -0.1911$$

Now, marching onto panel 4 from panel 2, we note that the net turning angle is again $30^{\circ }$. The P-M angle for panel 2 is $\nu _2 \approx 18^{\circ }$, which gives the P-M angle for panel 4 to be $\nu _4 \approx 48^{\circ }$. From P-M tables, we get $M_4 \approx 2.9$. The isentropic conditions between panels 2 and 4 can be used to establish static pressure, $p_4$

$$ \frac{p_4}{p_2} = \left( \frac{1+\left( \gamma -1 \right) M_2^2 /2}{1+\left( \gamma -1 \right) M_4^2 /2} \right) ^{\frac{\gamma }{\gamma -1}} \approx 0.15861 $$

We use chain rule to write

$$p_4/p_{\infty } = \left( p_4/p_2 \right) \left( p_2/p_{\infty } \right) =0.15861 \cdot 3.126=0.4958$$

Therefore, the pressure coefficient follows as

$$C_p = \frac{2}{\gamma M_{\infty }^2} \left[ \frac{p}{p_{\infty }} - 1 \right] $$

or

$$C_{p_{4}} \approx -0.1251$$

The normal force coefficient is:

$$ c_n = 0.5 \left( C_{p_{2}} + C_{p_{4}} - C_{p_{1}} - C_{p_{3}} \right) $$

Therefore, we get the normal force coefficient as $c_n \approx 0.1968$. To calculate the axial force, we first establish the airfoil thickness-to-chord ratio as $t/c = \tan 15^{\circ }$. This gives $t/c \approx 0.2678$ (a rather thick profile for supersonic flow!). Therefore, the axial force coefficient may be written as:

$$c_a = \frac{1}{2} \left( C_{p_{1}} + C_{p_{2}} - C_{p_{3}} - C_{p_{4}} \right) \frac{t}{c} $$

Upon substitution, we get $c_a \approx 0.1397$.

We use Eq. (4.37) to get lift and drag coefficients from the normal and axial force coefficients and the angle of attack, $\alpha$, namely $c_l \approx 0.1839$ and $c_d \approx 0.1563$. Finally, lift-to-drag ratio is $L^{'}/D^{'} = 0.1839/0.1563 \sim 1.18$.

4.3 Bi-Convex Airfoil

Fig. 4.22

Definition sketch of a general (asymmetrical) bi-convex airfoil in supersonic flow at angle of attack

We are now ready to apply the shock-expansion theory to bi-convex airfoils, which represent a broad class of supersonic airfoil shapes, e.g., circular-arc airfoils. The general characteristics of these supersonic airfoils are sharp leading and trailing edges, small thickness and small camber. These are indeed the qualities that lead to a reduced wave drag. The definition sketch in Fig. 4.22 shows a general asymmetrical bi-convex airfoil. The upper and lower surfaces are identified by their own functions, $y_\text {u} \left( x \right)$ and $y_\text {l} \left( x \right)$ respectively. The nose angles on the upper and lower surfaces, $\theta _\text {nu}$ and $\theta _\text {nl}$, are also identified. The chordwise position of the maximum thickness, $x_{t_{\text {max}}}$, and the maximum thickness itself, $t_{\text {max}}$, are shown. The upstream supersonic flow is at angle $\alpha$ (angle of attack) with respect to chord.

First, we need to identify the waves that are needed to turn the flow on the upper and lower surfaces at the leading edge (nose). Depending on the relative values of the upper surface nose angle and the angle of attack, we may have an oblique shock or a Prandtl-Meyer expansion fan at the leading edge on the upper surface. This is easily established by the following rules at the leading edge (for the case of positive angle of attack, i.e., $\alpha >0$):  

Case 1: $\alpha < \theta _\text {nu}$ :

The wave is an oblique shock, which turns the supersonic flow by $\left( \theta _\text {nu} - \alpha \right)$ angle (note that the flow turns into itself).

Case 2: $\alpha = \theta _\text {nu}$ :

A Mach wave appears at the leading edge on the upper surface with zero net turning (since $\alpha = \theta _\text {nu}$) imposed across the Mach wave.

Case 3: $\alpha > \theta _\text {nu}$ :

A Prandtl-Meyer expansion fan will be formed at the leading edge on the upper surface, with a net turning of ($\alpha - \theta _\text {nu}$). Note that the flow turns out of itself.

 

Fig. 4.23

The wave pattern on a bi-convex airfoil at positive angle of attack, $\alpha$. a Case 1: with $\alpha < \theta _\text {nu}$ (an oblique shock will form at the leading edge on the upper surface). b Case 2: with $\alpha = \theta _\text {nu}$ (upstream flow is aligned with the upper surface at the leading edge). c Case 3: with $\alpha > \theta _\text {nu}$ (Prandtl-Meyer expansion fan is formed at the leading edge on the upper surface)

The leading edge of the lower surface will encounter an oblique shock, which turns the flow by ($\theta _\text {nl} + \alpha$) angle. The identification of the leading-edge waves will allow us to march across the waves and onto the upper and lower surfaces of the bi-convex airfoil. The waves that are formed on the airfoil beyond the leading edge are all expansion Mach waves that cause the flow to accelerate and the static pressure to drop. In the vicinity of the airfoil, we will have the wave patterns corresponding to Case-1, Case-2 and Case-3, shown in Fig. 4.23 where solid lines show oblique shocks and the dashed lines are expansion Mach waves.

The waves at the trailing edge (on the upper and lower surfaces) are formed to equalize the static pressure and flow direction downstream of the airfoil. In supersonic airfoil theory, we define the slipstream as a (free) vortex sheet that is formed at the trailing edge of the airfoil that separates the upper and lower flows with flow directions on two sides coinciding with the slipstream and the static pressure being continuous across the slipstream. Although the flows on two sides of the slipstream are parallel, they are not equal in magnitude, i.e., $V_4 \ne V_5$. It is indeed the jump in tangential velocity across the slipstream that is called the vortex sheet strength. Also, since the airfoil is in the zone of silence of the waves at the trailing edge, the flow on the airfoil is not affected by the waves at the trailing edge. However, these waves impact the formation of slipstream (i.e., vortex sheet), which may impact aircraft components downstream of the wing (e.g., engine inlet). We show the slipstream downstream of a supersonic airfoil at high angle of attack and its two boundary conditions in a definition sketch in Fig. 4.24. We have (arbitrarily) chosen to show an oblique shock wave as well as a Prandtl-Meyer expansion fan at the trailing edge of the airfoil in Fig. 4.24. However, it is possible to have two oblique shock waves at the trailing edge, if the flow environment downstream of the airfoil (i.e., static pressure and flow direction) demands it. Indeed, the nature of these waves at the trailing edge is established by the two boundary conditions on the slipstream, namely the equality of static pressure $\left( p_4 = p_5 \right)$, which is known as the pressure equilibrium condition, and the flow angles $\left( \theta _4 = \theta _5 \right)$.

Fig. 4.24

The slipstream downstream of a supersonic airfoil at angle of attack with possible trailing edge waves

Once, we established the nature of the waves at the leading edge, we may proceed to calculate the pressure and Mach number immediately downstream of the leading edge on the upper and lower surfaces of the airfoil. The flow downstream of the leading edge on a bi-convex airfoil undergoes a continuous acceleration on both upper and lower surfaces (as shown in Figs. 4.23 and 4.24). Thus, the flow expands isentropically on both sides of the airfoil downstream of the leading edge where oblique shocks may appear. The difference between the upper and lower surfaces is therefore in the initial conditions that are set at the nose by the type and the strength of the waves that are formed at the leading edge. Otherwise, the solution methodology is identical on the upper and lower surfaces of the bi-convex airfoil.

