/*! 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 62 13.62 Coal with a mass analysis ... [FREE SOLUTION] | 91Ó°ÊÓ

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13.62 Coal with a mass analysis of \(88 \% \mathrm{C}, 6 \% \mathrm{H}, 4 \% \mathrm{O}, 1 \% \mathrm{~N}\), 1\% S burns completely with the theoretical amount of air. Determine (a) the amount of \(\mathrm{SO}_{2}\) produced, in \(\mathrm{kg}\) per \(\mathrm{kg}\) of coal. (b) the air-fuel ratio on a mass basis. (c) For environmental reasons, it is desired to separate the \(\mathrm{SO}_{2}\) from the combustion products by supplying the products at \(340 \mathrm{~K}, 1\) atm to a device operating isothermally at \(340 \mathrm{~K}\). At steady state, a stream of \(\mathrm{SO}_{2}\) and a stream of the remaining gases exit, each at \(340 \mathrm{~K}, 1 \mathrm{~atm}\). If the coal is burned at a rate of \(10 \mathrm{~kg} / \mathrm{s}\), determine the minimum theoretical power input required by the device, in \(\mathrm{kW}\). Heat transfer with the surroundings occurs, but kinetic and potential energy effects can be ignored.

Short Answer

Expert verified
The \(\mathrm{SO_2}\) produced is 0.02 kg per kg of coal. The air-fuel ratio is 11.22 kg-air/kg-coal. The theoretical minimum power required is zero.

Step by step solution

01

Analyze composition of coal

The coal composition is given as: \(88 \, \text{wt}\text{\text{%}} \text{C}, 6 \, \text{wt}\text{\text{%}} \text{H}, 4 \, \text{wt}\text{\text{%}} \text{O}, 1 \, \text{wt}\text{\text{%}} \text{N}, 1 \, \text{wt}\text{\text{%}} \text{S}\).
02

Determine the amount of \(\mathrm{SO_2}\) produced

Since 1% of the coal's mass is sulfur, for 1 kg of coal burned, we have 0.01 kg of sulfur. The reaction of sulfur during combustion is: \( \mathrm{S} + \mathrm{O}_{2} \rightarrow \mathrm{SO}_{2} \). Thus, the mass of \(\mathrm{SO_2}\) produced per kg of coal is calculated as: \( \text{mass of } \mathrm{SO_2} = 0.01 \, \text{kg of S} \times \dfrac{64 \, \text{g/mol}}{32 \, \text{g/mol}}= 0.02 \, \text{kg} \).
03

Calculate the air-fuel ratio

For complete combustion of carbon, hydrogen and sulfur in 1 kg of coal, the required air is determined as: \( \text{C:} \, 0.88 \text{ kg of C} \times \dfrac{32 \, \text{g/mol}}{12 \, \text{g/mol}} = 2.33 \text{ kg of } \mathrm{O_2} \) \( \text{H:} \, 0.06 \text{ kg of H} \times \dfrac{8 \, \text{g/mol}}{2 \, \text{g/mol}} = 0.24 \text{ kg of } \mathrm{O_2} \) \( \text{S:} \, 0.01 \text{ kg of S} \times \dfrac{32 \text{ g}}{32 \text{ g}} = 0.01 \text{ kg of } \mathrm{O_2} \) Total \( \mathrm{O_2} = 2.33 + 0.24 + 0.01 = 2.58 \text{ kg} \) of \( \mathrm{O_2} \). Since air contains 23% oxygen by mass: \( \mathrm{Air-fuel \, ratio} = \dfrac{2.58 \, \text{kg}}{\dfrac{23}{100}} = 11.22 \, \text{kg/kg fuel} \)
04

Determine minimum theoretical power for separating \(\mathrm{SO_2}\)

The power required for isothermal separation involves calculating the Gibbs free energy change. The change \( \Delta G\) is given by the chemical potential \( \mu \)of \( \mathrm{SO_2}\) at given conditions, calculated as: \( \Delta G_{m} = RT \, \ln \dfrac{p}{p_{0}} \) where \( T = 340 \, \text{K}, \ p = 1 \, \text{atm} \) Plugging in the values: \( \, \Delta G_{m} = (8.314 \, \text{J/mol K})(340 \, \text{K}) \, \ln (1) = 0 \, \text{J/mol} \). As there is no change in free energy, the theoretical minimum power is zero.

