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Sample Certificate in Mine Environmental Control Exam Practice Questions
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1A vertical downcast shaft has a collar elevation at surface where barometric pressure is 85.0 kPa. If the shaft depth is 2 000 m and the mean air density down the shaft is 1.15 kg/m³, what is the barometric pressure at the shaft bottom? (Assume g = 9.81 m/s²)
A.107.56 kPa
B.96.28 kPa
C.114.42 kPa
D.102.10 kPa
Explanation: The barometric pressure increase due to depth (static head) is calculated as ΔP = ρ · g · z = 1.15 kg/m³ × 9.81 m/s² × 2 000 m = 22 563 Pa = 22.563 kPa. Adding this to the surface pressure gives P_bottom = 85.0 + 22.563 = 107.56 kPa.
2In mine psychrometrics, why is Sigma Heat Content (S) preferred over total enthalpy (h) when analyzing adiabatic moisture exchange along mine airways?
A.Sigma heat content remains strictly constant during pure sensible and latent heat exchanges where liquid water evaporates at the wet-bulb temperature
B.Sigma heat content includes the sensible heat of liquid water added to or removed from the airway, whereas enthalpy excludes it
C.Sigma heat content is independent of barometric pressure, making it uniform across all mine depths
D.Sigma heat content eliminates the need to measure dry-bulb temperature
Explanation: Sigma heat content (S = c_pa · t + w · h_fg) represents the heat content of moist air per unit mass of dry air above a liquid reference state at wet-bulb temperature. During adiabatic evaporation of water at the wet-bulb temperature, Sigma heat content remains constant, whereas total enthalpy varies slightly due to liquid enthalpy terms.
3Calculate the density of dry mine air at a barometric pressure of 100.0 kPa and a dry-bulb temperature of 27.0 °C (300.15 K). Take the gas constant for dry air as R = 287.05 J/(kg·K).
A.1.161 kg/m³
B.1.293 kg/m³
C.1.050 kg/m³
D.1.225 kg/m³
Explanation: Using the ideal gas equation ρ = P / (R · T), where P = 100 000 Pa, R = 287.05 J/(kg·K), and T = 27.0 + 273.15 = 300.15 K: ρ = 100 000 / (287.05 × 300.15) = 100 000 / 86158 = 1.1606 kg/m³ ≈ 1.161 kg/m³.
4A deep gold mine has a downcast shaft with a mean air density of 1.22 kg/m³ and an upcast shaft with a mean air density of 1.10 kg/m³. If the vertical shaft depth is 1 800 m, calculate the Natural Ventilation Pressure (NVP) generated by the density difference. (g = 9.81 m/s²)
A.2 118.96 Pa
B.1 765.80 Pa
C.2 450.00 Pa
D.1 420.50 Pa
Explanation: Natural Ventilation Pressure is given by NVP = g · z · (ρ_down - ρ_up). Substituting values: NVP = 9.81 m/s² × 1 800 m × (1.22 - 1.10 kg/m³) = 9.81 × 1 800 × 0.12 = 2 118.96 Pa.
5According to the Atkinson equation for square-law fluid flow in mine airways (H_f = R · Q²), what happens to the frictional pressure drop (H_f) if the airflow quantity (Q) through the airway is doubled?
A.Frictional pressure drop increases by a factor of 4
B.Frictional pressure drop increases by a factor of 2
C.Frictional pressure drop increases by a factor of 8
D.Frictional pressure drop remains unchanged because Atkinson resistance R is constant
Explanation: The Atkinson equation states that frictional pressure drop H_f is proportional to the square of the volumetric flow rate Q (H_f = R · Q²). Doubling Q (2Q) results in H_f' = R · (2Q)² = 4 · R · Q², increasing the pressure drop by a factor of 4.
6Calculate the Atkinson resistance (R) of a rectangular mine airway 1 000 m long, with a perimeter of 16 m and a cross-sectional area of 16 m². The friction factor k is 0.010 N·s²/m⁴.
A.0.0391 N·s²/m⁸
B.0.6250 N·s²/m⁸
C.0.0024 N·s²/m⁸
D.0.1600 N·s²/m⁸
Explanation: The formula for Atkinson resistance is R = (k · C · L) / A³. Given k = 0.010 N·s²/m⁴, C = 16 m, L = 1 000 m, and A = 16 m²: R = (0.010 × 16 × 1000) / (16³) = 160 / 4096 = 0.0390625 N·s²/m⁸ ≈ 0.0391 N·s²/m⁸.
7A mine ventilation network passes a total airflow of 100 m³/s under a total fan pressure of 1 000 Pa. Calculate the Equivalent Orifice (A_e) of the mine.
A.1.23 m²
B.3.89 m²
C.0.39 m²
D.2.45 m²
Explanation: The standard Murgue / South African formula for Equivalent Orifice is A_e = (0.389 · Q) / √(H_p), where Q is in m³/s and H_p is in Pa. Substituting: A_e = (0.389 × 100) / √(1000) = 38.9 / 31.6227 = 1.230 m².
8A main axial-flow fan operates at 600 RPM and consumes 500 kW of shaft power. If the fan rotational speed is increased by 20% to 720 RPM while air density remains constant, what is the new power requirement?
A.864.0 kW
B.600.0 kW
C.720.0 kW
D.1 036.8 kW
Explanation: According to the Fan Laws, power is proportional to the cube of rotational speed: P₂ = P₁ · (N₂ / N₁)³. Here N₂ / N₁ = 720 / 600 = 1.20. P₂ = 500 kW × (1.20)³ = 500 × 1.728 = 864.0 kW.
9Two parallel airways have Atkinson resistances of R₁ = 0.16 N·s²/m⁸ and R₂ = 0.36 N·s²/m⁸. Calculate the equivalent resistance (R_eq) of the parallel combination.
A.0.0576 N·s²/m⁸
B.0.5200 N·s²/m⁸
C.0.1000 N·s²/m⁸
D.0.2600 N·s²/m⁸
Explanation: For parallel airways: 1 / √(R_eq) = 1 / √(R₁) + 1 / √(R₂). Here √(R₁) = √0.16 = 0.40 and √(R₂) = √0.36 = 0.60. 1 / √(R_eq) = 1 / 0.40 + 1 / 0.60 = 2.50 + 1.6667 = 4.1667. Therefore √(R_eq) = 1 / 4.1667 = 0.24, so R_eq = (0.24)² = 0.0576 N·s²/m⁸.
10Two split airways branch from a junction under a total available pressure drop of 500 Pa. Split A requires 30 m³/s and has a natural resistance of R_A = 0.20 N·s²/m⁸. Split B requires 20 m³/s and has R_B = 0.50 N·s²/m⁸. What pressure drop must be absorbed by a regulator placed in Split A?
A.320 Pa
B.180 Pa
C.200 Pa
D.500 Pa
Explanation: Natural pressure drop in Split A at 30 m³/s is H_A = R_A · Q_A² = 0.20 × (30)² = 180 Pa. Natural pressure drop in Split B at 20 m³/s is H_B = R_B · Q_B² = 0.50 × (20)² = 200 Pa. However, the total junction pressure differential must match the highest branch requirement or available head (500 Pa). To absorb excess head in Split A down from 500 Pa when operating at 30 m³/s, the regulator in A must drop H_reg = 500 - 180 = 320 Pa.
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