Solution Methodology

We start at the leading edge, calculating the nose angle on the upper surface, according to:

$$\begin{aligned} \theta _\text {nu} = \left. \tan ^{-1} \left( \text {d}y_\text {u} / \text {d}x \right) \right| _{x=0} \end{aligned}$$

(4.42)

Depending on the angle of attack, $\alpha$, we may have one of the three cases that we described earlier as determining the wave at the leading edge on the upper surface. We then march across the wave at the leading edge and establish:

• 1. $M_\text {nu}$, Mach number of the flow immediately downstream of the nose on the upper surface

• 2. $p_\text {nu}$, static pressure immediately downstream of the nose on the upper surface

We then propose to march downstream on the upper surface by creating a table of values for the surface pressure (and Mach number). For this purpose, we may divide the airfoil into $\left( n+1\right)$-chordwise calculation stations, from leading edge to trailing edge, similar to the graph shown in Fig. 4.25. Here, we used the chord length, $c$, as the normalizing length scale in the problem and graphed $y/c$ versus $x/c$, as the non-dimensional coordinates. Counting the leading and trailing edges as calculation stations, we have (arbitrarily) chosen 11 positions along the airfoil upper surface in Fig. 4.25, which divides the airfoil into ten segments/panels along the chord, as shown.

Fig. 4.25

The upper surface of a general bi-convex airfoil divided into 11 equally-spaced calculation stations and 10 panels

Next, we calculate the surface (inclination) angle at every calculation station, namely at $n=2$ to 11 [note that we have already calculated the leading edge angle, i.e., $n=1$ in Eq. (4.42)]. For the $i$th station, we use:

$$\begin{aligned} \theta _{\text {u} \left( i \right) } = \left. \tan ^{-1} \left( \text {d}y_\text {u} / \text {d}x \right) \right| _{x = x_i} \end{aligned}$$

(4.43)

The Mach number immediately downstream of the nose that we calculated earlier is now used to determine the Prandtl-Meyer angle at the nose, i.e., $\nu _n$ or $\nu _1$. Since the flow is continuously expanding on the bi-convex airfoil, the Prandtl-Meyer angle continually increases on the upper surface, from one station to the next, by the net flow turning angle between those stations, namely

$$\begin{aligned} \nu _{\text {u} \left( i+1 \right) } = \nu _{\text {u} \left( i \right) } + \left| \theta _{\text {u} \left( i+1 \right) } - \theta _{\text {u} \left( i \right) } \right| \end{aligned}$$

(4.44)

The absolute value in Eq. (4.44) shows that the Prandtl-Meyer angle increases as we march towards the trailing edge on the bi-convex airfoil. Now, we can determine the local Mach numbers, $M_{\text {u} \left( i \right) }$ based on the Prandtl-Meyer angles, $\nu _{\text {u} \left( i \right) }$. The isentropic flow condition on the upper (and lower) surface can be used to calculate the static pressure according to isentropic rule:

$$\begin{aligned} \frac{p_{\text {u} \left( i+1 \right) }}{p_{\text {u} \left( i \right) }} = \left[ \frac{1+\left( \frac{\gamma -1}{2}\right) M_{u(i)}^2}{1+\left( \frac{\gamma -1}{2}\right) M_{u(i+1)}^2} \right] ^{\frac{\gamma }{\gamma -1}} \end{aligned}$$

(4.45)

Fig. 4.26

Definition sketch of the effective panel for the $i$th calculation station

We have now created the pressure distribution on the upper surface of the airfoil at discrete points along the airfoil, from the leading edge to the trailing edge. By defining the effective panel areas for each calculated pressure as the half-way points between the adjacent calculation stations (or panels), we can proceed with the calculation of the incremental force, per effective panel area. A definition sketch is shown in Fig. 4.26 as an aid to understand the effective panel area and the calculation of the incremental force and pitching moment based on $p_{\text {u} \left( i \right) }$. The increment of normal components of force (per unit span) on the upper and lower surfaces due to the static pressures $p_{\text {u} \left( i \right) }$ and $p_{\text {l} \left( i \right) }$ are:

$$\begin{aligned} \varDelta N_{\text {u} \left( i \right) }^{'}&= -p_{\text {u} \left( i \right) } \left[ x_{i+1/2} - x_{i-1/2} \right] \end{aligned}$$

(4.46a)

$$\begin{aligned} \varDelta N_{\text {l} \left( i \right) }^{'}&= +p_{\text {l} \left( i \right) } \left[ x_{i+1/2} - x_{i-1/2} \right] \end{aligned}$$

(4.46b)

The increments of axial component of force (per unit span) on the upper and lower surface due to static pressures $p_{\text {u} \left( i \right) }$ and $p_{\text {l} \left( i \right) }$ are:

$$\begin{aligned} \varDelta A_{\text {u} \left( i \right) }^{'}&= +p_{\text {u} \left( i \right) } \left[ y_{u \left( i+1/2 \right) } - y_{u \left( i-1/2 \right) } \right] \end{aligned}$$

(4.47a)

$$\begin{aligned} \varDelta A_{\text {l} \left( i \right) }^{'}&= -p_{\text {l} \left( i \right) } \left[ y_{l \left( i+1/2 \right) } - y_{l \left( i-1/2 \right) } \right] \end{aligned}$$

(4.47b)

The increments of pitching moment about the leading edge (per unit span) on the upper and lower surfaces are:

$$\begin{aligned} \varDelta M_{\text {u} \left( i \right) }^{'}&= p_{\text {u} \left( i \right) } \left[ x_{i+1/2} - x_{i-1/2} \right] x_i + p_{\text {u} \left( i \right) } \left[ y_{u \left( i+1/2 \right) } - y_{u \left( i-1/2 \right) } \right] y_{\text {u} \left( i \right) } \end{aligned}$$

(4.48a)

$$\begin{aligned} \varDelta M_{\text {l} \left( i \right) }^{'}&= -p_{\text {l} \left( i \right) } \left[ x_{i+1/2} - x_{i-1/2} \right] x_i - p_{\text {l} \left( i \right) } \left[ y_{l \left( i+1/2 \right) } - y_{l \left( i-1/2 \right) } \right] y_{l \left( i \right) } \end{aligned}$$

(4.48b)

Finally, we add all the increments to get the normal and axial force components (per unit span) on the upper surface and proceed similarly on the lower surface. Equation (4.37) will then be used to express lift and wave drag (per unit span) in terms of the normal and axial force and the angle of attack, $\alpha$. We will now demonstrate the shock-expansion theory applied to an asymmetric, parabolic profile bi-convex airfoil with negative camber in the following example.

Example 4.6

Consider a 5 % thick bi-convex airfoil of parabolic arc upper and lower profiles with negative camber, as shown in the schematic drawing. The upper surface contributes 2 % to thickness at 50 % chord and is thus defined by:

$$ y_\text {u} \left( x \right) = 0.08x \left( 1-x \right) $$

The lower surface contributes 3 % thickness to the airfoil at mid chord and is thus defined by:

$$ y_\text {l} \left( x \right) = -0.12x \left( 1-x \right) $$

By virtue of negative camber, a stronger oblique shock appears at the leading edge on the lower surface than the oblique shock on the upper surface. Therefore, we expect this airfoil to produce lift (albeit small) even at zero angle of attack. Use shock-expansion theory to first calculate the pressure coefficient on the upper and lower surfaces, then calculate the lift, wave drag and pitching moment (about leading edge) coefficients for this airfoil at the freestream Mach number of $M_{\infty }=1.4$ and at zero angle of attack. Graph the airfoil and its upper and lower surface pressure distributions along the chord.

