Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2013 Article ID 269594 9 pageshttpdxdoiorg1011552013269594
Research ArticleSteering Parameters for Rock Grouting
Gunnar Gustafson Johan Claesson and Aringsa Fransson
School of Civil Engineering Chalmers University of Technology 41296 Gothenburg Sweden
Correspondence should be addressed to Asa Fransson asafranssonchalmersse
Received 31 May 2013 Accepted 3 September 2013
Academic Editor Ga Zhang
Copyright copy 2013 Gunnar Gustafson et al This is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work is properlycited
In Swedish tunnel grouting practice normally a fan of boreholes is drilled ahead of the tunnel front where cement grout is injectedin order to create a low permeability zone around the tunnel Demands on tunnel tightness have increased substantially in Swedenand this has led to a drastic increase of grouting costs Based on the flow equations for a Bingham fluid the penetration of grout asa function of grouting time is calculated This shows that the time scale of grouting in a borehole is only determined by groutingoverpressure and the rheological properties of the grout thus parameters that the grouter can choose Pressure grout propertiesand the fracture aperture determine themaximumpenetration of the groutThe smallest fracture aperture that requires to be sealedthus also governs the effective borehole distance Based on the identified parameters that define the grouting time-scale and groutpenetration an effective design of grouting operations can be set up The solution for time as a function of penetration depth isobtained in a closed form for parallel and pipe flow The new more intricate solution for the radial case is presented
1 Introduction
In Swedish tunnelling pregrouting is normally used whenconsidered necessary for the reduction of groundwaterinflows Cement grout occasionally with plasticisers addedis preferred for economical and environmental reasonsRecently the increased demands on tunnel tightness have ledto an approach to pregrouting where the whole tunnel issystematically pregrouted according to a few predeterminedstandard strategies This has led to a massive increase ofperformed grouting and subsequently there is a strong needfor effective design methods and steering parameters for thegrouting activities
In pregrouting a fan of boreholes is drilled around thetunnel periphery ahead of the tunnel front grout is injectedthrough the boreholes in order to create a low permeabilityzone around the tunnel and finally the tunnel is excavated bythe drill and blast method within the zone until the next cyclestarts with drilling of the grouting fan Normally groutingboreholes 15ndash18m long are used which give 3-4 blastingrounds per cycle
Figure 1 shows the grouting fan and some fractures as abackground for the design problem Through the borehole
grout is injected which spreads through the fractures At anytime the grout has penetrated a distance 119868 from the boreholewhich is individual for each fracture For a successful grout-ing the penetration between the boreholes should bridge thedistance between the boreholes119871 forwater-bearing fractureshaving a transmissivity 119879 above a critical value determinedby their frequency and the demands on tunnel tightnessRecent investigations of the transmissivity distributions offractures in Swedish Precambrian crystalline rocks [1ndash3] haveshown that only a small portion of the fractures and joints5ndash15 at a threshold level of 119879 = 10
minus9m2s are pervious andthat the statistical distribution of the transmissivities of theconductive fractures is approximately lognormal
The transmissivity is coupled to the hydraulic aperture ofthe fracture by the cubic law [4 5]
119879 =120588119908
1198921198873
12120583119908
(1)
where 120583119908
is the viscosity 120588119908
is the density of water and 119887 isthe so-called hydraulic aperture of the fractureThe hydraulicaperture determined by the cubic law has shown to be a goodestimate for the grouting aperture [6 7]
2 Journal of Applied Mathematics
Grouting borehole
Penetratinggrout
Fractures of different size and aperture
L
I
Figure 1 Grouting fan and grout penetration Borehole distance 119871grout penetration 119868
Grout Groundwater
Velocity profile
Stiff plug
2Z
120591
120591
pg
g
b
x
pw
I
Figure 2 Grout penetrating a fracture
From this it follows that in a borehole to be grouted onlya few fractures are pervious and only a small number of thesecontribute significantly to the groundwater flow through therock because of the large skewness of the transmissivitydistribution
The normally used cement grouts can reasonably well becharacterised as Bingham fluids [8ndash10] They are thus char-acterised by a yield strength 120591
0
and a plastic viscosity 120583119892
From the Bingham model it follows that flow can only takeplace in the parts of the fluid where the internal shear stressesexceed the yield strength This means that a stiff plug isformed in the centre of the flow channel surrounded byplastic flow zones see Figure 2 The advance of the groutfront ceases when the shear stresses at the walls of the fractureequal the yield strength of the grout A simple force balanceof the difference between the grouting and the resisting waterpressures Δ119901 = 119901
119892
minus 119901119908
and the shear stress gives themaximum grout penetration 119868max for a fracture of aperture119887 (eg [9 11])
119868max =Δ119901 sdot 119887
21205910
(2)
The relevant design question is thus how to make sure thatthe penetration length is long enough to bridge the distancebetween the grouting boreholes for the critical fractures andthe length of time it takes to reach the maximum penetrationor a significant portion of it
In order to obtain an analytical solution the problem hasto be simplified In particular it is assumed that the aperture isconstant not varying along the fractureThe grout propertiesare assumed to be constant in time These limitations shouldbe kept in mind when these analytical solutions are used
2 Derivation of EquationsResults and Discussion
21 Grout Penetration Let 119868(119905) be the position of the groutfront at time 119905 Figure 2 The velocity of grout 119889119868119889119905 movingin a horizontal facture of aperture 119887 can according to Hassler[9] be calculated as
Assuming parallel flow and a viscosity of the grout muchhigher than for water the pressure gradient can be simplifiedto be
119889119901
119889119909= minus
Δ119901
119868 (5)
Equations (4) (5) and (2) give 2119885119887 = 119868119868max The equationfor the relative penetration depth 119868
119863
= 119868119868max becomes from(3) after simplifications
119889119868119863
119889119905=
(1205910
)2
6120583119892
Δ119901sdot2 minus 3119868
119863
+ (119868119863
)3
119868119863
119868119863
=119868
119868max=2119885
119887
(6)
We define the characteristic time 1199050
and the dimensionlesstime 119905
119863
1199050
=
6120583119892
Δ119901
(1205910
)2
119905119863
=119905
1199050
(7)
Equation (6) gives the derivative 119889119868119863
119889119905119863
The derivative of119905119863
as a function of 119868119863
is
119889119905119863
119889119868119863
=119868119863
2 minus 3119868119863
+ (119868119863
)3
=119868119863
(2 + 119868119863
) (1 minus 119868119863
)2
(8)
The right-hand function of 119868119863
is the ratio between twopolynomials which may be expanded in partial fractionsThese are readily integrated We obtain the following explicitequation for the 119905
119863
as a function of 119868119863
119905119863
= 1198651
(119868119863
) 1198651
(119904) =119904
3 (1 minus 119904)+2
9sdot ln [2 (1 minus s)
2 + s]
(9)
It is straightforward to verify that derivative of (9) is given by(8) and that 119868
119863
= 0 for 119905119863
= 0A plot of 119868
119863
= 119868119868max as a function of 119905119863
= 119905(6120583119892
Δ1199011205912
0
)
is shown in Figure 3
Journal of Applied Mathematics 3
0010203040506070809
1
000001 00001 0001 001 01 1 10 100tD = tt0
I D=II m
ax
Figure 3 Relative penetration length as a function of dimensionlesstime in horizontal fracture
From (8) and Figure 3 some interesting observations canbe drawn
(i) The relative penetration is not a function of thefracture aperture 119887 This means that the penetrationprocess has the same time scale for all fractures withdifferent apertures penetrated by a borehole
(ii) The time scale is only a function of the groutingpressureΔ119901 and the grout properties120583
119892
and 1205910
Thusthe parameters are decided by choice of the grouter
(iii) The time scale is determined by 1199050
= 6120583119892
Δ1199011205912
0
sothat at this grouting time about 80 of the possiblepenetration length is reached in all fractures and after51199050
about 95 is reached After that the growth is veryslow and the economy of continued injection could beput in doubt
22 Experimental Verification A series of grouting exper-iments were published by Hakansson [10] He used thinplastic pipes instead of a parallel slot for his experimentsand several constitutive grout flowmodels were tested againstexperimental data As could be expected more complexmodels could give better fit to data but the Bingham modelgave adequate results especially in the light of its simplicity
The velocity of grout moving in a pipe of radius 1199030
Inserting (5) and (10) observing that 119889119909119889119905 = 119889119868119889119905 andusing the relative penetration depth 119868
119863119901
= 119868119868max119901 give aftersimplifications
119889119868119863119901
119889119905=
(1205910
)2
6120583119892
Δ119901sdot
3 minus 4119868119863119901
+ (119868119863119901
)4
119868119863119901
119868119863119901
=119868
119868max119901
(12)
Inserting 119905119863
= 119905(6120583119892
Δ1199011205912
0
) the previous equation gives thederivative 119889119868
119863119903
119889119905119863
The derivative of 119905119863
as a function of 119868119863119901
is
119889119905119863
119889119868119863119901
=
119868119863119901
3 minus 4119868119863119901
+ (119868119863119901
)4
=
119868119863119901
[1 minus 119868119863119901
]2
[3 + 2119868119863119901
+ (119868119863119901
)2
]
(121015840
)
This equation may with some difficulty be integrated Weobtain the following explicit equation for the 119905
119863
as a functionof 119868119863119901
119905119863
= 119865119901
(119868119863119901
)
119865119901
(119904) =119904
6 (1 minus 119904)+1
36sdot ln[ 3(1 minus s)2
3 + 2119904 + 1199042]
minus5radic2
36sdot arctan( 119904radic2
119904 + 3)
(13)
A long but straightforward calculation shows that the deriva-tive satisfies (12) It is easy to see that 119905
119863
= 0 for 119868119863119901
= 119904 = 0In Hakansson [10] two grouting experiments in 3 and
4mm pipes are reported In Table 1 the relevant parametersfor the experiments are shown based on the reported data InFigure 4 a direct comparison between the function 119868
119863119901
(119905119863
)
and experimental data is shownThe experimental data follow the theoretical function
extremely well up to a value of 119905119863
asymp 2 It shall also be borne inmind that the grout properties were taken directly from lab-oratory tests and no curve fitting was made Hakansson [10]who assumed them to be a result from differences betweenlaboratory values and experiment conditions also identifiedthe differences at the end of the curves As predicted the119868119863119903
minus 119905119863
-curves are almost identical for the two experimentsAnother striking fact is that more than 90 of the predictedpenetration is reached for 119905
119863
asymp 2
4 Journal of Applied Mathematics
0
02
04
06
08
1
12
000001 00001 0001 001 01 1 10 100
Pipe solution4 mm pipe3 mm pipe
tD = tt0
I D=II m
axr
Figure 4 Comparison of grout penetration function in a pipe withexperimental data from Hakansson [10]
Grout
Borehole
Fracture plane
pw
Q pg
I
r
Figure 5 Radial penetration of grout in a fracture
23 Penetration in a Two-Dimensional Fracture A morerealistic model of a fracture to grout is perhaps a pseudo-planewith a system of conductive areas and flow channels [5]If the transmissivity of the fracture is reasonably constant aparallel platemodelwith constant aperture b can approximateit If it is grouted through a borehole there will be a radialtwo-dimensional flow of grout out from the borehole seeFigure 5 In reality however the flow will as for flow of waterfrom a borehole be something in between a system of one-dimensional channels and radial flow [12]
Equations (3) and (4) give the grout flow in the plane caseThe grout flow velocity is constant (in 119909) and equal to thefront velocity 119889119868119889119905 In the radial case we replace 119909 by 119903 Thegrout flow velocity V
119892
(ms) decreases as 1119903 [16] Let 119903119887
bethe radius of the injection borehole and let 119903
119887
+119868 be the radiusof the grout injection front at any particular time 119905 We have
The injection excess pressure is Δ119901 We have the boundarycondition
119901 (119903119887
) minus 119901 (119903119887
+ 119868) = Δ119901 (19)
Here we neglect a pressure fall in the ground water since theviscosity of grout is much larger than that of water
The solution 119901(119903) of (18)-(19) has the front position 119868
as parameter The value of 119876 has to be adjusted so that thepressure difference Δ119901 is obtained in accordance with (19)The front position 119868 = 119868(119905) increases with time The flowvelocity at the grout front 119903 = 119903
119887
+ 119868(119905) is equal to the timederivative of 119868(119905) We have from (16)
This equation determines the motion of the grout front Itdepends on the required grout injection rate 119876(119868) which isobtained from the solution of (18)-(19) for each front position119868
The solution for radial grout flow is much more compli-cated than for the plain case and the pipe case We must firstsolve the implicit differential equation for 119901(119903) This involvesthe solution of a cubic equation in order to get the derivative119889119901119889119903 and an intricate integration in order to get 119901(119903) Fromthe solution we get the required grout flux for any frontposition 119868
With known function 119876(119868) we may determine themotion of the grout front from (20) by integration
The front position 119868 increases from zero at 119905 = 0 to amaximum value for infinite time Then the flux 119876 must bezero Equation (18) gives 119876 = 0 for 119904 = 1 Then we have alinear pressure variation
We get the same value (2) as in the plain caseThe complete solution in the radial case involves the
following constants
119868max =119887Δ119901
21205910
120574 =119868max119903119887
=119887Δ119901
2119903119887
1205910
1199050
=
6120583119892
Δ119901
(1205910
)2
1198760
=6120587119887(119868max)
2
1199050
=1205871198873
Δ119901
4120583119892
(23)
24 Solution for the Pressure In the dimensionless solutionfor the pressure we use the borehole radius as scaling length
1199031015840
=119903
119903119887
1198681015840
=119868
119903119887
119903119887
le 119903 le 119903119887
+ 119868 lArrrArr 1 le 1199031015840
le 1 + 1198681015840
(24)
The pressure is scaled by Δ119901120574 The variable 119904 for the deriv-ative of the pressure in (18) becomes
1199011015840
=120574 sdot (119901 minus 119901
119908
)
Δ119901997904rArr 119904 =
119887
21205910
sdot (minus119889119901
119889119903)
= minus119887Δ119901120574
21205910
119903119887
sdot1198891199011015840
1198891199031015840= minus
1198891199011015840
1198891199031015840
(25)
The dimensionless form of (18)-(19) becomes after somerecalculations
1199031015840
= 1198761015840
sdot 119892 (minus1198891199011015840
1198891199031015840) 119892 (119904) =
31199042
21199043 minus 31199042 + 1
1198761015840
=
2120583119892
119876
12058711988721205910
119903119887
1199011015840
(1) minus 1199011015840
(1 + 1198681015840
) = 120574
1 le 1199031015840
le 1 + 1198681015840
(26)
This is the basic equation to solve for the pressure distribu-tion It is to be solved for 0 lt 1198681015840 lt 120574 for positive values of theparameter 120574
The solution is derived in detail in [14] A brief derivationis presented in the appendix The dimensionless pressure isgiven by
1199011015840
(1199031015840
) = 120574 minus 1198761015840
sdot [119866 (1198761015840
) minus 119866(1198761015840
1199031015840)] 1 le 119903
1015840
le 1 + 1198681015840
(27)
The composite function 119866(119902) which is used for 119902 = 1198761015840 and
The function 119904(119902) is the root to the cubic equation 119902 sdot 119892(119904) = 1for 119904 gt 1 The function 119866(119904) is an integral of 119904 sdot 119889119892119889119904
The value of the factor1198761015840 has to be chosen so that the totalpressure difference corresponds to the injection pressure(26) This gives
120574 = 1198761015840
sdot [119866 (1198761015840
) minus 119866(1198761015840
1 + 1198681015840)] (29)
This equation determines 1198761015840 as a function of 1198681015840 and 120574
1198761015840
= 1198911015840
(1198681015840
120574) 0 le 1198681015840
le 120574 120574 gt 0 (30)
The value of1198761015840 for 1198681015840 = 120574 is zero in accordance with (21)-(22)1198911015840
(120574 120574) = 0
25 Motion of Grout Front In the dimensionless formulationof the equation for the motion of the grout front we use 119868maxas scaling length We also use 119876
0
and 1199050
from (23)
119868119863
=119868
119868max 119868
1015840
= 120574119868119863
119876119863
=119876
1198760
119905119863
=119905
1199050
(31)
The grout flux becomes from (23) and (26)
1198760
120574=1205871198872
1205910
119903119887
2120583119892
997904rArr 119876 =1198760
120574sdot 1198911015840
(1198681015840
120574) = 1198760
sdot 119876119863
(119868119863
120574)
(32)
The dimensionless grout flux is then
119876119863
(119868119863
120574) =1198911015840
(120574119868119863
120574)
120574 0 le 119868
119863
le 1 (33)
The dimensionless equation for the front motion is now from(32) (20) (31) and (23)
1198760
120574sdot 1198911015840
(120574119868119863
120574) = 2120587119887 sdot(119868max)
2
1199050
sdot (1
120574+ 119868119863
) sdot119889119868119863
119889119905119863
or 119889119905119863
119889119868119863
=120574
3sdot1120574 + 119868
119863
119891 (120574119868119863
120574)
(34)
6 Journal of Applied Mathematics
tD = tt0
I D=II m
ax
0010203040506070809
1
0000001 00001 001 1 100
120574 = 20
120574 = 50
120574 = 100
120574 = 200
120574 = 500
120574 = 1000
Figure 6 Grout penetration function 119868119863
= 119868119863
(119905119863
120574) for radial flow
By integration we get the time 119905119863
= 1199051199050
as an integral in 119868119863
119905119863
=1
3sdot int
119868119863
0
1 + 1205741198681015840
119863
119891 (1205741198681015840
119863
120574)1198891198681015840
119863
0 le 119868119863
lt 1 (35)
We get 119905119863
as a function of the grout front position 119868119863
Also in this case the inverse function describes the relativepenetration as a function of the dimensionless grouting timeFigure 6 shows this relation for a few 120574-values
A comparison of Figures 3 4 and 6 shows that the curvesfor 119868119863
(119905119863
) are similar for the three flow cases The main dif-ference to parallel flow is that penetration is somewhat slowerfor the radial case Around 80 of maximum penetrationis reached after 3119905
0
and to reach 90 takes about 71199050
Theprinciple is however the same and the curves could be usedin the same way
26 Injected Volume of Grout The injected volume of groutas a function of time is of interest The volume is
119881119892
(119905) = 120587119887 [(119903119887
