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Chapter 9: Problem 65

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]?

Short Answer

Expert verified
Required molar flow rates: methane = 15.1 kmol/h, air = 701.8 kmol/h. Volumetric flow rates: The effluent streams would depend on the composition of the combustion products. In the presence of CO in the stack gas, there would likely be a decrease in temperature in the combustion products and possibly changed volumetric flow rates due to changes in the molar flows of the components.

Step by step solution

01

Calculation of the Required Molar Flow Rates of Methane and Combustion Air

Given: methane is burnt with 40% excess air, volume flow rate of air (\(Q_{air}\)) = \(1.57 \times 10^{4} \, m^{3}(STP)/h\).\nUsing excess air percentage, the stoichiometric combustion equation can be formulated as:\[ CH_4 + (1+0.40)(2O_2+7.52N_2) \rightarrow CO_2 + 2H_2O + 9.024N_2\]Taking into account the molar volume (\(V_{m} = 22.4 \, m^{3}\)/kmol) at standard temperature and pressure (STP), the molar flowrate of air (\(N_{air}\)) can be calculated:\[ Q_{air} = N_{air}V_{m} \rightarrow N_{air} = Q_{air} / V_{m}\]Substituting values, we get \(N_{air} = \(1.57 \times 10^{4} \, m^{3}/h)/\(22.4 \, m^{3}/kmol) = 701.8 \, kmol/h\)
02

Calculation of Volumetric Flow Rates of Effluent Streams

For calculation of effluent streams, use the reaction stoichiometry. The sum of the molar flows of the products gives the total molar flow, which can be multiplied by the molar volume at the given temperature and pressure.\nThe effluent streams consist of Carbon Dioxide (CO2), Water vapour (H2O), Oxygen (O2) & Nitrogen (N2). Total molar flow from the combustion process (\(N_{total}\)) is sum of their individual molar flows.\nThe volumetric flow rate of effluent can be calculated by multiplying molar flow with the molar volume.
03

Explanation of Possible Changes in Temperatures and Volumetric Flow Rates of the Effluent Streams in Presence of CO

In presence of Carbon Monoxide (CO) in the stack effluent, this means methane isn鈥檛 combusting completely. When methane combustion isn鈥檛 complete, less heat will be generated leading to decrease in temp of combustion products & also decreasing the temperature of air in the spray dryer.\nIncomplete combustion would also change the molar flows of the components, resulting in the changes to the volumetric flow rates. Incomplete combustion means less CO2 and water vapor, but more CO and potentially some unburnt methane, makes the calculations complex. Therefore, without full computational details, exact changes to volumetric flow rates can鈥檛 be predicted.

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Key Concepts

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

Understanding Stoichiometry in Chemical Processes
Stoichiometry is the branch of chemistry that deals with the quantitative relationships of the reactants and products in a chemical reaction. It provides the proportions in which chemicals react and the yields of products we can expect from a reaction under specified conditions.

For example, in a combustion reaction where methane (CH4) is the fuel, stoichiometry helps us determine how much oxygen (O2) is needed for complete combustion. With a given amount of methane, stoichiometry can also predict the amount of carbon dioxide (CO2), water (H2O), and heat that will be produced. This is crucial in chemical process calculations where engineers need to ensure the optimal amount of reactants to avoid waste or unreacted excess.

Furthermore, stoichiometry isn't simply a theoretical concept; it has practical implications in designing combustion systems like furnaces or engines, where exact ratios determine both efficiency and environmental impact. For any chemical engineering student or professional, a firm grasp on stoichiometric principles is a must for accurate process analysis and design.
The Role of Combustion Reactions in Energy Production
Combustion reactions are exothermic processes where a fuel reacts with an oxidant, releasing energy in the form of heat or light. In industrial applications, such as power generation and heating systems, combustion reactions are harnessed for their capability to produce vast amounts of energy efficiently.

The combustion of methane, a common fuel, involves methane reacting with oxygen to produce carbon dioxide and water vapor. However, real-life combustion processes often require excess air to ensure complete combustion, as seen in the provided exercise. Moreover, variations like incomplete combustion can lead to the production of unwanted byproducts like carbon monoxide (CO), signaling inefficiencies or safety concerns within the process.

