Page 42 - Aerodynamics for Engineering Students
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Basic concepts and definitions  25

               This is very close to the given natural frequency of the tailplane, and there is thus a consider-
               able danger that the eddies might excite sympathetic vibration of the tailplane, possibly leading
               to structural failure of that component. Thus the shedding of eddies at this frequency is very
               dangerous to the aircraft.
               Example 1.3  An aircraft flies at a Mach number of 0.85 at 18300m where the pressure is
               7160Nm-2 and the temperature is -56.5  "C. A model of l/lOth scale is to be tested in a high-
               speed wind-tunnel. Calculate the total pressure of the tunnel stream necessary to give dynamic
               similarity, if the total temperature is 50 "C. It may be assumed that the dynamic viscosity is
               related to the temperature as follows:
                                             E= (6)
                                                      314
                                             Po

               where TO = 273°C  and  po = 1.71 x   kgm-ls-l
               (i)  Full-scale aircraft
                                M  = 0.85,  a = 20.05(273 - 56.5)'12 = 297ms-'
                                        V = 0.85 x 297 = 252m s-'

                                   p=-=  P    7160    = 0.1151 kgm-3
                                      RT   287.3 x  216.5

                                      e=($&) = 1.19
                                                 314
                                       P
                                      171
                                    = - 10-5 =   1.44 x   kgm-ls-l
                                      1.19
               Consider a dimension that, on the aircraft, has a length of 10 m. Then, basing the Reynolds
               number on this dimension:

                                            252 x 10 x 0.1151
                                 Ref=-= Vdp               = 20.2 x  lo6
                                       P      1.44 x  10-5
               (ii)  Model
               Total temperature Ts = 273 + 50 = 323 K
               Therefore at M  = 0.85:
                                        Ts     1
                                        -=  1 +-(OM)'=   1.1445
                                        T      5
                                        T = 282K
              Therefore
                                      a = 20.05 x  (282)'12 = 337m s-l
                                     -- r$)314=
                                      V = 0.85 x 337 = 287m s-'

                                     PO   -        1.0246
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