+ 119868 (119905))2
minus (119903119887
)2
]
= 120587119887119868(119905)2
sdot [1 +2119903119887
119868 (119905)]
(36)
Let 119881119892max be maximum injection volume and 119881
119863
the dimen-sionless volume of injected grout
119881119863
=
119881119892
119881119892max
119881119892max = 120587119887(119868max)
2
sdot [1 +2
120574] (36
1015840
)
Then we get using (31) (24) (23) and the relation (35)between 119868
119863
and 119905119863
119881119863
(119905119863
120574) = (119868119863
)2
sdot1 + 2 (120574119868
119863
)
1 + 2120574 119868119863
= 119868119863
(119905119863
120574) (37)
Equations presented in this paper have been used inGustafson and Stille [15] when considering stop criteria forgrouting Grouting projects where estimates of penetration
length have been made are for example [13 15 16] Penetra-tion length has also been a key to presenting a concept forestimation of deformation and stiffness of fractures based ongrouting data [13] In addition to grouting of tunnels theorieshave also been applied for grouting of dams [18]
3 Conclusions
The theoretical investigation of grout spread in one-dimensional conduits and radial spread in plane parallel frac-tures have shown very similar behavior for all the investigatedcases The penetration 119868 can be described as a product ofthe maximum penetration 119868max = Δ119901 sdot 120591
0
2119887 and a time-dependent scaling factor 119868
119863
(119905119863
) the relative penetrationlength HereΔ119901 is the driving pressure 120591
0
is the yield strengthof the grout and 119887 is the aperture of the penetrated fractureThe time factor or dimensionless grouting time 119905
119863
= 1199051199050
is the ratio between the actual grouting time 119905 and a timescaling factor 119905
0
= 6120583119892
Δ1199011205910
2 the characteristic groutingtime Here 120583
119892
is the Bingham viscosity of the grout Therelative penetration depth has a value of 70ndash90 for 119905 = 119905
0
and reaches a value of more than 90 for 119905 gt 71199050
for allfractures
From this a number of important conclusions can bedrawn
(i) The relative penetration is the same in all fracturesthat a grouted borehole cuts This means that giventhe same grout and pressure the grouting time shouldbe the same in high and low yielding boreholes inorder to get the same degree of tightening of allfractures This means that the tendency in practice togrout for a shorter time in tight boreholes will givepoor results for sealing of fine fractures
(ii) Themaximumpenetration is governed by the fractureaperture and pressure and yield strength of the groutThe latter are at the choice of the grouter
(iii) The relative penetration which governs much of thefinal result is determined by the grouting time
(iv) The pressure and the grout properties determine thedesired grouting time These are the choice of thegrouter alone
(v) It is poor economy to grout for a longer time thanabout 5119905
0
since the growth of the penetration is veryslow for a time longer than that On the other handif the borehole takes significant amounts of groutafter 5119905
0
there is reason to stop since it indicatesan unrestricted outflow of grout somewhere in thesystem
The significance of this for grouting design is as follows
(i) The conventional stop criteria based on volume orgrout flow can be replaced by a minimum time cri-terion based only on the parameters that the groutercan chose that is grouting pressure and yield strengthof the grout
Journal of Applied Mathematics 7
(ii) Based on an assessment of how fine fractures itis necessary to seal a maximum effective boreholedistance can be predicted given the pressure and theproperties of the grout
(iii) The time needed for effective grouting operations canbe estimated with better accuracy
(iv) In order to avoid unrestricted grout pumping also amaximum grouting time can be given where furtherinjection of grout will be unnecessary
Appendix
Derivation of the Solution for the Pressure
We seek the solution 1199011015840(1199031015840) to (26)
1199031015840
= 1198761015840
sdot 119892 (minus1198891199011015840
1198891199031015840) 1 le 119903
1015840
le 1 + 1198681015840
119892 (119904) =31199042
21199043 minus 31199042 + 1 0 le 119868
1015840
le 120574
(A1)
Here 1 + 1198681015840 is the position of the grout front The parameter120574 is positive Taking zero pressure at the grout front theboundary conditions for the dimensionless pressure become
1199011015840
(1) = 120574 1199011015840
(1 + 1198681015840
) = 0 (A2)
The dimensionless grout flux 1198761015840 is to be chosen so that theprevious boundary conditions are fulfilled The value of 1198761015840will depend on the front position 1198681015840
Solution in Parameter Form In order to see more directly thecharacter of the equation we make the following change ofnotation
119909 larrrarr 1199031015840
119910 larrrarr minus1199011015840
119891 (119904) = 1198761015840
sdot 119892 (119904) (A3)
The equation is then of the following type
119909 = 119891(119889119910
119889119909) (A4)
There is a general solution in a certain parameter form tothis type of implicit ordinary differential equation [19] Thesolution is
119909 (119904) = 119891 (119904) 119910 (119904) = 119904 sdot 119891 (119904) minus int
119904
119891 (1199041015840
) 1198891199041015840
(A5)
We have to show that this is indeed the solution We have
Theboundary condition (A2) at 1199031015840 = 1 is fulfilled for a certainchoice of 119870 The explicit solution is
1199011015840
(1199031015840
) = 120574 minus 1198761015840
sdot [119866 (1198761015840
) minus 119866(1198761015840
1199031015840)] 1 le 119903
1015840
le 1 + 1198681015840
(A13)
The other boundary condition (A2) at 1199031015840 = 1 + 1198681015840 is fulfilledwhen 1198761015840 satisfies the equation
120574 = 1198761015840
sdot [119866 (1198761015840
) minus 119866(1198761015840
1 + 1198681015840)] (A14)
We note that the derivative minus11988911990110158401198891199031015840 is given by 119904
119904 = minus1198891199011015840
1198891199031015840 (A15)
The pressure derivative is equal to ndash1 for zero flux (21) and(25) in the final stagnant position 119868
1015840
= 120574 The magnitudeof this derivative is larger than 1 for all preceding positions1198681015840
lt 120574 This means that 119904 is larger than (or equal to) 1 in thesolution
The Function 119866(119904) The solution (A13) and the compositefunction (A12) involve the function 119866(119904) defined in (A10)
8 Journal of Applied Mathematics
and (A1) The integral of 119892(119904) is obtained from an expansionin partial fractions We have
119892 (119904) =31199042
21199043 minus 31199042 + 1=
31199042
(2119904 + 1) (119904 minus 1)2
=1
3sdot
1
2119904 + 1+4
3sdot
1
119904 minus 1+
1
(119904 minus 1)2
(A16)
The integral of 119892(119904) is readily determined The function 119866(119904)becomes from (A10) and (A16)
119866 (119904) =1
6sdot ln (2119904 + 1) + 4
3sdot ln (119904 minus 1) minus 1
119904 minus 1
minus31199043
21199043 minus 31199042 + 1 119904 gt 1
(A17)
We will use the function for 1 lt 119904 lt infin
The Inverse 119904(119902) The inverse (A9) is for any 119902 ge 0 thesolution of the cubic equation
21199043
minus 31199042
+ 1 = 31199021199042
(A18)
The solution is reported in detail in [14] The cubic equationhas three real-valued solutions for positive 119902-values one ofwhich is larger than 1 (for 119902 = 0 there is a double root 119904 = 1
and a third root 119904 = minus05 (A16)) We need the solution 119904 gt 1It is given by
119904 (119902) =1
2radic1 + 119902 sdot sin (13) sdot arcsin [(1 + 119902)minus15] 119902 ge 0
(A19)
A plot shows that 119904(119902) is an increasing function from 119904(0) = 1
for 119902 ge 0 It has the asymptote 15 sdot (1 + 119902) for large 119902We will show that (A19) is the inverse We use the
notations
119904 (119902) = 119904 =1
2radic1 + 119902 sdot sin (1206013)
120601 = arcsin [(1 + 119902)minus15] (A20)
In (A18) we put 31199021199042 on the left-hand side divide by 1199043 andinsert 119904 = 119904(119902) from (A20) Then we have
On the third line we use a well-known trigonometric formularelating sin(1206013) to sin(120601) We have shown that (A19) is theinverse
Symbols and Units
119887 (m) Fracture aperture119868 (m) Penetration length of injected grout119868max (m) Maximum penetration length of grout119868max119901 (m) Maximum penetration length of grout in a
pipe1198681015840 (mdash) Ratio between penetration and borehole
(mdash) Relative penetration length in a pipe119871 (m) Distance between grouting boreholes119901 (Pa) Pressure119901119863
(mdash) Dimensionless pressure119901119892
(Pa) Grout pressure119901119908
(Pa) Water pressure119876 (m3s) Grout injection flow rate119903 (m) Pipe radius radial distance from borehole
centre119903119887
(m) Borehole radius119903119863
(mdash) Dimensionless radius119903119901
(m) Grout plug radius1199030
(m) Pipe radius1199031015840 (mdash) Ratio between distance from borehole centre
and borehole radius119879 (m2s) Transmissivity119905 (s) Grouting time1199050
(s) Characteristic grouting time119905119863
(mdash) Dimensionless grouting time119881119892
(m3) Injected volume of grout119881max (m
3) Maximum grout volume in a fracture119881119863
(mdash) Dimensionless grout volume119909 (m) Length coordinate119885 (mdash) Bingham half-plug thickness120574 (mdash) Ratio between maximum penetration and
borehole radiusΔ119901 (Pa) Driving pressure for grout120583119892
(Pas) Plastic viscosity of grout120583119908
(Pas) Viscosity of water120588119908
(kgm3) Density of water1205910
(Pa) Yield strength of grout
Acknowledgments
The authors would like to acknowledge the effort of GunnarGustafson who deceased during the study
References
[1] J D Osnes A Winberg and J E Andersson ldquoAnalysis of welltest datamdashapplication of probabilistic models to infer hydraulicproperties of fracturesrdquo Topical Report RSI-0338 RESPECRapid City Dakota 1988
Journal of Applied Mathematics 9
[2] C L Axelsson E K Jonsson J Geier and W DershowitzldquoDiscrete facture modellingrdquo SKB Progress Report 25-89-21SKB Stockholm Sweden 1990
[3] A Fransson ldquoNonparametric method for transmissivity distri-butions along boreholesrdquo Ground Water vol 40 no 2 pp 201ndash204 2002
[4] D T Snow ldquoThe frequency and apertures of fractures in rockrdquoInternational Journal of RockMechanics andMining Sciences andGeomechanics Abstracts vol 7 no 1 pp 23ndash40 1970
[5] R W Zimmerman and G S Bodvarsson ldquoHydraulic conduc-tivity of rock fracturesrdquo Transport in Porous Media vol 23 no1 pp 1ndash30 1996
[6] A Fransson ldquoCharacterisation of a fractured rock mass for agrouting field testrdquo Tunnelling and Underground Space Technol-ogy vol 16 no 4 pp 331ndash339 2001
[7] M Eriksson ldquoGrouting field experiment at the Aspo Hard RockLaboratoryrdquoTunnelling andUnderground Space Technology vol17 no 3 pp 287ndash293 2002
[8] M Wallner Propagation of Sedimentation Stable Cement Pastesin Jointed Rock University of Aachen Rock Mechanics andWaterways Construction BRD 1976
[9] L Hassler Grouting of rockmdashsimulation and classification [the-sis] Department of Soil and RockMechanics KTH StockholmSweden 1991
[10] U Hakansson Rheology of fresh cement-based grouts [thesis]Department of Soil and Rock Mechanics KTH StockholmSweden 1993
[11] G Gustafson and H Stille ldquoPrediction of groutability fromgrout properties and hydrogeological datardquo Tunnelling andUnderground Space Technology vol 11 no 3 pp 325ndash332 1996
[12] J A Barker ldquoA generalized radial flowmodel for hydraulic testsin fractured rockrdquoWater Resources Research vol 24 no 10 pp1796ndash1804 1988
[13] A Fransson C-F Tsang J Rutqvist and G Gustafson ldquoEsti-mation of deformation and stiffness of fractures close to tunnelsusing data from single-hole hydraulic testing and groutingrdquoInternational Journal of Rock Mechanics and Mining Sciencesvol 47 no 6 pp 887ndash893 2010
[14] J Claesson ldquoRadial bingham flow background report of adetailed solutionrdquo Tech Rep Department of Building Tech-nology Chalmers University of Technology Goteborg Sweden2003
[15] G Gustafson and H Stille ldquoStop criteria for cement groutingrdquoFelsbau vol 23 no 3 pp 62ndash68 2005
[16] L Hernqvist A Fransson G Gustafson A Emmelin MEriksson and H Stille ldquoAnalyses of the grouting results for asection of the APSE tunnel at Aspo Hard Rock LaboratoryrdquoInternational Journal of Rock Mechanics and Mining Sciencesvol 46 no 3 pp 439ndash449 2009
[17] C Butron G Gustafson A Fransson and J Funehag ldquoDripsealing of tunnels in hard rock a new concept for the designand evaluation of permeation groutingrdquo Tunnelling and Under-ground Space Technology vol 25 no 2 pp 114ndash121 2010
[18] H Stille G Gustafson and L Hassler ldquoApplication of newtheories and technology for grouting of dams and foundationson rockrdquoGeotechnical and Geological Engineering vol 30 no 3pp 603ndash624 2012
Figure 1 Grouting fan and grout penetration Borehole distance 119871grout penetration 119868
Grout Groundwater
Velocity profile
Stiff plug
2Z
120591
120591
pg
g
b
x
pw
I
Figure 2 Grout penetrating a fracture
From this it follows that in a borehole to be grouted onlya few fractures are pervious and only a small number of thesecontribute significantly to the groundwater flow through therock because of the large skewness of the transmissivitydistribution
The normally used cement grouts can reasonably well becharacterised as Bingham fluids [8ndash10] They are thus char-acterised by a yield strength 120591
0
and a plastic viscosity 120583119892
From the Bingham model it follows that flow can only takeplace in the parts of the fluid where the internal shear stressesexceed the yield strength This means that a stiff plug isformed in the centre of the flow channel surrounded byplastic flow zones see Figure 2 The advance of the groutfront ceases when the shear stresses at the walls of the fractureequal the yield strength of the grout A simple force balanceof the difference between the grouting and the resisting waterpressures Δ119901 = 119901
119892
minus 119901119908
and the shear stress gives themaximum grout penetration 119868max for a fracture of aperture119887 (eg [9 11])
119868max =Δ119901 sdot 119887
21205910
(2)
The relevant design question is thus how to make sure thatthe penetration length is long enough to bridge the distancebetween the grouting boreholes for the critical fractures andthe length of time it takes to reach the maximum penetrationor a significant portion of it
In order to obtain an analytical solution the problem hasto be simplified In particular it is assumed that the aperture isconstant not varying along the fractureThe grout propertiesare assumed to be constant in time These limitations shouldbe kept in mind when these analytical solutions are used
2 Derivation of EquationsResults and Discussion
21 Grout Penetration Let 119868(119905) be the position of the groutfront at time 119905 Figure 2 The velocity of grout 119889119868119889119905 movingin a horizontal facture of aperture 119887 can according to Hassler[9] be calculated as
Assuming parallel flow and a viscosity of the grout muchhigher than for water the pressure gradient can be simplifiedto be
119889119901
119889119909= minus
Δ119901
119868 (5)
Equations (4) (5) and (2) give 2119885119887 = 119868119868max The equationfor the relative penetration depth 119868
119863
= 119868119868max becomes from(3) after simplifications
119889119868119863
119889119905=
(1205910
)2
6120583119892
Δ119901sdot2 minus 3119868
119863
+ (119868119863
)3
119868119863
119868119863
=119868
119868max=2119885
119887
(6)
We define the characteristic time 1199050
and the dimensionlesstime 119905
119863
1199050
=
6120583119892
Δ119901
(1205910
)2
119905119863
=119905
1199050
(7)
Equation (6) gives the derivative 119889119868119863
119889119905119863
The derivative of119905119863
as a function of 119868119863
is
119889119905119863
119889119868119863
=119868119863
2 minus 3119868119863
+ (119868119863
)3
=119868119863
(2 + 119868119863
) (1 minus 119868119863
)2
(8)
The right-hand function of 119868119863
is the ratio between twopolynomials which may be expanded in partial fractionsThese are readily integrated We obtain the following explicitequation for the 119905
119863
as a function of 119868119863
119905119863
= 1198651
(119868119863
) 1198651
(119904) =119904
3 (1 minus 119904)+2
9sdot ln [2 (1 minus s)
2 + s]
(9)
It is straightforward to verify that derivative of (9) is given by(8) and that 119868
119863
= 0 for 119905119863
= 0A plot of 119868
119863
= 119868119868max as a function of 119905119863
= 119905(6120583119892
Δ1199011205912
0
)
is shown in Figure 3
Journal of Applied Mathematics 3
0010203040506070809
1
000001 00001 0001 001 01 1 10 100tD = tt0
I D=II m
ax
Figure 3 Relative penetration length as a function of dimensionlesstime in horizontal fracture
From (8) and Figure 3 some interesting observations canbe drawn
(i) The relative penetration is not a function of thefracture aperture 119887 This means that the penetrationprocess has the same time scale for all fractures withdifferent apertures penetrated by a borehole
(ii) The time scale is only a function of the groutingpressureΔ119901 and the grout properties120583
119892
and 1205910
Thusthe parameters are decided by choice of the grouter
(iii) The time scale is determined by 1199050
= 6120583119892
Δ1199011205912
0
sothat at this grouting time about 80 of the possiblepenetration length is reached in all fractures and after51199050
about 95 is reached After that the growth is veryslow and the economy of continued injection could beput in doubt
22 Experimental Verification A series of grouting exper-iments were published by Hakansson [10] He used thinplastic pipes instead of a parallel slot for his experimentsand several constitutive grout flowmodels were tested againstexperimental data As could be expected more complexmodels could give better fit to data but the Bingham modelgave adequate results especially in the light of its simplicity
The velocity of grout moving in a pipe of radius 1199030
Inserting (5) and (10) observing that 119889119909119889119905 = 119889119868119889119905 andusing the relative penetration depth 119868
119863119901
= 119868119868max119901 give aftersimplifications
119889119868119863119901
119889119905=
(1205910
)2
6120583119892
Δ119901sdot
3 minus 4119868119863119901
+ (119868119863119901
)4
119868119863119901
119868119863119901
=119868
119868max119901
(12)
Inserting 119905119863
= 119905(6120583119892
Δ1199011205912
0
) the previous equation gives thederivative 119889119868
119863119903
119889119905119863
The derivative of 119905119863
as a function of 119868119863119901
is
119889119905119863
119889119868119863119901
=
119868119863119901
3 minus 4119868119863119901
+ (119868119863119901
)4
=
119868119863119901
[1 minus 119868119863119901
]2
[3 + 2119868119863119901
+ (119868119863119901
)2
]
(121015840
)
This equation may with some difficulty be integrated Weobtain the following explicit equation for the 119905
119863
as a functionof 119868119863119901
119905119863
= 119865119901
(119868119863119901
)
119865119901
(119904) =119904
6 (1 minus 119904)+1
36sdot ln[ 3(1 minus s)2
3 + 2119904 + 1199042]
minus5radic2
36sdot arctan( 119904radic2
119904 + 3)
(13)
A long but straightforward calculation shows that the deriva-tive satisfies (12) It is easy to see that 119905
119863
= 0 for 119868119863119901
= 119904 = 0In Hakansson [10] two grouting experiments in 3 and
