E = 0.0000128


Example 8.8 A wastewater containing 25 mg/L of phenol and having the characteristic breakthrough of the previous example is to be treated by adsorption onto an activated carbon bed. The flow rate during the breakthrough experiment is 0.11 m /s; this is equivalent to a surficial velocity of 0.0088 m/s. The X/M ratio of the bed for the desired effluent of 0.06 mg/L is 0.02 kg solute per kg carbon. If the flow rate for design is also 0.11 m /s, design the absorption column. Assume the influent is introduced at the top of the bed. The packed density of the carbon bed is 721.58 kg/m3.

Solution: The design will include the determination of the amount of activated carbon needed, the dimensions of the column, and the interval of activated carbon replacement.

Amount of phenol to be removed = 0.11(0.025 - 0.00006)(60)(60)(24) = 237 kg/d Amount of activated carbon needed = 237/0.02 = 11,850 kg/d Ans

From the previous example, adsorbate retained in ¿length of test column at exhaustion = 2(0.002 - 0.001)(0.025) - 0.0000128 = 0.000037 kg. For As of the test column = 0.00051 m and S = 0.0051 m, adsorbate retained per unit volume in S = 14.23 kg.

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