Question

Hydrogen can cause fire hazards, and hydrogen gas leaking into surrounding air can lead to spontaneous ignition with extremely hot flames. Even at very low leakage rate, hydrogen can sustain combustion causing extended fire damages. Hydrogen gas is lighter than air, so if a leakage occurs it accumulates under roofs and forms explosive hazards. To prevent such hazards, buildings containing source of hydrogen must have adequate ventilation system and hydrogen sensors. Consider a metal spherical vessel, with an inner diameter of \(5 \mathrm{~m}\) and a thickness of \(3 \mathrm{~mm}\), containing hydrogen gas at \(2000 \mathrm{kPa}\). The vessel is situated in a room with atmospheric air at 1 atm. The ventilation system for the room is capable of keeping the air fresh, provided that the rate of hydrogen leakage is below $5 \mu \mathrm{g} / \mathrm{s}$. If the diffusion coefficient and solubility of hydrogen gas in the metal vessel are \(1.5 \times\) \(10^{-12} \mathrm{~m}^{2} / \mathrm{s}\) and \(0.005 \mathrm{kmol} / \mathrm{m}^{3}\)-bar, respectively, determine whether or not the vessel is safely containing the hydrogen gas.

Short answer

Based on the given information, we calculated the hydrogen gas leakage rate through the metal spherical vessel's wall as -3.177 x 10^3 μg/s. Comparing this to the maximum safe leakage rate of 5 μg/s, we can conclude that the vessel is not safely containing the hydrogen gas as the leakage rate is higher than the safe limit.

Step-by-step solution

Calculate the driving force

We are given that the hydrogen gas pressure inside the vessel is \(2000 \mathrm{kPa}\), and the atmospheric air pressure is \(1 \mathrm{~atm}\). We need to find the partial pressure difference which is the driving force for the diffusion of hydrogen through the vessel wall (inside pressure to outside pressure). Inside pressure of hydrogen gas (\(P_{i}\)): \(2000 \mathrm{kPa}\) 1 atmospheric pressure (\(P_{o}\)) in kPa: \(1 \mathrm{atm} = 101.325 \mathrm{kPa}\) Driving force for diffusion: \(P_{i} - P_{o} = 2000 - 101.325 = 1898.675 \mathrm{kPa}\)

Convert solubility into SI Units

The solubility of hydrogen gas \(\left( C \right)\) is given in \(\mathrm{kmol} / \mathrm{m}^{3}\)-bar, but the driving force we calculated is in kPa. We need to convert solubility from bar to kPa. 1 bar = 100 kPa \(C = 0.005 \cdot 100 = 0.5 \mathrm{kmol} / \mathrm{m}^{3} \mathrm{kPa}\)

Calculate leakage rate using Fick's 1st law

Now, we will use Fick's 1st law of diffusion to find the leakage rate of hydrogen gas through the vessel wall. Fick's 1st law: \(J = -D \frac{dC}{dx}\) Where \(J\) is the leakage rate, \(D\) is the diffusion coefficient, and \(\frac{dC}{dx}\) is the concentration gradient across the wall. The concentration gradient can be written as: \(\frac{dC}{dx} = \frac{C(P_{i}−P_{o})}{l}\) Here, \(l\) is the thickness of the vessel. Leakage rate (\(J\)): \(J = -D \frac{C(P_{i}−P_{o})}{l} = - \left( 1.5 \times 10^{-12} \right) \frac{0.5(1898.675)}{3 \times 10^{-3}}\)

Convert leakage rate to the required units

The answer obtained in step 3, J, is in the units of \(\mathrm{kmol} / \mathrm{m}^{2} \mathrm{s}\). We need to convert it to \(\mu \mathrm{g} / \mathrm{s}\). First, we need to convert kmol to grams. The molar mass of hydrogen is \(2 \mathrm{g/mol}\). So, 1 kmol is equal to 2 \(\times 10^{3}\) grams. J (in g/s): \(J \cdot 2 \times 10^3 = - \left( 1.5 \times 10^{-12} \right) \frac{0.5(1898.675)}{3 \times 10^{-3}} \cdot 2 \times 10^3\) Next, we'll convert grams to micrograms. 1 gram = \(10^6\) micrograms J (in \(\mu \mathrm{g} / \mathrm{s}\)): \(J \cdot 2 \times 10^{3} \cdot 10^{6} = - \left( 1.5 \times 10^{-12} \right) \frac{0.5(1898.675)}{3 \times 10^{-3}} \cdot 2 \times 10^{3} \cdot 10^{6}\)

Compare leakage rate with the maximum safe leakage rate

Now, we'll compare the calculated leakage rate with the maximum safe leakage rate for the ventilation system. Calculated leakage rate: \(J = -3.177 \times 10^3 \mu \mathrm{g} / \mathrm{s}\) Maximum safe leakage rate: \(5 \mu \mathrm{g} / \mathrm{s}\) Since the calculated leakage rate is higher than the maximum safe leakage rate, the vessel is not safely containing the hydrogen gas.