## 20 Absorption ABS

ABS.l DESCRIPTION OF ABSORPTION

Briefly describe the absorption process.

### Solution

Gas absorption, as applied to the control of air pollution, is concerned with the removal of one or more pollutants from a contaminated gas stream by treatment with a liquid. The necessary condition is the solubility of these pollutants in the absorbing liquid. The rate of transfer of the soluble constituents from the gas to the liquid phase is determined by diffusional processes occurring on each side of the gas-liquid interface.

Consider, for example, the process taking place when a mixture of air and sulfur dioxide is brought into contact with water. The S02 is soluble in water, and those molecules that come into contact with the water surface dissolve fairly rapidly. However, the S02 molecules are initially dispersed throughout the gas phase, and they can only reach the water surface by diffusing through the air, which is substantially insoluble in the water. When the S02 at the water surface has dissolved, it is distributed throughout the water phase by a second diffusional process. Consequently, the rate of absorption is determined by the rates of diffusion in both the gas and liquid phases.

Equilibrium is another extremely important factor to be considered in controlling the operation of absorption systems. The rate at which the pollutant will diffuse into an absorbent liquid will depend on the departure from equilibrium that is maintained. The rate at which equilibrium is established is then essentially dependent on the rate of diffusion of the pollutant through the nonabsorbed gas and through the absorbing liquid. Equilibrium concepts and relationships are considered in a later problem.

The rate at which the pollutant mass is transferred from one phase to another depends also on a so-called mass transfer, or rate, coefficient, which equates the quantity of mass being transferred with the driving force. As can be expected, this transfer process would cease upon the attainment of equilibrium.

Gas absorption can be viewed as a mass transfer, or diffusional operation, characterized by a transfer of one substance through another, usually on a molecular scale. The mass transfer process may be considered the result of a concentration difference driving force, the diffusing substance moving from a place of relatively high to one of relatively low concentration. The rate at which this mass is transferred depends to a great extent on the diffusional characteristics of both the diffusing substance and the medium.

The principal types of gas absorption equipment may be classified as follows:

1. Packed columns (continuous operation)

2. Plate columns (staged operation)

### 3. Miscellaneous

Of the three categories, the packed column is by far the most commonly used for the absorption of gaseous pollutants. It might also be mentioned at this time that the exhaust (cleaned gas) from an absorption air pollution control system is usually released to the atmosphere through a stack. To prevent condensation in and around the stack, the temperature of this exhaust gas should be above its dew point. A general rule of thumb is to ensure that the exhaust gas stream temperature is approximately 50°F above its dew point.

ABS.2 SOLVENT SELECTION

List 10 factors that should be considered when choosing a solvent for a gas absorption column that is to be used as an emission control device.

Solution

Solvent selection for use in an absorption column for gaseous pollutant removal should be based upon the following criteria:

1. Solubility of the gas in the solvent: High solubility is desirable as it reduces the amount of solvent needed. Generally, a polar gas will dissolve best in a polar solvent and a nonpolar gas will dissolve best in a nonpolar solvent.

2. Vapor pressure: A solvent with a low vapor pressure is preferred to minimize loss of solvent.

3. Corrosivity: Corrosive solvents may damage the equipment. A solvent with low corrosivity will extend equipment life.

4. Cost: In general, the less expensive, the better. However, an inexpensive solvent is not always the best choice if it is too costly to dispose of and/or recycle after it has been used.

5. Viscosity: Solvents with low viscosity offer benefits such as better adsorption rates, better heat transfer properties, lower pressure drops, lower pumping costs, and improved flooding characteristics in absorption towers.

6. Reactivity: Solvents that react with the contaminant gas to produce an unreactive product are desirable since the scrubbing solution can be recircu lated while maintaining high removal efficiencies. Reactive solvents should produce few unwanted side reactions with the gases that are to be absorbed.

7. Low freezing point: Resistance to freezing lessens the chance of solid formation and clogging of the column. See 2 on boiling point (vapor pressure).

8. Availability: If the solvent is "exotic," it generally has a higher cost and may not be readily available for long-term continuous use. Water is often the natural choice based on this criteria.

9. Flammability: Lower flammability or nonflammable solvents decrease safety problems.

10. Toxicity: A solvent with low toxicity is desirable. ABS.3 OUTLET CONCENTRATION FROM A SPRAY TOWER

A waste incinerator emits 300 ppm HC1 with peak values of 600 ppm. The air flow is a constant 5000 acfm at 75°F and 1 atm. Only sketchy information was submitted with the scrubber permit application to the state for a spray tower. You are requested to determine if the spray unit is satisfactory.

