1. Find: c(x,t) [mg/L] of chloride
continuous source of
chloride contamination 10 25 50
100
i=0.002 Cl mg L
co=725 x
Find the concentration of chloride, in milligrams per liter. [pause] In this problem, ---
x=15 [m] m s m2 s
η=0.23
K=3*10-5
D=1*10-9
t=1 [yr]
αx=0.83*(log(x))2.414 [m]

2. Find: c(x,t) [mg/L] of chloride
continuous source of
chloride contamination 10 25 50
100
i=0.002 Cl mg L
co=725 x
a continuous source of chloride contamination, with the initial concentration of 725 milligrams per liter, ---
x=15 [m] m s m2 s
η=0.23
K=3*10-5
D=1*10-9
t=1 [yr]
αx=0.83*(log(x))2.414 [m]

3. Find: c(x,t) [mg/L] of chloride
continuous source of
chloride contamination 10 25 50
100
i=0.002 Cl mg L
co=725 x
travels in a one-dimensional flow field, in the x direction. [pause] The problem asks to find the concentration, ---
x=15 [m] m s m2 s
η=0.23
K=3*10-5
D=1*10-9
t=1 [yr]
αx=0.83*(log(x))2.414 [m]

4. Find: c(x,t) [mg/L] of chloride
continuous source of
chloride contamination 10 25 50
100
i=0.002 Cl mg L
co=725 x
c sub x, t, at a distance x, 15 meters away from the source, and at a time t, 1 year after contamination begins.
x=15 [m] m s m2 s
η=0.23
K=3*10-5
D=1*10-9
t=1 [yr]
αx=0.83*(log(x))2.414 [m]

5. Find: c(x,t) [mg/L] of chloride
continuous source of
chloride contamination 10 25 50
100
i=0.002 Cl mg L
co=725 x
The problem also provides other parameters shown here. For a continuous source of contamination, ---
x=15 [m] m s m2 s
η=0.23
K=3*10-5
D=1*10-9
t=1 [yr]
αx=0.83*(log(x))2.414 [m]

6. Find: c(x,t) [mg/L] of chloride
x=15 [m]
co=725 mg L Cl
i=0.002 m s
the concentration at a downgrade distance, x, ---
K=3*10-5
t=1 [yr] m2 s
η=0.23
D=1*10-9 x
αx=0.83*(log(x))2.414 [m]

7. Find: c(x,t) [mg/L] of chlorideco
x-vx*t
c(x,t)=
erfc * 2
2 Dx*t
x+vx*t
vx*x
+exp
* erfc Dx
2 Dx*t
x=15 [m]
co=725 mg L Cl
i=0.002 m s
after a time duration, t, equals one half the initial concentration times times an expression which involves the ---
K=3*10-5
t=1 [yr] m2 s
η=0.23
D=1*10-9 x
αx=0.83*(log(x))2.414 [m]

8. Find: c(x,t) [mg/L] of chloride
complimentary co
x-vx*t
error function
c(x,t)=
erfc * 2
2 Dx*t
x+vx*t
vx*x
+exp
* erfc Dx
2 Dx*t
x=15 [m]
co=725 mg L Cl
i=0.002 m s
complimentary error function,
K=3*10-5
t=1 [yr] m2 s
η=0.23
D=1*10-9 x
αx=0.83*(log(x))2.414 [m]

9. Find: c(x,t) [mg/L] of chloride
complimentary co
x-vx*t
error function
c(x,t)=
erfc * 2
2 Dx*t
distance
x+vx*t
vx*x
+exp
* erfc Dx
2 Dx*t
x=15 [m]
co=725 mg L Cl
i=0.002 m s
the distance, ---
K=3*10-5
t=1 [yr] m2 s
η=0.23
D=1*10-9 x
αx=0.83*(log(x))2.414 [m]

