Page 720 - Automotive Engineering Powertrain Chassis System and Vehicle Body
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Interior noise: Assessment and control    C HAPTER 21.1

             Now, taking the linearised inviscid Euler equation  Substituting equation (E21.1.10) into equation (E21.1.9)
           (derived in Appendix 21.1G)

                                                                          Ae ikL  þ  x 1  Ae  ikL
                                                                                  xþ1
              vp     vu                                         Z MO
                 ¼ r 0                            (E21.1.2)           ¼ Y
              vx      vt                                        r c 0 S     ikL   x 1     ikL
                                                                 0
                                                                          Ae           Ae
                                                                                  xþ1
           From this:
                         ð                                      Dividing both numerator and denominator by A
                        1 vp
             uðx; tÞ¼       dt                    (E21.1.3)
                       r 0  vx                                        e ikL  þ  x 1  ikL
                                                                              e
                                                                Y ¼         xþ1                      (E21.1.11)
           So, a relationship is found between pressure gradient and  e ikL     x 1  e ikL
           particle velocity.                                               xþ1
             For harmonic waves the integral with respect to time
           is given by Morse and Ingard (1968),  1  .           Multiplying both numerator and denominator of
                                           iu                 equation (E21.1.11) by (xþ 1)
           So,
                                                                     e ikL x þ e ikL  þ e  ikL x   e  ikL
                      1   vp                                    Y ¼  e ikL x þ e ikL    e  ikL x þ e  ikL  (E21.1 12)
             uðx; tÞ¼                             (E21.1.4)
                     iur vx
                        0
                                                                Now, there are these two standard relationships (see
             vp             i½utþkðL xފ         i½ut kðL xފ
                ¼ ikðL   xÞAe          ikðL   xÞBe            Weltner et al. [1986] for example)
             vx
                                                                 ix
                                                  (E21.1.5)     e   e  ix
                                                                         ¼ sin x                     (E21.1.13)
                                                                   2i
           As k ¼ uc
                                                                 ix
                                                                e þe  ix
                      1                                            2    ¼ cos x                     (E21.1.13a)
             uðx; tÞ¼    Ae i½utþkðL xފ    Be i½ut kðL xފ
                     r c 0                                      Substituting equations (E21.1.13) and (E21.1.13a)
                      0
                                                  (E21.1.6)   into equation (E21.1.12)
           Now
                                                                     x e ikL  þ e  ikL  þ e ikL    e  ikL
                   Force                                        Y ¼
             Z ML ¼                               (E21.1.7)           x e ikL    e  ikL  þ e ikL  þ e  ikL
                     u
                                                                     2x cos kL þ i2 sin kL
                                                                Y ¼                                  (E21.1.14)
                                                                     2 cos kL þ i2x sin kL
             Force ¼ PðL; tÞ: S
                                                                Divide both the numerator and the denominator of
                         A þ B                                equation (E21.1.14) by 2 cos kL to get
             Z ML ¼ r c 0 S                       (E21.1.8)
                    0
                         A   B
                                                                      x þ i tan kL
             The input mechanical impedance is given by:        Y ¼  1 þ ix tan kL
                         Ae ikL  þ Be  ikL                               Z ML  þ i tan kL
             Z MO ¼ r c 0 S                       (E21.1.9)      Z MO  ¼  r 0 c 0 S   ½Kinsler et al:; 1982Š
                     0
                         Ae ikL    Be  ikL                      r c 0 S  1 þ i r 0 c 0 S  tan kL
                                                                             Z ML
                                                                 0
             The two equations may be combined to eliminate both                                     (E21.1.15)
           A and B from equation (E21.1.8)
                                                                      Z ML
                                                                Now       is a complex term
                                                                      0
                 A þ B     Z ML                                      r c 0 S
             x ¼                ¼ x
                 A   B    r c 0 S                                Z ML
                           0
             xA   xB ¼ A þ B                                    r c 0 S  ¼ r þ ix                    (E21.1.16)
                                                                 0
             xA   A ¼ xB þ B
                 ðx   1Þ                                        r ¼ acoustic (flow) resistance
             B ¼       A                         (E21.1.10)
                  x þ 1                                         x ¼ auoustic (flow) resistance
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