/*! 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 44 A fuel gas containing \(86 \%\) ... [FREE SOLUTION] | 91影视

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Chapter 5: Problem 44

A fuel gas containing \(86 \%\) methane, \(8 \%\) ethane, and \(6 \%\) propane by volume flows to a furnace at a rate of \(1450 \mathrm{m}^{3} / \mathrm{h}\) at \(15^{\circ} \mathrm{C}\) and \(150 \mathrm{kPa}\) (gauge), where it is burned with \(8 \%\) excess air. Calculate the required flow rate of air in SCMH (standard cubic meters per hour).

Short Answer

Expert verified
The required flow rate of air is approximately 17175.29 standard cubic meters per hour.

Step by step solution

01

Balanced Combustion Reactions

The first step is to write the balanced combustion reactions for each component of the mixed fuel. For methane: \(CH_{4} + 2O_{2} \rightarrow CO_{2} + 2H_{2}O\). For ethane: \(2C_{2}H_{6} + 7O_{2} \rightarrow 4CO_{2} + 6H_{2}O\). For propane: \(C_{3}H_{8} + 5O_{2} \rightarrow 3CO_{2} + 4H_{2}O\).
02

Mole Balances and Stoichiometric Coefficients

The stoichiometric coefficients from the balanced reactions represent the mole ratio in which the reactants react. The total moles of oxygen needed can be calculated by multiplying the stoichiometric coefficient of oxygen in each combustion reaction by the volume percent of the corresponding fuel component, and then summing up the results for all the fuel components. Therefore, for methane: \(2 \times 0.86 = 1.72\), for ethane: \(7/2 \times 0.08 = 0.28\), for propane: \(5 \times 0.06 = 0.30\). Summing these gives a total of \(2.3\) moles of \(O_{2}\) needed per mole of fuel.
03

Excess Air Calculation

The problem statement mentions that 8% excess air is used. Air consists of 21% oxygen by volume. As a result, if stoichiometric conditions were to be maintained, each mole of \(O_{2}\) would correspond to approximately 4.76 moles of air. Taking into account the excess air, the total moles of air becomes \(2.3 \times 4.76 \times 1.08 = 11.85\).
04

Flow Rate Calculation

The flow rate of the air can be calculated by multiplying the total moles of air needed per mole of fuel with the volumetric flow rate of the fuel. Using the conditions of standard temperature and pressure (STP), and assuming ideal gas behavior, 1 mole of any gas at STP occupies 22.4 liters. However, since the flow rate of the fuel is given in cubic meters per hour and considering that 1 cubic meter = 1000 liters, it is required to convert the flow rate of the fuel from cubic meters per hour to moles per hour. Therefore, the flow rate of the fuel in moles per hour is \(1450/22.4 = 64.73\) moles per hour. The flow rate of the air in moles per hour is then \(64.73 \times 11.85 = 767.29\) moles per hour. Converting this back to cubic meters per hour gives \(767.29 \times 22.4 = 17175.29\) SCMH.

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Key Concepts

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

Mole Balances
In any chemical reaction, understanding the concept of mole balances is crucial. It helps you calculate the amounts of reactants and products involved in a reaction. Mole balances use the principle that atoms are not created or destroyed in a chemical reaction.
Moles provide a useful means of converting between atomic-level and macroscopic quantities. This is particularly helpful in combustion reactions where fuel reacts with oxygen. Consider the combustion of methane, as described in the problem:
    \( CH_{4} + 2O_{2} \rightarrow CO_{2} + 2H_{2}O \).
Here, the moles of oxygen needed are twice the moles of methane. For other components like ethane and propane, the stoichiometry is different, affecting the total oxygen requirement.
To perform a proper mole balance, multiply the stoichiometric coefficients by the proportion of each component in the fuel. In this problem, you do this for methane, ethane, and propane, resulting in the total moles of oxygen needed for complete combustion. This mole balancing ensures each fuel component receives the correct amount of oxygen for full conversion into combustion products.
Excess Air Calculation
Excess air in combustion processes is simply air supplied beyond the stoichiometric requirement. It's used to ensure complete combustion of the fuel, which is crucial to avoid unburnt fuel loss and to ensure environmental compliance.
In this problem, 8% excess air means that 8% more air than the exact requirement is provided. Air is composed of about 21% oxygen by volume, which impacts the total air used. In stoichiometric terms, each mole of \(O_2\) corresponds to approximately 4.76 moles of air. Thus, for determining the actual air supplied, the total stoichiometrically needed air is increased by 8%. This can be calculated as:
    \[ (2.3 \, \text{moles of } O_2) \times (4.76 \, \text{moles of air per mole } O_2) \times (1.08 \, \text{for excess air}) = 11.85 \, \text{moles of air per mole of fuel} \].
Excess air ensures that all fuel molecules encounter sufficient oxygen atoms to fully combust. Without this, efficiency drops and emissions can increase, underscoring why engineers carefully calculate excess air.
Gas Flow Rate
Calculating the flow rate of gases in combustion scenarios allows engineers to ensure systems are operating efficiently. To find the air flow rate, you multiply the total moles of air needed per mole of fuel by the fuel flow rate.
Start by converting the volume of gas flow into moles using ideal gas laws. Given conditions or standard conditions like STP simplify this step. At STP, one mole of gas occupies 22.4 liters, but when dealing with flow rates, it's important to measure in cubic meters for standardization.For the given problem, converting the fuel flow rate involves:
    \[ 1450 \, \text{m}^3/ ext{h} \div 22.4 \, \text{L/mole} = 64.73 \, \text{moles per hour} \].
With 11.85 moles of air required per mole of fuel, the air flow rate is:
    \[ 64.73 \, \text{moles per hour} \times 11.85 = 767.29 \, \text{moles of air per hour} \]
Converting this back to standard cubic meters gives you:
    \[ 767.29 \, \text{moles of air per hour} \times 22.4 \, \text{L/mole} = 17175.29 \, \text{SCMH} \].
Calculating gas flow rates helps in designing systems for optimal fuel and air mixing, ensuring efficient and complete combustion.

