Surface Environment, Mine Ventilation and Underground Hazards · Mine Ventilation · 2 marks · Numerical answer
A 2×2.5 m face in a 1000 m drive uses 20 kg explosive producing 2200 cm³ fumes/kg. The duct outlet is 10 m from the face. Find airflow Q in m³/s reducing fumes to 5 ppm in 5 min. Use $$t=\frac{2.303V_m}{Q}\log_{10}\frac q{V_mc}+\frac{V-V_m}{Q},$$ where Vm is mixing-zone volume, V tunnel volume, q fume volume, c final fraction.
Tunnel volume V=5000 m³ and mixing volume Vm=50 m³. Fumes q=20×2200×10⁻⁶=0.044 m³ and c=5×10⁻⁶. With t=300 s, rearrange Q=[2.303×50 log₁₀(0.044/(50×5×10⁻⁶))+4950]/300. This gives about 17.36 m³/s. Both mixing removal and displacement along the remaining tunnel are included.
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