/*! 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 77 Methane is bumed with \(25 \%\) ... [FREE SOLUTION] | 91Ó°ÊÓ

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Methane is bumed with \(25 \%\) excess air in a continuous adiabatic reactor. The methane enters the reactor at \(25^{\circ} \mathrm{C}\) and 1.10 atm at a rate of \(550 \mathrm{L} / \mathrm{s}\), and the entering air is at \(150^{\circ} \mathrm{C}\) and 1.1 atm. Combustion in the reactor is complete, and the reactor effluent gas emerges at 1.05 atm. (a) Calculate the temperature and the degrees of superheat of the reactor effluent. (Consider water to be the only condensable species in the effluent.) (b) Suppose only 15\% excess air is supplied. Without doing any additional calculations, state how the temperature and degrees of superheat of the reactor effluent would be affected lincrease, decrease, remain the same, cannot tell without more information] and explain your reasoning. What risk is involved in lowering the percent excess air?

Short Answer

Expert verified
The effluent temperature and the degree of superheat could be calculated using principles of thermodynamics and energy balance, considering the adiabatic nature of the reactor. The reduction in excess air percentage would increase the temperature and degree of superheat of the reactor effluent, with risks of producing incomplete combustion products.

Step by step solution

01

Balancing The Reaction Equation For Complete Combustion

By the stoichiometry of the complete combustion reaction of methane with air (primarily oxygen), we know \(CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O\). As per the problem, there's 25% excess air, which means the total air required is 125% of the stoichiometric air, so the overall reaction becomes \(CH_4 + 2.5O_2 \rightarrow CO_2 + 2H_2O + 0.5O_2\).
02

Calculation of Molar Flow Rates

Methane is entering at a rate of 550 L/s at 1.10 atm and 25°C. Using the ideal gas law we can convert this flow rate to a molar flow rate. \(PV = nRT\), where P is pressure (1.10 atm), V is volume (550 L/s), R is the ideal gas constant (0.0821 L.atm/(K.mol)) and T is temperature (25°C = 298 K). Solving for n (the number of moles), we have \(n = PV/RT\).
03

Calculating The Combustion Temperature

Determining the exit temperature of the combustion requires the application of an energy balance around the reactor and the use of heat capacity data for the various species present. Since the reactor is adiabatic, there is no heat exchange with the surroundings, so all the heat generated by combustion is retained in the products. The temperature rise of the products is found by equating the heat generated from the combustion reaction (which can be determined from standard heat of combustion data and the molar flow rate of methane) with the increase in the enthalpy of the products (which is determined by integrating the heat capacities of the product species over the temperature range).
04

Calculate The Degree Of Superheat

The degree of superheat can be calculated by subtracting the saturation temperature at the pressure given from the temperature calculated in Step 3. The saturation temperature can be found from steam tables for the given pressure of the reactor effluent.
05

Discuss The Effects Of Reducing Excess Air

Reducing the amount of excess air would cause the exit temperature to increase, because there would be less nitrogen present to absorb some of the heat of the reaction. The degree of superheat would also increase for the same reason. One risk involved in lowering the percent excess air is incomplete combustion, which might lead to the formation of hazardous products like carbon monoxide (CO), or even unburnt methane (which is a significant greenhouse gas).

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Key Concepts

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

Combustion Reaction
A combustion reaction, often seen in chemical and mechanical engineering applications, involves the burning of a fuel in the presence of oxygen, releasing energy in the form of heat and light. Methane, a common fuel, reacts with oxygen in a 1:2 ratio for complete combustion, producing carbon dioxide and water vapor. The excess air ensures that there is more than enough oxygen to completely react with the fuel. Understanding the combustion reaction is crucial for optimizing the use of fuels and managing energy outputs efficiently.

In the given exercise, complete combustion of methane is assumed, which is an ideal scenario in practical applications. Industrial combustion systems aim for complete combustion to maximize energy generation while minimizing harmful emissions.
Stoichiometry
Stoichiometry is the branch of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. In an engineering context, it is used to calculate the amounts of inputs and outputs from a chemical process. For methane combustion, stoichiometry dictates the exact ratio of methane to oxygen required for complete combustion.

