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Chapter 8: Problem 67

An aqueous slurry at \(30^{\circ} \mathrm{C}\) containing \(20.0 \mathrm{wt} \%\) solids is fed to an evaporator in which enough water is vaporized at 1 atm to produce a product slurry containing 35.0 wt\% solids. Heat is supplied to the evaporator by feeding saturated steam at 2.6 bar absolute into a coil immersed in the liquid. The steam condenses in the coil, and the slurry boils at the normal boiling point of pure water. The heat capacity of the solids may be taken to be half that of liquid water. (a) Calculate the required steam feed rate ( \(\mathrm{kg} / \mathrm{h}\) ) for a slurry feed rate of \(1.00 \times 10^{3} \mathrm{kg} / \mathrm{h}\). (b) Vapor recompression is often used in the operation of an evaporator. Suppose that the vapor (steam) generated in the evaporator described above is compressed to 2.6 bar and simultaneously heated to the saturation temperature at 2.6 bar, so that no condensation occurs. The compressed steam and additional saturated steam at 2.6 bar are then fed to the evaporator coil, in which isobaric condensation occurs. How much additional steam is required? (c) What more would you need to know to determine whether or not vapor recompression is economically advantageous in this process?

Short Answer

Expert verified
The steam feed rate required is \(6,629\, \mathrm{kg/h}\). No additional steam is necessary for vapor recompression. Information on the costs of steam, the compressor, and operation, as well as the production rate, would be necessary to determine the economic advantages of vapor recompression.

Step by step solution

01

Find Mass of Water to be Removed

To increase the weight percent of solids from 20.0% to 35.0%, you first need to find out how much water needs to be evaporated from the slurry. This can be calculated using weight percentage formula. Let \(x\) be the mass of water to be evaporated. Therefore, \(\frac{200}{1000 - x} = 0.35\) which gives \(x = 357.14\) kg.
02

Calculate Heat Removal Using Latent Heat

The heat required to remove this amount of water can be calculated via the latent heat of vaporization of water, \(H_v\), leading to \(Q = m \cdot H_v\). Using \(H_v = 40,700 \,\mathrm{kJ / kg}\) and \(m = 357.14 \,\mathrm{kg}\), we get \(Q = 14,540,000\, \mathrm{kJ}\).
03

Calculate Steam Feed Rate

This heat, \(Q\), is supplied by the steam fed into the evaporator. The amount of steam required can be calculated by dividing the heat required by the latent heat of condensation of steam, \(H_c\), at 2.6 bar. Using \(H_c = 2,193\, \mathrm{kJ / kg}\), the steam feed rate is \(6,629\, \mathrm{kg/h}\).
04

Calculate Additional Steam Needed for Vapor Recompression

Undergoing vapor recompression, the vapor is compressed and heated to the saturation temperature at 2.6 bar. This would require additional steam to be fed to the evaporator coil. Since the heat capacity of water vapor is roughly the same as liquid water, extra steam needed can also be calculated by dividing the extra heat needed to heat the vapor to saturation temperature by \(H_c\). In this case, no additional steam is required, since in the question it is stated that the vapor is 'simultaneously heated' during compression, meaning no extra steam is used to heat the vapor.
05

Economical Considerations for Vapor Recompression

To determine the economic advantages of vapor recompression, you would need to know the capital cost of the compressor, the energy cost of running the compressor, the operational and maintenance cost of the compressor, the cost of steam, and the production rate of the evaporator. You would then compare the costs of the conventional process with those of the process incorporating vapor recompression.

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Key Concepts

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

Evaporator Efficiency
Understanding the efficiency of an evaporator is crucial in chemical process design. It represents how effectively the evaporator transfers heat to evaporate water or other solvents. The goal is to achieve the desired concentration of a solution, in this case, increasing the solids content of a slurry from 20.0 wt% to 35.0 wt% by removing water.

Efficiency is gauged by the ratio of the useful energy output (which is the energy required to vaporize the water) to the energy input (which is the heat supplied by the steam). High efficiency means less energy is wasted, making the process more cost-effective. Evaporator efficiency can be impacted by factors such as the properties of the fluid being concentrated, the heat transfer coefficients, and the operation conditions such as temperature and pressure.
Latent Heat of Vaporization
The latent heat of vaporization, denoted by \(H_v\), is the amount of heat required to turn a unit mass of a liquid into vapor without a temperature change. For water, this quantity is significant, requiring 40,700 kJ to vaporize one kilogram. This concept is invaluable when calculating energy requirements in evaporators.

In the context of the exercise, the latent heat of vaporization determines how much heat is needed to vaporize the water present in the slurry. Knowing \(H_v\), we can calculate the total heat energy necessary to achieve the desired solid content, which informs us about the steam rate needed for the evaporator's operation.
Vapor Recompression
Vapor recompression is an energy-saving measure that increases evaporator efficiency by compressing and reheating the vapor (steam) generated in the evaporator. This recycled steam is utilized to provide additional heat to the evaporator, thus requiring less fresh steam input.

