/*! 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 58 An ultimate analysis of a coal i... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 58

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.

Short Answer

Expert verified
The Dulong formula does not need further derivation. The mass ratio of \(SO_2\) emitted per kJ of energy produced can be calculated from the mass fractions of the elements and the Dulong formula. The plant operators stopped blowing fresh air into the stack after the regulations changed because it no longer helped lower the mass ratio of \( SO_2\) that is emitted.

Step by step solution

01

Converting mass percentages into mass fractions

Before proceeding with the task, first, the mass percentages need to be converted into mass fraction. Mass fraction is the ratio of the mass of a component to the total mass of the mixture. Hence, for each element in the coal, its mass fraction would be its mass percentage divided by 100.
02

Deriving the higher heating value (HHV)

With the corresponding mass fractions of the elements, we can derive the Higher Heating Value (HHV) from the Dulong formula. The Dulong formula already states the heating value as: \[HHV (kJ/kg) = 33,801(C) + 144,158[(H)-0.125(O)] + 9413(S)\]. Since this is already in terms of C, H, O, and S, no further derivation is required.
03

Calculating the heating value using the Dulong formula

To calculate \(\phi\), first, determine the heating value of the coal using the Dulong formula, substituting the mass fractions of the elements obtained in step 1.
04

Computing the amount of \(SO_2\)

Since all Sulphur turns into \(SO_2\), for every kg of S, we can obtain 1 mole of \(SO_2\). By using the molar mass of \(SO_2\) (approximately 64 g/mol), we can calculate how many kg of \(SO_2\) a kg of coal produces.
05

Estimating \(\phi\)

The ratio \(\phi\) is the ratio of \(\frac{kg SO_2}{kJ of heating value of fuel}\). Divide the amount of \(SO_2\) produced (found in step 4) by the heating value of the fuel (found in step 3), to estimate \(\phi\).
06

Understanding environmental regulations' impact on plant operators' practice

For the last part of the exercise, we need to understand why some plant operators blow clear air into the stack while the furnace is operating. This helped dilute the \(SO_2\) in the stack gas, thereby decreasing set the mole fraction of \(SO_2\). When regulations changed and focused on the mass ratio instead of the molar ratio of \(SO_2\), this practice became ineffective, because just diluting the \(SO_2\) gas with air does not change the mass of \(SO_2\) emitted.

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.

Dulong Formula
The Dulong Formula is a classic way to estimate the higher heating value (HHV) of coal from its elemental composition. Dulong's formula provides a mathematical expression to calculate the energy produced when coal is burned, which is crucial for optimizing industrial energy usage. In the formula- \(HHV (kJ/kg) = 33,801(C) + 144,158[(H)-0.125(O)] + 9413(S)\),- \(C\) is the mass fraction of carbon,\(H\) is the mass fraction of hydrogen,\(O\) is the mass fraction of oxygen,\(S\) is the mass fraction of sulfur.The term \(0.125(O)\) adjusts for the oxygen forming water with hydrogen in the coal. It's crucial to realize this formula isn't an exact calorimetric measure, but rather an estimate based on coal's ultimate analysis. It's especially useful in situations where direct measurement isn't feasible.
Higher Heating Value
The Higher Heating Value (HHV) quantifies the total energy released in combustion, considering the latent heat of vaporization of water. HHV differs from Lower Heating Value (LHV), which excludes the energy from condensing water vapor. In the context of coal, understanding HHV is vital for assessing fuel efficiency. HHV provides insights into how much usable energy is obtained from burning a specific fuel mass. In Dulong's formula: - HHV incorporates corrections for water vapor, - Meaning the latent heat from water turning to vapor is included in energy output. This comprehensive understanding aids engineers and scientists in evaluating and comparing fuel efficiency, which is integral for power plants and thermal energy systems.
Environmental Regulations
Environmental regulations, set by authorities like the Environmental Protection Agency (EPA), aim to minimize pollution and its impacts on the environment. For coal-burning power plants, specific standards limit sulfur dioxide (SO鈧�) emissions. Historically, regulations have evolved: - Earlier, mole fraction limits for SO鈧� in stack gases were enforced. - Now, the focus is on the mass of SO鈧� per unit of energy produced. By changing the focus to the mass ratio rather than the mole ratio, the regulation pushed for actual reductions in sulfur emissions, rather than simply diluting them with additional air. This is crucial for ensuring that environmental standards increase air quality and protect ecosystems from acid rain and other harmful consequences of sulfur pollution.
Sulfur Emissions
Sulfur emissions from burning coal can result in the production of sulfur dioxide (SO鈧�), a significant environmental pollutant. When coal containing sulfur is burned, it reacts with oxygen to form SO鈧� gas, which can contribute to acid rain.Key points about sulfur emissions include:- For each mass of sulfur combusted, a specific mass of SO鈧� is produced.- Understanding the sulfur content in coal helps estimate potential SO鈧� emissions.- Managing sulfur emissions is essential for meeting environmental standards.By calculating the ratio \(\phi = \frac{kg \, of \, SO_2}{ kJ \, of \, heating \, value}\), power plants can assess compliance with emission limits, contributing to reduced environmental impact.
Ultimate Analysis
Ultimate analysis in coal analysis refers to the detailed breakdown of a fuel's elemental composition. It identifies the percentages of carbon, hydrogen, sulfur, nitrogen, and oxygen present in the coal. Understanding ultimate analysis helps in: - Estimating energy content, like through the Dulong Formula, - Predicting emissions from combustion, - Deciding which coals are most suitable for specific applications. Ultimate analysis is foundational for industries in selecting the right type of coal for energy production. It provides essential data to optimize combustion processes and adherence to environmental regulations.

