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Chapter 4: Problem 9

Inside a distillation column (see Problem 4.8), a downward-flowing liquid and an upward-flowing vapor maintain contact with each other. For reasons we will discuss in greater detail in Chapter \(6,\) the vapor stream becomes increasingly rich in the more volatile components of the mixture as it moves up the column, and the liquid stream is enriched in the less volatile components as it moves down. The vapor leaving the top of the column goes to a condenser. A portion of the condensate is taken off as a product (the overhead product), and the remainder (the reflux) is returned to the top of the column to begin its downward journey as the liquid stream. The condensation process can be represented as shown below: A distillation column is being used to separate a liquid mixture of ethanol (more volatile) and water (less volatile). A vapor mixture containing 89.0 mole \(\%\) ethanol and the balance water enters the overhead condenser at a rate of \(100 \mathrm{lb}\) -mole/h. The liquid condensate has a density of \(49.01 \mathrm{b}_{\mathrm{m}} / \mathrm{ft}^{3},\) and the reflux ratio is \(3 \mathrm{lb}_{\mathrm{m}}\) reflux/lb \(_{\mathrm{m}}\) overhead product. When the system is operating at steady state, the tank collecting the condensate is half full of liquid and the mean residence time in the tank (volume of liquid/volumetric flow rate of liquid) is 10.0 minutes. Determine the overhead product volumetric flow rate (ft \(^{3}\) /min) and the condenser tank volume (gal).

Short Answer

Expert verified
The overhead product volumetric flow rate is \(0.0085 \, ft^{3}/min\) and the condenser tank volume is \(2.54 \) gallons.

Step by step solution

01

Calculate Volumetric Flow Rate of the Condensate

First, convert the flow rate from lb-mole/h to lb_m/ft^3. utilizing the given density of the liquid condensate:\[ \frac{100 \, lb_{mole/h}}{49.01 \, lb_{m} / ft^{3}} = 2.04 \, ft^{3}/h \]Then, convert this to ft^3/min as:\[ \frac{2.04 \, ft^{3}}{60} = 0.034 \, ft^{3}/min \] This is the volumetric flow rate of the condensate.
02

Calculate Volumetric Flow Rate of Overhead Product

Use the given reflux ratio to find the overhead product flow rate. The reflux ratio is defined as lb_m reflux per lb_m overhead product, which is given as 3. Therefore, let's denote the flow rate of the overhead product as F_O, then\[ 0.034 \, ft^{3}/min = F_O + 3F_O \]Solving this we get:\[ F_O = 0.0085 \, ft^{3}/min \] This is the volumetric flow rate of the overhead product.
03

Calculate the Condenser Tank Volume

Use the given mean residence time in the tank (volume of liquid/volumetric flow rate of liquid) to find the tank volume. By rearranging the formula, we get\[ Volume = Residence \, time \times Flow \, rate\]Substitute the given values:\[ Volume = 10 \, min \times 0.034 \, ft^{3}/min = 0.34 \, ft^{3}\]In gallons (as 1ft^3 = 7.481 gal):\[ Volume = 0.34 \, ft^{3} \times 7.481 \, gal/ft^{3} = 2.54 \, gal \] This is the volume of the condenser tank.

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Key Concepts

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

Volumetric Flow Rate Calculation
Understanding the volumetric flow rate is crucial when analyzing distillation column operations. It indicates the volume of fluid passing through a given area in a specific period. The solution provided illustrates a straightforward method for calculating the flow rate: first by converting the mass flow rate using the fluid density, and then by adjusting the units of time for practical use.

When calculating flow rate from a given mass flow rate, you need to use the density of the liquid to find its volume, then scale the units as needed, in this case from hours to minutes. If the column's productivity were to change, or if you needed to design a new system, these basic calculations would lay the groundwork for scaling up or down the equipment and processes involved.
Reflux Ratio
The reflux ratio is a key parameter in the operation of a distillation column, representing the ratio of liquid sent back into the column compared to the liquid taken out as the product. A higher reflux ratio usually means better separation but also higher energy costs, as more liquid needs to be reheated.

