1. Find: hmax [m] L 17.6 17.8 18.0 18.2 hmax h1 h2 L = 525 [m]
w = [m/day]
hmax h1 h2
Find the maximum water height in meters. [pause] In this problem, L
L = 525 [m] m
day
h1=17.6 [m]
K=14.5
h2=15.3 [m]

2. Find: hmax [m] L 17.6 17.8 18.0 18.2 hmax h1 h2 L = 525 [m]
w = [m/day]
hmax h1 h2
two bodies of water, having different water surface elevations, are separated by a known distance. L
L = 525 [m] m
day
h1=17.6 [m]
K=14.5
h2=15.3 [m]

3. Find: hmax [m] L 17.6 17.8 18.0 18.2 hmax h1 h2 L = 525 [m]
w = [m/day]
hmax h1 h2
The dividing land mass has a known hydraulic conductivity and experiences a constant recharge. L
L = 525 [m] m
day
h1=17.6 [m]
K=14.5
h2=15.3 [m]

4. Find: hmax [m] L Q=-K * i * A hmax h1 h2 flowrate L = 525 [m]
w = [m/day]
flowrate
hmax h1 h2
Darcy’s law states that flowrate through a soil is equal to --- L
L = 525 [m] m
day
h1=17.6 [m]
K=14.5
h2=15.3 [m]

5. Find: hmax [m] L Q=-K * i * A hmax h1 h2 flowrate hydraulic
w = [m/day]
flowrate
hydraulic
conductivity
hmax h1 h2
the hydraulic conductivity, --- L
L = 525 [m] m
day
h1=17.6 [m]
K=14.5
h2=15.3 [m]

6. Find: hmax [m] L Q=-K * i * A hmax h1 h2 gradient flowrate hydraulic
w = [m/day]
flowrate
hydraulic
conductivity
hmax h1 h2
times the gradient, --- L
L = 525 [m] m
day
h1=17.6 [m]
K=14.5
h2=15.3 [m]

7. Find: hmax [m] L Q=-K * i * A hmax h1 h2 gradient total area flowrate
w = [m/day]
flowrate
hydraulic
conductivity
hmax h1 h2
times the total area. [pause] From a one-dimensional perspective, --- L
L = 525 [m] m
day
h1=17.6 [m]
K=14.5
h2=15.3 [m]

8. Find: hmax [m] L Q=-K * i * A Q hmax w= h1 h2 A gradient total area
w = [m/day]
flowrate
hydraulic
conductivity Q
hmax w= h1 h2 A
the average recharge rate, w, is equal to the flowrate, divided by the area. L
L = 525 [m] m
day
h1=17.6 [m]
K=14.5
h2=15.3 [m]

9. Find: hmax [m] L Q=-K * i * A w=-K * i Q hmax w= h1 h2 A L = 525 [m]
w = [m/day]
w=-K * i Q
hmax w= h1 h2 A
This term is then substituted into Darcy’s equation, --- L
L = 525 [m] m
day
h1=17.6 [m]
K=14.5
h2=15.3 [m]

10. Find: hmax [m] L Q=-K * i * A w=-K * i w - = i K hmax h1 h2
w = [m/day]
w=-K * i w -
= i K
hmax h1 h2
both sides are divided K, the hydraulic conductivity. [pause] After skipping a few steps --- L
L = 525 [m] m
day
h1=17.6 [m]
K=14.5
h2=15.3 [m]

11. Find: hmax [m] + L Q=-K * i * A w=-K * i w - = i K w hmax - = h1 h2 K
w = [m/day]
w=-K * i w -
= i K w
δ2h2
δ2h2 1
hmax - = + h1 * h2 K 2
δx2
δy2
the gradient is rewritten as the sum of the partial gradients in the vertical and horizontal directions. L
L = 525 [m] m
day
h1=17.6 [m]
K=14.5
h2=15.3 [m]

12. Find: hmax [m] + L Q=-K * i * A w=-K * i w - = i K w hmax - = h1 h2 K
w = [m/day]
w=-K * i w -
= i K w
δ2h2
δ2h2 1
hmax - = + h1 * h2 K 2
δx2
δy2
Assuming flow at the soil water interface is purely horizontal, --- L
L = 525 [m] m
day
h1=17.6 [m]
K=14.5
h2=15.3 [m]

13. Find: hmax [m] + L Q=-K * i * A w=-K * i w - = i K w hmax - = h1 h2 K
w = [m/day]
w=-K * i w -
= i K w
δ2h2
δ2h2 1
hmax - = + h1 * h2 K 2
δx2
δy2
the partial differential equation is simplified by excluding the vertical component. L
L = 525 [m] m
day
h1=17.6 [m]
K=14.5
h2=15.3 [m]

14. Find: hmax [m] + - L Q=-K * i * A w=-K * i w - = i K w hmax - = h1 h2
w = [m/day]
w=-K * i w -
= i K w
δ2h2
δ2h2 1
hmax - = + h1 * h2 K 2
δx2
δy2
The expression is then integrated with respect to x ---
2*w
d2(h2) - L = K
δx2
L = 525 [m] m
day
h1=17.6 [m]
K=14.5
h2=15.3 [m]