4mm pipes are reported In Table 1 the relevant parametersfor the experiments are shown based on the reported data InFigure 4 a direct comparison between the function 119868
119863119901
(119905119863
)
and experimental data is shownThe experimental data follow the theoretical function
extremely well up to a value of 119905119863
asymp 2 It shall also be borne inmind that the grout properties were taken directly from lab-oratory tests and no curve fitting was made Hakansson [10]who assumed them to be a result from differences betweenlaboratory values and experiment conditions also identifiedthe differences at the end of the curves As predicted the119868119863119903
minus 119905119863
-curves are almost identical for the two experimentsAnother striking fact is that more than 90 of the predictedpenetration is reached for 119905
119863
asymp 2
4 Journal of Applied Mathematics
0
02
04
06
08
1
12
000001 00001 0001 001 01 1 10 100
Pipe solution4 mm pipe3 mm pipe
tD = tt0
I D=II m
axr
Figure 4 Comparison of grout penetration function in a pipe withexperimental data from Hakansson [10]
Grout
Borehole
Fracture plane
pw
Q pg
I
r
Figure 5 Radial penetration of grout in a fracture
23 Penetration in a Two-Dimensional Fracture A morerealistic model of a fracture to grout is perhaps a pseudo-planewith a system of conductive areas and flow channels [5]If the transmissivity of the fracture is reasonably constant aparallel platemodelwith constant aperture b can approximateit If it is grouted through a borehole there will be a radialtwo-dimensional flow of grout out from the borehole seeFigure 5 In reality however the flow will as for flow of waterfrom a borehole be something in between a system of one-dimensional channels and radial flow [12]
Equations (3) and (4) give the grout flow in the plane caseThe grout flow velocity is constant (in 119909) and equal to thefront velocity 119889119868119889119905 In the radial case we replace 119909 by 119903 Thegrout flow velocity V
119892
(ms) decreases as 1119903 [16] Let 119903119887
bethe radius of the injection borehole and let 119903
119887
+119868 be the radiusof the grout injection front at any particular time 119905 We have
The injection excess pressure is Δ119901 We have the boundarycondition
119901 (119903119887
) minus 119901 (119903119887
+ 119868) = Δ119901 (19)
Here we neglect a pressure fall in the ground water since theviscosity of grout is much larger than that of water
The solution 119901(119903) of (18)-(19) has the front position 119868
as parameter The value of 119876 has to be adjusted so that thepressure difference Δ119901 is obtained in accordance with (19)The front position 119868 = 119868(119905) increases with time The flowvelocity at the grout front 119903 = 119903
119887
+ 119868(119905) is equal to the timederivative of 119868(119905) We have from (16)
This equation determines the motion of the grout front Itdepends on the required grout injection rate 119876(119868) which isobtained from the solution of (18)-(19) for each front position119868
The solution for radial grout flow is much more compli-cated than for the plain case and the pipe case We must firstsolve the implicit differential equation for 119901(119903) This involvesthe solution of a cubic equation in order to get the derivative119889119901119889119903 and an intricate integration in order to get 119901(119903) Fromthe solution we get the required grout flux for any frontposition 119868
With known function 119876(119868) we may determine themotion of the grout front from (20) by integration
The front position 119868 increases from zero at 119905 = 0 to amaximum value for infinite time Then the flux 119876 must bezero Equation (18) gives 119876 = 0 for 119904 = 1 Then we have alinear pressure variation
We get the same value (2) as in the plain caseThe complete solution in the radial case involves the
following constants
119868max =119887Δ119901
21205910
120574 =119868max119903119887
=119887Δ119901
2119903119887
1205910
1199050
=
6120583119892
Δ119901
(1205910
)2
1198760
=6120587119887(119868max)
2
1199050
=1205871198873
Δ119901
4120583119892
(23)
24 Solution for the Pressure In the dimensionless solutionfor the pressure we use the borehole radius as scaling length
1199031015840
=119903
119903119887
1198681015840
=119868
119903119887
119903119887
le 119903 le 119903119887
+ 119868 lArrrArr 1 le 1199031015840
le 1 + 1198681015840
(24)
The pressure is scaled by Δ119901120574 The variable 119904 for the deriv-ative of the pressure in (18) becomes
1199011015840
=120574 sdot (119901 minus 119901
119908
)
Δ119901997904rArr 119904 =
119887
21205910
sdot (minus119889119901
119889119903)
= minus119887Δ119901120574
21205910
119903119887
sdot1198891199011015840
1198891199031015840= minus
1198891199011015840
1198891199031015840
(25)
The dimensionless form of (18)-(19) becomes after somerecalculations
1199031015840
= 1198761015840
sdot 119892 (minus1198891199011015840
1198891199031015840) 119892 (119904) =
31199042
21199043 minus 31199042 + 1
1198761015840
=
2120583119892
119876
12058711988721205910
119903119887
1199011015840
(1) minus 1199011015840
(1 + 1198681015840
) = 120574
1 le 1199031015840
le 1 + 1198681015840
(26)
This is the basic equation to solve for the pressure distribu-tion It is to be solved for 0 lt 1198681015840 lt 120574 for positive values of theparameter 120574
The solution is derived in detail in [14] A brief derivationis presented in the appendix The dimensionless pressure isgiven by
1199011015840
(1199031015840
) = 120574 minus 1198761015840
sdot [119866 (1198761015840
) minus 119866(1198761015840
1199031015840)] 1 le 119903
1015840
le 1 + 1198681015840
(27)
The composite function 119866(119902) which is used for 119902 = 1198761015840 and
The function 119904(119902) is the root to the cubic equation 119902 sdot 119892(119904) = 1for 119904 gt 1 The function 119866(119904) is an integral of 119904 sdot 119889119892119889119904
The value of the factor1198761015840 has to be chosen so that the totalpressure difference corresponds to the injection pressure(26) This gives
120574 = 1198761015840
sdot [119866 (1198761015840
) minus 119866(1198761015840
1 + 1198681015840)] (29)
This equation determines 1198761015840 as a function of 1198681015840 and 120574
1198761015840
= 1198911015840
(1198681015840
120574) 0 le 1198681015840
le 120574 120574 gt 0 (30)
The value of1198761015840 for 1198681015840 = 120574 is zero in accordance with (21)-(22)1198911015840
(120574 120574) = 0
25 Motion of Grout Front In the dimensionless formulationof the equation for the motion of the grout front we use 119868maxas scaling length We also use 119876
0
and 1199050
from (23)
119868119863
=119868
119868max 119868
1015840
= 120574119868119863
119876119863
=119876
1198760
119905119863
=119905
1199050
(31)
The grout flux becomes from (23) and (26)
1198760
120574=1205871198872
1205910
119903119887
2120583119892
997904rArr 119876 =1198760
120574sdot 1198911015840
(1198681015840
120574) = 1198760
sdot 119876119863
(119868119863
120574)
(32)
The dimensionless grout flux is then
119876119863
(119868119863
120574) =1198911015840
(120574119868119863
120574)
120574 0 le 119868
119863
le 1 (33)
The dimensionless equation for the front motion is now from(32) (20) (31) and (23)
1198760
120574sdot 1198911015840
(120574119868119863
120574) = 2120587119887 sdot(119868max)
2
1199050
sdot (1
120574+ 119868119863
) sdot119889119868119863
119889119905119863
or 119889119905119863
119889119868119863
=120574
3sdot1120574 + 119868
119863
119891 (120574119868119863
120574)
(34)
6 Journal of Applied Mathematics
tD = tt0
I D=II m
ax
0010203040506070809
1
0000001 00001 001 1 100
120574 = 20
120574 = 50
120574 = 100
120574 = 200
120574 = 500
120574 = 1000
Figure 6 Grout penetration function 119868119863
= 119868119863
(119905119863
120574) for radial flow
By integration we get the time 119905119863
= 1199051199050
as an integral in 119868119863
119905119863
=1
3sdot int
119868119863
0
1 + 1205741198681015840
119863
119891 (1205741198681015840
119863
120574)1198891198681015840
119863
0 le 119868119863
lt 1 (35)
We get 119905119863
as a function of the grout front position 119868119863
Also in this case the inverse function describes the relativepenetration as a function of the dimensionless grouting timeFigure 6 shows this relation for a few 120574-values
A comparison of Figures 3 4 and 6 shows that the curvesfor 119868119863
(119905119863
) are similar for the three flow cases The main dif-ference to parallel flow is that penetration is somewhat slowerfor the radial case Around 80 of maximum penetrationis reached after 3119905
0
and to reach 90 takes about 71199050
Theprinciple is however the same and the curves could be usedin the same way
26 Injected Volume of Grout The injected volume of groutas a function of time is of interest The volume is
119881119892
(119905) = 120587119887 [(119903119887
+ 119868 (119905))2
minus (119903119887
)2
]
= 120587119887119868(119905)2
sdot [1 +2119903119887
119868 (119905)]
(36)
Let 119881119892max be maximum injection volume and 119881
119863
the dimen-sionless volume of injected grout
119881119863
=
119881119892
119881119892max
119881119892max = 120587119887(119868max)
2
sdot [1 +2
120574] (36
1015840
)
Then we get using (31) (24) (23) and the relation (35)between 119868
119863
and 119905119863
119881119863
(119905119863
120574) = (119868119863
)2
sdot1 + 2 (120574119868
119863
)
1 + 2120574 119868119863
= 119868119863
(119905119863
120574) (37)
Equations presented in this paper have been used inGustafson and Stille [15] when considering stop criteria forgrouting Grouting projects where estimates of penetration
length have been made are for example [13 15 16] Penetra-tion length has also been a key to presenting a concept forestimation of deformation and stiffness of fractures based ongrouting data [13] In addition to grouting of tunnels theorieshave also been applied for grouting of dams [18]
3 Conclusions
The theoretical investigation of grout spread in one-dimensional conduits and radial spread in plane parallel frac-tures have shown very similar behavior for all the investigatedcases The penetration 119868 can be described as a product ofthe maximum penetration 119868max = Δ119901 sdot 120591
0
2119887 and a time-dependent scaling factor 119868
119863
(119905119863
) the relative penetrationlength HereΔ119901 is the driving pressure 120591
0
is the yield strengthof the grout and 119887 is the aperture of the penetrated fractureThe time factor or dimensionless grouting time 119905
119863
= 1199051199050
is the ratio between the actual grouting time 119905 and a timescaling factor 119905
0
= 6120583119892
Δ1199011205910
2 the characteristic groutingtime Here 120583
119892
is the Bingham viscosity of the grout Therelative penetration depth has a value of 70ndash90 for 119905 = 119905
0
and reaches a value of more than 90 for 119905 gt 71199050
for allfractures
From this a number of important conclusions can bedrawn
(i) The relative penetration is the same in all fracturesthat a grouted borehole cuts This means that giventhe same grout and pressure the grouting time shouldbe the same in high and low yielding boreholes inorder to get the same degree of tightening of allfractures This means that the tendency in practice togrout for a shorter time in tight boreholes will givepoor results for sealing of fine fractures
(ii) Themaximumpenetration is governed by the fractureaperture and pressure and yield strength of the groutThe latter are at the choice of the grouter
(iii) The relative penetration which governs much of thefinal result is determined by the grouting time
(iv) The pressure and the grout properties determine thedesired grouting time These are the choice of thegrouter alone
(v) It is poor economy to grout for a longer time thanabout 5119905
0
since the growth of the penetration is veryslow for a time longer than that On the other handif the borehole takes significant amounts of groutafter 5119905
0
there is reason to stop since it indicatesan unrestricted outflow of grout somewhere in thesystem
The significance of this for grouting design is as follows
(i) The conventional stop criteria based on volume orgrout flow can be replaced by a minimum time cri-terion based only on the parameters that the groutercan chose that is grouting pressure and yield strengthof the grout
Journal of Applied Mathematics 7
(ii) Based on an assessment of how fine fractures itis necessary to seal a maximum effective boreholedistance can be predicted given the pressure and theproperties of the grout
(iii) The time needed for effective grouting operations canbe estimated with better accuracy
(iv) In order to avoid unrestricted grout pumping also amaximum grouting time can be given where furtherinjection of grout will be unnecessary
Appendix
Derivation of the Solution for the Pressure
We seek the solution 1199011015840(1199031015840) to (26)
1199031015840
= 1198761015840
sdot 119892 (minus1198891199011015840
1198891199031015840) 1 le 119903
1015840
le 1 + 1198681015840
119892 (119904) =31199042
21199043 minus 31199042 + 1 0 le 119868
1015840
le 120574
(A1)
Here 1 + 1198681015840 is the position of the grout front The parameter120574 is positive Taking zero pressure at the grout front theboundary conditions for the dimensionless pressure become
1199011015840
(1) = 120574 1199011015840
(1 + 1198681015840
) = 0 (A2)
The dimensionless grout flux 1198761015840 is to be chosen so that theprevious boundary conditions are fulfilled The value of 1198761015840will depend on the front position 1198681015840
Solution in Parameter Form In order to see more directly thecharacter of the equation we make the following change ofnotation
119909 larrrarr 1199031015840
119910 larrrarr minus1199011015840
119891 (119904) = 1198761015840
sdot 119892 (119904) (A3)
The equation is then of the following type
119909 = 119891(119889119910
119889119909) (A4)
There is a general solution in a certain parameter form tothis type of implicit ordinary differential equation [19] Thesolution is
119909 (119904) = 119891 (119904) 119910 (119904) = 119904 sdot 119891 (119904) minus int
119904
119891 (1199041015840
) 1198891199041015840
(A5)
We have to show that this is indeed the solution We have
Theboundary condition (A2) at 1199031015840 = 1 is fulfilled for a certainchoice of 119870 The explicit solution is
1199011015840
(1199031015840
) = 120574 minus 1198761015840
sdot [119866 (1198761015840
) minus 119866(1198761015840
1199031015840)] 1 le 119903
1015840
le 1 + 1198681015840
(A13)
The other boundary condition (A2) at 1199031015840 = 1 + 1198681015840 is fulfilledwhen 1198761015840 satisfies the equation
120574 = 1198761015840
sdot [119866 (1198761015840
) minus 119866(1198761015840
1 + 1198681015840)] (A14)
We note that the derivative minus11988911990110158401198891199031015840 is given by 119904
119904 = minus1198891199011015840
1198891199031015840 (A15)
The pressure derivative is equal to ndash1 for zero flux (21) and(25) in the final stagnant position 119868
1015840
= 120574 The magnitudeof this derivative is larger than 1 for all preceding positions1198681015840
lt 120574 This means that 119904 is larger than (or equal to) 1 in thesolution
The Function 119866(119904) The solution (A13) and the compositefunction (A12) involve the function 119866(119904) defined in (A10)
8 Journal of Applied Mathematics
and (A1) The integral of 119892(119904) is obtained from an expansionin partial fractions We have
119892 (119904) =31199042
21199043 minus 31199042 + 1=
31199042
(2119904 + 1) (119904 minus 1)2
=1
3sdot
1
2119904 + 1+4
3sdot
1
119904 minus 1+
1
(119904 minus 1)2
(A16)
The integral of 119892(119904) is readily determined The function 119866(119904)becomes from (A10) and (A16)
119866 (119904) =1
6sdot ln (2119904 + 1) + 4
3sdot ln (119904 minus 1) minus 1
119904 minus 1
minus31199043
21199043 minus 31199042 + 1 119904 gt 1
(A17)
We will use the function for 1 lt 119904 lt infin
The Inverse 119904(119902) The inverse (A9) is for any 119902 ge 0 thesolution of the cubic equation
21199043
minus 31199042
+ 1 = 31199021199042
(A18)
The solution is reported in detail in [14] The cubic equationhas three real-valued solutions for positive 119902-values one ofwhich is larger than 1 (for 119902 = 0 there is a double root 119904 = 1
and a third root 119904 = minus05 (A16)) We need the solution 119904 gt 1It is given by
119904 (119902) =1
2radic1 + 119902 sdot sin (13) sdot arcsin [(1 + 119902)minus15] 119902 ge 0
(A19)
A plot shows that 119904(119902) is an increasing function from 119904(0) = 1
for 119902 ge 0 It has the asymptote 15 sdot (1 + 119902) for large 119902We will show that (A19) is the inverse We use the
notations
119904 (119902) = 119904 =1
2radic1 + 119902 sdot sin (1206013)
120601 = arcsin [(1 + 119902)minus15] (A20)
In (A18) we put 31199021199042 on the left-hand side divide by 1199043 andinsert 119904 = 119904(119902) from (A20) Then we have
On the third line we use a well-known trigonometric formularelating sin(1206013) to sin(120601) We have shown that (A19) is theinverse
Symbols and Units
119887 (m) Fracture aperture119868 (m) Penetration length of injected grout119868max (m) Maximum penetration length of grout119868max119901 (m) Maximum penetration length of grout in a
pipe1198681015840 (mdash) Ratio between penetration and borehole
(mdash) Relative penetration length in a pipe119871 (m) Distance between grouting boreholes119901 (Pa) Pressure119901119863
(mdash) Dimensionless pressure119901119892
(Pa) Grout pressure119901119908
(Pa) Water pressure119876 (m3s) Grout injection flow rate119903 (m) Pipe radius radial distance from borehole
centre119903119887
(m) Borehole radius119903119863
(mdash) Dimensionless radius119903119901
(m) Grout plug radius1199030
(m) Pipe radius1199031015840 (mdash) Ratio between distance from borehole centre
and borehole radius119879 (m2s) Transmissivity119905 (s) Grouting time1199050
(s) Characteristic grouting time119905119863
(mdash) Dimensionless grouting time119881119892
(m3) Injected volume of grout119881max (m
3) Maximum grout volume in a fracture119881119863
(mdash) Dimensionless grout volume119909 (m) Length coordinate119885 (mdash) Bingham half-plug thickness120574 (mdash) Ratio between maximum penetration and
borehole radiusΔ119901 (Pa) Driving pressure for grout120583119892
(Pas) Plastic viscosity of grout120583119908
(Pas) Viscosity of water120588119908
(kgm3) Density of water1205910
(Pa) Yield strength of grout
Acknowledgments
The authors would like to acknowledge the effort of GunnarGustafson who deceased during the study
References
[1] J D Osnes A Winberg and J E Andersson ldquoAnalysis of welltest datamdashapplication of probabilistic models to infer hydraulicproperties of fracturesrdquo Topical Report RSI-0338 RESPECRapid City Dakota 1988
Journal of Applied Mathematics 9
[2] C L Axelsson E K Jonsson J Geier and W DershowitzldquoDiscrete facture modellingrdquo SKB Progress Report 25-89-21SKB Stockholm Sweden 1990
[3] A Fransson ldquoNonparametric method for transmissivity distri-butions along boreholesrdquo Ground Water vol 40 no 2 pp 201ndash204 2002
[4] D T Snow ldquoThe frequency and apertures of fractures in rockrdquoInternational Journal of RockMechanics andMining Sciences andGeomechanics Abstracts vol 7 no 1 pp 23ndash40 1970
[5] R W Zimmerman and G S Bodvarsson ldquoHydraulic conduc-tivity of rock fracturesrdquo Transport in Porous Media vol 23 no1 pp 1ndash30 1996
[6] A Fransson ldquoCharacterisation of a fractured rock mass for agrouting field testrdquo Tunnelling and Underground Space Technol-ogy vol 16 no 4 pp 331ndash339 2001
