/*! 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 34 You were recently hired as a pro... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 34

You were recently hired as a process engineer by a pulp and paper manufacturing firm. Your new boss calls you in and tells you about a pulp dryer designed to reduce the moisture content of \(1500 \mathrm{kg} / \mathrm{min}\) of wet pulp from \(0.9 \mathrm{kg} \mathrm{H}_{2} \mathrm{O} / \mathrm{kg}\) dry pulp to \(0.15 \mathrm{wt} \% \mathrm{H}_{2} \mathrm{O}\). The design called for drawing atmospheric air at \(90 \%\) relative humidity, \(25^{\circ} \mathrm{C}, 760 \mathrm{mm}\) Hg into a blower that forces the air through a heater and into the dryer. When the operation was put into service, weather conditions were exactly as assumed in the design, and measurements showed that the air leaving the dryer was at \(80^{\circ} \mathrm{C}\) and a gauge pressure of \(10 \mathrm{mm}\) Hg. However, there was no way to check the operation of the blower to see if it was delivering the specified volumetric flow rate of air. Your boss wants to check that value and asks you to devise a method for doing so. You go back to your office, sketch the process, and determine that you can estimate the air flow rate from the given information if you also know the moisture content of the air leaving the dryer.(a) Propose a method to estimate the moisture content of the exit air. (b) Suppose your measurement is carried out and you learn that the exit air at \(10 \mathrm{mm}\) Hg gauge has a dew point of \(40^{\circ} \mathrm{C}\). Use that information and the mass of water removed from the wet pulp to determine the volumetric flow rate ( \(\mathrm{m}^{3} / \mathrm{min}\) ) of air entering the system.

Short Answer

Expert verified
The volumetric flow rate of air entering the system is approximately 2559.725 \(m^3/min\).

Step by step solution

01

The Moisture Content Estimation Method

The moisture content of the air leaving the dryer can be estimated using a moisture analyzer or a hygrometer, which measures the amount of water vapor in the air, thus giving an understanding of the moisture content.
02

Mass of Water Removal Calculation

The initial moisture per dry pulp is 0.9 kg while the final is 0.0015 kg because the percentage is given by mass. Therefore, the mass of water removed per kg of dry pulp is \(0.9 - 0.0015 = 0.8985\) kg. With an input of 1500 kg wet pulp per minute, the moisture removal rate is calculated as \((1500 kg/min ) * (0.8985 kg/ kg dry pulp)= 1347.75 kg/min\).
03

Exit Air Moisture Content Calculation

From the dew point of air, we find the moisture content because dew point temperature indicates the amount of water vapor in the air. In the \(40^{\circ}C\) dew point at \(10 mm Hg\) gauge, corresponds to \(0.5 kg H2O/ kg dry air\). This is obtained from steam tables or Psychrometric chart.
04

Volumetric Flow Rate Calculation

The volumetric flow rate can be estimated using the mass balance around the dryer. The air and moisture entering the system equals the dry air and moisture leaving. Since dry air leaves the system with each kg carrying 0.5 kg of water, for every 1 kg of water removed, 2 kg of dry air is required. This leads to a dry air flow rate of \((1347.75 kg/min) * (2 kg dry air/ kg H2O) = 2695.5 kg/min \). The specific volume of dry air at exit conditions is about \(0.95 m3/kg\) from the ideal gas law, leading to a volumetric flow rate of \( (2695.5 kg/min) * (0.95 m3/kg) = 2559.725 m3/min\).

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Key Concepts

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

Pulp Dryer Operation
In the pulp and paper industry, the pulp dryer is a fundamental component used to reduce the moisture content of pulp. This process ensures pulp is at an optimal moisture level for further processing or shipping. To understand pulp dryer operation, it's vital to recognize how it removes excess water from the wet pulp.

The process typically involves passing hot air through the moist pulp, which helps evaporate water. This hot air absorbs the moisture from the pulp, then exits the system carrying water vapor away.

Key points of the operation include:
  • The initial moisture content of the pulp, which in some scenarios can be as high as 90% compared to the dry pulp weight.
  • The desired final moisture content, usually specified as a percentage or in terms of weight by a quality control requirement.
  • Control of temperature and flow rate of the drying air to meet the specified reduction in moisture content.
  • Efficient moisture removal ensures reduced energy consumption and prevents any damage to the pulp fibers.
Understanding and monitoring these aspects are crucial for efficient pulp dryer operation.
Volumetric Flow Rate Calculation
Calculating the volumetric flow rate of air in a drying process helps in determining how effectively the system is operating. This calculation is often required when commissioning dryers or troubleshooting inefficiencies in operation.

The first step involves using the mass balance principle. The air entering the system, along with its enthalpy, should match the air and moisture leaving the system. Once we know the moisture content at the exit of the dryer, we can relate it to the dry air required to remove this moisture.

