Mine Ventilation: Mine Atmosphere, Heat and Cooling, Atkinson’s Law, Networks, Natural Ventilation, Fans, Auxiliary Ventilation and Surveys
1. The underground atmosphere, heat load and air cooling
Fresh air is about 78% nitrogen, 21% oxygen (20.95% by volume in dry air), 0.93% argon and 0.04% carbon dioxide. Underground it loses oxygen to the oxidation of coal, timber and sulphides and to breathing and combustion, and gains carbon dioxide, methane, carbon monoxide, blasting fumes, dust, heat and moisture. Indian coal-mine regulations require the general body of air where people work to carry not less than 19% oxygen and not more than 0.5% carbon dioxide; the limits for flammable and toxic gases are in the next chapter. The job of ventilation is to supply that air, dilute and remove contaminants below their limits, and keep the climate workable.
| Source | Mechanism |
|---|---|
| Autocompression | air descending a shaft is compressed and warms at about g/cp ≈ 9.81/1005 = 9.76 °C per 1000 m (dry adiabatic), regardless of the rock |
| Strata heat | virgin rock temperature rises with depth by the geothermal gradient, and exposed rock gives up heat to the air, most in new workings |
| Machinery | almost all the electrical or diesel power used underground ends as heat; diesels also add moisture and fumes |
| Other | oxidation of coal and sulphides, explosives, human metabolism, hot groundwater, lights |
The body loses heat by convection, radiation and, in hot mines above all, by the evaporation of sweat, so humidity matters as much as temperature. The wet-bulb temperature is the single most useful index of heat stress; others are the effective temperature (from dry bulb, wet bulb and air velocity), the air cooling power in W/m² (the heat the air can remove from a sweating body) and the kata thermometer. Raising air velocity helps as long as the air is cooler than skin. When more air cannot be brought in, it is cooled: spot coolers at the face, bulk air cooling of intake air on the surface or at shaft bottom, chilled-water circuits, and ice plants in very deep mines.
2. Mechanics of airflow: Atkinson’s equation, series and parallel airways
Air moves from high to low pressure, and the frictional pressure drop along an airway in turbulent flow is given by Atkinson’s equation p = k C L V²/A = k C L Q²/A³, i.e. p = R Q² with the airway resistance R = k C L/A³. In consistent SI units p is in Pa, the friction factor k in N s²/m⁴ (equivalently kg/m³), the perimeter C and length L in m, the area A in m², V in m/s and Q in m³/s, which puts R in N s²/m⁸. An airway 1000 m long, 12 m² in area and 14 m in perimeter with k = 0.012 N s²/m⁴ has R = 0.012 × 14 × 1000/12³ = 0.0972 N s²/m⁸, and passing 30 m³/s costs 0.0972 × 900 = 87.5 Pa. The friction factor is proportional to air density, so tabulated values at standard density (1.2 kg/m³) are scaled to site density.
Airways combine like non-linear resistors. In series the same Q flows through each and the pressures add, so R = R₁ + R₂ + …. In parallel the same p acts across each and the flows add; since Qᵢ = √(p/Rᵢ), 1/√R = 1/√R₁ + 1/√R₂ + …, and each branch takes Qᵢ = Q √(R/Rᵢ). For branches of 0.9 and 0.4 N s²/m⁸, 1/√R = 1.054 + 1.581 = 2.635, so R = 0.144; in series with a further 0.256 the total is 0.400 N s²/m⁸, and 50 m³/s needs 0.4 × 2500 = 1000 Pa, with 50 × √(0.144/0.9) = 20 m³/s in the first branch and 30 in the second. Shock losses at bends, junctions and changes of section are added as equivalent lengths or as multiples of the velocity pressure ρV²/2.
A mine’s overall resistance is often expressed as its equivalent orifice, the area of a sharp-edged hole that would pass the same Q at the same pressure: A = 1.19 Q/√p (A in m², Q in m³/s, p in Pa). A mine passing 60 m³/s at 900 Pa has an equivalent orifice of 1.19 × 60/30 = 2.38 m² — a larger orifice means an easier mine to ventilate.
