/*! 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 13 Methane is generated via the ana... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 13

Methane is generated via the anaerobic decomposition (biological degradation in the absence of oxygen) of solid waste in landfills. Collecting the methane for use as a fuel rather than allowing it to disperse into the atmosphere provides a useful supplement to natural gas as an energy source. If a batch of waste with mass \(M\) (tonnes) is deposited in a landfill at \(t=0,\) the rate of methane generation at time \(t\) is given by $$\dot{V}_{\mathrm{CH}_{4}}(t)=k L_{0} M_{\text {waste }} e^{-k t}$$ where \(\dot{V}_{\mathrm{CH}_{4}}\) is the rate at which methane is generated in standard cubic meters per year, \(k\) is a rate constant, \(L_{0}\) is the total potential yield of landfill gas in standard cubic meters per tonne of waste, and \(M_{\text {watte is the tonnes of waste in the landfill at } t=0}\). (a) Starting with Equation 1, derive an expression for the mass generation rate of methane, \(\dot{M}_{\mathrm{CH}_{4}}(t)\) Without doing any calculations, sketch the shape of a plot of \(M_{\mathrm{CH}, \text { versus } t \text { from } t=0 \text { to } t=3 \mathrm{y},}\) and graphically show on the plot the total masses of methane generated in Years \(1,2,\) and \(3 .\) Then derive an expression for \(M_{\mathrm{CH}_{4}}(t),\) the total mass of methane (tonnes) generated from \(t=0\) to a time \(t\) (b) A new landfill has a yield potential \(L_{0}=100\) SCM CH \(_{4}\) /tonne waste and a rate constant \(k=0.04 \mathrm{y}^{-1} .\) At the beginning of the first year, 48,000 tonnes of waste are deposited in the landfill. Calculate the tonnes of methane generated from this deposit over a three-year period. (c) A colleague solving the problem of Part (b) calculates the methane produced in three years from the \(4.8 \times 10^{4}\) tonnes of waste as $$M_{\mathrm{CH}_{4}}(t=3)=\dot{M}_{\mathrm{CH}_{4}}(t=0) \times 1 \mathrm{y}+\dot{M}_{\mathrm{CH}_{4}}(t=1) \times 1 \mathrm{y}+\dot{M}_{\mathrm{CH}_{4}}(t=2) \times 1 \mathrm{y}$$ where \(\dot{M}_{\mathrm{CH}_{4}}\) is the first expression derived in Part (a). Briefly state what has been assumed about the rate of methane generation. Calculate the value determined with this method and the percentage error in the calculation. Show graphically what the calculated value corresponds to on another sketch of \(M_{\mathrm{CH}_{4}}\) versus \(t\) (d) The following amounts of waste are deposited in the landfill on January 1 in each of three consecutive years. Exploratory Exercises - Research and Discover (e) Explain in your own words the benefits of reducing the release of methane from landfills and of using the methane as a fuel instead of natural gas. (f) One way to avoid the environmental hazard of methane generation is to incinerate the waste before it has a chance to decompose. What problems might this alternative process introduce?

Short Answer

Expert verified
Methane mass generation rate equation is \(\dot{M}_{\mathrm{CH}_{4}}(t) = 16 \times k L_{0} M_{\text {waste }} e^{-kt}\) and integral methane mass equation is \(M_{\mathrm{CH}_{4}}(t) = 16L_{0}M_{\text{waste}}(1 - e^{-kt})\). Calculated methane generation over 3 years depends on specific landfill parameters, while the colleague's simplified method leads to overestimation due to it presuming constant generation rate. Benefits of capturing and utilizing methane include greenhouse effect mitigation and additional fuel source. Potential problems with incineration include air pollution and difficulties in residual waste disposal.

Step by step solution

01

Deriving the Methane Mass Generation Rate

To derive the methane mass generation rate, \(\dot{M}_{\mathrm{CH}_{4}}(t)\), multiply the given methane generation rate equation by the molecular mass of methane, which is 16 g/mol. This gives: \(\dot{M}_{\mathrm{CH}_{4}}(t) = 16 \times k L_{0} M_{\text {waste }} e^{-kt}\).
02

Sketching a Plot

Since the equation is an exponential decay function, sketch a curve that starts at a point (0, \(16kL_{0}M_{\text {waste }}\)) and decreases as \(t\) increases, asymptotically approaching the x-axis but never reaching it.
03

Deriving Methane Mass

To find total mass of methane generated up to time \(t\), integrate \(\dot{M}_{\mathrm{CH}_{4}}(t)\) from 0 to \(t\): \(M_{\mathrm{CH}_{4}}(t) = \int_{0}^{t} 16 \times k L_{0} M_{\text {waste }} e^{-kt} dt\). Solving this integral gives: \(M_{\mathrm{CH}_{4}}(t) = 16L_{0}M_{\text{waste}}(1 - e^{-kt})\).
04

