(ii) Based on an assessment of how fine fractures it
is necessary to seal, a maximum effective borehole
distancecanbepredictedgiventhepressureandthe
properties of the grout.
(iii) The time needed for effective grouting operations can
be estimated with better accuracy.
(iv) In order to avoid unrestricted grout pumping also a
maximum grouting time can be given, where further
injection of grout will be unnecessary.Appendix
Derivation of the Solution for the Pressure
We s e e k t h e s o l u t i o n푝耠(푟耠)to ( 26 ):
푟耠=푄耠⋅푔(−
푑푝耠
푑푟耠
), 1≤푟耠≤1+퐼耠,
푔(푠)=
3푠^2
2푠^3 −3푠^2 +1
,0≤퐼耠≤훾.
(A.1)
Here,1+퐼耠is the position of the grout front. The parameter
훾is positive. Taking zero pressure at the grout front, the
boundary conditions for the dimensionless pressure become
푝耠( 1 )=훾, 푝耠(1 + 퐼耠)=0. (A.2)The dimensionless grout flux푄耠is to be chosen so that the
previous boundary conditions are fulfilled. The value of푄耠
will depend on the front position퐼耠.
Solution in Parameter Form.In order to see more directly the
character of the equation, we make the following change of
notation:
푥←→푟耠,푦←→−푝耠,푓(푠)=푄耠⋅푔(푠). (A.3)The equation is then of the following type:
푥=푓(
푑푦
푑푥
). (A.4)
There is a general solution in a certain parameter form to
this type of implicit ordinary differential equation [ 19 ]. The
solution is
푥(푠)=푓(푠),푦(푠)=푠⋅푓(푠)−∫
푠
푓(푠耠)푑푠耠. (A.5)We have to show that this is indeed the solution. We have
푑푥
푑푠=
푑푓
푑푠
,
푑푦
푑푠
=1⋅푓(푠)+푠⋅
푑푓
푑푠
−푓(푠)=푠⋅
푑푓
푑푠
.
(A.6)
The ratio between these equations gives that푠is equal to the
derivative푑푦/푑푥.Wehave
푑푦
푑푥=
푑푦/푑푠
푑푥/푑푠
=푠㨐⇒푓(
푑푦
푑푥
)=푓(푠)=푥. (A.7)
The right-hand equation shows that (A.5)isthesolutionto
(A.4).Explicit Solution.Applying this technique to (A.1), we get the
solution푟耠=푄耠⋅푔(푠),−푝耠(푠)=푠⋅푄耠⋅푔(푠)−푄耠⋅∫
푠
푔(푠耠)푑푠耠.(A.8)
We introduce the inverse to푔(푠)in the following way:1=푞⋅푔(푠)⇐⇒ 푠 = 푔−1(
1
푞
)=̃푠(푞)⇐⇒ 1 = 푞 ⋅ 푔(̃푠(푞)).
(A.9)
The pressure with a free constant퐾for the pressure level may
now be written as푝耠(푠)=푄耠⋅퐺(푠)+퐾, 퐺(푠)=∫
푠
푔(푠耠)푑푠耠−푠⋅푔(푠).
(A.10)The solution is then from (A.8)–(A.10)(with푞=푄耠/푟耠)푝耠(푠)=푄耠⋅퐺(푠)+퐾, 푠=푠(̃
푄耠
푟耠
) (A.11)
or, introducing the composite function퐺(푞)̃ ,퐺(푞)=퐺(̃ ̃푠(푞)), 푝耠(푟耠)=푄耠⋅퐺(̃ 푄
耠푟耠)+퐾.(A.12)
The boundary condition (A.2)at푟耠=1is fulfilled for a certain
choice of퐾. The explicit solution is푝耠(푟耠)=훾−푄耠⋅[퐺(푄̃ 耠)−퐺(̃
푄耠
푟耠
)], 1 ≤ 푟耠≤1+퐼耠.
(A.13)
The other boundary condition (A.2)at푟耠=1+퐼耠is fulfilled
when푄耠satisfies the equation훾=푄耠⋅[퐺(푄̃ 耠)−퐺(̃
푄耠
1+퐼耠
)]. (A.14)
We note that the derivative−푑푝耠/푑푟耠is given by푠:푠=−
푑푝耠
푑푟耠
. (A.15)
The pressure derivative is equal to –1 for zero flux, ( 21 )and
( 25 ), in the final stagnant position퐼耠=훾.Themagnitude
of this derivative is larger than 1 for all preceding positions
퐼耠<훾.Thismeansthat푠is larger than (or equal to) 1 in the
solution.The Function퐺(푠).The solution (A.13)andthecomposite
function (A.12)involvethefunction퐺(푠)defined in (A.10)