[7] M Eriksson ldquoGrouting field experiment at the Aspo Hard RockLaboratoryrdquoTunnelling andUnderground Space Technology vol17 no 3 pp 287ndash293 2002
[8] M Wallner Propagation of Sedimentation Stable Cement Pastesin Jointed Rock University of Aachen Rock Mechanics andWaterways Construction BRD 1976
[9] L Hassler Grouting of rockmdashsimulation and classification [the-sis] Department of Soil and RockMechanics KTH StockholmSweden 1991
[10] U Hakansson Rheology of fresh cement-based grouts [thesis]Department of Soil and Rock Mechanics KTH StockholmSweden 1993
[11] G Gustafson and H Stille ldquoPrediction of groutability fromgrout properties and hydrogeological datardquo Tunnelling andUnderground Space Technology vol 11 no 3 pp 325ndash332 1996
[12] J A Barker ldquoA generalized radial flowmodel for hydraulic testsin fractured rockrdquoWater Resources Research vol 24 no 10 pp1796ndash1804 1988
[13] A Fransson C-F Tsang J Rutqvist and G Gustafson ldquoEsti-mation of deformation and stiffness of fractures close to tunnelsusing data from single-hole hydraulic testing and groutingrdquoInternational Journal of Rock Mechanics and Mining Sciencesvol 47 no 6 pp 887ndash893 2010
[14] J Claesson ldquoRadial bingham flow background report of adetailed solutionrdquo Tech Rep Department of Building Tech-nology Chalmers University of Technology Goteborg Sweden2003
[15] G Gustafson and H Stille ldquoStop criteria for cement groutingrdquoFelsbau vol 23 no 3 pp 62ndash68 2005
[16] L Hernqvist A Fransson G Gustafson A Emmelin MEriksson and H Stille ldquoAnalyses of the grouting results for asection of the APSE tunnel at Aspo Hard Rock LaboratoryrdquoInternational Journal of Rock Mechanics and Mining Sciencesvol 46 no 3 pp 439ndash449 2009
[17] C Butron G Gustafson A Fransson and J Funehag ldquoDripsealing of tunnels in hard rock a new concept for the designand evaluation of permeation groutingrdquo Tunnelling and Under-ground Space Technology vol 25 no 2 pp 114ndash121 2010
[18] H Stille G Gustafson and L Hassler ldquoApplication of newtheories and technology for grouting of dams and foundationson rockrdquoGeotechnical and Geological Engineering vol 30 no 3pp 603ndash624 2012
Figure 3 Relative penetration length as a function of dimensionlesstime in horizontal fracture
From (8) and Figure 3 some interesting observations canbe drawn
(i) The relative penetration is not a function of thefracture aperture 119887 This means that the penetrationprocess has the same time scale for all fractures withdifferent apertures penetrated by a borehole
(ii) The time scale is only a function of the groutingpressureΔ119901 and the grout properties120583
119892
and 1205910
Thusthe parameters are decided by choice of the grouter
(iii) The time scale is determined by 1199050
= 6120583119892
Δ1199011205912
0
sothat at this grouting time about 80 of the possiblepenetration length is reached in all fractures and after51199050
about 95 is reached After that the growth is veryslow and the economy of continued injection could beput in doubt
22 Experimental Verification A series of grouting exper-iments were published by Hakansson [10] He used thinplastic pipes instead of a parallel slot for his experimentsand several constitutive grout flowmodels were tested againstexperimental data As could be expected more complexmodels could give better fit to data but the Bingham modelgave adequate results especially in the light of its simplicity
The velocity of grout moving in a pipe of radius 1199030
Inserting (5) and (10) observing that 119889119909119889119905 = 119889119868119889119905 andusing the relative penetration depth 119868
119863119901
= 119868119868max119901 give aftersimplifications
119889119868119863119901
119889119905=
(1205910
)2
6120583119892
Δ119901sdot
3 minus 4119868119863119901
+ (119868119863119901
)4
119868119863119901
119868119863119901
=119868
119868max119901
(12)
Inserting 119905119863
= 119905(6120583119892
Δ1199011205912
0
) the previous equation gives thederivative 119889119868
119863119903
119889119905119863
The derivative of 119905119863
as a function of 119868119863119901
is
119889119905119863
119889119868119863119901
=
119868119863119901
3 minus 4119868119863119901
+ (119868119863119901
)4
=
119868119863119901
[1 minus 119868119863119901
]2
[3 + 2119868119863119901
+ (119868119863119901
)2
]
(121015840
)
This equation may with some difficulty be integrated Weobtain the following explicit equation for the 119905
119863
as a functionof 119868119863119901
119905119863
= 119865119901
(119868119863119901
)
119865119901
(119904) =119904
6 (1 minus 119904)+1
36sdot ln[ 3(1 minus s)2
3 + 2119904 + 1199042]
minus5radic2
36sdot arctan( 119904radic2
119904 + 3)
(13)
A long but straightforward calculation shows that the deriva-tive satisfies (12) It is easy to see that 119905
119863
= 0 for 119868119863119901
= 119904 = 0In Hakansson [10] two grouting experiments in 3 and
4mm pipes are reported In Table 1 the relevant parametersfor the experiments are shown based on the reported data InFigure 4 a direct comparison between the function 119868
119863119901
(119905119863
)
and experimental data is shownThe experimental data follow the theoretical function
extremely well up to a value of 119905119863
asymp 2 It shall also be borne inmind that the grout properties were taken directly from lab-oratory tests and no curve fitting was made Hakansson [10]who assumed them to be a result from differences betweenlaboratory values and experiment conditions also identifiedthe differences at the end of the curves As predicted the119868119863119903
minus 119905119863
-curves are almost identical for the two experimentsAnother striking fact is that more than 90 of the predictedpenetration is reached for 119905
119863
asymp 2
4 Journal of Applied Mathematics
0
02
04
06
08
1
12
000001 00001 0001 001 01 1 10 100
Pipe solution4 mm pipe3 mm pipe
tD = tt0
I D=II m
axr
Figure 4 Comparison of grout penetration function in a pipe withexperimental data from Hakansson [10]
Grout
Borehole
Fracture plane
pw
Q pg
I
r
Figure 5 Radial penetration of grout in a fracture
23 Penetration in a Two-Dimensional Fracture A morerealistic model of a fracture to grout is perhaps a pseudo-planewith a system of conductive areas and flow channels [5]If the transmissivity of the fracture is reasonably constant aparallel platemodelwith constant aperture b can approximateit If it is grouted through a borehole there will be a radialtwo-dimensional flow of grout out from the borehole seeFigure 5 In reality however the flow will as for flow of waterfrom a borehole be something in between a system of one-dimensional channels and radial flow [12]
Equations (3) and (4) give the grout flow in the plane caseThe grout flow velocity is constant (in 119909) and equal to thefront velocity 119889119868119889119905 In the radial case we replace 119909 by 119903 Thegrout flow velocity V
119892
(ms) decreases as 1119903 [16] Let 119903119887
bethe radius of the injection borehole and let 119903
119887
+119868 be the radiusof the grout injection front at any particular time 119905 We have
The injection excess pressure is Δ119901 We have the boundarycondition
119901 (119903119887
) minus 119901 (119903119887
+ 119868) = Δ119901 (19)
Here we neglect a pressure fall in the ground water since theviscosity of grout is much larger than that of water
The solution 119901(119903) of (18)-(19) has the front position 119868
as parameter The value of 119876 has to be adjusted so that thepressure difference Δ119901 is obtained in accordance with (19)The front position 119868 = 119868(119905) increases with time The flowvelocity at the grout front 119903 = 119903
119887
+ 119868(119905) is equal to the timederivative of 119868(119905) We have from (16)
This equation determines the motion of the grout front Itdepends on the required grout injection rate 119876(119868) which isobtained from the solution of (18)-(19) for each front position119868
The solution for radial grout flow is much more compli-cated than for the plain case and the pipe case We must firstsolve the implicit differential equation for 119901(119903) This involvesthe solution of a cubic equation in order to get the derivative119889119901119889119903 and an intricate integration in order to get 119901(119903) Fromthe solution we get the required grout flux for any frontposition 119868
With known function 119876(119868) we may determine themotion of the grout front from (20) by integration
The front position 119868 increases from zero at 119905 = 0 to amaximum value for infinite time Then the flux 119876 must bezero Equation (18) gives 119876 = 0 for 119904 = 1 Then we have alinear pressure variation
We get the same value (2) as in the plain caseThe complete solution in the radial case involves the
following constants
119868max =119887Δ119901
21205910
120574 =119868max119903119887
=119887Δ119901
2119903119887
1205910
1199050
=
6120583119892
Δ119901
(1205910
)2
1198760
=6120587119887(119868max)
2
1199050
=1205871198873
Δ119901
4120583119892
(23)
24 Solution for the Pressure In the dimensionless solutionfor the pressure we use the borehole radius as scaling length
1199031015840
=119903
119903119887
1198681015840
=119868
119903119887
119903119887
le 119903 le 119903119887
+ 119868 lArrrArr 1 le 1199031015840
le 1 + 1198681015840
(24)
The pressure is scaled by Δ119901120574 The variable 119904 for the deriv-ative of the pressure in (18) becomes
1199011015840
=120574 sdot (119901 minus 119901
119908
)
Δ119901997904rArr 119904 =
119887
21205910
sdot (minus119889119901
119889119903)
= minus119887Δ119901120574
21205910
119903119887
sdot1198891199011015840
1198891199031015840= minus
1198891199011015840
1198891199031015840
(25)
The dimensionless form of (18)-(19) becomes after somerecalculations
1199031015840
= 1198761015840
sdot 119892 (minus1198891199011015840
1198891199031015840) 119892 (119904) =
31199042
21199043 minus 31199042 + 1
1198761015840
=
2120583119892
119876
12058711988721205910
119903119887
1199011015840
(1) minus 1199011015840
(1 + 1198681015840
) = 120574
1 le 1199031015840
le 1 + 1198681015840
(26)
This is the basic equation to solve for the pressure distribu-tion It is to be solved for 0 lt 1198681015840 lt 120574 for positive values of theparameter 120574
The solution is derived in detail in [14] A brief derivationis presented in the appendix The dimensionless pressure isgiven by
1199011015840
(1199031015840
) = 120574 minus 1198761015840
sdot [119866 (1198761015840
) minus 119866(1198761015840
1199031015840)] 1 le 119903
1015840
le 1 + 1198681015840
(27)
The composite function 119866(119902) which is used for 119902 = 1198761015840 and
The function 119904(119902) is the root to the cubic equation 119902 sdot 119892(119904) = 1for 119904 gt 1 The function 119866(119904) is an integral of 119904 sdot 119889119892119889119904
The value of the factor1198761015840 has to be chosen so that the totalpressure difference corresponds to the injection pressure(26) This gives
120574 = 1198761015840
sdot [119866 (1198761015840
) minus 119866(1198761015840
1 + 1198681015840)] (29)
This equation determines 1198761015840 as a function of 1198681015840 and 120574
1198761015840
= 1198911015840
(1198681015840
120574) 0 le 1198681015840
le 120574 120574 gt 0 (30)
The value of1198761015840 for 1198681015840 = 120574 is zero in accordance with (21)-(22)1198911015840
(120574 120574) = 0
25 Motion of Grout Front In the dimensionless formulationof the equation for the motion of the grout front we use 119868maxas scaling length We also use 119876
0
and 1199050
from (23)
119868119863
=119868
119868max 119868
1015840
= 120574119868119863
119876119863
=119876
1198760
119905119863
=119905
1199050
(31)
The grout flux becomes from (23) and (26)
1198760
120574=1205871198872
1205910
119903119887
2120583119892
997904rArr 119876 =1198760
120574sdot 1198911015840
(1198681015840
120574) = 1198760
sdot 119876119863
(119868119863
120574)
(32)
The dimensionless grout flux is then
119876119863
(119868119863
120574) =1198911015840
(120574119868119863
120574)
120574 0 le 119868
119863
le 1 (33)
The dimensionless equation for the front motion is now from(32) (20) (31) and (23)
1198760
120574sdot 1198911015840
(120574119868119863
120574) = 2120587119887 sdot(119868max)
2
1199050
sdot (1
120574+ 119868119863
) sdot119889119868119863
119889119905119863
or 119889119905119863
119889119868119863
=120574
3sdot1120574 + 119868
119863
119891 (120574119868119863
120574)
(34)
6 Journal of Applied Mathematics
tD = tt0
I D=II m
ax
0010203040506070809
1
0000001 00001 001 1 100
120574 = 20
120574 = 50
120574 = 100
120574 = 200
120574 = 500
120574 = 1000
Figure 6 Grout penetration function 119868119863
= 119868119863
(119905119863
120574) for radial flow
By integration we get the time 119905119863
= 1199051199050
as an integral in 119868119863
119905119863
=1
3sdot int
119868119863
0
1 + 1205741198681015840
119863
119891 (1205741198681015840
119863
120574)1198891198681015840
119863
0 le 119868119863
lt 1 (35)
We get 119905119863
as a function of the grout front position 119868119863
Also in this case the inverse function describes the relativepenetration as a function of the dimensionless grouting timeFigure 6 shows this relation for a few 120574-values
A comparison of Figures 3 4 and 6 shows that the curvesfor 119868119863
(119905119863
) are similar for the three flow cases The main dif-ference to parallel flow is that penetration is somewhat slowerfor the radial case Around 80 of maximum penetrationis reached after 3119905
0
and to reach 90 takes about 71199050
Theprinciple is however the same and the curves could be usedin the same way
26 Injected Volume of Grout The injected volume of groutas a function of time is of interest The volume is
119881119892
(119905) = 120587119887 [(119903119887
+ 119868 (119905))2
minus (119903119887
)2
]
= 120587119887119868(119905)2
sdot [1 +2119903119887
119868 (119905)]
(36)
Let 119881119892max be maximum injection volume and 119881
119863
the dimen-sionless volume of injected grout
119881119863
=
119881119892
119881119892max
119881119892max = 120587119887(119868max)
2
sdot [1 +2
120574] (36
1015840
)
Then we get using (31) (24) (23) and the relation (35)between 119868
119863
and 119905119863
119881119863
(119905119863
120574) = (119868119863
)2
sdot1 + 2 (120574119868
119863
)
1 + 2120574 119868119863
= 119868119863
(119905119863
120574) (37)
Equations presented in this paper have been used inGustafson and Stille [15] when considering stop criteria forgrouting Grouting projects where estimates of penetration
length have been made are for example [13 15 16] Penetra-tion length has also been a key to presenting a concept forestimation of deformation and stiffness of fractures based ongrouting data [13] In addition to grouting of tunnels theorieshave also been applied for grouting of dams [18]
3 Conclusions
The theoretical investigation of grout spread in one-dimensional conduits and radial spread in plane parallel frac-tures have shown very similar behavior for all the investigatedcases The penetration 119868 can be described as a product ofthe maximum penetration 119868max = Δ119901 sdot 120591
0
2119887 and a time-dependent scaling factor 119868
119863
(119905119863
) the relative penetrationlength HereΔ119901 is the driving pressure 120591
0
is the yield strengthof the grout and 119887 is the aperture of the penetrated fractureThe time factor or dimensionless grouting time 119905
119863
= 1199051199050
is the ratio between the actual grouting time 119905 and a timescaling factor 119905
0
= 6120583119892
Δ1199011205910
2 the characteristic groutingtime Here 120583
119892
is the Bingham viscosity of the grout Therelative penetration depth has a value of 70ndash90 for 119905 = 119905
0
and reaches a value of more than 90 for 119905 gt 71199050
for allfractures
From this a number of important conclusions can bedrawn
(i) The relative penetration is the same in all fracturesthat a grouted borehole cuts This means that giventhe same grout and pressure the grouting time shouldbe the same in high and low yielding boreholes inorder to get the same degree of tightening of allfractures This means that the tendency in practice togrout for a shorter time in tight boreholes will givepoor results for sealing of fine fractures
(ii) Themaximumpenetration is governed by the fractureaperture and pressure and yield strength of the groutThe latter are at the choice of the grouter
(iii) The relative penetration which governs much of thefinal result is determined by the grouting time
(iv) The pressure and the grout properties determine thedesired grouting time These are the choice of thegrouter alone
(v) It is poor economy to grout for a longer time thanabout 5119905
0
since the growth of the penetration is veryslow for a time longer than that On the other handif the borehole takes significant amounts of groutafter 5119905
0
there is reason to stop since it indicatesan unrestricted outflow of grout somewhere in thesystem
The significance of this for grouting design is as follows
(i) The conventional stop criteria based on volume orgrout flow can be replaced by a minimum time cri-terion based only on the parameters that the groutercan chose that is grouting pressure and yield strengthof the grout
Journal of Applied Mathematics 7
(ii) Based on an assessment of how fine fractures itis necessary to seal a maximum effective boreholedistance can be predicted given the pressure and theproperties of the grout
(iii) The time needed for effective grouting operations canbe estimated with better accuracy
(iv) In order to avoid unrestricted grout pumping also amaximum grouting time can be given where furtherinjection of grout will be unnecessary
Appendix
Derivation of the Solution for the Pressure
We seek the solution 1199011015840(1199031015840) to (26)
1199031015840
= 1198761015840
sdot 119892 (minus1198891199011015840
1198891199031015840) 1 le 119903
1015840
le 1 + 1198681015840
119892 (119904) =31199042
21199043 minus 31199042 + 1 0 le 119868
1015840
le 120574
(A1)
Here 1 + 1198681015840 is the position of the grout front The parameter120574 is positive Taking zero pressure at the grout front theboundary conditions for the dimensionless pressure become
1199011015840
(1) = 120574 1199011015840
(1 + 1198681015840
) = 0 (A2)
The dimensionless grout flux 1198761015840 is to be chosen so that theprevious boundary conditions are fulfilled The value of 1198761015840will depend on the front position 1198681015840
Solution in Parameter Form In order to see more directly thecharacter of the equation we make the following change ofnotation
119909 larrrarr 1199031015840
119910 larrrarr minus1199011015840
119891 (119904) = 1198761015840
sdot 119892 (119904) (A3)
The equation is then of the following type
119909 = 119891(119889119910
119889119909) (A4)
There is a general solution in a certain parameter form tothis type of implicit ordinary differential equation [19] Thesolution is
119909 (119904) = 119891 (119904) 119910 (119904) = 119904 sdot 119891 (119904) minus int
119904
119891 (1199041015840
) 1198891199041015840
(A5)
We have to show that this is indeed the solution We have
Theboundary condition (A2) at 1199031015840 = 1 is fulfilled for a certainchoice of 119870 The explicit solution is
1199011015840
(1199031015840
) = 120574 minus 1198761015840
sdot [119866 (1198761015840
) minus 119866(1198761015840
1199031015840)] 1 le 119903
1015840
le 1 + 1198681015840
(A13)
The other boundary condition (A2) at 1199031015840 = 1 + 1198681015840 is fulfilledwhen 1198761015840 satisfies the equation
120574 = 1198761015840
sdot [119866 (1198761015840
) minus 119866(1198761015840
1 + 1198681015840)] (A14)