Calculating Methane Over Three Years

Insert values into the derived equation from Step 3 for \(L_{0}=100\), \(k=0.04\) and \(M_{\text{waste}}=480000\), then calculate for \(t=3\).
05

Analyzing Assumption

The colleague's method assumes that the rate of methane generation remains constant for each year. However, as our derived equation indicates, this rate actually decreases exponentially with time.
06

Calculating Error

Calculate the methane mass for each year using the equation from step 5 and the colleague's method, and find the percentage difference between both methods.
07

Visual Representation of Error

On the graphical plot, represent the accurate methane amount as result from our equation, and the approximated methane amount as obtained from colleague's method, for each separate year.
08

Discussing Benefits

The exercise questions the advantages of methane utilization instead of its direct contribution to greenhouse effect, and benefits are to be listed.
09

Discussing Incineration Issues

The final step questions the problems related to alternate waste processing method - incineration.

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 91影视!

Key Concepts

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

Chemical Kinetics
Chemical kinetics is the study of rates of chemical processes, and understanding this concept is crucial when delving into methane generation in landfills. The exercise provided introduces a fundamental aspect of kinetics: the rate equation for methane generation, \( \dot{V}_{\mathrm{CH}_{4}}(t) = k L_{0} M_{\text {waste }} e^{-k t} \). This equation represents how the rate of methane production decreases over time due to the exponential decay term, \( e^{-kt} \).

The rate constant, \( k \), is indicative of how quickly methane production declines. A larger value of \( k \), typically inherent to the nature of the waste and conditions within the landfill, means that the methane generation rate will decrease more rapidly. By integrating the rate equation with respect to time, we can calculate the total mass of methane produced over a specific period. This provides a more comprehensive understanding of the methane's lifespan within the landfill environment and underscores the importance of chemical kinetics in predicting and managing methane generation in waste sites.
Methane as Sustainable Fuel
Methane, a potent greenhouse gas, when released into the atmosphere contributes to climate change; however, its capture and use as a sustainable fuel can mitigate these environmental impacts. Harvesting landfill methane for energy repurposes a waste byproduct into a valuable resource. Compared to other non-renewable energy sources, like coal or oil, methane burns more cleanly, releasing fewer pollutants such as nitrogen oxides and particulates.

Methane's high energy content makes it an excellent choice for both heat and power generation, and its use displaces the demand for fossil fuels, thereby conserving natural resources and reducing our carbon footprint. Moreover, using methane from landfills aligns with the circular economy principles, adding value to what otherwise would be a harmful and wasted byproduct.
Environmental Impact of Landfill Gases
Landfill gases, primarily composed of methane and carbon dioxide, have far-reaching environmental impacts. Methane is of particular concern due to its high global warming potential, which is approximately 28-36 times greater than CO2 over a 100-year period. When landfills are not properly managed, methane escapes into the atmosphere, exacerbating greenhouse gas effects such as global warming.

In addition to climate change implications, these gases can harm local ecosystems, infiltrate and deteriorate the air quality, and pose a risk of explosions or fires. Addressing the release of landfill gases is therefore critical for both environmental protection and public safety.
Waste Management Strategies
Effective waste management strategies are imperative for reducing the generation of methane in landfills. Techniques such as recycling, composting, and waste-to-energy technologies can significantly decrease the volume of organic waste that ends up in landfills.

Landfill gas collection systems also play an essential role by capturing methane produced from waste decomposition and converting it into energy. Further, policies encouraging reduced waste generation and the segregation of organic matter from the waste stream also cut down on the quantity of methane produced.

A progressive approach to waste management includes both preventing the production of landfill gases and utilizing them as energy, reflecting a shift towards sustainability that benefits both the environment and the economy.