Here's a simplified approach:
  • Calculate the moisture removed per minute from the starting and target moisture levels and the throughput of the pulp.
  • Determine the mass of dry air needed per kilogram of water removed, using the moisture distribution data in the outgoing air.
  • Apply these values in a mass flow equation to find the dry air rate.
  • Lastly, use the specific volume of the air at the exit conditions to convert the mass flow into a volumetric flow rate.
This volumetric flow metric gives insight into the amount of air (in cubic meters per minute) needed to effectively remove the designated amount of water and helps in assessing the dryer鈥檚 performance.
Dew Point Temperature
Dew point temperature is a critical parameter in the drying process, as it defines the temperature at which air becomes saturated with moisture, leading to condensation. In our scenario, knowing the dew point of the exit air offers clear insights into the moisture content of the outgoing air stream.

This temperature is determined under the specific pressure conditions prevailing in the system. At a gauge pressure, for example of 10 mm Hg, the dew point helps understand how much moisture is being carried away by the air.

Important aspects include:
  • The dew point temperature helps us understand the saturation level of moisture in the air.
  • By referring to steam tables or psychrometric charts, engineers can determine how much water vapor the air can hold at specific conditions.
  • A higher dew point implies more moisture in the air, which can be a measure of dryer efficiency.
  • Monitoring changes in dew point temperature can signal potential issues or adjustments needed in the drying process.
Therefore, understanding and utilizing dew point temperature is invaluable in regulating moisture content within air drying applications.

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Most popular questions from this chapter

Acetaldehyde is synthesized by the catalytic dehydrogenation of ethanol:$$ \mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH}\rightarrow\mathrm{CH}_{3}\mathrm{CHO}+\mathrm{H}_{2}.$$ Fresh feed (pure ethanol) is blended with a recycle stream (95 mole\% ethanol and 5\% acetaldehyde), and the combined stream is heated and vaporized, entering the reactor at \(280^{\circ} \mathrm{C}\). Gases leaving the reactor are cooled to \(-40^{\circ} \mathrm{C}\) to condense the acetaldehyde and unreacted ethanol. Off-gas from the condenser is sent to a scrubber, where the uncondensed organic compounds are removed and hydrogen is recovered as a by- product. The condensate from the condenser, which is 45 mole\% ethanol, is sent to a distillation column that produces a distillate containing 99 mole\% acetaldehyde and a bottoms product that constitutes the recycle blended with fresh feed to the process. The production rate of the distillate is \(1000 \mathrm{kg} / \mathrm{h}\). The pressure throughout the process may be taken as 1 atm absolute. (a) Calculate the molar flow rates ( \(\mathrm{kmol} / \mathrm{h}\) ) of the fresh feed, the recycle stream, and the hydrogen in the off-gas. Also determine the volumetric flow rate \(\left(\mathrm{m}^{3} / \mathrm{h}\right)\) of the feed to the reactor. (Suggestion:Use Raoult's law in the analysis of the condenser.)(b) Estimate (i) the overall and single-pass conversions of ethanol and (ii) the rates ( \(\mathrm{kmol} / \mathrm{h}\) ) at which ethanol and acetaldehyde are sent to the scrubber.

An aqueous solution of urea \((\mathrm{MW}=60.06)\) freezes at \(-4.6^{\circ} \mathrm{C}\) and 1 atm. Estimate the normal boiling point of the solution; then calculate the mass of urea (grams) that would have to be added to \(1.00 \mathrm{kg}\) of solution to raise the normal boiling point by \(3^{\circ} \mathrm{C}\).