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 standard heat of the reaction $$4 \mathrm{NH}_{3}(\mathrm{g})+5 \mathrm{O}_{2}(\mathrm{g}) \rightarrow 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})$$ $$\text { is } \Delta H_{r}^{\prime}=-904.7 \mathrm{kJ}$$ (a) Briefly explain what that means. Your explanation may take the form "When ___ (specify quantities of reactant species and their physical states) react to form ___ (quantities of product species and their physical state), the change in enthalpy is ___ . (b) Is the reaction exothermic or endothermic at \(25^{\circ} \mathrm{C}\) ? Would you have to heat or cool the reactor to keep the temperature constant? What would the temperature do if the reactor ran adiabatically? What can you infer about the energy required to break the molecular bonds of the reactants and that released when the product bonds form? (c) What is \(\Delta H_{r}\) for $$2 \mathrm{NH}_{3}(\mathrm{g})+\frac{5}{2} \mathrm{O}_{2} \rightarrow 2 \mathrm{NO}(\mathrm{g})+3 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})$$ (d) What is \(\Delta H_{r}\) for $$\mathrm{NO}(\mathrm{g})+\frac{3}{2} \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightarrow \mathrm{NH}_{3}(\mathrm{g})+\frac{5}{4} \mathrm{O}_{2}$$ (e) Estimate the enthalpy change associated with the consumption of \(340 \mathrm{g} \mathrm{NH}_{2} / \mathrm{s}\) if the reactants and products are all at \(25^{\circ} \mathrm{C}\). (See Example \(9.1-1 .\) ) What have you assumed about the reactor pressure? You don't have to assume that it equals 1 atm.) (f) The values of \(\Delta H_{\mathrm{r}}\) given in this problem apply to water vapor at \(25^{\circ} \mathrm{C}\) and 1 atm, and yet the normal boiling point of water is \(100^{\circ} \mathrm{C}\). Can water exist as a vapor at \(25^{\circ} \mathrm{C}\) and a total pressure of \(1 \mathrm{atm} ?\) Explain your answer.

