Question 3 of 6: Oxygen Depletion by Inerting and Liquid Discharge Through a Crack

Question 3: Oxygen Depletion by Inerting and Liquid Discharge Through a Crack (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) Nitrogen purge volume to reach 19.5% and 16% oxygen

Part (a) —

Given. Tank volume $V=1000\ \text{ft}^3$; initial O₂ = 21% by volume; N₂ purge gas added at the tank's own conditions (77°F, 1 atm), well mixed, with the tank vented so total pressure stays at 1 atm as gas is added.

Find. Volume of additional N₂ needed to bring the O₂ concentration down to (i) 19.5% and (ii) 16%.

Approach. For a well-mixed vessel purged by adding inert gas while venting to hold total pressure constant, the oxygen mole fraction decays exponentially with the volume of purge gas added (a standard inerting/purging result — each increment of added N₂ displaces the same fraction of whatever gas is currently in the tank, including the N₂ just added): $$\frac{y_{O_2}}{y_{O_2,0}}=\exp\!\left(-\frac{V_{N_2}}{V_{tank}}\right)$$

1. Rearrange for the purge volume. $$V_{N_2}=-V_{tank}\,\ln\!\left(\frac{y_{O_2}}{y_{O_2,0}}\right)$$
2. Target (i): 19.5% O₂. $$V_{N_2}=-1000\,\ln\!\left(\frac{0.195}{0.21}\right)=\boxed{74.1\ \text{ft}^3}$$
3. Target (ii): 16% O₂. $$V_{N_2}=-1000\,\ln\!\left(\frac{0.16}{0.21}\right)=\boxed{271.9\ \text{ft}^3}$$

Because the mole-fraction decay is exponential, roughly 3.7× more N₂ is needed to push the concentration all the way down to the 16% distress threshold than to the 19.5% no-SCBA threshold — the last few percentage points of oxygen are the most expensive to remove, which is also why an atmosphere can look "almost normal" by smell/feel while already being dangerously oxygen-deficient.

(b) Mass flow rate of liquid ammonia through a 0.1-inch crack

Part (b) —

Given. Ammonia stored at $P_g=2000$ psig, $T=80\ ^\circ\text{F}$; crack diameter $d=0.1$ in; discharge to atmosphere ($P_{atm}=14.7$ psia).

Find. The mass flow rate through the crack.

Approach. Ammonia's saturation (vapour) pressure at 80°F is only about 154 psig, far below the stated 2000 psig storage pressure, and 2000 psig (2014.7 psia) exceeds ammonia's own critical pressure (1657 psia) while the storage temperature (80°F) is far below its critical temperature (270°F) — so the material inside the vessel is a single-phase, compressed (sub-cooled) liquid, not a saturated vapour/liquid mixture. The crack is therefore modelled with the standard liquid-discharge-through-a-hole source model, not a choked-gas-flow model: $$\dot m=C_o\,A\,\sqrt{2\,g_c\,\rho_L\,(P_g-P_{atm,g})}$$ with a sharp-edged-orifice discharge coefficient $C_o\approx0.61$ and liquid ammonia density at 80°F, $\rho_L\approx37.5\ \text{lb}_m/\text{ft}^3$.

1. Crack area. $$A=\frac{\pi}{4}d^2=\frac{\pi}{4}\left(\frac{0.1}{12}\right)^2=5.5e-05\ \text{ft}^2$$
2. Driving pressure. Discharging to atmosphere, the driving $\Delta P$ is simply the gauge pressure inside the vessel: $$\Delta P = 2000\ \text{psig} = 2000\times144=288{,}000\ \text{lb}_f/\text{ft}^2$$
3. Mass flow rate. $$\dot m=0.61\times5.5e-05\times\sqrt{2(32.174)(37.5)(288{,}000)}=\boxed{0.877\ \text{lb}_m/\text{s}}$$ equivalently $52.6\ \text{lb}_m/\text{min}$ ($0.398\ \text{kg/s}$).

A sub-millimetre crack at this driving pressure is enough to discharge nearly 24 kilograms of ammonia per minute — the consequence-analysis distance in a full release scenario would then follow from this steady mass release rate feeding a downwind dispersion calculation of exactly the type performed in Question 4.