/*! 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 86 In a condenser to recover solven... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In a condenser to recover solvent vapors, \(9 \mathrm{kmol}\) of steam (species 1) are condensed for each \(\mathrm{kmol}\) of solvent (species 2). At the condensate surface \(x_{2, s}=0.85 .\) The total condensation rate of steam and solvent vapors is \(0.002 \mathrm{kmol} / \mathrm{m}^{2} \mathrm{~s}\). Using a one-dimensional model, find the mole fractions \(x_{1}\) and \(x_{2}\) as functions of distance from the condensate surface. The total pressure is 3 atm, \(T_{s}=320 \mathrm{~K}\), and \(\mathscr{D}_{12}=10 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\).

Short Answer

Expert verified
Mole fractions: \(x_2(z) = -200z + 0.15\), \(x_1(z) = 200z + 0.85\).

Step by step solution

01

Define the problem and known values

We are tasked with finding the mole fractions \(x_1\) and \(x_2\) as functions of distance from the condensate surface. Knowing that steam (species 1) and solvent (species 2) condense, we have:- Composition at condensate surface: \(x_{2,s} = 0.85\)- Total condensation rate: \(0.002 \text{ kmol/m}^2 \text{s}\)- Total pressure: \(P = 3 \text{ atm}\)- Temperature at surface: \(T_s = 320 \text{ K}\)- Binary diffusion coefficient: \(\mathscr{D}_{12} = 10 \times 10^{-6} \text{ m}^2/\text{s}\)- Condensation ratio: \( rac{9 \text{ kmol steam}}{1 \text{ kmol solvent}}\)
02

Use the condensation ratio to find \(x_{1,s}\)

The condensation rate describes the ratio between steam and solvent:\[ \frac{9 \text{ kmol steam}}{1 \text{ kmol solvent}} = \frac{y_1}{y_2} \]At the surface:\[ x_1 + x_2 = 1 \rightarrow x_1 + 0.85 = 1 \rightarrow x_1 = 0.15 \]
03

Use Fick's Law to relate mole fractions and distance

Fick's first law for constant diffusion and steady state is:\[ J = -\mathscr{D}_{12} \frac{d \alpha}{dz} \]where \(J\) is the condensation rate of the vapor. Here, \(\alpha\) stands for \(y_2\) (mole fraction of the solvent in vapor), and \(\mathscr{D}_{12}\) is the diffusion coefficient. Since \(J = 0.002\), you can express it as:\[ 0.002 = -10 \times 10^{-6} \frac{dy}{dz} \]Rearrange it as:\[ \frac{dy}{dz} = -200\]
04

Integrate to find the mole fraction profile

Integrate the equation:\[ \frac{dy}{dz} = -200 \]This gives:\[ y(z) = -200z + C \]Use the boundary condition at \(z = 0\) (at the surface \(x_{2,s} = 0.85\)):\[ y(0) = 0.15 \]Thus:\[ 0.15 = -200(0) + C \rightarrow C = 0.15 \]So:\[ y(z) = -200z + 0.15 \]
05

Determine \(x_1(z)\) from \(x_2(z)\)

Since \(x_1 + x_2 = 1\), we have:\[ x_1(z) = 1 - x_2(z) \]Therefore:\[ x_1(z) = 1 - (-200z + 0.15) \]\[ x_1(z) = 200z + 0.85 \]

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 91Ó°ÊÓ!

Key Concepts

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

Mole Fraction
The concept of mole fraction is crucial in understanding how components in a mixture are proportioned. In a two-component system like the one we are dealing with, the mole fraction indicates how many moles of a particular species are present relative to the total moles of all components in the mixture.

For our scenario, if we look at the condensate surface, we have the mole fractions denoted as \(x_1\) for steam and \(x_2\) for the solvent. These fractions can be mathematically expressed as:
- \(x_1 = \frac{n_1}{n_1 + n_2}\)
- \(x_2 = \frac{n_2}{n_1 + n_2}\)

Here, \(n_1\) represents the moles of steam and \(n_2\) the moles of the solvent. Thus, the sum of mole fractions must always equal one, \(x_1 + x_2 = 1\). Understanding this relationship helps in determining the proportions of each component throughout the distance from the surface.
Diffusion Coefficient
The diffusion coefficient, denoted by \(\mathscr{D}_{12}\), is a term that represents how rapidly a substance spreads through another. In our case, it describes the rate at which steam and solvent vapors are diffusing within the condenser.

The diffusion coefficient is affected by factors such as temperature and pressure, but for the given problem, it is considered constant. It has units of \(\text{m}^2/\text{s}\), indicating how much a substance can diffuse over a given area per second. The given value in the problem is \(\mathscr{D}_{12} = 10 \times 10^{-6} \, \text{m}^2/\text{s}\).