 

Solution:

First, we calculate the leading edge angles for the upper and lower surfaces.

$$ \left. \frac{\text {d}y_\text {u}}{ \text {d}x} \right| _{x=0} = \left. \left( 0.08 - 0.16 x \right) \right| _{x=0} = 0.08 $$

Therefore, the nose angle on the upper surface is $\theta _{\text {nu}} \approx 4.57^{\circ }$. Similarly, we calculate the leading edge angle for the lower surface,

$$ \left. \frac{\text {d}y_\text {l}}{ \text {d}x} \right| _{x=0} = \left. \left( -0.12 + 0.24 x \right) \right| _{x=0} = -0.12 \quad \rightarrow \quad \theta _\text {nl} \approx -6.84^{\circ }$$

In both cases, the supersonic flow turns into itself and thus requires an oblique shock to form at the leading edge on both upper and lower surfaces. Using oblique shock charts (Fig. 4.13), we get:

$$\begin{aligned}&M_{\infty } = 1.4 \text {~and~} \theta _{\text {nu}} \approx 4.6^{\circ } \quad \xrightarrow {\text {OS}} \quad \beta _\text {u} \approx 52.2^{\circ } \\&M_{\infty } \sin \beta \approx 1.11 \quad \xrightarrow {\text {NS}} \quad M_{\text {nu}_{\text {nose}}} \approx 0.907 \text {~and~} p_{n,u}/p_{\infty } \approx 1.261 \\&M_{\text {u}_{\text {nose}}} = M_{\text {nu}_{\text {nose}}} / \sin \left( \beta - \theta \right) \approx 1.228 \end{aligned}$$

The Prandtl-Meyer angle immediately downstream of the leading edge on the upper surface corresponds to $M_{\text {u}_{\text {nose}}}=1.228$. We can use Eq. (4.35) to get:

$$\nu _{\text {u}_{\text {nose}}} \approx 4.25^{\circ }$$

Now, from the equation of the slope of the upper surface, $\frac{\text {d}y_\text {u}}{\text {d}x} = 0.08 - 0.16 x$, we calculate the surface angle at increments of 10 % chord. This procedure is easily implemented in a spreadsheet program (e.g., Excel). The results are shown in Table 4.1.

Table 4.1 Geometric and flow calculations on a bi-convex airfoil using Shock-expansion theory

Table 4.2 Geometric and flow calculations on a bi-convex airfoil using Shock-expansion theory

The lower surface is then treated in the same way, i.e., we first establish the oblique shock wave angle at the leading edge for the lower surface, i.e.,

$$\begin{aligned}&M_{\infty } = 1.4 \text {~and~} \theta _{\text {nl}} \approx -6.84^{\circ } \quad \xrightarrow {OS} \quad \beta _\text {l} \approx 56.2^{\circ } \\&M_{\infty } \sin \beta \approx 1.16 \quad \xrightarrow {NS} \quad M_{\text {nl}_{\text {nose}}} \approx 0.866 \text {~and~} p_{l}/p_{\infty } \approx 1.4124 \\&M_{\text {l}_{\text {nose}}} = M_{\text {nl}_{\text {nose}}} / \sin \left( \beta - \theta \right) \approx 1.141 \end{aligned}$$

The Prandtl-Meyer angle immediately downstream of the leading edge on the lower surface corresponds to $M_{\text {l}_{\text {nose}}} =1.141$. We can use Eq. (4.35) to get:

$$\theta _{\text {l}_{\text {nose}}} \approx 2.185^{\circ }$$

Now, from the equation of the slope of the lower surface, $\frac{\text {d}y_\text {l}}{\text {d}x} = -0.12 + 0.24 x$, we calculate the surface angle at increments of 10 % chord. This procedure is easily implemented in a spreadsheet program (e.g., Excel). Table 4.1 shows the spreadsheet results pertaining to the geometry, Prandtl-Meyer angle and Mach numbers on the upper and lower surfaces of the airfoil. Tau is the thickness ratio for the upper and lower surfaces, 0.02 and 0.03 respectively.

From local Mach number calculations, we determine the static pressure and thus static pressure coefficient on the airfoil using isentropic relation (Eq. 4.45). The increments of force and pitching moment are also calculated using Eqs. (4.46)–(4.48). These calculations are performed using a spreadsheet program (see Table 4.2).

Fig. 4.27

Application of shock-expansion theory to a 5 % thick bi-convex airfoil with negative camber. a The 5 % thick parabolic bi-convex airfoil with negative camber (2 % upper thickness and 3 % lower thickness). b Pressure distribution on the upper and lower surfaces

The normal force coefficient is the sum of $\text {d}c_n$ column (in Table 4.2): $c_n \approx 0.00233$. The axial force coefficient is the sum of $\text {d}c_{a,\, u}$ and $\text {d}c_{a,\, l}$ columns (in Table 4.2): $c_a \approx 0.01453$. The pitching moment coefficient about the leading edge is the sum of $\text {d}c_{m,\, u}$ and $\text {d}c_{m,\, l}$ (in Table 4.2): $c_{m,\, {\textit{LE}}} \approx 0.0132$.

Since the airfoil is at zero angle of attack, the normal and axial force directions coincide with the lift and drag directions respectively. The shape of the airfoil, including the mean camber line is shown in Fig. 4.27a and the pressure distribution is shown in Fig. 4.27b, from the Excel files.

4.4 Axisymmetric and Slender Bodies

Flow around axisymmetric bodies in supersonic flow involves conical shocks and expansion Mach cones. In the absence of angle of attack, a pointed axisymmetric body that supports a conical shock may be analyzed using the conical shock charts in Fig. 4.17. However when the body is in an angle of attack, the axis of the conical shock will no longer coincide with the body axis. Figure 4.28 shows an example of a cone at an angle of attack and zero yaw. The axis of the cone is along x-direction, as shown. The cone cross section is therefore in the y-z plane. We define the meridian angle, $\varphi$ in the y-z plane with $\varPhi =0$ at the top and $180^{\circ }$ at the bottom of the cone, as shown. Therefore the windward side of the cone in positive angle of attack is at $\varPhi =180^{\circ }$ and the leeward side of the cone is at $\varPhi =0^{\circ }$, as indicated. The meridian angle becomes important in asymmetrical flows as it describes the rays along the cone surface where flow attains unique properties.

Fig. 4.28

Cone at an angle of attack (and zero yaw) in a supersonic flow showing an asymmetrical disposition of a conical shock attached to the cone at the vertex. a An axisymmetric cone at an angle of attack. b Meridian angle, $\varPhi$ at the base of the cone

We expect the windward side of the cone to experience higher static pressures and the leeward side of the cone to experience lower static pressures. The same arguments apply to a cone in sideslip or yaw. Therefore, in general we expect the static pressure gradient that is set up (in the meridian plane) around a cone with angle of attack and yaw to create a spiraling flow around the cone, as shown in Fig. 4.29.

Fig. 4.29

Spiraling streamlines on a cone in supersonic flow at angle of attack and yaw

An approximate method of calculation for slender bodies of revolution is based on the method of local cones. Consider a pointed body of revolution at zero angle of attack as shown in Fig. 4.30. The semi-vertex angle, $\delta _0$ is assumed to be less than the maximum turning angle, $\delta _{\text {max}}$, corresponding to a cone at the given flight Mach number, $M_{\infty }$. Therefore we guarantee that the conical shock at the vertex is attached to the axisymmetric body. Any other point on the body, e.g., point A, makes an angle $\delta$ with respect to the axis, or freestream direction. Since $\delta < \delta _0$, we expect the static pressure at point A to be less than the static pressure at the vertex. The method of local cones suggests that the reduced pressure at point A is due to a smaller angle cone that is tangent to the body at point A had it been placed in freestream Mach number of $M_{\infty }$. This method then predicts a pressure coefficient of zero at point B on the shoulder of the axisymmetric body where the angle is zero with respect to the body axis. In the aft portion of the body, i.e., past point B, the method of local cones does not apply. The aft region is known as the shadow zone.

