acf24527-231b-477b-ad7f-4602190d9704 13 4 5 decimal

&[DATE] &[TIME] - &[FILENAME]

SOLUTION:

L 1000 m * :

Δh 3 m * - :

K 3.5 m day / * :

α 0.35 :

Δt = ?

v K Δh L / * - : m day / 0.0105

v.α v α / : m day / 0.03

Δt L v.α / : yr 91.2636

Textbook approach: Note that the notation for porosity, now referred to as effective porosity, is not longer α, but .

n.e

For a differential of length, dx, along the flow, the head drop is -dh, and, with constant Darcy velocity, and homogeneous hydraulic conductivity K, we have (using h=0, at x =0):

K K 11 line :

1 h 0 h int v K / (- x 0 x int ≡

v K dh dx / * - ≡

dh v K / dx * - ≡

h v K / x * - ≡

K 3.5 m day / * :

For

x L : 1000 m *

and

h 3 m * - :

v K h x / * - : m day / 0.0105

v.α v α / : m day / 0.03

Δt L v.α / : yr 91.2636

-->

SOLUTION:

For this case, the elevation of

the water table can be used also as the thickness of the flow, i.e., b = h. The Dupuit assumption consists of taking the flow velocities to be horizontal, so that Darcy's law, applied to an increment, dx, of distance along the flow direction, with constant flow Q in width W of aquifer is:

Q K W * h * dh dx / * - ≡

h dh * Q K W * / dx * - ≡

h h h.0 h int q K / - x 0 x int ≡

Using q = Q/W,

12 / h 2^h.0 2^- (* q K / x * - ≡

h 2^h.0 2^q 2 K * / x * - ≡

h h.0 2^q 2 K * / x * - sqrt ≡

q K 2 x * / h.0 2^h 2^- (* ≡

K 101 -^cm s / * :

h.0 6.5 m * :

h 4.0 m * :

x 150 m * :

q K 2 x * / h.0 2^h 2^- (* : m 2^day / 7.56

Base flow to a stream

Symmetric streams, with heads h.a at the streams, x measured from the

C.L., stream at x = a. Uniform rain W (L/T), producing a discharge, at section x, equal to

q W x * ≡

[1]

At section x, applying Dupuit's assumption for an unconfined aquifer, we would have:

q K h * dh dx / * - ≡

[2]

K h * dh dx / * - W x * ≡

Making [1] = [2],

K h h h.a h int * - W x x a x int * ≡

K 2 / h 2^h.a 2^- (* - W 2 / x 2^a 2^- (* ≡

h 2^h.a 2^W K / a 2^x 2^- (* + ≡

Each stream receives a flow

W a *

being fed from each half of the surrounding dividing zone.

Thus, the total recharge per unit length of channel is

Q.b

2 a * W *

This is the base flow.

EXAMPLE -

Suppose that, for the situation depicted in the figure immediately above, the

hydraulic conductivity is 0.0018 m/day, the streams are separated by 125 m, and a well located at the mid point of the dividing zone shows a piezometric head of 5.2 m, when the water depth in the streams, measured from the same datum as the head, is 3.2 m. Determine the recharge rate W (cm/hr), and the base flow Qa (m^2/day).

K 2 m day / * :

a 125 m * 2 / : 62.5 m *

SOLUTION:

h 5.2 m * :

x 0.0 m * :

and

h.a 3.2 m * :

W K h 2^h.a 2^- a 2^x 2^- / (* : cm hr / 0.0358

h 2^h.a 2^W K / a 2^x 2^- (* + ≡

From

Q.a 2 a * W * : m 2^day / 1.0752

-------------------------------------------------------------------------------------------

Unconfined aquifer between two river

Consider the case depicted in the figure. The position x is measured from River B, where the water depth is h1. River A is located at x = L from River B. The aquifer is recharged at a rate W.

