Engineering Rock Mechanics

Question and answers: permeability 149

equation for the head at an interior node is

Hj =

ci,j

I

(see Section 9.4 of ERM 1, where we show how this equation is derived

from the Kirchoff relation for the net flow at a node). Expanding this

equation for node 4 results in

H4 = Cl,4H1 + c2,4H2 + c5,4H5 + C12,4H12

cl,4 + c2,4 + ‘3.4 + cl2.4

Rearranging this to put all of the known heads on the left-hand side of

-c1,4HI - c2,4H2 = - (cl.4 + c2,4 + c5,4 + ~124) H4 + ~5,4H5 + ~12.4H12.

Continuing for the remaining nodes gives the following relations

the equation gives

Node 0 = c4,5H4 - (c4,5 + c6.5 + c9,5)H5 + c6,5H6 + c9,5H9

Nod? (^6) -c3,6H3 = c5,6HS - (c3.6 + c5.6 + c7.6 f c10,6)H6 + c7,6H7 + CIO,~H~O
Node 7 -c8,7H8 = C6,7H6 - (C6,7 + c8,7 + c10,7)H7 + Cl0,7H10
Node 10 0 = c6,I0H6 + ~7.10H7 + c9,10H9 -
Node 12 -~11,12H11 = c4,12H4 - (c4.12 + ~11.12 + ~13,12)H12 + C13.12H13
Node (^9 0) = c5,9H5 - (c5.9 + c10.9 + c13,9)H9 + cl0,9HlO + C13.9H13
(c6.10 -k c7.10 + c9.10 + Cl5,lO)HlO + c15,1OHl5
Node (^13) -C17,13H17 = c9,13H9 + C12.13H12 -
(c9.13 f c12.13 f cl4.13 + c17,13)H13 + C14,13H14
Node 14 -C18,14H18 = C13.14H13 - (c13,~ + Ci5.14 f Ci8,14)H14 f C15.14H15
Node (^15) -C16,15H16 = cl0.15HlO + C14,15H14 - (clO,l5 + c14,15 f c16,15)H15.
These equations are now assembled into a system of ten simultaneous
equations. It is not convenient to solve such a large system by hand,
and so a matrix solution using a computer or appropriate hand-held
calculator should be sought. With this technique, we build and solve the
following matrix equation (note that for the sake of brevity the left-hand