Question

A natural gas (methane, \(\mathrm{CH}_{4}\) ) storage facility uses 3 -cm- diameter by 6 -m-long vent tubes on its storage tanks to keep the pressure in these tanks at atmospheric value. If the diffusion coefficient for methane in air is \(0.2 \times 10^{-4} \mathrm{~m}^{2} / \mathrm{s}\) and the temperature of the tank and environment is \(300 \mathrm{~K}\), the rate at which natural gas is lost from a tank through one vent tube is (a) \(13 \times 10^{-5} \mathrm{~kg} /\) day (b) \(3.2 \times 10^{-5} \mathrm{~kg} / \mathrm{day}\) (c) \(8.7 \times 10^{-5} \mathrm{~kg} / \mathrm{day}\) (d) \(5.3 \times 10^{-5} \mathrm{~kg} / \mathrm{day}\) (e) \(0.12 \times 10^{-5} \mathrm{~kg} / \mathrm{day}\)

Short answer

Based on the given information and calculations, the rate at which natural gas (methane) is lost through one vent tube per day is approximately: (a) \(13 \times 10^{-5} ~kg/day\)

Step-by-step solution

Calculate the molar flow rate through the vent tube

The molar flow rate, \(N\), can be determined using the formula: $$ N = -D \frac{A \Delta c}{L} $$ where: \(D\) is the diffusion coefficient for methane in air \((0.2 \times 10^{-4} ~m^2/s)\), \(A\) is the cross-sectional area of the vent tube, \(\Delta c\) is the concentration difference of methane across the vent tube, \(L\) is the length of the vent tube, \(R\) is the gas constant \((8.31 ~J/mol~K)\). The molar flow rate is inversely related to the length of vent tube (longer vent tube, lesser the loss of methane). We need to calculate the cross-sectional area and concentration difference of the vent tube.

Calculate the cross-sectional area of the vent tube

The diameter of the vent tube is given as \(3~cm\). Convert this to meters for consistency: $$ 3 ~cm = 0.03 ~m $$ Now, calculate the cross-sectional area \(A\): $$ A = \frac{\pi d^2}{4} = \frac{\pi \times (0.03)^2}{4} = 7.069 \times 10^{-4}~m^2 $$

Calculate the concentration difference of methane across the vent tube

Given the temperature of the environment and the tank is \(300~K\), we can find the concentration difference, \(\Delta c\), using the equation: $$ \Delta c = \frac{P}{RT} $$ where: \(P\) is the pressure difference across the vent tube, which is assumed to be atmospheric \((1 ~atm)\), \(R\) is the gas constant \((8.31 ~J/mol~K)\), \(T\) is the given temperature of \(300~K\). Converting pressure difference to SI unit (Pascal): $$ 1 ~atm = 101,325 ~Pa $$ Now calculate the concentration difference, \(\Delta c\): $$ \Delta c = \frac{101,325}{8.31 \times 300} = 4.065 ~mol/m^3 $$

Calculate the molar flow rate, \(N\)

Now that we have the cross-sectional area \(A\), concentration difference \(\Delta c\), and the length of the vent tube \(L\), we can calculate the molar flow rate \(N\): $$ N = -D \frac{A \Delta c}{L} = - (0.2 \times 10^{-4}) \times \frac{(7.069 \times 10^{-4})(4.065)}{6} = - 9.54 \times 10^{-9} ~mol/s $$ The negative sign indicates the direction of flow, but we are only interested in the magnitude.

Convert molar flow rate to mass flow rate

To find the mass flow rate, we'll multiply the molar flow rate by the molar mass of methane: $$ \text{Mass flow rate} = |N| \times M_{CH_4} $$ where \(M_{CH_4}\) is the molar mass of methane \((16~g/mol)\). Convert this to kilograms: $$ M_{CH_4} = 16 ~g/mol \times \frac{1~ kg}{1000 ~g} = 0.016 ~kg/mol $$ Now calculate the mass flow rate: $$ \text{Mass flow rate} = 9.54 \times 10^{-9} ~mol/s \times 0.016 ~kg/mol = 1.5264 \times 10^{-10} ~kg/s $$

Convert mass flow rate to mass loss rate per day

To find the rate at which natural gas is lost from a tank through one vent tube per day, multiply the mass flow rate by the number of seconds in a day: $$ \text{Mass loss rate per day} = 1.5264 \times 10^{-10} ~kg/s \times 24 \times 60 \times 60 ~s/day = 1.319 \times 10^{-5} ~kg/day $$ This value is closest to option (a) \(13 \times 10^{-5}~ kg/day\). Therefore, the correct answer is (a) \(13 \times 10^{-5} ~kg/day\).

Key concepts

Diffusion Coefficient

The diffusion coefficient, represented by the symbol D, is a significant factor in the transport of substances through a medium. It measures how fast a substance like methane can spread out from a region of high concentration to an area of low concentration within a carrier medium like air. The coefficient is affected by the properties of both the substance and the medium, as well as temperature and pressure conditions. In our textbook example, a diffusion coefficient of \(0.2 \times 10^{-4} \mathrm{~m}^{2} / \mathrm{s}\) for methane in air is a given constant for the conditions of the exercise.

When considering vent tubes in a natural gas storage facility, understanding the diffusion coefficient is key to predicting the rate at which gas may escape. The correct calculation and application of this coefficient can lead to more accurate predictions of material losses or infiltration rates. The concept can also be applied in various fields such as environmental engineering, chemical process design, and more. An improved understanding of how diffusion coefficients are determined and affect mass transfer in vent tubes can prove invaluable in these disciplines.

Molar Flow Rate Calculation

The molar flow rate represents the number of moles of a substance passing through a cross-section per unit time and is denoted as \(N\). To calculate the molar flow rate for methane escaping from a vent tube, we apply the formula: \[N = -D \frac{A \Delta c}{L}\] where \(A\) stands for the cross-sectional area of the vent tube, \(\Delta c\) is the concentration difference across the tube, and \(L\) represents the vent tube length. This equation is a simplification of Fick's first law of diffusion, emphasizing the inverse relationship between molar flow rate and the length of the vent tube; a longer tube generally results in less material loss. Molar flow rate calculations are not just used for calculating losses; they are essential for the design and operation of a wide array of chemical processes in industry.

In the given problem, after calculating the respective values for \(A\) and \(\Delta c\), the molar flow rate is carefully determined. Understanding how to compute each variable and applying them in the equation enriches students' practical skills and aids accurate process control in real-world applications.

Mass Flow Rate Conversion

After obtaining the molar flow rate in our example, the next step is to convert it into a mass flow rate for practical applications. The mass flow rate is the amount of mass flowing through a cross-section per unit time and is crucial for process engineers to design and monitor systems. This conversion uses the molar mass of the substance involved, in this case, methane (\(CH_4\)).

The formula used for conversion in the example is straightforward: \[\text{Mass flow rate} = |N| \times M_{CH_4}\] where \(M_{CH_4}\) is the molar mass of methane. We then convert the mass flow rate from moles per second to kilograms per day, accommodating standard operating or reporting time frames. By understanding how to convert molar flow rate to mass flow rate, students can bridge the gap between theoretical calculations and practical measurement, which often involves weighing mass instead of counting molecules in industrial settings.