Since there are no interruptions like testing or sample recovery involved in blasthole drilling, the aim is to drill the hole at the fastest penetration rate without causing excessive wear to the bit or drill string components.

To achieve this, apart from other influencing factors such as rotary speed, thrust on the bit, percussion blow count and energy, it is equally important to choose ideal flushing conditions. To make this choice, one must consider the pros and cons of compressed air flushing.

Today, blasthole drilling is done exclusively by rotary or rotary percussion drilling methods. The interaction between the bit and the formation creates cuttings at the bottom of the hole. A flowing medium is essential to remove these chips. Often compressed air is chosen as the flushing medium in both rotary and rotary percussion drilling methods. It is efficient in cleaning the hole due to its self-expanding property.

From the air compressor, the flushing air is taken to the drill string through tubings. The drill string components' central bore ensures that compressed air reaches the bottom of the hole. While flowing out of the hole through the annular space between blasthole and drill string, it carries the formation chips.

To effectively flush the blasthole, the compressor must supply a sufficient volume of compressed air at an adequate pressure. Therefore, it is necessary to first determine the volume and then the pressure needed at the compressor.

Blastholes are drilled in all the directions in a vertical plane. When a hole is drilled upward at 180 degrees, gravity makes the cuttings fall eliminating the need for flushing or a flushing medium. Even at an angle of (180 - R), where R is the angle of repose for the fragments of the rock, a flushing medium may be unnecessary. The particles are likely to roll down through the annular space. A flushing medium must be present when the angle is less than (180 - R).

In general, the need for flushing increases as the drilling direction changes from vertically upward to vertically downward. To flush vertically downward holes, the flushing medium must reach maximum upward velocity, which requires a maximum air discharge rate of the medium.

Table 1 - Sphericity of Shapes

Particle Shape	Sphericity
Sphere	1
Octahedron	0.85
Cube	0.81
Prism L×L×2L	0.77
Prism L×2L×3L	0.73
Cylinder H = R/15	0.25
Cylinder H = R/3	0.59
Cylinder H = R	0.83
Cylinder H = 2	0.87
Cylinder H = 20R	0.58

When a solid body falls through a medium, three forces act on it: buoyant force, drag force and gravitational force. While the gravitational force creates downward movement to the body, buoyant and drag forces oppose such movement. The gravitational and buoyant forces are constant. The drag force depends upon many factors and increases with the body's velocity. Therefore, the body's acceleration continues to decrease with increasing velocity and eventually reduces to zero. Thereafter, the body falls at a constant velocity, called terminal velocity.

For the falling body, the gravitational, buoyant and drag forces are:

4šR³D_mG/3, 4šR³D_fG/3 and 6šηRV, respectively.

By equating these components we get:

V = 2(D_m - D_f )R²/9η

Where R = particle radius, D_m = particle density, D_f = density of compressed air, η = viscosity of compressed air, G = gravitational constant, and V = terminal velocity of the falling particle.

This equation is known as Stoke's Law. This form of Stoke's Law is applicable under a certain pattern of fluid flow around the particle. Rather than diving deep into the subject of fluid mechanics and associated complex mathematics, it can be said that for spherical bodies of solid material falling through still air, the terminal velocity V_t in meters per second is given by the equation:

V_t = (3.1GDD_mD/Df)^½

Where G = 9.80665 sq. meter per second, Dm = particle density in kg per cu meter, D = particle diameter in meters and Df = air density in kg per cu meter.

The above formula assumes that spherical, smooth-surfaced bodies falling with long spacing in still air.

For typical rock cutting the air density is 1.225 kg per cu meter, particle density is 2,700 kg per cu meter and the diameter is 5 mm or 0.005 meters. By substituting the values, we get the terminal velocity as 18.30 meters per second (1,098 meters per minute or 3,600 ft per minute). If air is moving upwards at velocity V_a equal to terminal velocity V_t and if the cross section of the annular space is large, then the particles will float in the air without gravitational fall. If V_a is greater than V_t then the particle will be thrown upward at a speed of V_a - V_t.

The upward velocity of air in the annular space is called bailing velocity. For any particular case the bailing velocity V_ba can be easily calculated by the equation:

V_ba = 4•10⁶•Q/(š(D²-d²))

Where V_ba = bailing velocity in meters per minute, Q = discharge of compressor in cu meters per minute, D = drill hole diameter in mm, and d = drill pipe outside diameter in mm.