Fig. 4.30

A (pointed) slender body of revolution in supersonic flow at zero angle of attack. a Slender body of revolution at zero angle of attack. b Local cone (tangent) to point A has semi-vertex angle, $\delta$

Example 4.7

Consider a slender body of revolution at zero angle of attack in supersonic flow with $M_{\infty }=2.0$. The semi-vertex angle is $\delta _0=30^{\circ }$. Calculate the pressure coefficient at the vertex and estimate the pressure coefficient at point A, where the body angle is $\delta =10^{\circ }$ using the method of local cones.

 

Solution:

From conical shock charts (Fig. 4.17), for Mach 2.0 and a cone of $30^{\circ }$ semi-vertex angle, we get the conical shock angle at the vertex to be $\sigma _0 \approx 48^{\circ }$. The normal component of flow to the shock is therefore:

$$\begin{aligned} M_{\infty } \sin \sigma _0 \approx 2 \sin \left( 48^{\circ } \right) = 1.486 \end{aligned}$$

Therefore the static pressure ratio at the nose is calculated from

$$\begin{aligned} \frac{p_N}{p_{\infty }} = 1 + \frac{2 \gamma }{\gamma +1} \left( M_{\infty }^2 \sin ^2 \sigma _0 - 1 \right) \end{aligned}$$

To be $p_N/p_{\infty } = 2.4106$. Assuming the pressure at point A corresponds to the cone pressure of $10^{\circ }$ semi-vertex angle in Mach 2 flow, we use Fig. 4.17b (the conical shock charts) to get the ratio of the cone surface static pressure to the freestream stagnation pressure (from Fig. 4.17b) to be: $p_c/p_{t \infty } \approx 0.17$. For Mach 2, $p_{t \infty }/p_{\infty } = 7.824$ and therefore

$$\begin{aligned} \frac{p_A}{p_{\infty }} = \frac{p_A}{p_{t \infty }} \frac{p_{t \infty }}{p_{\infty }} \approx \left( 0.17\right) \left( 7.824\right) \approx 1.330 \end{aligned}$$

We note the static pressure at point A is significantly less than the pressure at the nose, as expected. The static pressure at point A is ${\sim }55\,\%$ of the static pressure at the nose.

Fig. 4.31

Definition sketch used in the method of local cones for a slender body in angle of attack

The same approximation applies to a slender axisymmetric body in angle of attack, $\alpha$. For a general axisymmetric body at an angle of attack, we define an equivalent cone angle, $\delta _c$ that is tangent to point A and the freestream direction is the axis of the equivalent cone, as shown in Fig. 4.31. The body angle at A is $\delta$, the nose angle is $\delta _0$ and the equivalent cone angle is $\delta _c$. The meridian angle is also shown in Fig. 4.31, which affects the equivalent cone angle according to (4.49).

The relationship between the angles $\delta _c$, $\delta$, $\varphi$ and $\alpha$ (in radian) is:

$$\begin{aligned} \delta _c = \delta - \alpha \cos \varPhi + \frac{\alpha ^2}{2} \cot \delta \cdot \sin ^2 \varPhi \end{aligned}$$

(4.49)

Example 4.8

Consider a slender axisymmetric body, similar to the body shown in Fig. 4.31, with the semi-vertex angle, $\delta _0=20^{\circ }$, in a supersonic flow with $M_{\infty }=2$ and at 10 degrees angle of attack, i.e., $\alpha =10^{\circ }$. Apply the method of local cones to point A on the body where the local body angle is $10^{\circ }$; and A is on the windward side of the body with $\varPhi =180^{\circ }$. Estimate the pressure coefficient at A using the method of local cones.

Solution:

The nose angle with respect to the flow direction on the windward side is the sum of $20^{\circ }$ for the nose and $10^{\circ }$ for the angle of attack (this may also be deduced from Eq. (4.49)). Therefore, the conical shock angle corresponding to $M_{\infty }=2$ and equivalent cone angle of $30^{\circ }$ is (the same as Example 4.7): $\sigma _0 \approx 48^{\circ }$ on the windward side. The rest of the calculations immediately downstream of the nose on the windward side follow the Example 4.7, namely

$$M_{\infty } \sin \sigma _0 \approx 2 \sin \left( 48^{\circ } \right) = 1.486$$

and the static pressure ratio at the nose (on the windward side) is calculated to be:

$$\left. \frac{p_N}{p_{\infty }} \right| _{\varphi =180^{\circ }} = 1 + \frac{2 \gamma }{\gamma + 1} \left( M_{\infty }^2 \sin ^2 \sigma _0 - 1 \right) \approx 2.411 $$

To get the equivalent cone angle for point A, we substitute for the body angle at A, namely $\delta = 10^{\circ }$ and the meridian angle, $\varPhi = 180^{\circ }$ as well as the angle of attack, $\alpha = 10^{\circ }$ in Eq. (4.49). We get:

$$\delta _c = \delta - \alpha \cos \varPhi + \frac{\alpha ^2}{2} \cot \delta \cdot \sin ^2 \varPhi \quad \rightarrow \quad \delta _c = 20^{\circ }$$

Therefore the corresponding (equivalent) conical shock produces the ratio of the cone surface pressure to freestream stagnation pressure of $p_c/p_{t \infty } \approx 0.25$ at $M_{\infty }=2$ (from Fig. 4.17b). From isentropic table, we get $p_{t \infty }/p_{\infty } = 7.824$ for Mach 2, and therefore we estimate $p_A/p_{\infty }$ to be:

$$\frac{p_A}{p_{\infty }} = \frac{p_A}{p_{t \infty }} \frac{p_{t \infty }}{p_{\infty }} \approx \left( 0.25\right) \left( 7.824\right) \approx 1.956$$

Fig. 4.32

The effect of angle of attack and body shape on nose wave drag (after Ref. [3])

In Example 4.8, we assumed that the body surface pressure was the same as the equivalent cone surface pressure. However, within the framework of method of local cones; there are two hypotheses that are advanced on the flow conditions, i.e., pressure and velocity, at point A as compared to an equivalent cone that is tangent to A, namely:  

Hypothesis 1::

pressure at A is identical as on the equivalent cone with $\delta _c$, good for high velocities

Hypothesis 2::

velocity at A is identical as on the equivalent cone, which is better for low velocities

 

We will examine these hypotheses more closely in the next section (Examples and Applications). In addition, the base pressure is assumed to be equal to the freestream pressure, i.e., $p_b=p_{\infty }$, which is a reasonably accurate assumption for most applications of truncated bodies in supersonic flow. Once the pressure distribution on the slender axisymmetric body is obtained [using (4.49) for an equivalent cone] and further assuming that the base pressure is equal to the ambient pressure, we may approximate the aerodynamic forces and moments on the body, namely, lift, wave drag, side force and three moment coefficients for the pointed slender body in supersonic flow. The effect of angle of attack and body shape (i.e., nose bluntness) on the nose wave drag coefficient is shown through experimental data in Fig. 4.32 (from Krasnov 1970). Body number 1 is a flat-head cylinder, whereas the bodies designated as number 2 and 3 represent a streamlined nose of 1.5 and 2.5 (nose) fineness ratios, respectively. Note that the lower-end range of freestream Mach numbers dips into the transonic zone. Here, the effect of angle of attack is seen as an upward shift in $C_D$.

