Question 2 of 6: OSHA Incidence Rates, Risk Management Steps, and Vacuum Purging

Question 2: OSHA Incidence Rates, Risk Management Steps, and Vacuum Purging (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) OSHA incidence rates — injuries and lost workdays

Part (a) —

Given. 1,200 full-time employees; 38 lost-time injuries in the year; 274 lost workdays resulting from those injuries.

Find. The OSHA incidence rate based on (i) injuries and (ii) lost workdays.

Approach. OSHA's standard incidence-rate formula normalises the count of a chosen event to a base of 100 full-time-equivalent workers, using the statistical convention that 100 FTE work 200,000 hours per year (100 workers × 2,000 h/yr each).

1. Total hours worked. $$H=1200\times2000=\boxed{2{,}400{,}000\ \text{hours}}$$
2. Injury incidence rate. $$IR_{inj}=\frac{N_{inj}\times200{,}000}{H}=\frac{38\times200{,}000}{2{,}400{,}000}=\boxed{3.17\ \text{injuries per 100 FTE-yr}}$$
3. Lost-workday rate. $$IR_{LWD}=\frac{N_{LWD}\times200{,}000}{H}=\frac{274\times200{,}000}{2{,}400{,}000}=\boxed{22.83\ \text{lost workdays per 100 FTE-yr}}$$

Both rates are reported per 100 full-time-equivalent workers so that facilities of different sizes can be benchmarked on a common basis; the lost-workday rate is the more severity-sensitive metric, since a single serious injury can contribute dozens of lost days while the injury count itself only increments by one.

(b) Main steps in risk assessment and management

Part (b) — Risk assessment and management, taken together, follow a five-step cyclical process:

1. Hazard identification. Systematically identify every hazard present in the operation (process hazard reviews, HAZOP, job safety analyses, incident/near-miss history).
2. Risk assessment. For each hazard, estimate the likelihood and severity of harm and rank hazards by the resulting risk level, so effort is directed at the highest-risk items first.
3. Control selection and design. Choose controls for the ranked risks following the hierarchy of controls — elimination and substitution preferred over engineering controls, then administrative controls, with PPE as the last line of defence.
4. Implementation. Put the chosen controls into practice through written procedures, engineering changes, and worker training.
5. Monitoring and review. Periodically verify that controls remain effective and that no new hazards have appeared, feeding findings back into a fresh round of hazard identification.

The cycle is deliberately continuous rather than a one-time exercise: a mine ventilation and occupational hygiene program has to keep re-assessing as workings advance, equipment changes, and new substances are introduced, so step 5 always loops back to step 1.

(c) Vacuum purge cycles — O₂ 21% to 1%

Part (c) —

Given. Initial vessel O₂ concentration $y_0=21\%$; target $y_n\le1\%$; inert purge gas contains 9000 ppm O₂ ($y_i=0.9\%$).

Find. The number of vacuum-purge cycles $n$ required.

Approach. No vacuum-pump pressures were stated in the source (only the concentrations); per the exam's own note 1 ("in case of doubt the student is allowed to make assumption... clear statement of any assumptions made"), this solution assumes a typical rough-vacuum pump pulling the vessel down to $P_{min}=20\ \text{mmHg}$ absolute before each refill to atmospheric $P_{max}=760\ \text{mmHg}$ — the standard textbook figure for this class of problem. Each vacuum-purge cycle then follows the repeated-dilution relation for an inert gas of non-zero purity: $$y_n=y_i+(y_0-y_i)\left(\frac{P_{min}}{P_{max}}\right)^n$$

1. Pressure ratio per cycle. $$r=\frac{P_{min}}{P_{max}}=\frac{20}{760}=0.02632$$
2. Solve for $n$. Setting $y_n=0.01$ and rearranging: $$n=\frac{\ln\!\left(\dfrac{y_n-y_i}{y_0-y_i}\right)}{\ln r}=\frac{\ln\!\left(\dfrac{0.01-0.009}{0.21-0.009}\right)}{\ln(0.02632)}=1.46$$ Since a fractional purge cycle is not physically meaningful, round up: $\boxed{n=2}$.
3. Check. After 1 cycle, $y_1=0.009+(0.201)(0.02632)^1=1.43\%$ — still above target. After 2 cycles, $$y_2=0.009+(0.201)(0.02632)^2=\boxed{0.914\%}$$ which is below the 1% target.

Two vacuum-purge cycles bring the vessel comfortably below the 1% target, with room to spare (0.91% vs. 1%) — a useful margin given the inert gas itself already carries 0.9% O₂, which sets an absolute floor no number of cycles can go below.