/*! 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 78 Methane and \(30 \%\) excess air... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Methane and \(30 \%\) excess air are to be fed to a combustion reactor. An inexperienced technician mistakes his instructions and charges the gases together in the required proportion into an evacuated closed tank. (The gases were supposed to be fed directly into the reactor.) The contents of the charged tank are at \(25^{\circ} \mathrm{C}\) and 4.00 atm absolute. (a) Calculate the standard internal energy of combustion of the methane combustion reaction. \(\Delta \hat{U}_{c}^{\circ}(\mathrm{kJ} / \mathrm{mol}),\) taking \(\mathrm{CO}_{2}(\mathrm{g})\) and \(\mathrm{H}_{2} \mathrm{O}(\mathrm{v})\) as the presumed products. Then prove that if the constant-pressure heat capacity of an ideal-gas species is independent of temperature, the specific internal energy of that species at temperature \(T\left(^{\circ} \mathrm{C}\right)\) relative to the same species at \(25^{\circ} \mathrm{C}\) is given by the expression $$\hat{U}=\left(C_{p}-R\right)\left(T-25^{\circ} \mathrm{C}\right)$$ where \(R\) is the gas constant. Use this formula in the next part of the problem. (b) You wish to calculate the maximum temperature, \(T_{\max }\left(^{\circ} \mathrm{C}\right),\) and corresponding pressure, \(P_{\max }(\text { atm }),\) that the tank would have to withstand if the mixture it contains were to be accidentally ignited. Taking molecular species at \(25^{\circ} \mathrm{C}\) as references and treating all species as ideal gases, prepare an inlet-outlet internal energy table for the closed system combustion process. In deriving expressions for each \(\dot{U}_{i}\) at the final reactor condition \(\left(T_{\max }, P_{\max }\right),\) use the following approximate values for \(C_{p_{i}}\left[\mathrm{k} J /\left(\mathrm{mol} \cdot^{\circ} \mathrm{C}\right)\right]: 0.033 \mathrm{for} \mathrm{O}_{2}, 0.032\) for \(\mathrm{N}_{2}, 0.052 \mathrm{for} \mathrm{CO}_{2},\) and \(0.040 \mathrm{for} \mathrm{H}_{2} \mathrm{O}(\mathrm{v}) .\) Then use an energy balance and the ideal-gas equation of state to perform the required calculations. (c) Why would the actual temperature and pressure attained in a real tank be less than the values calculated in Part (a)? (State several reasons.) (d) Think of ways that the tank contents might be accidentally ignited. The list should suggest why accepted plant safety regulations prohibit the storage of combustible vapor mixtures.

Short Answer

Expert verified
First compute the balanced chemical equation for the combustion of methane with a 30\% excess of air and subsequently the standard inner energy of combustion using enthalpies of formation. Compute the specific inner energy of an ideal gas species following the formula \( \hat{U} = (C_{p}-R) (T-25^{\circ} \mathrm{C})\). After preparing an internal energy table for the combustion process and computing the vortex quantities for each species, the energy balance and ideal gas equation of state are used to calculate the maximum temperature and pressure possible in the tank. Realistic values are usually lower due to factors such as heat losses to the surroundings, non-ideal behavior of gases, and incomplete combustion. One of the potential ignition sources and reasons for the prohibition of the storage of flammable gas mixtures includes sources of heat, sparks, and flames in the environment, as well as the electricity released during the combustion process.

Step by step solution

01

Calculate the combustion reaction

Determine the balanced chemical equation for methane combustion with \(30 \%\) excess air. The combustion equation is \(CH_{4} + 2O_{2} \rightarrow CO_{2} + 2H_{2}O\). Considering \(30 \%\) excess air, the reaction becomes: \(CH_{4} + 2.6O_{2} \rightarrow CO_{2} + 2H_{2}O + 0.6O_{2}\).
02

Calculate the standard internal energy of combustion

Use standard enthalpies of formation to calculate the standard internal energy change for the reaction using the formula \( \Delta U = \Delta H - T \Delta S\) and data from literature or textbooks. Enthalpies and entropies of formation are usually given at standard temperature and pressure (STP), i.e., \(25^{\circ} \mathrm{C}\) and 1 atm.
03

Derive the specific internal energy formula

Demonstrate that the specific internal energy of an ideal gas species at a certain temperature relative to \(25^{\circ} \mathrm{C}\) equals \(\hat{U} = (C_{p}-R) (T-25^{\circ} \mathrm{C})\). This comes from the definition of specific heat capacity at constant pressure \(C_{p}\), and the fact that for an ideal gas, the change in enthalpy equals the change in internal energy plus the product of pressure and change in volume, which leads to a difference in the constant volume and constant pressure heat capacities equal to the gas constant \(R\).
04

