This introduction is the same as for most of the other options and can be skipped if you are familiar with the gear.
If you wish you can let the simulation work right through the grouting of the hole without you being required to do any thinking! But, instead if you wish, you can do it by yourself .
This is a simple layout of the grouting equipment. The grout is mixed in the mixer at left, and then sent to the agitator, from which a pump is used to pump it along the circulation line to the grout hole. Grout not used there is pumped back to the agitator. .
In this sketch the grouting has progressed to the stage where the cracks have been partly filled with grout, but you will, of course, be starting with a fresh hole.
In the simulation the gear on the surface is not shown in further sketches. For explanations of the terminology
and criteria used consult the accompanying programmes and the Guide.
And now we come to running the
simulation. The pretend grout hole is 20 feet deep [6 metres, say] throughout the simulation and is vertical.
The grout does not contain additives.
Assume that the cracking system reached by this hole has the following properties:
1. The water test before grouting gave a lugeon value in the 'low' range. This range might be
regarded as a test result below about 10 lugeons.
2. Cracks are generally wider than 0.020 inch (0.5 mm, 500 microns) and less than about 0.30
inch [0.8 mm] open and are in sufficient number to provide some degree of travel for the grout through the
cracking system.
A suitable starting water:cement ratio for these could be 3:1.
The starting pressure, measured at the surface, should be 15 psi [1 bar] for the first few minutes.
Assume that no problems develop, such as leaks, connections, or rock movements, and that the grouting proceeds
smoothly. The pressure can therefore be gradually increasted to the recommended limit for sound rock, which is
25 psi [2 bar] for this depth of hole.
At 15 minutes assume that the amount of grout taken down the hole, so far, is 1.75 cubic feet [47 litres].
After that the hole continues to take grout but at a decreasing rate. At the end of 30 minutes we can pretend that the rate has decreased, in the manner in which it does in many jobs with fine cracking, and the grout taken is now only 1.00 cu ft [27 litres] It is then helpful in this simulation, to graph the grout behaviour:-
The plot shows the RATE of grout taken. The rate is measured in cubic feet per 15 minutes or litres per 15 minutes and is shown here in bar form to match readings taken at 15 minute intervals.
To simplify things, the cement injected is specified, in this simulator, in bags. These are 94 pound bags where traditional units are pretended to be in use, or 40 kilogramme bags for metric. (which are near enough the same for the purposes here).The amount of cement injected so far is 0.79 bags contained in 2.75 cu ft [74 litres] of grout.
SUMMARY OF APPLICATION SO FAR - at 30 minutes
Permeability before grouting is below 10 lugeons (low)
Water:cement ratio = 3:1. Pressure = 25 psi [2 bar].
Total cement injected so far = 0.50 bags in the first 15 minutes + 0.29 bags in the second 15 minutes =
0.79 bags total
It is necessary to decide what to do next. The choice is from one of the following:
Or if you have already worked through the simulation and wish to return to some of the steps again, CLICK HERE.for a list of them.