Repeat Exercise 15.21 for water containing 0.324 mg/L of benzene (B) and 0.630 mg/L of m-xylene (X).

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Repeat Exercise 15.21 for water containing 0.324 mg/L of benzene (B) and 0.630 mg/L of m-xylene (X). Adsorption isotherms at 25OC for these low concentrations are essentially independent and are given by

0.428 = 32 CB 0.333 9x = 125 c333 (1) (2)

The feed concentrations of the SOCs in the feed are to be reduced to 0.002 mg/Leach.

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Separation process principles

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Repeat Exercise 15.21 for water containing 0.324 mg/L of benzene (B) and 0.630 mg/L of m-xylene (X).

Question:

0.428 = 32 CB 0.333 9x = 125 c333 (1) (2)

Step by Step Answer:

a The minimum amount of adsorbent corresponds to equilibrium with a solution of the effluent concentration of 0002 mg of B or X with the other solute at a lower concentration Must determine which solu...View the full answer

Separation process principles

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