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Chapter 5: Problem 43

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?

Short Answer

Expert verified
a) The bottom of the column presents the greatest risk of liquid entrainment. b) The minimum column diameter is 1.56 ft. c) The vapor velocities at the column top and bottom are 4.05 ft/s and the vapor molar flow rate can be increased up to 31.01 lb-mole/min without causing liquid entrainment. d) To double the vapor molar flow rate without risking the liquid entrainment, the pressure inside the column must be doubled.

Step by step solution

01

Part (a) - Location of greatest risk of liquid entrainment

The greatest risk of liquid entrainment is at the bottom part of the distillation column. This is because that鈥檚 where the vapor is generated and starts its upward journey. The higher density of vapor at the bottom consequently increases the risk of entraining liquid droplets.
02

Part (b) - Calculation of the minimum column diameter

Using the given maximum vapor velocity of 5.0 ft/s and the molar flow rate of 25.0 lb-mole/min, we can calculate the cross-sectional area needed using the formula \(A = \dfrac{Q}{V}\), where Q is the molar flow rate and V is the velocity, which gives us \(A = \dfrac{25.0 \text{ lb-mole/min}}{5.0 \text{ ft/s}}\), and converting lb-mole/min to ft鲁/s because \(1 \text{ lb-mole/min} = 0.37948 \text{ ft鲁/s}\), we obtain \(A = \dfrac{25.0 \times 0.37948 \text{ ft鲁/s}}{5.0 \text{ ft/s}} = 1.898 \text{ ft虏}\). The column diameter can be then calculated from the area using the relation \(d = \sqrt{4A/\pi}\). Thus \(d = \sqrt{\dfrac{4\times1.898 \text{ ft虏}}{\pi}}\) which results in \(d = 1.56 \text{ ft}\), rounded to the nearest two decimal places.
03

Part (c) - Analysis with diameter \(10\%\) greater and estimation of vapor molar flow rate increase

If the column diameter is \(10\%\) greater than calculated in part (b), the new diameter will be \(1.10 \times d = 1.10 \times 1.56 \text{ ft} = 1.716 \text{ ft}\). With this new diameter, the vapor velocity both at the top and the bottom of the column is recalculated with \(v = \dfrac{Q}{A} = \dfrac{25.0\times0.37948 \text{ ft鲁/s}}{\pi \times (d/2)^2}\) and this gives us \(v = 4.05 \text{ ft/s}\). So, the vapor molar flow rate can be increased up to \(5.0 \text{ ft/s}\), therefore new flowrate \(Q' = A\times V' = \pi \times (d/2)^2 \times 5.0\), substituting the values and solving gives us \(Q' = 31.01 \text{ lb-mole/min}\), rounded to the nearest two decimal places.
04

Part (d) - Determination of the new pressure with vapor molar flow rate doubled

To keep the vapor velocity under 5.0 ft/s while doubling the vapor molar flow rate, the new pressure would need to increase proportionally as well, so it would essentially be double the original pressure. The exact value can't be determined as the original pressure isn't given in the problem but it will be double whatever that initial pressure was.

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Key Concepts

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

Column Diameter Calculation
In distillation processes, one of the critical aspects is ensuring the distillation column has an appropriate diameter. This is important for maintaining the desired vapor velocity and prevents liquid entrainment, which is when liquid droplets are suspended and carried away by the vapor stream. To calculate the diameter, we start with the formula for cross-sectional area \[ A = \frac{Q}{V} \]where \(Q\) is the molar flow rate in cubic feet per second, and \(V\) is the vapor velocity in feet per second. After calculating the area, the diameter \(d\) is derived from \[ d = \sqrt{\frac{4A}{\pi}}. \]This calculation ensures that with a vapor flow rate of 25 lb-mole/min and a maximum vapor velocity of 5 ft/s, the column is designed to handle the flow without risking liquid entrainment. A well-sized column helps maintain efficient separation processes and operational safety in the distillation system.
Vapor Velocity
Vapor velocity within a column is a critical factor in the design and operation of a distillation column. The velocity determines how efficiently the vapor can rise without entraining liquid droplets. The target here is to keep the vapor velocity below a threshold 鈥� 5 feet per second in this case 鈥� to minimize entrainment. If the velocity is too high, it increases the likelihood of carrying liquid droplets upward with the vapor, thus interfering with the separation process and reducing product purity.
Understanding vapor velocity not only involves measuring the speed of vapor but also requires careful adjustments based on calculated flow rates and column diameter. Maintaining optimal velocity assures the performance of the distillation column remains efficient and effective, aligning with desired outcomes such as purity levels and yield.
Liquid Entrainment
Liquid entrainment refers to the phenomenon where liquid droplets get carried away by the vapor stream because of high vapor velocities. This issue is predominantly a concern at the bottom of the distillation column where the vapor is generated, having a higher density and energy state. This density increases the risk of entraining the liquid, leading to inefficiencies in the separation process.
To mitigate entrainment, engineers calculate appropriate column diameters and vapor velocities, balancing them to ensure there's limited energy and space for liquid particles to be lifted by the vapor. Choosing the right internals and monitoring process conditions can further minimize entrainment risks, sustaining process integrity and improving output quality.
Separation Process
The separation process in a distillation column relies heavily on the principles of phase equilibrium. The purpose of this process is to separate components based on their volatilities, usually operating between varying temperatures from top to bottom of the column. For example, methanol and water are separated effectively due to their differing boiling points.
Ensuring efficient separation involves maintaining specific vapor velocities and column dimensions. Each of these factors contributes to the balance of phases, thus holding the components within the desired purity range. It's a precise operation that requires continuous adjustments and monitoring to accommodate flow rate changes and maintain consistency in separation performance.
Pressure Adjustment in Column
Adjusting the pressure within a distillation column allows for variations in vapor flow rate while maintaining control over vapor velocity. In scenarios where increased throughput is necessary, such as doubling the vapor flow rate, pressure adjustments can offset potential increases in vapor velocity that lead to liquid entrainment.
This pressure adjustment is directly linked to the ideal gas law, where increasing pressure can help to manage higher molar flow rates without exceeding velocity limits. Engineers must consider the current operating pressure and make proportional changes to accommodate new process demands while ensuring operational safety and efficiency.

