671017.pdf

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Pipe solution

4 mm pipe

3 mm pipe

tD=t/t 0

ID

=I/I

max

,r

Figure 4: Comparison of grout penetration function in a pipe with
experimental data from H ̊akansson [ 10 ].

Grout

Borehole

Fracture plane

pw

Q,pg

I

r

Figure 5: Radial penetration of grout in a fracture.

2.3.PenetrationinaTwo-DimensionalFracture.Amore
realistic model of a fracture to grout is perhaps a pseudo-
plane with a system of conductive areas and flow channels [ 5 ].
If the transmissivity of the fracture is reasonably constant, a
parallel plate model with constant aperture b can approximate
it. If it is grouted through a borehole, there will be a radial,
two-dimensional, flow of grout out from the borehole; see
Figure 5. In reality, however, the flow will as for flow of water
from a borehole be something in between a system of one-
dimensional channels and radial flow [ 12 ].
Equations ( 3 )and( 4 )givethegroutflowintheplanecase.
The grout flow velocity is constant (in푥)andequaltothe
front velocity푑퐼/푑푡. In the radial case we replace푥by푟.The
grout flow velocityV푔(m/s)decreases as1/푟,[ 16 ]. Let푟푏be
the radius of the injection borehole, and let푟푏+퐼be the radius
of the grout injection front at any particular time푡.Wehave

V푔=−

푑푝

푑푟

⋅

푏^2

12휇푔

[1 − 3 ⋅

푍

푏

+4⋅(

푍

푏

)

3

], 푟푏≤푟≤푟푏+퐼,

(14)

where

푍=휏 0 ⋅

儨儨

儨儨儨

儨儨

儨

푑푝

푑푟

儨儨

儨儨儨

儨儨

儨

−1

,푍<

푏

2

. (15)

Let the grout injection rate be푄(m^3 /s). The total grout flow

isthesameforall푟:

푄=2휋푟푏⋅V푔,푟푏≤푟≤푟푏+퐼. (16)

Combing ( 14 )and( 16 ), we get after some calculation the

following implicit differential equation for the pressure as a

function of the radius:

6휇푔푄

휋푏^2 휏 0

⋅

1

푟

=푠⋅[2−3⋅푠−1+푠−3],

푠=

푏

2푍

=

푏

2휏 0

⋅

儨儨儨

儨儨

儨儨儨

푑푝

푑푟

儨儨儨

儨儨

儨儨儨

(17)

or

푟=

2휇푔푄

휋푏^2 휏 0

⋅

3푠^2

2푠^3 −3푠^2 +1

,푠=−

푏

2휏 0

⋅

푑푝

푑푟

,

푟푏≤푟≤푟푏+퐼.

(18)

The injection excess pressure isΔ푝.Wehavetheboundary

condition

푝(푟푏)−푝(푟푏+퐼)=Δ푝. (19)

Here, we neglect a pressure fall in the ground water, since the

viscosity of grout is much larger than that of water.

The solution푝(푟)of ( 18 )-( 19 )hasthefrontposition퐼

as parameter. The value of푄hastobeadjustedsothatthe

pressure differenceΔ푝is obtained in accordance with ( 19 ).

The front position퐼=퐼(푡)increases with time. The flow

velocity at the grout front푟=푟푏+ 퐼(푡)is equal to the time

derivative of퐼(푡).Wehavefrom( 16 )

푄(퐼)=2휋푏⋅[푟푏+퐼(푡)]⋅

푑퐼

푑푡

,퐼( 0 )=0. (20)

This equation determines the motion of the grout front. It

depends on the required grout injection rate푄(퐼),whichis

obtained from the solution of ( 18 )-( 19 ) for each front position

퐼.

The solution for radial grout flow is much more compli-

cated than for the plain case and the pipe case. We must first

solve the implicit differential equation for푝(푟).Thisinvolves

the solution of a cubic equation in order to get the derivative

푑푝/푑푟and an intricate integration in order to get푝(푟).From

the solution, we get the required grout flux for any front

position퐼.

With known function 푄(퐼), we may determine the

motion of the grout front from ( 20 )byintegration.

The front position퐼increases from zero at푡=0to a

maximum value for infinite time. Then the flux푄must be

zero. Equation ( 18 )gives푄=0for푠=1.Thenwehavea

linear pressure variation:

푄=0, 푠=1㨐⇒−

푑푝

푑푟

=

2휏 0

푏

㨐⇒ 푝 = 퐾 −

2휏 0

푏

⋅푟.

(21)