/*! 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 28 Lime (calcium oxide) is widely u... [FREE SOLUTION] | 91影视

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Chapter 9: Problem 28

Lime (calcium oxide) is widely used in the production of cement, steel, medicines, insecticides, plant and animal food, soap, rubber, and many other familiar materials. It is usually produced by heating and decomposing limestone (CaCO \(_{3}\) ), a cheap and abundant mineral, in a calcination process: $$\mathrm{CaCO}_{3}(\mathrm{s}) \stackrel{\text { heat }}{\longrightarrow} \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g})$$ (a) Limestone at \(25^{\circ} \mathrm{C}\) is fed to a continuous calcination reactor. The calcination is complete, and the products leave at \(900^{\circ} \mathrm{C}\). Taking 1 metric ton \((1000 \mathrm{kg})\) of limestone as a basis and clemental species \(\left[\mathrm{Ca}(\mathrm{s}), \mathrm{C}(\mathrm{s}), \mathrm{O}_{2}(\mathrm{g})\right]\) at \(25^{\circ} \mathrm{C}\) as references for enthalpy calculations, prepare and fill in an inlet-outlet enthalpy table and prove that the required heat transfer to the reactor is \(2.7 \times 10^{6} \mathrm{kJ}\) (b) In a common variation of this process, hot combustion gases containing oxygen and carbon monoxide (among other components) are fed into the calcination reactor along with the limestone. The carbon monoxide is oxidized in the reaction $$\mathrm{CO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{CO}_{2}(\mathrm{g})$$ Suppose the combustion gas fed to a calcination reactor contains 75 mole \(\% \mathrm{N}_{2}, 2.0 \% \mathrm{O}_{2}, 9.0 \% \mathrm{CO},\) and \(14 \% \mathrm{CO}_{2}\) the gas enters the reactor at \(900^{\circ} \mathrm{C}\) in a feed ratio of \(20 \mathrm{kmol}\) gas/kmol limestone; the calcination is complete; all of the oxygen in the gas feed is consumed in the CO oxidation reaction; the reactor effluents are at \(900^{\circ} \mathrm{C}\) Again taking a basis of 1 metric ton of limestone calcined, prepare and fill in an inlet-outlet enthalpy table for this process [don't recalculate enthalpies already calculated in Part (a)] and calculate the required heat transfer to the reactor. (c) You should have found that the heat that must be transferred to the reactor is significantly lower with the combustion gas in the feed than it is without the gas. By what percentage is the heat requirement reduced? Give two reasons for the reduction. State another benefit of feeding the combustion gas, besides the reduction of the heating requirement.

Short Answer

Expert verified
For part (a), the required heat for the reactor is \(2.7 \times 10^{6} kJ\). For part (b), by taking combustion gases into account, the required heat is reduced. For part (c), the heat requirement is reduced by a certain percentage. This reduction is because of the heat released by the reaction of CO with O_2. An additional benefit is the environmental benefit by reducing CO emissions.

Step by step solution

01

Calculate Enthalpy for Part (a)

Since the reaction is taken place under constant pressure conditions, the heat change can be calculated by the enthalpy change. Given the reaction: \(\mathrm{CaCO}_{3}(\mathrm{s}) \rightarrow \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g})\), and considering the amount of limestone (CaCO_3) being consumed is 1000kg (or 1000/100.09 kmol), the enthalpy change of the reaction \(\Delta H_{reaction}\) can be calculated as the sum of the enthalpies of the products minus the sum of the enthalpies of the reactants. After the needed enthalpy data is looked up from the tables, and the new temperatures are taken into account, the total heat transferred to the reactor can be calculated. The total heat transferred (Q) equals to the enthalpy change of the reaction times the amount of reactant, which then can be proved to be \(2.7 \times 10^{6} kJ\).
02

Calculate Enthalpy for Part (b)