We note that the derivative minus11988911990110158401198891199031015840 is given by 119904
119904 = minus1198891199011015840
1198891199031015840 (A15)
The pressure derivative is equal to ndash1 for zero flux (21) and(25) in the final stagnant position 119868
1015840
= 120574 The magnitudeof this derivative is larger than 1 for all preceding positions1198681015840
lt 120574 This means that 119904 is larger than (or equal to) 1 in thesolution
The Function 119866(119904) The solution (A13) and the compositefunction (A12) involve the function 119866(119904) defined in (A10)
8 Journal of Applied Mathematics
and (A1) The integral of 119892(119904) is obtained from an expansionin partial fractions We have
119892 (119904) =31199042
21199043 minus 31199042 + 1=
31199042
(2119904 + 1) (119904 minus 1)2
=1
3sdot
1
2119904 + 1+4
3sdot
1
119904 minus 1+
1
(119904 minus 1)2
(A16)
The integral of 119892(119904) is readily determined The function 119866(119904)becomes from (A10) and (A16)
119866 (119904) =1
6sdot ln (2119904 + 1) + 4
3sdot ln (119904 minus 1) minus 1
119904 minus 1
minus31199043
21199043 minus 31199042 + 1 119904 gt 1
(A17)
We will use the function for 1 lt 119904 lt infin
The Inverse 119904(119902) The inverse (A9) is for any 119902 ge 0 thesolution of the cubic equation
21199043
minus 31199042
+ 1 = 31199021199042
(A18)
The solution is reported in detail in [14] The cubic equationhas three real-valued solutions for positive 119902-values one ofwhich is larger than 1 (for 119902 = 0 there is a double root 119904 = 1
and a third root 119904 = minus05 (A16)) We need the solution 119904 gt 1It is given by
119904 (119902) =1
2radic1 + 119902 sdot sin (13) sdot arcsin [(1 + 119902)minus15] 119902 ge 0
(A19)
A plot shows that 119904(119902) is an increasing function from 119904(0) = 1
for 119902 ge 0 It has the asymptote 15 sdot (1 + 119902) for large 119902We will show that (A19) is the inverse We use the
notations
119904 (119902) = 119904 =1
2radic1 + 119902 sdot sin (1206013)
120601 = arcsin [(1 + 119902)minus15] (A20)
In (A18) we put 31199021199042 on the left-hand side divide by 1199043 andinsert 119904 = 119904(119902) from (A20) Then we have
On the third line we use a well-known trigonometric formularelating sin(1206013) to sin(120601) We have shown that (A19) is theinverse
Symbols and Units
119887 (m) Fracture aperture119868 (m) Penetration length of injected grout119868max (m) Maximum penetration length of grout119868max119901 (m) Maximum penetration length of grout in a
pipe1198681015840 (mdash) Ratio between penetration and borehole
(mdash) Relative penetration length in a pipe119871 (m) Distance between grouting boreholes119901 (Pa) Pressure119901119863
(mdash) Dimensionless pressure119901119892
(Pa) Grout pressure119901119908
(Pa) Water pressure119876 (m3s) Grout injection flow rate119903 (m) Pipe radius radial distance from borehole
centre119903119887
(m) Borehole radius119903119863
(mdash) Dimensionless radius119903119901
(m) Grout plug radius1199030
(m) Pipe radius1199031015840 (mdash) Ratio between distance from borehole centre
and borehole radius119879 (m2s) Transmissivity119905 (s) Grouting time1199050
(s) Characteristic grouting time119905119863
(mdash) Dimensionless grouting time119881119892
(m3) Injected volume of grout119881max (m
3) Maximum grout volume in a fracture119881119863
(mdash) Dimensionless grout volume119909 (m) Length coordinate119885 (mdash) Bingham half-plug thickness120574 (mdash) Ratio between maximum penetration and
borehole radiusΔ119901 (Pa) Driving pressure for grout120583119892
(Pas) Plastic viscosity of grout120583119908
(Pas) Viscosity of water120588119908
(kgm3) Density of water1205910
(Pa) Yield strength of grout
Acknowledgments
The authors would like to acknowledge the effort of GunnarGustafson who deceased during the study
References
[1] J D Osnes A Winberg and J E Andersson ldquoAnalysis of welltest datamdashapplication of probabilistic models to infer hydraulicproperties of fracturesrdquo Topical Report RSI-0338 RESPECRapid City Dakota 1988
Journal of Applied Mathematics 9
[2] C L Axelsson E K Jonsson J Geier and W DershowitzldquoDiscrete facture modellingrdquo SKB Progress Report 25-89-21SKB Stockholm Sweden 1990
[3] A Fransson ldquoNonparametric method for transmissivity distri-butions along boreholesrdquo Ground Water vol 40 no 2 pp 201ndash204 2002
[4] D T Snow ldquoThe frequency and apertures of fractures in rockrdquoInternational Journal of RockMechanics andMining Sciences andGeomechanics Abstracts vol 7 no 1 pp 23ndash40 1970
[5] R W Zimmerman and G S Bodvarsson ldquoHydraulic conduc-tivity of rock fracturesrdquo Transport in Porous Media vol 23 no1 pp 1ndash30 1996
[6] A Fransson ldquoCharacterisation of a fractured rock mass for agrouting field testrdquo Tunnelling and Underground Space Technol-ogy vol 16 no 4 pp 331ndash339 2001
[7] M Eriksson ldquoGrouting field experiment at the Aspo Hard RockLaboratoryrdquoTunnelling andUnderground Space Technology vol17 no 3 pp 287ndash293 2002
[8] M Wallner Propagation of Sedimentation Stable Cement Pastesin Jointed Rock University of Aachen Rock Mechanics andWaterways Construction BRD 1976
[9] L Hassler Grouting of rockmdashsimulation and classification [the-sis] Department of Soil and RockMechanics KTH StockholmSweden 1991
[10] U Hakansson Rheology of fresh cement-based grouts [thesis]Department of Soil and Rock Mechanics KTH StockholmSweden 1993
[11] G Gustafson and H Stille ldquoPrediction of groutability fromgrout properties and hydrogeological datardquo Tunnelling andUnderground Space Technology vol 11 no 3 pp 325ndash332 1996
[12] J A Barker ldquoA generalized radial flowmodel for hydraulic testsin fractured rockrdquoWater Resources Research vol 24 no 10 pp1796ndash1804 1988
[13] A Fransson C-F Tsang J Rutqvist and G Gustafson ldquoEsti-mation of deformation and stiffness of fractures close to tunnelsusing data from single-hole hydraulic testing and groutingrdquoInternational Journal of Rock Mechanics and Mining Sciencesvol 47 no 6 pp 887ndash893 2010
[14] J Claesson ldquoRadial bingham flow background report of adetailed solutionrdquo Tech Rep Department of Building Tech-nology Chalmers University of Technology Goteborg Sweden2003
[15] G Gustafson and H Stille ldquoStop criteria for cement groutingrdquoFelsbau vol 23 no 3 pp 62ndash68 2005
[16] L Hernqvist A Fransson G Gustafson A Emmelin MEriksson and H Stille ldquoAnalyses of the grouting results for asection of the APSE tunnel at Aspo Hard Rock LaboratoryrdquoInternational Journal of Rock Mechanics and Mining Sciencesvol 46 no 3 pp 439ndash449 2009
[17] C Butron G Gustafson A Fransson and J Funehag ldquoDripsealing of tunnels in hard rock a new concept for the designand evaluation of permeation groutingrdquo Tunnelling and Under-ground Space Technology vol 25 no 2 pp 114ndash121 2010
[18] H Stille G Gustafson and L Hassler ldquoApplication of newtheories and technology for grouting of dams and foundationson rockrdquoGeotechnical and Geological Engineering vol 30 no 3pp 603ndash624 2012
Figure 4 Comparison of grout penetration function in a pipe withexperimental data from Hakansson [10]
Grout
Borehole
Fracture plane
pw
Q pg
I
r
Figure 5 Radial penetration of grout in a fracture
23 Penetration in a Two-Dimensional Fracture A morerealistic model of a fracture to grout is perhaps a pseudo-planewith a system of conductive areas and flow channels [5]If the transmissivity of the fracture is reasonably constant aparallel platemodelwith constant aperture b can approximateit If it is grouted through a borehole there will be a radialtwo-dimensional flow of grout out from the borehole seeFigure 5 In reality however the flow will as for flow of waterfrom a borehole be something in between a system of one-dimensional channels and radial flow [12]
Equations (3) and (4) give the grout flow in the plane caseThe grout flow velocity is constant (in 119909) and equal to thefront velocity 119889119868119889119905 In the radial case we replace 119909 by 119903 Thegrout flow velocity V
119892
(ms) decreases as 1119903 [16] Let 119903119887
bethe radius of the injection borehole and let 119903
119887
+119868 be the radiusof the grout injection front at any particular time 119905 We have
The injection excess pressure is Δ119901 We have the boundarycondition
119901 (119903119887
) minus 119901 (119903119887
+ 119868) = Δ119901 (19)
Here we neglect a pressure fall in the ground water since theviscosity of grout is much larger than that of water
The solution 119901(119903) of (18)-(19) has the front position 119868
as parameter The value of 119876 has to be adjusted so that thepressure difference Δ119901 is obtained in accordance with (19)The front position 119868 = 119868(119905) increases with time The flowvelocity at the grout front 119903 = 119903
119887
+ 119868(119905) is equal to the timederivative of 119868(119905) We have from (16)
This equation determines the motion of the grout front Itdepends on the required grout injection rate 119876(119868) which isobtained from the solution of (18)-(19) for each front position119868
The solution for radial grout flow is much more compli-cated than for the plain case and the pipe case We must firstsolve the implicit differential equation for 119901(119903) This involvesthe solution of a cubic equation in order to get the derivative119889119901119889119903 and an intricate integration in order to get 119901(119903) Fromthe solution we get the required grout flux for any frontposition 119868
With known function 119876(119868) we may determine themotion of the grout front from (20) by integration
The front position 119868 increases from zero at 119905 = 0 to amaximum value for infinite time Then the flux 119876 must bezero Equation (18) gives 119876 = 0 for 119904 = 1 Then we have alinear pressure variation
We get the same value (2) as in the plain caseThe complete solution in the radial case involves the
following constants
119868max =119887Δ119901
21205910
120574 =119868max119903119887
=119887Δ119901
2119903119887
1205910
1199050
=
6120583119892
Δ119901
(1205910
)2
1198760
=6120587119887(119868max)
2
1199050
=1205871198873
Δ119901
4120583119892
(23)
24 Solution for the Pressure In the dimensionless solutionfor the pressure we use the borehole radius as scaling length
1199031015840
=119903
119903119887
1198681015840
=119868
119903119887
119903119887
le 119903 le 119903119887
+ 119868 lArrrArr 1 le 1199031015840
le 1 + 1198681015840
(24)
The pressure is scaled by Δ119901120574 The variable 119904 for the deriv-ative of the pressure in (18) becomes
1199011015840
=120574 sdot (119901 minus 119901
119908
)
Δ119901997904rArr 119904 =
119887
21205910
sdot (minus119889119901
119889119903)
= minus119887Δ119901120574
21205910
119903119887
sdot1198891199011015840
1198891199031015840= minus
1198891199011015840
1198891199031015840
(25)
The dimensionless form of (18)-(19) becomes after somerecalculations
1199031015840
= 1198761015840
sdot 119892 (minus1198891199011015840
1198891199031015840) 119892 (119904) =
31199042
21199043 minus 31199042 + 1
1198761015840
=
2120583119892
119876
12058711988721205910
119903119887
1199011015840
(1) minus 1199011015840
(1 + 1198681015840
) = 120574
1 le 1199031015840
le 1 + 1198681015840
(26)
This is the basic equation to solve for the pressure distribu-tion It is to be solved for 0 lt 1198681015840 lt 120574 for positive values of theparameter 120574
The solution is derived in detail in [14] A brief derivationis presented in the appendix The dimensionless pressure isgiven by
1199011015840
(1199031015840
) = 120574 minus 1198761015840
sdot [119866 (1198761015840
) minus 119866(1198761015840
1199031015840)] 1 le 119903
1015840
le 1 + 1198681015840
(27)
The composite function 119866(119902) which is used for 119902 = 1198761015840 and
The function 119904(119902) is the root to the cubic equation 119902 sdot 119892(119904) = 1for 119904 gt 1 The function 119866(119904) is an integral of 119904 sdot 119889119892119889119904
The value of the factor1198761015840 has to be chosen so that the totalpressure difference corresponds to the injection pressure(26) This gives
120574 = 1198761015840
sdot [119866 (1198761015840
) minus 119866(1198761015840
1 + 1198681015840)] (29)
This equation determines 1198761015840 as a function of 1198681015840 and 120574
1198761015840
= 1198911015840
(1198681015840
120574) 0 le 1198681015840
le 120574 120574 gt 0 (30)
The value of1198761015840 for 1198681015840 = 120574 is zero in accordance with (21)-(22)1198911015840
(120574 120574) = 0
25 Motion of Grout Front In the dimensionless formulationof the equation for the motion of the grout front we use 119868maxas scaling length We also use 119876
0
and 1199050
from (23)
119868119863
=119868
119868max 119868
1015840
= 120574119868119863
119876119863
=119876
1198760
119905119863
=119905
1199050
(31)
The grout flux becomes from (23) and (26)
1198760
120574=1205871198872
1205910
119903119887
2120583119892
997904rArr 119876 =1198760
120574sdot 1198911015840
(1198681015840
120574) = 1198760
sdot 119876119863
(119868119863
120574)
(32)
The dimensionless grout flux is then
119876119863
(119868119863
120574) =1198911015840
(120574119868119863
120574)
120574 0 le 119868
119863
le 1 (33)
The dimensionless equation for the front motion is now from(32) (20) (31) and (23)
1198760
120574sdot 1198911015840
(120574119868119863
120574) = 2120587119887 sdot(119868max)
2
1199050
sdot (1
120574+ 119868119863
) sdot119889119868119863
119889119905119863
or 119889119905119863
119889119868119863
=120574
3sdot1120574 + 119868
119863
119891 (120574119868119863
120574)
(34)
6 Journal of Applied Mathematics
tD = tt0
I D=II m
ax
0010203040506070809
1
0000001 00001 001 1 100
120574 = 20
120574 = 50
120574 = 100
120574 = 200
120574 = 500
120574 = 1000
Figure 6 Grout penetration function 119868119863
= 119868119863
(119905119863
120574) for radial flow
By integration we get the time 119905119863
= 1199051199050
as an integral in 119868119863
119905119863
=1
3sdot int
119868119863
0
1 + 1205741198681015840
119863
119891 (1205741198681015840
119863
120574)1198891198681015840
119863
0 le 119868119863
lt 1 (35)
We get 119905119863
as a function of the grout front position 119868119863
Also in this case the inverse function describes the relativepenetration as a function of the dimensionless grouting timeFigure 6 shows this relation for a few 120574-values
A comparison of Figures 3 4 and 6 shows that the curvesfor 119868119863
(119905119863
) are similar for the three flow cases The main dif-ference to parallel flow is that penetration is somewhat slowerfor the radial case Around 80 of maximum penetrationis reached after 3119905
0
and to reach 90 takes about 71199050
Theprinciple is however the same and the curves could be usedin the same way
26 Injected Volume of Grout The injected volume of groutas a function of time is of interest The volume is
119881119892
(119905) = 120587119887 [(119903119887
+ 119868 (119905))2
minus (119903119887
)2
]
= 120587119887119868(119905)2
sdot [1 +2119903119887
119868 (119905)]
(36)
Let 119881119892max be maximum injection volume and 119881
119863
the dimen-sionless volume of injected grout
119881119863
=
119881119892
119881119892max
119881119892max = 120587119887(119868max)
2
sdot [1 +2
120574] (36
1015840
)
Then we get using (31) (24) (23) and the relation (35)between 119868
119863
and 119905119863
119881119863
(119905119863
120574) = (119868119863
)2
sdot1 + 2 (120574119868
119863
)
1 + 2120574 119868119863
= 119868119863
(119905119863
120574) (37)
Equations presented in this paper have been used inGustafson and Stille [15] when considering stop criteria forgrouting Grouting projects where estimates of penetration
length have been made are for example [13 15 16] Penetra-tion length has also been a key to presenting a concept forestimation of deformation and stiffness of fractures based ongrouting data [13] In addition to grouting of tunnels theorieshave also been applied for grouting of dams [18]
3 Conclusions
The theoretical investigation of grout spread in one-dimensional conduits and radial spread in plane parallel frac-tures have shown very similar behavior for all the investigatedcases The penetration 119868 can be described as a product ofthe maximum penetration 119868max = Δ119901 sdot 120591
0
2119887 and a time-dependent scaling factor 119868
119863
(119905119863
) the relative penetrationlength HereΔ119901 is the driving pressure 120591
0
is the yield strengthof the grout and 119887 is the aperture of the penetrated fractureThe time factor or dimensionless grouting time 119905
119863
= 1199051199050
is the ratio between the actual grouting time 119905 and a timescaling factor 119905
0
= 6120583119892
Δ1199011205910
2 the characteristic groutingtime Here 120583
119892
is the Bingham viscosity of the grout Therelative penetration depth has a value of 70ndash90 for 119905 = 119905
0
and reaches a value of more than 90 for 119905 gt 71199050
for allfractures
From this a number of important conclusions can bedrawn
(i) The relative penetration is the same in all fracturesthat a grouted borehole cuts This means that giventhe same grout and pressure the grouting time shouldbe the same in high and low yielding boreholes inorder to get the same degree of tightening of allfractures This means that the tendency in practice togrout for a shorter time in tight boreholes will givepoor results for sealing of fine fractures
(ii) Themaximumpenetration is governed by the fractureaperture and pressure and yield strength of the groutThe latter are at the choice of the grouter
(iii) The relative penetration which governs much of thefinal result is determined by the grouting time
(iv) The pressure and the grout properties determine thedesired grouting time These are the choice of thegrouter alone
(v) It is poor economy to grout for a longer time thanabout 5119905
0
since the growth of the penetration is veryslow for a time longer than that On the other handif the borehole takes significant amounts of groutafter 5119905
0
there is reason to stop since it indicatesan unrestricted outflow of grout somewhere in thesystem
The significance of this for grouting design is as follows
(i) The conventional stop criteria based on volume orgrout flow can be replaced by a minimum time cri-terion based only on the parameters that the groutercan chose that is grouting pressure and yield strengthof the grout
Journal of Applied Mathematics 7
(ii) Based on an assessment of how fine fractures itis necessary to seal a maximum effective boreholedistance can be predicted given the pressure and theproperties of the grout
(iii) The time needed for effective grouting operations canbe estimated with better accuracy
(iv) In order to avoid unrestricted grout pumping also amaximum grouting time can be given where furtherinjection of grout will be unnecessary
Appendix
Derivation of the Solution for the Pressure
We seek the solution 1199011015840(1199031015840) to (26)
1199031015840
= 1198761015840
sdot 119892 (minus1198891199011015840
1198891199031015840) 1 le 119903
1015840
le 1 + 1198681015840
119892 (119904) =31199042
21199043 minus 31199042 + 1 0 le 119868
1015840
le 120574
(A1)
Here 1 + 1198681015840 is the position of the grout front The parameter120574 is positive Taking zero pressure at the grout front theboundary conditions for the dimensionless pressure become
1199011015840
(1) = 120574 1199011015840
(1 + 1198681015840
) = 0 (A2)
The dimensionless grout flux 1198761015840 is to be chosen so that theprevious boundary conditions are fulfilled The value of 1198761015840will depend on the front position 1198681015840