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

The following chemical reactions take place in a liquid-phase batch reactor of constant volume \(V\). $$\begin{aligned} &\mathrm{A} \rightarrow 2 \mathrm{B} \quad r_{1}[\mathrm{mol} \mathrm{A} \text { consumed } /(\mathrm{L} \cdot \mathrm{s})]=0.100 C_{\mathrm{A}}\\\ &\mathbf{B} \rightarrow \mathbf{C} \quad r_{2}[\mathrm{mol} \mathbf{C} \text { generated } /(\mathbf{L} \cdot \mathbf{s})]=0.200 C_{\mathrm{B}}^{2} \end{aligned}$$ where the concentrations \(C_{\mathrm{A}}\) and \(C_{\mathrm{B}}\) are in mol/L. The reactor is initially charged with pure \(\mathrm{A}\) at a concentration of 1.00 mol/L. (a) Write expressions for ( \(i\) ) the rate of generation of \(\mathrm{B}\) in the first reaction and (ii) the rate of consumption of \(\mathrm{B}\) in the second reaction. (If this takes you more than about 10 seconds, you're missing the point.) (b) Write mole balances on A, B, and C, convert them into expressions for \(d C_{\mathrm{A}} / d t, d C_{\mathrm{B}} / d t\), and \(d C_{\mathrm{C}} / d t,\) and provide boundary conditions. (c) Without doing any calculations, sketch on a single graph the plots you would expect to obtain of \(C_{\mathrm{A}}\) versus \(t, C_{\mathrm{B}}\) versus \(t,\) and \(C_{\mathrm{C}}\) versus \(t .\) Clearly show the function values at \(t=0\) and \(t \rightarrow \infty\) and the curvature (concave up, concave down, or linear) in the vicinity of \(t=0 .\) Briefly explain your reasoning. (d) Solve the equations derived in Part (b) using a differential equation- solving program. On a single graph, show plots of \(C_{\mathrm{A} \text { versust }}, C_{\mathrm{B}}\) versus \(t,\) and \(C_{\mathrm{C}}\) versus \(t\) from \(t=0\) to \(t=50\) s. Verify that your predictions in Part (c) were correct. If they were not, change them and revise your explanation.

A steam coil is immersed in a stirred tank. Saturated steam at 7.50 bar condenses within the coil, and the condensate emerges at its saturation temperature. A solvent with a heat capacity of \(2.30 \mathrm{kJ} /\left(\mathrm{kg} \cdot^{\cdot} \mathrm{C}\right)\) is fed to the tank at a steady rate of \(12.0 \mathrm{kg} / \mathrm{min}\) and a temperature of \(25^{\circ} \mathrm{C},\) and the heated solvent is discharged at the same flow rate. The tank is initially filled with \(760 \mathrm{kg}\) of solvent at \(25^{\circ} \mathrm{C},\) at which point the flows of both steam and solvent are commenced. The rate at which heat is transferred from the steam coil to the solvent is given by the expression $$\dot{Q}=U A\left(T_{\mathrm{steam}}-T\right)$$ where \(U A\) (the product of a heat transfer coefficient and the coil surface area through which the heat is transferred) equals \(11.5 \mathrm{kJ} /\left(\min \cdot^{\circ} \mathrm{C}\right) .\) The tank is well stirred, so that the temperature of the contents is spatially uniform and equals the outlet temperature. (a) Prove that an energy balance on the tank contents reduces to the equation given below and supply an initial condition. \frac{d T}{d t}=1.50^{\circ} \mathrm{C} / \mathrm{min}-0.0224 T (b) Without integrating the equation, calculate the steady-state value of \(T\) and sketch the expected plot of \(T\) versus \(t,\) labeling the values of \(T_{\mathrm{b}}\) at \(t=0\) and \(t \rightarrow \infty\) (c) Integrate the balance equation to obtain an expression for \(T(t)\) and calculate the solvent temperature after 40 minutes. (d) The tank is shut down for routine maintenance, and a technician notices that a thin mineral scale has formed on the outside of the steam coil. The coil is treated with a mild acid that removes the scale and reinstalled in the tank. The process described above is run again with the same steam conditions, solvent flow rate, and mass of solvent charged to the tank, and the temperature after 40 minutes is \(55^{\circ} \mathrm{C}\) instead of the value calculated in Part (c). One of the system variables listed in the problem statement must have changed as a result of the change in the stirrer. Which variable would you guess it to be, and by what percentage of its initial value did it change?

A kettle containing 3.00 liters of water at a temperature of \(18^{\circ} \mathrm{C}\) is placed on an electric stove and begins to boil in three minutes. (a) Write an energy balance on the water and determine an expression for \(d T / d t,\) neglecting evaporation of water before the boiling point is reached, and provide an initial condition. Sketch a plot of \(T\) versus \(t\) from \(t=0\) to \(t=4\) minutes. (b) Calculate the average rate (W) at which heat is being added to the water. Then calculate the rate (g/s) at which water vaporizes once boiling begins. (c) The rate of heat output from the stove element differs significantly from the heating rate calculated in Part (b). In which direction, and why?

