Technical Report On The Design Of Mine Power System

Write a Technical Report on the Design of Mine Power System

There are various methods used in the mining sector. Notably, the major categories include shallow and deep mining; both having completely different requirements to facilitate the noble process of ensuring the minerals are brought to the surface in an efficient and safe fashion. Importantly, therefore, engineers are mostly required to design, develop, install, operate and maintain the power supply systems associated with mining. Now, there are many power supply system configurations based on different power sources; and selection of these systems would often be based on certain fundamental considerations (Tuck, no year). The two major sources of power that are commonly used include: the hydraulic powered system and compressed air system. The former is gaining momentum thanks to its greater working efficiency; however, the latter has traditionally been the major source of powering the drills due to its rather simpler configurations and cheaper maintenance and running costs (SME Inc, 1992). The focus of this report is on the compressed air supply system to power the rock drills which are located deep underneath. The next section provides an in-depth discussion on the proposed system. However, the case given for the purpose of the system design includes: A mine consisting of a 2-stoping section located at a depth of 2200m accessed by a shaft. One of the stopes houses 10 rock drills located at a distance of 1500m from the shaft and the other one is located at a distance of 1800m which houses 8 rock drills. Now, both sections are accessed from the same main drive by access drives that are 150m in length. It should be noted that there are no receivers as the compressed air supply is supposed to be continuous. Therefore, the aim of this report is to develop a workable mine power system design by considering the design criteria that is provided in the next section. Notably, however, the focus of the design is on deriving a system that if implemented will be more efficient, safe and easier to operate and maintain.

The functional elements constituting the system include: the prime mover (electrical motors); multistage centrifugal compressor; the main drive piping (steel pipes); access drive steel pipes and hose pipes 

The system operation would normally happen as follows: The prime mover (in this case the electric motor will power the multistage compressor which has multiple power strokes hence continuous supply of compressed air is made possible. The compressed air at a pressure that will sustain the drilling action is supplied to the rock drills situated in the two main stoping sections; this is done via a network of pipes mainly steel and hose pipes. It should be noted that due to losses during compressed air transmission (mainly due to friction and shock) the actual pressure supply to the drills will be less than the available pressure at the compressor outlet. The main drive shaft vertically houses the main steel pipe which then branches off into the two main access drives pipes to simultaneously access the two stopes having a total of 18 rock drills. The rock drills are then driven pneumatically to drill into the hard rocks in order to gain access to the required mineral ore. The action must be continuous hence the selection of multistage compressor with no receiver integrated in the system. However, there will be provision for intercooling between stages and will go up to three stages. 

Design Calculations

The centrifugal compressor was selected due to the following reasons:

• Low energy use for large capacities
• Quiet hence contributing system’s better acoustic performance
• The capacity is controllable via some linked control accessories

However,

• It is sensitive to dirt in air hence incorporation of air filtering and conditioning unit (such as dehumidifiers)
• There are normally relatively higher costs although it can be energy efficient if properly used (FOODSA, 2013)  

Density Calculations will be done to derive the proposed system:

The Rock drill design

According to the power point class notes, the rock drill pressure requirements is between 500 to 600kPa gauge as shown in table 1.

Table 1: Pressure requirements for different appliances

For the purpose of the design, let us work with the maximum hence 600kPa gauge fits the selection criteria

Now, the total pressure requirements for the drills= 18x600=10800kPa gauge

The air consumption per drill (as per table 2) assume it works like jack hummer hence consumption value at 0.06kg/s, again for design purpose, therefore the total air consumption for the 18 drills = 0.1x18= 1.08kg/s  

Table 2: Compressed consumption for different appliances

Now, the above will be the amount of air supplied from the main drive pipe (assuming zero leakages). However, design must cater for both frictional and shock losses hence compressor must work to deliver better compression.