Solution in Parameter Form In order to see more directly thecharacter of the equation we make the following change ofnotation
119909 larrrarr 1199031015840
119910 larrrarr minus1199011015840
119891 (119904) = 1198761015840
sdot 119892 (119904) (A3)
The equation is then of the following type
119909 = 119891(119889119910
119889119909) (A4)
There is a general solution in a certain parameter form tothis type of implicit ordinary differential equation [19] Thesolution is
119909 (119904) = 119891 (119904) 119910 (119904) = 119904 sdot 119891 (119904) minus int
119904
119891 (1199041015840
) 1198891199041015840
(A5)
We have to show that this is indeed the solution We have
Theboundary condition (A2) at 1199031015840 = 1 is fulfilled for a certainchoice of 119870 The explicit solution is
1199011015840
(1199031015840
) = 120574 minus 1198761015840
sdot [119866 (1198761015840
) minus 119866(1198761015840
1199031015840)] 1 le 119903
1015840
le 1 + 1198681015840
(A13)
The other boundary condition (A2) at 1199031015840 = 1 + 1198681015840 is fulfilledwhen 1198761015840 satisfies the equation
120574 = 1198761015840
sdot [119866 (1198761015840
) minus 119866(1198761015840
1 + 1198681015840)] (A14)
We note that the derivative minus11988911990110158401198891199031015840 is given by 119904
119904 = minus1198891199011015840
1198891199031015840 (A15)
The pressure derivative is equal to ndash1 for zero flux (21) and(25) in the final stagnant position 119868
1015840
= 120574 The magnitudeof this derivative is larger than 1 for all preceding positions1198681015840
lt 120574 This means that 119904 is larger than (or equal to) 1 in thesolution
The Function 119866(119904) The solution (A13) and the compositefunction (A12) involve the function 119866(119904) defined in (A10)
8 Journal of Applied Mathematics
and (A1) The integral of 119892(119904) is obtained from an expansionin partial fractions We have
119892 (119904) =31199042
21199043 minus 31199042 + 1=
31199042
(2119904 + 1) (119904 minus 1)2
=1
3sdot
1
2119904 + 1+4
3sdot
1
119904 minus 1+
1
(119904 minus 1)2
(A16)
The integral of 119892(119904) is readily determined The function 119866(119904)becomes from (A10) and (A16)
119866 (119904) =1
6sdot ln (2119904 + 1) + 4
3sdot ln (119904 minus 1) minus 1
119904 minus 1
minus31199043
21199043 minus 31199042 + 1 119904 gt 1
(A17)
We will use the function for 1 lt 119904 lt infin
The Inverse 119904(119902) The inverse (A9) is for any 119902 ge 0 thesolution of the cubic equation
21199043
minus 31199042
+ 1 = 31199021199042
(A18)
The solution is reported in detail in [14] The cubic equationhas three real-valued solutions for positive 119902-values one ofwhich is larger than 1 (for 119902 = 0 there is a double root 119904 = 1
and a third root 119904 = minus05 (A16)) We need the solution 119904 gt 1It is given by
119904 (119902) =1
2radic1 + 119902 sdot sin (13) sdot arcsin [(1 + 119902)minus15] 119902 ge 0
(A19)
A plot shows that 119904(119902) is an increasing function from 119904(0) = 1
for 119902 ge 0 It has the asymptote 15 sdot (1 + 119902) for large 119902We will show that (A19) is the inverse We use the
notations
119904 (119902) = 119904 =1
2radic1 + 119902 sdot sin (1206013)
120601 = arcsin [(1 + 119902)minus15] (A20)
In (A18) we put 31199021199042 on the left-hand side divide by 1199043 andinsert 119904 = 119904(119902) from (A20) Then we have
On the third line we use a well-known trigonometric formularelating sin(1206013) to sin(120601) We have shown that (A19) is theinverse
Symbols and Units
119887 (m) Fracture aperture119868 (m) Penetration length of injected grout119868max (m) Maximum penetration length of grout119868max119901 (m) Maximum penetration length of grout in a
pipe1198681015840 (mdash) Ratio between penetration and borehole
(mdash) Relative penetration length in a pipe119871 (m) Distance between grouting boreholes119901 (Pa) Pressure119901119863
(mdash) Dimensionless pressure119901119892
(Pa) Grout pressure119901119908
(Pa) Water pressure119876 (m3s) Grout injection flow rate119903 (m) Pipe radius radial distance from borehole
centre119903119887
(m) Borehole radius119903119863
(mdash) Dimensionless radius119903119901
(m) Grout plug radius1199030
(m) Pipe radius1199031015840 (mdash) Ratio between distance from borehole centre
and borehole radius119879 (m2s) Transmissivity119905 (s) Grouting time1199050
(s) Characteristic grouting time119905119863
(mdash) Dimensionless grouting time119881119892
(m3) Injected volume of grout119881max (m
3) Maximum grout volume in a fracture119881119863
(mdash) Dimensionless grout volume119909 (m) Length coordinate119885 (mdash) Bingham half-plug thickness120574 (mdash) Ratio between maximum penetration and
borehole radiusΔ119901 (Pa) Driving pressure for grout120583119892
(Pas) Plastic viscosity of grout120583119908
(Pas) Viscosity of water120588119908
(kgm3) Density of water1205910
(Pa) Yield strength of grout
Acknowledgments
The authors would like to acknowledge the effort of GunnarGustafson who deceased during the study
References
[1] J D Osnes A Winberg and J E Andersson ldquoAnalysis of welltest datamdashapplication of probabilistic models to infer hydraulicproperties of fracturesrdquo Topical Report RSI-0338 RESPECRapid City Dakota 1988
Journal of Applied Mathematics 9
[2] C L Axelsson E K Jonsson J Geier and W DershowitzldquoDiscrete facture modellingrdquo SKB Progress Report 25-89-21SKB Stockholm Sweden 1990
[3] A Fransson ldquoNonparametric method for transmissivity distri-butions along boreholesrdquo Ground Water vol 40 no 2 pp 201ndash204 2002
[4] D T Snow ldquoThe frequency and apertures of fractures in rockrdquoInternational Journal of RockMechanics andMining Sciences andGeomechanics Abstracts vol 7 no 1 pp 23ndash40 1970
[5] R W Zimmerman and G S Bodvarsson ldquoHydraulic conduc-tivity of rock fracturesrdquo Transport in Porous Media vol 23 no1 pp 1ndash30 1996
[6] A Fransson ldquoCharacterisation of a fractured rock mass for agrouting field testrdquo Tunnelling and Underground Space Technol-ogy vol 16 no 4 pp 331ndash339 2001
[7] M Eriksson ldquoGrouting field experiment at the Aspo Hard RockLaboratoryrdquoTunnelling andUnderground Space Technology vol17 no 3 pp 287ndash293 2002
[8] M Wallner Propagation of Sedimentation Stable Cement Pastesin Jointed Rock University of Aachen Rock Mechanics andWaterways Construction BRD 1976
[9] L Hassler Grouting of rockmdashsimulation and classification [the-sis] Department of Soil and RockMechanics KTH StockholmSweden 1991
[10] U Hakansson Rheology of fresh cement-based grouts [thesis]Department of Soil and Rock Mechanics KTH StockholmSweden 1993
[11] G Gustafson and H Stille ldquoPrediction of groutability fromgrout properties and hydrogeological datardquo Tunnelling andUnderground Space Technology vol 11 no 3 pp 325ndash332 1996
[12] J A Barker ldquoA generalized radial flowmodel for hydraulic testsin fractured rockrdquoWater Resources Research vol 24 no 10 pp1796ndash1804 1988
[13] A Fransson C-F Tsang J Rutqvist and G Gustafson ldquoEsti-mation of deformation and stiffness of fractures close to tunnelsusing data from single-hole hydraulic testing and groutingrdquoInternational Journal of Rock Mechanics and Mining Sciencesvol 47 no 6 pp 887ndash893 2010
[14] J Claesson ldquoRadial bingham flow background report of adetailed solutionrdquo Tech Rep Department of Building Tech-nology Chalmers University of Technology Goteborg Sweden2003
[15] G Gustafson and H Stille ldquoStop criteria for cement groutingrdquoFelsbau vol 23 no 3 pp 62ndash68 2005
[16] L Hernqvist A Fransson G Gustafson A Emmelin MEriksson and H Stille ldquoAnalyses of the grouting results for asection of the APSE tunnel at Aspo Hard Rock LaboratoryrdquoInternational Journal of Rock Mechanics and Mining Sciencesvol 46 no 3 pp 439ndash449 2009
[17] C Butron G Gustafson A Fransson and J Funehag ldquoDripsealing of tunnels in hard rock a new concept for the designand evaluation of permeation groutingrdquo Tunnelling and Under-ground Space Technology vol 25 no 2 pp 114ndash121 2010
[18] H Stille G Gustafson and L Hassler ldquoApplication of newtheories and technology for grouting of dams and foundationson rockrdquoGeotechnical and Geological Engineering vol 30 no 3pp 603ndash624 2012
We get the same value (2) as in the plain caseThe complete solution in the radial case involves the
following constants
119868max =119887Δ119901
21205910
120574 =119868max119903119887
=119887Δ119901
2119903119887
1205910
1199050
=
6120583119892
Δ119901
(1205910
)2
1198760
=6120587119887(119868max)
2
1199050
=1205871198873
Δ119901
4120583119892
(23)
24 Solution for the Pressure In the dimensionless solutionfor the pressure we use the borehole radius as scaling length
1199031015840
=119903
119903119887
1198681015840
=119868
119903119887
119903119887
le 119903 le 119903119887
+ 119868 lArrrArr 1 le 1199031015840
le 1 + 1198681015840
(24)
The pressure is scaled by Δ119901120574 The variable 119904 for the deriv-ative of the pressure in (18) becomes
1199011015840
=120574 sdot (119901 minus 119901
119908
)
Δ119901997904rArr 119904 =
119887
21205910
sdot (minus119889119901
119889119903)
= minus119887Δ119901120574
21205910
119903119887
sdot1198891199011015840
1198891199031015840= minus
1198891199011015840
1198891199031015840
(25)
The dimensionless form of (18)-(19) becomes after somerecalculations
1199031015840
= 1198761015840
sdot 119892 (minus1198891199011015840
1198891199031015840) 119892 (119904) =
31199042
21199043 minus 31199042 + 1
1198761015840
=
2120583119892
119876
12058711988721205910
119903119887
1199011015840
(1) minus 1199011015840
(1 + 1198681015840
) = 120574
1 le 1199031015840
le 1 + 1198681015840
(26)
This is the basic equation to solve for the pressure distribu-tion It is to be solved for 0 lt 1198681015840 lt 120574 for positive values of theparameter 120574
The solution is derived in detail in [14] A brief derivationis presented in the appendix The dimensionless pressure isgiven by
1199011015840
(1199031015840
) = 120574 minus 1198761015840
sdot [119866 (1198761015840
) minus 119866(1198761015840
1199031015840)] 1 le 119903
1015840
le 1 + 1198681015840
(27)
The composite function 119866(119902) which is used for 119902 = 1198761015840 and
The function 119904(119902) is the root to the cubic equation 119902 sdot 119892(119904) = 1for 119904 gt 1 The function 119866(119904) is an integral of 119904 sdot 119889119892119889119904
The value of the factor1198761015840 has to be chosen so that the totalpressure difference corresponds to the injection pressure(26) This gives
120574 = 1198761015840
sdot [119866 (1198761015840
) minus 119866(1198761015840
1 + 1198681015840)] (29)
This equation determines 1198761015840 as a function of 1198681015840 and 120574
1198761015840
= 1198911015840
(1198681015840
120574) 0 le 1198681015840
le 120574 120574 gt 0 (30)
The value of1198761015840 for 1198681015840 = 120574 is zero in accordance with (21)-(22)1198911015840
(120574 120574) = 0
25 Motion of Grout Front In the dimensionless formulationof the equation for the motion of the grout front we use 119868maxas scaling length We also use 119876
0
and 1199050
from (23)
119868119863
=119868
119868max 119868
1015840
= 120574119868119863
119876119863
=119876
1198760
119905119863
=119905
1199050
(31)
The grout flux becomes from (23) and (26)
1198760
120574=1205871198872
1205910
119903119887
2120583119892
997904rArr 119876 =1198760
120574sdot 1198911015840
(1198681015840
120574) = 1198760
sdot 119876119863
(119868119863
120574)
(32)
The dimensionless grout flux is then
119876119863
(119868119863
120574) =1198911015840
(120574119868119863
120574)
120574 0 le 119868
119863
le 1 (33)
The dimensionless equation for the front motion is now from(32) (20) (31) and (23)
1198760
120574sdot 1198911015840
(120574119868119863
120574) = 2120587119887 sdot(119868max)
2
1199050
sdot (1
120574+ 119868119863
) sdot119889119868119863
119889119905119863
or 119889119905119863
119889119868119863
=120574
3sdot1120574 + 119868
119863
119891 (120574119868119863
120574)
(34)
6 Journal of Applied Mathematics
tD = tt0
I D=II m
ax
0010203040506070809
1
0000001 00001 001 1 100
120574 = 20
120574 = 50
120574 = 100
120574 = 200
120574 = 500
120574 = 1000
Figure 6 Grout penetration function 119868119863
= 119868119863
(119905119863
120574) for radial flow
By integration we get the time 119905119863
= 1199051199050
as an integral in 119868119863
119905119863
=1
3sdot int
119868119863
0
1 + 1205741198681015840
119863
119891 (1205741198681015840
119863
120574)1198891198681015840
119863
0 le 119868119863
lt 1 (35)
We get 119905119863
as a function of the grout front position 119868119863
Also in this case the inverse function describes the relativepenetration as a function of the dimensionless grouting timeFigure 6 shows this relation for a few 120574-values
A comparison of Figures 3 4 and 6 shows that the curvesfor 119868119863
(119905119863
) are similar for the three flow cases The main dif-ference to parallel flow is that penetration is somewhat slowerfor the radial case Around 80 of maximum penetrationis reached after 3119905
0
and to reach 90 takes about 71199050
Theprinciple is however the same and the curves could be usedin the same way
26 Injected Volume of Grout The injected volume of groutas a function of time is of interest The volume is
119881119892
(119905) = 120587119887 [(119903119887
+ 119868 (119905))2
minus (119903119887
)2
]
= 120587119887119868(119905)2
sdot [1 +2119903119887
119868 (119905)]
(36)
Let 119881119892max be maximum injection volume and 119881
119863
the dimen-sionless volume of injected grout
119881119863
=
119881119892
119881119892max
119881119892max = 120587119887(119868max)
2
sdot [1 +2
120574] (36
1015840
)
Then we get using (31) (24) (23) and the relation (35)between 119868
119863
and 119905119863
119881119863
(119905119863
120574) = (119868119863
)2
sdot1 + 2 (120574119868
119863
)
1 + 2120574 119868119863
= 119868119863
(119905119863
120574) (37)
Equations presented in this paper have been used inGustafson and Stille [15] when considering stop criteria forgrouting Grouting projects where estimates of penetration
length have been made are for example [13 15 16] Penetra-tion length has also been a key to presenting a concept forestimation of deformation and stiffness of fractures based ongrouting data [13] In addition to grouting of tunnels theorieshave also been applied for grouting of dams [18]
3 Conclusions
The theoretical investigation of grout spread in one-dimensional conduits and radial spread in plane parallel frac-tures have shown very similar behavior for all the investigatedcases The penetration 119868 can be described as a product ofthe maximum penetration 119868max = Δ119901 sdot 120591
0
2119887 and a time-dependent scaling factor 119868
119863
(119905119863
) the relative penetrationlength HereΔ119901 is the driving pressure 120591
0
is the yield strengthof the grout and 119887 is the aperture of the penetrated fractureThe time factor or dimensionless grouting time 119905
119863
= 1199051199050
is the ratio between the actual grouting time 119905 and a timescaling factor 119905
0
= 6120583119892
Δ1199011205910
2 the characteristic groutingtime Here 120583
119892
is the Bingham viscosity of the grout Therelative penetration depth has a value of 70ndash90 for 119905 = 119905
0
and reaches a value of more than 90 for 119905 gt 71199050
for allfractures
From this a number of important conclusions can bedrawn
(i) The relative penetration is the same in all fracturesthat a grouted borehole cuts This means that giventhe same grout and pressure the grouting time shouldbe the same in high and low yielding boreholes inorder to get the same degree of tightening of allfractures This means that the tendency in practice togrout for a shorter time in tight boreholes will givepoor results for sealing of fine fractures
(ii) Themaximumpenetration is governed by the fractureaperture and pressure and yield strength of the groutThe latter are at the choice of the grouter
(iii) The relative penetration which governs much of thefinal result is determined by the grouting time
(iv) The pressure and the grout properties determine thedesired grouting time These are the choice of thegrouter alone
(v) It is poor economy to grout for a longer time thanabout 5119905
0
since the growth of the penetration is veryslow for a time longer than that On the other handif the borehole takes significant amounts of groutafter 5119905
0
there is reason to stop since it indicatesan unrestricted outflow of grout somewhere in thesystem
The significance of this for grouting design is as follows
(i) The conventional stop criteria based on volume orgrout flow can be replaced by a minimum time cri-terion based only on the parameters that the groutercan chose that is grouting pressure and yield strengthof the grout
Journal of Applied Mathematics 7
(ii) Based on an assessment of how fine fractures itis necessary to seal a maximum effective boreholedistance can be predicted given the pressure and theproperties of the grout
(iii) The time needed for effective grouting operations canbe estimated with better accuracy
(iv) In order to avoid unrestricted grout pumping also amaximum grouting time can be given where furtherinjection of grout will be unnecessary
Appendix
Derivation of the Solution for the Pressure
We seek the solution 1199011015840(1199031015840) to (26)
1199031015840
= 1198761015840
sdot 119892 (minus1198891199011015840
1198891199031015840) 1 le 119903
1015840
le 1 + 1198681015840
119892 (119904) =31199042
21199043 minus 31199042 + 1 0 le 119868
1015840
le 120574
(A1)
Here 1 + 1198681015840 is the position of the grout front The parameter120574 is positive Taking zero pressure at the grout front theboundary conditions for the dimensionless pressure become
1199011015840
(1) = 120574 1199011015840
(1 + 1198681015840
) = 0 (A2)
The dimensionless grout flux 1198761015840 is to be chosen so that theprevious boundary conditions are fulfilled The value of 1198761015840will depend on the front position 1198681015840
Solution in Parameter Form In order to see more directly thecharacter of the equation we make the following change ofnotation
119909 larrrarr 1199031015840
119910 larrrarr minus1199011015840
119891 (119904) = 1198761015840
sdot 119892 (119904) (A3)
The equation is then of the following type
119909 = 119891(119889119910
119889119909) (A4)
There is a general solution in a certain parameter form tothis type of implicit ordinary differential equation [19] Thesolution is
119909 (119904) = 119891 (119904) 119910 (119904) = 119904 sdot 119891 (119904) minus int
119904
119891 (1199041015840
) 1198891199041015840
(A5)
We have to show that this is indeed the solution We have
Theboundary condition (A2) at 1199031015840 = 1 is fulfilled for a certainchoice of 119870 The explicit solution is
1199011015840
(1199031015840
) = 120574 minus 1198761015840
sdot [119866 (1198761015840
) minus 119866(1198761015840
1199031015840)] 1 le 119903
1015840
le 1 + 1198681015840
(A13)
The other boundary condition (A2) at 1199031015840 = 1 + 1198681015840 is fulfilledwhen 1198761015840 satisfies the equation
120574 = 1198761015840
sdot [119866 (1198761015840
) minus 119866(1198761015840
1 + 1198681015840)] (A14)
We note that the derivative minus11988911990110158401198891199031015840 is given by 119904
119904 = minus1198891199011015840
1198891199031015840 (A15)
The pressure derivative is equal to ndash1 for zero flux (21) and(25) in the final stagnant position 119868
1015840
= 120574 The magnitudeof this derivative is larger than 1 for all preceding positions1198681015840
