SOUND WAVES IN A VISCOUS FLUID 



27 



In analogy with the result (86) for the simple case 

 of axial symmetry, 



^-(0,0) b{e,<t,) 



I(r,d,<t>) = ; (89) 



r- 



at a distance r from the source much greater than a 

 wavelength. 



The rate at which the projector emits energy in all 

 directions must be exhibited as a double integral in 

 this general case. The area element on the sphere of 

 radius r, intercepted between (5,0) and {d + dd, 

 </) + dip), is r^ cos dddd^; and the solid angle inter- 

 cepted by this area element is cos dddd<t>. Thus, in 

 view of equation (88), 



Emission through area element 



= F(0,0) b{e,<t>) cos eddd4> (90) 



and therefore 



Total rate of emission 



d(t> bie,<p) cos ede. (9i) 



-r J-W2 



Equation (91) can be put in the following form) 

 which is directly comparable to the law of emission 

 (59) of a point source : 



Rate of emission = 4x7^(0,0)5 (92) 



where 



I /» :r /• ir/2 



5 = — d<l)\ 5(6i,<^) COS ede. (93) 



itrJ-T J-t/2 



The factor 5 is a constant depending on the nature of 

 the source and may be called the directivity factor of 

 the source. For the point source of equation (59), this 

 directivity factor is 1 ; while for the double source of 

 equation (87) it is 3^. 



2.5 SOUND WAVES IN A VISCOUS FLUID 



In a homogeneous perfect fluid, the decrease of 

 sound intensity with increasing distance from the 

 source is due only to spreading according to the 

 inverse square law (58). However, sound intensity 

 measurements show clearly that the intensity loss in 

 the ocean tends to be much greater than the value pre- 

 dicted by equation (58). These extra losses above the 

 theoretical loss, (58), due to the fact that the ocean 

 is not a homogeneous perfect fluid, are called trans- 

 mission anomalies or, more loosely, attenuations. In 

 this section, we shall derive some results for sound 

 intensity in a viscous fluid and see how much of the 

 observed attenuation can be ascribed to fluid 

 viscosity. 



In the derivation of the wave equation (27) for a 

 perfect fluid, we used the equation of continuity (4), 

 the equations of motion (15), the equation of state 

 (18), and the law of forces (16). The equation of 

 continuity, the equations, of motion, and the equation 

 of state, it will be recalled, apply to any fluid, whether 

 it is perfect or viscous. The law of forces (16), how- 

 ever, is valid only for a perfect fluid. The exact law 

 of forces operating in a viscous fluid is quite difficult 

 to derive since it depends on the theory of viscous 

 fluid flow. It is sufficient to say that this complicated 

 law of forces, combined with equations (4), (18), and 

 (15), can be used to derive a general wave equation 

 for viscous fluids, analogous to equation (27). Under 

 the assumption that the resulting pressure distribu- 

 tion in the viscous medium depends only on the co- 

 ordinate X, this general equation reduces to' 



K d'^p 4m d'^p 

 po dx^ 3po dxdt 



(94) 



where m is the coefficient of shear viscosity. Equation 

 (94) is the plane wave equation for a viscous fluid. In 

 the absence of viscosity (m = 0), equation (94) re- 

 duces to equation (30). 



Let us see whether we can find a solution to equa- 

 tion (94) of the form 



By substituting this expression for p into equation 

 (94), we obtain 



z 



(95) 



m-= = 



- + -^2«/ 

 Po opo 



If c and a are defined by 



Po 



3 poc'' 



(96) 



(97) 



the relation (95) becomes 



m = ± ^ .— = + (- + ^.), (98) 



according to the binomial theorem, if we assiune that 

 a is so small that a^ and higher terms can be neglected. 

 In order to get the case of waves propagated in the 

 positive X direction, the negative sign of equation 

 (98) must be chosen and 2Tnmx becomes 



./ 



2inmx = —2in-x 

 c 



ax. 