Discharge volumes of compressed air are always given as free air discharge. This means that the air will occupy equivalent volume when the air pressure drops to that of standard atmospheric air. How much bailing velocity is appropriate for particular conditions of the blasthole is debatable.

Under certain blasthole conditions, cuttings may be very large but their quantum may less than 0.5%. In such cases it is inappropriate to have a huge compressor working all the time. It is certainly more economical to allow the chips to re-fragment at the bottom and move up the hole as their size reduces.

If the bailing velocity V_ba is 10% greater than the terminal velocity V_t, the chips will be taken out of the blasthole. The appropriate value of the desired bailing velocity for particular drilling conditions can be equated as:

V_bd = 1.1aV_t

Suppose the terminal velocity is 1,100 meters per minute, as calculated earlier. Therefore the desired V_ba is 1,210a in meters per second. Correlation factor a must consider all the deviations from the assumed conditions used while calculating a terminal velocity of 1,100 meters per minute.

Aspects to consider while arriving at the appropriate value of correlation factor a are the fragments' size, density, roughness and roundness. One also needs to consider the rate of fragmentation, annular space, inclination of the blasthole and the amount of water injection. In other words, the correlation factor can be equated as:

a = a_ra_daFLUSHING THE BLASTHOLE_roua_rnda_fra_aa_ia_w

Where,

• a_r = Fragment size factor
• a_d = Fragment density factor
• a_rou = Fragment roundness factor
• a_rnd = Fragment roughness factor
• a_fr = Fragmentation rate factor
• a_a = Available annular space factor
• a_i = Blasthole inclination factor
• a_w = Presence of water injection factor

Values of each of these correlation factors can be determined from the conditions experienced in drilling.

For smooth removal, the rock chips must be smaller than about 90% of the smallest airflow passage. Larger rock chips have to re-fragment and reduce to that size. In almost any type of drill bit, the smallest passage for airflow is the flushing groove. Studies of the geometry of different types of drill bits show that most of the bits have two to four flushing grooves, each with a mean dimension of about 7% to 8% of the bit diameter. It is, therefore, appropriate to assume that the size of largest cutting is about 1/15 of the bit diameter.

Field observations agree with the above correlation for small- and medium-size blastholes. In case of large blastholes with a diameter of about 300 mm, the largest cuttings were of 19 mm.

Since a bailing velocity of 1,100 meters per second has been calculated for particles of the size 5 mm (hole diameter of 75 mm), no correction will be necessary in this case and will have a value of 1. Further, as per the terminal velocity equation, the effect of particle size will be proportional to the square root of its diameter.

With this we can determine the relation of a_r with hole diameter D as:

a_r = ( D/75 )^½

Density D_m assumed while calculating value of 1,100 meters per minute is 2,700 kg per cu meter. In actual drilling practice the density of rock varies between about 2 to 5. Thus the relation between a_d and D_m can be written as:

a_d = (D_m/2,700 )^½

The bailing velocity of 1,100 meters per second was for perfectly smooth particles. In almost all the types of rocks the chips are very different from spherical. Igneous or metamorphic rock chips are closer to tetrahedron or parallelepiped, whereas sedimentary rock chips are close to flat disks. The coefficient of drag for nonspherical chips is much higher. Therefore, the bailing velocity required to lift such particles is less than that required for spherical particles.

Particle roundness or sphericity is the ratio of spherical surface area to that of a particle with equal volume. Table 1 presents sphericity coefficients of typical shapes. Based on the table, the value of the cuttings' sphericity will be about 0.75 for particles of igneous and metamorphic rocks and about 0.6 for sedimentary rocks. Very roughly, the drag coefficient can be equated to the sphericity. Therefore, value of a_rnd will be either 0.75 or 0.6.

Particle surface roughness affects terminal velocity. When the surface is rough the terminal velocity is less. The coarser the rock-forming grains, the higher the roughness. This is because fracture generally takes place at the joint between two particles. However, since the particles formed in the drilling process are small, their roughness does not affect the terminal velocity greatly. In the case of large chips, the reduction in terminal velocity from particle surface roughness can be as much as 10%.

Thus, the relationship between a_rou and D can be written as:

a_rou = (1+[ D/1,000 ]²)

B.V. Gokhale holds two engineering degrees and has worked in the drilling and mining field for more than 20 years. The research in this article comes from his recently published book “Blasthole Drilling Technology.”

Part two will appear in the February issue of Rock Products and will examine the remaining factors and verify the validity of the hypothesis through field observations. It also will examine air pressure requirements.