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Key Concepts

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

Combustion
Combustion, commonly known as burning, is a high-temperature exothermic reaction between a fuel and an oxidant. In this context, coal undergoes combustion in the presence of oxygen from the air. The complete combustion of coal typically yields carbon dioxide, water, nitrogen, and sulfur dioxide as products. The chemical reaction governing combustion is essential for understanding how different elements in the fuel, such as carbon, hydrogen, sulfur, and nitrogen, react with oxygen. Here, sulfur in the coal burns to form sulfur dioxide (SOâ‚‚): \( \mathrm{S} + \mathrm{O}_{2} \rightarrow \mathrm{SO}_{2} \). Knowing the precise composition of the coal helps predict the quantities of each combustion product accurately, crucial for engineering and environmental studies.
Air-Fuel Ratio
The air-fuel ratio is an important concept in thermodynamics and combustion engineering. It defines the amount of air required to completely combust a given amount of fuel. This ratio is imperative for achieving efficient burning with minimal pollutants. For coal with known percentages by mass of carbon, hydrogen, sulfur, and other elements, determining the air-fuel ratio allows engineers to design combustion systems that ensure complete oxidation of all elements in the fuel, minimizing unburnt components and possible pollutants. The calculation involves summing the required oxygen masses for each element and then using the fact that air is about 23% oxygen by mass. For example, the calculation for 1 kg of coal with the given composition results in an air-fuel ratio of 11.22 kg of air per kg of coal.
Thermodynamics in Engineering
Thermodynamics plays a pivotal role in engineering, especially in processes involving energy transfer, like combustion. The exercise involves principles such as energy balance, Gibbs free energy, and isothermal processes. When dealing with isothermal separation devices, the concept of Gibbs free energy (\(Delta G\)) is crucial. It gives an understanding of how spontaneous a reaction is under constant temperature and pressure conditions. In an isothermal separation of SOâ‚‚, the power input can be theoretically zero if there is no change in Gibbs free energy, as shown by the calculation for this system operating at 340K. Understanding thermodynamics helps engineers design systems that maximize efficiency while minimizing environmental impacts.
Sulfur Dioxide Emissions
Sulfur dioxide (SOâ‚‚) emissions are a major environmental concern associated with the combustion of sulfur-containing fuels like coal. SOâ‚‚ can lead to acid rain and respiratory problems in living beings. Hence, it is important to quantify and manage SOâ‚‚ emissions. In this problem, we'd determine the amount of SOâ‚‚ produced by burning coal, which was calculated to be 0.02 kg per kg of coal. This quantification is essential for complying with environmental regulations and designing emission control systems. One such control system is an isothermal device that separates SOâ‚‚ from other gases at high efficiency. The accurate design and power input calculations for these devices are critical to minimize pollution and protect the environment.

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Most popular questions from this chapter

13.58 Separate streams of hydrogen \(\left(\mathrm{H}_{2}\right)\) and oxygen \(\left(\mathrm{O}_{2}\right)\) at \(25^{\circ} \mathrm{C}, 1\) atm enter a fuel cell operating at steady state, and liquid water exits at \(25^{\circ} \mathrm{C}, 1 \mathrm{~atm}\). The hydrogen flow rate is \(2 \times 10^{-4}\) \(\mathrm{kmol} / \mathrm{s}\). If the fuel cell operates isothermally at \(25^{\circ} \mathrm{C}\), determine the maximum theoretical power it can develop and the accompanying rate of heat transfer, each in \(\mathrm{kW}\). Kinetic and potentially energy effects are negligible.

13\. Why is combustion inherently an irreversible process?

13.60 Streams of hydrogen \(\left(\mathrm{H}_{2}\right)\) and oxygen \(\left(\mathrm{O}_{2}\right)\), each at \(1 \mathrm{~atm}\), enter a fuel cell operating at steady state and water vapor exits at \(1 \mathrm{~atm}\). If the cell operates isothermally at (a) \(300 \mathrm{~K}\), (b) \(400 \mathrm{~K}\), and (c) \(500 \mathrm{~K}\), determine the maximum theoretical work that can be developed by the cell in each case, in kJ per kmol of hydrogen flowing, and comment. Heat transfer with the surroundings takes place at the cell temperature, and kinetic and potential energy effects can be ignored.

13.35 The energy required to vaporize the working fluid passing through the boiler of a simple vapor power plant is provided by the complete combustion of methane with \(110 \%\) of theoretical air. The fuel and air enter in separate streams at \(25^{\circ} \mathrm{C}, 1 \mathrm{~atm}\). Products of combustion exit the stack at \(150^{\circ} \mathrm{C}\), \(1 \mathrm{~atm}\). Plot the mass flow rate of fuel required, in \(\mathrm{kg} / \mathrm{h}\) per MW of power developed by the plant versus the plant thermal efficiency, \(\eta\). Consider \(\eta\) in the range \(30-40 \%\). Kinetic and potential energy effects are negligible.

13.52 Carbon monoxide (CO) at \(25^{\circ} \mathrm{C}, 1\) atm enters an insulated reactor operating at steady state and reacts completely with the theoretical amount of air entering in a separate stream at \(25^{\circ} \mathrm{C}, 1 \mathrm{~atm}\). The products of combustion exit as a mixture at \(1 \mathrm{~atm}\). For the reactor, determine the rate of entropy production, in \(\mathrm{kJ} / \mathrm{K}\) per kmol of CO entering. Neglect kinetic and potential energy effects.

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