/*! 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 24 Recovery and processing of vario... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Recovery and processing of various oils are important elements of the agricultural and food industries. For example, soybean hulls are removed from the beans, which are then flaked and contacted with hexane. The hexane extracts soybean oil and leaves very little oil in the residual solids. The solids are dried at an elevated temperature, and the dried solids are used to feed livestock or further processed to extract soy protein. The gas stream leaving the dryer is at \(80^{\circ} \mathrm{C}\) 1 atm absolute, and 50\% relative saturation.(a) To recover hexane, the gas leaving the dryer is fed to a condenser, which operates at 1 atm absolute. The gas leaving the condenser contains 5.00 mole \(\%\) hexane, and the hexane condensate is recovered at a rate of \(1.50 \mathrm{kmol} / \mathrm{min}\). (b) In an altemative arrangement, the gas leaving the dryer is compressed to 10.0 atm and the temperature simultancously is increased so that the relative saturation remains at \(50 \% .\) The gas then is cooled at constant pressure to produce a stream containing 5.00 mole \(\%\) hexane. Calculate the final gas temperature and the ratio of volumetric flow rates of the gas streams leaving and entering the condenser. State any assumptions you make.(c) What would you need to know to determine which of processes (a) and (b) is more cost- effective?

Short Answer

Expert verified
For part (a), the initial mole fraction of hexane in the gas is 0.0158. For part (b), the final gas temperature is 800°C and the volumetric flow rates ratio is 1/10. For part (c), a cost analysis including energy consumption for process (b) and hexane recovery cost for process (a) is needed.

Step by step solution

01

Calculating the initial mole fraction of hexane

For part (a), we first need to find the initial mole fraction of hexane in the gas. We know that the gas leaving the condenser contains 5.00 mole percent of hexane and the condensate is recovered at a rate of \(1.50 \mathrm{kmol} / \mathrm{min}\). The mole fraction of hexane is thus \( \frac{1.50 \mathrm{kmol}}{100-5} = 0.0158 \).
02

Calculating the final temperature and volumetric flow rate for scenario (b)

For part (b), the gas is compressed and heated in a way that the relative saturation remains at \(50 \%\). This implies that the mole fraction of hexane remains the same, \(0.0158\). The gas is then cooled at constant pressure. Since the mole fraction is constant, using ideal gas law at initial and final points, we find that the ratio of final to initial volume is the same as the ratio of initial to final temperature. The final temperature is thus \(80 \times 10 = 800 \degree C\). The ratio of volumetric flow rates is 1/10 since the final pressure is 10 times initial.
03

Analyzing cost-effectiveness of process (a) and (b)

For part (c), to determine the cost-effectiveness of processes (a) and (b), we would need to know the costs related to energy consumption in heating the gas to 800 degrees Celsius in the second scenario, and costs of compressing the gas to 10 atm. We would also need to know the costs related to hexane recovery in the first scenario and compare those costs.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Oil Recovery
Oil recovery is a crucial process in industries like agriculture and food. It is the technique used to extract oil from various materials, like soybeans. Imagine soybeans being processed: they are hulled, flaked, and then mixed with a solvent such as hexane.
The hexane serves as a medium to draw out nearly all the oil contained in the soybeans. This process enhances oil recovery because it ensures minimal oil remains in the waste solids.
These solids are then dried and can be put to use in different ways: for example, as feed for livestock or further processing to extract soy protein.
  • **Solvent-Based Extraction:** This method is chosen because of its efficiency in separating oil from solids.
  • **Post-Processing Use:** The residual cake, once oil-free, is not discarded but further processed for value generation.
This systematic and efficient approach ensures maximum yield and resource efficiency in oil recovery.
Gas Condensation
In the context of extracting oil, gas condensation plays a vital role, especially when dealing with solvents like hexane. After oil extraction, the residual gas containing hexane is directed towards a condenser.
This process is carried out to recover the hexane, transforming it from a vapor to a liquid form.
  • **Role of Condensation:** Condenses hexane from the vapor, allowing it to be reused or properly processed.
  • **Condensation Parameters:** Typically involves pressure changes, maintaining mole percent of substances (like 5% hexane in our setup), and achieving desired temperatures.
Condensation not only recovers valuable solvents but also plays an essential part in environmental and economic efficiency by allowing recycling and reducing emissions.
Chemical Engineering Calculations
Chemical engineering often necessitates intricate calculations to optimize processes. These calculations ensure that both oil recovery and gas condensation are as efficient as possible.
In our example, calculations are used to adjust parameters such as temperature and pressure, especially when comparing different setups for hexane recovery.
  • **Ideal Gas Law Application:** Used to link variables like temperature and pressure, giving insight into volumetric changes.
  • **Efficiency Analysis:** Calculates factors like flow rate ratios, ensuring resource and cost effectiveness.
  • **Cost-Effectiveness Evaluation:** Analysis of different process arrangements (like changing pressure or temperature) guides economic decisions.
With these calculations, chemical engineers make informed decisions, ensuring the process remains not only effective but also cost-efficient and sustainable.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The vapor pressure of an organic solvent is \(50 \mathrm{mm}\) Hg at \(25^{\circ} \mathrm{C}\) and \(200 \mathrm{mm} \mathrm{Hg}\) at \(45^{\circ} \mathrm{C}\). The solvent is the only species in a closed flask at \(35^{\circ} \mathrm{C}\) and is present in both liquid and vapor states. The volume of gas above the liquid is \(150 \mathrm{mL}\). (a) Estimate the amount of the solvent \((\mathrm{mol})\)contained in the gas phase. (b) What assumptions did you make? How would your answer change if the species dimerized (one molecule results from two molecules of the species combining)?

