6. Groundwater flow equation - University of Calgarypeople.ucalgary.ca/~hayashi/glgy401/lectures/401f06_6.pdf · 6-1 6. Groundwater flow equation Mass conservation: Min – Mout =

6-1

6. Groundwater flow equation

Mass conservation: Min – Mout = ∆M/∆t

Darcy’s law ∆M-∆h relationship

Now we can put them all together.

A

An aquifer having a cross sectional area A. Top and bottom are confined by aquitards, and one end is connected to a lake.

∂∂

∂∂

∆=

∂∂

−∂∂

∆−=∂∂

∆−=

∂∂

∆+−≅

−=−

xhK

xxA

xhK

xxA

xqxA

xqxxqxqA

xqxqAMM

w

wx

w

xxxw

xxwoutin

ρ

ρρ

ρ

ρ

)()(

)]()([

00

10

qx(x0) qx(x1)

x = x0 x1

Darcy’s law:

xhKqx ∂

∂−=

yhKqy ∂

∂−=

In general, h and q are functions of (x, y, z, t).

In this case, h and q depends on (x, t) only.

∆x = x1 - x0

How is the water level in wells related to the lake-water level?

6-2

Mass change in a REV having volume V = A∆x.∆M = VρwSs∆ψ from Eq.[4-2]

At a fixed location,∆h = ∆(z + ψ) = ∆z + ∆ψ = 0 + ∆ψ∴ ∆h = ∆ψ

∴ ∆M = VρwSs∆h = A∆x ρwSs∆h

This is the one-dimensional groundwater flow equation. Note that [6-1] and [6-2] represent exactly the same thing. Partial differential equations are nothing more than a language to describe the simple conservation principle.

We can solve [6-2] in a number of ways. The next example shows a simple numerical technique to solve [6-2b].

∆x

A

thS

xhK

x

thSxA

xhK

xxA

tMMM

s

sww

outin

∂∂

=

∂∂

∂∂

∆∆

∆=

∂∂

∂∂

∆

∆∆

=−

ρρ

Putting them all together,

Making ∆t very small,

[6-1]

[6-2a]

If K is constant (i.e. material is uniform), we have:

thS

xhK s ∂

∂=

∂∂

2

2

[6-2b]

6-3

Propagation of a hydraulic pulse

Water level in two lakes has been kept at z = 0 for a long time. The water level in the left lake suddenly rises at t = 0.

We divide the aquifer into a number of cells and analyze the mass balance of each cell. cell-1 cell-2 cell-3

∆x

M12 M23

xhhhAK

xhhAK

xhhAK

MMMM

w

ww

outin

∆+−

=

∆−

+∆−

−=

−=−

321

2312

2312

2ρ

ρρ

For example, for cell-2:

thh

xSAt

hhSV

tM tt

swtt

sw ∆−

∆ρ=∆

−ρ=

∆∆ ∆+∆+ 2@22@2

thh

xSAx

hhhAK ttsww ∆

−∆ρ=

∆+−

ρ∴ ∆+ 2@2321 2

[6-3]

∆x

z = 01

)2()( 32122@2 hhh

xStKhh

stt +−

∆∆

+=∴ ∆+

6-4

Eq. [6-3] is called forward finite difference equation, which is just another way of writing the mass balance equation [6-1] and the partial differential equation [6-2].

All of these give us the increment of storage. To calculate the storage amount itself, we need to specify:

Initial condition: h = 0 for all x > x0 at t = 0

The two end cells (0 and 10) only have one neighbor, and we cannot apply [6-3]; i.e. there is no h-1 or h11. One way to deal with this problem is to specify:

Boundary conditions: h = 1 at x = x0, h = 0 at x = x10

In this case, we force the boundary cells to take a specified value, which reflects the physical constraint on the system. This is called specified head or 1st type boundary condition.