However, in real-life applications, excess air is introduced to ensure complete combustion, as seen in our problem statement which uses 25% excess air. Stoichiometry becomes more complex with excess reactants and must account for the additional volume and mass flow rates, affecting overall reactor calculations.
Adiabatic Reactor
An adiabatic reactor is a system where no heat exchange occurs with its surroundings. This condition is essential for certain chemical processes where maintaining a specific temperature range is crucial for reaction efficiency and safety. In adiabatic conditions, all the heat generated from the combustion reaction stays within the system, raising the temperature of the products.

The calculation of temperature within an adiabatic reactor is critical and requires an understanding of energy conservation. Engineers must design such reactors to withstand high temperatures without releasing heat, ensuring efficiency and preventing energy loss.
Ideal Gas Law
The ideal gas law is an equation of state for a hypothetical ideal gas. It is a good approximation for behavior of real gases under many conditions, and it relates the pressure (P), volume (V), temperature (T), and the amount in moles (n) of the gas, with the universal gas constant (R). Expressed as the formula: \(PV = nRT\).

For the solution provided, the ideal gas law allows us to convert the volumetric flow rate of methane and air into a molar flow rate, which is a necessary step in calculating the energy balance and determining the final temperature and composition of products.
Energy Balance
Energy balance is a fundamental concept in engineering, stating that energy cannot be created or destroyed, only transformed or transferred. In the context of a reactor, this means that the energy input, in the form of fuel, must equal the energy output, which comes out as heat and work alongside the chemical products. The adiabatic reactor used in the exercise is an excellent example where all heat generated from the combustion is used to increase the enthalpy of the products, as there is no heat loss to the environment.

To calculate the temperature of the effluent, engineers use the energy balance, alongside specific heat capacity data, to determine how much energy is retained in the system and raises the temperature.
Degree of Superheat
The degree of superheat refers to how much a vapor, such as steam, is heated above its saturation temperature at a particular pressure. It is a critical parameter for efficiency in power generation and refrigeration. A higher degree of superheat indicates a higher temperature and thus more thermal energy in the vapor phase.

In our exercise scenario, calculating the degree of superheat involves determining the saturation temperature from steam tables for water vapor at the given pressure and subtracting it from the calculated effluent temperature. Understanding the degree of superheat helps engineers assess the performance and safety of heat transfer systems.

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Most popular questions from this chapter

