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A stream consisting of 44.6 mole \(\%\) benzene and \(55.4 \%\) toluene is fed at a constant rate to a process unit that produces two product streams, one a vapor and the other a liquid. The vapor flow rate is initially zero and asymptotically approaches half of the molar flow rate of the feed stream. Throughout this entire period, no material accumulates in the unit. When the vapor flow rate has become constant, the liquid is analyzed and found to be 28.0 mole\% benzene. (a) Sketch a plot of liquid and vapor flow rates versus time from startup to when the flow rates become constant. (b) Is this process batch or continuous? Is it transient or steady-state before the vapor flow rate reaches its asymptotic limit? What about after it becomes constant? (c) For a feed rate of 100 mol/min, draw and fully label a flowchart for the process after the vapor flow rate has reached its limiting value, and then use balances to calculate the molar flow rate of the liquid and the composition of the vapor in mole fractions.

Short Answer

Expert verified
The process is continuous and moves from transient to steady state. The vapor flow rate of benzene can be calculated using mass balance equations. The mole fraction of toluene in vapor and liquid streams is calculated as the difference between 1 and the mole fraction of benzene in the respective streams.

Step by step solution

01

Identify the nature of the process

The process starts with a single input feed and divides it into two product streams. One stream is vapor and the other is liquid. There are no losses, conversion or accumulation which meets the criteria of a splitting process.
02

Understanding of batch & continuous process

In a batch process, material is processed in definite batches and at the end of processing, the entire batch is removed from the equipment. Also, batch processing typically involves time variation. On the other hand, in a continuous process, the feed and products are continuously added and taken from the unit respectively. Here, since the operation is performed at a constant feed rate, the process is continuous.
03

Understanding of transient & steady-state process

In a transient state, the process variables change with time until the steady state is reached. Once steady state is achieved, variables like flow rate, concentration, temperature do not change with time. Here, the vapor flow rate initially varies and then eventually becomes constant indicating that the process moves from transient to steady state.
04

Draw a plot of liquid and vapor flow rates versus time

At startup, the vapor flow rate is zero and it increases over time eventually becoming a constant value, which is half of the molar feed rate. On the other hand, since the total matter does not accumulate in the system, any increase in the vapor flow rate must be counteracted by a reduction in the liquid flow rate. Therefore, at startup, the liquid flow rate is equal to the feed rate and it decreases until it reaches its steady state value, which is also half of the molar feed rate.
05

Mass balance calculation

Perform a mass balance calculation for both benzene and toluene to determine the molar flow rate and composition of the vapor product. For benzene, it is given by \( 0.446 * Feed Rate = Vapor Flow * Vapor mole fraction + Liquid Flow * Liquid mole fraction \). Similarly, a balance for toluene can be calculated. Substituting feed rate as 100 mol/min, vapor flow rate as 50 mol/min and liquid mole fraction of benzene as 0.280 in the above mentioned equation enables us to solve for the vapor mole fraction of benzene.
06

Calculation of toluene in vapor and liquid streams

As the process involves only benzene and toluene, the sum of mole fractions in every stream should be equal to 1. Therefore, the mole fraction of toluene in vapor and liquid streams can be calculated by subtracting the respective mole fractions of benzene from 1.