In practice, this ratio is used to balance between operational costs and the desired purity of the end product. The exercise's step-by-step solution demonstrates how to use this ratio to determine the flow rate of the overhead product by setting up a simple linear equation. This example underscores the importance of the reflux ratio in managing the efficiency and economics of a separation process.
Condenser Design
The condenser design is a crucial aspect of a distillation column, affecting both product quality and energy efficiency. It cools and condenses the vapor from the top of the column into a liquid form that can be either collected as a product or returned to the column as reflux. The design needs to ensure sufficient surface area for heat exchange, and the capacity to handle the calculated volumetric flow rate of the vapor.

In the given problem, the condenser must accommodate the flow rate dictated by the reflux ratio. The size and effectiveness of the condenser determine the steady state operation, directly influencing the mean residence time and ultimate product consistency. Well-designed condensers are vital for both the economic and technical success of distillation processes.
Mean Residence Time
The mean residence time indicates how long, on average, a liquid stays in a system. It is a critical factor for designing and operating any kind of tank or reservoir within a chemical process, including the tank in this distillation column exercise. It influences not just the size of the tank but also product quality.

By multiplying the mean residence time by the flow rate of the liquid, operators can calculate the necessary volume of the tank, as shown in the solution steps. In industrial scenarios, an optimal residence time is essential for maintaining a balance between equipment size and the throughput of the distillation process.
Chemical Processes
Distillation is one of the most widely used chemical processes for separating mixtures based on differences in the volatilities of the mixture components. The process鈥檚 efficiency depends on a multitude of factors including but not limited to flow rates, reflux ratios, and equipment design. Each of these elements must be finely tuned to ensure the desired separation is achieved in the most cost-effective manner.

Chemical process principles such as those demonstrated in the distillation column operation also apply to other separation techniques and are fundamental knowledge for anyone involved in chemical engineering or process design. An intimate understanding of these principles allows for optimization and troubleshooting in various stages of the chemical process industry.

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Most popular questions from this chapter

The popularity of orange juice, especially as a breakfast drink, makes this beverage an important factor in the economy of orange-growing regions. Most marketed juice is concentrated and frozen and then reconstituted before consumption, and some is "not-from-concentrate." Although concentrated juices are less popular in the United States than they were at one time, they still have a major segment of the market for orange juice. The approaches to concentrating orange juice include evaporation, freeze concentration, and reverse osmosis. Here we examine the evaporation process by focusing only on two constituents in the juice: solids and water. Fresh orange juice contains approximately 10 wt\% solids (sugar, citric acid, and other ingredients) and frozen concentrate contains approximately 42 wt\% solids. The frozen concentrate is obtained by evaporating water from the fresh juice to produce a mixture that is approximately 65 wt\% solids. However, so that the flavor of the concentrate will closely approximate that of fresh juice, the concentrate from the evaporator is blended with fresh orange juice (and other additives) to produce a final concentrate that is approximately 42 wt\% solids. (a) Draw and label a flowchart of this process, neglecting the vaporization of everything in the juice but water. First prove that the subsystem containing the point where the bypass stream splits off from the evaporator feed has one degree of freedom. (If you think it has zero degrees, try determining the unknown variables associated with this system.) Then perform the degree- offreedom analysis for the overall system, the evaporator, and the bypass- evaporator product mixing point, and write in order the equations you would solve to determine all unknown stream variables. In each equation, circle the variable for which you would solve, but don't do any calculations. (b) Calculate the amount of product (42\% concentrate) produced per 100 kg fresh juice fed to the process and the fraction of the feed that bypasses the evaporator. (c) Most of the volatile ingredients that provide the taste of the concentrate are contained in the fresh juice that bypasses the evaporator. You could get more of these ingredients in the final product by evaporating to (say) 90\% solids instead of 65\%; you could then bypass a greater fraction of the fresh juice and thereby obtain an even better tasting product. Suggest possible drawbacks to this proposal.

A process is carried out in which a mixture containing 25.0 wt\% methanol, \(42.5 \%\) ethanol, and the balance water is separated into two fractions. A technician draws and analyzes samples of both product streams and reports that one stream contains \(39.8 \%\) methanol and \(31.5 \%\) ethanol and the other contains 19.7\% methanol and 41.2\% ethanol. You examine the reported figures and tell the technician that they must be wrong and that stream analyses should be carried out again. (a) Prove your statement. (b) How many streams do you ask the technician to analyze? Explain.

