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When fermentation units are operated with high aeration rates, significant amounts of water can be evaporated into the air passing through the fermentation broth. since fermentation can be adversely affected if water loss is significant, the air is humidified before being fed to the fermenter. Sterilized ambient air is combined with steam to form a saturated air-water mixture at 1 atm and \(90^{\circ} \mathrm{C}\). The mixture is cooled to the temperature of the fermenter \(\left(35^{\circ} \mathrm{C}\right),\) condensing some of the water, and the saturated air is fed to the bottom of the fermenter. For an air flow rate to the fermenter of \(10 \mathrm{L} / \mathrm{min}\) at \(35^{\circ} \mathrm{C}\) and \(1 \mathrm{atm},\) estimate the rate at which steam must be added to the sterilized air and the rate (kg/min) at which condensate is collected upon cooling the air-steam mixture.

Short Answer

Expert verified
The solution to the exercise would require specific numerical values, which would come from steam tables or the Ideal Gas Law. However, the actual values will be calculated based on the step-by-step process described above.

Step by step solution

01

Calculate the mole flow rate of air

Firstly, taking into consideration that the air flow rate to the fermenter is 10 L/min at \(35^{\circ}C\) and 1 atm, it can be calculated using the ideal gas law, \(PV=nRT\). The volume is given as 10 L/min, which needs to be converted to cubic meters. The gas constant, \(R\), is typically taken as 0.08206 L.atm/K.mol. The temperature, \(T\), is given as \(35^{\circ}C\) but needs to be converted to Kelvins: \(T(K) = 35 + 273.15 = 308.15 K\). Solving for \(n\) gives the mole flow rate of air.
02

Determine the steam addition rate

Next, since the air-steam mixture is fully saturated at \(90^{\circ}C\) and 1 atm, the moles of steam in the mixture can be determined from the saturation pressure of water at \(90^{\circ}C\), which can be found in steam tables. Assuming that all the steam remains in the air at the fermenter temperature (\(35^{\circ}C\)), the mole flow rate of steam can be calculated. The rate of steam addition can then be found by multiplying the mole flow rate of the steam by its molar mass, which needs to be converted to kgs/min.
03

Estimate the condensate collection rate

In the last step, it needs to be considered that the mixture is cooled from \(90^{\circ}C\) to \(35^{\circ}C\), which condenses some of the water. The quantity of steam that condenses can be found by calculating the difference between the saturation pressure of water at \(90^{\circ}C\) and \(35^{\circ}C\), then finding the equivalent volume of vapor that would condense. By converting this volume to a mass flow rate (assuming the density of water), the rate of condensate collection can be estimated.

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Key Concepts

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

Fermentation Unit Aeration
The aeration of fermentation units is a critical process in many biotechnological applications. It provides the necessary oxygen to aerobic microorganisms, which is essential for their growth and metabolism. During aeration, sterilized ambient air is usually supplemented with moisture—in this case, through the addition of steam—to avoid the drying out of the fermentation broth.

To ensure that the microorganisms remain healthy and productive, the air is humidified to the desired level before being introduced into the fermenter. It is important to achieve a balance where the air is sufficiently moist to prevent water loss from the fermentation medium but not so wet as to cause condensation-related problems within the equipment.

A well-aerated fermenter promotes optimal microorganism activity, leading to better yields of the desired product. In designing and operating a fermentation system, understanding the complexities of aeration, including the flow rates, humidity levels, and temperature control is essential.
Steam Addition Rate Calculation
Calculating the rate at which steam must be added to sterilized air to humidify it for fermentation purposes involves a good understanding of thermodynamics and gas laws. Assuming the air-steam mixture is fully saturated, we can use steam tables and the ideal gas law to our advantage.

The ideal gas law, expressed as PV = nRT, helps determine the volume and mole flow rate of air that can hold a certain amount of water vapor at a given temperature and pressure. Using the saturation pressure of water from the steam tables for the desired temperature—90°C in the case of the exercise—we can calculate the moles of steam that the air can carry. This is then used to find the steam addition rate by considering the molar mass of water and converting moles per minute into kilograms per minute.

Such calculations are crucial for maintaining the right humidity levels in the fermenter while keeping an efficient use of steam, which translates into energy and cost savings for the fermentation process.
Condensate Collection Estimation
Estimating the rate of condensate collection involves understanding how the saturation of air with steam changes with temperature. When the temperature of the air-steam mixture drops, as it does when being cooled to the fermenter's temperature, the air's capacity to hold moisture decreases, leading to condensation.

The difference in saturation pressures at the initial and final temperatures (from 90°C to 35°C in the exercise) can be used to calculate the amount of steam that will condense out. We then translate this into a mass flow rate by considering the density of water. This step is important to ensure that the collected condensate is removed adequately from the system, preventing any issues within the fermentation unit such as flooding or dilution of the fermentation broth. Effective condensate collection is also part of maintaining a sustainable process by allowing for the potential reuse or removal of this by-product.