10. Find: c(x,t) [mg/L] of chloride
complimentary co
x-vx*t
error function
c(x,t)=
erfc * 2
2 Dx*t
distance
x+vx*t
vx*x
+exp
velocity
* erfc Dx
2 Dx*t
x=15 [m]
co=725 mg L Cl
i=0.002 m s
the velocity, ---
K=3*10-5
t=1 [yr] m2 s
η=0.23
D=1*10-9 x
αx=0.83*(log(x))2.414 [m]

11. Find: c(x,t) [mg/L] of chloride
complimentary co
x-vx*t
error function
c(x,t)=
erfc * 2
2 Dx*t
distance
x+vx*t
vx*x
+exp
velocity
* erfc Dx
2 Dx*t
hydrodynamic
dispersion
x=15 [m]
co=725 mg L Cl
i=0.002
coefficient m s
the hydrodynamic dispersion coefficient, ---
K=3*10-5
t=1 [yr] m2 s
η=0.23
D=1*10-9 x
αx=0.83*(log(x))2.414 [m]

12. Find: c(x,t) [mg/L] of chloride
complimentary co
x-vx*t
error function
c(x,t)=
erfc * 2
time
2 Dx*t
distance
x+vx*t
vx*x
+exp
velocity
* erfc Dx
2 Dx*t
hydrodynamic
dispersion
x=15 [m]
co=725 mg L Cl
i=0.002
coefficient m s
and the time. [pause]
K=3*10-5
t=1 [yr] m2 s
η=0.23
D=1*10-9 x
αx=0.83*(log(x))2.414 [m]

13. Find: c(x,t) [mg/L] of chlorideco
x-vx*t
c(x,t)=
erfc * 2
2 Dx*t
x+vx*t
vx*x
+exp
* erfc Dx
2 Dx*t
x=15 [m]
co=725 mg L Cl
i=0.002 m s
We already know the initial concentration, c sub not, the distance, x, and the time, t, ---
K=3*10-5
t=1 [yr] m2 s
η=0.23
D=1*10-9 x
αx=0.83*(log(x))2.414 [m]

14. Find: c(x,t) [mg/L] of chlorideco
x-vx*t
c(x,t)=
erfc * 2
2 Dx*t
velocity
x+vx*t
vx*x
+exp
* erfc Dx
2 Dx*t
hydrodynamic
dispersion
x=15 [m]
co=725 mg L Cl
i=0.002
coefficient m s
but we still need to determine the velocity and the hydrodynamic dispersion coefficient.
K=3*10-5
t=1 [yr] m2 s
η=0.23
D=1*10-9 x
αx=0.83*(log(x))2.414 [m]

15. Find: c(x,t) [mg/L] of chlorideco
x-vx*t
c(x,t)=
erfc * 2
2 Dx*t
velocity
x+vx*t
vx*x
+exp
* erfc Dx
2 Dx*t
hydrodynamic
dispersion
x=15 [m]
co=725 mg L Cl
i=0.002
coefficient m s
Let’s first solve for the velocity. [pause]
K=3*10-5
t=1 [yr] m2 s
η=0.23
D=1*10-9 x
αx=0.83*(log(x))2.414 [m]

16. Find: c(x,t) [mg/L] of chloride
velocity dh 1
v=k * * η dL
x=15 [m]
co=725 mg L Cl
i=0.002 m s
The actual velocity of water through a porous medium ---
K=3*10-5
t=1 [yr] m2 s
η=0.23
D=1*10-9 x
αx=0.83*(log(x))2.414 [m]

17. Find: c(x,t) [mg/L] of chloride
velocity dh 1
v=k * * η dL
Darcy
velocity
x=15 [m]
co=725 mg L Cl
i=0.002 m s
equals the Darcy velocity, ---
K=3*10-5
t=1 [yr] m2 s
η=0.23
D=1*10-9 x
αx=0.83*(log(x))2.414 [m]

18. Find: c(x,t) [mg/L] of chloride
velocity dh 1
v=k * * η dL
porosity
Darcy
velocity
x=15 [m]
co=725 mg L Cl
i=0.002 m s
divided by the porosity of the soil.
K=3*10-5
t=1 [yr] m2 s
η=0.23
D=1*10-9 x
αx=0.83*(log(x))2.414 [m]

