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Chapter 8: Problem 48

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.

Short Answer

Expert verified
a) To keep the process isothermal, the heat that needs to be removed from the adsorber can be found by calculating the heat released as a result of the adsorption of the carbon tetrachloride. The time taken to saturate the bed is related to the amount of carbon tetrachloride the activated carbon bed can adsorb, and the rate at which carbon tetrachloride enters the carbon bed. \n b) The surface-to-volume ratio for spherical particles is \(3/r\). The smaller the average diameter of the particles in the carbon bed, the more efficient the adsorption process is likely to be, due to the increased surface area available for adsorption. However, this ratio might not be as significant in terms of overall efficiency, since the main area for adsorption lies within the pores of the activated carbon particles.

Step by step solution

01

Determining mass of carbon tetrachloride in the gas

First, quantify how much carbon tetrachloride (\(CCl_{4}\)) is in the gas. Given nitrogen stream is fed at a rate of \(10.0 \, \mathrm{mol} / \mathrm{min}\) and the relative saturation of carbon tetrachloride is \(15 \% \), we can multiply these two values to find the molar amount of carbon tetrachloride in the incoming gas, then convert this to mass using the molar mass of carbon tetrachloride.
02

Calculating saturation time

To calculate the time it would take to saturate the carbon bed, we need to know the maximum amount of carbon tetrachloride the carbon bed can adsorb. Given the carbon bed weighs 6 kg and can adsorb 40% of its mass in carbon tetrachloride, we multiply the mass of the carbon bed by 0.40. Then we divide this maximum adsorbable mass by the mass flow rate of \(CCl_{4}\) calculated in Step 1. The result is the saturation time of the bed in minutes.
03

Calculating rate of heat removal

To determine the rate at which heat must be removed from the process, we must calculate the heat associated with the adsorption process. This can be found by multiplying the mass flow rate of \(CCl_{4}\) by the heat of condensation of \(CCl_{4}\), which is the energy released when one gram of \(CCl_{4}\) is adsorbed (i.e., changes from a gas to a condensed layer on the activated carbon). This will give the heat released in kJ/min. Because this is a constant temperature (isothermal) process, this is the heat that must be removed from the process.
04

Deriving the formula for the surface-to-volume ratio

For any sphere, the volume \(V = 4/3 \蟺 r^{3}\) and the surface area \(S = 4 \蟺 r^{2}\). Therefore, the surface-to-volume ratio is \(S/V = 3/r\).
05

Explaining the effect of particle size on adsorption efficiency

Smaller particles have a larger surface area for a given volume, hence a larger surface-to-volume ratio. This means that there are more surface sites available for the gas molecules to be adsorbed, improving the efficiency of the adsorption process. Particles with a small average diameter will have larger surface area accessible, resulting in a higher adsorption rate.
06

Discussing the significance of the surface to volume ratio

The surface-to-volume ratio as calculated by the above formula might not be that important in the overall efficiency of the adsorption process because most of the adsorption takes place not on the outside surface of the particles, but within the pores of the activated carbon particles. Hence, pore structure and porosity of the carbon particles could have a bigger impact on the efficiency of the adsorption process than the external surface-to-volume ratio of the carbon particles.

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Key Concepts

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

Activated Carbon
Activated carbon is a highly porous form of carbon with a large surface area, which makes it extremely effective for gas adsorption processes. The porous nature of activated carbon provides numerous sites for gas molecules such as carbon tetrachloride ( CCl_4). These molecules are adsorbed onto the surfaces within these pores, effectively removing them from a gas mixture.
Activated carbon is often used due to its ability to adsorb various gases and impurities, which helps in purifying air or separating components of gas mixtures.
  • High porosity and surface area allow for greater adsorption capacity.
  • It can be regenerated after saturation, which extends its usability.
  • Widely used in gas and liquid purification processes.
Understanding its structure and kinetics is key to optimizing adsorption processes in industrial applications.
Surface-to-Volume Ratio
In the context of adsorption, the surface-to-volume ratio is an important factor when considering the efficiency of adsorbent materials like activated carbon. This ratio indicates how much surface area is available for adsorption relative to the volume of the material.
The formula for the surface-to-volume ratio of a sphere is given by the expression \( \frac{3}{r} \), where \( r \) is the radius of the sphere. Hence, small particles with a smaller radius have a larger surface area for adsorption relative to their volume:
  • A higher surface-to-volume ratio increases the available adsorption sites.
  • For smaller average diameter particles, the larger the surface area that might enhance adsorption efficiency, particularly at the outer surface.
However, in activated carbon, not all adsorption occurs on the external surface. The abundant pores in the particles form an extensive internal surface where much of the adsorption takes place, making pore structure more crucial than this simple ratio suggests.
Saturation Time
Saturation time is the period required for a given amount of adsorbent, such as activated carbon, to become fully saturated with the adsorbate, like carbon tetrachloride ( CCl_4).
This is a key parameter when designing adsorption processes, as it influences how often the adsorbent must be regenerated or replaced:
  • Saturation time depends on the rate of adsorption and the adsorption capacity of the material.
  • In practice, knowing the saturation time helps optimize the maintenance schedule.
  • To determine the saturation time, consider the mass flow rate and maximum adsorbable mass.
In a stable process, the balance between adsorption rate and mass capacity allows for precise control over how frequently adsorption beds need servicing.
Heat of Adsorption
The heat of adsorption refers to the energy change that occurs when a gas is adsorbed onto a solid surface like activated carbon. It's an important factor in understanding and managing the adsorption process, particularly in isothermal systems where temperature regulation is crucial.
This energy release needs to be managed by removing the equivalent heat to maintain isothermal conditions. The heat of adsorption provides insights such as:
  • Energy is released when a gas condenses on a solid, similar to condensation heat.
  • For carbon tetrachloride, this entails the heat released when adsorption occurs, requiring cooling mechanisms to remove the heat.
  • Understanding this process helps design efficient cooling systems to maintain stability and efficiency in adsorption processes.
Overall, proper management of heat of adsorption is vital to ensure the thorough operation of an isothermal adsorption system.