Ammonia scrubbing is one of many processes for removing sulfur dioxide from flue gases. The gases are bubbled through an aqueous solution of ammonium sulfite, and the SO_reacts to form ammonium bisulfite: $$\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{3}(\mathrm{aq})+\mathrm{SO}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow 2 \mathrm{NH}_{4} \mathrm{HSO}_{3}(\mathrm{aq})$$ Subsequent process steps yield concentrated SO \(_{2}\) and regenerate ammonium sulfite, which is recycled to the scrubber. The sulfur dioxide is either oxidized and absorbed in water to form sulfuric acid or reduced to elemental sulfur. Flue gas from a power-plant boiler containing \(0.30 \% \mathrm{SO}_{2}\) by volume enters a scrubber at a rate of \(50,000 \mathrm{mol} / \mathrm{h}\) at \(50^{\circ} \mathrm{C} .\) The gas is bubbled through an aqueous solution containing \(10.0 \mathrm{mole} \%\) ammonium sulfite that enters the scrubber at \(25^{\circ} \mathrm{C}\). The gas and liquid effluents from the scrubber both emerge at \(35^{\circ} \mathrm{C}\). The scrubber removes \(90 \%\) of the \(S O_{2}\) entering with the flue gas. The effluent liquid is analyzed and is found to contain 1.5 moles \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{3}\) per mole of \(\mathrm{NH}_{4} \mathrm{HSO}_{3}\). The heat of formation of \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{3}(\mathrm{aq})\) at \(25^{\circ} \mathrm{C}\) is \(-890.0 \mathrm{kJ} / \mathrm{mol},\) and that of \(\mathrm{NH}_{4} \mathrm{HSO}_{3}(\mathrm{aq})\) is \(-760 \mathrm{kJ} / \mathrm{mol} .\) The heat capacities of all liquid solutions may be taken to be \(4.0 \mathrm{J} /\left(\mathrm{g} \cdot^{\circ} \mathrm{C}\right)\) and that of the flue gas may be taken to be that of nitrogen. Evaporation of water may be neglected. Calculate the required rate of heat transfer to or from the scrubber ( \(\mathrm{kW}\) ).

A natural gas containing 82.0 mole \(\% \mathrm{CH}_{4}\) and the balance \(\mathrm{C}_{2} \mathrm{H}_{6}\) is burned with \(20 \%\) excess air in a boiler furnace. The fuel gas enters the furnace at \(298 \mathrm{K}\), and the air is preheated to 423 \(\mathrm{K}\). The heat capacities of the stack-gas components may be assumed to have the following constant values: $$\begin{aligned}\mathrm{CO}_{2}: & C_{p}=50.0 \mathrm{J} /(\mathrm{mol} \cdot \mathrm{K}) \\ \mathrm{H}_{2} \mathrm{O}(\mathrm{v}): & C_{p}=38.5 \mathrm{J} /(\mathrm{mol} \cdot \mathrm{K}) \\\\\mathrm{O}_{2}: & C_{p}=33.1 \mathrm{J} /(\mathrm{mol} \cdot \mathrm{K}) \\ \mathrm{N}_{2}: & C_{p}=31.3 \mathrm{J} /(\mathrm{mol} \cdot \mathrm{K})\end{aligned}$$ (a) Assuming complete combustion of the fuel, calculate the adiabatic flame temperature. (b) How would the flame temperature change if the percent excess air were increased? How would it change if the percentage of methane in the fuel increased? Briefly explain both of your answers.

