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Chapter 5: Problem 92

The concentration of oxygen in a 5000 -liter tank containing air at 1 atm is to be reduced by pressure purging prior to charging a fuel into the tank. The tank is charged with nitrogen up to a high pressure and then vented back down to atmospheric pressure. The process is repeated as many times as required to bring the oxygen concentration below 10 ppm (i.c., to bring the mole fraction of \(\mathrm{O}_{2}\) below \(10.0 \times 10^{-6}\) ). Assume that the temperature is \(25^{\circ} \mathrm{C}\) at the beginning and end of each charging cycle. When doing \(P V T\) calculations in Parts (b) and (c), use the generalized compressibility chart if possible for the fully charged tank and assume that the tank contains pure nitrogen. (a) Speculate on why the tank is being purged. (b) Estimate the gauge pressure (atm) to which the tank must be charged if the purge is to be done in one charge-vent cycle. Then estimate the mass of nitrogen (kg) used in the process. (For this part, if you can't find the tank condition on the compressibility chart, assume ideal-gas behavior and state whether the resulting estimate of the pressure is too high or too low.) (c) Suppose nitrogen at 700 kPa gauge is used for the charging. Calculate the number of charge-vent cycles required and the total mass of nitrogen used. (d) Use your results to explain why multiple cycles at a lower gas pressure are preferable to a single cycle. What is a probable disadvantage of multiple cycles?

Short Answer

Expert verified
For purging the tank, lower pressures in multiple cycles are preferable because they would require less nitrogen and carry less risk of a hazardous high-pressure situation. However, the downside of this is that purging in multiple cycles would take more time.

Step by step solution

01

Reason for Purging

Considering that a fuel is to be charged into the tank, the purging is required to avoid the possibility of any explosive reaction with oxygen and to maintain the chemical stability of the fuel.
02

Estimate Pressure and Mass for Purging

Purging in one cycle means removal of all oxygen. Since we are starting with air at 1 atm which contains 21% oxygen, we want to reduce it to 10 ppm, which is like reducing it to 0.0001 %. Apply the ideal gas law to calculate the new pressure after purging by the formula \(P' = P \times (1 - X)\), where P' is the new pressure, P is the initial pressure (1 atm) and X is the fraction of oxygen (0.21 as 21%). To calculate the amount of nitrogen required, we need to calculate the moles of nitrogen needed to replace the oxygen. Each mole of oxygen replaced by nitrogen reduces its mole fraction by its initial value. So, moles of nitrogen = initial moles of air 脳 initial mole fraction of oxygen 脳 reduction factor. Mass of nitrogen can then be calculated using the molar mass of nitrogen.
03

Number of Cycles Required and Total Nitrogen Used

Given that nitrogen is charged at 700 kPa (approximately 6.9 atm), we need to calculate how much oxygen is left after one purge. Apply the same formula as in Step 2 for the reduction in oxygen fraction after one purge. Then, find out how many cycles required to reach 10 ppm. Recall from formula, number of cycles \(n\) can be derived from \(X = X_0(1 - X_N)^n \) where \(X_0\) is initial oxygen concentration and \(X_N\) is desired oxygen concentration. Total mass of nitrogen used is then the mass per cycle times the number of cycles
04

Purging Cycles Analysis

Based on results from previous steps, state why multiple cycles at lower gas pressure are preferable to a single cycle (since less nitrogen is needed and the pressure is safer), however also note that multiple cycles would take more time, which could be a disadvantage.

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Key Concepts

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

Ideal Gas Law Applications
The Ideal Gas Law, represented by the equation PV=nRT, where P is pressure, V is volume, n is the number of moles of gas, R is the ideal gas constant, and T is temperature in Kelvin, is a fundamental equation in the study of gases. In chemical engineering, it is especially useful for calculating the behavior of gases under varying conditions of pressure, volume, and temperature.

When applying this law to real-life scenarios such as pressure purging in a chemical process, we assume that the gases involved behave ideally, meaning their particles are non-attracting and occupy no volume. This assumption simplifies the calculation, but it's important to note when real gases might deviate from this behavior, taking into consideration the conditions at which the gas is being manipulated. In the given exercise, it is used to estimate the pressure and mass of nitrogen required to purge a tank of oxygen to a safe level for the introduction of fuel.