Eq. [6-3] indicates that the future state of cell-2 is dependent on the present state of itself and its neighbors. We can write [6-3] for all cells and solve them simultaneously using a spread sheet (lab exercise).

t = 0x 0 x 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 x 10

1 0 0 0 0 0 0 0 0 0 01 01 01 0∆t

6-5Next, suppose that the aquifer is “plugged” at x = x10. The mass balance equation for cell-10 is:

010,910 −=

∆∆ M

tM

Therefore, we do not have to include h11 in our calculation. This is another way of imposing a physical constraint on the system, and called specified flux or 2nd type boundary condition. We can also write this as:

0=∂∂xh

at x = x10

0

0.2

0.4

0.6

0.8

1

0 20 40 60 80 100x (m)

h (m

)

t = 120 min

t = 30 min

As t → ∞, the graph becomes a straight line. (1) The mass storage does not change any more.(2) Specific discharge is constant with respect to both t and x.

Solution with 1st type boundary conditions (BC’s)

This graphs show the solution of [6-2b] with the 1st type BC’s;h = 1 m at x = 0 and h = 0 at x = 100 m

6-6

0

0.2

0.4

0.6

0.8

1

0 20 40 60 80 100x (m)

h (m

)

Solution with 1st and 2nd type BC’s

Suppose we have the following BC’s to go with [6-2b].h = 1 m at x = 0 and ∂h/∂x = 0 at x = 100 m

At steady state, the solution looks different from the last one.Differential equation ↔ Governing processesBoundary conditions ↔ Physical constraints

Steady-state solution

∆S/∆t = 0 at steady state, which means ∂h/∂t = 0 and h is a function of x only.

Eq.[6-2b] is now written as: or02

2

=dx

hdK 02

2

=dx

hd

It is easy to show that the steady-state solution is given by:h = C1x + C2

where C1 and C2 are constants that are dependent on BC’s.

6-7

Effects of hydraulic conductivity

Let’s go back to the example with 1st type BC’s. The graph shows the h(x) profile at t = 120 min. In this case, K= 10-5 m/s.If K = 10-6 m/s, how would the profile look like at t = 120 min?How about K = 10-4 m/s?

0

0.5

1

0 20 40 60 80 100x (m)

h(m

)

h = 0h = 1

0

0.5

1

0 20 40 60 80 100x (m)

h(m

)

h = 0h = 1 K1 K1K2

Suppose the middle part of aquifer is filled with a material having a smaller K, say K2 = K1/2. At steady state, how will the h(x) profile look like?

6-8

Bulk hydraulic conductivity

Water is flowing through a box at Q (m3/s). We are not given the information of the material. We use Darcy’s law to assign the bulk hydraulic conductivity(Kb) of the box.

LhAK

LhhAKAqQ b

inoutb

∆−=

−−==

Q

L

A1

A2

A3

K1

K2

K3

(1) Parallel layers

AAK

AAK

AAKKb

332211 ++=∴

L

Q

Area A

h = hin h = hout

Q

hAQLKb ∆

−=∴

Kb is dependent on the property of sediments and how they are arranged in the box. Let’s examine two important cases.

LhKA

LhKA

LhKAQ

∆−

∆−

∆−=

33

2211

Kb is given by a weighted arithmetic average of each material. The layer thickness serves as a weighting factor. Thicker layers have a heavier influence on Kb than thinner layers.

6-9

1

3

3

2

2

1

1

−

++=∴

LKL

LKL

LKLKb

3

33

2

22

1

11 L

hKLhK

LhKq ∆

−=∆

−=∆

−=

hhhh ∆=∆+∆+∆∴ 321

Head drop in each layer must add up to the total head drop ∆h.

L1 L2 L3

K1 K2 K3

h(2) Serial layers

In this case, we need to have the same value of q in all layers.

Kb is given by a weighted harmonic average of each material.

Example

SupposeA1 = A2 = A3 = A/3 L1 = L2 = L3 = L/3K1 = K3, K2 = 0.01 K1

What is the Kb for parallel and serial cases?