Design of the pipe system:

Diameter and length of the pipe sections are determined as follows:

The pressure drop criterion can be used as the basis for determining the size of the pipes, namely: main drive, access drive and hose pipes:

For the main drive pipe, the total length is 2200m

Now, given the equation:  Pressure drop, P’= RfM²Lx10^-3/w…….(1)

From equation (1) above we will be able to determine the resistance factor Rf which will then be used to estimate the size of the pipe; however, the following must be calculated first:

Density of air by calculating values of w at different locations in the system:

From the surface to 2200mm, w= Px 10³/RT= (600+100) x10³/287x(273+26)

= 700 000/85813

= 8.1573= 8.16kg/m³

At the base where atmospheric pressure is 105kPa, w= (600+105)x 10³/287x(273+26)

=705000/85813

= 8.22kg/m³

And getting the average Wav= (8.22+8.16)/2= 8.19kg/m³

Note also that the pressure drop can be given in terms of mmHg by applying equation (2) below:

m’= C(wh)^0.5……..(2)

Now, assume a circular orifice is used to regulate the flow with a coefficient C lying between 0.61 to 0.64 hence take average 0.625 and substituting the values:

1.08= 0.625(hx8.19)^0.5

1.08/0.625= (8.19h)^0.5

Hx8.19= 2.9864, h= 0.3646mmHg 

The pressure drop, p’= h’?g= 9.81x 1000x 0.3646= 3.577kPa=3.6kPa

And substituting in equation (1):

Rf= 10x1000x8.19/(1.08²x2200) = 31.92

Using table 3 below to select the considerable pipe diameter:

Table 3: Resistance factor pipe selection criteria

Hence diameter of the main drive pipe will be fixed at 457mm and will be made of steel material

Design of access drives and hose pipes will follow the same criterion hence:

Now, since there is uniform supply of air to both stopes, the average air supply= m₁’= m’/2= 1.08/2= 0.54kg/s

Substituting in equation (2):

0.54= 0.625x(hx8.22)^0.5

Hx8.22= 0.864²=0.7465, h= 0.0908

The pressure drop, p’= 0.0908x9.81x1000= 0.891kPa

And substituting (when L=1500m) to get Rf= 0.891x8.22/(0.9²x1500)=0.006028

And when L=1800mm, Rf= 0.891x8.22/(0.9²x1800) = 0.0050233 

The total pressure that must be available at the compressor outlet= 10800+10+2.5x2= 10815kPa

This can only be achieved by installation of a 3-stage compressor configuration with intercooling between stages such that it is complete that is, the inlet temperature at every successive stage must be equal 

Other parameters that the selection must consider include: compressor pressure ratio (at stage level), airmass flow rate, and inlet temperatures

This was catered for during design of the pipes. However, shock losses were assumed negligible since if compared with frictional losses, it is nothing close to it hence the basis for size selection was based purely on frictional losses in pipes during transmission.

This is normally a phenomenon that is experienced when compressed air is moved from higher to lower elevation resulting into involuntary pressure increases which can affect the system performance if not considered in the design and will be determined as follows:

This can be done partially:

Exp2gH/RT= exp{2x9.81x2200/287x299}

         = exp{43164/85813}= exp(0.5030007)

= 1.6537

RfR²T²/(P²sgx10⁶)= 11.489x287²x299²/(100x10³)²x9.81x10⁶= 8.624X10^-7

Exp(2gH/RT-1)M²L/H= (1.6537-1)x1.08²x2200/2200= 0.7625

Pe= (100x10³)x(1.6537-8.624x10^-7x0.7625)^0.5

= 100x10³x(1.6537)^0.5= 128.6x10³=129kPa 

Conclusion

The design of the power supply system provided above has succinctly provided the technical path towards the configuration, implementation and operation of the said system. It should be noted that the system should be put under several tests before implementation to monitor its performance prior to the actual implementation so as to minimize on the project cost. Notably, there are modern simulation and modelling tools that can be used for this purpose. However, the proposed design provides an impetus from which further work on the same can be developed. As mentioned earlier, the design focused on improving efficiency, system safety and ease of operation. Therefore, it is highly recommended for implementation after further developments. 

Reference

Tuck, M.A. (no year). Compressed Air. University of Ballarat, Australia. Available from: https://air.ingersoll-rand.com [Accessed: 19/07/2017]

SME Inc. (1992). Compressed Air Power. 3^rd Edition, Colorado. Available from: https://foodsouthaustralia.com.au/wpcontent/uploads/2013/05/foodsa_eeetoolkit_compressedair.pdf [accessed: 19/07/2017]

FOODSA. (2013). Compressed Air: Energy Efficient Equipment Toolkit. Available from: https://foodsouthaustralia.com.au/wp-content/uploads/2013/05/foodsa_eeetoolkit_compressedair.pdf [accessed: 20/07/2017]

De La Vergne, J.N. (2003). The Hard Rock Miners Handbook. 3^rd Edition Mc Intosh Engineering, Inc. Available from: www.mcintoshengineering.com