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Chapter 5: Problem 45

Ethane at \(25^{\circ} \mathrm{C}\) and 1.1 atm (abs) flowing at a rate of \(100 \mathrm{mol} / \mathrm{s}\) is burned with \(20 \%\) excess oxygen at \(175^{\circ} \mathrm{C}\) and 1.1 atm \((\text { abs }) .\) The combustion products leave the furnace at \(800^{\circ} \mathrm{C}\) and 1 atm. (a) What is the volumetric flow rate of oxygen (L/s) fed to the furnace? (b) What should the volumetric flow rate of the combustion products be? State all assumptions you make. (c) The volumetric flow rate of the combustion products is measured and found to be different from the value calculated in Part (b). Assuming that no mistakes were made in the calculation, what could be going on that could lead to the discrepancy? Consider assumptions made in the calculations and things that can go wrong in a real system.

Short Answer

Expert verified
(a) The volumetric flow rate of oxygen fed into the furnace is approximately 15771 L/s. (b) The calculated volumetric flow rate of the combustion products should be roughly 44116 L/s. (c) Discrepancies can be due to non-ideal gas behaviour, losses during combustion, measurement errors, or system leaks.

Step by step solution

01

Write the balanced combustion reaction

The combustion reaction of ethane (C2H6) with oxygen (O2) can be written as: \[ \mathrm{C2H6 + \frac{7}{2}O2 \rightarrow 2CO2 + 3H2O }\]
02

Calculate the stoichiometric requirement of \(\mathrm{O2}\)

Given that ethane is burned with 20% excess \(\mathrm{O2}\), the molar flow rate of \(\mathrm{O2}\) required for the combustion is \(1.2 \times \frac{7}{2} = 4.2\) mol \(\mathrm{O2}\) for every 1 mol of \(\mathrm{C2H6}\). Therefore, the total molar flow rate of \(\mathrm{O2}\) is \(4.2 \times 100 = 420 \) mol/s.
03

Apply the ideal gas law

The ideal gas law is given as \[ PV=nRT \]. In this context, V is the volume of the gas, n is the number of moles, R is the gas constant and T is the temperature in Kelvin. Solving for V and substituting the given values (\(n=420 \) mol/s, \(R=0.0821\) L.atm/mol.K \(T=175+273= 448K\) and \(P=1.1\) atm), we get the volumetric flow rate of \(\mathrm{O2}\) as: \[ V = \frac{nRT}{P} = \frac{420 \times 0.0821 \times 448}{1.1} = 15770.5636 \] L/s.
04

Calculate the volumetric flow rate of the combustion products

According to the balanced reaction, 1 mol of ethane produces 2 mol of \(\mathrm{CO2}\) and 3 mol of \(\mathrm{H2O}\), that is, a total of 5 moles of gaseous products. So, in this case the total molar flow rate of the combustion products is \(5 \times 100 = 500\) mol/s. Substituting \(n=500 \) mol/s, \(R=0.0821\) L.atm/mol.K, \(T=800+273= 1073K\) and \(P=1\) atm into the ideal gas law equation, we get the volumetric flow rate of the combustion products as: \[ V = \frac{nRT}{P} = \frac{500 \times 0.0821 \times 1073}{1} = 44115.515 \] L/s.
05

Identify possible reasons for any discrepancy

The calculated values could be different from the real-world measurements due to several reasons such as deviations from the assumed ideal gas behaviour, loss of combustion products during the process, inaccurate measurement of initial conditions, or leaks in the system.

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Key Concepts

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

Ideal Gas Law
The ideal gas law is a fundamental principle in chemistry and physics that relates the pressure, volume, temperature, and the amount of an ideal gas. The relationship is given by the equation \( PV = nRT \), where \( P \) represents pressure in atmospheres, \( V \) is the volume in liters, \( n \) is the number of moles, \( R \) is the ideal gas constant (0.0821 L.atm/mol.K), and \( T \) is the temperature in Kelvin.

When dealing with real gases, especially at high temperatures and low pressures, they tend to behave similarly to an ideal gas. This assumption simplifies the complex interactions between particles in a gas, making it easier to predict and calculate properties such as volumetric flow rate in chemical processes like combustion. In the provided exercise, the ideal gas law is used to calculate the volumetric flow rate of both the oxygen fed to the furnace and the combustion products exiting the furnace.
Volumetric Flow Rate Calculation
The volumetric flow rate is a measure of the volume of fluid (in this case, a gas) that passes through a given surface per unit time. It is typically expressed in liters per second (L/s). For gases, the calculation of the volumetric flow rate often incorporates the ideal gas law, as changes in pressure, temperature, and the number of moles of gas affect the volume it occupies.

