We are going to give an example on blast pattern design for the case of underground drift (4 x 3m) with rectangular cross section. Parameters that we are using are:

Rock:

                Tensile strength = 6 MPa

                Poisson’s ratio = 0.25

Explosive:

                Density 1.1 g/cm3

                VOD = 4.1 km/s

                Patron diameter = 50mm

Drilling:

                Hole diameters = 51mm

                Hole length = 3m

                

For this purpose we will use burden calculator to determine burden of explosive charge and results are:

  • Burden (m): 1.45
  • r4 (m): 2.05
  • r8 (m): 1.02
  • r16 (m): 0.51
  • r32 (m): 0.26

 

As the first step we will design cut and we are going to illustrate spiral cut design in this article. The first, empty, hole can be with same or larger diameter, however we are using 51mm diameter for all holes.

Distance between first charge and empty hole is determined as:

$${{r}_{ch}}=5d=0.255m<{{r}_{32}}=0.26m$$

 

Where:     

                ${{r}_{ch}}$ - distance between empty hole and first charge

                $d$ - diameter of empty hole

 

After first charge detonates it creates new void for the following charge in sequence where blasted rock can move. If size of the create void is approximately S=0.3m distance between newly created void and next charge is determined as follows:

${{B}_{1}}=2.5\cdot S=0.75m$

 

In order to be sure that burden of new charge is not higher than it should be it is necessary that following criterion is satisfied:

${{B}_{1}}\le 0.98{{r}_{16}}$

 

Since 0.75m > 0.5m we need to reduce the distance using following expression:

${{B}_{1}}=1.2\cdot S=0.36m<0.92{{r}_{8}}=0.94m$

 

This distance is satisfactory and we are proceeding to the position of the next charge that is placed perpendicular to the newly created void (S=0.39m):

${{B}_{1}}=1.2\cdot S=0.46m<0.92\cdot {{r}_{8}}=0.85m$

 

Since this charge position satisfies, we proceed to the next one where the size of the void increased to S=0.49m:

${{B}_{1}}=1.2\cdot S=0.59m<0.92\cdot {{r}_{8}}=0.85m$

 

New void size is S=0.9m and considering this we have:

${{B}_{1}}=1.2\cdot S=1.08m>0.92\cdot {{r}_{8}}=0.85m$

 

Since this distance is not satisfactory we will make its correction as follows:

${{B}_{1}}=0.5\cdot S=0.45m<B=1.45m$

 

Figure 1 illustrates design of the spiral cut that is used in this example.

Figure 1 Spiral cut design

 

It is common that cut is placed centrally, but there are occasions when cut is dislocated from the center. After placing the cut other explosive charges are placed following few simple principles.

Main principle to follow is that distance between explosive charge and the void (free surface) must not exceed its calculated burden. Therefore, it is important to take care of the fracturing sequence. Therefore,  it is important to select proper delay time between charges, in other words we need to provide enough time for the explosive charge to do its work and create void for next charges.

And finally Figures 2 and 3 provide blasting pattern with dimensions while Figure 4 provides sequence of fracturing and initiation.

Figure 2 Blasting pattern

 

 

Figure 3 Spiral cut dimensions

 

Figure 4 Fracturing and initiation sequence

 

It is desirable that blasting pattern is simple in order to be easily drilled at site. Therefore, it is best practice to align boreholes in rows or to follow alignment as much as possible.

After design it is necessary to determine basic blasting parameters as follows.

If we are using drilling jumbo for drilling 3m long, 51mm diameter boreholes and 50mm explosive patrone (0.75kg, 382mm).

 

Stemming length is calculated as:

$0.7\cdot B=1m$

 

That means that charge length is 2m which implies number of patron per each borehole:

$$n=\frac{2000mm}{382mm}=5.2\to 5 patrons$$

 

By having 5 patrons of 382mm stemming length is slightly more than 1m. amount of explosive per each borehole is 3.75kg, with total number of charged holes is 27 and therefore total amount of explosive for the face is 101.25kg.

 

Specific quantity of explosives is:

$q=\cfrac{101.25kg}{3m\cdot 4m\cdot 3m}=2.81kg/{{m}^{3}}$

 

Blasting pattern summarized:

  1. 27+1 borehole
  2. 101.25kg of explosives
  3. 2.81kg/m3

Animated fracturing sequence

 

 

Add comment


Security code
Refresh