1. Suppose you are adding activated carbon to a water stream to remove a contami

Question

1. Suppose you are adding activated carbon to a water stream to remove a contaminant. The contaminant is adsorbed according to a linear isotherm with K = 0.5 L/g. The influent 5 concentration (Ci) is 5 mg/L. The target effluent concentration is 1.5 mg/L.

(a) What is the solid phase concentration (x/m) when the activated carbon is at equilibrium with the target concentration? Answer: 7.5x10-4 g/g

(b) What mass of contaminant must be removed from 1.0 million liters of water to reach the target concentration? Answer: 3500 g

(c) How many kg of activated carbon would be needed to treat 1.0 million liters of contaminated water down to the target concentration? Answer: 4667 kg

Explanation / Answer

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A a] The solid phase concentration (x/m) when the activated carbon is at equilibrium with the target concentration is equal to,

= K * target effluent concentration

= 0.5 L/g * 1.5 mg/L

= 0.75 mg/g

= 7.5 * 10(-4) g/g

A b] mass of contaminant must be removed from 1.0 million liters of water to reach the target concentration can be calculated as the difference between the influent concentration and the target effluent cpncentration which is,

5 mg/L - 1.5 mg/L

= 3.5 mg/L

For 1 million litres of water ,

= 3.5 * 1000000 /1000 g

= 3500 g

A c]

The total amount of activated carbon would be needed to treat 1.0 million liters of contaminated water down to the target concentration is equal to the ratio of mass of contaminant that must be removed from 1.0 million liters of water to the solid phase concentration (x/m) when the activated carbon is at equilibrium with the target concentration.

i.e, = 3500 g / (7.5 * 10-4) g/g

= 4666666.66 g

= 4667 kg.

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