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Chapter 5: Problem 47

The bacteria acetobacter aceti convert ethanol to acetic acid in the presence of oxygen according to the reaction $$\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}+\mathrm{O}_{2} \rightarrow \mathrm{CH}_{3} \mathrm{COOH}+\mathrm{H}_{2} \mathrm{O}$$ In a continuous fermentation process, ethanol enters the top of the fermenter at a rate of \(145 \mathrm{kg} / \mathrm{h}\), and the air fed to the bottom of the fermenter is \(25 \%\) in excess of the amount required to consume all of the ethanol. A gas stream containing nitrogen and unreacted oxygen leaves the top of the fermenter, and a liquid stream containing acetic acid, water, and \(10 \%\) of the entering ethanol leaves the bottom. Assume that none of the ethanol, water, and acetic acid in the reactor is vaporized. The fermenter operates at \(30^{\circ} \mathrm{C},\) maintains a liquid \((\mathrm{SG}=0.95)\) height of \(4.5 \mathrm{m},\) and is open to the atmosphere (i.e., the pressure at the top of the fermenter is 1 atm). (a) What is the volumetric flow rate of air as it enters the bottom of the fermenter? What is the volumetric flow rate of gas leaving the top of the fermenter? (b) Assume a linear relationship between the fraction of oxygen reacted and the position of gas bubbles rising through the liquid in the fermenter: for example, half of the oxygen reacted is consumed in the bottom half of the fermenter. At the vertical midpoint of the fermenter, the average bubble diameter is \(1.5 \mathrm{mm}\). What is the average bubble diameter at the entry point of the air and as the gas leaves the liquid at the top of the fermenter?

Short Answer

Expert verified
The volumetric flow rates of air at the bottom and of gas leaving at the top of the fermenter as well as the average bubble diameter at the entry point of the air and as the gas leaves the liquid at the top can be determined using stoichiometric relationships, mass balance principles and gas-liquid interactions in the fermenter. The specific values depend on the physical properties and operational conditions of the system.

Step by step solution

01

Determine stoichiometry

Given the reaction \(C_{2} H_{5} OH + O_{2} \rightarrow CH_{3} COOH + H_{2} O\), we can see that 1 mole of \(C_{2} H_{5} OH\) (ethanol) reacts with 1 mole of \(O_{2}\) (oxygen) to produce 1 mole of \(CH_{3} COOH\) (acetic acid) and 1 mole of \(H_{2} O\). So, mole to mole ratio is 1:1:1:1.
02

Find the air inflow rate

First find the molar flow rate of ethanol entering the system, then calculate the molar flow rate of oxygen needed for the reaction. Given that the air is 25% in excess of the required quantity, we can obtain the total molar flow rate of air entering the system. Convert this to volumetric flow rate using ideal gas law.
03

Find the gas outflow rate

Use the stoichiometry, air inlet rate and given information that all the ethanol is consumed to calculate the molar flow rate of oxygen leaving the fermenter. Convert this into volumetric flow rate using the ideal gas law.
04

Calculate bubble diameter at entry point, midpoint and top of the fermenter

Assuming a linear relationship between the fraction of oxygen reacted and the position of gas bubbles rising through the liquid, we determine the relationship between bubble diameter and its location. It's given that the average bubble diameter at the midpoint is 1.5 mm, therefore, the average bubble diameter at the entry point and at the top of the fermenter can be deduced.

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Key Concepts

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

Stoichiometry
Understanding stoichiometry is crucial for professionals in chemical engineering. It's the fundamental concept that deals with the quantitative relationships between the reactants and products in a chemical reaction. In the context of the bacterial conversion of ethanol to acetic acid, stoichiometry helps us determine how much oxygen is required to react with a given amount of ethanol, and subsequently, how much acetic acid and water will be produced.