Titanium dioxide \(\left(\mathrm{Ti} \mathrm{O}_{2}\right)\) is used extensively as a white pigment. It is produced from an ore that contains ilmenite \(\left(\mathrm{FeTiO}_{3}\right)\) and ferric oxide \(\left(\mathrm{Fe}_{2} \mathrm{O}_{3}\right) .\) The ore is digested with an aqueous sulfuric acid solution to produce an aqueous solution of titanyl sulfate \(\left[(\mathrm{TiO}) \mathrm{SO}_{4}\right]\) and ferrous sulfate (FeSO \(_{4}\) ). Water is added to hydrolyze the titanyl sulfate to \(\mathrm{H}_{2} \mathrm{TiO}_{3},\) which precipitates, and \(\mathrm{H}_{2} \mathrm{SO}_{4} .\) The precipitate is then roasted, driving off water and leaving a residue of pure titanium dioxide. (Several steps to remove iron from the intermediate solutions as iron sulfate have been omitted from this description.) Suppose an ore containing \(24.3 \%\) Ti by mass is digested with an \(80 \% \mathrm{H}_{2} \mathrm{SO}_{4}\) solution, supplied in \(50 \%\) excess of the amount needed to convert all the ilmenite to titanyl sulfate and all the ferric oxide to ferric sulfate \(\left[\mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}\right] .\) Further suppose that \(89 \%\) of the ilmenite actually decomposes. Calculate the masses (kg) of ore and 80\% sulfuric acid solution that must be fed to produce \(1000 \mathrm{kg}\) of pure \(\mathrm{TiO}_{2}\)

An unknown paraffinic hydrocarbon is defined by the chemical formula \(\mathrm{C}_{x} \mathrm{H}_{2 x+2}\). The paraffin is burned with air, and there is no CO in the combustion products. (a) Use a degree-of-freedom analysis to determine how many variables must be specified to determine the flow rates of all components entering and leaving the combustion unit. Express the fraction excess air as \(y\) and write elemental balances in terms of \(x, y,\) and the molar flow rate of \(C_{x} \mathrm{H}_{2 x+2}\) (b) Calculate the molar composition of the combustion product gas in terms of \(x\) for each of the following three cases: (i) theoretical air supplied ( \(y=0\) ), \(100 \%\) conversion of the paraffin; (ii) \(20 \%\) excess air \((y=0.2), 100 \%\) conversion of the paraffin; (iii) \(20 \%\) excess air, \(90 \%\) conversion of the paraffin. (c) Suppose \(x=3\) (i.e. the paraffin is propane). Assuming complete combustion of the hydrocarbon, what is the ratio of \(\mathrm{CO}_{2}\) to \(\mathrm{H}_{2} \mathrm{O}\) in the product gas? Use this result to suggest a method for determining the molecular formula of the paraffin.

Fermentation of sugars obtained from hydrolysis of starch or cellulosic biomass is an alternative to using petrochemicals as the feedstock in production of ethanol. One of the many commercial processes to do this \(^{16}\) uses an enzyme to hydrolyze starch in corn to maltose (a disaccharide consisting of two glucose units) and oligomers consisting of several glucose units. A yeast culture then converts the maltose to ethyl alcohol and carbon dioxide: $$\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}+\mathrm{H}_{2} \mathrm{O}(+\text { yeast }) \rightarrow 4 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}+4 \mathrm{CO}_{2}\left(+\text { yeast }+\mathrm{H}_{2} \mathrm{O}\right)$$ As the yeast grows, \(0.0794 \mathrm{kg}\) of yeast is produced for every \(\mathrm{kg}\) ethyl alcohol formed, and \(0.291 \mathrm{kg}\) water is produced for every kg of yeast formed. For use as a fuel, the product from such a process must be around 99.5 wt\% ethyl alcohol. Corn fed to the process is 72.0 wt\% starch on a moisture-free basis and contains 15.5 wt\% moisture. It is estimated that 101.2 bushels of corn can be harvested from an acre of com, that each bushel is equivalent to \(25.4 \mathrm{lb}_{\mathrm{m}}\) of corn, and that \(6.7 \mathrm{kg}\) of ethanol can be obtained from a bushel of corn. What acreage of farmland is required to produce 100,000 kg of ethanol product? What factors (economic and environmental) must be considered in comparing production of ethanol by this route with other routes involving petrochemical feedstocks?
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