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Most popular questions from this chapter

Steam reforming is an important technology for converting refined natural gas, which we take here to be methane, into a synthesis gas that can be used to produce a varicty of other chemical compounds. For example, consider a reformer to which natural gas and steam are fed in a ratio of 3.5 moles of steam per mole of methane. The reformer operates at 18 atm, and the reaction products leave the reformer in chemical equilibrium at \(875^{\circ} \mathrm{C}\). The steam reforming reaction is $$\mathrm{CH}_{4}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{CO}+3 \mathrm{H}_{2}$$ and the water-gas shift reaction also occurs in the reformer. $$\mathrm{CO}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{CO}_{2}+\mathrm{H}_{2}$$ The equilibrium constants for these two reactions are given by the expressions At \(875^{\circ} \mathrm{C}, K_{\mathrm{R}}=872.9 \mathrm{atm}^{2}\) and \(K \mathrm{w} \mathrm{G}=0.2482 .\) The process is to produce \(100.0 \mathrm{kmol} / \mathrm{h}\) of hydrogen. Calculate the feed rates (kmol/h) of methane and steam and the volumetric flow rate \(\left(\mathrm{m}^{3} / \mathrm{min}\right)\) of gas leaving the reformer.

The current global reliance on fossil fuels for heating, transportation, and electric power generation raises concems regarding the release of \(\mathrm{CO}_{2}\) and \(\mathrm{CH}_{4},\) which are greenhouse gases thought to lead to climate change, and NO, which contributes to smog. One potential solution to these problems is to produce transportation fuels from renewable biomass. You have been asked to evaluate a proposed process for converting forest residues to alcohols that may be used as transportation fuels. In the first stage of the process, steam and dry wood from hybrid poplar trees (which grow between five and eight feet a year and can be harvested roughly every five years) are fed to a gasifier in which the biomass is converted to light gases in the following reactions: $$\begin{aligned} \mathrm{C}+\mathrm{H}_{2} \mathrm{O} & \rightarrow \mathrm{CO}+\mathrm{H}_{2} \\\ \mathrm{CO}+\mathrm{H}_{2} \mathrm{O} & \rightarrow \mathrm{CO}_{2}+\mathrm{H}_{2} \\ \mathrm{C}+\mathrm{CO}_{2} & \rightarrow 2 \mathrm{CO} \\ \mathrm{C}+2 \mathrm{H}_{2} & \rightarrow \mathrm{CH}_{4} \\ \mathrm{CH}_{4}+\mathrm{H}_{2} \mathrm{O} & \rightarrow \mathrm{CO}+3 \mathrm{H}_{2} \end{aligned}$$ The effluents from the reactor are a gas stream containing \(\mathrm{H}_{2}, \mathrm{CO}, \mathrm{CO}_{2}, \mathrm{CH}_{4},\) and \(\mathrm{H}_{2} \mathrm{O},\) and a solid char stream that contains only carbon and hydrogen. The char is discarded and the gases go through additional steps in which the hydrogen and carbon monoxide are converted to mixed alcohols. This problem only concerns the gasifier. \(\cdot\) Elemental composition of biomass: 51.9 mass \(\%\) C \(, 6.3 \%\) H, and \(41.8 \%\) O \(\cdot\) Pressure and temperature of entering steam: \(155^{\circ} \mathrm{C}, 4.4 \mathrm{atm}\) \(\cdot\) Feed ratio of steam to biomass: 1.1 kg steam/kg biomass \(\cdot\) Yield and dry-basis composition of product gas: 1.35 kg dry gas/kg biomass at \(700^{\circ} \mathrm{C}, 1.2\) atm; 50.7 mol\% \(\mathrm{H}_{2}, 23.8 \%\) CO, \(18.0 \% \mathrm{CO}_{2}, 7.5 \% \mathrm{CH}_{4}\) (a) Taking a basis of \(100 \mathrm{kg}\) of biomass fed, draw and completely label a flowchart for the gasifier incorporating the given data, labeling the volumes of the steam fed and the gases produced. Perform a degree-of-freedom analysis. (b) Calculate the mass and mass composition of the char and the volumes of the steam feed and product gas streams. (c) List advantages and possible drawbacks of using biomass rather than petroleum as a fuel source.

