1. Introduction
This section is devoted to “7 bar” compressed air systems for mine application and does not deal with low-pressure (0.5 -2 bar) compressed air supplied by “blowers” in the mill for flotation cells or process tanks.
Compressed Air Inefficiency
A compressed air system is the traditional means used to supply energy for underground mining equipment. Converting electrical power into mechanical power by compressing air is the most expensive and least efficient utility in the mining industry. The actual efficiency may be near 20% compared to approximately 40% for diesel engines and 95% for electric motors. Because air has no smell or color, leaks are not readily detected and a large amount of compressor capacity is lost through waste and neglect. The efficiency of compressed air distribution systems is made worse because mines do not normally dry air beyond removing precipitation from primary cooling at the source. The air leaving a compressor is typically in the order of 2750 F (1350 C). The after cooler lowers it to approximately 1050 F (400 C) at which point it is 100% saturated.
The result of the cooling is to precipitate approximately 2/3 of the moisture contained in the air stream. This water is normally automatically removed from a separator at the after cooler and a surface receiver removes more. Unless the mine air is very warm, further cooling will take place underground precipitating water so rust develops on the inside walls of pipe lines, which eventually increases the resistance to the compressed air flow. The increased friction and rust build-up as a result of corrosion is one reason that underground mines experience a gradual decrease in air pressure over a prolonged period of time. After a number of years, the corrosion will progress to the point that steel pipelines require replacement. When compressed air travels downward into the mine (shaft column), it experiences warming due to auto-compression, but this is normally insufficient to prevent precipitation.
When the compressed air reaches its final destination, it expands and cools rapidly often resulting in icing that can completely freeze an underground machine. Another problem is noise, particularly noise produced from the exhaust ports of drilling equipment. The noise level for a jackleg drill operator is approximately 110 decibels, and for a jumbo operator (further away from the drill) the level is near 100. Hearing damage is believed most likely to occur at noise levels above 85 decibels between 1,200 Hz and 4,800Hz. In this frequency range, earplugs will lower the noise level to the ear of the operator by about 25 decibels and earmuffs by about 40. Hearing protection is less effective at lower frequencies.
Compressed Air Replacement
The introduction of hydraulic drills, replacement of slushers with LHD units, and advent of onboard compressors for mobile equipment have decreased the requirements for a compressed air circuit underground. A few mines (particularly in arctic regions) have completely eliminated the requirement for a stationary compressed air plant on surface. One mine in South Africa, Northair Platinum, has replaced compressed air with water from a shaft column. The drills are designed to run on hydropower instead of compressed air or hydraulic oil.
Most operating mines still find that a compressed air reticulation system in the mine is necessary. It is found useful for blowing muck; operating hand held drills and shop equipment, dislodging hangups, clearing pipelines, dispersing sediment, applying shotcrete, cleaning equipment, spot cooling, air lifts, diffusing stench gas, connection to refuge stations, and ventilating raise headings.
Compressor Capacity
Compressor capacity is expressed in terms of the volumetric rate of inlet air processed. Capacity refers to performance at sea level at a specified compression ratio or outlet pressure. In Imperial units, the outlet pressure of the compressed air is expressed in pounds per square inch over the atmospheric pressure (psi or psig). In metric terms, the pressure is correctly expressed in bars or kilopascals (kPa). A bar was once equivalent to standard atmospheric pressure (1.03 kg/cm2 or 14.7 psi).
Today, it is defined as being equal to 100 kilopascals. 1 bar = 1.02 kg/cm2 = 14.50 psi =100 kPa In empirical terms, the capacity is expressed in cubic feet per minute (cfm). In metric terms, it may be expressed in liters per second (l/s) or cubic meters per minute (m3/min).
1 m3/min = 16.7 l/s = 35.3 cfm
1 l/s = 2.12 cfm
Rules of Thumb
As a result of the decreasing requirement underground, certain traditional rules of thumb related to the required compressor plant capacity are no longer valid; therefore, these are not included in the list that follows.