The synthesis of cthyl chloride is accomplished by reacting ethylene with hydrogen chloride in the presence of an aluminum chloride catalyst: $$\mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{g})+\mathrm{HCl}(\mathrm{g}) \stackrel{\text { catallyst }}{\longrightarrow} \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}(\mathrm{g}) ; \quad \Delta H_{\mathrm{r}}\left(0^{\circ} \mathrm{C}\right)=-64.5 \mathrm{kJ}$$ Process data and a simplified schematic flowchart are given here. Data Reactor: adiabatic, outlet temperature \(=50^{\circ} \mathrm{C}\) Feed A: \(100 \% \mathrm{HCl}(\mathrm{g}), 0^{\circ} \mathrm{C}\) Feed \(\mathrm{B}: 93\) mole \(\% \mathrm{C}_{2} \mathrm{H}_{4}, 7 \% \mathrm{C}_{2} \mathrm{H}_{6}, 0^{\circ} \mathrm{C}\) Reactor: adiabatic, outlet temperature \(=50^{\circ} \mathrm{C}\) Feed A: 100\% HCl(g), 0"C Feed B: 93 mole\% C_H_4, 7\% C_H_0, 0"C Product C: Consists of 1.5\% of the HCl, 1.5\% of the C_2 \(\mathrm{H}_{4}\), and all of the \(\mathrm{C}_{2} \mathrm{H}_{6}\) that enter the reactor Product D: \(1600 \mathrm{kg} \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}(\mathrm{l}) / \mathrm{h}, 0^{\circ} \mathrm{C}\) Recycle to reactor: \(\mathbf{C}_{2} \mathrm{H}_{5} \mathrm{Cl}(\mathrm{l}), 0^{\circ} \mathrm{C}\) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}: \Delta \hat{H}_{\mathrm{y}}=24.7 \mathrm{kJ} / \mathrm{mol}\) (assume independent of \(T\) ) \(\left(C_{p}\right)_{C_{2} H_{3} C(v)}\left[\mathrm{kJ} /\left(\mathrm{mol} \cdot^{\circ} \mathrm{C}\right)\right]=0.052+8.7 \times 10^{-5} T\left(^{\circ} \mathrm{C}\right)\) The reaction is exothermic, and if the heat of reaction is not removed in some way, the reactor temperature could increase to an undesirably high level. To avoid this occurrence, the reaction is carried out with the catalyst suspended in liquid cthyl chloride. As the reaction proceeds, most of the heat liberated goes to vaporize the liquid, making it possible to keep the reaction temperature at or below 50^'C. The stream leaving the reactor contains cthyl chloride formed by reaction and that vaporized in the reactor. This stream passes through a heat exchanger where it is cooled to \(0^{\circ} \mathrm{C},\) condensing essentially all of the cthyl chloride and leaving only unreacted \(\mathrm{C}_{2} \mathrm{H}_{4}, \mathrm{HCl}\), and \(\mathrm{C}_{2} \mathrm{H}_{6}\) in the gas phase. A portion of the liquid condensate is recycled to the reactor at a rate equal to the rate at which ethyl chloride is vaporized, and the rest is taken off as product. At the process conditions, heats of mixing and the influence of pressure on enthalpy may be neglected. (a) At what rates (kmol/h) do the two feed streams enter the process? (b) Calculate the composition (component mole fractions) and molar flow rate of product stream \(\mathrm{C}\). (c) Write an energy balance around the reactor and use it to determine the rate at which ethyl chloride must be recycled. (d) A number of simplifying assumptions were made in the process description and the analysis of this process system, so the results obtained using a more realistic simulation would differ considerably from those you should have obtained in Parts (a)-(c). List as many of these assumptions as you can think of.

In a surface-coating operation, a polymer (plastic) dissolved in liquid acetone is sprayed on a solid surface and a stream of hot air is then blown over the surface, vaporizing the acetone and leaving a residual polymer film of uniform thickness. Because environmental standards do not allow discharging acetone into the atmosphere, a proposal to incinerate the stream is to be evaluated. The proposed process uses two parallel columns containing beds of solid particles. The air-acetone stream, which contains acetone and oxygen in stoichiometric proportion, enters one of the beds at \(1500 \mathrm{mm} \mathrm{Hg}\) absolute at a rate of 1410 standard cubic meters per minute. The particles in the bed have been preheated and transfer heat to the gas. The mixture ignites when its temperature reaches \(562^{\circ} \mathrm{C}\), and combustion takes place rapidly and adiabatically. The combustion products then pass through and heat the particles in the second bed, cooling down to \(350^{\circ} \mathrm{C}\) in the process. Periodically the flow is switched so that the heated outlet bed becomes the feed gas preheater/combustion reactor and vice versa. Use the following average values for \(C_{p}\left[\mathrm{kJ} /\left(\mathrm{mol} \cdot^{\circ} \mathrm{C}\right)\right]\) in solving the problems to be given: 0.126 for \(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}, 0.033\) for \(\mathrm{O}_{2}, 0.032\) for \(\mathrm{N}_{2}, 0.052\) for \(\mathrm{CO}_{2},\) and 0.040 for \(\mathrm{H}_{2} \mathrm{O}(\mathrm{v})\) (a) If the relative saturation of acetone in the feed stream is \(12.2 \%,\) what is the stream temperature? (b) Determine the composition of the gas after combustion, assuming that all of the acetone is converted to \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O},\) and estimate the temperature of this stream. (c) Estimate the rates ( \(\mathrm{kW}\) ) at which heat is transferred from the inlet bed particles to the feed gas prior to combustion and from the combustion gases to the outlet bed particles. Suggest an alternative to the two-bed feed switching arrangement that would achieve the same purpose.
See all solutions

Recommended explanations on Chemistry Textbooks

Organic Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Nuclear Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Making Measurements

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.