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Interior noise: Assessment and control    C HAPTER 21.1

           So                                                   p ¼ p 0 þ p 0                        (G21.1.1)
                                                                r ¼ r þ r 0                          (G21.1.2)
                                                                     0
             vr TOT        vðr TOT uÞ  dxS        (F21.1.2)     T ¼ T 0 þ T 0                        (G21.1.3)
              vt  dxS ¼       vx
             vr TOT  vðr TOT uÞ  ¼ 0              (F21.1.3)     The mass of air between the two plates is given by (r 0 þ
                                                               0
              vt  þ    vx                                     r )dx (Morse and Ingard, 1968). The net force acting on
                                                              this mass is p(x)   p(x þ dx). Using Newton’s second law,
             This equation of mass conservation may be linearised
           thus:                                              this must be equal to the mass times the acceleration of
                                                              the fluid.
             r TOT u ¼ðr þ rÞu                    (F21.1.4)
                       0
             r TOT u ¼ r u þ ru                                 pðxÞ  pðx þ dxÞ¼   vp dx ¼  du ðr þ r Þdx
                                                                                                   0
                       0
                                                                                   vx      dt  0
             The ru term is the product of two small quantities and                                  (G21.1.4)
           one might choose to neglect it and so
                                                                Now the total differential  du  may be expressed in
                                                                                         dx
             r TOT u z r u                        (F21.1.5)   its partial differential form where u ¼ f(x, t)and the
                      0
                                                              total change in u is given by the sum of the partial
             As r 0 is not a function of x:
                                                              changes:
             vðr TOT uÞ   vu
                vx   z r 0 vx                     (F21.1.6)         vu      vu
                                                              du ¼    dx þ    dtðWeltner et al: ð1986Þ for exampleÞ
             As r 0 is not a function of t either, so               vx      vt
             vr TOT  vr                           (F21.1.7)
              vt   ¼  vt                                        So, dividing both sides by dt in the limit dt/ 0

           and the linearised mass conservation equation becomes
                                                                du    vu dx  vu dt
                                                                   ¼    $  þ   $
             vr    vu                                            dt   vx dt   vt dt
               þ r 0  ¼ 0                         (F21.1.8)
             vt    vx                                           du  ¼ u vu  þ  vu
                                                                 dt    vx   vt
             Re-arranging
                                                                So, from equation (G21.1.4)
             vu     1 vr
                ¼                                 (F21.1.9)
             vx     r vt                                          vp        vu  vu
                     0
                                                                                         0
                                                                    dx ¼ u    þ    ðr þ r Þdx
                                                                                     0
                     1      1 vp                                  vx        vx   vt
             Now       ¼            from                                                             (G21.1.5)
                     r 0    k 0 vr                                vp  ¼ðr þ r Þ  vu  þ  vu
                                                                              0
                                                 (F21.1.10)       vx     0      vt  vx
                      vp  vu      1 vp    vr
             k ¼ r 0         ¼                                  This is the non-linear inviscid Euler equation (Kinsler
              0
                      vr  vx      k 0 vr  vt
                                                              et al., 1982).
                         r 0
           So finally the linerised mass conservation equation is:        vu  vu                        r 0
                                                                Now if u  vx     vt  and the condensation ðs ¼  r 0 Þ  1,
             vu     1 vp                                      the non-linear inviscid Euler equation reduces to its
                ¼                                (F21.1.11)
             vx     k 0 vt                                    linearised form, being
                                                                  vp     vu
                                                                     ¼ r 0                           (G21.1.6)
           Appendix 21.1G: The derivation of                      vx      vt
           the non-linear (and linearised)                      From this
           inviscid Euler equation                                         1  ð  vp
                                                                uðx; tÞ¼        dt                   (G21.1.7)
                                                                          r 0  vx
           Take two plane surfaces – one at x and the other at
           x þ dx, each one with unit area. There is an acoustic wave  So, a relationship between pressure gradient and
           causing:                                           particle velocity is found.


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