Prepare energy table

Calculate the vortex quantities for the combustion process using the formula \(U = n (C_{p}-R) (T-25^{\circ} \mathrm{C})\). Use given values for molar heat capacities and the vortex quantities from the balanced chemical equation. Solve the expressions for each molecule at the final reactor condition \(\left(T_{\max }, P_{\max }\right)\) using the provided values for each species’ specific heat capacities.
05

Perform energy balance calculation

Apply energy conservation to calculate the max temperature and pressure. The energy released by combustion equals the increase in internal energy of the combustion products. Then use ideal-gas equation of state \(PV=nRT\) to solve for pressure.
06

Explaining fluctuation in realistic values

Explain why the real temperature and pressure might be less than the calculated values, such as heat loss to the environment, non-ideality of gases, incomplete combustion, and others.
07

Identify potential ignition sources

List sources of ignition which include static electricity, external heat, sparks or open flames, among other things.

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.

Combustion Reactions
Combustion reactions are chemical processes in which a fuel reacts with an oxidant, typically oxygen, to release energy. This energy is often in the form of heat and light. In the case of methane combustion, the balanced chemical equation is:
  • CHâ‚„ + 2Oâ‚‚ → COâ‚‚ + 2Hâ‚‚O.
When there is a 30% excess of air, extra oxygen molecules are present, making the equation:
  • CHâ‚„ + 2.6Oâ‚‚ → COâ‚‚ + 2Hâ‚‚O + 0.6Oâ‚‚.
This reaction is exothermic, meaning it releases energy. The energy from combustion can be harnessed in engines and power plants.
To find out how much energy a reaction releases or absorbs, chemists calculate the reaction's internal energy change (∆U). This involves using the energy stored in chemical bonds and determining how much changes as new bonds form.
Understanding combustion reactions is crucial in designing safe and efficient systems for energy production.
Ideal Gas Law
The Ideal Gas Law is a fundamental equation that describes the behavior of an ideal gas. It is expressed as:
  • PV = nRT
Here:
  • P represents pressure,
  • V is volume,
  • n denotes the number of moles,
  • R is the universal gas constant, and
  • T stands for temperature in Kelvin.
This equation helps predict how a gas will respond to changes in pressure, volume, or temperature when other conditions remain constant.
In our scenario, calculating the maximum pressure ( P_{ ext{max}}) involved using the Ideal Gas Law after combustion in a closed tank, starting with conditions like 25°C and 4.00 atm, is critical.
Understanding these calculations is essential in determining whether a system can withstand certain pressures and temperatures during chemical reactions.
Internal Energy
Internal energy is the total energy contained within a system. It includes kinetic energy of molecules and the energy stored in chemical bonds.
For a reaction, such as methane combustion, internal energy change (∆U) can be calculated using the formula:
  • ∆U = ∆H - T∆S.
Here:
  • ∆H is the enthalpy change, and
  • T∆S accounts for energy affecting entropy at temperature T.
Through combustion, the internal energy of reactants decreases as energy is released.
When analyzing ideal gases, specific internal energy changes can be calculated using \[\hat{U} = (C_p - R)(T - 25^{\circ}C)\]This equation indicates how internal energy varies with temperature, factoring in heat capacity (C_p) and the constant R. Understanding changes in internal energy helps predict how energy is transferred during reactions and the resulting temperature variations.
Heat Capacity
Heat capacity is the amount of heat energy required to raise the temperature of a substance by one degree Celsius. It is an intrinsic property of matter that determines how substances respond to added or removed heat. In the context of ideal gases, the heat capacity at constant pressure ( C_p) is critical.
For our calculation, specific heat capacities ( C_p) for gases involved in the reaction are necessary. These values are typically provided in scientific problems and can vary for each gas. For example:
  • Oâ‚‚ has a C_p of 0.033 kJ/(mol•°C),
  • Nâ‚‚ has a C_p of 0.032 kJ/(mol•°C),
  • COâ‚‚ has a C_p of 0.052 kJ/(mol•°C), and
  • Hâ‚‚O(v) has a C_p of 0.040 kJ/(mol•°C).
Using these values helps calculate temperature changes in the system and the energy required or released during reactions. Having a deeper understanding of heat capacity allows engineers and scientists to design systems that effectively manage thermal energy.
Safety Regulations
Safety regulations are critical in preventing accidents in industrial and laboratory settings. They provide guidelines to mitigate risks associated with processes like combustion, where flammable gases are involved.
In our scenario, storing combustible mixtures in tanks can be hazardous. Ignition sources, such as static electricity or open flames, can accidentally ignite combustible vapors, causing explosions.
Following safety regulations means proper handling and storing of materials, using equipment designed to withstand foreseeable pressures and temperatures, and employing sensors to detect and manage leaks or ignition risks.
These protocols aim to protect workers and the environment from dangerous chemical reactions and their potential consequences. Understanding and adhering to safety regulations is as crucial as understanding the chemistry of reactions themselves.