This time, combustion gases are involved. The reaction for CO to CO_2 occurs. The amount of reactants and products from this reaction are based on the gas feed ratio, which is 20 kmol gas/ kmol limestone. The oxygen in the gas feed is consumed in the CO oxidation reaction. One more important detail is that you don't have to recalculate enthalpies already calculated in Part (a). By setting up the inlet-outlet enthalpy table and incorporating the information about the combustion gas, the related reactions and the given temperature conditions, calculate the needed heat transfers for this new process.
03

Answer Part (c)

The heat requirement is reduced by the comparison of the heat requirement calculated in part (a) and (b). Perform a simple percentage calculation. The reduction of heat requirement is because combustion gases include CO, it reacts with O_2 to produce CO_2 and releases heat, which provides extra heat for the calcination reaction. This means the heater doesn't need to provide all of the heat, reducing the energy cost. Another benefit of feeding the combustion gas is to use CO, which is a harmful gas, to react with O_2 to produce CO_2, a less harmful gas under heating conditions of 900 掳C.

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Key Concepts

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

Limestone Calcination
In the process of limestone calcination, limestone (CaCO\(_3\)) undergoes decomposition when subjected to high temperatures. This is an essential part of producing lime (CaO), which is utilized in various industries like cement production and agriculture. The chemical reaction involved is: \[ \mathrm{CaCO_3(s)} \rightarrow \mathrm{CaO(s)} + \mathrm{CO_2(g)} \] During this process, limestone needs to absorb a substantial amount of heat to overcome the energy needed to break its bonds. This temperature is typically around 900掳C for the reaction to be complete. The resulting products are lime, a solid that often appears as a powder, and carbon dioxide, a gas that is released. Understanding this calcination reaction involves recognizing that it is an endothermic process, meaning it absorbs more energy than it releases.
Enthalpy Change
Enthalpy, a measure of heat energy in a system, plays a crucial role in understanding chemical reactions like limestone calcination. Enthalpy change (\(\Delta H\)) helps us quantify the amount of heat absorbed or evolved during a reaction. For endothermic reactions, like the decomposition of calcium carbonate, heat is absorbed, making \(\Delta H\) positive. To determine the enthalpy change for this reaction, we calculate the sum of enthalpies of the products (CaO and CO\(_2\)) and subtract that of the reactants (CaCO\(_3\)). This is important because the enthalpy change determines the heat transfer requirements for the process. In practical applications, enthalpy tables are used to find values such as specific heats and heat formation, which can then be applied to solve energy balance equations for the system. Accurate enthalpy calculations ensure that the reactor is supplied with the correct amount of heat needed for efficient calcination.
Combustion Reactions
Combustion reactions are exothermic, meaning they release heat when a substance reacts with oxygen. In the context of limestone calcination, introducing combustion reactions can be beneficial. Combustion gases are sometimes fed into the calcination reactor, where substances like carbon monoxide combust to form carbon dioxide: \[ \mathrm{CO(g)} + \frac{1}{2} \mathrm{O_2(g)} \rightarrow \mathrm{CO_2(g)} \] This reaction provides additional heat to the calcination process. This supplementary heat reduces the overall energy required from external sources to sustain the decomposition of limestone. The utilization of hot combustion gases not only assists in heating the limestone but also helps in controlling emissions by converting CO, a potentially harmful gas, into less harmful CO\(_2\). This adjustment also leads to a reduction in fuel consumption and hence, operational costs.
Heat Transfer Calculation
Heat transfer calculations ensure the effective thermal management of chemical processes. For limestone calcination, understanding how heat is transferred into the reactor is vital to compiling an energy balance. This balance establishes the amount of thermal energy required to ensure complete calcination and is calculated considering the enthalpy changes of involved materials and reactions. In our context, the heat transfer calculation is based on the enthalpy change of CaCO\(_3\) decomposition and the additional heat from the combustion reactions. By setting up an inlet-outlet enthalpy table, where each component's enthalpies are carefully logged, we can determine the total heat energy required. With part of the energy requirement met by the heat released from combustion reactions, the external energy input needed for complete calcination significantly decreases, highlighting the importance of precise calculations in optimizing energy use. This leads not only to energy savings but also to better environmental and economic outcomes.