An accurate measurement of the involved reactants and products allows for the optimization of the process, minimizing fuel consumption and reducing the emission of pollutants. Thus, understanding the mechanics of combustion reactions not only enhances efficiency but is also integral to environmental conservation efforts.
Mass and Energy Balances in Chemical Process Engineering
Mass and energy balances are foundational principles in chemical engineering, enabling professionals to design, analyze, and optimize various industrial processes. A mass balance ensures that the mass in a system remains conserved; in other words, what goes into a process must come out in some form. This applies to all components, including reactants, products, and byproducts.

An energy balance, on the other hand, verifies the conservation of energy within a process. It accounts for energy inputs such as fuels and energy losses like waste heat. In an adiabatic system, like the one mentioned in the exercise, there is no exchange of heat with the surroundings, implying that all the heat produced from combustion should account for the heat content in the effluent streams and the heated air for the spray dryer.

When we talk about balance, it not only means ensuring the right proportions but also troubleshooting issues, just as in the example where the presence of CO in the stack gas indicates incomplete combustion. Thus, by applying principles of mass and energy balances, engineers can simulate, adapt, and improve chemical processes to achieve operational goals efficiently and economically.

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Most popular questions from this chapter

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.

The equilibrium constant for the ethane dehydrogenation reaction, $$\mathrm{C}_{2} \mathrm{H}_{6}(\mathrm{g}) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g})$$ is defined as $$K_{p}(\mathrm{atm})=\frac{y_{\mathrm{C}_{2} \mathrm{H}_{4}} y_{\mathrm{H}_{2}}}{y_{\mathrm{C}_{2} \mathrm{H}_{6}}} P$$ where \(P(\text { atm })\) is the total pressure and \(y_{i}\) is the mole fraction of the ith substance in an equilibrium mixture. The equilibrium constant has been found experimentally to vary with temperature according to the formula $$K_{p}(T)=7.28 \times 10^{6} \exp [-17,000 / T(\mathrm{K})]$$ The heat of reaction at \(1273 \mathrm{K}\) is \(+145.6 \mathrm{kJ}\), and the heat capacities of the reactive species may be approximated by the formulas $$\left.\begin{array}{rl}\left(C_{p}\right)_{\mathrm{C}_{2} \mathrm{H}_{4}} & =9.419+0.1147 T(\mathrm{K}) \\\\\left(C_{p}\right)_{\mathrm{H}_{2}} & =26.90+4.167 \times 10^{-3} T(\mathrm{K}) \\ \left(C_{p}\right)_{\mathrm{C}_{2} \mathrm{H}_{6}} & =11.35+0.1392 T(\mathrm{K}) \end{array}\right\\}[\mathrm{J} /(\mathrm{mol} \cdot \mathrm{K})]$$ Suppose pure cthane is fed to a continuous constant-pressure adiabatic reactor at \(1273 \mathrm{K}\) and pressure \(P(\text { atm }),\) the products emerge at \(T_{\mathrm{f}}(\mathrm{K})\) and \(P(\mathrm{atm}),\) and the residence time of the reaction mixture in the reactor is large enough for the outlet stream to be considered an equilibrium mixture of ethane, ethylene, and hydrogen. (a) Prove that the fractional conversion of ethane in the reactor is $$f=\left(\frac{K_{p}}{P+K_{p}}\right)^{1 / 2}$$ (b) Write an energy balance on the reactor, and use it to prove that $$f=\frac{1}{1+\phi\left(T_{\mathrm{f}}\right)}$$ where Finally, substitute for \(\Delta H_{\mathrm{r}}\) and the heat capacities in Equation 4 to derive an explicit expression for \(\phi\left(T_{\mathrm{f}}\right)\) (c) We now have two expressions for the fractional conversion \(f\) : Equation 2 and Equation 3 . If these expressions are equated, \(K_{p}\) is replaced by the expression of Equation \(1,\) and \(\phi\left(T_{\mathrm{f}}\right)\) is replaced by the expression derived in Part (b), the result is one equation in one unknown, \(T_{\mathrm{f}}\). Derive this equation, and transpose the right side to obtain an expression of the form $$\psi\left(T_{\mathrm{f}}\right)=0$$ (d) Prepare a spreadsheet to take \(P\) as input, solve Equation 5 for \(T_{\mathrm{f}}\) (use Goal Seek or Solver), and determine the final fractional conversion, \(\left.f \text { . (Suggestion: Set up columns for } P, T_{\mathrm{f}}, f, K_{p}, \phi, \text { and } \psi .\right)\) Run the program for \(P(\text { atm })=0.01,0.05,0.10,0.50,1.0,5.0,\) and \(10.0 .\) Plot \(T_{\mathrm{f}}\) versus \(P\) and \(f\) versus \(P,\) using a logarithmic coordinate scale for \(P\).

