Question

A wastewater containing 150 mg/l chlorobenzene is treated in a laboratory adsorption unit using a PVC column, 1.0 inch internal diameter, to an effluent concentration of 15 mg/l . Service times, and throughput volumes at specified depths and flowrates associated with a breakthrough concentration of 15.0 mg/l are given in table 1.

table1 : result of adsorption column experiment

Loading rate,gpm/ft²   Bed depth,ft Throughput volume, gal Time, hr

loading rate gpm/ft2	bed depth ft	throughput volume, gal	time, hr
2.5	3.0	810	980
	5.0	1750	2230
	7.0	2910	3440
5.0	3.0	605	420
	5.0	1495	1000
	9.0	3180	2185
7.5	5.0	1183	452
	9.0	2781	1075
	12.0	4000	1564

1) is the attainable effluent concentration satisfactory from a regulatory standpoint?

2) determine the Bohart-Adams constant ( K,N₀ and x₀) for each hydraulic loading.

3)base on data derived above design an adsorption column 2.0 ft internal diameter to treat a wastewater flow 5,000 gal/d containing 150 mg/l of CB. The attainable effluent concentration is 15 mg/l and it is desired to operate the column for 90 days(8 hourslday,7 days/week) before reching exhaustion.

4)calculate the yearly carbon requirements in cubic feet.

what kind of information do you need? these are all what i got from paper

Answer 1

1) The attainable effluent concentration of 15 mg/l for chlorobenzene is satisfactory. As per toxicity characteristice the regulatory level for chlorobenzene is 100 mg/l

2) The Bohart-Adam equation is given as:

t = (N₀/C₀V)x-1/C₀kln(C₀/C_b-1)

where,

t = time

N₀ = column adsorption capacity

C₀ = concentration of solute in feed solution

C_b = breakthrough concentration

V = linear feed velocity of the feed to bed

k = adsorption rate constant

x = bed depth

A plot of t vs x at different loading rates gives a staright line with:

slope = N₀/C₀V

Intercept = 1/C₀kln(C₀/C_b-1)

The graph is shown below:

1) For loading rate 2.5 gpm/ft2: Blue line

y = 615x-858.33

slope = 615 = N₀/C₀V

C0 = 150 mg/l

Cb = 15 mg/l

intercept = 858.33 = 1/k150ln(150/15 - 1)

k = 1.71 * 10^-5

Inorder to calculate N0, information on V is required

x0 = minimum bed depth. This can be ontained from the graph as the intercept on the abscissa at t= 0

x0 = 800

2) For loading rate 5 gpm/ft2: red line

y = 294.46 x - 466.96

intercept = 466.96 = 1/k150ln(150/15-1)

k = 3.14 * 10^-5

x0 = 500

3) For loading rate = 7.5gpm/ft: green line

y = 120.74x - 64.429

k = 2.27*10^-4

x0 = 100