For a length dx of the unconfined aquifer along the direction of the groundwater flow, we have

q.x K h * dh dx / * - ≡

[1]

dq.x dx / W ≡

Also, the discharge per unit width of aquifer, is

[2]. Combining [1] and [2]:

[2']

K h * dh dx / * - (x diff W ≡

K h * dh dx / * - W x * C.1 + ≡

. Integrating once,

, then

K h * dh * - W x * C.1 + (dx * ≡

. Integrate

K h 2^2 / * - W x 2^2 / * C.1 x * + C.2 + ≡

again,

[3]. Now, we introduce the boundary condition, start with

K h 2^2 / * - W x 2^2 / * C.1 x * + K 2 / h.1 2^* - ≡

K 2 / h.1 2^* - C.2 ≡

h h.1 ≡

at x = 0, then:

, thus, [3] becomes:

[4]

The second boundary conditions is:

h h.2 ≡

at x = L, so that [4] becomes

K 2 / h.2 2^* - W 2 / L 2^* C.1 L * + K 2 / h.1 2^* - ≡

C.1 K 2 L * / h.1 2^h.2 2^- (* W 2 / L * - ≡

[5]. With [5] into [4],

K 2 / h 2^* - W x 2^2 / * K 2 L * / h.1 2^h.2 2^- (* x * + W L * 2 / x * - K 2 / h.1 2^* - ≡

h 2^h.1 2^h.1 2^h.2 2^- L / (x * - W K / L x - (* x * + ≡

h 2^h.1 2^h.1 2^h.2 2^- L / (x * - W K / L x - (* x * + ≡

, or

[6]

2 h * dh dx / * h.1 2^h.2 2^- L / (- 2 W K / * L 2 / x - (* + ≡

From [6], using implicit derivatives:

. From which,

h dh dx / * h.1 2^h.2 2^- 2 L * / (- W K / L 2 / x - (* + ≡

. Result, which can now be introduced into [1], to calculate

q.x

q.x K h.1 2^h.2 2^- (* 2 L * / W L 2 / x - (* - ≡

[7]

To determine location of the crest in the water table, i.e., the value of x where h is a maximum, we make dh/dx = 0. Since, qx = h*(dh/dx), then, the crest is located where qx = 0. Let qx = 0 at x = d, then, from [7]

K h.1 2^h.2 2^- (* 2 L * / W L 2 / d - (* - 0 ≡

d L 2 / K W / h.1 2^h.2 2^- 2 L * / (* - ≡

[8]

[8], then

With this value of d, we can calculate the crest at the water table, from [6], by making x=d:

h.max h.1 2^h.1 2^h.2 2^- L / (x * - W K / L d - (* d * + sqrt ≡

[9]

--------------------------------------------------------------------------------------------

K 102 -^cm s / * :

W 1.6 m yr / * :

SOLUTION:

h.2 8.5 m * :

h.1 10 m * :

L 460 m * :

(from Figure)

n.e 0.35 :

(a) Location of the GW divide:

d L 2 / K W / h.1 2^h.2 2^- 2 L * / (* - : 170.5092 m *

measured from River B

L d - 289.4908 m *

measured from River A

The maximum elevation of the water table is:

h.max h.1 2^h.1 2^h.2 2^- L / (d * - W K / L d - (* d * + sqrt : 10.7117 m *

(b) To estimate travel time through the aquifer to each of the rivers, we use an approximate pore velocity (Darcy velocity divided by porosity) for each river:

i.A h.max h.2 - L d - / : 0.0076

v.A K i.A * n.e / : m day / 0.1886

* Flow towards river A:

Δt.A L d - v.A / : yr 4.2026

i.B h.max h.1 - d / : 0.0042

v.B K i.B * n.e / : m day / 0.103

* Flow towards river B:

Δt.B d v.B / : yr 4.5307

q.A K h.1 2^h.2 2^- (* 2 L * / W L 2 / L - (* - : m 3^day / km / 1268.1593

* For River A:

q.B K h.1 2^h.2 2^- (* 2 L * / W L 2 / 0 - (* - : m 3^day / km / 746.9419 -

* For River B: