Here,퐾is a constant. The boundary condition ( 19 ) deter-
mines the maximum value of퐼:
푝(푟푏)−푝(푟푏+퐼max)=
2휏 0
푏
⋅(−푟푏+푟푏+퐼max)=Δ푝㨐⇒퐼max=푏Δ푝
2휏 0
.
(22)
We g e t t h e s a m e v a l u e ( 2 )asintheplaincase.
The complete solution in the radial case involves the
following constants:
퐼max=푏Δ푝
2휏 0
,훾=
퐼max
푟푏=
푏Δ푝
2푟푏휏 0
,
푡 0 =
6휇푔Δ푝
(휏 0 )
2 ,푄^0 =
6휋푏(퐼max)^2
푡 0=
휋푏^3 Δ푝
4휇푔
.
(23)
2.4. Solution for the Pressure.In the dimensionless solution
forthepressure,weusetheboreholeradiusasscalinglength:
푟耠=
푟
푟푏
,퐼耠=
퐼
푟푏
,푟푏≤푟≤푟푏+퐼⇐⇒1≤푟耠≤1+퐼耠.
(24)
The pressure is scaled byΔ푝/훾.Thevariable푠for the deriv-
ative of the pressure in ( 18 )becomes
푝耠=
훾⋅(푝−푝푤)
Δ푝
㨐⇒ 푠 =
푏
2휏 0
⋅(−
푑푝
푑푟
)
=−
푏Δ푝/훾
2휏 0 푟푏
⋅
푑푝耠
푑푟耠
=−
푑푝耠
푑푟耠
.
(25)
The dimensionless form of ( 18 )-( 19 ) becomes after some
recalculations
푟耠=푄耠⋅푔(−
푑푝耠
푑푟耠
), 푔(푠)=
3푠^2
2푠^3 −3푠^2 +1
,
푄耠=
2휇푔푄
휋푏^2 휏 0 푟푏
,푝耠( 1 )−푝耠(1 + 퐼耠)=훾,
1≤푟耠≤1+퐼耠.
(26)
This is the basic equation to solve for the pressure distribu-
tion.Itistobesolvedfor0<퐼耠<훾for positive values of the
parameter훾.
The solution is derived in detail in [ 14 ]. A brief derivation
is presented in the appendix. The dimensionless pressure is
given by
푝耠(푟耠)=훾−푄耠⋅[퐺(푄̃ 耠)−퐺(̃
푄耠
푟耠
)], 1 ≤ 푟耠≤1+퐼耠.
(27)
The composite function퐺(푞)̃ ,whichisusedfor푞=푄耠and
푞=푄耠/푟耠, is defined by퐺(푞)=퐺(̃ ̃푠(푞)),
̃푠(푞)=
1
(^2) √1+푞⋅sin{(1/3)⋅arcsin[(1 + 푞)−1.5]}
,
퐺(푠)=
4
3
⋅ln(푠−1)+1
6
⋅ln(2푠 + 1)−1
푠−1
−
3푠^3
(2푠 + 1)(푠−1)^2
.
(28)
The functioñ푠(푞)is the root to the cubic equation푞⋅푔(푠)=1
for푠>1.Thefunction퐺(푠)is an integral of푠⋅푑푔/푑푠.
The value of the factor푄耠hastobechosensothatthetotal
pressure difference corresponds to the injection pressure,
( 26 ). This gives훾=푄耠⋅[퐺(푄̃ 耠)−퐺(̃
푄耠
1+퐼耠
)]. (29)
This equation determines푄耠as a function of퐼耠and훾:푄耠=푓耠(퐼耠,훾), 0≤퐼耠≤훾, 훾>0. (30)
The value of푄耠for퐼耠=훾is zero in accordance with ( 21 )-( 22 ):
푓耠(훾,훾) = 0.2.5. Motion of Grout Front.In the dimensionless formulation
oftheequationforthemotionofthegroutfront,weuse퐼max
as scaling length. We also use푄 0 and푡 0 from ( 23 )퐼퐷=
퐼
퐼max,퐼耠=훾퐼퐷,
푄퐷=
푄
푄 0
,푡퐷=
푡
푡 0
.
(31)
The grout flux becomes from ( 23 )and( 26 )푄 0
훾
=
휋푏^2 휏 0 푟푏
2휇푔
㨐⇒ 푄 =
푄 0
훾
⋅푓耠(퐼耠,훾)=푄 0 ⋅푄퐷(퐼퐷,훾).
(32)
The dimensionless grout flux is then푄퐷(퐼퐷,훾)=
푓耠(훾퐼퐷,훾)
훾
,0≤퐼퐷≤1. (33)
The dimensionless equation for the front motion is now from
( 32 ), ( 20 ), ( 31 ), and ( 23 )푄 0
훾
⋅푓耠(훾퐼퐷,훾)=2휋푏⋅
(퐼max)2푡 0⋅(
1
훾
+퐼퐷)⋅
푑퐼퐷
푑푡퐷
or