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Next: Vectors and Vector Fields Up: Two-Dimensional Compressible Inviscid Flow Previous: Flat Lifting Wings


  1. Show that Equation (15.12) can be written in the form

    $\displaystyle X^{\,3} + b\,X^{\,2}+ c\,X+ d = 0,$ (15.207)


    $\displaystyle X$ $\displaystyle = \sin^2\beta,$    
    $\displaystyle b$ $\displaystyle =-\left(\frac{{\rm Ma}_1^{\,2}+2}{{\rm Ma}_1^{\,2}}+\gamma\,\sin^2\theta\right),$    
    $\displaystyle c$ $\displaystyle = \frac{2\,{\rm Ma}_1^{\,2}+1}{{\rm Ma}_1^{\,4}}+\left[\left(\frac{\gamma+1}{2}\right)^2+\frac{\gamma-1}{{\rm Ma}_1^{\,2}}\right]\sin^2\theta,$    
    $\displaystyle d$ $\displaystyle = -\frac{\cos^2\theta}{{\rm Ma}_1^{\,4}}.$    


    $\displaystyle Q$ $\displaystyle = \frac{1}{3}\,c - \frac{1}{9}\,b^{\,2},$    
    $\displaystyle R$ $\displaystyle = \frac{1}{6}\,(b\,c-3\,d)-\frac{1}{27}\,b^{\,3},$    
    $\displaystyle D$ $\displaystyle = Q^{\,3}+R^{\,2},$    

    Demonstrate that the oblique shock solution only exists for $ D<0$ [i.e., when Equation (15.207) possesses three real roots.] Show that the strong shock solution, $ \beta_s$ , and the weak shock solution, $ \beta_w$ , are given by

    $\displaystyle \beta_s$ $\displaystyle = \tan^{-1}\left(\sqrt{\frac{\chi_s}{1-\chi_s}}\,\right),$    
    $\displaystyle \beta_w$ $\displaystyle = \tan^{-1}\left(\sqrt{\frac{\chi_w}{1-\chi_w}}\,\right),$    


    $\displaystyle \chi_s$ $\displaystyle = -\frac{1}{3}\,b+2\sqrt{-Q}\,\cos\phi,$    
    $\displaystyle \chi_w$ $\displaystyle = -\frac{1}{3}\,b-\sqrt{-Q}\left(\cos\phi-\sqrt{3}\,\sin\phi\right),$    
    $\displaystyle \phi$ $\displaystyle = \frac{1}{3}\left[\tan^{-1}\left(\frac{\sqrt{-D}}{R}\right)+{\mit\Delta}\right].$    

    Here, $ {\mit\Delta}=0$ if $ R\geq 0$ , and $ {\mit\Delta}=\pi$ if $ R<0$ .

  2. Assuming that information propagates with respect to a two-dimensional supersonic flow pattern at the local sound speed, show that, in order for the flow at some point $ P$ to affect the flow at some other point $ Q$ , the latter point must lie between the $ (+)$ and $ (-)$ characteristics that pass through $ P$ .

  3. Show that for a weak oblique shock with $ \beta\simeq \mu$ ,

    $\displaystyle \frac{{\mit\Delta}\rho}{\rho_1}$ $\displaystyle \simeq \frac{1}{\gamma}\,\frac{{\mit\Delta}p}{p_1},$    
    $\displaystyle \frac{{\mit\Delta}T}{T_1}$ $\displaystyle \simeq \left(\frac{\gamma-1}{\gamma}\right)\frac{{\mit\Delta}p}{p_1},$    
    $\displaystyle \frac{{\mit\Delta}{\cal S}}{{\cal R}}$ $\displaystyle \simeq \frac{(\gamma+1)}{12\,\gamma^{\,2}}\left(\frac{{\mit\Delta}p}{p_1}\right)^3,$    

    where $ {\mit\Delta}\rho =\rho_2-\rho_1$ , et cetera, and

    $\displaystyle \frac{{\mit\Delta}p}{p_1}\simeq \left(\frac{\gamma\,{\rm Ma}_1^{\,2}}{\sqrt{{\rm Ma}_1^{\,2}-1}}\right)\theta.

