Page 257 - 48Fundamentals of Compressible Fluid Mechanics
P. 257

13.4. SOLUTION OF MACH ANGLE                                        219

         Substituting the values of  and  equations (13.37)(13.38) into equation (13.28)
         provides the equation to be solved for c .


                                 "




                   < E  <




                      "
                                   .
               	    	 .
                                                                          (13.39)

                     "




                                             )  E
                                         9

         This author is not aware of any analytical demonstration in the literature which





                                                   16 . Nevertheless, this identity can



                                                                          . Table

         (13.6) is provided for the following demonstration. Substitution of all the above
         showing that the solution is identity zero for c

                                       .

              Utilizing the symmetry and antisymmetry of the qualities of the    and    for

         values into (13.28) results in
                                     regardless to Mach number. Hence, the physical
         be demonstrated by checking several points for example,
         interpretation of this fact that either that no shock can exist and the flow is without

              demonstrates that
                                             17
         any discontinuity or a normal shock exist . Note, in the previous case, positive
         c

         large deflection angle, there was transition from one kind of discontinuity to an-


         other.
                                , the
         question whether it is possible           coefficients
         for the oblique shock to exist?
         In the range where c
         The answer according to this

         analysis and stability analysis is       1.0          ,  ,'	 ,  .
                                                                -3
                                                                    -1
         not. And according to this anal-
         ysis no Mach wave can be gen-             2.0          3   0    -.
         erated from the wall with zero
         deflection. In other words, the                         -1  0
         wall doesn’t emit any signal to                               -
         the flow (assuming zero viscos-
         ity) which contradicts the com-
                                      Fig. 13.6: The various coefficients of three different
         mon approach. Nevertheless,
                                                                            is zero
         in the literature, there are sev-    Mach number to demonstrate that
                                                                              18
         eral papers suggesting zero strength Mach wave, other suggest singular point .
         The question of singular point or zero Mach wave strength are only of mathematical
         interest.
          16
           A mathematical challenge for those who like to work it out.
          17 There are several papers that attempted to prove this point in the past. Once this analytical solution
         was published, this proof became trivial. But for non ideal gas (real gas) this solution is only indication.
          18 See for example, paper by Rosles, Tabak, “Caustics of weak shock waves,” 206 Phys. Fluids 10 (1)
         , January 1998.
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