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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.

Many references give the specific gravity of gases with reference to air. For example, the specific gravity of carbon dioxide is 1.53 relative to air at the same temperature and pressure. Show that this value is correct as long as the ideal-gas equation of state applies.

The van der Waals equation of state (Equation \(5.3-7\) ) is to be used to estimate the specific molar volume \(\hat{V}(\mathrm{L} / \mathrm{mol})\) of air at specified values of \(T(\mathrm{K})\) and \(P(\mathrm{atm}) .\) The van der Waals constants for air are \(a=1.33 \mathrm{atm} \cdot \mathrm{L}^{2} / \mathrm{mol}^{2}\) and \(b=0.0366 \mathrm{L} / \mathrm{mol}\) (a) Show why the van der Waals equation is classified as a cubic equation of state by expressing it in the form $$f(\hat{V})=c_{3} \hat{V}^{3}+c_{2} \hat{V}^{2}+c_{1} \hat{V}+c_{0}=0$$ where the coefficients \(c_{3}, c_{2}, c_{1},\) and \(c_{0}\) involve \(P, R, T, a,\) and \(b .\) Calculate the values of these coefficients for air at \(223 \mathrm{K}\) and 50.0 atm. (Include the units when giving the values.) (b) What would the value of \(\hat{V}\) be if the ideal-gas equation of state were used for the calculation? Use this value as an initial estimate of \(\tilde{V}\) for air at \(223 \mathrm{K}\) and 50.0 atm and solve the van der Waals equation using Goal Seek or Solver in Excel. What percentage error results from the use of the ideal-gas equation of state, taking the van der Waals estimate to be correct? (c) Set up a spreadsheet to carry out the calculations of Part (b) for air at \(223 \mathrm{K}\) and several pressures. The spreadsheet should appear as follows: The polynomial expression for \(\hat{V}\left(f=c_{3} \hat{V}^{3}+c_{2} \hat{V}^{2}+\cdots\right)\) should be entered in the \(f(V)\) column, and the value in the \(V\) column should be determined using Goal Seek or Solver in Excel.

A stream of air enters a \(7.50-\mathrm{cm}\) ID pipe with a velocity of \(60.0 \mathrm{m} / \mathrm{s}\) at \(27^{\circ} \mathrm{C}\) and 1.80 bar (gauge). At a point downstream, the air flows through a \(5.00 \mathrm{cm}\) ID pipe at \(60^{\circ} \mathrm{C}\) and 1.53 bar (gauge). What is the average velocity of the gas at this point.

A 150 -liter cylinder of carbon monoxide is stored in a \(30.7-\mathrm{m}^{3}\) room. The pressure gauge on the tank reads 2500 psi when the tank is delivered. Sixty hours later the gauge reads 2245 psi. The Threshold Limit Value Ceiling (TLV-C) molar concentration of CO-that is, the concentration considered unsafe for even instantaneous human exposure-is \(200 \mathrm{ppm}\left(200 \times 10^{-6} \mathrm{mol} \mathrm{CO} / \mathrm{mol} \text { room air). }^{24}\right.\) The temperature of the room is constant at \(27^{\circ} \mathrm{C}\). (a) The decrease in pressure is a source of concern, but it may have resulted from a reduction in the temperature of the tank as it was transported from the loading dock to the air-conditioned laboratory. Without assuming that the gas behaves ideally, show that this is unlikely to be the case. (b) Having determined that the pressure decrease must be due to a leak, estimate the average leak rate (mol CO/h), again without assuming that the gas behaves ideally. (c) Calculate \(t_{\min }(\mathrm{h}),\) the minimum time from delivery at which the average concentration of \(\mathrm{CO}\) in the room could have reached the TLV-C concentration. Explain why the actual time to reach this concentration would be greater. (d) Why could it be disastrous to enter the room at any time without wearing proper personal protective equipment, even at a time \(t
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