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.

Short Answer

Expert verified
Answer: The mass flow rate of hydrogen gas lost to the atmosphere through the charging valve during the initial stages of the process is approximately \(3.08 \times 10^{-8} \mathrm{kg/s}\).

Step by step solution

01

Identify relevant information from the problem statement

The important information given in the problem is as follows: - Temperature of the system is \(25^{\circ} \mathrm{C}\), which is equal to \((273.15 + 25) K = 298.15 K\). - Pressure of the gas inside the tank is \(90 \mathrm{kPa}\). - Internal diameter of the charging valve is \(3 \mathrm{~cm}\); hence the radius is \(1.5 \mathrm{~cm}\) or \(0.015 \mathrm{~m}\). - The charging valve extends \(8 \mathrm{~cm}\) above the tank. - The equimolar counterdiffusion occurs through a \(10 \mathrm{~cm}\) long passageway. - Our objective is to determine the mass flow rate of hydrogen lost to the atmosphere.
02

Calculate the area of the valve opening for diffusion to occur

To find the area of the valve opening, we can use the formula for the area of a circle with the given radius: \(A = \pi r^2\) where \(A\) is the area, and \(r\) is the radius (0.015 m). Plugging in the value for the radius, we get: \(A = \pi (0.015 \mathrm{m})^2 = 7.07 \times 10^{-4} \mathrm{m}^2\)
03

Determine the molar mass of hydrogen gas and air

For the equimolar counterdiffusion, we need to know the molar masses of hydrogen gas (H\(_2\)) and air. We have: - Molar mass of hydrogen gas, \(M_\mathrm{H_2} = 2.016 \mathrm{g/mol} = 2.016 \times 10^{-3} \mathrm{kg/mol}\) - Molar mass of air, \(M_\mathrm{air} \approx 28.97 \mathrm{g/mol} = 28.97 \times 10^{-3} \mathrm{kg/mol}\)
04

Calculate the diffusion coefficient for hydrogen gas and air

The diffusion coefficient can be calculated using the following formula (Chapman-Enskog Theory): \(D = \frac{3.03 \times 10^{-3} (T)^{3/2} (1/M_\mathrm{H_2} + 1/M_\mathrm{air})^{1/2}}{P(\Sigma_\mathrm{H_2}+\Sigma_\mathrm{air})^{2/3}}\) where \(D\) is the diffusion coefficient, \(T\) is the temperature (298.15 K), \(M_\mathrm{H_2}\) and \(M_\mathrm{air}\) are the molar masses, \(P\) is the pressure (90 kPa), and \(\Sigma_\mathrm{H_2}\) and \(\Sigma_\mathrm{air}\) are the collision diameter parameters of hydrogen and air. For simplicity and approximation purposes, we can assume that \(\Sigma_\mathrm{H_2}+\Sigma_\mathrm{air} = 6.00 \times 10^{-10} \mathrm{m}\). Plugging in the values, we get: \(D = \frac{3.03 \times 10^{-3} \times (298.15 \mathrm{K})^{3/2} \times (1/2.016 \times 10^{-3} + 1/28.97 \times 10^{-3})^{1/2}}{90 \times 10^3 \times (6.00 \times 10^{-10})^{2/3}} = 6.247 \times 10^{-5} \mathrm{m^2/s}\)
05

Calculate the mass flow rate of hydrogen lost to the atmosphere

Now, we can use Fick's law to determine the mass flow rate of hydrogen gas lost to the atmosphere: \(\dot{m}_\mathrm{H_2} = \rho_\mathrm{H_2} \times D \times A \times \frac{\Delta C}{L}\) \(\rho_\mathrm{H_2}\) is the density of hydrogen gas, calculated using the ideal gas law: \(\rho_\mathrm{H_2} = \frac{P M_\mathrm{H_2}}{R_\mathrm{gas} T}\), where \(R_\mathrm{gas} = 8.314 \mathrm{J/(mol \cdot K)}\). Thus, we find: \(\rho_\mathrm{H_2} = \frac{90 \times 10^3 \times 2.016 \times 10^{-3}}{8.314 \times 298.15} = 0.0655 \mathrm{kg/m^3}\) We're assuming that the concentration difference is large enough that the system behaves roughly as if the concentration of hydrogen is constant over the 10 cm of the charging valve (\(\Delta C \approx 1\)). The length of the diffusion passageway is given as \(L=10 \mathrm{~cm} = 0.1 \mathrm{~m}\). Now, we can plug in the values to find the mass flow rate: \(\dot{m}_\mathrm{H_2} = 0.0655 \mathrm{kg/m^3} \times 6.247 \times 10^{-5} \mathrm{m^2/s} \times 7.07 \times 10^{-4} \mathrm{m}^2 \times \frac{1}{0.1 \mathrm{~m}}\) And we get: \(\dot{m}_\mathrm{H_2} = 3.076 \times 10^{-8} \mathrm{kg/s}\) So, the mass flow rate of hydrogen gas lost to the atmosphere through the valve during the initial stages of the process is approximately \(3.08 \times 10^{-8} \mathrm{kg/s}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Fick's Law of Diffusion
The principles of diffusion are foundational to understanding how gases mix or separate under specific conditions. One of the cornerstone principles in this topic is Fick's Law, which is crucial for estimating the rate at which particles, such as atoms or molecules, diffuse through a medium.

