91Ó°ÊÓ

A liquid mixture of acetone and water contains 35 mole\% acetone. The mixture is to be partially evaporated to produce a vapor that is 75 mole \(\%\) acetone and leave a residual liquid that is 18.7 mole \(\%\) (a) Suppose the process is to be carried out continuously and at steady state with a feed rate of 10.0 kmol/h. Let \(\dot{n}_{\mathrm{v}}\) and \(\dot{n}_{1}\) be the flow rates of the vapor and liquid product streams, respectively. Draw and label a process flowchart, then write and solve balances on total moles and on acetone to determine the values of \(\dot{n}_{\mathrm{v}}\) and \(\dot{n}_{\mathrm{l}}\). For each balance, state which terms in the general balance equation (accumulation \(=\)input \(+\)generation \(-\)output\(-\)consumption ) can be discarded and why. (See Example 4.2-2.) (b) Now suppose the process is to be carried out in a closed container that initially contains 10.0 kmol of the liquid mixture. Let \(n_{\mathrm{v}}\) and \(n_{1}\) be the moles of final vapor and liquid phases, respectively. Draw and label a process flowchart, then write and solve integral balances on total moles and on acetone. For each balance, state which terms of the general balance equation can be discarded and why. (c) Returning to the continuous process, suppose the vaporization unit is built and started and the product stream flow rates and compositions are measured. The measured acetone content of the vapor stream is 75 mole \(\%\) acetone, and the product stream flow rates have the values calculated in Part (a). However, the liquid product stream is found to contain 22.3 mole \(\%\) acetone. It is possible that there is an error in the measured composition of the liquid stream, but give at least five other reasons for the discrepancy. [Think about assumptions made in obtaining the solution of Part (a).]

Step by step solution

01

Determine Flow Rates for Steady State Continuous Process

First note that in a steady state process, there is no accumulation. Also, there is no generation or consumption of acetone - it is merely being redistributed between the vapor and liquid phases. So the balance equation simplifies to input = output. Let \( x_{f} = 0.35 \) be the mole fraction of acetone in the feed, \( x_{v} = 0.75 \) in the vapor, and \( x_{l} = 0.187 \) in the liquid. Write an overall mole balance for the entire process and an acetone balance. Then solve these two equations to find \( \dot{n}_v \) and \( \dot{n}_l \).

02

Determine Mole Fractions for Batch Process

Now consider this process occurs in a closed container, which means \( n_v \) and \( n_l \) are dependent upon the initial amount of mixture in the container. Therefore, the principle of the mole balance still applies but now the general balance equation simplifies to initial = final. Draw a similar process flowchart as in step 1 and set up integral balances on total moles and on acetone, discarding the terms for input and output. Then solve these equations to find \( n_v \) and \( n_l \).

03

Discuss Possible Discrepancies

Given a scenario where the vaporization unit is built and the results are inconsistent with calculations performed in part (a), identify potential sources for discrepancies. Factors to consider include: measurement errors, assumption of accurate feed composition, neglect of potential reactions, assumption of ideal phases, equipment inefficiencies, and potential presence of other components in the mixture.

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Key Concepts

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

Mole Fraction

In chemistry and process engineering, understanding the concept of mole fraction is crucial because it helps describe the composition of mixtures. The mole fraction of a component in a mixture is defined as the ratio of the number of moles of that component to the total number of moles of all components in the mixture. It's a dimensionless quantity often represented by the symbol \( x \), like \( x_f \) for feed, \( x_v \) for vapor, and \( x_l \) for liquid, in our exercise.

To calculate the mole fraction, you use the formula:
\[ x_i = \frac{n_i}{ ext{total moles}} \]
where \( n_i \) is the moles of the component of interest.

In our example, the feed mixture starts with 35 mole \( \% \) acetone, which means the mole fraction of acetone in the feed is 0.35. Mole fractions are useful because they remain constant across various conditions like pressure and temperature, making them especially important in analyses of chemical processes like vaporization.

Steady State

Steady state is a concept used to describe a condition where the variables (e.g., chemical concentrations, temperatures) remain constant over time in a system, even though there might be the flow of material through the system. In a steady state process, such as in part (a) of our exercise, there is no accumulation of mass within the system because input exactly equals output.

In terms of a balance equation, this means:
\[ ext{accumulation} = ext{input} - ext{output} + ext{generation} - ext{consumption} = 0 \]

Because we have no generation or consumption of acetone, it's about redistribution into vapor and liquid streams without any loss or gain within the system.

• No change in the amount stored, so steady conditions prevail.
• Simplifies calculations as we can focus directly on input equals output for each component.

Applying this to our exercise, both total mole balance and acetone mole balance simplify, helping us find flow rates efficiently.

Vaporization Process

Vaporization is the process where a liquid is converted into vapor. A common example is boiling water to produce steam. In industry, vaporization processes are used to separate components based on their volatility. The vaporization process in the exercise involves acetone and water, where control over the composition of vapor and residual liquid is necessary.

Critical factors include:

• Volatility: The more volatile component will occupy a higher mole fraction in the vapor phase.
• Energy Input: Adequate heat must be supplied to enable the phase change.
• Process Design: Equipment like distillation columns can enhance separation efficiency.

In the exercise, the vapor should contain 75 mole \( \% \) acetone, showcasing selective evaporation. Understanding vapor-liquid equilibrium can optimize such processes, indicating the extent to which components will separate under given conditions.

Batch Process

In a batch process, the operation is executed in a closed system where materials are added at the start, and the process runs without adding or removing material, except through the operation itself until the completion. It's characterized by pre-determined processing time and volume.

This differs from continuous processes in configuration and operation:

• No Flow: Since flow in or out isn't continuous, the system behaves as a self-contained unit.
• Initial to Final: You must determine the state change from the initial condition to the final state. It's often expressed as "initial = final" in mole balances.
• Flexibility: Easier to adjust conditions for small-scale production.

In part (b) of the exercise, this involves recalculating for a static mass within the container, making integral (rather than differential) balance equations suitable. The approach requires understanding the total closed system dynamics, ensuring proper redistribution during the vaporization phase.