    Here, $ \beta $ is the wave angle, $ \mu$ the Mach angle, $ \theta\ll 1$ the deflection angle, $ \gamma$ the ratio of specific heats, $ {\cal R}$ the specific gas constant, and $ {\rm Ma}_1$ the upstream Mach number. Furthermore, $ p_1$ , $ \rho_1$ , $ T_1$ , and $ {\cal S}_1$ are the upstream pressure, density, temperature, and specific entropy, respectively, whereas $ p_2$ , $ \rho_2$ , $ T_2$ , and $ {\cal S}_2$ are the corresponding downstream quantities. Show, also, that

    $\displaystyle \frac{{\mit\Delta}{\rm Ma}}{{\rm Ma}_1}\simeq -\left[\frac{1+[(\gamma-1)/2]\,{\rm Ma}_1^{\,2}}{\sqrt{{\rm Ma}_1^{\,2}-1}}\right]\theta.

  4. Show that for a weak oblique shock

    $\displaystyle \beta$ $\displaystyle \simeq \mu_1+\epsilon,$    
    $\displaystyle \beta-\theta$ $\displaystyle \simeq \mu_2-\epsilon,$    

    where $ \epsilon\ll 1$ . Here, $ \mu_1$ and $ \mu_2$ are the Mach angles upstream and downstream of the shock front, respectively. Moreover, $ \beta $ is the wave angle, and $ \theta\ll 1$ the deflection angle. Hence, deduce that the shock front subtends the same angle, $ \epsilon$ , with the Mach lines upstream and downstream of it. In other words, the shock position is the ``average'' of the Mach line positions on either side of it. Consider supersonic flow incident on a wedge of small nose angle, with an afterbody, as illustrated in Figure 15.9(a). Assume that the shock front is attached to the apex of the wedge, and that the flow downstream of the shock is supersonic. Use the result just proved to show that the shape of the shock front in the region of attenuation by expansion (i.e., in the region in which the shock front is intersected by Mach lines emanating from the shoulder) is parabolic. [Hint: Use the well-known optical result that a parabolic mirror perfectly focuses a parallel beam of light rays.] (Leipmann & Roshko 1957.)

  5. Show that, to second order in the deflection angle, $ \theta $ , the relative change in pressure across a weak oblique shock is written

    $\displaystyle \frac{{\mit\Delta}p}{p_1} \simeq \left(\frac{\gamma\,{\rm Ma}_1^{...
...\left[(\gamma+1)\,{\rm Ma}_1^{\,4}-4\,({\rm Ma}_1^{\,2}-1)\right]\theta^{\,2},

    where $ {\mit\Delta}p=p_2-p_1$ . Here, $ p_1$ is the upstream pressure, $ p_2$ the downstream pressure, $ {\rm Ma}_1$ the upstream Mach number, and $ \gamma$ the ratio of specific heats. (Leipmann & Roshko 1957.)

    Figure: Flow of Mach number $ {\rm Ma}_1$ over convex corner of deflection angle $ \vert\theta _2\vert$ .
\epsfysize =2.25in

  6. An ideal gas of pressure, density, temperature, and Mach number $ p_1$ , $ \rho_1$ , $ T_1$ , and $ {\rm Ma}_1$ , respectively, flows over a convex corner that turns through an angle $ \vert\theta _2\vert$ , as shown in Figure 15.16. Consider a particular Mach line in the Prandtl-Mayer expansion fan that subtends an angle $ \eta$ with the continuation of the upstream wall, as shown in the figure. Let $ p$ , $ \rho$ , $ T$ , $ {\rm Ma}$ , $ \vert\theta\vert$ , $ \mu$ , and $ \nu$ be the pressure, density, temperature, Mach number, magnitude of the deflection angle, Mach angle, and Prandtl-Mayer function, respectively, on the Mach line in question. Furthermore, let

