SECONDARY PRESSURE WAVES 363 



where a is the minunum radius of the gas sphere at the time for which 

 da/dt = 0. At this time, the pressure on the surface of the sphere must 

 equal the gas pressure P{d), and hence 



(9.10) Pn.ax - Po = "" P(a) 



r 



if Po is neglected in comparison with P{a) . At the time of the minimum, 

 the excess pressure in the water thus falls off inversely as the distance r 

 from the value P{a) on the gas sphere. 



In order to compute the value of Pmax in a given case, it is necessary 

 to evaluate P{a) and a for the explosive products. This is conveniently 

 done by noting that when a = a, all of the available energy Y is in the 

 form of internal energy E{a) of compression of the gas. For product 

 gases obeying the law PaV^ = k, where V is the volume per gram of 

 explosive and Pa is in dynes/cm.^, we therefore have 



E{a) = Y = -^^ = -^^[&0.5y^a''] 

 7-1 7-lL 3 J 



if a is the minimum radius in feet. Solving for Pa and substituting in 

 Eq. (9.10) gives 



(9.11) ^--^-i^3(T-l)5-- 



The value of d~^ can also be determined from the condition E{a)/Y 

 = 1. The expression for E{a)/Y in terms of a has been worked out in 

 section 8.9, and substitution of the value of a obtained by setting 

 E(a)/Y = 1 gives 



As written, this equation is expressed in the units of P(a) (dynes/cm.^), 

 r is the fraction of the detonation energy Q expressed in cal./gm. of ex- 

 plosive, W is the charge weight in pounds, and r is in feet. Converting 

 the equation to express pressure in the more generally employed units 

 of lb./in.2 gives 



(9.12) Pn.ax - Po= 2.26-10-6/c('4.19-10VQ^^^^ ^iV +^(:;3T) 



r 