6-10

Majority of shallow unconsolidated sediments have horizontal strata, and their hydraulic conductivity is higher in horizontaldirection (Kx and Ky) than in vertical direction (Kz).

zhKq

yhKq

xhKq zzyyxx ∂

∂−=

∂∂

−=∂∂

−=

Anisotropy of hydraulic conductivity

In layered sediments, the value of Kb varies with direction. This is called the anisotropy of hydraulic conductivity.Note that heterogeneity at a small scale appears as anisotropy at a larger scale.

Darcy’s law in anisotropic material is given by:

6-11

V

R2R1

V

R2

R1

I = V/R R = L/(AKE) KE: electrical conductivity

dxdhKq −=

dxdTKq HH −=

Electricity and groundwater obey the same form of equation.

Darcy’s law Ohm’s law Fourier’s law

dxdVKi E−=

i : current density qH : heat flux density

Laws like Darcy’s law are called constituitive equations or phenomenological laws. In physical sciences, we combine them with the conservation equation:

This concept is the foundation of many scientific disciplines.

Similarity between Darcy’s and Ohm’s Law

We saw that the Kb is the arithmetic average for the parallel case, and the harmonic average for the serial case. Does this remind us of something?

tSQQ outin ∆

∆=−

6-12Two-dimensional flow equation

Suppose a confined aquifer having a constant thickness (b). We can analyze the mass balance of a box in this aquifer in a way similar to the analysis of one-dimensional flow.

We saw that Min - Mout in x-direction is given by (see p.6-1):

Similarly, Min - Mout in y-direction is given by:

[ ]

∂∂

∂∂

∆∆=

∂∂

∂∂

∆=−xhK

xxyb

xhK

xxAMM xwxwxoutin ρρ

[ ]

∂∂

∂∂

∆∆=

∂∂

∂∂

∆=−yhK

yyxb

yhK

yyAMM ywywyoutin ρρ

qx

y0

y1

qy

x1x0

∆x = x1 - x0 ∆y = y1 - y0

xy

b

Storage change is (see p.6-2):∆M = Vbox ρwSs∆h = b∆x∆y ρwSs∆h

Putting them all together into Min - Mout = ∆M/∆t,

thSyxb

yhK

yyxb

xhK

xxyb swywxw ∆

∆∆∆=

∂∂

∂∂

∆∆+

∂∂

∂∂

∆∆ ρρρ

[Min – Mout]x [Min – Mout]y ∆M/∆t

6-13

This is the 2-D groundwater flow equation. At steady state, Eq. [6-4] becomes:

0=

∂∂

∂∂

+

∂∂

∂∂

yhK

yxhK

x yx [6-5]

thS

yhK

yxhK

x syx ∂∂

=

∂∂

∂∂

+

∂∂

∂∂

[6-4]

Dividing both sides by b∆x∆yρw

Plan view

Cross sectionSuppose a two-dimensional, confined aquifer of homogeneous, isotropic sediments. North and south boundaries of the aquifer are impermeable, and east and west boundaries are connected to rivers. Water levels in rivers are constant.

x

y h=

15 m

h=

10 m

0 100 m0

50 m

Graphical solution of 2-D flow equation

How does hydraulic head change from the west to east?

6-14

14 13 12 11 1015

Observe:

(1) Flow lines are normal to contour lines.

(2) Contours meets the impermeable boundaries at 90º.

These are common features of the steady-state flow in isotropic aquifers.

Water level in groundwater wells indicate hydraulic head in the confined aquifer. Joining the well water levels, we can define an imaginary surface of hydraulic head. The contours in the flow diagram above show the shape of the surface.

6-15Equipotential and flow line

h = 1080 m

h = 1075 m82

8177

78

74

73

x

ypotentiometric surface

impermeable clay

sand

The potentiometric surface is an imaginary surface defined by the water level in wells (hydraulic head, h) in a single aquifer. Like any other surface, we can draw contours of constant h, called equipotentials.