Methanol is produced by reacting carbon monoxide and hydrogen at \(644 \mathrm{K}\) over a \(\mathrm{ZnO}-\mathrm{Cr}_{2} \mathrm{O}_{3}\) catalyst. A mixture of \(\mathrm{CO}\) and \(\mathrm{H}_{2}\) in a ratio \(2 \mathrm{mol} \mathrm{H}_{2} / \mathrm{mol}\) CO is compressed and fed to the catalyst bed at \(644 \mathrm{K}\) and 34.5 MPa absolute. A single-pass conversion of 25\% is obtained. The space velocity, or ratio of the volumetric flow rate of the feed gas to the volume of the catalyst bed, is The product gases are passed through a condenser, in which the methanol is liquefied. (a) You are designing a reactor to produce \(54.5 \mathrm{kmol} \mathrm{CH}_{3} \mathrm{OH} / \mathrm{h}\). Estimate (i) the volumetric flow rate that the compressor must be capable of delivering if no gases are recycled, and (ii) the required volume of the catalyst bed. (Use Kay's rule for pressure-volume calculations.) (b) If (as is done in practice) the gases from the condenser are recycled to the reactor, the compressor is then required to deliver only the fresh feed. What volumetric flow rate must it deliver assuming that the methanol produced is completely recovered in the condenser? (In practice it is not; moreover, a purge stream must be taken off to prevent the buildup of impurities in the system.)

A tank in a room at \(19^{\circ} \mathrm{C}\) is initially open to the atmosphere on a day when the barometric pressure is 102 kPa. A block of dry ice (solid \(\mathrm{CO}_{2}\) ) with a mass of \(15.7 \mathrm{kg}\) is dropped into the tank, which is then sealed. The reading on the tank pressure gauge initially rises very quickly, then much more slowly, eventually reaching a value of 3.27 MPa. Assume \(T_{\text {final }}=19^{\circ} \mathrm{C}\) (a) How many moles of air were in the tank initially? Neglect the volume occupied by \(\mathrm{CO}_{2}\) in the solid state, and assume that a negligible amount of \(\mathrm{CO}_{2}\) escapes prior to the sealing of the tank. (b) Estimate the percentage error made by neglecting the volume of the block of dry ice placed in the tank. (The specific gravity of solid carbon dioxide is approximately 1.56 .) (c) What is the final density (g/L) of the gas in the tank? (d) Explain the observed variation of pressure with time. More specifically, what is happening in the tank during the initial rapid pressure increase and during the later slow pressure increase?

A stream of hot dry nitrogen flows through a process unit that contains liquid acetone. A substantial portion of the acetone vaporizes and is carried off by the nitrogen. The combined gases leave the recovery unit at \(205^{\circ} \mathrm{C}\) and 1.1 bar and enter a condenser in which a portion of the acetone is liquefied. The remaining gas leaves the condenser at \(10^{\circ} \mathrm{C}\) and 40 bar. The partial pressure of acetone in the feed to the condenser is 0.100 bar, and that in the effluent gas from the condenser is 0.379 bar. Assume ideal-gas behavior. (a) Calculate for a basis of \(1 \mathrm{m}^{3}\) of gas fed to the condenser the mass of acetone condensed ( \(\mathrm{kg}\) ) and the volume of gas leaving the condenser \(\left(\mathrm{m}^{3}\right)\) (b) Suppose the volumetric flow rate of the gas leaving the condenser is \(20.0 \mathrm{m}^{3} / \mathrm{h}\). Calculate the rate (kg/h) at which acetone is vaporized in the solvent recovery unit.
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