/*! 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

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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.

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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.

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Most popular questions from this chapter

A gas mixture containing 85.0 mole \(\% \mathrm{N}_{2}\) and the balance \(n\) -hexane flows through a pipe at a rate of \(100.0 \mathrm{m}^{3} / \mathrm{h} .\) The pressure is 2.00 atm absolute and the temperature is \(100^{\circ} \mathrm{C}\). (a) What is the molar flow rate of the gas in \(\mathrm{kmol} / \mathrm{h}\) ? (b) Is the gas saturated? If not, to what temperature ( \(^{C} C\) ) would it have to be cooled at constant pressure in order to begin condensing hexane? (c) To what temperature ( \(C\) ) would the gas have to be cooled at constant pressure in order to condense \(80 \%\) of the hexane?

The vapor pressure of ethylene glycol at several temperatures is given below:$$\begin{array}{|l|r|r|r|r|r|r|}\hline T\left(^{\circ} \mathrm{C}\right) & 79.7 & 105.8 & 120.0 & 141.8 & 178.5 & 197.3 \\\\\hline p^{*}(\mathrm{mm} \mathrm{Hg}) & 5.0 & 20.0 & 40.0 & 100.0 & 400.0 & 760.0 \\\\\hline\end{array}$$ a semilog plot e vapor-pressure data and determine a linear expression for \(\ln p^{*}\) function of \(1 / T(\mathrm{K}) .\) Use the results to estimate the heat of vaporization \((\mathrm{kJ} / \mathrm{mol})\) of ethylene glycol, and then use that value in the Clausius-Clapeyron equation to estimate the vapor pressures at each of the temperatures given in the table.(b) Repeat Part (a) using the Slope and Intercept functions of APEx to obtain the expression for \(\ln p^{*}\) vs. \(1 / T(\mathrm{K})\).(c) Use the results from Part (b) to estimate vapor pressures of ethylene glycol at \(50^{\circ} \mathrm{C}, 80^{\circ} \mathrm{C},\) and \(110^{\circ} \mathrm{C} .\) Also estimate the boiling point of this substance at system pressures of \(760 \mathrm{mm} \mathrm{Hg}\) and 2000 mm Hg. Compare all five results with those obtained directly using APEx functions. In which of the estimates at the given temperatures and pressures would you have the least confidence? Explain your reasoning.

Pure chlorobenzene is contained in a flask attached to an open-end mercury manometer. When the flask contents are at \(58.3^{\circ} \mathrm{C}\), the height of the mercury in the arm of the manometer connected to the flask is \(747 \mathrm{mm}\) and that in the arm open to the atmosphere is \(52 \mathrm{mm} . \mathrm{At} 110^{\circ} \mathrm{C},\) the mercury level is \(577 \mathrm{mm}\) in the arm connected to the flask and \(222 \mathrm{mm}\) in the other arm. Atmospheric pressure is \(755 \mathrm{mm} \mathrm{Hg}\). (a) Extrapolate the data using the Clausius-Clapeyron equation to estimate the vapor pressure of chlorobenzene at \(130^{\circ} \mathrm{C}\). (b) Air saturated with chlorobenzene at \(130^{\circ} \mathrm{C}\) and \(101.3 \mathrm{kPa}\) is cooled to \(58.3^{\circ} \mathrm{C}\) at constant pressure. Estimate the percentage of the chlorobenzene originally in the vapor that condenses. (See Example 6.3-2.)(c) Summarize the assumptions you made in doing the calculation of Part (b).

A fuel gas containing methane and ethane is burned with air in a furnace, producing a stack gas at \(300^{\circ} \mathrm{C}\) and \(105 \mathrm{kPa}\) (absolute). You analyze the stack gas and find that it contains no unburned hydrocarbons, oxygen, or carbon monoxide. You also determine the dew-point temperature.(a) Estimate the range of possible dew-point temperatures by determining the dew points when the feed is either pure methane or pure ethane. (b) Estimate the fraction of the feed that is methane if the measured dew- point temperature is \(59.5^{\circ} \mathrm{C}\). (c) What range of measured dew point temperatures would lead to calculated methane mole fractions within 5\% of the value determined in Part (b)?

Nitric acid is a chemical intermediate primarily used in the synthesis of ammonium nitrate, which is used in the manufacture of fertilizers. The acid also is important in the production of other nitrates and in the separation of metals from ores. Nitric acid may be produced by oxidizing ammonia to nitric oxide over a platinum-rhodium catalyst, then oxidizing the nitric oxide to nitrogen dioxide in a separate unit where it is absorbed in water to form an aqueous solution of nitric acid.The reaction sequence is as follows:$$\begin{aligned} 4 \mathrm{NH}_{3}+5 \mathrm{O}_{2} & \rightarrow 4 \mathrm{NO}+6 \mathrm{H}_{2} \mathrm{O} \\\4 \mathrm{NO}+2 \mathrm{O}_{2} & \rightarrow 4 \mathrm{NO}_{2} \\\4 \mathrm{NO}_{2}+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})+\mathrm{O}_{2} & \rightarrow 4 \mathrm{HNO}_{3}(\mathrm{aq}) \end{aligned}$$.Ammonia vapor produced by vaporizing pure liquid ammonia at 820 kPa absolute is mixed with air, and the combined stream enters the ammonia oxidation unit. Air at \(30^{\circ} \mathrm{C}, 1\) atm absolute, and \(50 \%\) relative humidity is compressed and fed to the process. A fraction of the air is sent to the cooling and hydration units, while the remainder is passed through a heat exchanger and mixed with the ammonia. The total oxygen fed to the process is the amount stoichiometrically required to convert all of the ammonia to HNO \(_{3},\) while the fraction sent to the ammonia oxidizer corresponds to the stoichiometric amount required to convert ammonia to NO.The ammonia reacts completely in the oxidizer, with \(97 \%\) forming NO and the rest forming \(\mathrm{N}_{2}\). Only a negligible amount of \(\mathrm{NO}_{2}\) is formed in the oxidizer. However, the gas leaving the oxidizer is subjected to a series of cooling and hydration steps in which the NO is completely oxidized to \(\mathrm{NO}_{2}\) which in turn combines with water (some of which is present in the gas from the oxidizer and the rest is added) to form a 55 wt\% aqueous solution of nitric acid. The product gas from the process may be taken to contain only \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\). (a) Taking a basis of \(100 \mathrm{kmol}\) of ammonia fed to the process, calculate (i) the volumes \(\left(\mathrm{m}^{3}\right)\) of the ammonia vapor and air fed to the process using the compressibility-factor equation of state; (ii) the amount (kmol) and composition (in mole fractions) of the gas leaving the oxidation unit; (iii) the required volume of liquid water \(\left(\mathrm{m}^{3}\right)\) that must be fed to the cooling and hydration units; and (iv) the fraction of the air fed to the ammonia oxidizer. (b) Scale the results from Part (a) to a new basis of 100 metric tons per hour of 55\% nitric acid solution.(c) Nitrogen oxides (collectively referred to as \(\mathrm{NO}_{x}\) ) are a category of pollutants that are formed in many ways, including processes like that described in this problem. List the annual emission rates of the three largest sources of \(\mathrm{NO}_{x}\) emissions in your home region. What are the effects of exposure to excessive concentrations of \(\mathrm{NO}_{x} ?\) (d) A platinum-rhodium catalyst is used in ammonia oxidation. Fxplain the function of the catalyst, describe its structure, and explain the relationship of the structure to the function.

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