The reaction \( C_{2}H_{5}OH + O_{2} \rightarrow CH_{3}COOH + H_{2}O \) is balanced, meaning one mole of ethanol reacts with one mole of oxygen to produce one mole of acetic acid and one mole of water. To further clarify for students, a 'mole' is a standard scientific unit for measuring large quantities of very small entities such as atoms, molecules, or other specified particles. The equation also tells us that for every kilogram of ethanol consumed, there will be a proportional kilogram of acetic acid produced, since their molar ratios are equal. This is a key concept in designing and operating a fermentation process effectively and efficiently.
Volumetric Flow Rate
The volumetric flow rate is a measure of the volume of fluid that passes through a given surface per unit time. It's usually expressed in units like liters per second or cubic meters per hour. In chemical engineering education, grasping the concept of the volumetric flow rate is necessary because it impacts the design and scaling of reactors and other process equipment.

In our exercise, we first calculate the volumetric flow rate of air needed for the fermentation process. Air is being fed in excess (25%), so we must account for this surplus when calculating the volumetric flow rate. Using the stoichiometric relationship established earlier, we convert the molar flow rate of oxygen to the molar flow rate of air by accounting for the excess and then use the ideal gas law (PV = nRT) to find the volumetric flow rate, where P is pressure, V is volume, n is number of moles, R is the gas constant, and T is temperature. Similarly, the volumetric flow rate of the gas leaving the fermenter is calculated by considering the molar flow rate of the remaining oxygen, as all ethanol should be consumed. This concept ensures that there is sufficient reaction with ethanol to produce acetic acid while also being mindful of economic and environmental concerns.
Fermentation Process
The fermentation process is a metabolic pathway that converts sugar to acids, gases, or alcohol. It occurs in yeast and bacteria, and also in oxygen-starved muscle cells. In chemical engineering, fermentation is a key operation, particularly in the production of alcohols, solvents, and organic acids. Our focus lies on the production of acetic acid from ethanol by the bacteria acetobacter aceti.

Different types of fermentations, such as continuous fermentation shown in this exercise, are critical for the efficient production of chemicals. Continuous fermentation involves the constant addition of nutrients and removal of products, which is beneficial for the production of acetic acid since the process does not have to be stopped to harvest the product. This style of fermentation can be more efficient than batch processes, especially for large-scale production. The maintenance of a stable environment, such as temperature, pH, and oxygen availability, which are implied in the exercise, is also an integral part of the fermentation process. This ensures that bacteria will continuously convert ethanol into acetic acid, maintaining the reaction's efficiency.

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Most popular questions from this chapter

Spray drying is a process in which a liquid containing dissolved or suspended solids is injected into a chamber through a spray nozzle or centrifugal disk atomizer. The resulting mist is contacted with hot air, which evaporates most or all of the liquid, leaving the dried solids to fall to a conveyor belt at the bottom of the chamber. Powdered milk is produced in a spray dryer \(6 \mathrm{m}\) in diameter by \(6 \mathrm{m}\) high. Air enters at \(167^{\circ} \mathrm{C}\) and \(-40 \mathrm{cm} \mathrm{H}_{2} \mathrm{O}\). The milk fed to the atomizer contains \(70 \%\) water by mass, all of which evaporates. The outlet gas contains 12 mole \(\%\) water and leaves the chamber at \(83^{\circ} \mathrm{C}\) and \(1 \mathrm{atm}\) (absolute) at a rate of \(311 \mathrm{m}^{3} / \mathrm{min}\). (a) Calculate the production rate of dried milk and the volumetric flow rate of the inlet air. Estimate the upward velocity of air (m/s) at the bottom of the dryer. (b) Engineers often face the challenge of what to do to a process when demand for a product increases (or decreases). Suppose in the present case production must be doubled. (i) Why is it unlikely that the flow rates of feed and air can simply be increased to achieve the new production rate? (ii) An obvious option is to buy another dryer like the existing one and operate the two in parallel. Give two advantages and two disadvantages of this option. (iii) Still another possibility is to buy a larger dryer to replace the original unit. Give two advantages and two disadvantages of doing so. Estimate the approximate dimensions of the larger unit.