4.5 Examples and Applications

4.5.1 The Shape and Geometric Parameters of an (Axisymmetric) Ogive Nose

Consider a pointed axisymmetric body of nose length, $l_N$, and a base diameter of $d$. Tangent ogive describes such a pointed body where the body contour, in the meridian plane, is a circular arc with zero slope at the base. Figure 4.33 shows a definition sketch of a tangent ogive in the meridian plane.

Fig. 4.33

An axisymmetric tangent ogive in the $\left( r, x\right)$ plane

We define the fineness ratio for the nose as the ratio of nose length-to base diameter, following Chap. 3 notation:

$$\begin{aligned} F_N \equiv \frac{l_N}{d} \end{aligned}$$

(4.50)

The non-dimensional coordinates describing the body are:

$$\begin{aligned} \overline{r}&\equiv \frac{r}{d/2} \end{aligned}$$

(4.51a)

$$\begin{aligned} \overline{x}&\equiv \frac{x}{l_N} \end{aligned}$$

(4.51b)

$$\begin{aligned} \overline{R}&\equiv \frac{R}{d} \end{aligned}$$

(4.51c)

where $R$ is the radius of curvature of the body (i.e., a constant for a tangent ogive). In terms of the non-dimensional coordinates, the equation for the (circular arc) ogive is described by:

$$\begin{aligned} \overline{r} = 1 - 2 \overline{R} \left\{ 1- \left[ 1- \frac{F_N^2}{\overline{R}^2} \left( 1- \overline{x} \right) ^2 \right] ^{1/2} \right\} \end{aligned}$$

(4.52)

The nose fineness ratio may be related to the non-dimensional radius of curvature via:

$$\begin{aligned} F_N = \sqrt{\overline{R} - 1/4} \end{aligned}$$

(4.53)

The local body slope is obtained by differentiating $r \left( x \right)$ with respect to $x$, i.e.,

$$\begin{aligned} \tan \delta = \frac{\text {d}r}{\text {d}x} = \frac{1}{2 F_N} \frac{\text {d}\overline{r}}{\text {d}\overline{x}} = \frac{F_N}{\overline{R}} \left( 1-\overline{x} \right) \left[ 1 - \frac{F_N^2}{\overline{R}^2} \left( 1-\overline{x} \right) ^2 \right] ^{-1/2} \end{aligned}$$

(4.54)

The nose angle $\delta _0$ is related to the nose fineness ratio via:

$$\begin{aligned} \sin \delta _0 = \frac{F_N}{\overline{R}} = \frac{F_N}{F_N^2 + 1/4} \end{aligned}$$

(4.55)

Based on the method of local cones described in Sect. 4.4, we propose to examine the two hypotheses as they apply to a tangent ogive, namely:

• 1. the pressure on the body is equal to the pressure on the equivalent cone, of the same local angle $\delta$, and the stagnation pressure corresponding to $\delta _0$ cone (at the nose)

• 2. the velocity on the body is equal to the velocity on an equivalent cone, of the same local angle, $\delta$ and the stagnation pressure corresponding to a cone of $\delta$, semi-vertex angle (i.e., for higher total pressure recovery than the first hypothesis)

The first hypothesis assumes the same pressure results in a pressure coefficient, $C_{pp}$ and the second method assumes the same surface velocity produces $C_{pv}$. These pressure coefficients are related to each other via [3]:

$$\begin{aligned} C_{pv} = C_{pp} \overline{\nu } + \frac{2 \left( \overline{\nu } -1 \right) }{\gamma M_{\infty }^2} \end{aligned}$$

(4.56)

where,

$$\begin{aligned} \overline{\nu }&\equiv \frac{\nu _0}{\nu } \end{aligned}$$

(4.57)

$$\begin{aligned} \nu _0&\equiv \frac{\left. p_t \right| _{\delta _{0}}}{p_{t \infty }}; \quad \text {Stagnation pressure ratio on a cone with the nose angle} ~ \delta _0 \end{aligned}$$

(4.58)

$$\begin{aligned} \nu&\equiv \frac{\left. p_t \right| _{\delta }}{p_{t \infty }}; \quad \text {Stagnation pressure ratio on a cone with the nose angle} ~ \delta \end{aligned}$$

(4.59)

Fig. 4.34

Pressure coefficient on a tangent ogive based on competing hypotheses [3]

The second hypothesis, i.e., $C_{pv}$ proves to be a more accurate estimate of the local pressure coefficient than the $C_{pp}$. The $C_{pv}$ leads to a negative pressure coefficient on the aft sections of the ogive, i.e., near the shoulder and the shadow region that is borne by both the Method of Characteristics as well as the experimental data. The graphs of $C_{pp}$ and $C_{pv}$ on tangent ogives at Mach numbers 2 and 5 are shown in Figs. 4.34 and 4.35 respectively (from Ref. [3]). Note that the pressure coefficient $C_{pv}$ near the shoulder region of the tangent ogive where $\delta \approx 0$, dips into the negative territory for tan $\delta _0>0.35$, or semi-apex angle of ${>}19.3^{\circ }$.

There is a semi-empirical correlation for the wave drag coefficient of a tangent ogive in supersonic flow that is given in Ref. [3], which is:

$$\begin{aligned} C_D = C_{pc} \left[ 1 - \frac{196 F_N^2 - 16}{14 \left( M_{\infty } + 18 \right) F_N^2} \right] \end{aligned}$$

(4.60)

This is valid for: $1.5 \le M_{\infty } \le 3.5$ and $10^{\circ } \le \delta _0 \le 45^{\circ }$.

In Eq. (4.60), $C_D$ is the pressure or wave drag coefficient of the tangent ogive (nose) and $C_{pc}$ is the pressure coefficient on a cone of the (semi-) apex angle $\delta _0$.

Fig. 4.35

Pressure coefficient on a tangent ogive based on competing hypotheses [3]

Example 4.9

A tangent ogive has a nose fineness ratio of $F_N=2.825$. Calculate:

• (a) the non-dimensional radius of curvature, $\overline{R}$ for the nose

• (b) the semi-vertex angle, $\delta _0$ in degrees

• (c) the pressure coefficients $C_{pp}$ and $C_{pv}$ at $M_{\infty }=2$ where the body angle is $11.2^{\circ }$

• (d) the wave drag coefficient of this tangent ogive (nose) at $M_{\infty }=2$

Solution:

From Eq. (4.53), we establish the non-dimensional radius of curvature of the nose to be:

$$\overline{R} = F_N^2 + 1/4 = 8.23$$

From Eq. (4.55) we calculate the semi-vertex angle at the nose:

$$ \delta _0 = \sin ^{-1} \left( F_N / \overline{R} \right) \approx 20^{\circ }$$

For the body angle of $11.2^{\circ }$ and the semi-apex angle of $20^{\circ }$ we calculate the tangents to be: $\tan \delta \approx 0.20$ and $\tan \delta _0 \approx 0.365$. We use these values in Fig. 4.34 to get the two pressure coefficients: $C_{pp} \approx 0.13$ and $C_{pv} \approx 0.12$ Finally, the pressure drag coefficient of the tangent ogive is estimated from the semi-empirical correlation (4.60). But first, we need the pressure coefficient on the cone of the semi-apex angle $20^{\circ }$, in Mach-2 flow (from Fig. 4.17f, noting that the cone pressure coefficient and the wave drag coefficients are the same since the base pressure is assumed to be equal to $p_{\infty }$), $C_{pc} \approx 0.325$

$$C_D = C_{pc} \left[ 1 - \frac{196 F_N^2 - 16}{14 \left( M_{\infty } + 18 \right) F_N^2} \right] \approx 0.0998 \approx 0.1$$