Various amino acids have utility as food additives and in medical applications. They are often synthesized by fermentation using a specific microorganism to convert a substrate (e.g., a sugar) into the desired product. Small quantities of other species also may be formed and must be removed to meet product specifications. For example, isoleucine (Ile), which has a molecular weight of \(131.2,\) is an essential amino acid \(^{16}\) produced by fermentation, and other amino acids such as leucine and valine also are found in the fermentation broth. The broth is subjected to several processing steps to remove these and other impurities, but final processing by crystallization is required to meet stringent specifications on purity. The strategy is to crystallize the hydrated acid form of Ile (Ile. \(\mathrm{HCl} \cdot \mathrm{H}_{2} \mathrm{O}\) ), whose crystals exclude other amino acids, and then to redissolve, neutralize, and crystallize the final Ile product. In a batch process designed to manufacture \(2500 \mathrm{kg}\) of Ile per batch, an aqueous feed solution containing 35 g Ile/dL and much lower concentrations of leucine and valine is fed to the final purification stages. The pH of the solution is 1.1 and its specific gravity is 1.02. The solution is heated to \(60^{\circ} \mathrm{C}\) and 35-wt\% HCl solution is added in a ratio of 0.4 kg per kg of feed. The addition of HCl causes the formation of crystals of Ile\cdotHCl\cdot \(\mathrm{H}_{2} \mathrm{O},\) and the production of these crystals is further increased by slowly lowering the temperature to \(20^{\circ} \mathrm{C}\). At the final crystallizer conditions the Ile solubility is \(5 \mathrm{g}\) Ile/ \(100 \mathrm{g}\) solution. The resulting slurry is sent to a centrifuge where the crystals are separated from the liquid solution and the crystal cake is washed with water. The solids leaving the centrifuge contain \(12 \%\) free water (i.e., not part of the crystal structure) and \(88 \%\) pure crystals of Ile\(\cdot \mathrm{HCl} \cdot \mathrm{H}_{2} \mathrm{O}\). \(\mathrm{H}_{2} \mathrm{O}\).The washed crystals "water to form a solution that is 4.0 g Ile/dL with gravity of 1.1. The solution is sent to an ion exchange unit where HCl is removed. Upon leaving the ion exchange unit the solution has a pH of about \(5.5 .\) It is sent to a second crystallizer where the temperature is gradually reduced to \(10^{\circ} \mathrm{C}\) and the Ile solubility is \(3.4 \mathrm{g} \mathrm{Ile} / 100 \mathrm{g} \mathrm{H}_{2} \mathrm{O}\). The crystals are separated from the slurry by centrifugation, washed with pure water, and sent to a dryer for final processing. (a) Construct a labeled flowchart for the process. (b) Choosing a basis of 1 kg of feed solution, estimate (i) the mass of HCl solution added to the system, (ii) the water added to redissolve the Ile.HCI. \(\mathrm{H}_{2} \mathrm{O}\) crystals, (iii) the mass of \(\mathrm{HCl}\) removed in the ion exchange unit, and (iv) the mass of final Ile product. (c) Scale the quantities calculated in Part (b) to the production rate of 2500 kg Ile/batch. (d) Estimate the active volume (in liters) of each of the crystallizers. (e) Amino acids are amphoteric, which means they can either donate or accept a proton \(\left(\mathrm{H}^{+}\right) .\) At low pH they tend to accept a proton and become acidic while at high pH they tend to donate a proton and become basic. They also are known as zwitterions because their ends are oppositely charged, even though the overall molecule is neutral. Isoleucine is reported to have an isoelectric point (pI) of 6.02 and \(\mathrm{pK}_{\mathrm{a}}\) values of 2.36 and \(9.60 .\) Look up the meaning of these terms and prepare a plot showing how these values are used in plotting the distribution of Ile between acid, zwitterionic (neutral), and basic forms as a function of pH. Explain why such a distribution is important in carrying out the separations described in the process.

Sulfur trioxide (SO \(_{3}\) ) dissolves in and reacts with water to form an aqueous solution of sulfuric acid \(\left(\mathrm{H}_{2} \mathrm{SO}_{4}\right) .\) The vapor in equilibrium with the solution contains both \(\mathrm{SO}_{3}\) and \(\mathrm{H}_{2} \mathrm{O}\). If enough \(\mathrm{SO}_{3}\) is added, all of the water reacts and the solution becomes pure \(\mathrm{H}_{2} \mathrm{SO}_{4}\). If still more \(\mathrm{SO}_{3}\) is added, it dissolves to form a solution of \(\mathrm{SO}_{3}\) in \(\mathrm{H}_{2} \mathrm{SO}_{4}\), called oleum or fuming sulfuric acid. The vapor in equilibrium with oleum is pure \(\mathrm{SO}_{3}\). Twenty percent oleum by definition contains \(20 \mathrm{kg}\) of dissolved \(\mathrm{SO}_{3}\) and \(80 \mathrm{kg}\) of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) per hundred kilograms of solution. Alternatively, the oleum composition can be expressed as \(\% \mathrm{SO}_{3}\) by mass, with the constituents of the oleum considered to be \(\mathrm{SO}_{3}\) and \(\mathrm{H}_{2} \mathrm{O}\). (a) Prove that a \(15.0 \%\) oleum contains \(84.4 \% \mathrm{SO}_{3}\) (b) Suppose a gas stream at \(40^{\circ} \mathrm{C}\) and 1.2 atm containing 90 mole \(\% \mathrm{SO}_{3}\) and \(10 \% \mathrm{N}_{2}\) contacts a liquid stream of 98 wt\% \(\mathrm{H}_{2} \mathrm{SO}_{4}\) (aq), producing \(15 \%\) oleum. Tabulated equilibrium data indicate that the partial pressure of \(S O_{3}\) in equilibrium with this oleum is 1.15 mm Hg. Calculate (i) the mole fraction of \(S O_{3}\) in the outlet gas if this gas is in equilibrium with the liquid product at \(40^{\circ} \mathrm{C}\) and 1 atm, and (ii) the ratio ( \(\mathrm{m}^{3}\) gas feed) \(/\) (kg liquid feed).

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.
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