Chap. 10] 



ELECTRICAL METHODS 



<Jx = 



3o-2 + 2t'i(<ri — 0^2) 

 3<ri — V\{a\ — 0-2) 



• a\. 



635 

 (10-76). 



Assuming now that v\ = r (or 2v\ = Vx , a porosity of 50 per cent), we 



obtain from eq. (10-76): ax = -^ — , and therefore, for the ratio, 



5<ri + (72 



Ox 



ai 



0-2 



2^^ + 4 



5^^ + l 



<T2 



(10-7c) 



«5 



From this it follows that, for a conductivity of the pore filling medium very 

 much greater than that of the grains, or for ci )S> 0-2, we have (Tx == \<ti , 

 and for <t^^ ai , <Xx =f 4o-i . 

 Fig. 10-1 is a graph of eq. 

 (10-7c). For a porosity 

 of 50 per cent, the conduc- 

 tivity of the aggregate in- 

 creases almost in direct 

 proportion to the conduc- 

 tivity of the medium fill- 

 ing the pores (assuming 

 all pores to be filled with 

 the medium of the con- 

 ductivity o-i). For virtu- 

 ally all porous rocks we 

 are, therefore, justified in 

 disregarding the conduc- 

 tivity of the mineral grains 

 and operating only with 



the conductivity of the medium filling the pores 

 formula (10-76), we get 



Fig. 10-1. Rock resistivity as a function of re- 

 sistivity of pore-filling medium, for 50 per cent 

 porosity (after Hummel). 



Thus, letting (72 = in 



2vi 3 - yi 

 — -tri or px = —7i Pi. 



3 - vi 



2vi 



(10-7rf) 



Eq. (10-7d) can be further simpUfied for porosities less than 25 per cent 

 by assuming that the pore volume is equivalent to a system of cylindrical 

 tubes of identical radius r, not touching ohe another, traversing the sub- 

 stances in the three directions. If there are n tubes per square centimeter, 

 their area on one side of a cube (or one-third of the pore volume) is v/Z = 