This coefficient plays a pivotal role in calculating the vapor concentration profiles, through a link with Fick's Law of Diffusion, which will be explained further below.
Fick's Law
Fick's Law is fundamental in understanding how diffusion occurs. It provides a relationship between diffusion flow rate and concentration gradient. Fick’s first law specifically relates the diffusion flux \(J\) to the gradient of concentration \(\alpha\), where \(J\) is represented by the equation:
\[ J = -\mathscr{D}_{12} \frac{d\alpha}{dz} \]
For steady-state diffusion where the diffusion coefficient \(\mathscr{D}_{12}\) remains constant, we can determine the rate at which \(y_2\) — the mole fraction of the solvent in the vapor phase — changes as a function of distance \(z\) from the condensate surface.

In the problem setup, this allows us to connect the tangible rate of condensation \(J=0.002\) to the gradient of mole fractions. By rearranging the law, we obtain \(\frac{dy}{dz} = -200\), which can be solved by integration to find the concentration profile \(y(z)\). This resulting profile is pivotal in understanding how the vapor concentration changes with distance in the condenser.
Binary Diffusion
Binary diffusion involves the movement of two species as they exchange places due to microscopic collisions. In our context, this means the distribution of steam and solvent vapors through the condenser system.

When considering binary diffusion, it’s essential to view how these vapor species interact. Each species impacts the other’s movement while maintaining the overall composition determined by mole fractions.

The diffusion coefficient is key here, as it provides insight into how easily these species mix together. During the diffusion process, each vapor’s concentration alters spatially based on the opposing concentration gradients, temperature, pressure, and of course, the diffusion coefficient.
  • Streamline understanding of vapor transport.
  • Separation of components in mixtures.
  • Enhance controlled recovery as seen in condensation processes.
Hence, binary diffusion is not only critical in this problem, but it also plays an extensive role in many industrial applications where separation and purification of substances are needed.

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

Plot a graph of the mass fraction of water vapor in saturated water vapor-air mixtures at 1 atm pressure for \(273.15 \mathrm{~K}

Experimental oxidation data for a titanium alloy, used to sheath hypersonic vehicle nose cones, indicates that the total weight gain can be described by a parabolic law of the form \(w^{2}=C t\), where \(C=480 \exp \left(-E_{a} / \mathscr{R T}\right) \mathrm{g}^{2} / \mathrm{cm}^{4} \mathrm{~s}\), and \(E_{a}=61,800\) \(\mathrm{cal} / \mathrm{mol}\). Examination of oxidized samples shows that about \(20 \%\) of the oxygen goes to form a rutile \(\left(\mathrm{Ti} \mathrm{O}_{2}\right)\) surface scale. Assuming that the diffusion coefficient for oxygen in alpha titanium can be expressed as \(\mathscr{D}_{12}=\mathscr{D}_{0} \exp \left(-E_{a} / \mathscr{R} T\right)\), with \(E_{a}\) again \(61,800 \mathrm{cal} / \mathrm{mol}\), estimate \(\mathscr{D}_{0}\). Show also that the parabolic oxidation law implies diffusion across the rutile scale characterized by a linear concentration gradient. Use an alloy density of \(4400 \mathrm{~kg} / \mathrm{m}^{3}\).

Air at \(315 \mathrm{~K}, 1 \mathrm{~atm}\), and \(20 \%\) RH flows at \(12 \mathrm{~m} / \mathrm{s}\) along a \(30 \mathrm{~cm}-\) square copper plate in a wind tunnel. The plate has a porous surface, and water is supplied to maintain a completely wet surface. The walls of the wind tunnel are at \(10^{\circ} \mathrm{C}\). Estimate the equilibrium temperature of the plate if the backside of the plate is insulated and the water supply temperature is controlled to be equal to the plate surface temperature. Take \(\varepsilon=0.92\) for the wetted surface. Use a transition Reynolds number of \(3 \times 10^{5}\).

(i) A mixture of noble gases contains equal mole fractions of helium, argon, and xenon. What is the composition in terms of mass fractions? (ii) If the mixture contains equal mass fractions of \(\mathrm{He}, \mathrm{Ar}\), and \(\mathrm{Xe}\), what are the corresponding mole fractions?

On a clear night the air temperature is \(5^{\circ} \mathrm{C}\) and has a relative humidity of \(60 \%\). The effective sky temperature as a blackbody is \(-23^{\circ} \mathrm{C}\). Water at \(15^{\circ} \mathrm{C}\) is poured into a shallow, well-insulated pan, giving a layer \(1 \mathrm{~mm}\) deep. The pan is in full view of the sky and is exposed to a light breeze, giving an estimated convective heat transfer coefficient of \(10 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\) on the water surface. (i) Show that it is possible for ice to form. (ii) How long after the first ice appears will the water be completely frozen? The enthalpy of fusion of water is \(335 \mathrm{~kJ} / \mathrm{kg}\). (Hint: Perform a surface energy balance at \(0^{\circ} \mathrm{C}\).)

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.