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Most popular questions from this chapter

The heat capacity at constant pressure of a gas is determined experimentally at several temperatures, with the following results: $$\begin{array}{|l|c|c|c|c|c|c|c|} \hline T\left(^{\circ} \mathrm{C}\right) & 0 & 100 & 200 & 300 & 400 & 500 & 600 \\ \hline C_{p}\left[\mathrm{J} /\left(\mathrm{mol} \cdot^{\circ} \mathrm{C}\right)\right] & 33.5 & 35.1 & 36.7 & 38.4 & 40.2 & 42.0 & 43.9 \\ \hline \end{array}$$ (a) Calculate the heat (kW) required to raise 150 mol/s of the gas from 0^ C to 600^'C, using Simpson's rule (Appendix A.3) to integrate the tabulated heat capacities. (b) Use the method of least squares (Appendix A.1) to derive a linear expression for \(C_{p}(T)\) in the range \(0^{\circ} \mathrm{C}\) to \(600^{\circ} \mathrm{C},\) and use this expression to estimate once again the heat ( \(\mathrm{kW}\) ) required to raise 150 mol/s of the gas from 0 ^ C to 600^'C. If the estimates differ, in which one would you have more confidence, and why?

Ever wonder why espresso costs much more per cup than regular drip coffee? Part of the reason is the expensive equipment needed to brew a proper espresso. A high-powered burr grinder first shears the coffee beans to a fine powder without producing too much heat. (Heating the coffee in the grinding stage prematurely releases the volatile oils that give espresso its rich flavor and aroma.) The ground coffee is put into a cylindrical container called a gruppa and tamped down firmly to provide an even flow of water through it. An electrically heated boiler inside the espresso machine maintains water in a reservoir at 1.4 bar and \(109^{\circ} \mathrm{C}\). An electric pump takes cold water at \(15^{\circ} \mathrm{C}\) and 1 bar, raises its pressure to slightly above 9 bar, and feeds it into a heating coil that passes through the reservoir. Heat transferred from the reservoir through the coil wall raises the water temperature to \(96^{\circ} \mathrm{C}\). The heated water flows into the top of the gruppa at \(96^{\circ} \mathrm{C}\) and 9 bar, passes slowly through the tightly packed ground beans, and dissolves the oils and some of the solids in the beans to become espresso, which decompresses to 1 atm as it exits the machine. The water temperature and uniform flow through the bed of packed coffee in the gruppa lead to the more intense flavor of espresso relative to normal drip coffee. Water drawn directly from the reservoir is expanded to atmospheric pressure where it forms steam, which is used to heat and froth milk for lattes and cappuccinos. (a) Sketch this process, using blocks to represent the pump, reservoir, and gruppa. Label all heat and work flows in the process, including electrical energy. (b) To make a 14-oz latte, you would steam 12 ounces of cold milk (3^'C) until it reaches 71 ^ C and pour it over 2 ounces of espresso. Assume that the steam cools but none of it condenses as it bubbles through the milk. For each latte made, the heating element that maintains the reservoir temperature must supply enough energy to heat the espresso water plus enough to heat the milk, plus additional energy. Assuming $$ \left(C_{p}\right)_{\operatorname{milk}}=3.93 \frac{\mathrm{J}}{\mathrm{g} \cdot^{\circ} \mathrm{C}}, \quad \mathrm{SG}_{\mathrm{milk}}=1.03 $$ calculate the quantity of electrical energy that must be provided to the heating element to accomplish those two functions. Why would more energy than what you calculate be required? (There are several reasons.) (c) Coffee beans contain a considerable amount of trapped carbon dioxide, not all of which is released when the beans are ground. When the hot pressurized water percolates through the ground beans, some of the carbon dioxide is absorbed in the liquid. When the liquid is then dispensed at atmospheric pressure, fine \(\mathrm{CO}_{2}\) bubbles come out of solution. In addition, one of the chemical compounds formed when the coffee beans are roasted and extracted into the espresso is melanoidin, a surfactant. Surfactant molecules are asymmetrical, with one end being hydrophilic (drawn to water) and the other end hydrophobic (repelled by water). When the bubbles (thin water films containing \(\mathrm{CO}_{2}\) ) pass through the espresso liquid, the hydrophilic ends of the melanoidin molecules attach to the bubbles and the dissolved bean oils in turn attach to the hydrophobic ends. The result is that the bubbles emerge coated with the oils to form the crema , the familiar reddish brown stable foam at the surface of good espresso. Speculate on why you don't see crema in normal drip coffee. (Hint: Henry's law should show up in your explanation.) Note: All soaps and shampoos contain at least one surfactant species. (A common one is sodium lauryl sulfate.) Its presence explains why if you have greasy hands, washing with plain water may leave the grease untouched but washing with soap removes the grease. (d) Explain in your own words (i) how espresso is made, (ii) why espresso has a more intense flavor than regular drip coffee, (iii) what the crema in espresso is, how it forms, and why it doesn't appear in regular drip coffee, and (iv) why washing with plain water does not remove grease but washing with soap does. (Note: Many people automatically assume that all chemical engineers are extraordinarily intelligent. If you can explain those four things, you can help perpetuate that belief.)