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

An ultimate analysis of a coal is a series of operations that yields the percentages by mass of carbon, hydrogen, nitrogen, oxygen, and sulfur in the coal. The heating value of a coal is best determined in a calorimeter, but it may be estimated with reasonable accuracy from the ultimate analysis using the Dulong formula: $$H H V(\mathrm{k} J / \mathrm{kg})=33,801(\mathrm{C})+144,158[(\mathrm{H})-0.125(\mathrm{O})]+9413(\mathrm{S})$$ where (C), (H), (O), and (S) are the mass fractions of the corresponding elements. The 0.125(O) term accounts for the hydrogen bound in the water contained in the coal. (a) Derive an expression for the higher heating value ( \(H H V\) ) of a coal in terms of \(\mathrm{C}, \mathrm{H}, \mathrm{O},\) and \(\mathrm{S},\) and compare your result with the Dulong formula. Suggest a reason for the difference. (b) A coal with an ultimate analysis of \(75.8 \mathrm{wt} \% \mathrm{C}, 5.1 \% \mathrm{H}, 8.2 \% \mathrm{O}, 1.5 \% \mathrm{N}, 1.6 \% \mathrm{S},\) and \(7.8 \%\) ash (noncombustible) is burned in a power-plant boiler fumace. All of the sulfur in the coal forms \(\mathrm{SO}_{2}\) The gas leaving the furnace is fed through a tall stack and discharged to the atmosphere. The ratio \(\phi\) (\(\mathrm{kg} \mathrm{SO}_{2}\) in the stack gas/kJ heating value of the fuel) must be below a specified value for the power plant to be in compliance with Environmental Protection Agency regulations regarding sulfur emissions. Estimate \(\phi\), using the Dulong formula for the heating value of the coal. (c) An earlier version of the EPA regulation specified that the mole fraction of \(\mathrm{SO}_{2}\) in the stack gas must be less than a specified amount to avoid a costly fine and the required installation of an expensive stack gas scrubbing unit. When this regulation was in force, a few unethical plant operators blew clear air into the base of the stack while the furnace was operating. Briefly explain why they did so and why they stopped this practice when the new regulation was introduced.

A culture of the fungus aspergillus niger is used industrially in the manufacture of citric acid and other organic species. Cells of the fungus have an ultimate analysis of \(\mathrm{CH}_{1,79} \mathrm{N}_{0.2} \mathrm{O}_{0.5}\), and the heat of formation of this species is necessary to approximate the heat duty for the bioreactor in which citric acid is to be produced. You collect a dried sample of the fungus and determine its heat of combustion to be \(-550 \mathrm{kJ} / \mathrm{mol} .\) Estimate the heat of formation \((\mathrm{kJ} / \mathrm{mol})\) of the dried fungus cells.

Methane is burned completely with 40\% excess air. The methane enters the combustion chamber at \(25^{\circ} \mathrm{C},\) the combustion air enters at \(150^{\circ} \mathrm{C},\) and the stack gas \(\left[\mathrm{CO}_{2}, \mathrm{H}_{2} \mathrm{O}(\mathrm{v}), \mathrm{O}_{2}, \mathrm{N}_{2}\right]\) exits at \(450^{\circ} \mathrm{C} .\) The chamber functions as a preheater for an air stream flowing in a pipe through the chamber to a spray dryer. The air enters the chamber at \(25^{\circ} \mathrm{C}\) at a rate of \(1.57 \times 10^{4} \mathrm{m}^{3}(\mathrm{STP}) / \mathrm{h}\) and is heated to \(181^{\circ} \mathrm{C}\). All of the heat generated by combustion is used to heat the combustion products and the air going to the spray dryer (i.e., the combustion chamber may be considered adiabatic). (a) Draw and completely label the process flow diagram and perform a degree- of-freedom analysis. (b) Calculate the required molar flow rates of methane and combustion air (kmol/h) and the volumetric flow rates \(\left(\mathrm{m}^{3} / \mathrm{h}\right)\) of the two effluent streams. State all assumptions you make. (c) When the system goes on line for the first time, environmental monitoring of the stack gas reveals a considerable quantity of CO, suggesting a problem with either the design or the operation of the combustion chamber. What changes from your calculated values would you expect to see in the temperatures and volumetric flow rates of the effluent streams [increase, decrease, cannot tell without doing the calculations]?

