Answer:
a) benzene = 910 days
b) toluene = 1612.67 days
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:
[tex]R=1+\frac{\rho * K_{d} }{n}, \rho = 2.6\\ R=1+\frac{2.6*1.8}{0.37} =13.65[/tex]
The time will take will be:
[tex]t=\frac{xR}{a} , x=12,a=0.18\\t=\frac{12*13.65}{0.18} =910days[/tex]
b) For toluene:
[tex]R=1+\frac{2.6*3.3 }{0.37} = 24.19[/tex]
[tex]t=\frac{12*24.19}{0.18} =1612.67days[/tex]
Answer:
(a) For benzene, t = 909.33 days or 2.49 years
(b) For toluene, t = 1612.67 days or 4.42 years
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