2. Rules of Thumb
Power
• The horsepower required for a stationary single-stage electric compressor is approximately 28% that of its capacity, expressed in cfm (sea level at 125 psig). Source: Lyman Scheel
• The horsepower required for a portable diesel air compressor is approximately 33% that of its capacity, expressed in cfm (sea level at 125 psig). Source: Franklin Matthias
• To increase the output pressure of a two-stage compressor from 100 to 120 psig requires a 10% increase in horsepower (1% for each 2 psig). Source: Ingersoll-Rand Air Intake
• The area of the intake duct should be not less than ½ the area of the low-pressure cylinder of a two-stage reciprocating compressor. Source: Lewis and Clark Cooling
• A water-cooled after-cooler will require approximately 3 USGPM per 100 cfm of air compressed to 100 psig. Source: Ingersoll-Rand Receiver
• The primary receiver capacity should be six times the compressor capacity per second of free air for automatic valve unloading. Source: Atlas Copco
• The difference between automatic valve unloading and loading pressure limits should not be less than 0.4 bar. Source: Atlas Copco Air Line Losses
• At 100 psi, a 6-inch diameter airline will carry 3,000 cfm one mile with a loss of approximately 12 psi. Source: Franklin Matthias
• At 100 psi, a 4-inch diameter airline will carry 1,000 cfm one mile with a loss of approximately 12 psi. Source: Franklin Matthias
• A line leak or cracked valve with an opening equivalent to 1/8-inch (3 mm) diameter will leak 25 cfm (42m3/min.) at 100 psig (7 bars). Source: Lanny Pasternack
• In a well-managed system, the air leaks should not exceed 15% of productive consumption. Source: Lanny Pasternack
• Many older mines waste as much as 70% of their compressed air capacity through leakage. Source: Robert McKellar
• Drilling requires a 25-psi air-drop across the bit for cooling to which must be added the circulation loss for bailing of cuttings in the borehole at a velocity of 5,000 fpm, or more. Source: Reed Tool
• Except in South Africa, pneumatic drills are usually designed to operate at 90 psig (6.2 bars). Their drilling speed will be reduced by 30% at 70 psig (4.8 bars). Source: Christopher Bise
• A line oiler reduces the air pressure by 5 psi. Source: Ingersoll-Rand
• An exhaust muffler can increase the required air pressure by 5 psi, or more. Source: Morris Medd
• A constant speed compressor designed to be fed at 60 cycles (hertz) will operate at 50 cycles, but experience a reduction in capacity of about 17%. Source: Jack de la Vergne
Altitude
• A constant speed compressor (or booster) underground will require 1% more horsepower for every 100m of depth below sea level. Source: Atlas Copco
• Auto-compression will increase the gage pressure of a column of air in a mineshaft by approximately 10% for each 3,000 feet of depth (11% for each 1,000m). Source: Jack de la Vergne
• The compressed air from a constant speed compressor will have 1% less capacity to do useful work for every 100m above sea level that it is located. Source: Atlas Copco
3. Tricks of the Trade
• A common misinterpretation of compressor capacity is to assume it refers to the outlet volume; in fact, it always refers to the inlet of ambient air. Source: Ingersoll Rand
• To maximize precipitation in a primary receiver, it should be placed in a cool location, but not outside in winter if the ambient air temperature falls much below freezing. Source: Jack de la Vergne
• A needle gauge (with spare needles) is an indispensable tool for determining pressure in a hose at the machine end. Source: Henry Lavigne
• “Air-less” mines are well served with skid-mounted portable compressors that can be towed as required around the workings by any unit of mobile equipment. Source: Tony Keene
• Its usually almost impossible to properly seal the bulkhead on a blind drift used to provide a large receiver underground. Source: Jim Redpath
• At altitudes less than 13,000 feet (4,000m), the textbook reduction factors for the capacity of an electric motor driving an air compressor may usually be ignored provided the motor has sufficient cooling. Source: George Greer
• Icing up of a unit of underground equipment can be avoided with a de-mister attachment or the application of tanner gas to the compressed air distribution lines. Source: Rudy Warren
• Employing plastic (instead of steel) pipe for distribution on mine levels can avert line pressure losses due to friction from corrosion inside the pipe. Source: Jim Tilley
• Line pressure losses due to shock can be diminished by the insertion of reducers at T and Y connections. Source: Khoa Mai
• Turbulent flow (as opposed to laminar or mixed flow) should be assumed when calculating line losses for a compressed air line at a mine. Source: Andy Pitz
4. Air Line Diameter
A myriad of methods exist to determine the diameter of an airline from simple rules of thumb to tables in handbooks to complicated formulae. The formulas are mainly derivations of D’Arcy’s elementary formula (below), taking some account for the compressibility of air.
H = fLd2/2g
The inside pipe diameter (d) required for a temporary installation (± one year) of new steel pipe or a permanent installation with a stainless steel or plastic pipe may be calculated by modifying the Weymouth formula to the following configuration.
The pipe diameter (d) required for a permanent installation of new steel pipe or a hose length before the lubricator (oiler) attachment may be calculated by the Simons formula.
In which, d= inside pipe diameter (inches)
L = length of air line (feet)
Q = Free air flow (cfm)
P1 and P2 are the absolute pressures at the inlet and the outlet (psia)
Example
Determine the minimum inside pipe diameter required for a 10-psi drop.
Facts:
1. L = 5,800 feet
2. Q = 10,000 cfm
3. Atmospheric pressure = 15 psia
4. Inlet pressure = 110 psig, so P2 = (110 +15) = 125 psia
5. Outlet pressure (10 psi drop), P1 = (125 - 10) = 115 psia
Solution:
1. For a temporary steel airline:
d=[10,0002 x 5800 /1736 (1252-1152)] 0.188 = 9.3 inches
2. For a permanent steel airline
d=[10,0002 x 5800 /2000 (1252-1152)] 0.2 = 10.4 inches
(Refer to Table 20-8 in Chapter 20.11 for standard steel pipe dimensions.)