During the purging process, it's crucial to account for the initial and final gas concentrations, ensuring the safety and efficacy of the procedure. As oxygen is removed and replaced by nitrogen, the application of the Ideal Gas Law helps us estimate the new pressure necessary for a complete purge in one cycle or multiple cycles, as well as the mass of nitrogen needed - both critical factors for the success of the operation.
Compressibility Chart Usage
Compressibility charts are valuable tools used to determine the deviation of real gas behavior from ideal gas laws under various pressures and temperatures. These charts provide the compressibility factor Z, which corrects the ideal gas law for real gas behavior. The compressibility factor is defined as Z=PV/(nRT), where a Z value of 1 indicates ideal behavior.

Chemical engineers use these charts to improve the accuracy of their PV calculations when dealing with high pressures or low temperatures where gases may not act ideally. In the exercise, the compressibility chart is referenced to calculate the properties of nitrogen in the fully charged tank to account for deviations from the Ideal Gas Law which assumes a Z value of 1. If the state of nitrogen is outside the range of the chart, or the chart is not readily accessible, the exercise improvement advice suggests assuming ideal-gas behavior while noting whether the estimate is too high or low. While this is a practical approach, it is essential to understand its limitations and the potential for error in the absence of compressibility data.
Chemical Engineering Safety
Safety is paramount in chemical engineering processes, especially when handling flammable or explosive substances, such as in the pressure purging process described in the problem. Purging is a crucial step to prevent any explosive reactions by removing oxygen that could otherwise react with the fuel being loaded into the tank.

Engineers must design safe procedures that lower the concentration of oxygen to non-flammable levels, adhering to strict safety regulations and guidelines. This is due to the potential risk that any residual oxygen presents when it comes in contact with highly reactive substances. The approach to use multiple charge-vent cycles at a lower gas pressure, rather than a single high-pressure purge, is an application of these safety practices. It minimizes risks associated with handling high-pressure gases and the potential for accidents. In the realm of safety, time is also a trade-off, as multiple cycles might require more time and might potentially increase exposure to risk. Nevertheless, this methodical approach generally aligns with the industry's best practices for safety.
Mole Fraction Calculations
Mole fraction is a way of expressing the concentration of a component in a mixture and is calculated as the moles of the component divided by the total moles of all components in the mixture. It's a dimensionless number that plays a key role in processes like purging because it helps in setting and achieving concentration targets. For example, reaching a mole fraction of oxygen below 10^-6 is critical for safety in the given exercise.

The exercise involves calculating the number of purging cycles required to dilute the oxygen concentration to a safe level. By using repeated calculations of the mole fraction after each cycle, engineers can determine how many cycles are needed to eventually reach the desired oxygen concentration. This process of calculation not only ensures the safety and effectiveness of the purging process but also allows for optimization of resource usage, in this case, the nitrogen used for purging.

Understanding mole fraction calculations is essential for engineers to accurately design and control chemical processes, preventing safety hazards and ensuring the proper functioning of equipment.

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Most popular questions from this chapter

