/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 21 Hydrogen at a pressure of 2 atm ... [FREE SOLUTION] | 91影视

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Chapter 14: Problem 21

Hydrogen at a pressure of 2 atm flows within a tube of diameter \(40 \mathrm{~mm}\) and wall thickness \(0.5 \mathrm{~mm}\). The outer surface is exposed to a gas stream for which the hydrogen partial pressure is \(0.1 \mathrm{~atm}\). The mass diffusivity and solubility of hydrogen in the tube material are \(1.8 \times 10^{-11} \mathrm{~m}^{2} / \mathrm{s}\) and \(160 \mathrm{kmol} / \mathrm{m}^{3}\) *atm, respectively. When the system is at \(500 \mathrm{~K}\), what is the rate of hydrogen transfer through the tube per unit length \((\mathrm{kg} / \mathrm{s} \cdot \mathrm{m})\) ?

Short Answer

Expert verified
The rate of hydrogen transfer through the tube per unit length is approximately \(6.79 \times 10^{-4}\,\mathrm{kg}/\mathrm{s} \cdot \mathrm{m}\).

Step by step solution

01

Calculate Concentration Difference

Since the flow is steady, the driving force is the concentration difference across the wall. We can calculate the concentration at both ends by using the solubility and the pressures given: \(C_{1} = S \cdot P_{1}\) \(C_{2} = S \cdot P_{2}\) Where \(C_1\) and \(C_2\) represent the concentration of hydrogen at the inner and outer sides of the tube, respectively, S is the solubility, and \(P_1\) and \(P_2\) are inner and outer pressures. Using the given values, we get: \(C_{1} = 160\,\mathrm{kmol}/\mathrm{m}^{3}\text{ }\mathrm{atm} \cdot 2\,\mathrm{atm} = 320\,\mathrm{kmol}/\mathrm{m}^{3}\) \(C_{2} = 160\,\mathrm{kmol}/\mathrm{m}^{3}\text{ }\mathrm{atm} \cdot 0.1\,\mathrm{atm} = 16\,\mathrm{kmol}/\mathrm{m}^{3}\)
02

Use Fick's Law of Diffusion

Now, we use Fick's law of diffusion to find the rate of hydrogen transfer through the tube per unit length: \(J = -D \frac{dC}{dx}\) where J is the molar flux, D is the mass diffusivity, and \(dC/dx\) is the concentration gradient of hydrogen across the wall. Rearranging the Fick's law for our case: \(J = D \frac{C_{1} - C_{2}}{\delta}\) where \(\delta\) is the wall thickness.
03

Calculate Molar Flux

Using the given values, we can calculate the molar flux: \(J = 1.8 \times 10^{-11}\,\mathrm{m}^{2}/\mathrm{s} \cdot \frac{320 - 16}{0.5 \times 10^{-3}\,\mathrm{m}} = 3.456 \times 10^{-6}\,\mathrm{kmol}/\mathrm{s} \cdot \mathrm{m}^{2}\)
04

Convert Molar Flux to Mass Flux

We will now convert the molar flux to mass flux by multiplying it with the molar mass of hydrogen: \(m' = J \times M_{H2}\) where \(m'\) is the mass flux, and \(M_{H2}\) is the molar mass of hydrogen. Using the molar mass of hydrogen, \(M_{H2} = 2\,\mathrm{kg}/\mathrm{kmol}\), we get: \(m' = 3.456 \times 10^{-6}\,\mathrm{kmol}/\mathrm{s} \cdot \mathrm{m}^{2} \cdot 2\,\mathrm{kg}/\mathrm{kmol}= 6.912 \times 10^{-6}\,\mathrm{kg}/\mathrm{s} \cdot \mathrm{m}^{2}\)
05

Calculate the Rate of Hydrogen Transfer per Unit Length

To get the rate of hydrogen transfer through the tube per unit length, we need to divide the mass flux by the tube's cross-sectional area: \(Q = \frac{m'}{A} = \frac{6.912 \times 10^{-6}\,\mathrm{kg}/\mathrm{s} \cdot \mathrm{m}^{2}}{\frac{\pi}{4} (0.04\,\mathrm{m})^{2} - \frac{\pi}{4}(0.04\,\mathrm{m} - 2 \times 0.0005\,\mathrm{m})^{2}}\) Calculating the rate, we get: \(Q \approx 6.79 \times 10^{-4}\,\mathrm{kg}/\mathrm{s} \cdot \mathrm{m}\) So, the rate of hydrogen transfer through the tube per unit length is approximately \(6.79 \times 10^{-4}\,\mathrm{kg}/\mathrm{s} \cdot \mathrm{m}\).

