/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 41 An important parameter in the de... [FREE SOLUTION] | 91Ó°ÊÓ

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An important parameter in the design of gas absorbers is the ratio of the flow rate of the feed liquid to that of the feed gas. The lower the value of this ratio, the lower the cost of the solvent required to process a given quantity of gas but the taller the absorber must be to achieve a specified separation.Propane is recovered from a 7 mole \(\%\) propane \(-93 \%\) nitrogen mixture by contacting the mixture with liquid \(n\) -decane. An insignificant amount of decane is vaporized in the process, and \(98.5 \%\) of the propane entering the unit is absorbed.(a) The highest possible propane mole fraction in the exiting liquid is that in equilibrium with the propane mole fraction in the feed gas (a condition requiring an infinitely tall column). Using Raoult's law to relate the mole fractions of propane in the feed gas and liquid, calculate the ratio \(\left(\dot{n}_{L_{1}} / \dot{n}_{G_{2}}\right)\) corresponding to this limiting condition.(b) Suppose the actual feed ratio \(\left(\dot{n}_{L_{1}} / \dot{n}_{G_{2}}\right)\) is 1.2 times the value calculated in Part (a) and the percentage of the entering propane absorbed is the same (98.5\%). Calculate the mole fraction of propane in the exiting liquid.(c) What are the costs and benefits associated with increasing \(\left(\dot{n}_{L_{1}} / \dot{n}_{G_{2}}\right)\) from its minimum value [the value calculated in Part (a)]? What would you have to know to determine the most cost-effective value of this ratio?

Short Answer

Expert verified
The calculation based on Raoult's law for part a provides the highest possible mole fraction \(x_{i}\) in the exiting liquid under equilibrium conditions. With the given actual feed ratio 1.2 times the value calculated in part a, we can calculate the mole fraction of propane in the exiting liquid for part b. In part c, costs and benefits would have to be evaluated in context of increased ratio of the flow rate of the feed liquid to that of the feed gas. For a comprehensive cost assessment, we need specific cost derivatives like solvent cost and construction costs.

Step by step solution

01

Apply Raoult's Law for Limiting Condition

Raoult's law is given by \(y_{i}P = x_{i}P_{i}^{sat}\), where \(y_{i}\) is the mole fraction of propane in the gas phase, \(P\) is the total pressure, \(x_{i}\) is the mole fraction in the liquid phase, and \(P_{i}^{sat}\) is the saturation pressure of propane. Given that the mole fraction in the feed gas is 7%, apply the law to solve for \(x_{i}\), the mole fraction in the liquid.
02

Calculate the Ratio

Given that all the propane in the absorbed gas (7% mole fraction) goes to the liquid phase and becomes \(x_{i}\) mole fraction after absorption, the actual feed ratio \(\left(\dot{n}_{L_{1}} / \dot{n}_{G_{2}}\right)\) can be calculated using the quantities of propane in the gas \(y_{i}\) (7%) and in the liquid \(x_{i}\).
03

Calculate the Mole Fraction for Part b

For Part (b), it is stated that the actual feed ratio is 1.2 times the value calculated in Part (a), with the same percentage of propane absorbed. To find the new mole fraction of propane in the exiting liquid, multiply the propane absorbed by 1.2.
04

Evaluate Costs and Benefits

In Part (c), we need to evaluate how changing the ratio from its minimum value affects the overall cost and efficiency. It is known that as the ratio increases, the height of the tank (and thus the cost) decreases. However, the cost of solvent also increases. To determine the most cost-effective value, we need information such as the cost of the solvent, the cost of tank construction and operation, and the market value of propane.