5. Air Lines Leaks
Older underground mines are plagued with air leaks. Actual measurements obtained by measuring the airflow at shutdown of underground operations at two mines produced the following results.
• At the first mine (30 years old), leakage consumed 52% of the installed compressor capacity.
• At the second (20 years old), leakage consumed 39% of installed compressor capacity.
At the second mine, 867 leaks were discovered of which 46% were involved grooved pipe mechanical couplings, 10% threaded pipe couplings, 23% faulty valves, 12% damaged hoses, and the remainder (9%) miscellaneous faults.
The leakage in all or any part of a pipeline network may be calculated by using the following process.
• Calculate the volume of air in the line or network to be tested
• Convert the volume to free air (atmospheric pressure)
• Fill the lines with air at normal operating pressure
• Close the lines at every end
• Time the fall in line pressure until it reaches zero
• Apply the Briggs formula: Q = 5V/2t
In which,
Q = leakage in cfm
V = volume of free air in the system
t = the time in minutes from shutdown until the gage pressure reaches zero
A procedure has evolved to deal with leaks when they become intolerable. A dedicated team is assembled who may spend six weeks tracing the entire pipe network in the underground mine. Leaks are mended when discovered. Piping to blind ends is permanently capped. As the work progresses, the line pressure increases, which in some cases precipitates leaks in steel pipelines that must then be replaced. At some mines, it is not practical to have one repair crew working in different beats, so the non-staff members of the dedicated team are rotated as required. Preventive measures against leaks include an employee awareness campaign and quality control over pipe alignment at couplings.
6. Equipment Air Requirements
Table 19-1 contains typical compressed air requirements for a selection of underground equipment.
Refer to Table 19-2 for consumption data on additional equipment.
Table 19-1 Typical Compressed Air Requirements
7. Compressed Air Plant Capacity
The following methods may be employed to determine the plant capacity required for a proposed mine.
• Formula
• Averaging data from actual installations at comparable mines
• Detailed spreadsheet analysis
Example
Calculate the compressed air capacity required for a proposed underground mine by each of the three methods above.
Facts:
1. The mine will be a trackless operation using diesel powered mobile equipment.
2. The mine production rate will be 3,000 short tpd
Solution:
1. Formula
C = 140 T0.5 (O’Hara)
T = Short tons of ore mined daily (3,000)
C = New plant capacity in cfm
C = 140 x (3,000)0.5 = 140 x 55 = 7,700 cfm
2. Comparable Mine Installations
3. Detailed Spreadsheet Analysis
The spreadsheet lists each item of equipment along with its nominal air consumption. The consumption figure is then rationalized to account for operating hours and utilization. A spreadsheet for this example and the resulting solution (8,000 cfm) are found in Table 19-2. Of the three solutions obtained (7,700, 9,800, 8,000), the middle one is approximately 25% higher than the other two. This may be expected since the value was derived mainly from installations at older mines. In this case, a plant capacity of 8,000 cfm would likely be selected for the new mine.
Table 19-2 Equipment Air Consumption
8. Altitude and Depth
The atmospheric pressure is less at high altitude than it is at sea level. The atmospheric pressure in a deep mine is greater. The same is true of a column of compressed air such as the airline in a mineshaft. The compressed air in a shaft column will gain more pressure with depth than the atmosphere and so the gage pressure will increase at the bottom of a mineshaft. (If the airflow is moving, this gain will be mitigated by friction.) The difference in atmospheric pressure and gage pressure due to change in altitude may be considered adiabatic and each calculated with the following formula.
P2 = P1[1 - g(z1-z2)/T1Cp]1/k
In which,
T1 = 200C = 2930K (atmosphere)
P2 = Final pressure, kPa T1 = 400C = 3130K (compressed air)
P1 = Initial pressure, kPa Cp = 1014 J/kgK (atmosphere)
g = 9.81 m2/s Cp = 1079 J/kgK (compressed air)
z1 = initial altitude, m 1/k = 3.500 (atmosphere)
z2 = final altitude, m 1/k = 3.759 (compressed air)
Example
Determine the atmospheric pressure and the static airline pressure for the following situation.
Facts:
1. The depth is 2,000m
2. The mine has a compressor at sea level
3. The atmospheric pressure is 101.3 kPa (14.7 psi)
4. The air is compressed to a gage pressure of 689 kPa (100 psi)
Solution:
1. Atmospheric pressure at depth
P2 = P1[1 - g(z1-z2)]/(T1Cp)1/k = 101.3[1 – 9.81(0-2,000)]/(293 x 1014)]3.500 = 127 kPa
2. Compressed air pressure at depth
P2 = P1[1 - g(z1-z2)/T1Cp]1/k = (689+101.3) [1 - 9.81(0-2,000)/(313 x 1079)]3.759 = 977 kPa
3. Gage pressure at depth = 977 -127 = 850 kPa (123.4 psi)
In Table 19-3, this sample problem is extended for different altitudes and depths of mining.
Table 19-3
Standard Atmosphere at Altitudes/Depths and Air Compressed to 100psig (689 kPa) at Sea Level
