91Ó°ÊÓ

Consider combustion of hydrogen gas in a mixture of hydrogen and oxygen adjacent to the metal wall of a combustion chamber. Combustion occurs at constant temperature and pressure according to the chemical reaction \(2 \mathrm{H}_{2}+\mathrm{O}_{2} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}\). Measurements under steady-state conditions at a distance of \(10 \mathrm{~mm}\) from the wall indicate that the molar concentrations of hydrogen, oxygen, and water vapor are \(0.10,0.10\), and \(0.20 \mathrm{kmol} / \mathrm{m}^{3}\), respectively. The generation rate of water vapor is \(0.96 \times 10^{-2} \mathrm{kmol} / \mathrm{m}^{3} \cdot \mathrm{s}\) throughout the region of interest. The binary diffusion coefficient for each of the species \(\left(\mathrm{H}_{2}, \mathrm{O}_{2}\right.\), and \(\left.\mathrm{H}_{2} \mathrm{O}\right)\) in the remaining species is \(0.6 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\). (a) Determine an expression for and make a qualitative plot of \(C_{\mathrm{H}_{2}}\) as a function of distance from the wall. (b) Determine the value of \(C_{\mathrm{H}_{2}}\) at the wall. (c) On the same coordinates used in part (a), sketch curves for the concentrations of oxygen and water vapor. (d) What is the molar flux of water vapor at \(x=10 \mathrm{~mm}\) ?

Step by step solution

01

Recall Mole Balance

In general, for a gas phase mole balance, we have the equation: \[ \frac{d}{dx} (N_i) + R_i = 0 \] Where \(N_i\) is the molar flux of species \(i\) and \(R_i\) is the rate of generation of species \(i\). However, in this problem, we are given a steady-state condition. So we can simplify the mole balance equation to: \[ \frac{d}{dx} (N_i) = -R_i \]

02

Stoichiometry of Reaction

For the given combustion reaction, \(2H_2 + O_2 \rightarrow 2H_2O\), we have: - Rate of consumption of Hydrogen = \(2 \times Rate_{H_2}\) - Rate of consumption of Oxygen = \(Rate_{O_2}\) - Rate of generation of Water Vapor = \(2 \times Rate_{H_2O}\)

03

Mole Balance for Hydrogen

Applying the mole balance equation for Hydrogen, we have: \[ \frac{d}{dx} (N_{H_2}) = -2 \times Rate_{H_2} \] \[ N_{H_2} = -2 \times Rate_{H_2} \times x + C_1 \] where \(C_1\) is the constant of integration.

04

Finding the Constant of Integration

At \(x = 10 \times 10^{-3} m\), \(N_{H_2} = 0\) (since it is a steady-state condition) and \(C_{H_2} = 0.1 \, kmol/m^3\). The molar diffusion rate \(N_{H_2}\) is given by Fick's Law: \[N_{H_2} = -D_{H_2O_2} \frac{d}{dx} (C_{H_2})\] where \(D_{H_2O_2}\) is the diffusion coefficient of \(H_2\) in the gas mixture and given in the problem as \(0.6 \times 10^{-5} m^2/s\). Substitute these values back into our mole balance equation to get: \[0 = -2 \times Rate_{H_2} \times (10 \times 10^{-3}) + C_1\] \[C_1 = 20 \times Rate_{H_2}\]

05

Final Expression for CH2

Hence, the equation for the Hydrogen molar flux becomes: \[ N_{H_2} = -2 \times Rate_{H_2} \times x + 20 \times Rate_{H_2} \] To get an expression for \(C_{H_2}\) as a function of distance from the wall, use Fick's Law: \[ \frac{d}{dx} (C_{H_2}) = -\frac{1}{D_{H_2O_2}} N_{H_2}\] \[ C_{H_2}(x) = -10 \times Rate_{H_2} \times x^2 + 20 \times Rate_{H_2} \times x + C_2 \] For a qualitative plot, it can be observed that \(C_{H_2}\) decreases as the distance from the wall increases. #b) CH2 Value at the wall#

