Surface Environment, Mine Ventilation and Underground Hazards · Mine Ventilation · 2 marks · Numerical answer
A main fan circulates 30 m³/s through parallel districts A and B. Resistances are RA=2.0 and RB=4.0 N s²/m⁸; the common intake and return each have resistance 0.2 N s²/m⁸. A booster in B must increase its flow by 20%, with main-fan pressure unchanged. Find booster pressure in Pa, to two decimals.
Initially 2QA²=4QB² and QA+QB=30, giving QA=17.57359 and QB=12.42641 m³/s. Main pressure is 0.4×30²+4QB²=977.66235 Pa. New QB is 14.91169. Solve 2QA′²+0.4(QA′+14.91169)²=977.66235, obtaining QA′=16.91771. The branch pressure available without the booster is 2QA′²; therefore booster pressure is 4×14.91169²−2×16.91771²=317.01575 Pa, or 317.02 Pa. A’s flow changes because the common-airway pressure loss changes.
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