15. Find: hmax [m] + - - L Q=-K * i * A w=-K * i w - = i K w hmax - = h1
w = [m/day]
w=-K * i w -
= i K w
δ2h2
δ2h2 1
hmax - = + h1 * h2 K 2
δx2
δy2
and the constants of integration are solved using known boundary conditions.
2*w
d2(h2) - L = K
δx2
w*x2 -
h2=
+c1*x + c2 K

16. Find: hmax [m] + - - L x Q=-K * i * A w=-K * i w - = i K w hmax - = h1
w = [m/day]
w=-K * i w -
= i K w
δ2h2
δ2h2 1
hmax - = + h1 * h2 K 2
δx2
δy2
If we set x equal to the horizontal distance from the left bank, in the direction of the right bank, ---
2*w
d2(h2) - L = K
δx2 x
w*x2 -
h2=
+c1*x + c2 K

17. Find: hmax [m] + - - L x Q=-K * i * A w=-K * i w - = i K w hmax - = h
w = [m/day]
w=-K * i w -
= i K w
δ2h2
δ2h2 1
hmax - = + h h1 * h2 K 2
δx2
δy2
and we set h, equal to the vertical distance up to the groundwater table, ---
2*w
d2(h2) - L = K
δx2 x
w*x2 -
h2=
+c1*x + c2 K

18. Find: hmax [m] + - - L x Q=-K * i * A w=-K * i w - = i K w hmax - = h
w = [m/day]
w=-K * i w -
= i K w
δ2h2
δ2h2 1
hmax - = + h h1 * h2 K 2
δx2
δy2
the first boundary condition will be x = 0, h = h1, and the second boundary condition will be ---
2*w
d2(h2) - L = K
δx2 x
w*x2
BC#1: x=0, h=h1 -
h2=
+c1*x + c2 K

19. Find: hmax [m] + - - L x w - = i K w hmax - = h h1 h2 K = K w*x2
w = [m/day] w -
= i K w
δ2h2
δ2h2 1
hmax - = + h h1 * h2 K 2
δx2
δy2
x = L, h = h2. Finally, the height of the groundwater table,
2*w
d2(h2) - L = K
δx2 x
w*x2
BC#1: x=0, h=h1 -
h2=
+c1*x + c2 K
BC#2: x=L, h=h2

20. Find: hmax [m] + - - + L x w - = i K w hmax - = h h1 h2 K = K w*x2
(h1 - h2 ) * x 2 2
*(L-x) * x
h = h1- +
[m/day] 2 K L w -
= i K w
δ2h2
δ2h2 1
hmax - = + h h1 * h2 K 2
δx2
δy2
as a function of x equals the square root of the quadratic, shown here. [pause] From the given data, ---
2*w
d2(h2) - L = K
δx2 x
w*x2
BC#1: x=0, h=h1 -
h2=
+c1*x + c2 K
BC#2: x=L, h=h2

21. Find: hmax [m] + - - + L x w - = i K w hmax - = h h1 h2 K = K w*x2 K w
(h1 - h2 ) * x 2 2
*(L-x) * x
h = h1- +
[m/day] 2 K L w -
= i K w
δ2h2
δ2h2 1
hmax - = + h h1 * h2 K 2
δx2
δy2
we know h1, h2, the recharge w, the hydraulic conductivity K, and the length L, ---
2*w
d2(h2) - L = K
δx2 x
L = 525 [m]
w*x2 m
day -
h1=17.6 [m]
h2=
+c1*x + c2
K=14.5 K
h2=15.3 [m]

22. Find: hmax [m] ? ? ? + L x hmax h h1 h2 w (h1 - h2 ) * x 2 2 h = h1- 2
*(L-x) * x
h = h1- +
[m/day] 2 K L ? ? ?
hmax h h1 h2
but the value of x is unknown. [pause] Since the problem asks to find the maximum height of the groundwater table, L x
L = 525 [m] m
day
h1=17.6 [m]
K=14.5
h2=15.3 [m]

23. Find: hmax [m] ? ? ? + L x hmax h h1 h2 w (h1 - h2 ) * x 2 2 h = h1- 2
*(L-x) * x
h = h1- +
[m/day] 2 K L ? ? ?
hmax h h1 h2
we need to find the value of x which corresponds to this maximum height. L x
L = 525 [m] m
day
h1=17.6 [m]
K=14.5
h2=15.3 [m]

24. Find: hmax [m] ? ? ? + L x hmax h h1 h2 w (h1 - h2 ) * x 2 2 h = h1- 2
*(L-x) * x
h = h1- +
[m/day] 2 K L ? ? ? dh =0 dx
hmax h h1 h2
At the groundwater divide, the slope of the groundwater table is zero, therefore, we can --- L x
L = 525 [m] m
day
h1=17.6 [m]
K=14.5
h2=15.3 [m]