4.5.2 Extension to Transonic Speeds

Fig. 4.36

Wave drag coefficient, $C_{D_p}$, at sonic flow (after Ref. [3]). a Cone. b Ogive

A slender body in a slightly supersonic flow, i.e., for $M_{\infty }= 1+\varepsilon$ creates a normal shock that stands ahead of the body. The flow downstream of the normal shock is subsonic with $M \approx 1-\varepsilon$. This is the result of the linear behavior of normal shocks in slightly supersonic flows, as we demonstrated earlier in this chapter. Therefore the local Mach number on a slender body, $M$, does not change appreciably when the freestream flow condition is near sonic. This argument, first forwarded by Liepmann and Bryson [4] can be expressed mathematically as:

$$\begin{aligned} \left. \frac{\partial M}{\partial M_{\infty }} \right| _{M_{\infty } = 1} = 0 \end{aligned}$$

(4.61)

Fig. 4.37

Experimental data on wave drag coefficient in transonic flow (after Ref. [3]). Note that $\lambda _N$ is the nose fineness ratio, $F_N$ in our notation. a Cone. b Ogive

This is the principle of stationary local Mach numbers near sonic freestream flow conditions on slender bodies. This condition can be used as an extrapolation tool to extend the results of force and moment coefficients from subsonic and supersonic sides through the transonic region. We will apply this principle to a slender cone and an ogive in transonic flow. First, the experimental results for the wave drag coefficient for cones and ogives are shown in Fig. 4.36a, b respectively at sonic freestream Mach number [3]. Note that the cone semi-apex angle of 90$^\circ$ is the case of a flat-head cylinder and the ogive with the semi-vertex angle of 90$^\circ$ is the case of hemisphere. Applying the principle of stationary Mach numbers to slender axisymmetric bodies, e.g., cone or ogive, where transonic similarity applies, produces:

$$\begin{aligned} C_D = \frac{\gamma M_{\infty }^2 + 1}{\left( \gamma +1\right) M_{\infty }^2} \left( C_D \right) _{M_{\infty }=1} + \frac{2 \left( M_{\infty }^2 - 1 \right) }{\left( \gamma +1 \right) M_{\infty }^2} \end{aligned}$$

(4.62)

This expression is valid for axisymmetric slender bodies that include both cones and ogives. Therefore the experimental data on $C_D$-cone at $M_{\infty }=1$, as shown in Fig. 4.36a substituted in Eq. (4.62) allows for the extension of wave drag in transonic range, i.e., from subsonic to supersonic. In particular the slope of the drag coefficient near sonic flow is determined. The proof of the match between the theory and the experimental results on a cone in the transonic range is shown in Fig. 4.37a [3]. The rising dashed lines through sonic Mach number are produced from the theoretical model, i.e., Eq. (4.62). Similarly, the wave drag coefficient on an ogive in the vicinity of sonic flow may be explored using Eq. (4.62) in conjunction with the ogive wave drag coefficient for sonic flow, i.e., Fig. 4.36b. Comparison with experimental results shown in Fig. 4.37b is in good agreement with the theoretical model.

4.6 Summary

Shock-expansion theory is a powerful tool in analyzing aerodynamic forces and moments on pointed bodies in steady supersonic flow where the oblique shock(s) at the leading edge are attached. This method produces exact results within the confines of inviscid flow theory. The method breaks down however when the shocks become detached where the flow at the leading edge exceeds the maximum turning angle supported by straight oblique shocks. The two-dimensional flow problems are easily analyzed using plane oblique shocks and expansion waves. The three-dimensional problems are analyzed using conical shocks. The 3-D expansion through conical Mach waves is best accomplished using the method of characteristics, i.e., the subject of Chap. 5. Approximate method of local cones is introduced to estimate the effect of 3-D expansion, using conical shocks associated with a smaller nose angle. The problem of asymmetrical flows is also analyzed through the method of local cones. The extension of aerodynamic forces through sonic velocity is accomplished through the principle of stationary local Mach numbers. The wave drag coefficient on cones and ogives in transonic flow are in good agreement with the theoretical predictions.

Problems

4.1

A two-dimensional projectile with a sharp nose is exposed to a Mach 3 flow, as shown. Assuming the base pressure is $p_{\infty }$, calculate:

• (a) $p_1/p_{\infty }$

• (b) $p_2/p_{\infty }$

• (c) wave drag coefficient referenced to chord, $C_d \equiv D^{'}/q_{\infty } \cdot c$.

 

4.2

Calculate the stagnation pressure measured by a Pitot tube on an inclined ramp in a supersonic flow, as shown.

 

4.3

A symmetrical half-diamond airfoil has a leading-edge angle of $5^{\circ }$. This airfoil is set at $5^{\circ }$ angle of attack, as shown. The airfoil is placed in a windtunnel with test section (T.S.) Mach number $M_{T.S.}=2.0$, $p_{t,T.S.}=100$ kPa and $T_{t,T.S.}=25\,^{\circ }\text {C}$. Assuming $\gamma =1.4$ and $c_p=1.004$ kJ/kg. K, use shock-expansion theory to calculate:

• (a) Static pressure, $p_1$ (in kPa)

• (b) Static pressure, $p_2$ (in kPa)

• (c) Static pressure, $p_3$ (in kPa).

 

4.4

A supersonic flow expands around a sharp corner, as shown. Calculate the following parameters:

• (a) Downstream Mach number, $M_2$

• (b) Flow area ratio, $A_2/A_1$

• (c) The angle of Prandtl-Meyer fan envelope, $\delta$.

 

4.5

A symmetrical half-diamond airfoil has a nose angle $\theta _{\text {nose}}=30^{\circ }$ and is exposed to a supersonic flow. A Pitot tube is installed on each of the three surfaces, as shown. Calculate the Pitot tube readings on the airfoil surfaces at zero angle of attack.

 

4.6

Calculate the wave drag coefficient of a 2-D sharp-nosed projectile, as shown, assuming the base static pressure is equal to ambient, i.e., $p_{\text {base}}=p_{\infty }$. The 2-D wave drag coefficient here is defined based on $b$, i.e., $C_d \equiv D^{'}/q_{\infty }\cdot b$. Assume the angle of attack is zero.

 

4.7

A Prandtl-Meyer centered expansion wave is visualized with the wave angles as shown. Calculate:

• (a) the flow turning angle, i.e., wall angle, $\theta _w$

• (b) velocity ratio across the expansion waves $V_2/V_1$.

 

4.8

A half-diamond airfoil with a nose angle of $15^{\circ }$ and a thickness-to-chord ratio of 10 % is in a supersonic flow, as shown. Calculate:

• (a) the chordwise location of the maximum thickness point (in % c)

• (b) the trailing-edge angle

• (c) the non-dimensional pressure on the three surfaces, $p_1/p_{\infty }$, $p_2/p_{\infty }$, $p_3/p_{\infty }$.

 

4.9

Calculate the 2-D lift and drag coefficients of a two-dimensional wave rider, as shown. Assume the base pressure is equal to the ambient, $p_4=p_1$. Shock wave angle is $\beta _1=45^{\circ }$. The upper surface is aligned with the flight direction (surface 2). Surface 3 is the lower surface and surface 4 is the base.

