/*! 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 fuel gas containing \(86 \%\) methane, \(8 \%\) ethane, and \(6 \%\) propane by volume flows to a furnace at a rate of \(1450 \mathrm{m}^{3} / \mathrm{h}\) at \(15^{\circ} \mathrm{C}\) and \(150 \mathrm{kPa}\) (gauge), where it is burned with \(8 \%\) excess air. Calculate the required flow rate of air in SCMH (standard cubic meters per hour).

Steam reforming is an important technology for converting refined natural gas, which we take here to be methane, into a synthesis gas that can be used to produce a varicty of other chemical compounds. For example, consider a reformer to which natural gas and steam are fed in a ratio of 3.5 moles of steam per mole of methane. The reformer operates at 18 atm, and the reaction products leave the reformer in chemical equilibrium at \(875^{\circ} \mathrm{C}\). The steam reforming reaction is $$\mathrm{CH}_{4}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{CO}+3 \mathrm{H}_{2}$$ and the water-gas shift reaction also occurs in the reformer. $$\mathrm{CO}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{CO}_{2}+\mathrm{H}_{2}$$ The equilibrium constants for these two reactions are given by the expressions At \(875^{\circ} \mathrm{C}, K_{\mathrm{R}}=872.9 \mathrm{atm}^{2}\) and \(K \mathrm{w} \mathrm{G}=0.2482 .\) The process is to produce \(100.0 \mathrm{kmol} / \mathrm{h}\) of hydrogen. Calculate the feed rates (kmol/h) of methane and steam and the volumetric flow rate \(\left(\mathrm{m}^{3} / \mathrm{min}\right)\) of gas leaving the reformer.

A process stream flowing at \(35 \mathrm{kmol} / \mathrm{h}\) contains 15 mole \(\%\) hydrogen and the remainder 1 -butene. The stream pressure is 10.0 atm absolute, the temperature is \(50^{\circ} \mathrm{C}\), and the velocity is \(150 \mathrm{m} / \mathrm{min}\). Determine the diameter (in \(\mathrm{cm}\) ) of the pipe transporting this stream, using Kay's rule in your calculations.

When a liquid or a gas occupies a volume, it may be assumed to fill the volume completely. On the other hand, when solid particles occupy a volume, there are always spaces (voids) among the particles. The porosity or void fraction of a bed of particles is the ratio (void volume)/(total bed volume). The bulk density of the solids is the ratio (mass of solids)/(total bed volume), and the absolute density of the solids has the usual definition (mass of solids)/(volume of solids). Suppose \(600.0 \mathrm{g}\) of a crushed ore is placed in a graduated cylinder, filling it to the \(184 \mathrm{cm}^{3}\) level. One hundred \(\mathrm{cm}^{3}\) of water is then added to the cylinder, whereupon the water level is observed to be at the \(233.5 \mathrm{cm}^{3}\) mark. Calculate the porosity of the dry particle bed, the bulk density of the ore in this bed, and the absolute density of the ore.

Chemicals are stored in a laboratory with volume \(V\left(\mathrm{m}^{3}\right) .\) As a consequence of poor laboratory practices, a hazardous species, A, enters the room air (from inside the room) at a constant rate \(\dot{m}_{\mathrm{A}}(\mathrm{g} \mathrm{A} / \mathrm{h})\) The room is ventilated with clean air flowing at a constant rate \(\dot{V}_{\text {air }}\left(\mathrm{m}^{3} / \mathrm{h}\right) .\) The average concentration of A in the room air builds up until it reaches a steady-state value \(C_{\mathrm{A}, \mathrm{r}}\left(\mathrm{g} \mathrm{A} / \mathrm{m}^{3}\right)\) (a) List at least four situations that could lead to A getting into the room air. (b) Assume that the A is perfectly mixed with the room air and derive the formula $$\dot{m}_{\mathrm{A}}=\dot{V}_{\mathrm{air}} C_{\mathrm{A}}$$ (c) The assumption of perfect mixing is never justificd when the enclosed space is a room (as opposed to, say, a stirred reactor). In practice, the concentration of A varies from one point in the room to another: it is relatively high near the point where A enters the room air and relatively low in regions far from that point, including the ventilator outlet duct. If we say that \(C_{\mathrm{A}, \text { duct }}=k C_{\mathrm{A}}\) where \(k < 1\) is a nonideal mixing factor (generally between 0.1 and \(0.5,\) with the lowest value corresponding to the poorest mixing), then the equation of Part (b) becomes $$\dot{m}_{\mathrm{A}}=k \dot{V}_{\mathrm{air}} C_{\mathrm{A}}$$ Use this equation and the ideal-gas equation of state to derive the following expression for the average mole fraction of \(A\) in the room air: $$y_{\mathrm{A}}=\frac{\dot{m}_{\mathrm{A}}}{k \dot{V}_{\mathrm{air}}} \frac{R T}{M_{\mathrm{A}} P}$$ where \(M_{\mathrm{A}}\) is the molecular weight of \(\mathrm{A}\) (d) The permissible exposure level (PEL) for styrene \((M=104.14\) ) defined by the U.S. Occupational Safcty and Health Administration is 50 ppm (molar basis). \(^{21}\) An open storage tank in a polymerization laboratory contains styrene. The evaporation rate from this tank is estimated to be \(9.0 \mathrm{g} / \mathrm{h}\). Room temperature is \(20^{\circ} \mathrm{C}\). Assuming that the laboratory air is reasonably well mixed (so that \(k=0.5\) ), calculate the minimum ventilation rate \(\left(\mathrm{m}^{3} / \mathrm{h}\right)\) required to keep the average styrene concentration at or below the PEL. Then give several reasons why working in the laboratory might still be hazardous if the calculated minimum ventilation rate is used. (e) Would the hazard level in the situation described in Part (d) increase or decrease if the temperature in the room were to increase? (Increase, decrease, no way to tell.) Explain your answer, citing at least two effects of temperature in your explanation.

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