In chemical vapor deposition (CVD), a semiconducting or insulating solid material is formed in a reaction between a gaseous species and a species adsorbed on the surface of silicon wafers (disks about \(10 \mathrm{cm}\) in diameter and \(1 \mathrm{mm}\) thick). The coated wafers are subjected to further processing to produce the microelectronic chips in computers and most other electronic devices in use today. In one such process, silicon dioxide (MW \(=60.06, \mathrm{SG}=2.67\) ) is formed in the reaction between gaseous dichlorosilane (DCS) and adsorbed nitrous oxide: $$\mathrm{SiH}_{2} \mathrm{Cl}_{2}(\mathrm{g})+2 \mathrm{N}_{2} \mathrm{O}(\mathrm{ads}) \rightarrow \mathrm{SiO}_{2}(\mathrm{s})+2 \mathrm{N}_{2}(\mathrm{g})+2 \mathrm{HCl}(\mathrm{g})$$ A mixture of DCS and \(\mathrm{N}_{2} \mathrm{O}\) flows through a "boat reactor" - a horizontal pipe in which 50 to 100 silicon wafers about \(12 \mathrm{cm}\) in diameter and \(1 \mathrm{mm}\) thick are set upright along the reactor length, with about \(20 \mathrm{mm}\) separation between each wafer. A side view of the reactor is shown below: The feed gas enters the reactor at a rate of 3.74 SCMM (standard cubic meters per minute) and contains 22.0 mole\% DCS and the balance \(\mathrm{N}_{2} \mathrm{O}\). In the reactor, the gas flows around the wafers, DCS and \(\mathrm{N}_{2} \mathrm{O}\) diffuse into the spaces between the wafers, \(\mathrm{N}_{2} \mathrm{O}\) is adsorbed on the wafer surfaces, and the adsorbed \(\mathrm{N}_{2} \mathrm{O}\) reacts with gascous DCS. The silicon dioxide formed remains on the surface, and the nitrogen and hydrogen chloride go into the gas phase and eventually leave the reactor with the unconsumed reactants. The temperature and absolute pressure in the reactor are constant at \(900^{\circ} \mathrm{C}\) and 604 millitorr. (a) The percentage conversion of DCS at a certain axial position (distance along the length of the reactor) is 60\%. Calculate the volumetric flow rate (m \(^{3} / \mathrm{min}\) ) of gas at this axial position. (b) The rate of deposition of silicon dioxide per unit area of wafer surface is given by the formula $$r\left(\frac{\mathrm{mol} \mathrm{SiO}_{2}}{\mathrm{m}^{2} \cdot \mathrm{s}}\right)=3.16 \times 10^{-8} \mathrm{p}_{\mathrm{DCS}} p_{\mathrm{N}_{2} \mathrm{O}}^{0.65}$$ where \(p_{\mathrm{DCS}}\) and \(p_{\mathrm{N}, \mathrm{o}}\) are the partial pressures of \(\mathrm{DCS}\) and \(\mathrm{N}_{2} \mathrm{O}\) in millitorr. What is \(r\) at the axial position in the reactor where the DCS conversion is \(60 \% ?\) (c) Consider a wafer located at the axial position determined in Part (b). How thick is the silicon dioxide layer on that wafer after two hours of reactor operation, assuming that gas diffusion is rapid enough at the low reactor pressure for the composition of the gas (and hence the component partial pressures) to be uniform over the wafer surface? Express your answer in angstroms, where 1 \(\AA=1.0 \times 10^{-10} \mathrm{m}\). (Hint: You can calculate the rate of growth of the \(\mathrm{Si} \mathrm{O}_{2}\) layer in \(\mathrm{A} / \mathrm{min}\) from \(r\) and propertics of \(\mathrm{SiO}_{2}\) given in the problem statement.) Would the thickness be greater or less than this value at an axial position closer to the reactor entrance? Briefly explain your answer.

An oxygen tank with a volume of \(2.5 \mathrm{ft}^{3}\) is kept in a room at \(50^{\circ} \mathrm{F}\). An engineer has used the idealgas equation of state to determine that if the tank is first evacuated and then charged with \(35.3 \mathrm{lb}_{\mathrm{m}}\) of pure oxygen, its rated maximum allowable working pressure (MAWP) will be attained. Operation at pressures above this value is considered unsafe. (a) What is the maximum allowable working pressure (psig) of the tank? (b) You suspect that at the conditions of the fully charged tank, ideal-gas behavior may not be a good assumption. Use the SRK equation of state to obtain a better estimate of the maximum mass of oxygen that may be charged into the tank. Did the ideal-gas assumption lead to a conservative estimate (on the safe side) or a nonconservative estimate of the amount of oxygen that could be charged? (c) Suppose the tank is charged and ruptures before the amount of oxygen calculated in Part (b) enters it. (It should have been able to withstand pressures up to four times the MAWP.) Think of at least five possible explanations for the failure of the tank below its rated pressure limit.

A natural gas contains 95 wt\% \(\mathrm{CH}_{4}\) and the balance \(\mathrm{C}_{2} \mathrm{H}_{6}\). Five hundred cubic meters per hour of this gas at \(40^{\circ} \mathrm{C}\) and 1.1 bar is to be burned with \(25 \%\) excess air. The air flowmeter is calibrated to read the volumetric flow rate at standard temperature and pressure. What should the meter read (in SCMH) when the flow rate is set to the desired value?