    $\displaystyle \eta$ $\displaystyle = \nu({\rm Ma}_1)+\frac{\pi}{2}-\left(\frac{\gamma+1}{\gamma-1}\right)^{1/2}z,$    
    $\displaystyle z_1$ $\displaystyle = \tan^{-1}\left[\left(\frac{\gamma-1}{\gamma+1}\right)^{1/2}({\rm Ma}_1^{\,2}-1)^{1/2}\right],$    

    where $ \gamma$ is the ratio of specific heats. Show that, inside the fan,

    $\displaystyle \mu$ $\displaystyle = \cot^{-1}\left[({\rm Ma}_1^{\,2}-1)^{1/2}\,\frac{\tan z}{\tan z_1}\right],$    
    $\displaystyle \vert\theta\vert$ $\displaystyle =\tan^{-1}\left[({\rm Ma}_1^{\,2}-1)^{1/2}\right] -\tan^{-1}\left...
...{\tan z}{\tan z_1}\right] +\left(\frac{\gamma+1}{\gamma-1}\right)^{1/2}(z-z_1),$    
    $\displaystyle {\rm Ma}^{\,2}$ $\displaystyle = 1+ ({\rm Ma}_1^{\,2}-1)\,\frac{\tan^{\,2} z}{\tan^{\,2} z_1},$    
    $\displaystyle \nu$ $\displaystyle = \nu({\rm Ma}_1) +\tan^{-1}\left[({\rm Ma}_1^{\,2}-1)^{1/2}\righ...
...tan^{-1}\left[({\rm Ma}_1^{\,2}-1)^{1/2}\frac{\tan z}{\tan z_1}\right]\nonumber$    
      $\displaystyle \phantom{=}+\left(\frac{\gamma+1}{\gamma-1}\right)^{1/2}(z-z_1),$    

    $\displaystyle p$ $\displaystyle = p_1\left(\frac{\cos z}{\cos z_1}\right)^{2\,\gamma/(\gamma-1)},$    
    $\displaystyle \rho$ $\displaystyle =\rho_1\left(\frac{\cos z}{\cos z_1}\right)^{2/(\gamma-1)},$    
    $\displaystyle T$ $\displaystyle =T_1\left(\frac{\cos z}{\cos z_1}\right)^{2},$    

    where $ z_1\leq z\leq z_2$ . Here, $ z_2$ is defined implicitly by $ \vert\theta\vert(z_2)=\vert\theta_2\vert$ . Demonstrate that the fan extends over the range of angles $ \eta_2\leq \eta\leq \eta_1$ , where

    $\displaystyle \eta_1$ $\displaystyle =\cot^{-1}\left[({\rm Ma}_1^{\,2}-1)^{1/2}\right],$    
    $\displaystyle \eta_2$ $\displaystyle = \eta_1-\left(\frac{\gamma+1}{\gamma-1}\right)^{1/2}(z_2-z_1).$    

    Figure: Reflection of oblique shock by wall. Here, $ {\rm Ma}_1$ , $ {\rm Ma}_2$ , et cetera, are Mach numbers.
\epsfysize =2.25in

    Show that $ \mu$ , $ p$ , $ \rho\rightarrow 0$ , $ {\rm Ma}\rightarrow \infty$ ,

    $\displaystyle \nu\rightarrow\nu_{\rm max} \equiv \frac{\pi}{2}\left[\left(\frac{\gamma+1}{\gamma-1}\right)^{1/2}-1\right],


    $\displaystyle \vert\theta\vert\rightarrow\vert\theta_{\rm max}\vert\equiv \nu_{\rm max}-\nu({\rm Ma}_1),

    in the limit that $ z\rightarrow\pi/2$ . Hence, deduce that if $ \vert\theta_2\vert>\vert\theta_{\rm max}\vert$ then the the fan only extends over the region $ -\vert\theta_{\rm max}\vert\leq \eta\leq \eta_1$ , and the region $ -\vert\theta_2\vert< \eta<-\vert\theta_{\rm max}\vert$ is occupied by a vacuum (i.e., a gas with zero pressure and density).

