THE SCIENCE AND APPLICATIONS OF ACOUSTICS - H. H. Arnold ...

8.4 Waves in Pipes: Junctions and Branches 155compliance C s per unit length. It also follows that the distribution of mass per unitlength of the pipe can be represented by m s = ρ 0 A. The acoustic inductance perunit length becomes M s = m s /A 2 = ρ 0 /A.We shall now find the mechanical stiffness per unit length. When the fluid is compressedadiabatically by a small linear displacement δ , then p = ρ 0 c 2 (δ / ), andthe force providing this impetus is pA. Hence the stiffness becomes S = pA/δ ;and the stiffness per unit length is S s = ρ 0 c 2 A. The mechanical compliance C mrelates to acoustic compliance C by C = A 2 C m , and on a per unit length basis, it followsthat C s = A/(ρ 0 c 2 ). By analogy the acoustic impedance of the pipe is given byZ =which corresponds to Equation (8.8).√M sC s= ρ 0cA8.4 Waves in Pipes: Junctions and BranchesConsider a sound wave traveling in the positive x-direction, represented byp i = ae i(ωt−kx) , (8.9)is incident upon a point x = 0, where the acoustic impedance changes from ρ 0 c/Ato some complex value Z 0 . At this point a reflected wavep r = be i(ωt+kx) (8.10)will be produced and travel in the negative x-direction. It is our task to findthe power reflection and transmission coefficients for this point. The acousticimpedance at any point in the pipe is given byZ = p i + p r= ρ 0c ae −ikx + be +ikx(8.11)U i + U r A ae −ikx − be +ikxwhich, at x = 0, reduces toZ 0 = ρ 0cAa + ba − b(8.12)Equation (8.12) can be rearranged tobZ 0 − ρ 0ca = AZ 0 + ρ (8.13)0cAThe sound power reflection coefficient R p , which yields the fraction of the incidentpower that is reflected, is given by(R p =b2 R 0 − ρ )0c 2+ X2 ∣ a ∣ =A 0(R 0 + ρ )0c 2(8.14)+ X2A 0