/*! 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 23 Spray drying is a process in whi... [FREE SOLUTION] | 91Ó°ÊÓ

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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.

Short Answer

Expert verified
The estimated production rate of dried milk, volumetric flow rate of the inlet air, and upward velocity at the bottom of the dryer are to be calculated in the mathematical part. In the analytical part, it is stated that while increasing the flow rates can lead to incomplete drying, using two dryers will take up more space and have higher cost, yet having a larger dryer might lead to inefficiencies in the process.

Step by step solution

01

Calculation of Outlet Air Flow Rate

Since the outlet gas contains 12 mole \% water, it can be inferred that it contains 88 mole \% dry air. From the ideal gas law, the outlet air flow rate \(Q_{out}\) can be calculated by using the formula \(Q_{out} = 311 m^{3/min} * (1-0.12)\)
02

Calculation of Air Flow Rate at Inlet and Milk Production Rate

Using the balance equation, flowrate of air at the inlet \(Q_{in}\) can be determined by subtracting the mass flowrate of water evaporated from the outlet air flowrate, \(Q_{in} = Q_{out} - (mass flow rate of water evaporated)\). Afterwards, the production rate of dried milk can be calculated by the formula \(Q_{milk} = flow rate of atomized milk - mass flow rate of water evaporated\).
03

Calculation of Upward Velocity of Air

The upward velocity of air at the bottom of the dryer can be calculated using the formula \(v = Q_{in} / Area\), where ‘Area’ is the cross-sectional area of the dryer.
04

Analyzing the Options for Increasing Production

Increasing the flow rates may lead to insufficient contact time between the hot air and the atomized milk, thus all the water may not evaporate. Operating two identical dryers could double the production, but the cost and space required are significant. Using a larger dryer is another solution, but it may require modifying the existing facilities and also lead to inefficiency, as the contact time could be more difficult to control with a larger size.
05

Estimation of the Size of the Larger Unit

The dimensions of a larger unit could be estimated by assuming that the volume should be doubled to achieve double the production rate, while maintaining the approximate physical shape of the dryer (a cylinder) for the sake of simplicity.

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Key Concepts

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

Chemical Engineering
Spray drying is commonly utilized in chemical engineering to produce dry powders from liquid substances. It involves atomizing a liquid feed into fine droplets and introducing it to a stream of hot gas. This technique rapidly removes moisture, resulting in a dry, powdery product.
The method is ideal for maintaining the integrity of heat-sensitive materials. Chemical engineers select this process due to its efficiency and ability to produce uniform particle sizes. Understanding the thermodynamics and kinetics of heat transfer and evaporation is crucial here.
Engineering considerations include choosing the correct nozzle or disk for atomization, optimizing gas flow rates, and controlling temperature for effective drying. This requires a strong grasp of the physical and chemical properties involved.
Mass Balance
Mass balance is a fundamental concept used in evaluating processes like spray drying. It ensures that the quantity of each component at the start and end of a process remains consistent.
In the context of the spray drying process, calculating mass balance involves determining the mass flow rates of water and air. By applying the law of conservation of mass, we calculate the mass rate of the dried product. The equation starts with the incoming mass of atomized milk and subtracts the mass of water that evaporates.
This balance helps understand and optimize the process, ensuring no discrepancies occur in the input and output mass. It's a critical step that allows engineers to predict and verify the production output efficiently.
Process Optimization
Process optimization in the spray drying operation involves fine-tuning the different variables involved to enhance efficiency.
Engineers must consider the condition of the air in terms of temperature and flow, as well as the feed rate of the liquid. An increase in production demand brings challenges, such as ensuring complete evaporation of water without simply increasing feed rates.
To double production, engineers might explore adding equipment or altering system parameters. Options include utilizing multiple dryers in parallel or investing in a single, larger dryer. These choices involve assessing costs, spatial requirements, and potential changes in drying efficiency.
Each modification must be assessed for its viability, the energy required, and whether it meets product moisture content criteria. The key here is achieving more output with minimal resource increase.
Drying Technology
Drying technology is a pivotal element in the spray drying process, focusing on the removal of moisture from the feed.
The technology must ensure efficient moisture evaporation so that the final product adheres to desired consistency and shelf-stability. Hot air temperatures and dryer design, such as height and diameter, play essential roles in ensuring optimal drying conditions.
When considering upgrades to the drying system, engineers evaluate how moisture will be managed. Upgrading to a larger or additional drying unit involves reassessing airflow pathways to ensure uniform drying.
This aspect of the technology requires expertise in thermodynamics, understanding material properties, and designing effective heat transfer systems. Innovations in this field continually enhance drying rates, energy efficiency, and product quality.