3. Natural and mechanical ventilation, fans and auxiliary ventilation
Natural ventilation pressure (NVP) arises when the air in the downcast is denser — colder — than in the upcast: NVP = g H (ρ̄d − ρ̄u), with ρ̄ the mean densities of the two columns over the depth H. A 400 m deep mine with a density difference of 0.05 kg/m³ has an NVP of 9.81 × 400 × 0.05 = 196.2 Pa. It changes with the seasons and the time of day, may reverse in summer where intakes are warm, and either assists or opposes the fan; it is never relied on alone in a large mine.
Main fans are centrifugal (air turns through 90°; backward-bladed designs have a stable, non-overloading characteristic) or axial (air passes straight through; blade pitch can be adjusted, and there is a stall region at high pressure where operation is unstable). A fan’s characteristic is its p–Q curve with its power and efficiency curves; the mine’s is the parabola p = RQ², and the operating point is where they cross. If a fan develops p = 2400 − 0.1Q² on a mine of R = 0.5, then 0.5Q² = 2400 − 0.1Q² gives Q = √4000 = 63.25 m³/s at 2000 Pa. The air power is p Q, and the shaft power p Q/η. Fans in series add pressures at the same quantity; fans in parallel add quantities at the same pressure. Booster fans underground raise the pressure in one district and must be interlocked and monitored to prevent recirculation.
| Quantity | Varies as | Speed 600 → 720 rpm (×1.2) |
|---|---|---|
| Air quantity Q | N | 100 → 120 m³/s |
| Fan pressure p | N² (and ρ) | 1500 → 2160 Pa |
| Power P | N³ (and ρ) | × 1.728 |
Auxiliary ventilation takes air from the main circuit into a dead-end heading through a small fan and ducting. A forcing system blows fresh air through the duct to the face, where its jet sweeps gas from the face — but returns the contaminated air down the heading past the men. An exhausting system draws air from the face through the duct, so the heading carries fresh air and dust is captured at source, but the suction reaches only a short distance from the duct end. An overlap system combines a forcing duct with a short exhausting duct and a dust filter. Ducts leak, so the fan must supply more than the face receives, and the auxiliary fan must be placed in intake air with enough through-flow in the main airway to prevent recirculation.
4. Ventilation survey, planning and networks
A ventilation survey measures quantities and pressures to find where the air goes and where it is lost. Quantity: Q = V A, with the mean velocity from a vane anemometer traversed across the airway (or smoke-tube timing at low speed) and the area measured at the same station. Pressure: the gauge-and-tube method runs a long tube between two stations to a differential gauge and reads the frictional drop directly; the barometer method reads absolute pressures and corrects them for elevation and air density. From p and Q each branch’s resistance follows, R = p/Q², and the survey calibrates the network model used for planning.
Planning fixes the air each district needs from the largest of the demands: the quantity per person and per kW of diesel power that the regulations and good practice require, the velocity range for dust and comfort, the heat to be removed, and the dilution of gas. Gas emitted at q m³/s into air arriving with concentration Cᵢ must be diluted below a limit C: Q = q/(C − Cᵢ), so 0.3 m³/s of methane with a 1% limit and 0.1% already in the intake needs 0.3/0.009 = 33.3 m³/s. A ventilation network is solved with Kirchhoff’s laws: the algebraic sum of flows at a junction is zero, and round any closed mesh the frictional drops sum to the fan pressure and NVP in that mesh. The Hardy Cross method iterates: guess flows that satisfy the junction law, then correct each mesh by ΔQ = −Σ(RQ|Q| − p_fan − NVP)/Σ 2R|Q| until the corrections vanish.
Key takeaways
- Atkinson: p = RQ² with R = kCL/A³; in SI, k in N s²/m⁴ gives R in N s²/m⁸ and p in Pa; halving A multiplies R by eight.
- Series: R = ΣRᵢ; parallel: 1/√R = Σ1/√Rᵢ and Qᵢ = Q√(R/Rᵢ); equivalent orifice A = 1.19Q/√p.