    Assuming that $ \vert\theta\vert\ll 1$ , show that

    $\displaystyle \vert\theta\vert$ $\displaystyle \simeq \frac{2}{\gamma-1}\left(\frac{\gamma-1}{\gamma+1}\right)^{1/2}\left(\frac{{\rm Ma}_1^{\,2}-1}{{\rm Ma}_1^{\,2}}\right)(z-z_1),$    
    $\displaystyle \eta_1-\eta$ $\displaystyle \simeq \frac{\gamma-1}{2}\left(\frac{{\rm Ma}_1^{\,2}}{{\rm Ma}_1^{\,2}-1}\right)\vert\theta\vert,$    


    $\displaystyle \frac{{\mit\Delta}p}{p_1}$ $\displaystyle \simeq -\left(\frac{\gamma\,{\rm Ma}_1^{\,2}}{\sqrt{{\rm Ma}_1^{\,2}-1}}\right)\vert\theta\vert,$    
    $\displaystyle \frac{{\mit\Delta}\rho}{\rho_1}$ $\displaystyle \simeq \frac{1}{\gamma}\,\frac{{\mit\Delta}p}{p_1},$    
    $\displaystyle \frac{{\mit\Delta}T}{T_1}$ $\displaystyle \simeq \frac{\gamma-1}{\gamma}\,\frac{{\mit\Delta} p}{p_1},$    
    $\displaystyle \frac{{\mit\Delta}{\rm Ma}}{{\rm Ma}_1}$ $\displaystyle \simeq -\left(\frac{1+[(\gamma-1)/2]\,{\rm Ma}_1^{\,2}}{\gamma\,{\rm Ma}_1^{\,2}}\right)\frac{{\mit\Delta}p}{p_1},$    

    where $ {\mit\Delta}p=p-p_1$ , et cetera. Of course, the previous four relations are the same as those for a weak shock. (See Exercise iii.) Why is this not surprising? (Hint: The jump in specific entropy across a weak shock is third order in the deflection angle.) Deduce that to second order in the deflection angle,

    $\displaystyle \frac{{\mit\Delta}p}{p_1} \simeq -\left(\frac{\gamma\,{\rm Ma}_1^...
...amma+1)\,{\rm Ma}_1^{\,4}-4\,({\rm Ma}_1^{\,2}-1)\right]\vert\theta\vert^{\,2}

    for a weak Prandtl-Mayer fan. (See Exercise v.)

  7. If an oblique shock is intercepted by a wall then it is reflected, as illustrated in Figure 15.17. Calculate $ \beta_1'-\beta_1$ , assuming that the shocks are sufficiently weak that the approximate expressions of Section 15.4 can be used. Demonstrate that, in this limit,

    $\displaystyle \beta_1'-\beta_1\simeq 2\left[\left(\frac{\gamma+1}{4}\right)\left(\frac{{\rm Ma}_1^{\,2}}{{\rm Ma}_1^{\,2}-1}\right)-1\right]\theta_1,

    where $ \theta_1\ll 1$ is the deflection angle of the incident shock, and $ \gamma$ the ratio of specific heats. Show that $ \beta_1'-\beta_1>0$ if $ 1<{\rm Ma}_1<
[4/(3-\gamma)]^{1/2}$ and $ \beta_1'-\beta_1<0$ if $ {\rm Ma}_1> [4/(3-\gamma)]^{1/2}$ . Demonstrate that $ {\rm Ma}_2\simeq {\rm Ma}_1\,(1-\delta)$ and $ {\rm Ma}_3\simeq {\rm Ma}_1\,(1-2\,\delta)$ , where

    $\displaystyle \delta = \left[\frac{1+[(\gamma-1)/2]\,{\rm Ma}_1^{\,2}}{\sqrt{{\rm Ma}_1^{\,2}-1}}\right]\theta_1.

    Figure: Merging of two oblique shocks of the same family produced by successive concave corners of deflection angles $ \theta _1$ and $ \theta _2$ . Here, $ {\rm Ma}_1$ , $ {\rm Ma}_2$ , et cetera, are Mach numbers.
\epsfysize =2.5in

  8. Figure 15.18 shows a situation in which two oblique shocks of the same family [in this case, the $ (-)$ family], produced by successive concave corners in a wall, merge together to form a single stronger shock [of the $ (-)$ family]. Assuming that the shocks are sufficiently weak that the approximate expressions of Section 15.4 can be used (which implies that $ \theta_1\ll 1$ and $ \theta_2\ll 1$ ), demonstrate that