lt 120574 This means that 119904 is larger than (or equal to) 1 in thesolution
The Function 119866(119904) The solution (A13) and the compositefunction (A12) involve the function 119866(119904) defined in (A10)
8 Journal of Applied Mathematics
and (A1) The integral of 119892(119904) is obtained from an expansionin partial fractions We have
119892 (119904) =31199042
21199043 minus 31199042 + 1=
31199042
(2119904 + 1) (119904 minus 1)2
=1
3sdot
1
2119904 + 1+4
3sdot
1
119904 minus 1+
1
(119904 minus 1)2
(A16)
The integral of 119892(119904) is readily determined The function 119866(119904)becomes from (A10) and (A16)
119866 (119904) =1
6sdot ln (2119904 + 1) + 4
3sdot ln (119904 minus 1) minus 1
119904 minus 1
minus31199043
21199043 minus 31199042 + 1 119904 gt 1
(A17)
We will use the function for 1 lt 119904 lt infin
The Inverse 119904(119902) The inverse (A9) is for any 119902 ge 0 thesolution of the cubic equation
21199043
minus 31199042
+ 1 = 31199021199042
(A18)
The solution is reported in detail in [14] The cubic equationhas three real-valued solutions for positive 119902-values one ofwhich is larger than 1 (for 119902 = 0 there is a double root 119904 = 1
and a third root 119904 = minus05 (A16)) We need the solution 119904 gt 1It is given by
119904 (119902) =1
2radic1 + 119902 sdot sin (13) sdot arcsin [(1 + 119902)minus15] 119902 ge 0
(A19)
A plot shows that 119904(119902) is an increasing function from 119904(0) = 1
for 119902 ge 0 It has the asymptote 15 sdot (1 + 119902) for large 119902We will show that (A19) is the inverse We use the
notations
119904 (119902) = 119904 =1
2radic1 + 119902 sdot sin (1206013)
120601 = arcsin [(1 + 119902)minus15] (A20)
In (A18) we put 31199021199042 on the left-hand side divide by 1199043 andinsert 119904 = 119904(119902) from (A20) Then we have
On the third line we use a well-known trigonometric formularelating sin(1206013) to sin(120601) We have shown that (A19) is theinverse
Symbols and Units
119887 (m) Fracture aperture119868 (m) Penetration length of injected grout119868max (m) Maximum penetration length of grout119868max119901 (m) Maximum penetration length of grout in a
pipe1198681015840 (mdash) Ratio between penetration and borehole
(mdash) Relative penetration length in a pipe119871 (m) Distance between grouting boreholes119901 (Pa) Pressure119901119863
(mdash) Dimensionless pressure119901119892
(Pa) Grout pressure119901119908
(Pa) Water pressure119876 (m3s) Grout injection flow rate119903 (m) Pipe radius radial distance from borehole
centre119903119887
(m) Borehole radius119903119863
(mdash) Dimensionless radius119903119901
(m) Grout plug radius1199030
(m) Pipe radius1199031015840 (mdash) Ratio between distance from borehole centre
and borehole radius119879 (m2s) Transmissivity119905 (s) Grouting time1199050
(s) Characteristic grouting time119905119863
(mdash) Dimensionless grouting time119881119892
(m3) Injected volume of grout119881max (m
3) Maximum grout volume in a fracture119881119863
(mdash) Dimensionless grout volume119909 (m) Length coordinate119885 (mdash) Bingham half-plug thickness120574 (mdash) Ratio between maximum penetration and
borehole radiusΔ119901 (Pa) Driving pressure for grout120583119892
(Pas) Plastic viscosity of grout120583119908
(Pas) Viscosity of water120588119908
(kgm3) Density of water1205910
(Pa) Yield strength of grout
Acknowledgments
The authors would like to acknowledge the effort of GunnarGustafson who deceased during the study
References
[1] J D Osnes A Winberg and J E Andersson ldquoAnalysis of welltest datamdashapplication of probabilistic models to infer hydraulicproperties of fracturesrdquo Topical Report RSI-0338 RESPECRapid City Dakota 1988
Journal of Applied Mathematics 9
[2] C L Axelsson E K Jonsson J Geier and W DershowitzldquoDiscrete facture modellingrdquo SKB Progress Report 25-89-21SKB Stockholm Sweden 1990
[3] A Fransson ldquoNonparametric method for transmissivity distri-butions along boreholesrdquo Ground Water vol 40 no 2 pp 201ndash204 2002
[4] D T Snow ldquoThe frequency and apertures of fractures in rockrdquoInternational Journal of RockMechanics andMining Sciences andGeomechanics Abstracts vol 7 no 1 pp 23ndash40 1970
[5] R W Zimmerman and G S Bodvarsson ldquoHydraulic conduc-tivity of rock fracturesrdquo Transport in Porous Media vol 23 no1 pp 1ndash30 1996
[6] A Fransson ldquoCharacterisation of a fractured rock mass for agrouting field testrdquo Tunnelling and Underground Space Technol-ogy vol 16 no 4 pp 331ndash339 2001
[7] M Eriksson ldquoGrouting field experiment at the Aspo Hard RockLaboratoryrdquoTunnelling andUnderground Space Technology vol17 no 3 pp 287ndash293 2002
[8] M Wallner Propagation of Sedimentation Stable Cement Pastesin Jointed Rock University of Aachen Rock Mechanics andWaterways Construction BRD 1976
[9] L Hassler Grouting of rockmdashsimulation and classification [the-sis] Department of Soil and RockMechanics KTH StockholmSweden 1991
[10] U Hakansson Rheology of fresh cement-based grouts [thesis]Department of Soil and Rock Mechanics KTH StockholmSweden 1993
[11] G Gustafson and H Stille ldquoPrediction of groutability fromgrout properties and hydrogeological datardquo Tunnelling andUnderground Space Technology vol 11 no 3 pp 325ndash332 1996
[12] J A Barker ldquoA generalized radial flowmodel for hydraulic testsin fractured rockrdquoWater Resources Research vol 24 no 10 pp1796ndash1804 1988
[13] A Fransson C-F Tsang J Rutqvist and G Gustafson ldquoEsti-mation of deformation and stiffness of fractures close to tunnelsusing data from single-hole hydraulic testing and groutingrdquoInternational Journal of Rock Mechanics and Mining Sciencesvol 47 no 6 pp 887ndash893 2010
[14] J Claesson ldquoRadial bingham flow background report of adetailed solutionrdquo Tech Rep Department of Building Tech-nology Chalmers University of Technology Goteborg Sweden2003
[15] G Gustafson and H Stille ldquoStop criteria for cement groutingrdquoFelsbau vol 23 no 3 pp 62ndash68 2005
[16] L Hernqvist A Fransson G Gustafson A Emmelin MEriksson and H Stille ldquoAnalyses of the grouting results for asection of the APSE tunnel at Aspo Hard Rock LaboratoryrdquoInternational Journal of Rock Mechanics and Mining Sciencesvol 46 no 3 pp 439ndash449 2009
[17] C Butron G Gustafson A Fransson and J Funehag ldquoDripsealing of tunnels in hard rock a new concept for the designand evaluation of permeation groutingrdquo Tunnelling and Under-ground Space Technology vol 25 no 2 pp 114ndash121 2010
[18] H Stille G Gustafson and L Hassler ldquoApplication of newtheories and technology for grouting of dams and foundationson rockrdquoGeotechnical and Geological Engineering vol 30 no 3pp 603ndash624 2012
as a function of the grout front position 119868119863
Also in this case the inverse function describes the relativepenetration as a function of the dimensionless grouting timeFigure 6 shows this relation for a few 120574-values
A comparison of Figures 3 4 and 6 shows that the curvesfor 119868119863
(119905119863
) are similar for the three flow cases The main dif-ference to parallel flow is that penetration is somewhat slowerfor the radial case Around 80 of maximum penetrationis reached after 3119905
0
and to reach 90 takes about 71199050
Theprinciple is however the same and the curves could be usedin the same way
26 Injected Volume of Grout The injected volume of groutas a function of time is of interest The volume is
119881119892
(119905) = 120587119887 [(119903119887
+ 119868 (119905))2
minus (119903119887
)2
]
= 120587119887119868(119905)2
sdot [1 +2119903119887
119868 (119905)]
(36)
Let 119881119892max be maximum injection volume and 119881
119863
the dimen-sionless volume of injected grout
119881119863
=
119881119892
119881119892max
119881119892max = 120587119887(119868max)
2
sdot [1 +2
120574] (36
1015840
)
Then we get using (31) (24) (23) and the relation (35)between 119868
119863
and 119905119863
119881119863
(119905119863
120574) = (119868119863
)2
sdot1 + 2 (120574119868
119863
)
1 + 2120574 119868119863
= 119868119863
(119905119863
120574) (37)
Equations presented in this paper have been used inGustafson and Stille [15] when considering stop criteria forgrouting Grouting projects where estimates of penetration
length have been made are for example [13 15 16] Penetra-tion length has also been a key to presenting a concept forestimation of deformation and stiffness of fractures based ongrouting data [13] In addition to grouting of tunnels theorieshave also been applied for grouting of dams [18]
3 Conclusions
The theoretical investigation of grout spread in one-dimensional conduits and radial spread in plane parallel frac-tures have shown very similar behavior for all the investigatedcases The penetration 119868 can be described as a product ofthe maximum penetration 119868max = Δ119901 sdot 120591
0
2119887 and a time-dependent scaling factor 119868
119863
(119905119863
) the relative penetrationlength HereΔ119901 is the driving pressure 120591
0
is the yield strengthof the grout and 119887 is the aperture of the penetrated fractureThe time factor or dimensionless grouting time 119905
119863
= 1199051199050
is the ratio between the actual grouting time 119905 and a timescaling factor 119905
0
= 6120583119892
Δ1199011205910
2 the characteristic groutingtime Here 120583
119892
is the Bingham viscosity of the grout Therelative penetration depth has a value of 70ndash90 for 119905 = 119905
0
and reaches a value of more than 90 for 119905 gt 71199050
for allfractures
From this a number of important conclusions can bedrawn
(i) The relative penetration is the same in all fracturesthat a grouted borehole cuts This means that giventhe same grout and pressure the grouting time shouldbe the same in high and low yielding boreholes inorder to get the same degree of tightening of allfractures This means that the tendency in practice togrout for a shorter time in tight boreholes will givepoor results for sealing of fine fractures
(ii) Themaximumpenetration is governed by the fractureaperture and pressure and yield strength of the groutThe latter are at the choice of the grouter
(iii) The relative penetration which governs much of thefinal result is determined by the grouting time
(iv) The pressure and the grout properties determine thedesired grouting time These are the choice of thegrouter alone
(v) It is poor economy to grout for a longer time thanabout 5119905
0
since the growth of the penetration is veryslow for a time longer than that On the other handif the borehole takes significant amounts of groutafter 5119905
0
there is reason to stop since it indicatesan unrestricted outflow of grout somewhere in thesystem
The significance of this for grouting design is as follows
(i) The conventional stop criteria based on volume orgrout flow can be replaced by a minimum time cri-terion based only on the parameters that the groutercan chose that is grouting pressure and yield strengthof the grout
Journal of Applied Mathematics 7
(ii) Based on an assessment of how fine fractures itis necessary to seal a maximum effective boreholedistance can be predicted given the pressure and theproperties of the grout
(iii) The time needed for effective grouting operations canbe estimated with better accuracy
(iv) In order to avoid unrestricted grout pumping also amaximum grouting time can be given where furtherinjection of grout will be unnecessary
Appendix
Derivation of the Solution for the Pressure
We seek the solution 1199011015840(1199031015840) to (26)
1199031015840
= 1198761015840
sdot 119892 (minus1198891199011015840
1198891199031015840) 1 le 119903
1015840
le 1 + 1198681015840
119892 (119904) =31199042
21199043 minus 31199042 + 1 0 le 119868
1015840
le 120574
(A1)
Here 1 + 1198681015840 is the position of the grout front The parameter120574 is positive Taking zero pressure at the grout front theboundary conditions for the dimensionless pressure become
1199011015840
(1) = 120574 1199011015840
(1 + 1198681015840
) = 0 (A2)
The dimensionless grout flux 1198761015840 is to be chosen so that theprevious boundary conditions are fulfilled The value of 1198761015840will depend on the front position 1198681015840
Solution in Parameter Form In order to see more directly thecharacter of the equation we make the following change ofnotation
119909 larrrarr 1199031015840
119910 larrrarr minus1199011015840
119891 (119904) = 1198761015840
sdot 119892 (119904) (A3)
The equation is then of the following type
119909 = 119891(119889119910
119889119909) (A4)
There is a general solution in a certain parameter form tothis type of implicit ordinary differential equation [19] Thesolution is
119909 (119904) = 119891 (119904) 119910 (119904) = 119904 sdot 119891 (119904) minus int
119904
119891 (1199041015840
) 1198891199041015840
(A5)
We have to show that this is indeed the solution We have
Theboundary condition (A2) at 1199031015840 = 1 is fulfilled for a certainchoice of 119870 The explicit solution is
1199011015840
(1199031015840
) = 120574 minus 1198761015840
sdot [119866 (1198761015840
) minus 119866(1198761015840
1199031015840)] 1 le 119903
1015840
le 1 + 1198681015840
(A13)
The other boundary condition (A2) at 1199031015840 = 1 + 1198681015840 is fulfilledwhen 1198761015840 satisfies the equation
120574 = 1198761015840
sdot [119866 (1198761015840
) minus 119866(1198761015840
1 + 1198681015840)] (A14)
We note that the derivative minus11988911990110158401198891199031015840 is given by 119904
119904 = minus1198891199011015840
1198891199031015840 (A15)
The pressure derivative is equal to ndash1 for zero flux (21) and(25) in the final stagnant position 119868
1015840
= 120574 The magnitudeof this derivative is larger than 1 for all preceding positions1198681015840
lt 120574 This means that 119904 is larger than (or equal to) 1 in thesolution
The Function 119866(119904) The solution (A13) and the compositefunction (A12) involve the function 119866(119904) defined in (A10)
8 Journal of Applied Mathematics
and (A1) The integral of 119892(119904) is obtained from an expansionin partial fractions We have
119892 (119904) =31199042
21199043 minus 31199042 + 1=
31199042
(2119904 + 1) (119904 minus 1)2
=1
3sdot
1
2119904 + 1+4
3sdot
1
119904 minus 1+
1
(119904 minus 1)2
(A16)
The integral of 119892(119904) is readily determined The function 119866(119904)becomes from (A10) and (A16)
119866 (119904) =1
6sdot ln (2119904 + 1) + 4
3sdot ln (119904 minus 1) minus 1
119904 minus 1
minus31199043
21199043 minus 31199042 + 1 119904 gt 1
(A17)
We will use the function for 1 lt 119904 lt infin
The Inverse 119904(119902) The inverse (A9) is for any 119902 ge 0 thesolution of the cubic equation
21199043
minus 31199042
+ 1 = 31199021199042
(A18)
The solution is reported in detail in [14] The cubic equationhas three real-valued solutions for positive 119902-values one ofwhich is larger than 1 (for 119902 = 0 there is a double root 119904 = 1
and a third root 119904 = minus05 (A16)) We need the solution 119904 gt 1It is given by
119904 (119902) =1
2radic1 + 119902 sdot sin (13) sdot arcsin [(1 + 119902)minus15] 119902 ge 0
(A19)
A plot shows that 119904(119902) is an increasing function from 119904(0) = 1
for 119902 ge 0 It has the asymptote 15 sdot (1 + 119902) for large 119902We will show that (A19) is the inverse We use the
notations
119904 (119902) = 119904 =1
2radic1 + 119902 sdot sin (1206013)
120601 = arcsin [(1 + 119902)minus15] (A20)
In (A18) we put 31199021199042 on the left-hand side divide by 1199043 andinsert 119904 = 119904(119902) from (A20) Then we have
On the third line we use a well-known trigonometric formularelating sin(1206013) to sin(120601) We have shown that (A19) is theinverse
Symbols and Units
119887 (m) Fracture aperture119868 (m) Penetration length of injected grout119868max (m) Maximum penetration length of grout119868max119901 (m) Maximum penetration length of grout in a
pipe1198681015840 (mdash) Ratio between penetration and borehole
(mdash) Relative penetration length in a pipe119871 (m) Distance between grouting boreholes119901 (Pa) Pressure119901119863
(mdash) Dimensionless pressure119901119892
(Pa) Grout pressure119901119908
(Pa) Water pressure119876 (m3s) Grout injection flow rate119903 (m) Pipe radius radial distance from borehole
centre119903119887
(m) Borehole radius119903119863
(mdash) Dimensionless radius119903119901
(m) Grout plug radius1199030
(m) Pipe radius1199031015840 (mdash) Ratio between distance from borehole centre
and borehole radius119879 (m2s) Transmissivity119905 (s) Grouting time1199050
(s) Characteristic grouting time119905119863
(mdash) Dimensionless grouting time119881119892
(m3) Injected volume of grout119881max (m
3) Maximum grout volume in a fracture119881119863
(mdash) Dimensionless grout volume119909 (m) Length coordinate119885 (mdash) Bingham half-plug thickness120574 (mdash) Ratio between maximum penetration and
borehole radiusΔ119901 (Pa) Driving pressure for grout120583119892
(Pas) Plastic viscosity of grout120583119908
(Pas) Viscosity of water120588119908
(kgm3) Density of water1205910
(Pa) Yield strength of grout
Acknowledgments
The authors would like to acknowledge the effort of GunnarGustafson who deceased during the study
References
[1] J D Osnes A Winberg and J E Andersson ldquoAnalysis of welltest datamdashapplication of probabilistic models to infer hydraulicproperties of fracturesrdquo Topical Report RSI-0338 RESPECRapid City Dakota 1988
Journal of Applied Mathematics 9
[2] C L Axelsson E K Jonsson J Geier and W DershowitzldquoDiscrete facture modellingrdquo SKB Progress Report 25-89-21SKB Stockholm Sweden 1990
[3] A Fransson ldquoNonparametric method for transmissivity distri-butions along boreholesrdquo Ground Water vol 40 no 2 pp 201ndash204 2002
[4] D T Snow ldquoThe frequency and apertures of fractures in rockrdquoInternational Journal of RockMechanics andMining Sciences andGeomechanics Abstracts vol 7 no 1 pp 23ndash40 1970
[5] R W Zimmerman and G S Bodvarsson ldquoHydraulic conduc-tivity of rock fracturesrdquo Transport in Porous Media vol 23 no1 pp 1ndash30 1996
[6] A Fransson ldquoCharacterisation of a fractured rock mass for agrouting field testrdquo Tunnelling and Underground Space Technol-ogy vol 16 no 4 pp 331ndash339 2001
[7] M Eriksson ldquoGrouting field experiment at the Aspo Hard RockLaboratoryrdquoTunnelling andUnderground Space Technology vol17 no 3 pp 287ndash293 2002
[8] M Wallner Propagation of Sedimentation Stable Cement Pastesin Jointed Rock University of Aachen Rock Mechanics andWaterways Construction BRD 1976
[9] L Hassler Grouting of rockmdashsimulation and classification [the-sis] Department of Soil and RockMechanics KTH StockholmSweden 1991
[10] U Hakansson Rheology of fresh cement-based grouts [thesis]Department of Soil and Rock Mechanics KTH StockholmSweden 1993
[11] G Gustafson and H Stille ldquoPrediction of groutability fromgrout properties and hydrogeological datardquo Tunnelling andUnderground Space Technology vol 11 no 3 pp 325ndash332 1996
[12] J A Barker ldquoA generalized radial flowmodel for hydraulic testsin fractured rockrdquoWater Resources Research vol 24 no 10 pp1796ndash1804 1988
[13] A Fransson C-F Tsang J Rutqvist and G Gustafson ldquoEsti-mation of deformation and stiffness of fractures close to tunnelsusing data from single-hole hydraulic testing and groutingrdquoInternational Journal of Rock Mechanics and Mining Sciencesvol 47 no 6 pp 887ndash893 2010
[14] J Claesson ldquoRadial bingham flow background report of adetailed solutionrdquo Tech Rep Department of Building Tech-nology Chalmers University of Technology Goteborg Sweden2003