Liquid \(n\) -pentane at \(25^{\circ} \mathrm{C}\) is burned with \(30 \%\) excess oxygen (not air) fed at \(75^{\circ} \mathrm{C}\). The adiabatic flame temperature is \(T_{\mathrm{ad}}\left(^{\circ} \mathrm{C}\right)\) (a) Take as a basis of calculation \(1.00 \mathrm{mol} \mathrm{C}_{5} \mathrm{H}_{12}(1)\) burned and use an energy balance on the adiabatic reactor to derive an equation of the form \(f\left(T_{\mathrm{ad}}\right)=0,\) where \(f\left(T_{\mathrm{ad}}\right)\) is a fourth-order polynomial \(\left[f\left(T_{\mathrm{ad}}\right)=c_{0}+c_{1} T_{\mathrm{ad}}+c_{2} T_{\mathrm{ad}}^{2}+c_{3} T_{\mathrm{ad}}^{3}+c_{4} T_{\mathrm{ad} \mathrm{d}}^{4}\right]\). If your derivation is correct, the ratio \(c_{0} / c_{4}\) should equal \(-6.892 \times 10^{14} .\) Use a spreadsheet program to determine \(T_{\mathrm{ad}}\) (b) Repeat the calculation of Part (a) using successively the first two terms, the first three terms, and the first four terms of the fourth-order polynomial equation. If the solution of Part (a) is taken to be exact, what percentage errors are associated with the linear (two-term), quadratic (three-term), and cubic (four-term) approximations? (c) Why is the fourth-order solution at best an approximation and quite possibly a poor one? (Hint: Examine the conditions of applicability of the heat capacity formulas in Table B.2.)

You are checking the performance of a reactor in which acetylene is produced from methane in the reaction $$2 \mathrm{CH}_{4}(\mathrm{g}) \rightarrow \mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g})$$ An undesired side reaction is the decomposition of acetylene: $$\mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{C}(\mathrm{s})+\mathrm{H}_{2}(\mathrm{g})$$ Methane is fed to the reactor at \(1500^{\circ} \mathrm{C}\) at a rate of \(10.0 \mathrm{mol} \mathrm{CH}_{4} / \mathrm{s}\). Heat is transferred to the reactor at a rate of \(975 \mathrm{kW}\). The product temperature is \(1500^{\circ} \mathrm{C}\) and the fractional conversion of methane is 0.600 . A flowchart of the process and an enthalpy table are shown below. (a) Using the heat capacitics given below for enthalpy calculations, write and solve material balances and an energy balance to determine the product component flow rates and the yield of acctylene (mol \(\mathbf{C}_{2} \mathbf{H}_{2}\) produced/mol \(\mathbf{C H}_{4}\) consumed). $$\begin{aligned}\mathrm{CH}_{4}(\mathrm{g}): & C_{p} \approx 0.079 \mathrm{kJ} /\left(\mathrm{mol} \cdot^{\circ} \mathrm{C}\right) \\ \mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{g}): & C_{p} \approx 0.052 \mathrm{kJ} /\left(\mathrm{mol} \cdot^{\circ} \mathrm{C}\right) \\ \mathrm{H}_{2}(\mathrm{g}): & C_{p} \approx 0.031 \mathrm{kJ} /\left(\mathrm{mol} \cdot^{\circ} \mathrm{C}\right) \\ \mathrm{C}(\mathrm{s}): & C_{p} \approx 0.022 \mathrm{kJ} /\left(\mathrm{mol} \cdot^{\circ} \mathrm{C}\right)\end{aligned}$$ For example, the specific enthalpy of methane at \(1500^{\circ} \mathrm{C}\) relative to methane at \(25^{\circ} \mathrm{C}\) is \(\left[0.079 \mathrm{kJ} /\left(\mathrm{mol} \cdot^{\circ} \mathrm{C}\right)\right]\left(1500^{\circ} \mathrm{C}-25^{\circ} \mathrm{C}\right)=116.5 \mathrm{kJ} / \mathrm{mol}\) (b) The reactor efficiency may be defined as the ratio (actual acetylene yield/acetylene yield with no side reaction). What is the reactor efficiency for this process? (c) The mean residence time in the reactor \([\tau(\mathrm{s})]\) is the average time gas molecules spend in the reactor in going from inlet to outlet. The more \(\tau\) increases, the greater the extent of reaction for every reaction occurring in the process. For a given feed rate, \(\tau\) is proportional to the reactor volume and inversely proportional to the feed stream flow rate. (i) If the mean residence time increases to infinity, what would you expect to find in the product stream? Explain. (ii) Someone proposes running the process with a much greater feed rate than the one used in Part (a), separating the products from the unconsumed reactants, and recycling the reactants. Why would you expect that process design to increase the reactor efficiency? What else would you need to know to determine whether the new design would be cost-effective?

See all solutions

Recommended explanations on Chemistry 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.