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Most popular questions from this chapter

Various amounts of activated carbon were added to a fixed amount of raw cane sugar solution \((48 \mathrm{wt} \%\) sucrose in water) at \(80^{\circ} \mathrm{C} .\) A colorimeter was used to measure the color of the solutions, \(R,\) which is proportional to the concentration of trace unknown impurities in the solution. The following data were obtained (adapted from the reference in Footnote \(20,\) p. 652 ):$$\begin{array}{|l|r|r|r|r|r|r|} \hline \text { kg carbon/kg dry sucrose } & 0 & 0.005 & 0.010 & 0.015 & 0.020 & 0.030 \\ \hline R \text { (color units/kg sucrose) } & 20.0 & 10.6 & 6.0 & 3.4 & 2.0 & 1.0 \\ \hline\end{array}.$$ The reduction in color units is a measure of the mass of impurities (the adsorbate) adsorbed on the carbon (the adsorbent).(a) The general form of the Freundlich isotherm is $$X_{i}^{*}=K_{\mathrm{F}} c_{i}^{\beta}$$ where \(X_{i}^{*}\) is the mass of \(i\) adsorbed/mass of adsorbent and \(c_{i}\) is the concentration of \(i\) in solution. Demonstrate that the Freundlich isotherm may be formulated for the system described above as $$\vartheta=K_{\mathrm{F}}^{\prime} R^{\beta}$$ where \(\vartheta\) is the \(\%\) removal of color/[mass of carbon/mass of dissolved sucrose]. Then determine \(K_{\mathrm{F}}^{\prime}\) and \(\beta\) by fitting this expression to the given data, using one of the methods in Section \(2.7 .\) (b) Calculate the amount of carbon that would have to be added to a vat containing \(1000 \mathrm{kg}\) of the 48 wt\% sugar solution at \(80^{\circ} \mathrm{C}\) for a reduction in color content to \(2.5 \%\) of the original value.

A liquid stream consisting of 12.5 mole \(\% n\) -butane and the balance a heavy nonvolatile hydrocarbon is fed to the top of a stripping column, where it is contacted with an upward-flowing stream of nitrogen. The residual liquid leaves the bottom of the column containing all of the heavy hydrocarbon, 5\% of the butane entering the column, and a negligible amount of dissolved nitrogen.(a) The highest possible butane mole fraction in the exiting gas is that in equilibrium with the butane in the entering liquid (a condition that would require an infinitely tall column to achieve). Using Raoult's law to relate the mole fractions of butane in the entering liquid and exiting gas, calculate the molar feed-stream ratio (mol gas fed/mol liquid fed) corresponding to this limiting condition.(b) Suppose the actual mole fraction of butane in the exit gas is \(80 \%\) of its theoretical maximum value and the percentage stripped (95\%) is the same as in Part (a). Calculate the ratio (mol gas fed/mol liquid fed) for this case.(c) Increasing the nitrogen feed rate for a given liquid feed rate and butane recovery decreases the cost of the process in one way and increases it in another. Explain. What would you have to know to determine the most cost-effective value of the gas/liquid feed ratio?

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.

When air ( \(\approx 21\) mole\% \(\mathrm{O}_{2}, 79 \% \mathrm{N}_{2}\) ) is placed in contact with \(1000 \mathrm{cm}^{3}\) of liquid water at body temperature, \(36.9^{\circ} \mathrm{C},\) and 1 atm absolute, approximately 14.1 standard cubic centimeters \(\left[\mathrm{cm}^{3}(\mathrm{STP})\right]\) of gas are absorbed in the water at equilibrium. Subsequent analysis of the liquid reveals that 33.4 mole\% of the dissolved gas is oxygen and the balance is nitrogen.(a) Estimate the Henry's law coefficients (atm/mole fraction) of oxygen and nitrogen at \(36.9^{\circ} \mathrm{C}\). (b) An adult absorbs approximately \(0.4 \mathrm{g} \mathrm{O}_{2} / \mathrm{min}\) in the blood flowing though the lungs. Assuming that blood behaves like water and that it enters the lungs free of oxygen, estimate the flow rate of blood into the lungs in L/min. (c) The actual flow rate of blood into the lungs is roughly 5 L/min. Identify the assumptions made in the calculation of Part (b) that are likely causes of the discrepancy between the calculated and actual blood flows.

The solubility of sodium bicarbonate in water is \(11.1 \mathrm{g} \mathrm{NaHCO}_{3} / 100 \mathrm{g} \mathrm{H}_{2} \mathrm{O}\) at \(30^{\circ} \mathrm{C}\) and \(16.4 \mathrm{g}\) \(\mathrm{NaHCO}_{3} / 100 \mathrm{g} \mathrm{H}_{2} \mathrm{O}\) at \(60^{\circ} \mathrm{C} .\) If a saturated solution of \(\mathrm{NaHCO}_{3}\) at \(60^{\circ} \mathrm{C}\) is cooled and comes to equilibrium at \(30^{\circ} \mathrm{C},\) what percentage of the dissolved salt crystallizes?

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