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.

In a coal gasification process, carbon (the primary constituent of coal) reacts with steam to produce carbon monoxide and hydrogen (synthesis gas). The gas may either be burned or subjected to further processing to produce any of a variety of chemicals. A coal contains 10.5 wt\% moisture (water) and 22.6 wt\% noncombustible ash. The remaining fraction of the coal contains 81.2 wife \(\mathrm{C}, 13.4 \%\) O, and \(5.4 \%\) H. A coal slurry containing \(2.00 \mathrm{kg}\) coal/kg water is fed at \(25^{\circ} \mathrm{C}\) to an adiabatic gasification reactor along with a stream of pure oxygen at the same temperature. The following reactions take place in the reactor: $$\begin{array}{l}\mathrm{C}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{v}) \rightarrow \mathrm{CO}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g}): \quad \Delta H_{\mathrm{r}}^{\circ}=+131.3 \mathrm{kJ} \\\\\mathrm{C}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{CO}_{2}(\mathrm{g}): \quad \Delta H_{\mathrm{r}}^{\circ}=-393.5 \mathrm{kJ} \\ 2 \mathrm{H}(\mathrm{in} \mathrm{coal})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{v}): \quad \Delta H_{\mathrm{r}}^{\circ} \approx-242 \mathrm{kJ}\end{array}$$ Gas and slag (molten ash) leave the reactor at \(2500^{\circ} \mathrm{C}\). The gas contains \(\mathrm{CO}, \mathrm{H}_{2}, \mathrm{CO}_{2},\) and \(\mathrm{H}_{2} \mathrm{O}^{14}\) (a) Feeding oxygen to the reactor lowers the yield of synthesis gas, but no gasifier ever operates without supplementary oxygen. Why does the oxygen lower the yield? Why it is nevertheless always supplied. (Hint: All the necessary information is contained in the first two stoichiometric equations and associated heats of reaction shown above.) (b) Suppose the oxygen gas fed to the reactor and the oxygen in the coal combine with all the hydrogen in the coal (Reaction 3) and with some of the carbon (Reaction 2), and the remainder of the carbon is consumed in Reaction 1. Taking a basis of 1.00 kg coal fed to the reactor and letting \(n_{0}\) equal the moles of \(\mathrm{O}_{2}\) fed, draw and label a flowchart. Then derive expressions for the molar flow rates of the four outlet gas species in terms of \(n_{0}\). [Partial solution: \(n_{\mathrm{H}_{2}}=\left(51.3-n_{0}\right)\) mol \(\mathrm{H}_{2} . \mathrm{J}\) (c) The standard heat of combustion of the coal has been determined to be -21,400 kJ/kg, taking \(\mathrm{CO}_{2}(\mathrm{g})\) and \(\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) to be the combustion products. Use this value and the given clemental composition of the coal to prove that the standard heat of formation of the coal is \(-1510 \mathrm{kJ} / \mathrm{kg}\). Then use an energy balance to calculate \(n_{0},\) using the following approximate heat capacities in your calculation: Take the heat of fusion of ash (the heat required to convert ash to slag) to be \(710 \mathrm{kJ} / \mathrm{kg}\).

A 2.00 mole \(\%\) sulfuric acid solution is neutralized with a 5.00 mole\% sodium hydroxide solution in a continuous reactor. All reactants enter at \(25^{\circ} \mathrm{C}\). The standard heat of solution of sodium sulfate is \(-1.17 \mathrm{kJ} / \mathrm{mol} \mathrm{Na}_{2} \mathrm{SO}_{4},\) and the heat capacities of all solutions may be taken to be that of pure liquid water [4.184 kJ/(kg.'C)]. (a) How much heat (kJ/kg acid solution fed) must be transferred to or from the reactor contents (state which it is) if the product solution emerges at \(40^{\circ} \mathrm{C} ?\) (b) Estimate the product solution temperature if the reactor is adiabatic, neglecting heat transferred between the reactor contents and the reactor wall.
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