A gas-phase decomposition reaction with stoichiometry \(2 \mathrm{A} \rightarrow 2 \mathrm{B}+\mathrm{C}\) follows a second-order rate law (see Problem 10.19): $$r_{\mathrm{d}}\left[\operatorname{mol} /\left(\mathrm{m}^{3} \cdot \mathrm{s}\right)\right]=k C_{\mathrm{A}}^{2}$$ where \(C_{\mathrm{A}}\) is the reactant concentration in \(\mathrm{mol} / \mathrm{m}^{3}\). The rate constant \(k\) varies with the reaction temperature according to the Arrhenius law $$k\left[\mathrm{m}^{3} /(\mathrm{mol} \cdot \mathrm{s})\right]=k_{0} \exp (-E / R T)$$ where \(k_{0}\left[\mathrm{m}^{3} /(\mathrm{mol} \cdot \mathrm{s}]\right)=\) the preexponential factor \(E(\mathrm{J} / \mathrm{mol})=\) the reaction activation energy \(R=\) the gas constant \(T(\mathrm{K})=\) the reaction temperature (a) Suppose the reaction is carried out in a batch reactor of constant volume \(V\left(\mathrm{m}^{3}\right)\) at a constant temperature \(T(\mathrm{K}),\) beginning with pure \(\mathrm{A}\) at a concentration \(C_{\mathrm{A} 0} .\) Write a differential balance on A and integrate it to obtain an expression for \(C_{\mathrm{A}}(t)\) in terms of \(C_{\mathrm{A} 0}\) and \(k\) (b) Let \(P_{0}(\text { atm })\) be the initial reactor pressure. Prove that \(t_{1 / 2}\), the time required to achieve a \(50 \%\) conversion of \(\mathrm{A}\) in the reactor, equals \(R T / k P_{0},\) and derive an expression for \(P_{1 / 2},\) the reactor pressure at this point, in terms of \(P_{0} .\) Assume ideal- gas behavior. (c) The decomposition of nitrous oxide \(\left(\mathrm{N}_{2} \mathrm{O}\right)\) to nitrogen and oxygen is carried out in a 5.00 -liter batch reactor at a constant temperature of \(1015 \mathrm{K},\) beginning with pure \(\mathrm{N}_{2} \mathrm{O}\) at several initial pressures. The reactor pressure \(P(t)\) is monitored, and the times \(\left(t_{1 / 2}\right)\) required to achieve \(50 \%\) conversion of \(\mathrm{N}_{2} \mathrm{O}\) are noted. $$\begin{array}{|c|c|c|c|c|} \hline P_{0}(\mathrm{atm}) & 0.135 & 0.286 & 0.416 & 0.683 \\ \hline t_{1 / 2}(\mathrm{s}) & 1060 & 500 & 344 & 209 \\ \hline \end{array}$$ Use these results to verify that the \(\mathrm{N}_{2} \mathrm{O}\) decomposition reaction is second-order and determine the value of \(k\) at \(T=1015 \mathrm{K}\) (d) The same experiment is performed at several other temperatures at a single initial pressure of 1.00 atm, with the following results: $$\begin{array}{|c|c|c|c|c|} \hline T(\mathrm{K}) & 900 & 950 & 1000 & 1050 \\ \hline t_{1 / 2}(\mathrm{s}) & 5464 & 1004 & 219 & 55 \\ \hline \end{array}$$ Use a graphical method to determine the Arrhenius law parameters ( \(k_{0}\) and \(E\) ) for the reaction. (e) Suppose the reaction is carried out in a batch reactor at \(T=980 \mathrm{K},\) beginning with a mixture at 1.20 atm containing 70 mole \(\%\) N \(_{2}\) O and the balance a chemically inert gas. How long (minutes) will it take to achieve a \(90 \%\) conversion of \(\mathrm{N}_{2} \mathrm{O} ?\)

Methanol is added to a storage tank at a rate of \(1200 \mathrm{kg} / \mathrm{h}\) and is simultaneously withdrawn at a rate \(\dot{m}_{w}(t)(\mathrm{kg} / \mathrm{h})\) that increases linearly with time. At \(t=0\) the tank contains \(750 \mathrm{kg}\) of the liquid and \(\dot{m}_{w}=750 \mathrm{kg} / \mathrm{h} .\) Five hours later \(\dot{m}_{\mathrm{w}}\) equals \(1000 \mathrm{kg} / \mathrm{h}\) (a) Calculate an expression for \(\dot{m}_{w}(t),\) letting \(t=0\) signify the time at which \(\dot{m}_{w}=750 \mathrm{kg} / \mathrm{h},\) and incorporate it into a differential methanol balance, letting \(M(\mathrm{kg})\) be the mass of methanol in the tank at any time. (b) Integrate the balance equation to obtain an expression for \(M(t)\) and check the solution two ways. (See Example 10.2-1.) For now, assume that the tank has an infinite capacity. (c) Calculate how long it will take for the mass of methanol in the tank to reach its maximum value, and calculate that value. Then calculate the time it will take to empty the tank. (d) Now suppose the tank volume is \(3.40 \mathrm{m}^{3}\). Draw a plot of \(M\) versus \(t\), covering the period from \(t=0\) to an hour after the tank is empty. Write expressions for \(M(t)\) in each time range when the function changes.
See all solutions

Recommended explanations on Chemistry Textbooks

Organic Chemistry

Read Explanation

Chemical Reactions

Read Explanation

The Earths Atmosphere

Read Explanation

Making Measurements

Read Explanation

Inorganic Chemistry

Read Explanation

Chemical Analysis

Read Explanation
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.