- NVP = gH(ρ̄d − ρ̄u); the operating point is where the fan curve meets p = RQ²; Q ∝ N, p ∝ N², P ∝ N³.
- Autocompression adds about 9.76 °C per 1000 m of descent; wet-bulb temperature is the key heat-stress index.
- Dilution air Q = q/(C − Cᵢ); Kirchhoff’s junction and mesh laws, iterated by Hardy Cross, solve a network.
Practice questions (15)
Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.
An airway is 1000 m long, 12 m² in cross-section and 14 m in perimeter, with an Atkinson friction factor of 0.012 N s²/m⁴. The pressure drop when it carries 30 m³/s, in Pa, to one decimal place, is ____.
Numerical answer — type the value.
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Answer: 87.5
R = kCL/A³ = 0.012 × 14 × 1000/1728 = 0.09722 N s²/m⁸, and p = RQ² = 0.09722 × 900 = 87.5 Pa. Using A² in place of A³ gives 1050 Pa, and p = RQ (not squared) gives 2.9 Pa.Two airways of resistance 0.9 and 0.4 N s²/m⁸ are connected in parallel. Their equivalent resistance, in N s²/m⁸, to three decimal places, is ____.
Numerical answer — type the value.
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Answer: 0.144
1/√R = 1/√0.9 + 1/√0.4 = 1.0541 + 1.5811 = 2.6352, so √R = 0.3795 and R = 0.144. The electrical-parallel formula 1/R = 1/0.9 + 1/0.4 gives 0.277, which is wrong because p varies as Q², not Q.The parallel pair of 0.9 and 0.4 N s²/m⁸ is in series with an airway of 0.256 N s²/m⁸. The pressure needed to pass 50 m³/s through the whole circuit, in Pa, is ____.
Numerical answer — type the value.
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Answer: 1000
The parallel pair is 0.144 N s²/m⁸, so the total is 0.144 + 0.256 = 0.400, and p = 0.4 × 50² = 1000 Pa. Adding 0.9 + 0.4 + 0.256 as if all were in series gives 1.556 and 3890 Pa.In the same circuit carrying 50 m³/s, the quantity flowing through the 0.9 N s²/m⁸ branch, in m³/s, is ____.
Numerical answer — type the value.
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Answer: 20
Q₁ = Q √(R_eq/R₁) = 50 × √(0.144/0.9) = 50 × 0.4 = 20 m³/s, and the 0.4 branch takes 30. Check: 0.9 × 20² = 360 Pa = 0.4 × 30². Splitting in inverse proportion to R gives 15.4 m³/s, the linear-resistor answer.A main fan running at 600 rpm develops 1500 Pa. If its speed is raised to 720 rpm on the same mine, the fan pressure, in Pa, becomes ____.
Numerical answer — type the value.
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Answer: 2160
p ∝ N², so p₂ = 1500 × (720/600)² = 1500 × 1.44 = 2160 Pa, while Q rises in proportion to N (×1.2) and power by 1.728. Scaling pressure linearly gives 1800 Pa, which is the quantity law misapplied.A mine is 400 m deep, and the mean air density in the downcast shaft exceeds that in the upcast by 0.05 kg/m³. Taking g = 9.81 m/s², the natural ventilation pressure, in Pa, to one decimal place, is ____.
Numerical answer — type the value.
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Answer: 196.2
NVP = g H Δρ = 9.81 × 400 × 0.05 = 196.2 Pa, acting to drive air down the denser column. Omitting g gives 20 kg/m², a mass per area rather than a pressure; the sign reverses if the upcast column becomes the colder one.A fan delivers 100 m³/s at a fan pressure of 1500 Pa with an efficiency of 75%. The shaft power input, in kW, is ____.
Numerical answer — type the value.
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Answer: 200
Air power = pQ = 1500 × 100 = 150 000 W = 150 kW, and shaft power = 150/0.75 = 200 kW. Quoting 150 kW stops at the air power, and 150 × 0.75 = 112.5 kW applies the efficiency the wrong way round.A mine passes 60 m³/s under a total mine pressure of 900 Pa. Its equivalent orifice, in m², using A = 1.19 Q/√p, to two decimal places, is ____.