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Key Concepts

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

Raoult's Law
Raoult's Law is a fundamental principle in thermodynamics that relates the partial pressure of a component in a gas phase to its mole fraction in a liquid phase. This law is crucial in processes such as gas absorption, where we separate components through contact with a liquid.
In essence, Raoult's Law expresses that for an ideal solution, the partial pressure of a component (\(y_i P\)) is equal to the mole fraction of that component in the liquid phase (\(x_i\)) multiplied by the component's saturation pressure (\(P_i^{sat}\)). This relationship forms the basis of calculating the distribution of propane between gaseous and liquid phases in our gas absorber exercise.
To apply Raoult's Law effectively, we first need to understand the composition of the starting gas mixture. In this exercise, we are dealing with a 7% propane mixture, which helps us determine \(x_i\) once we have \(y_i\) and \(P_i^{sat}\). This relationship helps calculate the limiting condition, where the column must be infinitely tall for completion.
A practical application: the higher the propane mole fraction we achieve in the liquid by applying Raoult's Law, the more efficient the absorber becomes.
Solvent Cost Evaluation
When considering gas absorbers, solvent cost evaluation becomes a key factor. Simply put, solvent cost evaluation is assessing the expenses associated with using a specific liquid to capture and separate gases.
This involves not just the direct purchase price of the solvent, but also indirect costs like transportation, storage, and disposal. Additionally, the choice of solvent impacts the overall operational costs, such as energy consumption for heating or cooling to achieve optimal absorption.
Incorporating solvent costs into our decision-making is essential for determining the most economical approach. While a more efficient solvent might improve absorber performance, it might also come at a high cost, affecting the project's overall budget.
  • Initial solvent purchase and replenishment
  • Energy costs for regulating absorber temperature
  • Maintenance and handling
Evaluating these costs against the rates of absorption and column efficiency gives a clearer picture of the best practice. The exercise hints at this by suggesting adjustments to the feed ratio, which could save costs but require a critical balance between solvent expense and column efficiency.
Choosing the right solvent and optimizing its use is a continuous puzzle of cost-effectiveness and performance.
Absorption Efficiency
Absorption efficiency in a gas absorber refers to the percentage of a target gas component that is successfully captured and removed from a gas mixture during the absorption process.
This efficiency is an indicator of how effective a particular absorber is in separating gases. In the provided exercise, 98.5% of the entering propane is absorbed, indicating high absorption efficiency.
Several factors directly influence absorption efficiency:
  • Solvent selection: The properties of the solvent, such as its affinity for the target gas, are critical.
  • Operating conditions: Temperature and pressure settings must be optimized for maximum efficiency.
  • Absorber design: The physical characteristics, such as height and surface area for contact, play a role in effectiveness.
Achieving high absorption efficiency often necessitates a trade-off between operational cost and the physical footprint of the absorber (i.e., height). A taller column may achieve better separation but at a higher cost due to materials and construction expenses.
Thus, balancing these aspects ensures that the overall performance of the gas absorbing system is optimized, with efficiency tallied against economic constraints.

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Most popular questions from this chapter

The feed to a distillation column (sketched below) is a 45.0 mole\% \(n\) -pentane- 55.0 mole\% n-hexane liquid mixture. The vapor stream leaving the top of the column, which contains 98.0 mole\% pentane and the balance hexane, goes to a total condenser (which means all the vapor is condensed). Half of the liquid condensate is returned to the top of the column as reflux and the rest is withdrawn as overhead product (distillate) at a rate of \(85.0 \mathrm{kmol} / \mathrm{h}\). The distillate contains \(95.0 \%\) of the pentane fed to the column. The liquid stream leaving the bottom of the column goes to a reboiler. Part of the stream is vaporized; the vapor is returned to the bottom of the column as boilup, and the residual liquid is withdrawn as bottoms product.(a) Calculate the molar flow rate of the feed stream and the molar flow rate and composition of the bottoms product stream. (b) Estimate the temperature of the vapor entering the condenser, assuming that it is saturated (at its dew point) at an absolute pressure of 1 atm and that Raoult's law applies to both pentane and hexane. Then estimate the volumetric flow rates of the vapor stream leaving the column and of the liquid distillate product. State any assumptions you make. (c) Estimate the temperature of the reboiler and the composition of the vapor boilup, again assuming operation at 1 atm.(d) Calculate the minimum diameter of the pipe connecting the column and the condenser if the maximum allowable vapor velocity in the pipe is \(10 \mathrm{m} / \mathrm{s}\). Then list all the assumptions underlying the calculation of that number.