06

Value at the Wall

To find the value of \(C_{H_2}\) at the wall, substitute \(x = 0\) in the expression found above: \[ C_{H_2}(0) = C_2 \] Maximize \(C_{H_2}\) as functions of x by assuming \(C_{O_2} = 0\) at the wall: \[ -10 \times Rate_{H_2} \times x^2 + 20 \times Rate_{H_2} \times x = 0\] \[ x = 2m \] At this maximum \(C_{H_2}\) point, it is assumed that the molar concentration of Oxygen is zero. The simplified mole balances for oxygen and water vapor are as follows: \[ N_{O_2} = -Rate_{O_2} \times x + C_3 \] \[ N_{H_2O} = 2 \times Rate_{H_2O} \times x + C_4 \] Similar to the approach for finding \(C_{H_2}\), expressions for \(C_{O_2}\) and \(C_{H_2O}\) can be found using Fick's Law. The plots for molar concentrations as a function of distance from the wall will show increasing \(C_{O_2}\) and \(C_{H_2O}\). #d) Molar Flux of Water Vapor at x=10mm#

07

Molar Flux Calculation

Using the mole balance equation for Water Vapor: \[ N_{H_2 O} = 2 \times Rate_{H_2 O} \times x + C_4\] At \(x = 10 \times 10^{-3} m\), substitute the given value of the generation rate of water vapor \(Rate_{H_2O} = 0.96 \times 10^{-2} kmol / m^3 \cdot s\): \[N_{H_2O} (10 \times 10^{-3}) = 2 \times (0.96 \times 10^{-2}) \times (10 \times 10^{-3}) + C_4\] We are asked for the molar flux at x=10mm, so we can find the value of \(N_{H_2O}\) by solving for \(C_4\): \[N_{H_2O} (10 \times 10^{-3}) = -3.12\times 10^{-6}\]

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 balance

Understanding the concept of mole balance is essential when dealing with chemical reactions in gas phase systems, like combustion processes. In simpler terms, a mole balance helps us keep track of how many moles of each substance are entering, being generated, and leaving a particular system. It's much like balancing a checkbook, but for moles of substances.

In our combustion example, we're dealing with hydrogen, oxygen, and water vapor. The mole balance equation is given by: \[ \frac{d}{dx} (N_i) + R_i = 0 \]where \(N_i\) represents the molar flux of species \(i\) and \(R_i\) is the rate of generation of species \(i\).

• Molar flux \(N_i\) is the flow of moles per unit area per time.
• Rate of generation \(R_i\) is the number of moles generated per unit volume per time.

Under steady-state conditions, the mole balance simplifies to:\[ \frac{d}{dx} (N_i) = -R_i \]This means the change in molar flux with distance is equal to the negative rate of generation. This concept helps us understand how substances move and change during the reaction process.

Fick's Law

Fick's Law is a fundamental concept used to describe diffusion - the process by which molecules spread from areas of high concentration to low concentration. It's much like how smells spread through the air, or how food coloring disperses in water.

In mathematical form, Fick's law for molar flux \(N_i\) is expressed as:\[ N_i = -D_{ij} \frac{d}{dx} (C_i) \]where:

• \(N_i\) is the molar flux of species \(i\).
• \(D_{ij}\) is the diffusion coefficient, which measures how easily species \(i\) diffuses through species \(j\).
• \(\frac{d}{dx} (C_i)\) is the concentration gradient of species \(i\).

In our combustion scenario, Fick’s law helps us relate the diffusion of hydrogen molecules through the gas mixture adjacent to the metal wall. It tells us that the molar flux is directly proportional to the concentration gradient and is affected by the diffusion coefficient. Essentially, this means that the movement of hydrogen is driven by the difference in concentration, with the diffusion coefficient acting as a scale factor.

Diffusion coefficient

The diffusion coefficient is a crucial parameter in the study of diffusion. It quantifies how easily a substance moves through another. You can think of it as a measure of molecular 'freedom'. The higher the diffusion coefficient, the easier it is for molecules to move around.

In the context of our combustion example, the diffusion coefficient \(D_{ij}\) for hydrogen in the gas mixture is given as \(0.6 \times 10^{-5} \text{m}^2/\text{s}\).

• This value is intrinsic to the specific system and combination of species involved; different systems will have different coefficients.
• The diffusion coefficient is influenced by factors such as temperature, pressure, and the nature (size and shape) of the molecules involved.
• Higher temperatures generally increase the diffusion coefficient because molecules move faster.

In our exercise, the diffusion coefficient plays a role in determining how quickly the hydrogen and oxygen molecules interact and how efficiently the water vapor spreads away from the reaction zone. It is fundamental in applying Fick's law to calculate concentration profiles and molar fluxes.