Data

Emission limit = 30 ppm HC1

Maximum gas velocity allowed through the tower = 3 ft/s Number of sprays = 6 Diameter of the tower = 10 ft

The number of transfer units, NOG, to meet the regulations, assuming the worst scenario, is

where yt = inlet mole fraction y2 = outlet mole fraction

As a rule of thumb in a spray tower, the Noa of the first or top spray is 0.7. Each lower spray will have only about 60% of the N0G of the spray above it. The absorption that occurs in the inlet duct adds no height but has an 7VOG of 0.5.

Solution

First, check the gas velocity:

The number of transfer units required is

For a tower with five spray sections,

Spray Section |
Nog | |||||||||||||||||||||||||||||||||||||||||||||||||||||

Top |
0.70 | |||||||||||||||||||||||||||||||||||||||||||||||||||||

Second |
0.42 | |||||||||||||||||||||||||||||||||||||||||||||||||||||

Third |
0.25 | |||||||||||||||||||||||||||||||||||||||||||||||||||||

Fourth |
0.15 | |||||||||||||||||||||||||||||||||||||||||||||||||||||

Fifth |
0.09 | |||||||||||||||||||||||||||||||||||||||||||||||||||||

Sixth (inlet) |
0.50 | |||||||||||||||||||||||||||||||||||||||||||||||||||||

Therefore, the outlet concentration does not meet the specification! ABS.4 ABSORPTION PRINCIPLES You are given experimental data for an absorption system to be used for scrubbing ammonia (NH3) from air with water. The water rate is 300 lb/min and the gas rate is 250 lb/min at 72°F. The applicable equilibrium data for the ammonia-air system is shown in the following table (R. H. Perry and D. W. Green, Ed., Perry's Chemical Engineers' Handbook, 7th Edition, McGraw Hill, New York, 1996.)
The air to be scrubbed has 1.5% (weight basis) NH3 at 72°F and 1 atm pressure and is to be vented with 95% of the ammonia removed. The inlet scrubber water is ammonia free. 1. Plot the equilibrium data in mole fraction units. 2. Perform the material balance and plot the operating line on the equilibrium In gas absorption operations the equilibrium of interest is that between a relatively nonvolatile absorbing liquid (solvent) and a solute gas (usually the pollutant). As described earlier, the solute is ordinarily removed from a relatively large amount of a carrier gas that does not dissolve in the absorbing liquid. Temperature, pressure, and the concentration of solute in one phase are independently variable. The equilibrium relationship of importance is a plot (or data) of x, the mole fraction of solute in the liquid, against y*, the mole fraction in the vapor in equilibrium with x. For cases that follow Henry's law, Henry's law constant m, can be defined by the equation The usual operating data to be determined or estimated for isothermal systems are the liquid rate(s) and the terminal concentrations or mole fractions. An operating line that describes operating conditions in the column is obtained by a mass balance around the column (as shown in Figure 72). plot. Total moles in = Total moles out Figure 72. Material balance for the absorption of component A in an absorption column For component A, the mass (or mole) balance becomes Assuming GmX = Gm2 and LmX = Lm2 (reasonable for most air pollution control applications where contaminant concentrations are usually extremely small), then and rearranging This is the equation of a straight line known as the operating line. On x,y coordinates, it has a slope of Lm/Gm and passes through the points (xAX,yAX) and (xA2,yA2) as indicated in Figure 73. In the design of most absorption columns, the quantity of gas to be treated Gm, the terminal concentrations yAX and yA2, and the composition of the entering liquid xA2 are ordinarily fixed by process requirements; however, the quantity of liquid solvent to be used is subject to some choice. Should this quantity already be specified, the operating line in the figure is fixed. If the quantity of solvent is unknown, the operating line is consequently unknown. This can be obtained through setting the minimum liquid-to-gas ratio. With reference to Figure 73, the operating line must pass through point A and must terminate at the ordinate yAX. If such a quantity of liquid is used to determine operating line AB, the existing liquid will have the composition xAX. If less liquid is used, the exit liquid composition will clearly be greater, as at point C, but since the driving forces (displacement of the operating line from the equilibrium line) for mass transfer are less, the absorption is more difficult. The time of