A vapor mixture of \(n\) -butane (B) and \(n\) -hexane (H) contains 50.0 mole\% butane at \(120^{\circ} \mathrm{C}\) and 1.0 atm. A stream of this mixture flowing at a rate of \(150.0 \mathrm{L} / \mathrm{s}\) is cooled and compressed, causing some but not all of the vapor to condense. (Treat this process as a single-unit operation.) Liquid and vapor product streams emerge from the process in equilibrium at \(T\left(^{\circ} \mathrm{C}\right)\) and \(1100 \mathrm{mm} \mathrm{Hg}\). The vapor product contains 60.0 mole\% butane.(a) Draw and label a flowchart. Perform a degree-of-freedom analysis to show that you have enough information to determine the required final temperature ( \(T\) ), the composition of the liquid product (component mole fractions), and the molar flow rates of the liquid and vapor products from the given information and Antoine expressions for the vapor pressures \(p_{\mathrm{B}}^{*}(T)\) and \(p_{\mathrm{H}}^{*}(T) .\) Just identify the equations - for example, mole balance on butane or Raoult's law for hexane-but don't write them yet.(b) Write in order the equations you would use to determine the quantities listed in Part (a) and also the fractional condensation of hexane (mol \(\mathrm{H}\) condensed/mol \(\mathrm{H}\) fed). In each equation, circle the variable for which you would solve. Do no algebra or calculations.(c) Complete the calculations either manually or with an equation-solving program.(d) State three assumptions you made that could lead to errors in the calculated quantities.

Air at \(25^{\circ} \mathrm{C}\) and 1 atm with a relative humidity of \(25 \%\) is to be dehumidified in an adsorption column packed with silica gel. The equilibrium adsorptivity of water on silica gel is given by the expression \(^{19}\).$$X^{*}(\mathrm{kg} \text { water/ } 100 \mathrm{kg} \text { silica gel })=12.5 \frac{p_{\mathrm{H}_{2} \mathrm{O}}}{p_{\mathrm{H}_{2} \mathrm{O}}^{*}}$$ where \(p_{\mathrm{H}_{2} \mathrm{O}}\) is the partial pressure of water in the gas contacting the silica gel and \(p_{\mathrm{H}, \mathrm{O}}^{*}\) is the vapor pressure of water at the system temperature. Air is fed to the column at a rate of 1.50 L/min until the silica gel is saturated (i.e., until it reaches equilibrium with the feed air), at which point the flow is stopped and the silica gel regenerated. (a) Calculate the minimum amount of silica gel needed in the column if regeneration is to take place no more frequently than every two hours. State any assumptions you make. (b) Briefly describe this process in terms that a high school student would have no trouble understanding. (What is the process designed to do, what happens within the column, and why is regeneration of the column packing necessary?)

The solubility coefficient of a gas may be defined as the number of cubic centimeters (STP) of the gas that dissolves in \(1 \mathrm{cm}^{3}\) of a solvent under a partial pressure of 1 atm. The solubility coefficient of \(\mathrm{CO}_{2}\) in water at \(20^{\circ} \mathrm{C}\) is \(0.0901 \mathrm{cm}^{3} \mathrm{CO}_{2}(\mathrm{STP}) / \mathrm{cm}^{3} \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\). (a) Calculate the Henry's law constant in atm/mole fraction for \(\mathrm{CO}_{2}\) in \(\mathrm{H}_{2} \mathrm{O}\) at \(20^{\circ} \mathrm{C}\) from the given solubility coefficient. (b) How many grams of \(\mathrm{CO}_{2}\) can be dissolved in a \(12-\mathrm{oz}\) bottle of soda at \(20^{\circ} \mathrm{C}\) if the gas above the soda is pure \(\mathrm{CO}_{2}\) at a gauge pressure of 2.5 atm ( 1 liter \(=33.8\) fluid ounces)? Assume the liquid properties are those of water. (c) What volume would the dissolved \(C O_{2}\) occupy if it were released from solution at body temperature and pressure \(-37^{\circ} \mathrm{C}\) and 1 atm?

Using Raoult's law or Henry's law for each substance (whichever one you think appropriate), calculate the pressure and gas-phase composition (mole fractions) in a system containing a liquid that is 0.3 mole \(\% \mathrm{N}_{2}\) and 99.7 mole \(\%\) water in equilibrium with nitrogen gas and water vapor at \(80^{\circ} \mathrm{C}\).

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