A small power plant produces \(500 \mathrm{MW}\) of electricity through combustion of coal that has the following composition on a dry basis: 76.2 wt\% carbon, \(5.6 \%\) hydrogen, \(3.5 \%\) sulfur, \(7.5 \%\) oxygen, and the remainder ash. The coal contains 4.0 wt\% water. The feed rate of coal is 183 tons/h, and it is burned with \(15 \%\) excess air at 1 atm, \(80^{\circ} \mathrm{F}\), and \(30.0 \%\) relative humidity. (a) Estimate the volumetric flow rate (ft \(^{3} / \mathrm{min}\) ) of air drawn into the furnace. (b) Effluent gases are discharged from the furnace at \(625^{\circ} \mathrm{F}\) and 1 atm. Estimate the molar (lb-mole/ min) and volumetric (ft \(^{3} / \mathrm{min}\) ) flow rates of gas leaving the furnace. (c) Injection of dry limestone ( \(\mathrm{CaCO}_{3}\) ) into the furnace is being considered as a means of reducing the \(\mathrm{SO}_{2}\) emitted from the plant. The technology calls for \(\mathrm{SO}_{2}\) to react with limestone: $$\mathrm{CaCO}_{3}+\mathrm{SO}_{2}+\frac{1}{2} \mathrm{O}_{2} \rightarrow \mathrm{CaSO}_{4}+\mathrm{CO}_{2}$$ Unfortunately, the process is expected to remove only \(75 \%\) of the \(\mathrm{SO}_{2}\) in the effluent gases, even though the limestone is fed at a rate 2.5 times the stoichiometric amount. What is the required feed rate of limestone? since some of the \(S O_{2}\) is removed from the furnace efflucnt [in contrast to Part (b)], recalculate the molar flow rate and composition of the effluent from the fumace. (d) The gas leaving the furnace passes through an electrostatic precipitator, where particulates from ash and limestone are removed, and then enters a stack (chimney) for release to the atmosphere. What is the gas velocity at a point in the stack where the stack diameter is \(25 \mathrm{ft}\) and the temperature is \(300^{\circ} \mathrm{F}\) ? Does the gas discharged from the stack meet the new Environmental Protection Agency standard that emissions from such power plants contain less than 75 parts of \(\mathrm{SO}_{2}\) per billion?

Phosgene (CCl, O) is a colorless gas that was used as an agent of chemical warfare in World War I. It has the odor of new-mown hay (which is a good warning if you know the smell of new-mown hay). Pete Brouillette, an innovative chemical engincering student, came up with what he believed was an effective new process that utilized phosgene as a starting material. He immediately set up a reactor and a system for analyzing the reaction mixture with a gas chromatograph. To calibrate the chromatograph (i.e., to determine its response to a known quantity of phosgene), he evacuated a 15.0 cm length of tubing with an outside diameter of \(0.635 \mathrm{cm}\) and a wall thickness of \(0.559 \mathrm{mm}\), and then connected the tube to the outlet valve of a cylinder containing pure phosgene. The idea was to crack the valve, fill the tube with phosgene, close the valve, feed the tube contents into the chromatograph, and observe the instrument response. What Pete hadn't thought about (among other things) was that the phosgene was stored in the cylinder at a pressure high enough for it to be a liquid. When he opened the cylinder valve, the liquid rapidly flowed into the tube and filled it. Now he was stuck with a tube full of liquid phosgene at a pressure the tube was not designed to support. Within a minute he was reminded of a tractor ride his father had once given him through a hayfield, and he knew that the phosgene was leaking. He quickly ran out of the lab, called campus security, and told them that a toxic leak had occurred, that the building had to be evacuated, and the tube removed and disposed of properly. Personnel in air masks shortly appeared, took care of the problem, and then began an investigation that is still continuing. (a) Show why one of the reasons phosgene was an effective weapon is that it would collect in low spots soldiers often mistakenly entered for protection. (b) Pete's intention was to let the tube equilibrate at room temperature ( \(23^{\circ} \mathrm{C}\) ) and atmospheric pressure. How many gram-moles of phosgene would have been contained in the sample fed to the chromatograph if his plan had worked? (c) The laboratory in which Pete was working had a volume of \(2200 \mathrm{ft}^{3}\), the specific gravity of liquid phosgene is \(1.37,\) and Pete had read somewhere that the maximum "safe" concentration of phosgene in air is \(0.1 \mathrm{ppm}\) \(\left(0.1 \times 10^{-6} \mathrm{mol} \mathrm{CCl}_{2} \mathrm{O} / \mathrm{mol}\) air) \right. Would the "safe" concentration have been exceeded if all the liquid phosgene in the tube had evaporated into the room? Even if the limit would not have been exceeded, give several reasons why the lab would still have been unsafe. (d) List several things Pete did (or failed to do) that made his experiment unnecessarily hazardous.
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