Answer :
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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