A gas containing water vapor has a dry-basis composition of 7.5 mole \(\%\) CO, \(11.5 \%\) CO \(_{2}, 0.5 \%.\) \(\mathrm{O}_{2},\) and \(80.5 \% \mathrm{N}_{2} .\) The gas leaves a catalyst regeneration unit at \(620^{\circ} \mathrm{C}\) and 1 atm with a dew point of \(57^{\circ} \mathrm{C}\) at a flow rate of \(28.5 \mathrm{SCMH}\left[\mathrm{m}^{3}(\mathrm{STP}) / \mathrm{h}\right] .\) Valuable solid catalyst particles entrained in the gas are to be recovered in an electrostatic precipitator, but the gas must first be cooled to \(425^{\circ} \mathrm{C}\) to prevent damage to the precipitator electrodes. The cooling is accomplished by spraying water at \(20^{\circ} \mathrm{C}\) into the gas. (a) Use simultaneous material and energy balances on the spray cooler to calculate the required water feed rate ( \(\mathrm{kg} / \mathrm{h}\) ). Treat the spray cooler as adiabatic and neglect the heat transferred from the entrained solid particles as they cool. (b) In terms that a high school senior could understand, explain the operation of the spray cooler in this problem. (What happens when the cold water contacts the hot gas?)

The specific internal energy of formaldehyde (HCHO) vapor at 1 atm and moderate temperatures is given by the formula $$\hat{U}(\mathrm{J} / \mathrm{mol})=25.96 T+0.02134 T^{2}$$ where \(T\) is in \(^{\circ} \mathrm{C}\) (a) Calculate the specific internal energies of formaldehyde vapor at \(0^{\circ} \mathrm{C}\) and \(200^{\circ} \mathrm{C}\). What reference temperature was used to generate the given expression for \(\hat{U} ?\) (b) The value of \(\hat{U}\) calculated for \(200^{\circ} \mathrm{C}\) is not the true value of the specific internal energy of formaldehyde vapor at this condition. Why not? (Hint: Refer back to Section 7.5a.) Briefly state the physical significance of the calculated quantity. (c) Use the closed system energy balance to calculate the heat (J) required to raise the temperature of 3.0 mol HCHO at constant volume from 0^0 C to 200^'C. List all of your assumptions. (d) From the definition of heat capacity at constant volume, derive a formula for \(C_{v}(T)\left[\mathrm{J} /\left(\mathrm{mol} \cdot^{\circ} \mathrm{C}\right)\right]\) Then use this formula and Equation \(8.3-6\) to calculate the heat \((\) J) required to raise the temperature of 3.0 mol of HCHO(v) at constant volume from 0^ C to 200^'C. [You should get the same result you got 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.
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