 

4.10

A half-diamond unsymmetrical airfoil is shown at $5^{\circ }$ angle-of-attack. Calculate:

• (a) thickness-to-chord ratio, $t/c$

• (b) $p_1/p_{\infty }$, $p_2/p_{\infty }$ and $p_3/p_{\infty }$

• (c) normal force coefficient, $C_n$

• (d) axial force coefficient, $C_a$.

 

4.11

A symmetrical diamond airfoil is shown. Free-stream Mach number is $M_1=5.0$. The angle of attack is $5^{\circ }$. Use shock-expansion theory to calculate:

• (a) thickness-to-chord ratio, $t/c$

• (b) pressure coefficients $C_{p_2}$, $C_{p_3}$, $C_{p_4}$, $C_{p_5}$

• (c) axial force coefficient, $C_a$.

 

4.12

Consider a 10 % thick bi-convex airfoil of parabolic arc upper and lower profiles with negative camber. The upper surface contributes 3 % to thickness at 50 % chord and the lower surface contributes 7 % thickness to the airfoil at mid chord. First, write the equations $y_u \left( x \right)$ and $y_l \left( x \right)$ for the upper and lower profiles. Next, use shock-expansion theory to calculate:

• (a) the pressure coefficient at the leading edge, and increments of 10 % chord until trailing edge

• (b) the lift coefficient

• (c) the (wave) drag coefficient

• (d) the pitching moment coefficient about the leading edge.

 

4.13

A hexagonal airfoil is a suitable shape for supersonic flow. Assuming the airfoil shown is 10 % thick and it is at zero angle of attack in Mach 2.4 flow, use shock-expansion theory to calculate:

• (a) pressure coefficients on all six panels

• (b) the airfoil wave drag coefficient.

 

4.14

Identify the waves and draw the wave pattern around the two-dimensional symmetrical body at points A, B and C, as shown. For expansion waves, calculate the wave angle of the head and tail waves and with the oblique shocks calculate the wave angle. Note that the body is symmetrical and is set at zero angle-of-attack.

 

4.15

A supersonic flow approaches an expansion corner, as shown. The static pressure ratio $p_2/p_1$ is 0.25. Calculate:

• (a) Mach number on the expansion ramp, $M_2$

• (b) Wall angle, $\theta _w$ (in degrees).

 

4.16

Two compression ramps in supersonic flow create two plane oblique shocks, as shown. Calculate:

• (a) shock wave angle, $\beta _1$ (in degrees)

• (b) ramp angle, $\theta _1$ (in degrees)

• (c) shock wave angle, $\beta _2$ (in degrees)

• (d) estimate the second ramp angle, $\theta _2$ (in degrees).

 

4.17

A sharp hexagonal airfoil is placed in a supersonic flow. The Mach numbers on its upper three surfaces are as indicated. Calculate:

• (a) The semi-nose angle, $\theta _{\text {LE}}$

• (b) Static pressure ratio, $p_2/p_1$

• (c) Static pressure ratio, $p_3/p_1$

• (d) The semi trailing-edge angle, $\theta _{\text {TE}}$.

 

4.18

Consider a symmetrical, bi-convex parabolic profile airfoil that is of 10 % thickness, as shown. The airfoil is in Mach 1.8 flow at zero angle of attack. Using shock-expansion theory, calculate:

• (a) Pressure distribution on the airfoil at $x/c =$0, 0.2, 0.4, 0.6, 0.8 and 1.0

• (b) Estimate the wave drag coefficient by numerical integration of the pressures in part (a).

 

4.19

Apply shock-expansion theory to the diamond airfoil in supersonic flow (as shown), to calculate:

• (a) upper and lower thickness-to-chord ratios, $t_1/c$ and $t_2/c$

• (b) the pressure coefficients, $C_{p_1}$, $C_{p_2}$, $C_{p_3}$, $C_{p_4}$

• (c) the normal force coefficient, $c_n$

• (d) the axial force coefficient, $c_a$

• (e) lift-to-drag ratio, $L^{'}/D^{'}$

• (f) the pitching moment coefficient about the leading edge, $c_{m, LE}$.

 

4.20

Consider a (two-dimensional) compression ramp in supersonic flow in air $\left( \gamma =1.4 \right)$ followed by a sharp corner that turns the flow back to horizontal, as shown. The expansion waves that are centered at the shoulder intersect the attached oblique shock at the nose and cause wave reflections (not shown). If we ignore the reflected waves at the shock, we can use the shock-expansion theory to calculate the ratio of static pressure downstream of the shoulder, $p_3$, to upstream of the shock, $p_1$.

 

Write a computer program or use the exact shock-expansion formulas in a spreadsheet, to calculate and graph $p_3/p_1$ for a range of Mach numbers, $M_1=1.5, 3, 5, 10$ and a range of ramp angles, $\theta$, from the corresponding wave angles, $\mu$, to near maximum turning angles. Do we recover the flow static pressure $p_1$ when we get to region 3?

4.21

An oblique shock is described by its upstream flow components ($u$ and $w$ are normal and parallel to the shock, respectively) as shown. Calculate:

• (a) Upstream speed of sound, $a_1$

• (b) Characteristic Mach number, $M_2^*$.

 

4.22

A flat plate is in Mach 1.2 flow at $2^{\circ }$ angle of attack. Use shock-expansion theory to predict its lift-to-wave drag ratio. The ratio of specific heats is $\gamma =1.4$.

4.23

A bi-convex airfoil has a nose angle, $\theta _{\text {nose}} = 10^{\circ }$. For an angle of attack of $5^{\circ }$ and freestream Mach number, $M_{\infty } = 2.4$, calculate:

• (a) the upper surface static pressure at the nose, $p_{\text {nose}}/p_{\infty }$

• (b) the static pressure at the mid-chord point, $p_{\text {A}}/p_{\infty }$.

 

4.24

Consider a symmetrical bi-convex airfoil with a half-nose angle of $10^{\circ }$. It is in a supersonic flow with $M_{\infty } =2.8$ and has an angle-of-attack of $2^{\circ }$. Use shock-expansion theory to calculate the static pressure difference between the lower and upper surfaces at 50 % chord (see figure), $\varDelta p/ p_{\infty }$.

 

4.25

Consider a slender body of revolution at zero angle of attack in supersonic flow with $M_{\infty }=3.0$. The semi-vertex angle is $\delta _0=40^{\circ }$. Calculate the pressure coefficient at the vertex and estimate the pressure coefficient at point A, where the body angle is $\delta =20^{\circ }$ using the method of local cones.

 

4.26

Consider a slender axisymmetric body, as shown, with the semi-vertex angle, $\delta _0=30^{\circ }$, in a supersonic flow with $M_{\infty }=2.4$ and at 5 degrees angle of attack, i.e., $\alpha =5^{\circ }$. Apply the method of local cones to point A on the body where the local body angle is $10^{\circ }$; and A is on the meridian plane with $\varPhi =120^{\circ }$. Estimate the pressure coefficient at A.

 

4.27

A tangent ogive has a nose fineness ratio of $F_N = 2.5$. Calculate:

• (a) the non-dimensional radius of curvature, $\overline{R}$ for the nose

• (b) the semi-vertex angle, $\delta _0$ in degrees

• (c) the pressure coefficients $C_{pp}$ and $C_{pv}$ at $M_{\infty }=5$ where the body angle is $10^{\circ }$

• (d) the wave drag coefficient of this tangent ogive (nose) at $M_{\infty }=5$.