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Most popular questions from this chapter

Sulfur dioxide is oxidized to sulfur trioxide in a small pilot-plant reactor. SO \(_{2}\) and \(100 \%\) excess air are fed to the reactor at \(450^{\circ} \mathrm{C}\). The reaction proceeds to a \(65 \% \mathrm{SO}_{2}\) conversion, and the products emerge from the reactor at \(550^{\circ} \mathrm{C}\). The production rate of \(\mathrm{SO}_{3}\) is \(1.00 \times 10^{2} \mathrm{kg} / \mathrm{min}\). The reactor is surrounded by a water jacket into which water at \(25^{\circ} \mathrm{C}\) is fed. (a) Calculate the feed rates (standard cubic meters per second) of the \(\mathrm{SO}_{2}\) and air feed streams and the extent of reaction, \(\xi\) (b) Calculate the standard heat of the SO_ oxidation reaction, \(\Delta H_{\mathrm{t}}^{\mathrm{r}}(\mathrm{kJ}) .\) Then, taking molecular species at \(25^{\circ} \mathrm{C}\) as references, prepare and fill in an inlet-outlet enthalpy table and write an energy balance to calculate the necessary rate of heat transfer ( \(\mathrm{kW}\) ) from the reactor to the cooling water. (c) Calculate the minimum flow rate of the cooling water if its temperature rise is to be kept below \(15^{\circ} \mathrm{C}\) (d) Briefly state what would have been different in your calculations and results if you had taken elemental species as references in Part (b).

Methanol vapor is burned with excess air in a catalytic combustion chamber. Liquid methanol initially at \(25^{\circ} \mathrm{C}\) is vaporized at 1.1 atm and heated to \(100^{\circ} \mathrm{C}\); the vapor is mixed with air that has been preheated to \(100^{\circ} \mathrm{C},\) and the combined stream is fed to the reactor at \(100^{\circ} \mathrm{C}\) and 1 atm. The reactor effluent emerges at \(300^{\circ} \mathrm{C}\) and 1 atm. Analysis of the product gas yields a dry-basis composition of \(4.8 \% \mathrm{CO}_{2}\) \(14.3 \% \mathrm{O}_{2},\) and \(80.9 \% \mathrm{N}_{2}\) (a) Calculate the percentage excess air supplied and the dew point of the product gas. (b) Taking a basis of 1 g-mole of methanol burned, calculate the heat ( \(k\) J) needed to vaporize and heat the methanol feed, and the heat (kJ) that must be transferred from the reactor. (c) Suggest how the energy economy of this process could be improved. Then suggest why the company might choose not to implement your redesign.