19. Find: c(x,t) [mg/L] of chloride
velocity dh 1
v=k * * η dL
porosity
Darcy
velocity
x=15 [m]
co=725 mg L Cl
i=0.002 m
Plugging the the appropriate values, the velocity in the x direction equals ---
K=3*10-5
t=1 [yr] s m2 s
η=0.23
D=1*10-9 x
αx=0.83*(log(x))2.414 [m]

20. Find: c(x,t) [mg/L] of chloride
velocity dh 1
v=k * * η dL
porosity
Darcy
velocity m
vx=2.61*10-7 s
x=15 [m]
co=725 mg L Cl
i=0.002 m
2.61 times 10 to the –7 meters per second. [pause] With the velocity solved, ----
K=3*10-5
t=1 [yr] s m2 s
η=0.23
D=1*10-9 x
αx=0.83*(log(x))2.414 [m]

21. Find: c(x,t) [mg/L] of chloride
hydrodynamic
dispersion
coefficient m s
vx=2.61*10-7
x=15 [m]
co=725 mg L Cl
i=0.002 m
we’ll next determine the hydrodynamic dispersion coefficient, ---
K=3*10-5 s
t=1 [yr] m2 s
η=0.23
D=1*10-9 x
αx=0.83*(log(x))2.414 [m]

22. Find: c(x,t) [mg/L] of chloride
hydrodynamic
Dx = Dx + D
dispersion
coefficient m s
vx=2.61*10-7
x=15 [m]
co=725 mg L Cl
i=0.002 m
in the x direction, D sub x, which is the sum of the ---
K=3*10-5 s
t=1 [yr] m2 s
η=0.23
D=1*10-9 x
αx=0.83*(log(x))2.414 [m]

23. Find: c(x,t) [mg/L] of chloride
diffusion
hydrodynamic
Dx = Dx + D
dispersion
coefficient
coefficient
mechanical
dispersion
coefficient m s
vx=2.61*10-7
x=15 [m]
co=725 mg L Cl
i=0.002 m
mechanical dispersion coefficient, and the diffusion coefficient.
K=3*10-5 s
t=1 [yr] m2 s
η=0.23
D=1*10-9 x
αx=0.83*(log(x))2.414 [m]

24. Find: c(x,t) [mg/L] of chloride
diffusion
hydrodynamic
Dx = Dx + D
dispersion
coefficient
coefficient
mechanical
dispersion
coefficient m s
vx=2.61*10-7
x=15 [m]
co=725 mg L Cl
i=0.002 m
Since the problem statement provides the diffusion coefficient,
K=3*10-5 s
t=1 [yr] m2 s
η=0.23
D=1*10-9 x
αx=0.83*(log(x))2.414 [m]

25. Find: c(x,t) [mg/L] of chloride
diffusion
hydrodynamic
Dx = Dx + D
dispersion
coefficient
coefficient
mechanical
dispersion
coefficient m s
vx=2.61*10-7
x=15 [m]
co=725 mg L Cl
i=0.002 m
we’ll solve for the the mechanical dispersion, which equals, ---
K=3*10-5 s
t=1 [yr] m2 s
η=0.23
D=1*10-9 x
αx=0.83*(log(x))2.414 [m]

26. Find: c(x,t) [mg/L] of chloride
diffusion
hydrodynamic
Dx = Dx + D
dispersion
coefficient
coefficient
velocity
Dx = αx * vx
dynamic
dispersivity m s
vx=2.61*10-7
x=15 [m]
co=725 mg L Cl
i=0.002 m
the dynamic dispersivity, times the velocity.
K=3*10-5 s
t=1 [yr] m2 s
η=0.23
D=1*10-9 x
αx=0.83*(log(x))2.414 [m]

