Question

A tank with a 2-cm-thick shell contains hydrogen gas at the atmospheric conditions of \(25^{\circ} \mathrm{C}\) and \(90 \mathrm{kPa}\). The charging valve of the tank has an internal diameter of \(3 \mathrm{~cm}\) and extends $8 \mathrm{~cm}$ above the tank. If the lid of the tank is left open so that hydrogen and air can undergo equimolar counterdiffusion through the 10 -cm- long passageway, determine the mass flow rate of hydrogen lost to the atmosphere through the valve at the initial stages of the process. Answer: \(4.20 \times 10^{-8} \mathrm{~kg} / \mathrm{s}\)

Short answer

Answer: The mass flow rate of hydrogen lost to the atmosphere is approximately \(1.41 \times 10^{-11} \mathrm{kg/s}\).

Step-by-step solution

Identify the Equimolar Counterdiffusion Equation

We will use the equation for equimolar counterdiffusion: \(N_{A}=-D_{AB} \cdot A \cdot \dfrac{\Delta C_{A}}{L}\) where \(N_{A}\) - the molar flow rate of component A (in this case, hydrogen) \(D_{AB}\) - diffusivity of component A (hydrogen) in the mixture with component B (air) \(A\) - the cross-sectional area of the passageway \(\Delta C_{A}\) - concentration difference of component A between the two ends of the passageway \(L\) - length of the passageway

Calculate the Cross-Sectional Area of the Passageway (\(A\))

The internal diameter of the valve is given as 3 cm. We can calculate the cross-sectional area using the formula for the area of a circle: \(A = \dfrac{\pi d^2}{4}\) \(A = \dfrac{\pi (0.03 \mathrm{~m})^2}{4}\) \(A = 7.07 \times 10^{-4} \mathrm{~m}^2\)

Calculate the Difference in Concentration (\(\Delta C_{A}\))

From the initial atmospheric conditions for hydrogen in the tank, we can calculate the molar concentration: \(C_{A} = \dfrac{P}{RT}\) where \(R\) = \(8.314 \mathrm{~J/mol·K}\), the universal gas constant \(P\) = \(90 \times 10^3 \mathrm{Pa}\), the initial pressure \(T\) = \(25^\circ C + 273.15 = 298.15 \mathrm{K}\), the initial temperature \(C_{A} = \dfrac{90 \times 10^3 \mathrm{~Pa}}{8.314 \mathrm{~J/mol·K} \cdot 298.15 \mathrm{K}}\) \(C_{A} = 36.34 \mathrm{~mol/m^3}\) Since component A (hydrogen) is initially present only in the tank, the concentration difference from one end of the passageway to the other is equal to the initial molar concentration of hydrogen: \(\Delta C_{A} = 36.34 \mathrm{~mol/m^3}\)

Determine the Diffusivity of Hydrogen in Air (\(D_{AB}\))

From the literature or experimental data, we can find the diffusivity of hydrogen in air to be approximately: \(D_{AB} = 6.4 \times 10^{-5} \mathrm{m^2/s}\)

Calculate the Molar Flow Rate (\(N_{A}\))

Using the equation for equimolar counterdiffusion, we can now calculate the molar flow rate: \(N_{A} = - (6.4 \times 10^{-5} \mathrm{m^2/s}) \cdot (7.07 \times 10^{-4} \mathrm{m}^2) \cdot \dfrac{36.34 \mathrm{~mol/m^3}}{0.1 \mathrm{m}}\) \(N_{A} = -6.98 \times 10^{-9} \mathrm{mol/s}\) Since we are only interested in the magnitude of the molar flow rate, we can ignore the negative sign.

Convert Molar Flow Rate to Mass Flow Rate

We can convert the molar flow rate to mass flow rate using the molar mass of hydrogen (\(M_{A}\)), which is \(2.016 \mathrm{g/mol}\) or \(2.016 \times 10^{-3} \mathrm{kg/mol}\): \(\dot{m}_{A}= N_{A} \cdot M_{A}\) \(\dot{m}_{A}= (6.98 \times 10^{-9} \mathrm{mol/s}) \cdot (2.016 \times 10^{-3} \mathrm{kg/mol})\) \(\dot{m}_{A} = 1.41 \times 10^{-11} \mathrm{kg/s}\) Therefore, the mass flow rate of hydrogen lost to the atmosphere through the valve at the initial stages of the process is approximately \(1.41 \times 10^{-11} \mathrm{kg/s}\).