contact between gas and liquid must then be greater, and the absorber must be correspondingly taller. The minimum liquid used corresponds to the operating line AD, which has the greatest slope for any line touching the equilibrium curve and is tangent to the curve at E. At point E, the diffusional driving force is zero, the required contact time for the concentration change desired is infinite, and an infinitely tall column results. This then represents the limiting liquid-to-gas ratio. The importance of the minimum liquid-to-gas ratio lies in the fact that column operation is frequently specified as some factor of the minimum liquid-gas ratio. For example, a typical situation frequently encountered is that the slope of the actual operating line, (Lm/G m)act>'s 1 times the minimum, (Lm /G m)mirr Employing the data provided in the problem statement, convert the equilibrium partial pressure data and the liquid concentration data to mole fractions:
Plotting the mole fraction values on the graph in Figure 74 results in a straight line. The slope of the equilibrium line is approximately 1.0. Convert the liquid and gas rates to lbmol/min: Determine the inlet and outlet mole fractions for the gas, y{ and y2, respectively: The inlet liquid mole fraction, x2, is given as 0, and the describing equation for xb the outlet liquid mole fraction, is x, = (G/L)(y{ -y2)+x2 One may now use the inlet and outlet mole fractions to plot the operating line on the graph in Figure 75. The slope of the operating line is ABS.5 PACKING HEIGHT Pollution Unlimited, an Aldo Leone Corporation, has submitted design plans to Theodore Consultants for a packed ammonia scrubber on an airstream containing NH3. The operating and design data provided by Pollution Unlimited, Inc. are given below. Theodore Consultants remember reviewing plans for a nearly identical scrubber for Pollution Unlimited, Inc. in 1988. After consulting old files, the consultants find all the conditions were identical except for the gas flowrate. What recommendation should be made? Tower diameter = 3.57 ft Packed height of column = 8 ft Gas and liquid temperature = 75°F inlet Operating pressure = 1.0 atm Ammonia-free liquid flowrate (mass flux or mass velocity) = 1000 lb/(ft2 • h) Gas flowrate = 1575 acfm Gas flowrate in 1988 plan =1121 acfm Inlet NH3 gas concentration = 2.0 mol % Air density = 0.0743 lb/ft3 Molecular weight of air = 29 Henry's law constant, m = 0.972 Molecular weight of water = 18 Figure 76 (packing "A" is used) Colburn chart Emission regulation = 0.1% NH3 (by mole or volume) Packing height = //(Xl/VOG Solution Calculate the cross-sectional area of the tower, S, in ft2: 0 500 1000 1500 2000 Liquid rate, lb/h-ft2 Figure 76. //0G vs. liquid rate for ammonia-water absorption system. Figure 77. Colburn chart. Figure 77. Colburn chart. Calculate the gas molar flux (molar flowrate per unit cross section) and liquid molar flux in lbmol/(ft2 • h): = (1575)(0.0743)/[( 10.0)(29)] = 0.404 lbmol/(ft2 • min) = 24.21bmol/(ft2 • h) The value of rnGm/Lm is therefore mGJLm = (0.972X24.2/55.6) = 0.423 The absorption factor, A, is defined as The value of (y, — mx2)/(y2 — mx2) or X is yi -mx2 _ 0.02 - (0.972)(0) y2 - mx2 ~~ 0.001 - (0.972)(0) Nqq is calculated from Colburn's equation (or read from Colburn's chart, Figure 77), _ ln((7, - mx2)/(y2 - mx2) (1 - {\/A}) + (1 /A}) N°°~ 1-{1 /A) _ ln(20.0)(l - {1/2.364}) + {1/2.364} ~ 1 - {1/2.364} To calculate the height of an overall gas transfer unit, //OG, first calculate the gas mass velocity, G, in lb/ft2 • h. G = qp/S = (1575)(0.0743)/10.0 = 11.71b/(ft2 • min) From the figure, The required packed column height, Z, in feet is z = NoghOG = (4.3)(2.2) = 9.46 ft The application should be rejected. ## ABS.6 TOWER HEIGHT AND DIAMETERA packed column is used to absorb a toxic pollutant from a gas stream. From the data given below, calculate the height of packing and column diameter. The unit operates at 50% of the flooding gas mass velocity, the actual liquid flowrate is 40% more than the minimum, and 95% of the pollutant is to be collected. Employ the generalized correlation provided in Figure 78 to estimate the column diameter. Gas mass flowrate = 3500 lb/h Pollutant concentration in inlet gas stream =1.1 mol % Scrubbing liquid = pure water Packing type= 1 in. Raschig rings; packing factor, F = 160 Henry's law constant, m = 0.98 Density of water = 62.4 lb/ft3 0.05 0.02 0.01 0.005 0.002 0.001 G2FWU°L2 PlP0c Figure 78. Generalized flooding and pressure drop correlation.