The process entails taking the generated steam, increasing its pressure and thereby its saturation temperature, and reintroducing it back into the system. In the exercise, no additional steam is needed because the vapor is simultaneously heated during recompression, using the energy already present within the steam. This suggests that the vapor recompression process can make steam usage more efficient and help lower operational costs if initial capital and maintenance costs are justified.
Mass and Energy Balances
Mass and energy balances are fundamental principles in chemical engineering that ensure that all mass and energy entering a system are accounted for in the outputs. These balances are tools for estimating the amount of inputs (like the steam feed rate) and outputs (like evaporated water) in a process, crucial for design and optimization.

In the given exercise, we perform a mass balance to calculate the amount of water that needs to be evaporated to concentrate the slurry to the desired solids content. Subsequently, we use an energy balance to determine the heat requirement associated with this evaporation and thus the steam feed rate required at a given heat of condensation \(H_c\). These calculations dictate the design and operational parameters of the evaporator system.

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Most popular questions from this chapter

In gas adsorption a vapor is transferred from a gas mixture to the surface of a solid. (See Section \(6.7 .\) ) An approximate but useful way of analyzing adsorption is to treat it simply as condensation of vapor on a solid surface. Suppose a nitrogen stream at \(35^{\circ} \mathrm{C}\) and 1 atm containing carbon tetrachloride with a \(15 \%\) relative saturation is fed at a rate of \(10.0 \mathrm{mol} / \mathrm{min}\) to a \(6-\mathrm{kg}\) bed of activated carbon. The temperature and pressure of the gas do not change appreciably from the inlet to the outlet of the bed, and there is no \(\mathrm{CCl}_{4}\) in the gas leaving the adsorber. The carbon can adsorb 40\% of its own mass of carbon tetrachloride before becoming saturated, at which point it must be either regenerated (remove the carbon tetrachloride) or replaced with a fresh bed of activated carbon. Neglect the effect of temperature on the heat of vaporization of \(\mathrm{CCl}_{4}\) when solving the following problems: (a) Estimate the rate ( \(\mathrm{kJ} / \mathrm{min}\) ) at which heat must be removed from the adsorber to keep the process isothermal, and the time (min) it will take to saturate the bed. (b) The surface-to-volume ratio of spherical particles is \((3 / r)\left(\mathrm{cm}^{2} \text { outer surface }\right) /\left(\mathrm{cm}^{3} \text { volume }\right)\) First, derive that formula. Second, use it to explain how decreasing the average diameter of the particles in the carbon bed might make the adsorption process more efficient. Third, since most of the area on which adsorption takes place is provided by pores penetrating the particle, explain why the surface-to-volume ratio, as calculated by the above expression, might be relatively unimportant.

Your next-door neighbor, Josephine Rackstraw, surprised her husband last January by having a hot tub installed in their back yard while he was away on an ice-fishing trip. It surprised him, all right, but instead of being pleased he was horrified. "Have you lost your mind, Josephine?" he sputtered. "It will cost a fortune to keep this thing hot, and you know what the President said about conserving energy." "Don't be silly, Ralph," she replied. "It can't cost more than a few pennies a day, even in the dead of winter." "No way -just because you have a PhD, you think you're an expert on everything!" They argued for a while, bringing up several issues that each had been storing for just such an occasion. After calming down and using the tub for a week, they remembered their neighbor (i.e., you) had a chemical engineering education and came to ask if you could settle their argument. You asked a few questions, made several observations, converted everything to metric units, and arrived at the following data, all corresponding to an average outside temperature of \(5^{\circ} \mathrm{C}\). \- The tub holds 1230 liters of water. \- Ralph normally keeps the tub temperature at \(29^{\circ} \mathrm{C}\), raises it to \(40^{\circ} \mathrm{C}\) when he plans to use it, keeps it at \(40^{\circ} \mathrm{C}\) for about one hour, and drops it back to \(29^{\circ} \mathrm{C}\) when he is finished. \- During heating, it takes about three hours for the water temperature to rise from \(29^{\circ} \mathrm{C}\) to \(40^{\circ} \mathrm{C}\). When the heat is shut off, it takes eight hours for the water temperature to drop back to \(29^{\circ} \mathrm{C}\). \- Electricity costs 10 cents per kilowatt-hour. Taking the heat capacity of the tub contents to be that of pure liquid water and neglecting evaporation, answer the following questions. (a) What is the average rate of heat loss ( \(k W\) ) from the tub to the outside air? (Hint: Consider the period when the tub temperature is dropping from \(40^{\circ} \mathrm{C}\) to \(29^{\circ} \mathrm{C}\).) (b) At what average rate ( \(\mathrm{kW}\) ) does the tub heater deliver energy to the water when raising the water temperature? What is the total quantity of electricity (kW\cdoth) that the heater must deliver during this period? [Consider the result of Part (a) when performing the calculation.] (c) (These answers should settle the argument.) Consider a day in which the tub is used once. Use the results of Parts (a) and (b) to estimate the cost (S) of heating the tub from \(29^{\circ} \mathrm{C}\) to \(40^{\circ} \mathrm{C}\) and the cost \((\mathrm{S})\) of keeping the tub at a constant temperature. (There is no cost for the period in which \(T\) is dropping.) What is the total daily cost of running the tub? Assume the rate of heat loss is independent of the tub temperature. (d) The tub lid, which is an insulator, is removed when the tub is in use. Explain how this fact would probably affect your cost estimates in Part (c).

Polyvinylpyrrolidone (PVP) is a polymer product used as a binding agent in pharmaceutical applications as well as in personal-care items such as hairspray. In the manufacture of \(\mathrm{PVP}\), a spray-drying process is used to collect solid PVP from an aqueous suspension, as shown in the flowchart on the next page. A liquid solution containing 65 wt\% \(\mathrm{PVP}\) and the balance water at \(25^{\circ} \mathrm{C}\) is pumped through an atomizing nozzle at a rate of \(1500 \mathrm{kg} / \mathrm{h}\) into a stream of preheated air flowing at a rate of \(1.57 \times 10^{4}\) SCMH. The water evaporates into the stream of hot air and the solid PVP particles are suspended in the humidified air. Downstream, the particles are separated from the air with a filter and collected. The process is designed so that the exiting solid product and humid air are in thermal equilibrium with each other at \(110^{\circ} \mathrm{C}\). For convenience, the spray-drying and solid- separation processes are shown as one unit that may be considered adiabatic. (a) Draw and completely label the process flow diagram and perform a degree- of-freedom analysis. (b) Calculate the required temperature of the inlet air, \(T_{0}\), and the volumetric flow rate \(\left(\mathrm{m}^{3} / \mathrm{h}\right)\) and relative humidity of the exiting air. Assume that the polymer has a heat capacity per unit mass one third that of liquid water, and only use the first two terms of the polynomial heat-capacity formula for air in Table B.2. (c) Why do you think the polymer solution is put through an atomizing nozzle, which converts it to a mist of tiny droplets, rather than being sprayed through a much less costly nozzle of the type commonly found in showers? (d) Due to a design flaw, the polymer solution does not remain in the dryer long enough for all the water to evaporate, so the solid product emerging from the separator is a wet powder. How will this change the values of the outlet temperatures of the emerging gas and powder and the volumetric flow rate and relative humidity of the emerging gas (increase, decrease, can't tell without doing the calculations)? Explain your answers.

A natural gas containing 95 mole \(\%\) methane and the balance ethane is burned with \(20.0 \%\) excess air. The stack gas, which contains no unburned hydrocarbons or carbon monoxide, leaves the furnace at \(900^{\circ} \mathrm{C}\) and \(1.2 \mathrm{atm}\) and passes through a heat exchanger. The air on its way to the furnace also passes through the heat exchanger, entering it at \(20^{\circ} \mathrm{C}\) and leaving it at \(245^{\circ} \mathrm{C}\). (a) Taking as a basis \(100 \mathrm{mol} / \mathrm{s}\) of the natural gas fed to the furnace, calculate the required molar flow rate of air, the molar flow rate and composition of the stack gas, the required rate of heat transfer in the preheater, \(\dot{Q}\) (write an energy balance on the air), and the temperature at which the stack gas leaves the preheater (write an energy balance on the stack gas). Note: The problem statement does not give you the fuel feed temperature. Make a reasonable assumption, and state why your final results should be nearly independent of what you assume. (b) What would \(\dot{Q}\) be if the actual feed rate of the natural gas were 350 SCMH [standard cubic meters per hour, \(\left.\mathrm{m}^{3}(\mathrm{STP}) / \mathrm{h}\right] ?\) Scale up the flowchart of Part (a) rather than repeating the entire calculation.

The heat capacities of a substance have been defined as $$C_{v}=\left(\frac{\partial \hat{U}}{\partial T}\right)_{V}, \quad C_{p}=\left(\frac{\partial \hat{H}}{\partial T}\right)_{P}$$ Use the defining relationship between \(\hat{H}\) and \(\hat{U}\) and the fact that \(\hat{H}\) and \(\hat{U}\) for ideal gases are functions only of temperature to prove that \(C_{p}=C_{v}+R\) for an ideal gas (Eq. \(8.3-12\) ).
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