The volumetric flow rate calculation provides important information in engineering and scientific applications such as identifying the capacity of a gas stream to enter a reactor, size pipelines, and ensure the right amount of reactants in industrial processes. In the context of the exercise, after determining the molar flow rate of oxygen required for ethane combustion, using the ideal gas law helps convert this into a real-world figure, the volumetric flow rate of oxygen, which is then used similarly to find the volumetric flow rate of the combustion products.
Chemical Process Assumptions
In solving chemical engineering problems, certain assumptions are made to simplify the calculations and make them more manageable. Common assumptions include treating gases as ideal, assuming constant temperature and pressure throughout the process, and neglecting losses due to incomplete combustion or physical leaks.

In real-world applications, these assumptions may not hold true. Factors such as non-ideal gas behavior, heat losses, pressure drops, and equipment inefficiencies can lead to discrepancies between calculated and actual values. In the exercise, if the measured volumetric flow rate of combustion products does not match the calculation, it could indicate such real-world effects. This highlights the importance of understanding the limitations of theoretical models and the potential need for adjusting them to better reflect the system being studied.

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Most popular questions from this chapter

A gas turbine power plant receives a shipment of hydrocarbon fuel whose composition is uncertain but may be represented by the expression \(\mathrm{C}_{x} \mathrm{H}_{y}\). The fuel is burned with excess air. An analysis of the product gas gives the following results on a moisture-free basis: \(10.5 \%(\mathrm{v} / \mathrm{v}) \mathrm{CO}_{2}, 5.3 \% \mathrm{O}_{2},\) and \(84.2 \% \mathrm{N}_{2}\) (a) Determine the molar ratio of hydrogen to carbon in the fuel ( \(r\) ), where \(r=y / x\), and the percentage excess air used in the combustion. (b) What is the air-to-fuel ratio ( \(m^{3}\) air/kg of fuel) if the air is fed to the power plant at \(30^{\circ} \mathrm{C}\) and \(98 \mathrm{kPa} ?\) (c) The specific gravity of the fuel (a petroleum product) is \(0.85 .\) Estimate the ratio standard cubic feet of gas fed to the turbine per barrel of fuel. (d) What are the issues associated with using oil as a fuel as opposed to natural gas? Consider two factors: (i) the complete composition of typical fuel oils and their resulting emissions, and (ii) the availability and global distribution of the two fuel sources.

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 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.

Propylene is hydrogenated in a batch reactor: $$\mathrm{C}_{3} \mathrm{H}_{6}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g}) \rightarrow \mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{g})$$ Equimolar amounts of propylene and hydrogen are fed into the reactor at \(25^{\circ} \mathrm{C}\) and a total absolute pressure of 32.0 atm. The reactor temperature is raised to \(235^{\circ} \mathrm{C}\) and held constant thereafter until the reaction is complete. The propylene conversion at the beginning of the isothermal period is \(53.2 \%\) You may assume ideal-gas behavior for this problem, although at the high pressures involved this assumption constitutes an approximation at best. (a) What is the final reactor pressure? (b) What is the percentage conversion of propylene when \(P=35.1\) atm? (c) Construct a graph of pressure versus fractional conversion of propylene covering the isothermal period of operation. Use the graph to confirm the results in Parts (a) and (b). (Suggestion: Use a spreadsheet.)

A distillation column is being used to separate methanol and water at atmospheric pressure. The column temperature varies from approximately \(65^{\circ} \mathrm{C}\) at the top to \(100^{\circ} \mathrm{C}\) at the bottom. Liquid enters the top of the column and flows down to the bottom; vapor is generated in a reboiler at the bottom of the column, flows upward, and leaves at the top. The molar flow rate of vapor up the column may be assumed to be constant from top to bottom. The vapor velocity is kept below \(5.0 \mathrm{ft} / \mathrm{s}\) to keep the vapor from entraining liquid (suspending and carrying away liquid droplets). (a) Where in the column is the greatest risk of liquid entrainment? Explain your answer. (b) Assuming that the liquid flowing down the column and the column internals (equipment inside the column) occupy a negligible fraction of the column cross-sectional area, estimate the minimum column diameter if the vapor flow rate is 25.0 lb-mole/min. (c) Suppose the column is constructed with a diameter \(10 \%\) greater than that determined in Part (b). What are the vapor velocities at the top and bottom of the column if the vapor molar flow rate in both locations is 25.0 ib-mole/min? How much can the vapor molar flow rate be increased without causing liquid entrainment? (d) There is a need to increase process throughput, which would require the vapor molar flow rate to be doubled. It has been suggested that increasing the pressure in the column would allow that to be done without risking excessive liquid entrainment. Again applying a vapor velocity limit of \(5 \mathrm{ft} / \mathrm{s}\) what would the new pressure be?
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