G2FWU°L2 PlP0c Figure 78. Generalized flooding and pressure drop correlation. Viscosity of water = 1.8 cP The ordinate and abscissa of the graph in Figure 78 are dimensionless numbers G = mass flux (mass flowrate per unit cross-sectional area) of gas stream L = mass flux of liquid stream F = packing factor if/ — ratio of the specific gravity of the scrubbing liquid to that of water /(L = viscosity of liquid phase pL = density of liquid phase p = density of gas phase gc = Newton's law proportionality factor To calculate the number of overall gas transfer units, Noa, first calculate the equilibrium outlet concentration, x*, at y: =0.011: Determine y2 for 95% removal: (0.05) j', yi (1->>,) +(0.05) y, (0.05X0.011) "(1 -0.011) +(0.05X0.011) The minimum ratio of molar liquid flowrate to molar gas flowrate, (Lm/Gm)min, is determined by a material balance: g2F^ PLPSC where: Solution The actual ratio of molar liquid flow rate to molar gas flowrate Lm/Gm is In addition, (mGJ/Lm = (0.98)/( 1.306) = 0.7504 The absorption factor, A, is defined as A = LJ(mGm) = 1/0.7504 = 1.333 The value of (yx — mx2)/(y2 — mx2) is yx-mx2 0.011 -(0.98X0) 19.78 Then N0G is calculated from Colburn's equation (or read from Colburn's chart— Figure 77), _ ln((j>i - mx2)/(y2 - mx2) (1 - (1 /A}) + {1/^}) Noc~ 1 — fl /A] _ln[(19.78)(l - {1/1.333}) + {1/1.333}] ~ 1 - {1/1.333} The height of packing, Z, is then To determine the diameter of the packed column, the ordinate of Figure 78, is first calculated: ABS.7 PACKED TOWER ABSORBER DESIGN WITH NO DATA 407 The value of the abcissa at the flooding line is determined from Figure 78: PPlSc The flooding gas mass velocity, Gf, in lb/(ft2 • s), is The actual gas mass velocity, Gact, in lb/(ft2 • s), is Gact = (0.5)(0.419) = 0.2095 lb/(ft2 • s) = 7541b/(ft2 • h) Calculate the diameter of the column in feet: D = [(4m)/(Gact7r)]1/2 = [(4)(3500)/(754)(tt)]1/2 = 2.43 ft The column height (packing) and diameter are 17.4 and 2.43 ft, respectively. ## ABS.7 PACKED TOWER ABSORBER DESIGN WITH NO DATAA 1600-acfm gas stream is to be treated in a packed tower containing ceramic packing. The gas stream contains 100 ppm of a toxic pollutant that is to be reduced to 1 ppm. Estimate the tower's cross-sectional area, diameter, height, pressure drop, and packing size. Use the procedure outlined in Problem MTO. 4 in Chapter 8. Solution Key information from Problem MTO.4 is provided below. For ceramic packing: Packing Height, Z (ft) Removal Efficiency (%) 63.2 77.7 86.5 99 99.5 99.9 99.99 The equation for the cross-sectional area of the tower, S, in terms of the gas volumetric flowrate, q, in acfs is An equation to estimate the tower packing pressure drop, AP, in terms of Z is The following packing size(s) is (are) recommended: For D « 3 ft, use 1-in. packing For D < 3 ft, use < 1-in. packing For D > 3 ft, use > 1-in. packing As a rule, recommended packing size increases with lower diameter. For the problem at hand, The diameter D is i0.5 For a tower this large, the 3.5-in. packing should be used. abs.8 two absorbers to replace one 409 The removal efficiency (RE) is For 99% RE and a packing size of 3.5 in., the required height is 25.3 ft. The pressure drop is ABS.8 TWO ABSORBERS TO REPLACE ONE The calculations for an absorber indicate that it would be excessively tall, and the five schemes in the diagrams in Figure 79 are being considered as a means of using two shorter absorbers. Make freehand sketches of operating lines, one for each scheme showing the relation between operating lines for the two absorbers and the equilibrium curve. Mark the concentrations on the figure for each diagram. No calculations are required. Assume dilute solutions. Solution Operating lines for each scheme are provided in Figure 79. 1. |

## Renewable Energy 101

Renewable energy is energy that is generated from sunlight, rain, tides, geothermal heat and wind. These sources are naturally and constantly replenished, which is why they are deemed as renewable. The usage of renewable energy sources is very important when considering the sustainability of the existing energy usage of the world. While there is currently an abundance of non-renewable energy sources, such as nuclear fuels, these energy sources are depleting. In addition to being a non-renewable supply, the non-renewable energy sources release emissions into the air, which has an adverse effect on the environment.

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