The wastewater treatment plant at the Ossabaw Paper Company paper mill generates about 24 tonnes of sludge per day. The consistency of the sludge is \(35 \%,\) meaning that the sludge contains 35 wt\% solids and the balance liquids. The mill currently spends \(\$ 40\) /tonne to dispose of the sludge in a landfill. The plant environmental cngincer has determined that if the sludge consistency could be increased to \(75 \%\) the sludge could be incinerated (burned) to generate useful energy and to eliminate the environmental problems associated with landfill disposal. A flowchart for a preliminary design of the proposed sludge-treatment process follows. For simplicity, we will assume that the liquid in the sludge is just water. Process description: The sludge from the wastewater treatment plant (Stream ? passes through a dryer where a portion of the water in the sludge is vaporized. The heat required for the vaporization comes from condensing saturated steam at 4.00 bar (Stream ?. The steam fed to the dryer is produced in the plant's oil-fired boiler from feedwater at \(20^{\circ} \mathrm{C}\) (Stream ? The heat required to produce the steam is transferred from the boiler furnace, where fuel oil (Stream ? from the boiler furnace, where fuel oil (Stream ? .The concentrated sludge coming from the dryer (Stream ? which has a consistency of \(75 \%,\) is fed to an incinerator. The heating value of the sludge is insufficient to keep the incinerator temperature high enough for complete combustion, so natural gas (Stream ? is used as a supplementary fuel. A stream of outside air at \(25^{\circ} \mathrm{C}\) (Stream ? Is heated to \(110^{\circ} \mathrm{C}\) and fed to the incinerator along with the concentrated sludge and natural gas. The waste gas from the incinerator is discharged to the atmosphere. Fuel oil: The oil is a low-sulfur No. 6 fuel oil. Its ultimate (elemental) analysis on a weight basis is \(87 \% \mathrm{C}, 10 \% \mathrm{H}, 0.84 \% \mathrm{S},\) and the balance oxygen, nitrogen, and nonvolatile ash. The higher heating value of the oil is \(3.75 \times 10^{4} \mathrm{kJ} / \mathrm{kg}\) and the heat capacity is \(C_{p}=1.8 \mathrm{kJ} /\left(\mathrm{kg} \cdot^{\circ} \mathrm{C}\right)\) Boiler: The boiler has an efficiency of \(62 \%,\) meaning that \(62 \%\) of the heating value of the fuel oil burned is used to produce saturated steam at 4.00 bar from boiler feedwater at 20^{\circ } \mathrm { C } \text { . Fuel oil at } 6 5 ^ { \circ } \mathrm { C } and dry air at \(125^{\circ} \mathrm{C}\) are fed to the boiler furnace. The air feed rate is \(25 \%\) in excess of the amount theoretically required for complete consumption of the fuel. Sludge: The sludge from the wastewater treatment plant contains \(35 \%\) w/w solids (S) and the balance liquids (which for the purposes of this problem may be treated as only water) and enters the dryer at \(22^{\circ} \mathrm{C} .\) The sludge includes a number of volatile organic species, some of which may be toxic, and has a terrible odor. The heat capacity of the solids is approximately constant at \(2.5 \mathrm{kJ} /\left(\mathrm{kg} \cdot^{\circ} \mathrm{C}\right)\). Dryer: The dryer has an efficiency of \(55 \%,\) meaning that the heat transferred to the sludge, \(Q_{2},\) is \(55 \%\) of the total heat lost by the condensing steam, and the remainder, \(Q_{3}\), is lost to the surroundings. The dryer operates at 1 atm, and the water vapor and concentrated sludge emerge at the corresponding saturation temperature. The steam condensate leaves the dryer as a liquid saturated at 4.00 bar. Incinerator: The concentrated sludge has a heating value of \(19,000 \mathrm{kJ} / \mathrm{kg}\) dry solids. For a feed sludge of \(75 \%\) consistency, the incinerator requires 195 SCM natural gas/tonne wet sludge \(\left[1 \mathrm{SCM}=1 \mathrm{m}^{3}(\mathrm{STP})\right]\) The theoretical air requirement for the sludge is 2.5 SCM air/ 10,000 kJ of heating value. Air is fed in \(100 \%\) excess of the amount theoretically required to burn the sludge and the natural gas. (a) Use material and energy balances to calculate the mass flow rates (tonnes/day) of Streams ? ? ? ? ? ? and ? and heat flows \(\dot{Q}_{0}, \dot{Q}_{1}, \ldots, \dot{Q}_{4}(\mathrm{kJ} / \text { day }) .\) Take the molecular weight of air to be \(29.0 .\) (Caution: Before you start doing lengthy and unnecessary energy balance calculations on the boiler furnace, remember the given furnace efficiency.) Exploratory Exercises - Research and Discover (b) The money saved by implementing this process will be the current cost of disposing of the wastewater plant sludge in a landfill. Two major costs of implementing the process are the installed costs of the new dryer and incinerator. What other costs must be taken into account when determining the economic feasibility of the process? Why might management decide to go ahead with the project even if it proves to be unprofitable? (c) What opportunities exist for improving the energy economy of the process? (Hint: Think about the need to preheat the fuel oil and the boiler and incinerator air streams and consider heat exchange possibilities.) (d) The driving force for the introduction of this process is to eliminate the environmental cost of sludge disposal. What is that cost- -that is, what environmental penalties and risks are associated with using landfills for hazardous waste disposal? What environmental problems might incineration introduce?

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.

Use Hess's law to calculate the standard heat of the water-gas shift reaction $$\mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{v}) \rightarrow \mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g})$$ from each of the two sets of data given here. (a) \(\mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g}): \quad \Delta H_{\mathrm{r}}^{\circ}=+1226 \mathrm{Btu}\) $$\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{v}): \quad \Delta \hat{H}_{\mathrm{v}}=+18,935 \mathrm{Btu} / \mathrm{lb}-\mathrm{mole}$$ $$\begin{aligned}&\text { (b) } \mathrm{CO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{CO}_{2}(\mathrm{g}): \quad \Delta H_{\mathrm{r}}^{\circ}=-121,740 \mathrm{Btu}\\\&\mathrm{H}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{v}): \quad \Delta H_{\mathrm{r}}^{\circ}=-104,040 \mathrm{Btu} \end{aligned}$$

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.

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.

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