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Most popular questions from this chapter

A slurry contains crystals of copper sulfate pentahydrate \(\left[\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}(\mathrm{s}), \text { specific gravity }=2.3\right]\) suspended in an aqueous copper sulfate solution (liquid SG \(=1.2\) ). A sensitive transducer is used to measure the pressure difference, \(\Delta P(\mathrm{Pa}),\) between two points in the sample container separated by a vertical distance of \(h\) meters. The reading is in turn used to determine the mass fraction of crystals in the slurry, \(x_{\mathrm{c}}(\mathrm{kg}\) crystals/kg slurry). (a) Derive an expression for the transducer reading, \(\Delta P(\mathrm{Pa}),\) in terms of the overall slurry density, \(\rho_{\mathrm{s}}\left(\mathrm{kg} / \mathrm{m}^{3}\right),\) assuming that the equation used to calculate the pressure head in Chapter 3 \(\left(P=P_{0}+\rho g h\right)\) is valid for this two-phase system. (b) Validate the following expression relating the overall slurry density to the liquid and solid crystal densities \(\left(\rho_{1} \text { and } \rho_{c}\right)\) and the mass fraction of crystals in the slurry: $$\frac{1}{\rho_{\mathrm{si}}}=\frac{x_{\mathrm{c}}}{\rho_{\mathrm{c}}}+\frac{\left(1-x_{\mathrm{c}}\right)}{\rho_{1}}$$ (c) Suppose \(175 \mathrm{kg}\) of the slurry is placed in the sample container with \(h=0.200 \mathrm{m}\) and a transducer reading \(\Delta P=2775\) Pa is obtained. Calculate \((\mathrm{i}) \rho_{\mathrm{s}_{\mathrm{s}},(\mathrm{ii})} x_{\mathrm{c}},\) (iii) the total slurry volume, (iv) the mass of crystals in the slurry, (v) the mass of anhydrous copper sulfate (CuSO \(_{4}\) without the water of hydration) in the crystals, (vi) the mass of liquid solution, and (vii) the volume of liquid solution. (d) Prepare a spreadsheet to generate a calibration curve of \(x_{c}\) versus \(\Delta P\) for this device. Take as inputs \(\rho_{\mathrm{c}}\left(\mathrm{kg} / \mathrm{m}^{3}\right), \rho_{1}\left(\mathrm{kg} / \mathrm{m}^{3}\right),\) and \(h(\mathrm{m}),\) and calculate \(\Delta P(\mathrm{Pa})\) for \(x_{\mathrm{c}}=0.0,0.05,0.10, \ldots, 0.60\) Run the program for the parameter values in this problem \(\left(\rho_{\mathrm{c}}=2300, \rho_{1}=1200, \text { and } h=0.200\right)\) Then plot \(x_{c}\) versus \(\Delta P\) (have the spreadsheet program do it, if possible), and verify that the value of \(x_{c}\) corresponding to \(\Delta P=2775\) Pa on the calibration curve corresponds to the value calculated in Part (c). (e) Derive the expression in Part (b). Take a basis of \(1 \mathrm{kg}\) of slurry \(\left[x_{\mathrm{c}}(\mathrm{kg}), V_{c}\left(\mathrm{m}^{3}\right)\right.\) crystals, \(\left.\left(1-x_{\mathrm{c}}\right)(\mathrm{kg}), V_{l}\left(\mathrm{m}^{3}\right) \text { liquid }\right],\) and use the fact that the volumes of the crystals and liquid are additive.

