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24-MMP-B4 Mine Ventilation and Occupational Hygiene · May 2013

Question 6 of 6: Dilution Ventilation Design and Risk Management

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Exams (BC), 09-MMP-B4 Occupational Health, Safety and Loss Management (Mine Ventilation and Occupational Hygiene), May 2013, 3 hours, open book with calculator permitted. Answer any five of the six questions; every question (1-6) is answered in full as a complete study resource.

Reference texts: Crowl & Louvar, Chemical Process Safety: Fundamentals with Applications, 4th ed.; ACGIH, TLVs and BEIs and Industrial Ventilation: A Manual of Recommended Practice; OSHA 29 CFR 1904 Recordkeeping; WorkSafeBC/BC Health, Safety and Reclamation Code for Mines.

Question 6: Dilution Ventilation Design and Risk Management (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

(a) Dilution ventilation rate for fugitive solvent RXP

Part (a) —

Given. Evaporation rate 3.0 gal per 8-hour shift; TLV-TWA = 50 ppm; $T=77\ ^\circ\text{F}$, $P=1$ atm; $SG=0.85$; $MW=145$; mixing factor $K=5$ ("average" ventilation effectiveness).

Find. The dilution air flow rate required to keep the area below the TLV.

Approach. Convert the liquid evaporation rate to a molar vapour generation rate, use the ideal gas law at the stated conditions to get the pure-vapour volumetric generation rate, then divide by the TLV (as a volume fraction) and multiply by the imperfect-mixing safety factor $K$ (ACGIH guidance: $K\approx1$–3 excellent, 3–5 good, 5–10 fair/average, >10 poor mixing — "average" here is taken as $K=5$).

  1. Liquid evaporation rate. $$ER=\frac{3.0\ \text{gal}}{8\times60\ \text{min}}=0.00625\ \text{gal/min}$$
  2. Mass and molar generation rate. Liquid density $\rho_L=SG\times8.34=0.85\times8.34=7.09\ \text{lb/gal}$: $$\dot m=ER\times\rho_L=0.00625\times7.09=0.0443\ \text{lb/min}$$ $$\dot n=\frac{\dot m}{MW}=\frac{0.0443}{145}=0.000306\ \text{lbmol/min}$$
  3. Vapour volumetric generation rate (ideal gas, 77°F, 1 atm). $$\dot V_{vap}=\frac{\dot nRT}{P}=\frac{(0.000306)(10.731)(537)}{14.696}=\boxed{0.12\ \text{ft}^3/\text{min pure vapour}}$$
  4. Required dilution air flow. $$Q=\frac{\dot V_{vap}}{TLV}\times K=\frac{0.12}{50\times10^{-6}}\times5=\boxed{11982\ \text{ft}^3/\text{min}}$$
QuantityValue
Vapour generation rate (pure RXP)0.12 ft³/min
Required dilution ventilation rate, $Q$ ($K=5$)≈ 11982 ft³/min (≈ 339 m³/h)

Roughly 12,000 cfm of dilution air is needed to hold this fugitive emission below its TLV even though the raw vapour generation rate is only about 0.12 ft³/min of pure solvent — the TLV of 50 ppm is a 20,000:1 dilution ratio on its own, and the $K=5$ imperfect-mixing factor multiplies that by another 5× because real room air never mixes perfectly with a point-source leak.

Check: "average" ventilation condition is taken as ACGIH mixing factor $K=5$ (the mid-point of the commonly cited fair/average band); a poorer-mixing facility (K=10, as used for the practice set below) would require twice this airflow for the identical leak.

(b) Risk Management steps

Part (b) — Risk management in a hazardous operation follows a cyclical, proactive sequence rather than a one-time calculation:

REACTIVE — derive preventive actions FROM accidentsAccidentoccursInvestigateIdentify rootcauseCorrectiveactionVerify &monitorPROACTIVE — develop preventive actions BEFORE accidentsHazardIDRiskassessmentControldesignTraining& SOPsAudit& review
Fig. 6b — The five-step proactive risk-management cycle, contrasted with a purely reactive (post-accident) response.
  1. Hazard identification. Systematically identify every hazard present in the operation (e.g. process hazard reviews, HAZOP, job safety analyses).
  2. Risk assessment. For each identified hazard, estimate the likelihood and severity of harm (risk = likelihood × consequence, exactly the definition established in Question 2(c)) and rank hazards by the resulting risk level.
  3. Control design. Select controls for the highest-ranked risks, following the hierarchy of controls (elimination/substitution preferred over engineering, administrative, and PPE measures — the same inherent-safety preference discussed in Question 2(b)).
  4. Training and SOPs. Implement the chosen controls through written procedures and worker training so they are actually followed in practice.
  5. Audit and review. Periodically verify that controls remain effective and that no new hazards have been introduced, feeding findings back into a fresh round of hazard identification — closing the loop.

Example. A mine considering underground diesel-equipment operation first identifies the hazard (diesel particulate matter and CO/NOx exposure in a confined underground atmosphere), assesses the risk (high likelihood of daily exposure, moderate-to-severe long-term respiratory consequence), designs controls following the hierarchy (substitute to battery-electric equipment where feasible; if not, engineering controls such as diesel particulate filters and forced ventilation; administrative controls such as exposure-duration limits and rotation; PPE such as respirators as the last line of defence), trains crews on the ventilation and exposure-limit procedures, and then audits airborne DPM concentrations on a fixed schedule — feeding any exceedance back into a reassessment of the controls, exactly as the domino/pentagon models in Questions 1(a) and 5(c) show that interrupting any single link in the chain prevents the loss.

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