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Key Concepts

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

Mass Balance
In chemical process engineering, performing a mass balance is critical for analyzing how stream compositions and flow rates interact in a system. Mass balance involves accounting for the intake, distribution, and exit of materials. In the exercise, a mass balance focuses on a feed stream being split into vapor and liquid streams. Applying this concept ensures that the sum of the input equals the sum of the outputs since no accumulation occurs within the unit. For example, the mass flow rate of benzene and toluene entering the system must match the combined rates leaving in vapor and liquid forms. A mass balance equation might look like this for benzene:
  • \( 0.446 \, \text{mole fraction} \times \text{Feed Rate} = \text{Vapor Flow Rate} \times \text{Vapor mole fraction} + \text{Liquid Flow Rate} \times \text{Liquid mole fraction} \).
This equation ensures all molecules in the process are accounted for and transitions the understanding toward the equilibrium and split behavior of these substances.
Transient State
The transient state in a chemical process refers to the period during which variables—such as flow rate or concentration—change with time. This phase occurs before reaching steady state, highlighting the system's gradual change towards equilibrium. In our exercise, the process starts with a newly introduced feed. Initially, no vapor is present, marking a transient state. Consequently, the vapor flow rate will rise from zero until it stabilizes. During this time, you observe shifting compositions and flow rates that create dynamic conditions. Understanding transient conditions is essential for predicting how a system reaches its final output or equilibrium state, accommodating any operational variability and ensuring system stability.
Steady State
Thriving at steady state is the goal for many continuous processes in chemical engineering. Once reached, steady state means that key variables remain constant over time. In the context of the exercise, once the vapor flow rate becomes constant, so do other variables such as composition and flow rates of the liquid and vapor outlets. At this point, one can consistently predict outputs, optimizing the process for efficiency and desired product quality. Steady state greatly simplifies calculations and control since introductions and removals balance out, leading to stable operation irrespective of the complex dynamics that may have preceded it.
Continuous Process
Continuous processes are characterized by the uninterrupted feeding of materials and extraction of products. In contrast to batch processes, which require stopping and starting, continuous processes maintain a state of equilibrium. In this exercise, the feed is introduced to the system at a constant rate, indicating a continuous process. By maintaining consistent input and output levels, downtime is minimized, allowing for efficient material use and consistent product quality. Continuous processes are ideal for large-scale production due to their ability to manage sustained operations with less intervention.
Vapor-Liquid Equilibrium
Vapor-liquid equilibrium (VLE) is a crucial concept for understanding the distribution of compound components between vapor and liquid phases at given conditions. In the context of the exercise, VLE principles help define how benzene and toluene distribute themselves between the vapor and liquid streams.
  • In a state of equilibrium, at a given temperature and pressure, the chemical potential of each component is equal in both phases.
  • This means that the concentrations in the liquid and vapor phases become stable.
Understanding VLE allows engineers to predict component behavior, facilitating the design of efficient separation and purification processes. The composition in each phase is important for calculating mole fractions of the liquid and vapor streams, making VLE analysis integral to process optimization.

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Most popular questions from this chapter

A mixture of 75 mole \(\%\) methane and 25 mole \(\%\) hydrogen is burned with \(25 \%\) excess air. Fractional conversions of \(90 \%\) of the methane and \(85 \%\) of the hydrogen are achieved; of the methane that reacts, \(95 \%\) reacts to form \(\mathrm{CO}_{2}\) and the balance reacts to form CO. The hot combustion product gas passes through a boiler in which heat transferred from the gas converts boiler feedwater into steam. (a) Calculate the concentration of \(\mathrm{CO}\) (ppm) in the stack gas. (b) The CO in the stack gas is a pollutant. Its concentration can be decreased by increasing the percent excess air fed to the furnace. Think of at least two costs of doing so. (Hint: The heat released by the combustion goes into heating the combustion products; the higher the combustion product temperature, the more steam is produced.)

