Question

The underground storage of a gas station has leaked gasoline into the ground. Among the constituents of gasoline are benzene, with a Kd value of 1.8 L/kg for the soil in that location, and toluene, with a Kd of 3.3 L/kg. The soil has a solids density of 2.6 kg/L and a hydraulic conductivity of 2.9 × 10^-5 m/s and is 37% pores (by volume). The site is to be remediated by pumping out and treating groundwater from a well 15 m downgradient from the site of the spill, where the water table is 0.4 m lower than at the leaking tank. How long will it take for:

a. benzene
b. toluene to reach the well?

Answer 1

(a) For benzene, t = 909.33 days or 2.49 years

(b) For toluene, t = 1612.67 days or 4.42 years

Step-by-step explanation:

Kd value of benzene = 1.8L/kg

Kd value of toluene = 3.3 L/kg

Density of soil (p) = 2.6 kg/L = 2600 kg/m3

Hydraulic conductivity K = 2.9 x 10-5 m/s

Porosity (n) = 0.37

Well depth(x) = 12m

To calculate the time for benzene and toluene to reach the well, we use the formula;

x = t u /R -----------------------------1

But velocity (u) = Ki/n

u= 2.9 x 10-5 x (0.4/15) / 0.37

= 2.09 x 10-6 m/s or 0.18 m/day

For benzene:

R = 1 + (p*Kd)/n

R= 1 + 2.6(1.8)/0.37

= 1 + 12.64

= 13.64

Substituting the values for R and u into equation 1, we have

x = t u /R

12 = t x 0.18 / 13.64

12 = 0.013t

t = 12/0.013

t = 909.33 days

or t = 909.33/365 days = 2.49 years

(b) For Toluene:

R = 1 + (p*Kd)/n

R= 1 + 2.6(3.3)/0.37

= 1 + 23.19

= 24.19

Substituting into equation 1, we have

x = t u/R

12 = t x 0.18 /24.19

12 = 0.00744 t

t = 12/0.00744

t = 1612.67 days

or 1612.67/365 = 4.42 years

Answer 2

a) benzene = 910 days

b) toluene = 1612.67 days

Step-by-step explanation:

Given:

Kd = 1.8 L/kg (benzene)

Kd = 3.3 L/kg (toluene)

psolid = solids density = 2.6 kg/L

K = 2.9x10⁻⁵m/s

pores = n = 0.37

water table = 0.4 m

ground water = 15 m

u = K/n = (2.9x10⁻⁵ * (0.4/15)) / 0.37 = 2.09x10⁻⁶m/s

a) For benzene:

`R=1+(\rho * K_(d) )/(n), \rho = 2.6\\ R=1+(2.6*1.8)/(0.37) =13.65`

The time will take will be:

`t=(xR)/(a) , x=12,a=0.18\\t=(12*13.65)/(0.18) =910days`

b) For toluene:

`R=1+(2.6*3.3 )/(0.37) = 24.19`

`t=(12*24.19)/(0.18) =1612.67days`