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18-Env-B5 Industrial & Hazardous Waste Management · December 2019

Question 7 of 10: Granular Activated Carbon Required for Xylene Removal

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Reference texts: LaGrega, Buckingham & Evans, Hazardous Waste Management, 2nd ed.; Nemerow & Dasgupta, Industrial and Hazardous Waste Treatment, 2nd ed.; Davis & Cornwell, Introduction to Environmental Engineering, 6th ed.; Metcalf & Eddy, Wastewater Engineering: Treatment and Resource Recovery, 5th ed.; Cooper & Alley, Air Pollution Control: A Design Approach; ACGIH, Industrial Ventilation: A Manual of Recommended Practice; Ontario Environmental Protection Act, R.S.O. 1990, c. E.19 and O. Reg. 347 (Waste Management – General); U.S. EPA SW-846 Method 1311 (Toxicity Characteristic Leaching Procedure).

Question 7: Granular Activated Carbon Required for Xylene Removal (10 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Flow $Q = 10{,}000\ \text{gal/day}$; influent xylenes $C_0 = 600\ \text{mg/L}$; required effluent $C_e = 10\ \text{mg/L}$; Freundlich isotherm $q = 51.3\,C_f^{0.187}$ (mg xylene per g carbon, $C_f$ in mg/L).

Find. The mass of granular activated carbon (GAC) required per day to achieve the target effluent concentration.

Approach. Compute the daily mass of xylenes that must be removed from a simple influent–effluent mass balance, then size the carbon from the Freundlich isotherm evaluated at the target effluent concentration $C_e$ — the standard equilibrium-adsorber sizing convention, since a well-run GAC bed is taken to breakthrough, at which point the carbon leaving service is essentially in equilibrium with the very effluent concentration the design targets.

  1. Convert flow and find the mass of xylenes removed per day. $$Q = 10{,}000\ \text{gal/day}\times3.785\ \tfrac{\text{L}}{\text{gal}} = 37{,}854\ \text{L/day}$$ $$\dot m_{\text{removed}} = Q(C_0-C_e) = 37{,}854\ \text{L/day}\times(600-10)\ \text{mg/L} = \boxed{22{,}334{,}000\ \text{mg/day}} = 22.33\ \text{kg/day}$$
  2. Carbon capacity at the target effluent concentration. Evaluating the Freundlich isotherm at $C_f = C_e = 10\ \text{mg/L}$: $$q_e = 51.3\,(10)^{0.187} = 51.3\times1.538 = \boxed{78.9\ \text{mg xylene/g carbon}}$$
  3. Carbon required. $$\dot m_{\text{carbon}} = \frac{\dot m_{\text{removed}}}{q_e} = \frac{22{,}334{,}000\ \text{mg/day}}{78.9\ \text{mg/g}} = \boxed{283{,}000\ \text{g/day}} = 283\ \text{kg/day}$$
Final results
QuantityValue
Xylenes removed22.33 kg/day
Carbon capacity at $C_e=10$ mg/L78.9 mg/g
Carbon required283 kg/day
Check: evaluating $q_e$ at the target EFFLUENT concentration (not the influent) is the standard, most-common convention for this class of exam problem — it represents the carbon at the point of breakthrough. Evaluating instead at $C_0=600$ mg/L (a fresh-bed, far-from-breakthrough state) would give a larger apparent capacity, $q=51.3(600)^{0.187}\approx137.9$ mg/g, and a proportionally smaller (non-conservative) carbon estimate of about 162 kg/day; the effluent-based figure above is the defensible design value.