4.28

Use the wave drag coefficient for a cone of semi-vertex angle of $45^{\circ }$ at $M_{\infty }=1.0$, from Fig. 4.36a to create the pressure drag behavior of the cone in the transonic range from $M_{\infty }=0.8$ to $M_{\infty }=1.5$. Graph your results and compare it to the experimental data in Fig. 4.37a.

4.29

Consider four cones with semi-vertex angles of $0^{\circ }$, $15^{\circ }$, $30^{\circ }$ and $40^{\circ }$, as shown. First for a flow Mach number of $M_{\infty }=3.0$, calculate the cones’ wave drag coefficients and graph them on a chart, (as shown). You may use conical shock charts of Shapiro (Fig. 4.17) for $\gamma =1.4$.

 

Finally, if we install a Pitot tube on the $30^{\circ }$ cone in the Mach 3 flow, calculate the stagnation pressure that the Pitot tube reads, as a fraction of flight stagnation pressure.

4.30

Use the wave drag coefficient for an ogive of fineness ratio 3.0 at $M_{\infty }=1.0$, from Fig. 4.36b. You first need to calculate the semi-vertex angle for a tangent ogive that has a fineness ratio of 3.0 before you can use Fig. 4.36b. Next, calculate the pressure drag behavior of the ogive in the transonic range from $M_{\infty }=1.0$ to $M_{\infty }=1.4$. Graph your results and compare it to the experimental data in Fig. 4.37b for the ogive with fineness ratio 3.0.

4.31

A cone of $30^{\circ }$ semi-vertex angle is in supersonic flow at $M_{\infty }=2.0$ and zero angle of attack. Use conical shock charts to find:

• (a) the conical shock angle, $\sigma$, in degrees

• (b) the Mach number on the surface of the cone, $M_c$

• (c) the pressure drag coefficient of the cone, $C_{D_p}$.

 

4.32

Consider a $30^{\circ }$ (2-D) ramp in a Mach 3 flow of air with $\gamma =1.4$. Also, there is a cone of semi-vertex angle of $30^{\circ }$ in the same supersonic flow condition. Calculate and compare:

• (a) The wave angles on the 2-D ramp and the cone

• (b) The surface pressure on the ramp and the cone

• (c) Explain the 3-D relieving effect as evidenced by this problem.

4.33

A two-dimensional wedge is in supersonic flight, as shown. The freestream velocity is parallel to the upper surface. Assuming that the base pressure, $p_3 = p_{\infty }$ and $\gamma =1.4$, calculate:

• (a) Pressure coefficients, $C_{p1}$ and $C_{p2}$

• (b) Wave drag coefficient, $c_d$, referenced to base area, i.e., $c_d=D^{'}/q_{\infty }b \left( 1 \right)$

• (c) Lift coefficient, $c_l$, referenced to planform area, i.e., $c_l=L^{'}/q_{\infty }c \left( 1 \right)$.

 

4.34

Supersonic flow approaches a sharp expansion corner as shown. Calculate:

• (a) Speed of gas upstream of the corner, $V_1$

• (b) Mach number downstream of the corner, $M_2$

• (c) Speed of sound downstream of the corner, $a_2$.

 

4.35

An oblique shock intersects a wall at a corner, as shown (not to scale). What kind of wave(s) will form at the corner to accommodate the turning of the flow? Calculate the Mach number in region 3.

 

4.36

A flat plate is placed in a supersonic flow, as shown. Use shock-expansion theory to calculate:

• (a) lift-to-(wave) drag ratio, $L^{'}/D^{'}$

• (b) the angle of the slipstream, $\varPhi$ (in degrees).

 

4.37

In the method of local cones, we introduced a function for the cone semi-apex angle, $\delta _c$ that was related to the angle of attack (in radians), local body angle and the meridian angle following:

$$\begin{aligned} \delta _c = \delta - \alpha \cos \varPhi + \frac{\alpha ^2}{2} \cot \delta \cdot \sin ^2 \varPhi \end{aligned}$$

We wish to graph $\delta _c$ as a function of $\overline{x}$ for a tangent ogive, for $\alpha =10^{\circ }$ and the local ogive angle $\delta$ varying between the nose angle, $\delta _0 =45^{\circ }$ to the shoulder value, where $\delta =0$ around the circumference of the ogive, i.e., with $\varPhi$ varying between $0^{\circ }$ and $360^{\circ }$. First make a table for $\delta$ versus $\overline{x}$ for a tangent ogive in increments of $\varDelta \overline{x} = 0.1$ along the axis of the ogive, i.e.,

$\overline{x}$	0	0.1	0.2	0.3	0.4	0.5	0.6	0.7	0.8	0.9	1
$\delta ~(^{\circ })$	45										0

Then, calculate (e.g., using a spreadsheet) and graph $\delta _c$ versus $\varPhi$ at each $\overline{x}$. Note that there is a singularity at the shoulder where $\delta =0$ [due to $\cot \left( \delta \right)$ term], that corresponds to $\overline{x} = 1$. To avoid the singularity at the shoulder; we limit $\overline{x}$ to 0.99. Identify the shadow region on the ogive at angle of attack, where $\delta _c$ is negative.

4.38

Consider a flat plate at angle of attack in supersonic flow. Use shock-expansion theory to calculate and graph the drag polar (i.e., graph of $C_l$ vs. $C_d$) for the airfoil in the range of $\alpha$ from 0 to $12^{\circ }$ for constant freestream Mach numbers of $M_{\infty }=2$, 3 and 5.

4.39

A symmetrical diamond airfoil has a semi-vertex angle of $20^{\circ }$ in a Mach 2 flow in air ($\gamma =1.4$).

• (a) At what angle of attack, will the leading-edge shock detach?

• (b) At what angle of attack will the flow downstream of the leading-edge shock be sonic?.

4.40

Consider a $15^{\circ }$ (2-D) ramp in a Mach-3 flow of air with $\gamma =1.4$. Also, there is a cone of semi-vertex angle of $15^{\circ }$ in the same supersonic flow condition. Calculate and compare:

• (a) surface Mach number on the ramp and the cone

• (b) the wave angles on the 2-D ramp and the cone

• (c) the surface pressure coefficient on the ramp and the cone

• (d) wave drag coefficient, referenced to base area for the ramp and the cone.

4.41

An over-expanded jet emerges from a symmetrical 2-D nozzle in a quiescent ambient at pressure $p_a$, as shown in Fig. 4.38. Due to symmetry, the centerline is a streamline, thus it behaves like a solid wall. The oblique shocks that are formed (at points A and B on the nozzle lip) to satisfy the pressure equilibrium condition on the free boundary, i.e., $p_2= p_a$, are reflected from the centerline (at point C) as oblique shocks. The reflected oblique shocks impinge on the free boundary (at points D and E) and are reflected as centered expansion waves (DF and DG) to maintain the free boundary at constant pressure, i.e., $p_4 = p_a$.

Fig. 4.38

Figure belonging to Problem 4.41

Calculate:

• (a) The nozzle pressure ratio, $p_{t1}/p_a$

• (b) The angle of the free boundary AD with respect to the x-axis

• (c) Mach number in field number 3, $M_3$

• (d) The ratio of static pressure in field number 3 to ambient pressure, $p_3/p_a$

• (e) The angle of the free boundary DH with respect to the x-axis.

4.42

An airfoil is in transonic flow. The freestream Mach number is $M_{\infty } = 0.90$ and the local Mach number at the peak suction point is $M_{\text {max}}=1.24$. Calculate:

• (a) the critical pressure coefficient, $C_{p,~\text {crit}}$

• (b) the pressure coefficient at peak suction point, $C_{p,~\text {min}}$

• (c) the pressure coefficient at the stagnation point, $C_{p,~\text {stag}}$.