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Key Concepts

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

Fick's Law of Diffusion
Fick's Law of Diffusion is a crucial principle in mass transfer that helps explain how particles like hydrogen move from areas of high concentration to areas of low concentration. Imagine a crowded party; people tend to move towards less crowded spots. Similarly, particles diffuse along their concentration gradient. Fick's law is mathematically expressed as: \[ J = -D \frac{dC}{dx} \] where:
  • \( J \) is the molar flux, the rate at which particles pass through a unit area.
  • \( D \) is the mass diffusivity, indicating how easily particles spread out.
  • \( \frac{dC}{dx} \) is the concentration gradient, the change in concentration over a distance.
In the exercise, Fick's Law was used to determine the rate of hydrogen transfer through a tube's wall. By calculating the concentration difference inside and outside the tube, due to varying pressures, the law helps set the foundation for finding how much hydrogen flows through the material over time.
Hydrogen Solubility
Hydrogen solubility refers to how much hydrogen can dissolve in a given material. It acts like a sponge's capacity to hold water. This property is vital when considering how gas molecules like hydrogen interact with, or pass through, solid materials. Solubility is influenced by factors such as pressure and the material's nature. In the tube example, it's given as \( 160 \ \mathrm{kmol}/\mathrm{m}^3 \mathrm{\text{ }} \mathrm{atm} \). This value tells us how much hydrogen can "fit" in the material surrounding the tube at a specific pressure. By knowing the solubility and pressure details, concentrations within the material can be calculated. In practical terms, understanding hydrogen solubility means ensuring that the right materials are used for applications where hydrogen containment is crucial, like in pipes or storage tanks.
Mass Diffusivity
Mass diffusivity is a measure of how well a substance, like hydrogen, can diffuse through another medium. It reflects the ease with which molecules spread out over time. Higher diffusivity means particles move more readily, similar to how faster winds spread scents quicker.The diffusivity for hydrogen given in the problem is \( 1.8 \times 10^{-11} \ \mathrm{m}^2/\mathrm{s} \). This relatively small number indicates hydrogen doesn鈥檛 easily diffuse through the tube wall compared to more permeable materials. Mass diffusivity plays a pivotal role when applying Fick's Law, as shown in the problem, directly affecting the calculation of hydrogen's transfer rate. By understanding this concept, engineers can predict and control how gases behave when in contact with different materials, which is essential for designing effective industrial processes involving gas exchange.

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Most popular questions from this chapter

An old-fashioned glass apothecary jar contains a patent medicine. The neck is closed with a rubber stopper that is \(20 \mathrm{~mm}\) tall, with a diameter of \(10 \mathrm{~mm}\) at the bottom end, widening to \(20 \mathrm{~mm}\) at the top end. The molar concentration of medicine vapor in the stopper is \(2 \times 10^{-3} \mathrm{kmol} / \mathrm{m}^{3}\) at the bottom surface and is negligible at the top surface. If the mass diffusivity of medicine vapor in rubber is \(0.2 \times 10^{-9} \mathrm{~m}^{2} / \mathrm{s}\), find the rate \((\mathrm{kmol} / \mathrm{s})\) at which vapor exits through the stopper.

A solar pond operates on the principle that heat losses from a shallow layer of water, which acts as a solar absorber, may be minimized by establishing a stable vertical salinity gradient in the water. In practice such a condition may be achieved by applying a layer of pure salt to the bottom and adding an overlying layer of pure water. The salt enters into solution at the bottom and is transferred through the water layer by diffusion, thereby establishing salt-stratified conditions. As a first approximation, the total mass density \(\rho\) and the diffusion coefficient for salt in water \(\left(D_{\mathrm{AB}}\right)\) may be assumed to be constant, with \(D_{\mathrm{AB}}=1.2 \times 10^{-9} \mathrm{~m}^{2} / \mathrm{s}\). (a) If a saturated density of \(\rho_{\mathrm{A}, s}\) is maintained for salt in solution at the bottom of the water layer of thickness \(L=1 \mathrm{~m}\), how long will it take for the mass density of salt at the top of the layer to reach \(25 \%\) of saturation? (b) In the time required to achieve \(25 \%\) of saturation at the top of the layer, how much salt is transferred from the bottom into the water per unit surface area \(\left(\mathrm{kg} / \mathrm{m}^{2}\right)\) ? The saturation density of salt in solution is \(\rho_{A, s}=380 \mathrm{~kg} / \mathrm{m}^{3}\). (c) If the bottom is depleted of salt at the time that the salt density reaches \(25 \%\) of saturation at the top, what is the final (steady-state) density of the salt at the bottom? What is the final density of the salt at the top?

Insulation degrades (experiences an increase in thermal conductivity) if it is subjected to water vapor condensation. The problem may occur in home insulation during cold periods, when vapor in a humidified room diffuses through the drywall (plaster board) and condenses in the adjoining insulation. Estimate the mass diffusion rate for a \(3 \mathrm{~m} \times 5 \mathrm{~m}\) wall, under conditions for which the vapor pressure is \(0.03\) bar in the room air and \(0.0\) bar in the insulation. The drywall is \(10 \mathrm{~mm}\) thick, and the solubility of water vapor in the wall material is approximately \(5 \times 10^{-3} \mathrm{kmol} / \mathrm{m}^{3}\) - bar. The binary diffusion coefficient for water vapor in the drywall is approximately \(10^{-9} \mathrm{~m}^{2} / \mathrm{s}\).

Consider the DVD of Problem 14.49, except now the reacting polymer is blended uniformly with the polycarbonate to reduce manufacturing costs. Assume that a first-order homogeneous chemical reaction takes place between the polymer and oxygen; the reaction rate is proportional to the oxygen molar concentration. (a) Write the governing equation, boundary conditions, and initial condition for the oxygen molar concentration after the DVD is removed from the oxygen- proof pouch, for a DVD of thickness \(2 L\). Do not solve. (b) The DVD will gradually become more opaque over time as the reaction proceeds. The ability to read the DVD will depend on how well the laser light can penetrate through the thickness of the DVD. Therefore, it is important to know the volume-averaged molar concentration of product, \(\bar{C}_{\text {prod }}\), as a function of time. Write an expression for \(\bar{C}_{\text {prod }}\) in terms of the oxygen molar concentration, assuming that every mole of oxygen that reacts with the polymer results in \(p\) moles of product.

A He-Xe mixture containing \(0.75\) mole fraction of helium is used for cooling of electronics in an avionics application. At a temperature of \(300 \mathrm{~K}\) and atmospheric pressure, calculate the mass fraction of helium and the mass density, molar concentration, and molecular weight of the mixture. If the cooling system capacity is \(10 \mathrm{~L}\), what is the mass of the coolant?
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