Lewis \(^{12}\) describes the hazards of breathing air containing appreciable amounts of an asphyxiant (a gas that has no specific toxicity but, when inhaled, excludes oxygen from the lungs). When the mole percent of the asphyxiant in the air reaches \(50 \%,\) marked symptoms of distress appear, and at \(75 \%\) death occurs in a matter of minutes. A small storage room whose dimensions are \(2 \mathrm{m} \times 1.5 \mathrm{m} \times 3 \mathrm{m}\) contains a number of expensive and dangerous chemicals. To prevent unauthorized entry, the door to the room is always locked and can be opened with a key from either side. A cylinder of liquid carbon dioxide is stored in the room. The valve on the cylinder is faulty and some of the contents have escaped over the weekend. The room temperature is \(25^{\circ} \mathrm{C}\). (a) If the concentration of \(\mathrm{CO}_{2}\) reaches the lethal 75 mole \(\%\) level, what would be the mole percent of \(\mathrm{O}_{2} ?\) (b) How much \(\mathrm{CO}_{2}(\mathrm{kg})\) is present in the room when the lethal concentration is reached? Why would more than that amount have to escape from the cylinder for this concentration to be reached? (c) Describe a set of events that could result in a fatality in the given situation. Suggest at least two measures that would reduce the hazards associated with storage of this scemingly harmless substance.

An ideal-gas mixture contains \(35 \%\) helium, \(20 \%\) methane, and \(45 \%\) nitrogen by volume at 2.00 atm absolute and \(90^{\circ} \mathrm{C}\). Calculate (a) the partial pressure of each component, (b) the mass fraction of methane, (c) the average molecular weight of the gas, and (d) the density of the gas in \(\mathrm{kg} / \mathrm{m}^{3}\).

The van der Waals equation of state (Equation \(5.3-7\) ) is to be used to estimate the specific molar volume \(\hat{V}(\mathrm{L} / \mathrm{mol})\) of air at specified values of \(T(\mathrm{K})\) and \(P(\mathrm{atm}) .\) The van der Waals constants for air are \(a=1.33 \mathrm{atm} \cdot \mathrm{L}^{2} / \mathrm{mol}^{2}\) and \(b=0.0366 \mathrm{L} / \mathrm{mol}\) (a) Show why the van der Waals equation is classified as a cubic equation of state by expressing it in the form $$f(\hat{V})=c_{3} \hat{V}^{3}+c_{2} \hat{V}^{2}+c_{1} \hat{V}+c_{0}=0$$ where the coefficients \(c_{3}, c_{2}, c_{1},\) and \(c_{0}\) involve \(P, R, T, a,\) and \(b .\) Calculate the values of these coefficients for air at \(223 \mathrm{K}\) and 50.0 atm. (Include the units when giving the values.) (b) What would the value of \(\hat{V}\) be if the ideal-gas equation of state were used for the calculation? Use this value as an initial estimate of \(\tilde{V}\) for air at \(223 \mathrm{K}\) and 50.0 atm and solve the van der Waals equation using Goal Seek or Solver in Excel. What percentage error results from the use of the ideal-gas equation of state, taking the van der Waals estimate to be correct? (c) Set up a spreadsheet to carry out the calculations of Part (b) for air at \(223 \mathrm{K}\) and several pressures. The spreadsheet should appear as follows: The polynomial expression for \(\hat{V}\left(f=c_{3} \hat{V}^{3}+c_{2} \hat{V}^{2}+\cdots\right)\) should be entered in the \(f(V)\) column, and the value in the \(V\) column should be determined using Goal Seek or Solver in Excel.

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?

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