    $\displaystyle \delta\simeq \epsilon_1+\epsilon_2,


    $\displaystyle \epsilon_1$ $\displaystyle =\left(\frac{\gamma+1}{4}\right)\left(\frac{{\rm Ma}_1^{\,2}}{{\rm Ma}_1^{\,2}-1}\right)\theta_1,$    
    $\displaystyle \epsilon_2$ $\displaystyle = \left(\frac{\gamma+1}{4}\right)\left(\frac{{\rm Ma}_1^{\,2}}{{\rm Ma}_1^{\,2}-1}\right)\theta_2,$    

    and $ \gamma$ is the ratio of specific heats. Show that

    $\displaystyle \beta_3\simeq \beta_1+\epsilon_2,

    and also that $ {\rm Ma}_2\simeq {\rm Ma}_1\,(1-\delta_1)$ and $ {\rm Ma}_3\simeq {\rm Ma}_4 \simeq {\rm Ma}_1\,(1-\delta_1-\delta_2)$ , where

    $\displaystyle \delta_1$ $\displaystyle = \left[\frac{1+[(\gamma-1)/2]\,{\rm Ma}_1^{\,2}}{\sqrt{{\rm Ma}_1^{\,2}-1}}\right]\theta_1,$    
    $\displaystyle \delta_2$ $\displaystyle = \left[\frac{1+[(\gamma-1)/2]\,{\rm Ma}_1^{\,2}}{\sqrt{{\rm Ma}_1^{\,2}-1}}\right]\theta_2.$    

    Demonstrate that the strength of the merged shock is approximately the sum of the strengths of the two component shocks, and, hence, that the pressures on either side of the slipstream shown in the figure are equal (at least, to first order in $ \theta _1$ and $ \theta _2$ ). Finally, show that

    $\displaystyle \frac{{\cal S}_4-{\cal S}_3}{{\cal R}}$ $\displaystyle \simeq \frac{\gamma\,(\gamma+1)}{4}\,\frac{{\rm Ma}_1^{\,6}}{({\rm Ma}_1^{\,2}-1)^{3/2}}\,\theta_1\,\theta_2\,(\theta_1+\theta_2),$    

    where $ {\cal S}$ is specific entropy, and $ {\cal R}$ the specific gas constant. It is, thus, clear that the specific entropy is not quite the same on either side of the slipstream.

    Figure: Crossing of two oblique shocks of different families. Here, $ {\rm Ma}_1$ , $ {\rm Ma}_2$ , et cetera, are Mach numbers, and $ \theta _1$ , $ \theta _2$ , et cetera, are deflection angles.
\epsfysize =2.5in

  9. If two shocks of opposite families intersect then they pass through one another, but are slightly bent in the process, as illustrated in Figure 15.19. Assuming that the shocks are sufficiently weak that the approximate expressions of Section 15.4 can be used (which implies that $ \theta_1\ll 1$ , $ \theta_2\ll 1$ , et cetera), show that

    $\displaystyle \frac{p_2-p_1}{p_1}$ $\displaystyle \simeq \left(\frac{\gamma\,{\rm Ma}_1^{\,2}}{\sqrt{{\rm Ma}_1^{\,2}-1}}\right)\theta_1,$    
    $\displaystyle \frac{p_3-p_1}{p_1}$ $\displaystyle \simeq \left(\frac{\gamma\,{\rm Ma}_1^{\,2}}{\sqrt{{\rm Ma}_1^{\,2}-1}}\right)(\theta_1+\theta_2),$    
    $\displaystyle \frac{p_2'-p_1}{p_1}$ $\displaystyle \simeq \left(\frac{\gamma\,{\rm Ma}_1^{\,2}}{\sqrt{{\rm Ma}_1^{\,2}-1}}\right)\theta_1',$    
    $\displaystyle \frac{p_3'-p_1}{p_1}$ $\displaystyle \simeq \left(\frac{\gamma\,{\rm Ma}_1^{\,2}}{\sqrt{{\rm Ma}_1^{\,2}-1}}\right)(\theta_1'+\theta_2'),$    

    where $ p$ denotes pressure. Hence, deduce that

    $\displaystyle \theta_2$ $\displaystyle \simeq \theta_1',$    
    $\displaystyle \theta_2'$ $\displaystyle \simeq \theta_1,$    
    $\displaystyle \delta$ $\displaystyle \simeq \theta_1'-\theta_1.$    

    Show, that the respective strengths of the two shocks are unaffected by the intersection (at least, to first order in the deflection angles).

