E = 6.78
9.6.4 Ammonia Stripping (or Absorption)
The reaction of NH3 with water can be represented by
From this reaction, raising the pH (as represented by the OH-) will drive the reaction to the left, increasing the concentration of NH3. This makes ammonia more easily removed by stripping. In practice, the pH is raised to about 10 to 11 using lime. Stripping is done by introducing the wastewater at the top of the column and allowing it to flow down countercurrent to the flow of air introduced at the bottom. The stripping medium inside the column may be composed of packings or fillings, such as Raschig rings and Berl saddles, or sieve trays and bubble caps. The liquid flows in thin sheets around the medium, thereby, allowing more intimate contact between liquid and the stripping air.
Let G be the moles per unit time of ammonia-free air, L be the moles per unit time of ammonia-free water, [F1] be the moles of ammonia per mole of ammonia-free air at the bottom, [F2] be the moles of ammonia per mole of ammonia-free air at the top, [XJ be the moles of ammonia per mole ammonia-free water at the bottom, and [X2] be the moles of ammonia per mole of ammonia-free water at the top. At any section between the bottom and the top of the column, the [X]s and [ y]s are simply [X] and [Y ]. By mass balance, the ammonia stripped from the wastewater is equal to the ammonia absorbed by the air. Thus,
Let the equilibrium concentrations corresponding to [X] and [Y ] be [X ] and
[ Y ], respectively. From the equilibrium between ammonia in water and ammonia
in air, the following equations for the relationships between [X ] and [Y ] at various temperatures and one atmosphere of total pressure may be derived (Metcalf and Eddy, 1991):
The coefficients of [X ] in these equations plot a straight line with the temperatures. Letting these coefficients be K produces the regression equation,
K = 0.0585 Tk - 16.31 with Tk in degrees Kelvin
Therefore, in general, the relationship between [X ] and [ Y ] using the temperature in degrees Kelvin, is
This is the equation of the equilibrium line for ammonia stripping or absorption. It may be expressed in terms of the respective mole fractions [ y f ] and [ x* ]. Performing the operation and solving for [ y*] produces
L, [Xj], [X2], and [Yj] are known values. Assuming the ammonia in the air leaving the top of the tower is in equilibrium with the ammonia content of the entering wastewater, [Y2] may be written as
[ Y 2 ] = [ Y * ] = (0.0585 Tk -16.31)[ X2 ] (9.59)
Thus, the ratio G/L, the moles of air needed for stripping per mole of water all in solute-free basis is
In terms of mole fractions,
l (^-[.x/1];i[(0.0i58-);r)--1--i.3i)(i-[-ii/-];-[|i/2]-(i-[x/:?])[y/1]] (. )
G is the theoretical amount of air needed to strip ammonia in the wastewater from a concentration of [X2] down to a concentration of [XJ. In practice, the minimum theoretical G is often multiplied by a factor of 1.5 to 2.0. Once G has been established, the cross-sectional area of the tower may be computed using the equation of continuity. The superficial velocity through the tower should be less than the velocity that will cause flooding or boiling up of the incoming wastewater. The method for estimating tower height was previously discussed.
Operating line for ammonia. The equation G([F] - [Fj]) = L([X] - [XJ) is the equation for the operating line for ammonia. Expressing in terms of the equivalent mole fractions [y], [y^], [xf], and [x^], respectively, and solving the resulting expression for [ yf],
G[yfJ(1 - [Xft])( 1 - [Xf]) + L[[Xf](1 - [yft])(1 - [xf J) - [xf1](1 - [yf1])(1 - [xf])]
Gbyjq - [Xf1])(1 - [Xf]) + L[[Xf](1 - [yfj)(1 - [x^]) - [XfJQ - [y^g - [x,])] (1-[y/1])(1-[x/1])(1 -[X/])
Example 9.13 A wastewater containing 25 mg/L of NH3-N is to be stripped. The temperature of operation is 15°C and the flow is 0.013 m /s. Determine the amount of air needed assuming a multiplying factor of 1.5 for the minimum theoretical G. Calculate the cross-sectional area of the tower. Assume a superficial velocity of 0.30 m/s.
Solution: To get the minimum G, assume the operation at the top of the tower is at equilibrium. Thus,
G = (^.^585^^™]- [^ 1L of water = 1000 g = 1,000,000 mg Therefore,
mol H2O (ammonia free) [ F1 ] = 0. Tk = 15 + 273 = 288 ° K
G = 1.5 (1.86)(0.722) = 2.01 kgmol/s = 1.74( 108) gmol/d = 1.74(108)(22.4= 4.11 (106) m3/d Ans
Overall cross-sectional area of towers = _ 4'11(1° L,.- = 158.56 m3
Using three towers, cross-sectional area = 1158-.56 = 52.85 m2 Ans
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