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Packed absorption and stripping columns. Prof. Dr. Marco Mazzotti - Institut für Verfahrenstechnik. 1. HETP - approach. Packed columns are continuous contacting devices that do not have the physically distinguishable stages found in trayed columns. .

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## Packed absorption and stripping columns

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**Packed absorption and stripping columns**Prof. Dr. Marco Mazzotti - Institut für Verfahrenstechnik**1. HETP - approach**Packed columns are continuous contacting devices that do not have the physically distinguishable stages found in trayed columns. In practice, packed columns are often analyzed on the basis of equivalent equilibrium stages using a Height Equivalent to a Theoretical Plate (HETP): Knowing the value of the HETP and the theoretical number of stages n of a trayed column, we can easily calculate the height H of the column : The HETP concept, unfortunately, has no theoretical basis. HETP values can only be calculated using experimental data from laboratory or commercial-size columns.**y2< y spec**G, y2 L, x2 T, p G, y1 L, x1 2. Absorption: Mass transfer approach (HTU, NTU) For packed columns, it is preferable to determine packed height from a more theoretically based method using mass transfer coefficients. The absorption problem is usually presented as follows. There is a polluted gas stream coming out from a process. The pollutant must be recovered in order to clean the gas. z = H At the bottom and the top of the column, the compositions of the entering and leaving streams are: y x Furthermore, we introduce the coordinate z, which describes the height of the column. z = 0 Process The green, upper envelope is needed for the operating line of the absorption column.**First, we need a material balance around the green, upper**envelope of the column. It is the operating line, going through the point (x2,y2): y y1 y* = m x Then we need the equilibrium condition: y2 x2 x1 x We can now draw the equilibrium and operating line into the diagram. From the operating line with the smallest slope (Lmin/G), we can get (L/G) with the known formula:**As a third equation, we need a mass transfer rate equation.**We take a small slice of the column. The material balance over the “gas side” of this slice gives: L G N L G S is the cross-sectional area of the tower. Please note that N, G and L are defined as fluxes and not as molar flow rates [mol/s]: Determination of the packed height of a column most commonly involves the overall gas-phase coefficientKy because the liquid usually has a strong affinity for the solute. Its driving force is the mole fraction difference (y-y*):**Dividing the mass transfer rate equation by S and z, we**get: Because we want a differential height of the slice, we let z 0. Introducing the definition of N: Separating variables and integration gives: Taking constant terms out of the integral and changing the integration limits: HOG NOG The right-hand side can be written as the product of the two terms HOG and NOG:**The term HOG is called the overall Height of a Transfer Unit**(HTU) based on the gas phase. Experimental data show that the HTU varies less with G than with Kya. The smaller the HTU,the more efficient is the contacting. The term NOG is called the overall Number of Transfer Units (NTU) based on the gas phase. It represents the overall change in solute mole fraction divided by the average mole fraction driving force. The larger the NTU, the greater is the extent of contacting required. Now we would like to solve the integral of NOG. Therefore we replace y* by equation (2): Solving (1) for x, knowing that A=L/(Gm): Introducing the result into the equation for NOG:**Integration of NOG gives:**Splitting the inner part of the logarithm into two parts: We already know the fraction of absorption: Introducing and doing some transformations, we finally get for NOG:**y**y y eq. line eq. line eq. line x x x 3. Comparison between HTU / NTU and HETP The height of the column can be calculated in two ways: The NTU and the HTU should not be confused with the HETP and the number of theoretical equilibrium stages n, which can be calculated with the Kremser Equation: When the operating and equilibrium lines are not only straight but also parallel, NTU = n and HTU = HETP. Otherwise, the NTU is greater than or less than n. op. line op. line op. line**When the operating and equilibrium lines are straight but**not parallel (NTU n), we need a formula to transform them. We can write: Replacing NOG and n by the formulas found earlier, we get for HETP: Doing the same calculation for NOG, we find: Finally we want to calculate the volumetric overall mass transfer coefficientKya. We know that: Solving for Kya, we find:**L, x2**G, y2 T, p G, y1 L, x1 4. Stripping: Mass transfer approach (HTU, NTU) Process Now we want to focus on a stripping problem, which is usually presented as follows. There is a polluted liquid stream coming out from a process. The pollutant must be recovered in order to clean the liquid. z = H First, we need a material balance around the green, upper envelope of the column. It is the operating line, going through the point (x1,y1): x y z = 0 Then we need the equilibrium condition:**L**G N L G We can now draw the equilibrium and operating line into the diagram. From the operating line with the largest slope (L/G)max, we can get (L/G) with the known formula: y y2 y* = m x y1 As a third equation, we need a mass transfer rate equation. We take a small slice of the column. The material balance over the “liquid side” of this slice gives: x1 x2 x The flux N involves the overall liquid-phase coefficientKx and the driving force (x-x*):**Dividing the mass transfer rate equation by S and z, we**get: We let z 0 and introduce the definition of N: Separating variables and integration gives: HOL NOL The term HOL is called the overall Height of a Transfer Unit (HTU) based on the liquid phase. The term NOL is called the overall Number of Transfer Units (NTU) based on the liquid phase.**We already know the fraction of strippingσ:**Furthermore, we know the stripping factor S: The solution of the integral of NOL can be found if one proceeds exactly as in the case of absorption: Finally, after some transformations, we find:

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