Magnesium sulfate has a number of uses, some of which are related to the ability of the anhydrate form to remove water from air and others based on the high solubility of the heptahydrate \(\left(\mathrm{MgSO}_{4} \cdot 7 \mathrm{H}_{2} \mathrm{O}\right)\) form, also known as Epsom salt. The densities of the anhydrate and heptahydrate crystalline forms are 2.66 and \(1.68 \mathrm{g} / \mathrm{mL},\) respectively. Suppose you wish to form a 20.0 wt\% \(\mathrm{MgSO}_{4}\) aqueous solution by simply pouring crystals of one of the forms into a tank of water while the temperature is held constant at \(30^{\circ} \mathrm{C}\). The specific gravity of the 20.0 wt\% solution at \(30^{\circ} \mathrm{C}\) is \(1.22 .\) Answer the following questions for both forms of the \(\mathrm{MgSO}_{4}\) crystals: (a) What volume of water should be in the tank before crystals are added if the final product is to be 1000 kg of the 20 wt\% solution? (b) Suppose the tank diameter is \(0.30 \mathrm{m}\). What is the height of liquid in the tank before the crystals are added? (c) What is the height of the water in the tank after addition of the crystals but before they begin to dissolve? (d) What is the height of liquid in the tank after all the MgSO \(_{4}\) has dissolved?

A fuel cell is an electrochemical device that reacts hydrogen with oxygen from the air to produce water and DC electricity. A proposed application is replacement of the gasoline-fueled internal combustion engine in an automobile with a \(100 \mathrm{kW}\) fuel cell. You are on a summer internship with a gas supplier planning to transport hydrogen to service stations for use in cars powered by fuel cells. The hydrogen is to be transported in tube trailers, each of which has 10 tubes of length \(10.5 \mathrm{m}\) and diameter \(0.56 \mathrm{m}\). Hydrogen in the tubes at 2600 psig and an average temperature of \(298 \mathrm{K}\) is discharged at service stations to a final pressure of 55 psig. Refueling cach fuel-cell-powered automobile is estimated to require 4.0 kg of hydrogen. (a) You and your office-mate- an intern from a different university - have been asked to estimate the number of automobiles that can be refueled by one tube-trailer load of hydrogen. He does a very quick calculation and comes up with a value of 95 cars. Speculate how he did it and provide support for your speculation. What was his mistake? (b) Do the calculation using the SRK equation of state. Instead of using Eqs. \(5.3-11\) and \(5.3-13\) for the parameter \(\alpha,\) use the following correlation developed specifically for hydrogen: \(^{23}\) \(\alpha=1.202 \exp \left(-0.3228 T_{\mathrm{r}}\right)\) (c) Do the calculation using the law of corresponding states. (d) In which of the three estimates would you have the greatest confidence, and why?

The volume of a dry box (a closed chamber with dry nitrogen flowing through it) is \(2.0 \mathrm{m}^{3}\). The dry box is maintained at a slight positive gauge pressure of \(10 \mathrm{cm} \mathrm{H}_{2} \mathrm{O}\) and room temperature \(\left(25^{\circ} \mathrm{C}\right) .\) If the contents of the box are to be replaced every five minutes, calculate the required mass flow rate of nitrogen in \(g / \min\) by (a) direct solution of the ideal-gas equation of state and (b) conversion from standard conditions. You may assume the gas in the dry box is well mixed.

The concentration of oxygen in a 5000 -liter tank containing air at 1 atm is to be reduced by pressure purging prior to charging a fuel into the tank. The tank is charged with nitrogen up to a high pressure and then vented back down to atmospheric pressure. The process is repeated as many times as required to bring the oxygen concentration below 10 ppm (i.c., to bring the mole fraction of \(\mathrm{O}_{2}\) below \(10.0 \times 10^{-6}\) ). Assume that the temperature is \(25^{\circ} \mathrm{C}\) at the beginning and end of each charging cycle. When doing \(P V T\) calculations in Parts (b) and (c), use the generalized compressibility chart if possible for the fully charged tank and assume that the tank contains pure nitrogen. (a) Speculate on why the tank is being purged. (b) Estimate the gauge pressure (atm) to which the tank must be charged if the purge is to be done in one charge-vent cycle. Then estimate the mass of nitrogen (kg) used in the process. (For this part, if you can't find the tank condition on the compressibility chart, assume ideal-gas behavior and state whether the resulting estimate of the pressure is too high or too low.) (c) Suppose nitrogen at 700 kPa gauge is used for the charging. Calculate the number of charge-vent cycles required and the total mass of nitrogen used. (d) Use your results to explain why multiple cycles at a lower gas pressure are preferable to a single cycle. What is a probable disadvantage of multiple cycles?
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