An air conditioner is designed to bring \(10,000 \mathrm{ft}^{3} / \mathrm{min}\) of outside air \(\left(90^{\circ} \mathrm{F}, 29.8 \text { inches } \mathrm{Hg} .88 \%\right.\) relative humidity) to \(40^{\circ} \mathrm{F}\), thereby condensing a portion of the water vapor, and then to reheat the air before releasing it into a room at \(65^{\circ} \mathrm{F}\). Calculate the rate of condensation (gallons \(\mathrm{H}_{2} \mathrm{O} / \mathrm{min}\) ) and the volumetric flow rate of the air delivered to the room. (Suggestion: On the flowchart, treat the coolingcondensation and the reheating as separate process steps.)

In-Hexane is used to extract oil from soybeans. (See Problem 6.24 .) The solid residue from the extraction unit, which contains 0.78 kg liquid hexane/kg dry solids, is contacted in a dryer with nitrogen that enters at \(85^{\circ} \mathrm{C}\). The solids leave the dryer containing \(0.05 \mathrm{kg}\) liquid hexane/kg dry solids, and the gas leaves the dryer at \(80^{\circ} \mathrm{C}\) and 1.0 atm with a relative saturation of \(70 \% .\) The gas is then fed to a condenser in which it is compressed to 5.0 atm and cooled to \(28^{\circ} \mathrm{C}\), enabling some of the hexane to be recovered as condensate.(a) Calculate the fractional recovery of hexane (kg condensed/kg fed in wet solids). (b) A proposal has been made to split the gas stream leaving the condenser, combining 90\% of it with fresh makeup nitrogen, heating the combined stream to \(85^{\circ} \mathrm{C},\) and recycling the heated stream to the dryer inlet. What fraction of the fresh nitrogen required in the process of Part (a) would be saved by introducing the recycle? What costs would be incurred by introducing the recycle?

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 \(80^{\circ} \mathrm{C}\).

The following diagram shows a staged absorption column in which \(n\) -hexane (H) is absorbed from a gas into a heavy oil.A gas feed stream containing 5.0 mole \(\%\) hexane vapor and the balance nitrogen enters at the bottom of an absorption column at a basis rate of \(100 \mathrm{mol} / \mathrm{s}\), and a nonvolatile oil enters the top of the column in a ratio 2 mol oil fed/mol gas fed. The absorber consists of a series of ideal stages (see Problem 6.66), arranged so that gas flows upward and liquid flows downward. The liquid and gas streams leaving each stage are in equilibrium with each other (by the definition of an ideal stage), with compositions related by Raoult's law. The absorber operates at an approximately constant temperature \(T\left(^{\circ} \mathrm{C}\right)\) and \(760 \mathrm{mm}\) Hg. Of the hexane entering the column, \(99.5 \%\) is absorbed and leaves in the liquid column effluent. At the given conditions it may be assumed that \(\mathrm{N}_{2}\) is insoluble in the oil and that none of the oil vaporizes.(a) Calculate the molar flow rates and mole fractions of hexane in the gas and liquid streams leaving the column. Then calculate the average values of the liquid and gas molar flow rates in the column, \(\dot{n}_{\mathrm{L}}(\mathrm{mol} / \mathrm{s})\) and \(\dot{n}_{\mathrm{G}}(\mathrm{mol} / \mathrm{s}) .\) For simplicity, in subsequent calculations use the average values for liquid and gas molar flow rates within the column, but the actual values for the corresponding flow rates entering and leaving the column.(b) Considering the bottom stage to be ideal, estimate the mole fractions of hexane in the gas leaving that stage \(\left(y_{N}\right)\) and in the liquid entering it \(\left(x_{N-1}\right)\) if the column temperature is \(50^{\circ} \mathrm{C}\). (c) Suppose that \(x_{i}\) and \(y_{i}\) are the mole fractions of hexane in the liquid and gas streams leaving stage \(i\) Derive the following equations from an equilibrium relationship and a mass balance around a section of the column encompassing stage \(i\) and the bottom of the column:$$\begin{array}{c}y_{i}=x_{i} p_{i}^{*}(T) / P \\ x_{i-1}=\left(x_{N} n_{L, N}+y_{i} \dot{n}_{\mathrm{G}}-y_{N+1} \dot{n}_{\mathrm{G}+1}\right) / \dot{n}_{\mathrm{L}}\end{array}$$Verify that these equations yield the answers you calculated in Part (b). (d) Examine the effect of operating temperature on the column by estimating the number of ideal stages necessary to achieve the desired separation. In the calculations, which will be done using a spreadsheet, take the pressure in the column to be constant at 760 torr, but consider three different operating temperatures: \(30^{\circ} \mathrm{C}\) \(50^{\circ} \mathrm{C},\) and \(70^{\circ} \mathrm{C} .\) The calculations will follow a stage-to-stage strategy beginning at the bottom of the column and repeatedly applying Equations (1) and (2) until the mole fraction of hexane in the vapor leaving the column is less than or equal to that calculated in Part (a). You may use APEx or the Antoine equation and Table B.4 to estimate the hexane vapor pressure. The calculations for the case of \(T=30^{\circ} \mathrm{C}\) illustrate how to proceed; for this case, \(y_{N-1} < y_{1}=0.00263\) after only two stages.(e) You can see that the number of stages required increases as the column temperature increases. In fact, there is a maximum temperature beyond which the required separation cannot be achieved. At that temperature, the entering gas and leaving liquid are approximately in equilibrium, so that \(x_{N} p^{*}(T)=y_{N+1} P .\) Use either APEx or the Antoine equation to estimate the maximum temperature at which the separation can be achieved.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.