Methanol is formed from carbon monoxide and hydrogen in the gas-phase reaction The mole fractions of the reactive species at equilibrium satisfy the relation where \(P\) is the total pressure (atm), \(K_{c}\) the reaction equilibrium constant (atm \(^{-2}\) ), and \(T\) the temperature (K). The equilibrium constant \(K_{c}\) equals 10.5 at 373 K, and \(2.316 \times 10^{-4}\) at \(573 \mathrm{K}\). A semilog plot of \(K_{\mathrm{c}}\) (logarithmic scale) versus 1/ \(T\) (rectangular scale) is approximately linear between \(T=300 \mathrm{K}\) and \(T=600 \mathrm{K}\) (a) Derive a formula for \(K_{\mathrm{c}}(T),\) and use it to show that \(K_{\mathrm{e}}(450 \mathrm{K})=0.0548 \mathrm{atm}^{-2}\) (b) Write expressions for \(n_{A}, n_{B},\) and \(n_{C}\) (gram-moles of each species), and then \(y_{A}, y_{B},\) and \(y_{C},\) in terms of \(n_{\mathrm{A} 0}, n_{\mathrm{B} 0}, n_{\mathrm{C} 0},\) and \(\xi,\) the extent of reaction. Then derive an equation involving only \(n_{\mathrm{A} 0}, n_{\mathrm{B} 0}, n_{\mathrm{C} 0}, P, T,\) and \(\xi_{e},\) where \(\xi_{e}\) is the extent of reaction at equilibrium. (c) Suppose you begin with equimolar quantities of CO and \(\mathrm{H}_{2}\) and no \(\mathrm{CH}_{3} \mathrm{OH}\), and the reaction proceeds to equilibrium at 423 K and 2.00 atm. Calculate the molar composition of the product ( \(y_{\mathrm{A}}\), \(\left.y_{\mathrm{B}}, \text { and } y_{\mathrm{C}}\right)\) and the fractional conversion of \(\mathrm{CO}\) (d) The conversion of CO and \(\mathrm{H}_{2}\) can be enhanced by removing methanol from the reactor while leaving unreacted CO and \(\mathrm{H}_{2}\) in the vessel. Review the equations you derived in solving Part (c) and determine any physical constraints on \(\xi_{c}\) associated with \(n_{\mathrm{A} 0}=n_{\mathrm{B} 0}=1\) mol. Now suppose that 90\% of the methanol is removed from the reactor as it is produced; in other words, only 10\% of the methanol formed remains in the reactor. Estimate the fractional conversion of CO and the total gram moles of methanol produced in the modified operation. (e) Repeat Part (d), but now assume that \(n_{\mathrm{B} 0}=2\) mol. Explain the significant increase in fractional conversion of CO. (f) Write a set of equations for \(y_{\mathrm{A}}, y_{\mathrm{B}}, y_{\mathrm{C}},\) and \(f_{\mathrm{A}}\) (the fractional conversion of \(\mathrm{CO}\) ) in terms of \(y_{\mathrm{A} 0}, y_{\mathrm{B} 0}, T,\) and \(P(\) the reactor temperature and pressure at equilibrium). Enter the equations in an equation-solving program. Check the program by running it for the conditions of Part (c), then use it to determine the effects on \(f_{\mathrm{A}}\) (increase, decrease, or no effect) of separately increasing, (i) the fraction of \(\mathrm{CH}_{3} \mathrm{OH}\) in the feed, (ii) temperature, and (iii) pressure.

A liquid mixture contains \(60.0 \mathrm{wt} \%\) ethanol \((\mathrm{E}), 5.0 \mathrm{wt} \%\) of a dissolved solute \((\mathrm{S}),\) and the balance water. A stream of this mixture is fed to a continuous distillation column operating at steady state. Product streams emerge at the top and bottom of the column. The column design calls for the product streams to have equal mass flow rates and for the top stream to contain 90.0 wt\% ethanol and no S. (a) Assume a basis of calculation, draw and fully label a process flowchart, do the degree-of-freedom analysis, and verify that all unknown stream flows and compositions can be calculated. (Don't do any calculations yet.) (b) Calculate (i) the mass fraction of \(S\) in the bottom stream and (ii) the fraction of the ethanol in the feed that leaves in the bottom product stream (i.e., \(\mathrm{kg} \mathrm{E}\) in bottom stream/kg \(\mathrm{E}\) in feed) if the process operates as designed. (c) An analyzer is available to determine the composition of ethanol-water mixtures. The calibration curve for the analyzer is a straight line on a plot on logarithmic axes of mass fraction of ethanol, \(x\) (kg E/kg mixture), versus analyzer reading, \(R\). The line passes through the points \((R=15, x=\) 0.100) and \((R=38, x=0.400)\). Derive an expression for \(x\) as a function of \(R(x=\cdots\) ) based on the calibration, and use it to determine the value of \(R\) that should be obtained if the top product stream from the distillation column is analyzed. (d) Suppose a sample of the top stream is taken and analyzed and the reading obtained is not the one calculated in Part (c). Assume that the calculation in Part (c) is correct and that the plant operator followed the correct procedure in doing the analysis. Give five significantly different possible causes for the deviation between \(R_{\text {measured and }} R_{\text {prediced }}\), including several assumptions made when writing the balances of Part (c). For each one, suggest something that the operator could do to check whether it is in fact the problem.