27. Find: c(x,t) [mg/L] of chloride
diffusion
hydrodynamic
Dx = Dx + D
dispersion
coefficient
coefficient
velocity
Dx = αx * vx
dynamic
dispersivity m s
vx=2.61*10-7
x=15 [m]
co=725 mg L Cl
i=0.002 m
We already solved for the velocity, ---
K=3*10-5 s
t=1 [yr] m2 s
η=0.23
D=1*10-9 x
αx=0.83*(log(x))2.414 [m]

28. Find: c(x,t) [mg/L] of chloride
diffusion
hydrodynamic
Dx = Dx + D
dispersion
coefficient
coefficient
velocity
Dx = αx * vx
dynamic
dispersivity
αx=0.83*(log(x))2.414 [m] m s
vx=2.61*10-7
x=15 [m]
co=725 mg L Cl
i=0.002 m
but we still need to figure out the dispersivity. When the distance of ---
K=3*10-5 s
t=1 [yr] m2 s
η=0.23
D=1*10-9 x
αx=0.83*(log(x))2.414 [m]

29. Find: c(x,t) [mg/L] of chloride
diffusion
hydrodynamic
Dx = Dx + D
dispersion
coefficient
coefficient
velocity
Dx = αx * vx
dynamic
dispersivity
αx=0.83*(log(x))2.414 [m] m s
vx=2.61*10-7
x=15 [m]
co=725 mg L Cl
i=0.002 m
15 meters is plugged into this equation,
K=3*10-5 s
t=1 [yr] m2 s
η=0.23
D=1*10-9 x
αx=0.83*(log(x))2.414 [m]

30. Find: c(x,t) [mg/L] of chloride
diffusion
hydrodynamic
Dx = Dx + D
dispersion
coefficient
coefficient
velocity
Dx = αx * vx
dynamic
dispersivity
αx=0.83*(log(x))2.414 [m]
αx=1.23 [m] m s
vx=2.61*10-7
x=15 [m]
co=725 mg L Cl
i=0.002 m
the dispersivity equals 1.23 meters. [pause] Multiplying this 1.23 meters by, ---
K=3*10-5 s
t=1 [yr] m2 s
η=0.23
D=1*10-9 x
αx=0.83*(log(x))2.414 [m]

31. Find: c(x,t) [mg/L] of chloride
diffusion
hydrodynamic
Dx = Dx + D
dispersion
coefficient
coefficient
Dx = αx * vx
αx=1.23 [m] m
vx=2.61*10-7
x=15 [m]
co=725 mg L Cl
i=0.002 s m
the velocity of 2.61 times 10 to the –7 meters per second, ---
K=3*10-5 s
t=1 [yr] m2 s
η=0.23
D=1*10-9 x
αx=0.83*(log(x))2.414 [m]

32. Find: c(x,t) [mg/L] of chloride
diffusion
hydrodynamic
Dx = Dx + D
dispersion
coefficient
coefficient
Dx = αx * vx
αx=1.23 [m] m2 s
Dx=3.21*10-7 m
vx=2.61*10-7
x=15 [m]
co=725 mg L Cl
i=0.002 s m
the mechanical dispersion coefficient equals 3.21 times 10 to the –7, meters squared per second.
K=3*10-5 s
t=1 [yr] m2 s
η=0.23
D=1*10-9 x
αx=0.83*(log(x))2.414 [m]

33. Find: c(x,t) [mg/L] of chloride
diffusion
hydrodynamic
Dx = Dx + D
dispersion
coefficient
coefficient m2 s
Dx=3.21*10-7 m
vx=2.61*10-7
x=15 [m]
co=725 mg L Cl
i=0.002 s m
With a little addition, we can solve for the hydrodynamic dispersion coefficient, which equals, ---
K=3*10-5 s
t=1 [yr] m2
η=0.23
D=1*10-9 s x
αx=0.83*(log(x))2.414 [m]

34. Find: c(x,t) [mg/L] of chloride
diffusion
hydrodynamic
Dx = Dx + D
dispersion
coefficient
coefficient m2 s
Dx=3.21*10-7 m2 s
Dx = 3.22*10-7 m
vx=2.61*10-7
x=15 [m]
co=725 mg L Cl
i=0.002 s m
3.22 times 10 to the –7, meters squared per second. [pause] Having solved for the ---
K=3*10-5 s
t=1 [yr] m2 s
η=0.23
D=1*10-9 x
αx=0.83*(log(x))2.414 [m]