  10. Show that Equation (14.66) can be written in the form

    $\displaystyle \sin^2\mu +\left(\frac{\gamma-1}{\gamma+1}\right)\cos^2\mu=\frac{c_1^{\,2}}{w^{\,2}},

    where $ \mu$ is the Mach angle, $ \gamma$ the ratio of specific heats, $ c_1$ the sound speed at the sonic point, and $ w$ the flow speed. Deduce that

    $\displaystyle \sqrt{{\rm Ma}^{\,2}-1}\,\frac{dw}{w}=\left(\frac{b^{\,2}-1}{b^{\,2}+\tan^2\mu}\right)d\mu,


    $\displaystyle b^{\,2}= \frac{\gamma-1}{\gamma+1},

    and, hence, that

    $\displaystyle \int \sqrt{{\rm Ma}^{\,2}-1}\,\frac{dw}{w} = \mu -\frac{1}{b}\,\tan^{-1}\left(\frac{1}{b}\,\tan\mu\right)+{\rm const.}

    Finally, demonstrate that

    $\displaystyle \nu({\rm Ma})\equiv \int_1^{\rm Ma}\sqrt{{\rm Ma}^{\,2}-1}\,\frac{dw}{w}$ $\displaystyle = \left(\frac{\gamma+1}{\gamma-1}\right)^{1/2}\tan^{-1}\left(\lef...
...\frac{\gamma-1}{\gamma+1}\right)({\rm Ma}^{\,2}-1)\right]^{1/2}\right)\nonumber$    
      $\displaystyle \phantom{=}-\tan^{-1}\left(\left[{\rm Ma}^{\,2}-1\right]^{1/2}\right).$    

    (Leipmann & Roshko 1957.)

  11. Show that for a thin, symmetrical airfoil, with zero angle of attack, whose profile is a lens defined by two circular arcs, the drag coefficient is

    $\displaystyle C_D = \frac{16}{3\sqrt{{\rm Ma}_1^{\,2}-1}}\left(\frac{t}{c}\right)^2,

    where $ {\rm Ma}_1$ is the upstream Mach number, $ t$ the maximum thickness, and $ c$ the chord-length. Demonstrate that for a given thickness ratio, $ t/c$ , the airfoil with the minimum drag is a symmetric diamond profile. (Leipmann & Roshko 1957.)

  12. Prove that on a supersonic swept-back wing of infinite span the thin-airfoil pressure coefficient, (15.50), is multiplied by the sweepback factor,

    $\displaystyle \frac{1}{\sqrt{1-n^{\,2}}},


    $\displaystyle n=\frac{\tan{\mit\Lambda}}{\sqrt{{\rm Ma}_1^{\,2}-1}}.

    Here, $ {\mit\Lambda}$ is the sweepback angle, and $ {\rm Ma}_1$ the upstream Mach number. [Hint: Resolve $ {\rm Ma}_1$ into components normal and parallel to the leading edge. The flow may then be studied in the plane normal to the leading edge using standard thin-airfoil theory.] (Leipmann & Roshko 1957.)

  13. Consider the problem of subsonic flow past a wave-shaped wall that was discussed in Section 15.13. Show that if the flow is bounded by a second wall that lies at $ y=b$ (where $ b>0$ ) then

    $\displaystyle \phi(x,y) = \frac{U\,\epsilon}{m}\,\cos(\alpha\,x)\,\frac{\cosh[m\,\alpha\,(b-y)]}{\sinh(m\,\alpha\,b)}.

    Show, on the other hand, that if the flow is bounded by a free surface at $ y=b$ (where $ p=p_\infty$ ) then

    $\displaystyle \phi(x,y) = \frac{U\,\epsilon}{m}\,\cos(\alpha\,x)\,\frac{\sinh[m\,\alpha\,(b-y)]}{\cosh(m\,\alpha\,b)}.

    (Leipmann & Roshko 1957.)

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Next: Vectors and Vector Fields Up: Two-Dimensional Compressible Inviscid Flow Previous: Flat Lifting Wings
Richard Fitzpatrick 2016-03-31