An evaporation-crystallization process of the type described in Example \(4.5-2\) is used to obtain solid potassium sulfate from an aqueous solution of this salt. The fresh feed to the process contains 19.6 wt\% \(\mathrm{K}_{2} \mathrm{SO}_{4}\). The wet filter cake consists of solid \(\mathrm{K}_{2} \mathrm{SO}_{4}\) crystals and a \(40.0 \mathrm{wt} \% \mathrm{K}_{2} \mathrm{SO}_{4}\) solution, in a ratio \(10 \mathrm{kg}\) crystals/kg solution. The filtrate, also a \(40.0 \%\) solution, is recycled to join the fresh feed. Of the water fed to the evaporator, 45.0\% is evaporated. The evaporator has a maximum capacity of 175 kg water evaporated/s. (a) Assume the process is operating at maximum capacity. Draw and label a flowchart and do the degree-of-freedom analysis for the overall system, the recycle-fresh feed mixing point, the evaporator, and the crystallizer. Then write in an efficient order (minimizing simultaneous equations) the equations you would solve to determine all unknown stream variables. In each equation, circle the variable for which you would solve, but don't do the calculations. (b) Calculate the maximum production rate of solid \(\mathrm{K}_{2} \mathrm{SO}_{4}\), the rate at which fresh feed must be supplied to achieve this production rate, and the ratio kg recycle/kg fresh feed. (c) Calculate the composition and feed rate of the stream entering the crystallizer if the process is scaled to 75\% of its maximum capacity. (d) The wet filter cake is subjected to another operation after leaving the filter. Suggest what it might be. Also, list what you think the principal operating costs for this process might be. (e) Use an equation-solving computer program to solve the equations derived in Part (a). Verify that you get the same solutions determined in Part (b).

\- An equimolar liquid mixture of benzene and toluene is separated into two product streams by distillation. A process flowchart and a somewhat oversimplified description of what happens in the process follow: Inside the column a liquid stream flows downward and a vapor stream rises. At each point in the column some of the liquid vaporizes and some of the vapor condenses. The vapor leaving the top of the column, which contains 97 mole\% benzene, is completely condensed and split into two equal fractions: one is taken off as the overhead product stream, and the other (the reflux) is recycled to the top of the column. The overhead product stream contains \(89.2 \%\) of the benzene fed to the column. The liquid leaving the bottom of the column is fed to a partial reboiler in which \(45 \%\) of it is vaporized. The vapor generated in the reboiler (the boilup) is recycled to become the rising vapor stream in the column, and the residual reboiler liquid is taken off as the bottom product stream. The compositions of the streams leaving the reboiler are governed by the relation $$\frac{y_{\mathrm{B}} /\left(1-y_{\mathrm{B}}\right)}{x_{\mathrm{B}} /\left(1-x_{\mathrm{B}}\right)}=2.25$$ where \(y_{\mathrm{B}}\) and \(x_{\mathrm{B}}\) are the mole fractions of benzene in the vapor and liquid streams, respectively. (a) Take a basis of 100 mol fed to the column. Draw and completely label a flowchart, and for each of four systems (overall process, column, condenser, and reboiler), do the degree-of-freedom analysis and identify a system with which the process analysis might appropriately begin (one with zero degrees of freedom). (b) Write in order the equations you would solve to determine all unknown variables on the flowchart, circling the variable for which you would solve in each equation. Do not do the calculations in this part. (c) Calculate the molar amounts of the overhead and bottoms products, the mole fraction of benzene in the bottoms product, and the percentage recovery of toluene in the bottoms product \((100 \times\) moles toluene in bottoms/mole toluene in feed).

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