∂∂

−∂∂

−yh

xh

What does this vector represent?,

yhKq

xhKq yyxx ∂

∂−=

∂∂

−=

From Darcy’s law:

In vector form, we can write,

(qx, qy) = (-Kx∂h/∂x, -Ky∂h/∂y)

If the material is isotropic (Kx = Ky), we can just use K and

(qx, qy) = K(-∂h/∂x, -∂h/∂y)

i.e. the flow direction is parallel to the gradient vector.

6-16

Hydraulic conductivity of anisotropic materials varies with direction, and flow lines may not be normal to equipotentials.

Groundwater flow direction is is normal to equipotentialswhen Kx = Ky.

6-17

Flownet construction

Remember the flow is at steady state.

Is q constant throughout the aquifer?Qin Qout

Plan view

Cross section

impermeable

impermeable

x

y h=

10 m

h=

7 m

0 100 m0

50 m

Let’s go back to the analysis of the confined aquifer. This time it has somewhat irregular shape.

Hydraulic head is 10 m in the west river and 7 m in the east river.

Can we draw equipotentials?

6-18

Equipotentials and are shown below. Note that they meet the impermeable boundary at 90º. Why?

Can we draw a few flow lines?

(1) Flow lines are normal to equipotentials.

(2) Spacing between flow lines?9.

5

9.0

8.5

8.0

7.510 7.0

7.5

7.25

l1

b1

l2

b2

The strip between two flow lines is called a “flow tube”. It is customary to draw flow lines so that each flow tube has the same Q. Suppose a horizontal layer in the aquifer having a thickness (normal to the page) of 1 m. The flow rate through a 1-m thick tube is given by:

If we chose the spacing so that bi = li, then Q in each tube is equal to K∆h.

Flow nets constructed this way provide a useful tool for the analysis of steady-state flow in homogeneous and isotropicmaterials.

i

iii l

bhKdldhKbQ ∆=−= )1( where ∆h = 0.25 m

6-19

9.5

9.0

8.5

8.0

7.510 7.0

Above is an example of a properly constructed flow net. Suppose that K = 10-4 m/s. We can say that:(1) There are four and half flow tubes.(2) Each tube carries Q = 0.25 × 10-4 m3/s. (3) 1-m thick layer has a total flow rate of 1.1 × 10-4m3/s.(4) If the aquifer thickness is 5 m, the total flow rate will be

5.5 × 10-4 m3/s.

Effects of heterogeneity

Can we construct a flow net for this case?

x

y

h=

15 m

h=

10 m

0 100 m0

50 mK1 = 10-3 m/s

K1 = 10-3 m/sK 2= 0.

3 ×10

-3 m/s

6-20Flow lines and equipotentials are still normal to each other. However, heterogeneity creates some features that were not seen in the homogeneous case.(1) Flow lines refract at the zone boundaries.

(2) Equipotentials are much denser in the low-K zone.

Groundwater takes the “paths of least resistance” by taking the shortest path through the low-K material. The shortest path through the low-K zone is achieved by flowing straight across the zone.

x

y

h=

15 m

h=

10 m

0 100 m0

50 mK1 = 10-3 m/s

K2 = 5 × 10-3 m/s

How will the flow net look like in this case?

14 13

12 11 1015

6-21In this case, the paths of least resistance are achieved by channeling the flow through the high-K zone.

K1 < K2

Note that, at steady state, the flow volume (Q = Aq) crossing the interface is equal on both sides.

Let’s say q1 and q2 are the specific discharges along flow directions. From mass balance principle,

A1q1 = A2q2

where ∆h is the head difference between the two contours. Note that contours are continuous across the interface.

K2 is greater than K1, which means θ2 must be greater than θ1. The flow lines refract.

222

111 L

hKALhKA ∆

=∆

∴

2

1

22

11

2

1

tantan

//

θθ

==∴ALAL

KK

Tangent law (optional)

θ2

q1

A2

θ1

q2

L2

L1A1 h

h + ∆h

K1

K2

14 13 12 11 1015