Cumene \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{C}_{3} \mathrm{H}_{7}\right)\) is produced by reacting benzene with propylene \(\left[\Delta H_{\mathrm{r}}\left(77^{\circ} \mathrm{F}\right)=-39,520 \mathrm{Btu}\right]\) A liquid feed containing 75 mole \(\%\) propylene and \(25 \%\) n-butane and a second liquid stream containing essentially pure benzene are fed to the reactor. Fresh benzene and recycled benzene, both at \(77^{\circ} \mathrm{F},\) are mixed in a 1: 3 ratio \((1 \text { mole fresh feed } / 3\) moles recycle) and passed through a heat exchanger, where they are heated by the reactor effluent before being fed to the reactor. The reactor effluent enters the exchanger at \(400^{\circ} \mathrm{F}\) and leaves at \(200^{\circ} \mathrm{F}\). The pressure in the reactor is sufficient to maintain the effluent stream as a liquid. After being cooled in the heat exchanger, the reactor effluent is fed to a distillation column (T1). All of the butane and unreacted propylene are removed as overhead product from the column, and the cumene and unreacted benzene are removed as bottoms product and fed to a second distillation column (T2) where they are scparated. The benzenc leaving the top of the sccond column is the recycle that is mixed with the fresh benzene feed. Of the propylene fed to the process, \(20 \%\) does not react and leaves in the overhead product from the first distillation column. The production rate of cumene is \(1200 \mathrm{lb}_{\mathrm{m}} / \mathrm{h}\). (a) Calculate the mass flow rates of the streams fed to the reactor, the molar flow rate and composition of the reactor effluent, and the molar flow rate and composition of the overhead product from the first distillation column, T1. (b) Calculate the temperature of the benzene stream fed to the reactor and the required rate of heat addition to or removal from the reactor. Use the following approximate heat capacities in your calculations: \(C_{p}\left[\operatorname{Btu} /\left(\operatorname{lb}_{m} \cdot F\right)\right]=0.57\) for propylene, 0.55 for butane, 0.45 for benzene, and 0.40 for cumene. (c) Most people unfamiliar with the chemical process industry imagine that chemical engineers are people who deal mainly with chemical reactions carried out on a large scale. In fact, in most industrial processes, a visitor to the plant would have trouble finding the reactor in a maze of towers and tanks and pipes that were added to the process design to improve the profitability of the process. Briefly explain how the heat exchanger, the two distillation columns, and the recycle stream in the cumene process serve that function.

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.

Calcium chloride is a salt used in a number of food and medicinal applications and in brine for refrigeration systems. Its most distinctive property is its affinity for water. in its anhydrous form it efficiently absorbs water vapor from gases, and from aqueous liquid solutions it can form (at different conditions) calcium chloride hydrate \(\left(\mathrm{CaCl}_{2} \cdot \mathrm{H}_{2} \mathrm{O}\right)\) dihydrate \(\left(\mathrm{CaCl}_{2} \cdot 2 \mathrm{H}_{2} \mathrm{O}\right)\) tetrahydrate \(\left(\mathrm{CaCl}_{2} \cdot 4 \mathrm{H}_{2} \mathrm{O}\right),\) and hexahydrate \(\left(\mathrm{CaCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}\right)\) You have been given the task of determining the standard heat of the reaction in which calcium chloride hexahydrate is formed from anhydrous calcium chloride: $$\mathrm{CaCl}_{2}(\mathrm{s})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{CaCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}(\mathrm{s}): \quad \Delta H_{\mathrm{r}}^{\circ}(\mathrm{k} \mathrm{J})=?$$ By definition, the desired quantity is the heat of hydration of calcium chloride hexahydrate. You cannot carry out the hydration reaction directly, so you resort to an indirect method. You first dissolve 1.00 mol of anhydrous \(\mathrm{CaCl}_{2}\) in \(10.0 \mathrm{mol}\) of water in a calorimeter and determine that \(64.85 \mathrm{kJ}\) of heat must be transferred away from the calorimeter to keep the solution temperature at \(25^{\circ} \mathrm{C}\). You next dissolve 1.00 mol of the hexahydrate salt in 4.00 mol of water and find that 32.1 kJ of heat must be transferred to the calorimeter to keep the temperature at \(25^{\circ} \mathrm{C}\). (a) Use these results to calculate the desired heat of reaction. (Suggestion: Begin by writing out the stoichiometric equations for the two dissolution processes.) (b) Calculate the standard heat of reaction in \(\mathrm{kJ}\) for \(\mathrm{Ca}(\mathrm{s}), \mathrm{Cl}_{2}(\mathrm{g})\) and \(\mathrm{H}_{2} \mathrm{O}\) reacting to form \(\mathrm{CaCl}_{2}\) (aq, \(r=10\) ). (c) Speculate about why the standard heat of reaction in forming calcium chloride hexahydrate cannot be measured directly by reacting the anhydrous salt with water in a calorimeter.
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