35. Find: c(x,t) [mg/L] of chloride
velocity
hydrodynamic
dispersion m
vx=2.61*10-7
coefficient s m2
Dx = 3.22*10-7
x=15 [m]
co=725 mg L Cl
i=0.002 s m s
velocity and dispersion coefficient, we can now plug these values into ---
K=3*10-5
t=1 [yr] m2 s
η=0.23
D=1*10-9 x
αx=0.83*(log(x))2.414 [m]

36. Find: c(x,t) [mg/L] of chlorideco
x-vx*t
c(x,t)=
erfc * 2
2 Dx*t
x+vx*t
vx*x
+exp
* erfc m
vx=2.61*10-7 Dx
2 Dx*t s m2
Dx = 3.22*10-7
x=15 [m]
co=725 mg L Cl
i=0.002 s m s
the original equation for the concentration, as well as the variables for, ---
K=3*10-5
t=1 [yr] m2 s
η=0.23
D=1*10-9 x
αx=0.83*(log(x))2.414 [m]

37. Find: c(x,t) [mg/L] of chlorideco
x-vx*t
c(x,t)=
erfc * 2
2 Dx*t
x+vx*t
vx*x
+exp
* erfc m
vx=2.61*10-7 Dx
2 Dx*t s m2
Dx = 3.22*10-7 s
i=0.002 m s
the initial concentration, the distance, and the time. The equation simplifies to ---
K=3*10-5
t=1 [yr] Cl mg m2 s
η=0.23
co=725
D=1*10-9 L x
αx=0.83*(log(x))2.414 [m]
x=15 [m]

38. Find: c(x,t) [mg/L] of chloride
erfc(1.06) * L
+exp(12.16) * erfc(3.64) m
vx=2.61*10-7 s m2
Dx = 3.22*10-7 s
i=0.002 m s
362.5 milligrams per liter, times the quantity, the complimentary error function of 1.06, plus e raised to times the complimentary error of [pause]
K=3*10-5
t=1 [yr] Cl mg m2 s
η=0.23
co=725
D=1*10-9 L x
αx=0.83*(log(x))2.414 [m]
x=15 [m]

39. Find: c(x,t) [mg/L] of chloride
erfc(1.06) * L
+exp(12.16) * erfc(3.64) m
vx=2.61*10-7 s m2
Dx = 3.22*10-7 s
i=0.002 m s
The complimentary error function, of a variable, beta, is equal to ---
K=3*10-5
t=1 [yr] Cl mg m2 s
η=0.23
co=725
D=1*10-9 L x
αx=0.83*(log(x))2.414 [m]
x=15 [m]

40. Find: c(x,t) [mg/L] of chloride
erfc(1.06) * L
+exp(12.16) * erfc(3.64)
erfc(β)=1-erf(β)
i=0.002 m s
1 minus the error function, of beta, where the error function of beta equals, ---
K=3*10-5
t=1 [yr] Cl mg m2 s
η=0.23
co=725
D=1*10-9 L x
αx=0.83*(log(x))2.414 [m]
x=15 [m]

41. Find: c(x,t) [mg/L] of chloride
erfc(1.06) * L
+exp(12.16) * erfc(3.64)
erfc(β)=1-erf(β) β 2 2
erf(β)=
e-u du π
i=0.002 m s
2 over root PI time the integral of e to the negative u squared du evaluated from 0 to beta. [pause] Error function values are typically looked up in tables.
K=3*10-5
t=1 [yr] Cl mg m2 s
η=0.23
co=725
D=1*10-9 L x
αx=0.83*(log(x))2.414 [m]
x=15 [m]