Fick's Law expresses the flux of a diffusing substance across an area per unit time due to a concentration gradient. In simple terms, it tells us how quickly a gas, like hydrogen in the tank example, will spread out. The law is mathematically formulated as: \[\begin{equation} J = -D \frac{dC}{dx} \end{equation}\]Here, \( J \) represents the diffusion flux, indicating the amount of substance that will flow through a unit area during a unit time. The constant \( D \) is the diffusion coefficient, and \(-\frac{dC}{dx}\) denotes the concentration gradient across the distance \( x \). In essence, the flux goes from regions of higher concentration to lower concentration, and the 'minus' sign denotes this direction.

When applying Fick's Law to the tank problem to determine the mass flow rate of hydrogen, it is crucial to understand that it depends on both the area available for diffusion and the concentration difference across the valve. By focusing on these factors and properly calculating the diffusion coefficient, students can effectively estimate the rate at which hydrogen escapes the tank.
Equimolar Counterdiffusion in Gas Mixtures
The concept of equimolar counterdiffusion is particularly relevant when two gases are diffusing through one another without an accompanying net flow of mass. This phenomenon occurs when the molar flux of one gas in one direction is equal to the molar flux of another gas in the opposite direction. Imagine two types of molecules, one dancing to the left and another to the right at the same pace across a dance floor; they swap places but maintain an equal rhythm.

In the exercise with the hydrogen tank, we are dealing with equimolar counterdiffusion between hydrogen and air. The calculation becomes a tad more complex as both gases are moving through each other in opposite directions. This point is nuanced but critical, as it affects the final mass flow rate calculation. To resolve any difficulties that may arise from this concept, consider both gases separately, assess their individual diffusion rates, and ensure that the molar flux of hydrogen going out is equal to the molar flux of air coming in. With this balanced view, students can avoid common errors and better grasp the subtleties of mass transfer in mixed gas scenarios.
The Role of the Ideal Gas Law in Mass Transfer Calculations
The Ideal Gas Law is a powerful tool in various scientific fields, and it finds a crucial place when calculating mass transfer rates in gaseous systems. The equation, given by \(PV = nRT\), establishes a direct relationship among the pressure \(P\), volume \(V\), and temperature \(T\) of an ideal gas with the amount of substance \(n\) and the ideal gas constant \(R\).

When dealing with the flow of gases, like the hydrogen escaping from the tank, it is necessary to determine the density of the escaping gas. This point is where the ideal gas law comes into play by allowing us to calculate the density \( \rho \) with the formula:\(\rho = \frac{P M}{RT}\)In this expression, \( M \) is the molar mass of the gas. In our tank scenario, applying the ideal gas law reveals the density of the hydrogen gas, which, in turn, is vital to determine the mass flow rate using Fick's law.

It's essential for students to grasp that the ideal gas law is an approximation that works well under standard conditions. Although real gases have different behaviors, especially under high pressure or low temperature, understanding the ideal gas law allows students to handle a broad range of problems effectively and serves as a stepping stone to more complex thermodynamic concepts.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A glass of milk left on top of a counter in the kitchen at \(15^{\circ} \mathrm{C}, 88 \mathrm{kPa}\), and 50 percent relative humidity is tightly sealed by a sheet of \(0.009-\mathrm{mm}\)-thick aluminum foil whose permeance is \(2.9 \times 10^{-12} \mathrm{~kg} / \mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{Pa}\). The inner diameter of the glass is \(12 \mathrm{~cm}\). Assuming the air in the glass to be saturated at all times, determine how much the level of the milk in the glass will recede in \(12 \mathrm{~h}\). Answer: \(0.0011 \mathrm{~mm}\)

The diffusion coefficient of carbon in steel is given as $$ D_{A B}=2.67 \times 10^{-5} \exp (-17,400 / T) \quad\left(\mathrm{m}^{2} / \mathrm{s}\right) $$ where \(T\) is in \(\mathrm{K}\). Determine the diffusion coefficient from \(300 \mathrm{~K}\) to \(1500 \mathrm{~K}\) in \(100 \mathrm{~K}\) increments and plot the results.

Consider a nickel wall separating hydrogen gas that is maintained on one side at \(5 \mathrm{~atm}\) and on the opposite at \(3 \mathrm{~atm}\). If the temperature is constant at \(85^{\circ} \mathrm{C}\), determine \((a)\) the mass densities of hydrogen gas in the nickel wall on both sides and \((b)\) the mass densities of hydrogen outside the nickel wall on both sides.

What is the relation \((f / 2) \mathrm{Re}=\mathrm{Nu}=\mathrm{Sh}\) known as? Under what conditions is it valid? What is the practical importance of it? \(\mathrm{St}_{\text {mass }} \mathrm{Sc}^{2 / 3}\) and what are the names of the variables in it? Under what conditions is it valid? What is the importance of it in engineering?

Consider a 30-cm-diameter pan filled with water at \(15^{\circ} \mathrm{C}\) in a room at \(20^{\circ} \mathrm{C}, 1 \mathrm{~atm}\), and 30 percent relative humidity. Determine \((a)\) the rate of heat transfer by convection, (b) the rate of evaporation of water, and \((c)\) the rate of heat transfer to the water needed to maintain its temperature at \(15^{\circ} \mathrm{C}\). Disregard any radiation effects.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free