[15] G Gustafson and H Stille ldquoStop criteria for cement groutingrdquoFelsbau vol 23 no 3 pp 62ndash68 2005
[16] L Hernqvist A Fransson G Gustafson A Emmelin MEriksson and H Stille ldquoAnalyses of the grouting results for asection of the APSE tunnel at Aspo Hard Rock LaboratoryrdquoInternational Journal of Rock Mechanics and Mining Sciencesvol 46 no 3 pp 439ndash449 2009
[17] C Butron G Gustafson A Fransson and J Funehag ldquoDripsealing of tunnels in hard rock a new concept for the designand evaluation of permeation groutingrdquo Tunnelling and Under-ground Space Technology vol 25 no 2 pp 114ndash121 2010
[18] H Stille G Gustafson and L Hassler ldquoApplication of newtheories and technology for grouting of dams and foundationson rockrdquoGeotechnical and Geological Engineering vol 30 no 3pp 603ndash624 2012
(ii) Based on an assessment of how fine fractures itis necessary to seal a maximum effective boreholedistance can be predicted given the pressure and theproperties of the grout
(iii) The time needed for effective grouting operations canbe estimated with better accuracy
(iv) In order to avoid unrestricted grout pumping also amaximum grouting time can be given where furtherinjection of grout will be unnecessary
Appendix
Derivation of the Solution for the Pressure
We seek the solution 1199011015840(1199031015840) to (26)
1199031015840
= 1198761015840
sdot 119892 (minus1198891199011015840
1198891199031015840) 1 le 119903
1015840
le 1 + 1198681015840
119892 (119904) =31199042
21199043 minus 31199042 + 1 0 le 119868
1015840
le 120574
(A1)
Here 1 + 1198681015840 is the position of the grout front The parameter120574 is positive Taking zero pressure at the grout front theboundary conditions for the dimensionless pressure become
1199011015840
(1) = 120574 1199011015840
(1 + 1198681015840
) = 0 (A2)
The dimensionless grout flux 1198761015840 is to be chosen so that theprevious boundary conditions are fulfilled The value of 1198761015840will depend on the front position 1198681015840
Solution in Parameter Form In order to see more directly thecharacter of the equation we make the following change ofnotation
119909 larrrarr 1199031015840
119910 larrrarr minus1199011015840
119891 (119904) = 1198761015840
sdot 119892 (119904) (A3)
The equation is then of the following type
119909 = 119891(119889119910
119889119909) (A4)
There is a general solution in a certain parameter form tothis type of implicit ordinary differential equation [19] Thesolution is
119909 (119904) = 119891 (119904) 119910 (119904) = 119904 sdot 119891 (119904) minus int
119904
119891 (1199041015840
) 1198891199041015840
(A5)
We have to show that this is indeed the solution We have
Theboundary condition (A2) at 1199031015840 = 1 is fulfilled for a certainchoice of 119870 The explicit solution is
1199011015840
(1199031015840
) = 120574 minus 1198761015840
sdot [119866 (1198761015840
) minus 119866(1198761015840
1199031015840)] 1 le 119903
1015840
le 1 + 1198681015840
(A13)
The other boundary condition (A2) at 1199031015840 = 1 + 1198681015840 is fulfilledwhen 1198761015840 satisfies the equation
120574 = 1198761015840
sdot [119866 (1198761015840
) minus 119866(1198761015840
1 + 1198681015840)] (A14)
We note that the derivative minus11988911990110158401198891199031015840 is given by 119904
119904 = minus1198891199011015840
1198891199031015840 (A15)
The pressure derivative is equal to ndash1 for zero flux (21) and(25) in the final stagnant position 119868
1015840
= 120574 The magnitudeof this derivative is larger than 1 for all preceding positions1198681015840
lt 120574 This means that 119904 is larger than (or equal to) 1 in thesolution
The Function 119866(119904) The solution (A13) and the compositefunction (A12) involve the function 119866(119904) defined in (A10)
8 Journal of Applied Mathematics
and (A1) The integral of 119892(119904) is obtained from an expansionin partial fractions We have
119892 (119904) =31199042
21199043 minus 31199042 + 1=
31199042
(2119904 + 1) (119904 minus 1)2
=1
3sdot
1
2119904 + 1+4
3sdot
1
119904 minus 1+
1
(119904 minus 1)2
(A16)
The integral of 119892(119904) is readily determined The function 119866(119904)becomes from (A10) and (A16)
119866 (119904) =1
6sdot ln (2119904 + 1) + 4
3sdot ln (119904 minus 1) minus 1
119904 minus 1
minus31199043
21199043 minus 31199042 + 1 119904 gt 1
(A17)
We will use the function for 1 lt 119904 lt infin
The Inverse 119904(119902) The inverse (A9) is for any 119902 ge 0 thesolution of the cubic equation
21199043
minus 31199042
+ 1 = 31199021199042
(A18)
The solution is reported in detail in [14] The cubic equationhas three real-valued solutions for positive 119902-values one ofwhich is larger than 1 (for 119902 = 0 there is a double root 119904 = 1
and a third root 119904 = minus05 (A16)) We need the solution 119904 gt 1It is given by
119904 (119902) =1
2radic1 + 119902 sdot sin (13) sdot arcsin [(1 + 119902)minus15] 119902 ge 0
(A19)
A plot shows that 119904(119902) is an increasing function from 119904(0) = 1
for 119902 ge 0 It has the asymptote 15 sdot (1 + 119902) for large 119902We will show that (A19) is the inverse We use the
notations
119904 (119902) = 119904 =1
2radic1 + 119902 sdot sin (1206013)
120601 = arcsin [(1 + 119902)minus15] (A20)
In (A18) we put 31199021199042 on the left-hand side divide by 1199043 andinsert 119904 = 119904(119902) from (A20) Then we have
On the third line we use a well-known trigonometric formularelating sin(1206013) to sin(120601) We have shown that (A19) is theinverse
Symbols and Units
119887 (m) Fracture aperture119868 (m) Penetration length of injected grout119868max (m) Maximum penetration length of grout119868max119901 (m) Maximum penetration length of grout in a
pipe1198681015840 (mdash) Ratio between penetration and borehole
(mdash) Relative penetration length in a pipe119871 (m) Distance between grouting boreholes119901 (Pa) Pressure119901119863
(mdash) Dimensionless pressure119901119892
(Pa) Grout pressure119901119908
(Pa) Water pressure119876 (m3s) Grout injection flow rate119903 (m) Pipe radius radial distance from borehole
centre119903119887
(m) Borehole radius119903119863
(mdash) Dimensionless radius119903119901
(m) Grout plug radius1199030
(m) Pipe radius1199031015840 (mdash) Ratio between distance from borehole centre
and borehole radius119879 (m2s) Transmissivity119905 (s) Grouting time1199050
(s) Characteristic grouting time119905119863
(mdash) Dimensionless grouting time119881119892
(m3) Injected volume of grout119881max (m
3) Maximum grout volume in a fracture119881119863
(mdash) Dimensionless grout volume119909 (m) Length coordinate119885 (mdash) Bingham half-plug thickness120574 (mdash) Ratio between maximum penetration and
borehole radiusΔ119901 (Pa) Driving pressure for grout120583119892
(Pas) Plastic viscosity of grout120583119908
(Pas) Viscosity of water120588119908
(kgm3) Density of water1205910
(Pa) Yield strength of grout
Acknowledgments
The authors would like to acknowledge the effort of GunnarGustafson who deceased during the study
References
[1] J D Osnes A Winberg and J E Andersson ldquoAnalysis of welltest datamdashapplication of probabilistic models to infer hydraulicproperties of fracturesrdquo Topical Report RSI-0338 RESPECRapid City Dakota 1988
Journal of Applied Mathematics 9
[2] C L Axelsson E K Jonsson J Geier and W DershowitzldquoDiscrete facture modellingrdquo SKB Progress Report 25-89-21SKB Stockholm Sweden 1990
[3] A Fransson ldquoNonparametric method for transmissivity distri-butions along boreholesrdquo Ground Water vol 40 no 2 pp 201ndash204 2002
[4] D T Snow ldquoThe frequency and apertures of fractures in rockrdquoInternational Journal of RockMechanics andMining Sciences andGeomechanics Abstracts vol 7 no 1 pp 23ndash40 1970
[5] R W Zimmerman and G S Bodvarsson ldquoHydraulic conduc-tivity of rock fracturesrdquo Transport in Porous Media vol 23 no1 pp 1ndash30 1996
[6] A Fransson ldquoCharacterisation of a fractured rock mass for agrouting field testrdquo Tunnelling and Underground Space Technol-ogy vol 16 no 4 pp 331ndash339 2001
[7] M Eriksson ldquoGrouting field experiment at the Aspo Hard RockLaboratoryrdquoTunnelling andUnderground Space Technology vol17 no 3 pp 287ndash293 2002
[8] M Wallner Propagation of Sedimentation Stable Cement Pastesin Jointed Rock University of Aachen Rock Mechanics andWaterways Construction BRD 1976
[9] L Hassler Grouting of rockmdashsimulation and classification [the-sis] Department of Soil and RockMechanics KTH StockholmSweden 1991
[10] U Hakansson Rheology of fresh cement-based grouts [thesis]Department of Soil and Rock Mechanics KTH StockholmSweden 1993
[11] G Gustafson and H Stille ldquoPrediction of groutability fromgrout properties and hydrogeological datardquo Tunnelling andUnderground Space Technology vol 11 no 3 pp 325ndash332 1996
[12] J A Barker ldquoA generalized radial flowmodel for hydraulic testsin fractured rockrdquoWater Resources Research vol 24 no 10 pp1796ndash1804 1988
[13] A Fransson C-F Tsang J Rutqvist and G Gustafson ldquoEsti-mation of deformation and stiffness of fractures close to tunnelsusing data from single-hole hydraulic testing and groutingrdquoInternational Journal of Rock Mechanics and Mining Sciencesvol 47 no 6 pp 887ndash893 2010
[14] J Claesson ldquoRadial bingham flow background report of adetailed solutionrdquo Tech Rep Department of Building Tech-nology Chalmers University of Technology Goteborg Sweden2003
[15] G Gustafson and H Stille ldquoStop criteria for cement groutingrdquoFelsbau vol 23 no 3 pp 62ndash68 2005
[16] L Hernqvist A Fransson G Gustafson A Emmelin MEriksson and H Stille ldquoAnalyses of the grouting results for asection of the APSE tunnel at Aspo Hard Rock LaboratoryrdquoInternational Journal of Rock Mechanics and Mining Sciencesvol 46 no 3 pp 439ndash449 2009
[17] C Butron G Gustafson A Fransson and J Funehag ldquoDripsealing of tunnels in hard rock a new concept for the designand evaluation of permeation groutingrdquo Tunnelling and Under-ground Space Technology vol 25 no 2 pp 114ndash121 2010
[18] H Stille G Gustafson and L Hassler ldquoApplication of newtheories and technology for grouting of dams and foundationson rockrdquoGeotechnical and Geological Engineering vol 30 no 3pp 603ndash624 2012
and (A1) The integral of 119892(119904) is obtained from an expansionin partial fractions We have
119892 (119904) =31199042
21199043 minus 31199042 + 1=
31199042
(2119904 + 1) (119904 minus 1)2
=1
3sdot
1
2119904 + 1+4
3sdot
1
119904 minus 1+
1
(119904 minus 1)2
(A16)
The integral of 119892(119904) is readily determined The function 119866(119904)becomes from (A10) and (A16)
119866 (119904) =1
6sdot ln (2119904 + 1) + 4
3sdot ln (119904 minus 1) minus 1
119904 minus 1
minus31199043
21199043 minus 31199042 + 1 119904 gt 1
(A17)
We will use the function for 1 lt 119904 lt infin
The Inverse 119904(119902) The inverse (A9) is for any 119902 ge 0 thesolution of the cubic equation
21199043
minus 31199042
+ 1 = 31199021199042
(A18)
The solution is reported in detail in [14] The cubic equationhas three real-valued solutions for positive 119902-values one ofwhich is larger than 1 (for 119902 = 0 there is a double root 119904 = 1
and a third root 119904 = minus05 (A16)) We need the solution 119904 gt 1It is given by
119904 (119902) =1
2radic1 + 119902 sdot sin (13) sdot arcsin [(1 + 119902)minus15] 119902 ge 0
(A19)
A plot shows that 119904(119902) is an increasing function from 119904(0) = 1
for 119902 ge 0 It has the asymptote 15 sdot (1 + 119902) for large 119902We will show that (A19) is the inverse We use the
notations
119904 (119902) = 119904 =1
2radic1 + 119902 sdot sin (1206013)
120601 = arcsin [(1 + 119902)minus15] (A20)
In (A18) we put 31199021199042 on the left-hand side divide by 1199043 andinsert 119904 = 119904(119902) from (A20) Then we have
On the third line we use a well-known trigonometric formularelating sin(1206013) to sin(120601) We have shown that (A19) is theinverse
Symbols and Units
119887 (m) Fracture aperture119868 (m) Penetration length of injected grout119868max (m) Maximum penetration length of grout119868max119901 (m) Maximum penetration length of grout in a
pipe1198681015840 (mdash) Ratio between penetration and borehole
(mdash) Relative penetration length in a pipe119871 (m) Distance between grouting boreholes119901 (Pa) Pressure119901119863
(mdash) Dimensionless pressure119901119892
(Pa) Grout pressure119901119908
(Pa) Water pressure119876 (m3s) Grout injection flow rate119903 (m) Pipe radius radial distance from borehole
centre119903119887
(m) Borehole radius119903119863
(mdash) Dimensionless radius119903119901
(m) Grout plug radius1199030
(m) Pipe radius1199031015840 (mdash) Ratio between distance from borehole centre
and borehole radius119879 (m2s) Transmissivity119905 (s) Grouting time1199050
(s) Characteristic grouting time119905119863
(mdash) Dimensionless grouting time119881119892
(m3) Injected volume of grout119881max (m
3) Maximum grout volume in a fracture119881119863
(mdash) Dimensionless grout volume119909 (m) Length coordinate119885 (mdash) Bingham half-plug thickness120574 (mdash) Ratio between maximum penetration and
borehole radiusΔ119901 (Pa) Driving pressure for grout120583119892
(Pas) Plastic viscosity of grout120583119908
(Pas) Viscosity of water120588119908
(kgm3) Density of water1205910
(Pa) Yield strength of grout
Acknowledgments
The authors would like to acknowledge the effort of GunnarGustafson who deceased during the study
References
[1] J D Osnes A Winberg and J E Andersson ldquoAnalysis of welltest datamdashapplication of probabilistic models to infer hydraulicproperties of fracturesrdquo Topical Report RSI-0338 RESPECRapid City Dakota 1988
Journal of Applied Mathematics 9
[2] C L Axelsson E K Jonsson J Geier and W DershowitzldquoDiscrete facture modellingrdquo SKB Progress Report 25-89-21SKB Stockholm Sweden 1990
[3] A Fransson ldquoNonparametric method for transmissivity distri-butions along boreholesrdquo Ground Water vol 40 no 2 pp 201ndash204 2002
[4] D T Snow ldquoThe frequency and apertures of fractures in rockrdquoInternational Journal of RockMechanics andMining Sciences andGeomechanics Abstracts vol 7 no 1 pp 23ndash40 1970
[5] R W Zimmerman and G S Bodvarsson ldquoHydraulic conduc-tivity of rock fracturesrdquo Transport in Porous Media vol 23 no1 pp 1ndash30 1996
[6] A Fransson ldquoCharacterisation of a fractured rock mass for agrouting field testrdquo Tunnelling and Underground Space Technol-ogy vol 16 no 4 pp 331ndash339 2001
[7] M Eriksson ldquoGrouting field experiment at the Aspo Hard RockLaboratoryrdquoTunnelling andUnderground Space Technology vol17 no 3 pp 287ndash293 2002
[8] M Wallner Propagation of Sedimentation Stable Cement Pastesin Jointed Rock University of Aachen Rock Mechanics andWaterways Construction BRD 1976
[9] L Hassler Grouting of rockmdashsimulation and classification [the-sis] Department of Soil and RockMechanics KTH StockholmSweden 1991
[10] U Hakansson Rheology of fresh cement-based grouts [thesis]Department of Soil and Rock Mechanics KTH StockholmSweden 1993
[11] G Gustafson and H Stille ldquoPrediction of groutability fromgrout properties and hydrogeological datardquo Tunnelling andUnderground Space Technology vol 11 no 3 pp 325ndash332 1996
[12] J A Barker ldquoA generalized radial flowmodel for hydraulic testsin fractured rockrdquoWater Resources Research vol 24 no 10 pp1796ndash1804 1988
[13] A Fransson C-F Tsang J Rutqvist and G Gustafson ldquoEsti-mation of deformation and stiffness of fractures close to tunnelsusing data from single-hole hydraulic testing and groutingrdquoInternational Journal of Rock Mechanics and Mining Sciencesvol 47 no 6 pp 887ndash893 2010
[14] J Claesson ldquoRadial bingham flow background report of adetailed solutionrdquo Tech Rep Department of Building Tech-nology Chalmers University of Technology Goteborg Sweden2003
[15] G Gustafson and H Stille ldquoStop criteria for cement groutingrdquoFelsbau vol 23 no 3 pp 62ndash68 2005
[16] L Hernqvist A Fransson G Gustafson A Emmelin MEriksson and H Stille ldquoAnalyses of the grouting results for asection of the APSE tunnel at Aspo Hard Rock LaboratoryrdquoInternational Journal of Rock Mechanics and Mining Sciencesvol 46 no 3 pp 439ndash449 2009
[17] C Butron G Gustafson A Fransson and J Funehag ldquoDripsealing of tunnels in hard rock a new concept for the designand evaluation of permeation groutingrdquo Tunnelling and Under-ground Space Technology vol 25 no 2 pp 114ndash121 2010
[18] H Stille G Gustafson and L Hassler ldquoApplication of newtheories and technology for grouting of dams and foundationson rockrdquoGeotechnical and Geological Engineering vol 30 no 3pp 603ndash624 2012
[2] C L Axelsson E K Jonsson J Geier and W DershowitzldquoDiscrete facture modellingrdquo SKB Progress Report 25-89-21SKB Stockholm Sweden 1990
[3] A Fransson ldquoNonparametric method for transmissivity distri-butions along boreholesrdquo Ground Water vol 40 no 2 pp 201ndash204 2002
[4] D T Snow ldquoThe frequency and apertures of fractures in rockrdquoInternational Journal of RockMechanics andMining Sciences andGeomechanics Abstracts vol 7 no 1 pp 23ndash40 1970
[5] R W Zimmerman and G S Bodvarsson ldquoHydraulic conduc-tivity of rock fracturesrdquo Transport in Porous Media vol 23 no1 pp 1ndash30 1996
[6] A Fransson ldquoCharacterisation of a fractured rock mass for agrouting field testrdquo Tunnelling and Underground Space Technol-ogy vol 16 no 4 pp 331ndash339 2001
[7] M Eriksson ldquoGrouting field experiment at the Aspo Hard RockLaboratoryrdquoTunnelling andUnderground Space Technology vol17 no 3 pp 287ndash293 2002
[8] M Wallner Propagation of Sedimentation Stable Cement Pastesin Jointed Rock University of Aachen Rock Mechanics andWaterways Construction BRD 1976
[9] L Hassler Grouting of rockmdashsimulation and classification [the-sis] Department of Soil and RockMechanics KTH StockholmSweden 1991
[10] U Hakansson Rheology of fresh cement-based grouts [thesis]Department of Soil and Rock Mechanics KTH StockholmSweden 1993
[11] G Gustafson and H Stille ldquoPrediction of groutability fromgrout properties and hydrogeological datardquo Tunnelling andUnderground Space Technology vol 11 no 3 pp 325ndash332 1996
[12] J A Barker ldquoA generalized radial flowmodel for hydraulic testsin fractured rockrdquoWater Resources Research vol 24 no 10 pp1796ndash1804 1988
[13] A Fransson C-F Tsang J Rutqvist and G Gustafson ldquoEsti-mation of deformation and stiffness of fractures close to tunnelsusing data from single-hole hydraulic testing and groutingrdquoInternational Journal of Rock Mechanics and Mining Sciencesvol 47 no 6 pp 887ndash893 2010
[14] J Claesson ldquoRadial bingham flow background report of adetailed solutionrdquo Tech Rep Department of Building Tech-nology Chalmers University of Technology Goteborg Sweden2003
[15] G Gustafson and H Stille ldquoStop criteria for cement groutingrdquoFelsbau vol 23 no 3 pp 62ndash68 2005
[16] L Hernqvist A Fransson G Gustafson A Emmelin MEriksson and H Stille ldquoAnalyses of the grouting results for asection of the APSE tunnel at Aspo Hard Rock LaboratoryrdquoInternational Journal of Rock Mechanics and Mining Sciencesvol 46 no 3 pp 439ndash449 2009
[17] C Butron G Gustafson A Fransson and J Funehag ldquoDripsealing of tunnels in hard rock a new concept for the designand evaluation of permeation groutingrdquo Tunnelling and Under-ground Space Technology vol 25 no 2 pp 114ndash121 2010
[18] H Stille G Gustafson and L Hassler ldquoApplication of newtheories and technology for grouting of dams and foundationson rockrdquoGeotechnical and Geological Engineering vol 30 no 3pp 603ndash624 2012