42. Find: c(x,t) [mg/L] of chloride
erfc(1.06) * L
+exp(12.16) * erfc(3.64)
erfc(1.0) =
erfc(1.1) =
i=0.002 m s
For example, complimentary error function values are looked up for 1.0 and 1.1, and the value corresponding to
K=3*10-5
t=1 [yr] Cl mg m2 s
η=0.23
co=725
D=1*10-9 L x
αx=0.83*(log(x))2.414 [m]
x=15 [m]

43. Find: c(x,t) [mg/L] of chloride
0.1348 mg
c(x,t)=362.5
erfc(1.06) * L
+exp(12.16) * erfc(3.64)
erfc(1.0) =
interpolate
erfc(1.1) =
erfc(1.06) =
i=0.002 m s
is determined by interpolation, and equals The error function of 3.64,
K=3*10-5
t=1 [yr] Cl mg m2 s
η=0.23
co=725
D=1*10-9 L x
αx=0.83*(log(x))2.414 [m]
x=15 [m]

44. Find: c(x,t) [mg/L] of chloride
0.1348 mg
c(x,t)=362.5
erfc(1.06) * L
+exp(12.16) * erfc(3.64)
erf(3.64) ~1.0 ~
i=0.002 m s
is practically 1, therefore, its compliment is considered to be zero, ---
K=3*10-5
t=1 [yr] Cl mg m2 s
η=0.23
co=725
D=1*10-9 L x
αx=0.83*(log(x))2.414 [m]
x=15 [m]

45. Find: c(x,t) [mg/L] of chloride
0.1348 mg
c(x,t)=362.5
erfc(1.06) * L
+exp(12.16) * erfc(3.64)
erf(3.64) ~1.0 ~
erfc(3.64) ~ 0.0 ~
i=0.002 m s
and this last term cancels out of the equation. [pause] Solving for the concentration then becomes ---
K=3*10-5
t=1 [yr] Cl mg m2 s
η=0.23
co=725
D=1*10-9 L x
αx=0.83*(log(x))2.414 [m]
x=15 [m]

46. Find: c(x,t) [mg/L] of chloride
0.1348 * L
i=0.002 m s
362.5 milligrams per liter, times , which equals, ----
K=3*10-5
t=1 [yr] Cl mg m2 s
η=0.23
co=725
D=1*10-9 L x
αx=0.83*(log(x))2.414 [m]
x=15 [m]

47. Find: c(x,t) [mg/L] of chloride
0.1348 * L mg
c(x,t)=48.9 L
i=0.002 m s
48.9 milligrams per liter.
K=3*10-5
t=1 [yr] Cl mg m2 s
η=0.23
co=725
D=1*10-9 L x
αx=0.83*(log(x))2.414 [m]
x=15 [m]

48. Find: c(x,t) [mg/L] of chloride10 25 50
100 mg
c(x,t)=362.5
0.1348 * L mg
c(x,t)=48.9 L
i=0.002 m s
Looking over the possible solutions, ---
K=3*10-5
t=1 [yr] Cl mg m2 s
η=0.23
co=725
D=1*10-9 L x
αx=0.83*(log(x))2.414 [m]
x=15 [m]

49. Find: c(x,t) [mg/L] of chloride10 25 50
100 mg
c(x,t)=362.5
0.1348 * L mg
c(x,t)=48.9 L
AnswerC
i=0.002 m s
the answer is C.
K=3*10-5
t=1 [yr] Cl mg m2 s
η=0.23
co=725
D=1*10-9 L x
αx=0.83*(log(x))2.414 [m]
x=15 [m]

50. ? Index σ’v = Σ γ d γT=100 [lb/ft3] +γclay dclay 1
Find: σ’v at d = 30 feet
(1+wc)*γw
wc+(1/SG)
σ’v = Σ γ d d
Sand
10 ft
γT=100 [lb/ft3]
100 [lb/ft3]
10 [ft]
20 ft
Clay
= γsand dsand
+γclay dclay A W S
V [ft3]
W [lb]
40 ft
text
wc = 37% ? Δh
20 [ft]
(5 [cm])2 * π/4