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Your consulting firm is asked to design an activated carbon adsorption system to treat contaminated groundwater containing 10 mg/L of benzene with a treatment goal of 99% reduction in benzene concentration before injection into groundwater. The extraction well pumps groundwater at 100 gpm. The activated carbon removal capacity is expressed as the following Freundlich equation: 9 (mg/kg) = 0.2 CO.2, where Cis concentration (mg/L) of benzene after adsorption equilibrium. (a) What is the amount of activated carbon (kg) needed per day? (b) If the packing density of activated carbon is 1.2 g/L, how many 55-gallon tank it would fill? Approximate to nearest number exceeding the amount of activated carbon required.

Sagot :

the daily need for activated carbon in kg is 4319.327 kg of carbon. It would fill a 55-gallon tank, which is 18.

groundwater containing 10 mg/L of benzene

treatment goal of 99%

Flow rate of ground water =100 Gpm

                                            =378.541 Liter/min

now weight of benzene in water

  =378.541 Liter/min*10 mg/lit

  = 3.78541 gram/min

now per day weight of benzene

=3.78541 g/min*(24*60)min=5450.9904 gram

now frewndlich equation derives

q(mg/g)=2 c[tex]^{0.2}[/tex]

c=0.01*10=0.1 mg/L

q=2*[tex](0.1)^{0.2}[/tex] = 1.262

q=1.262 (mg/g)

it means 1.262 mg of benzene adsorps on one gram of activated carbon

a)now per day mass=5450.9904*10³ mg

1.262 mg needs 1g of carbon

1mg needs 1/1.262 g of carbon

5450.9904*10³  mg needs

((1/1.262)*5450.9904*10³ ) grams

=4319.327 Kg of carbon /day

b) packing density of activated carbon = 1.2 g/L

tank size =55 gallon

now 1.2g fills in 1 lit

1g fills in (1/1.2)litre

4319.327*10³ g fills in(1/1.2)*4319.327*10³ liter

                                   =3600 liter volume needed

1 tank volume =55gallon

                        =208.2 lit

number of tanks needed =3600 liter/208.2

                                          =17.29≅18 tanks

∴18 such tanks are needed

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