25. Find: hmax [m] + L x hmax h h1 h2 w (h1 - h2 ) * x 2 2 h = h1- 2 K L
*(L-x) * x
h = h1- +
[m/day] 2 K L dh =0 dx d
hmax h h1 h2 dx
differentiate our equation for the height of the groundwater table, with respect to x, --- L x
L = 525 [m] m
day
h1=17.6 [m]
K=14.5
h2=15.3 [m]

26. Find: hmax [m] + L x hmax h h1 h2 w (h1 - h2 ) * x 2 2 h = h1- 2 K L
*(L-x) * x
h = h1- +
[m/day] 2 K L dh =0 dx d =0
hmax h h1 h2 dx
set that value equal to zero, and solve for x. Doing so will calculate --- L x
L = 525 [m] m
day
h1=17.6 [m]
K=14.5
h2=15.3 [m]

27. Find: hmax [m] - + L x hmax h h1 h2 w (h1 - h2 ) * x 2 2 h = h1- 2 K L
*(L-x) * x
h = h1- +
[m/day] 2 K L dh =0 dx d =0
hmax h h1 h2 dx
the value of x, corresponding to the maximum groundwater height, h max. L K
(h1 - h2 ) 2 2 - L
xhmax= * w 2
2*L x
L = 525 [m] m
day
h1=17.6 [m]
K=14.5
h2=15.3 [m]

28. Find: hmax [m] - + L x hmax h h1 h2 w (h1 - h2 ) * x 2 2 h = h1- 2 K L
*(L-x) * x
h = h1- +
[m/day] 2 K L dh =0 dx d =0
hmax h h1 h2 dx
The known values of L, K, w, h1 and h2 are plugged in, and we compute the horizontal distance to from the left bank --- L K
(h1 - h2 ) 2 2 - L
xhmax= * w 2
2*L x
L = 525 [m] m
h1=17.6 [m]
K=14.5
day
h2=15.3 [m]

29. Find: hmax [m] - + L x hmax h h1 h2 w (h1 - h2 ) * x 2 2 h = h1- 2 K L
*(L-x) * x
h = h1- +
[m/day] 2 K L dh =0 dx d =0
hmax h h1 h2 dx
to the groundwater divide, to be meters. [pause] Now that we know the appropriate value for x, --- L K
(h1 - h2 ) 2 2 - L * w 2
2*L x
L = 525 [m] m
xhmax=113.2 [m]
h1=17.6 [m]
K=14.5
day
h2=15.3 [m]

30. Find: hmax [m] - + L x hmax h h1 h2 w (h1 - h2 ) * x 2 2 h = h1- 2 K L
*(L-x) * x
h = h1- +
[m/day] 2 K L dh =0 dx d =0
hmax h h1 h2 dx
it can be plugged into the equation for the height of the groundwater table, --- L K
(h1 - h2 ) 2 2 - L * w 2
2*L x
L = 525 [m] m
xhmax=113.2 [m]
h1=17.6 [m]
K=14.5
day
h2=15.3 [m]

31. Find: hmax [m] - + L x hmax h h1 h2 w (h1 - h2 ) * x 2 2 h = h1- 2 K L
*(L-x) * x
h = h1- +
[m/day] 2 K L dh =0 dx d =0
hmax h h1 h2 dx
in addition to the other known variables, and the maximum groundwater height --- L K
(h1 - h2 ) 2 2 - L * w 2
2*L x
L = 525 [m] m
xhmax=113.2 [m]
h1=17.6 [m]
K=14.5
day
h2=15.3 [m]

32. Find: hmax [m] - + L x hmax=17.78 [m] h1 h2 w (h1 - h2 ) * x 2 2
*(L-x) * x
h = h1- +
[m/day] 2 K L d =0
hmax=17.78 [m] h1 h2 dx
equals meters. [pause] L K
(h1 - h2 ) 2 2 - L * w 2
2*L x
L = 525 [m] m
xhmax=113.2 [m]
h1=17.6 [m]
K=14.5
day
h2=15.3 [m]

33. Find: hmax [m] - + L x 17.6 17.8 18.0 hmax=17.78 [m] 18.2 h1 h2 w
(h1 - h2 ) * x 2 2
*(L-x) * x
h = h1- +
[m/day] 2 K L
17.6
17.8
18.0
18.2 d =0
hmax=17.78 [m] h1 h2 dx
Looking back at the possible solutions, L K
(h1 - h2 ) 2 2 - L * w 2
2*L x
L = 525 [m] m
xhmax=113.2 [m]
h1=17.6 [m]
K=14.5
day
h2=15.3 [m]

34. Find: hmax [m] - + AnswerB L x 17.6 17.8 18.0 hmax=17.78 [m] 18.2 h1
(h1 - h2 ) * x 2 2
*(L-x) * x
h = h1- +
[m/day] 2 K L
17.6
17.8
18.0
18.2 d =0
hmax=17.78 [m] h1 h2 dx
AnswerB
the answer is B. L K
(h1 - h2 ) 2 2 - L * w 2
2*L x
L = 525 [m] m
xhmax=113.2 [m]
h1=17.6 [m]
K=14.5
day
h2=15.3 [m]

35. ? 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