Numerical answer — type the value.
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Answer: 2.38
A = 1.19 × 60/√900 = 71.4/30 = 2.38 m². The constant 1.19 applies with p in Pa; with the pressure in mm of water gauge the constant is 0.38, and mixing the two gives an orifice wrong by a factor of about 3.A development district emits 0.3 m³/s of methane. The intake air arriving at it already contains 0.1% methane, and the return must not exceed 1%. The minimum quantity of air required, in m³/s, to two decimal places, is ____.
Numerical answer — type the value.
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Answer: 33.33
Q = q/(C − Cᵢ) = 0.3/(0.010 − 0.001) = 0.3/0.009 = 33.33 m³/s. Ignoring the methane already in the intake gives 30 m³/s, and using percentages as whole numbers (0.3/0.9) gives 0.33, a hundredfold error.Taking g = 9.81 m/s² and cp = 1005 J/(kg·K) for dry air, the rise in dry-bulb temperature by autocompression of air descending a 1000 m shaft, in °C, to two decimal places, is ____.
Numerical answer — type the value.
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Answer: 9.76
ΔT = g Δz/cp = 9.81 × 1000/1005 = 9.76 °C, about 1 °C per 100 m. It is independent of the rock: it is the potential energy of the air turned into enthalpy. Evaporation in a wet shaft makes the observed dry-bulb rise smaller.A fan’s characteristic is p = 2400 − 0.1Q² (p in Pa, Q in m³/s) and the mine resistance is 0.5 N s²/m⁸. The air quantity at the operating point, in m³/s, to two decimal places, is ____.
Numerical answer — type the value.
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Answer: 63.25
At the operating point 0.5Q² = 2400 − 0.1Q², so 0.6Q² = 2400, Q² = 4000 and Q = 63.25 m³/s, at p = 0.5 × 4000 = 2000 Pa. Putting the fan’s shut-off pressure straight onto the mine curve gives √(2400/0.5) = 69.28, which ignores the fan’s falling characteristic.Compared with an exhausting auxiliary system, a forcing auxiliary system in a development heading:
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Answer: B — delivers a jet of fresh air that sweeps gas from the face, but returns the contaminated air along the heading
A forcing duct throws a jet well beyond its end and scours the face, which suits gassy headings, but everything it picks up travels back along the heading. The exhausting system is the one that keeps the heading fresh and captures dust at source through the duct.Which of the following statements about mine fans are correct? (Select all that apply.)
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Answer: B — Operation of an axial fan in its stall region is unstable; C — Fans in series add their pressures at the same quantity; D — The blade pitch of an axial fan can be adjusted to change its duty
In series the same air passes through both fans and each adds pressure; in parallel they share one pressure and add quantities, so the first statement describes series, not parallel. An axial fan’s characteristic has a stall dip where two flows share a pressure and the flow hunts, and adjustable pitch is how axial fans are matched to a changing mine.Kirchhoff’s second law applied to a mine ventilation network states that round any closed mesh:
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Answer: B — the algebraic sum of the frictional pressure drops equals the fan pressure and natural ventilation pressure acting in that mesh
The mesh law is a pressure balance: going round a loop the pressure must return to its starting value, so the drops RQ|Q| are balanced by whatever fans and NVP act in the loop. The sum of quantities being zero is the first (junction) law, and resistances add only for branches in series.For turbulent flow in a mine airway obeying Atkinson’s law, which of the following are correct? (Select all that apply.)
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Answer: A — Resistance is proportional to perimeter and length and inversely proportional to the cube of the area; B — Doubling the quantity through a fixed airway quadruples the pressure drop; D — Doubling the quantity through a fixed airway raises the air power eightfold
R = kCL/A³; p = RQ² quadruples when Q doubles, and the air power pQ = RQ³ rises eightfold — the reason big quantities are expensive. The friction factor is proportional to air density, so k must be corrected from the standard 1.2 kg/m³ to the density in the airway.