Principles of groundwater flow

Darcy's Law, Duputi Equation, Flow Net

1. By : Putika Ashfar .K
May 2016
Principles of Groundwater Flow

2. References
Todd, D. K., 1980, Groundwater hydrology, 2d ed . : New
York, John Wiley, 535 p
Bear, J., Beljin, M. S., Rose, R. Fundamentals of ground
water Modelling, EPA Ground Water Issue, 1992.
Harbaugh, A. W. MODFLOW-2005, The U.S. Geological
Survey Modular ground water Model - The ground water
Flow Process: U.S. Geological Survey Techniques and
Methods 6-A16, 2005.

3. Governing equation of groundwater f low
Darcy’s Law
Darcy’s law is the most fundamental equation, govern
groundwater movement



K = hydraulic conductivity (m/d)
Q = water discharge (m
3
/d)
dh/dl = hydraulic gradient

4. Equation of groundwater flow
Confined Aquifer

5. Equation of groundwater flow
Unconfined Aquifer

6. Tensiometer (for unconfined aquifer)
Steady state unsaturated flow
Gradient
due to
capillary
force
Gradient
due to gravity
Land surface
Capillary fringe
Water table
datum
Ht
hp
z
b
hc hc

7. A confined aquifer, has a source of recharge , K = 50 m/day, n = 0,2.
The piezometric head is 1000 m apart , and water level in each
piezometric are 55 m and 50 m respectively from a common datum.
The average thickness of aquifer is 30 m. The average width flow is 5
km
Calculate :
1.The darcy seepage velocity in aquifer
2.The average time of travel, from the head of aquifer to 4 km
downstream

8. Direction of flow
55 m
50 m
1000 m
5 m
30 m
Bed rocks
Confining
unit

9. Cross-sectional area of aquifer (A)
b= 30 m
w = 5 km x 1000 = 5000 m
A = b . w = 150.000 m
2
Hydraulic gradient (HL)
H
L=Δh/L = (55 m – 50 m)/ 1000 m = 0,005
Rate of flow through aquifer (Q)
Q = K.A. H
L
Q = 50 m/ day. 150.000 m
2
. 0,005
Q= 37.500 m
3
/day

10. Darcy Velocity (v)
Q = A.v
v= Q/A = (37.500 m3/day)/(150.0000 m2)
v= 0,25 m/day
Seepage velocity (Vs)
Vs= v/n
Vs= 0,25 m/day /2 = 1,25 m/day
Time to travel 4 km downstream
v = S/T
T = S/v = (4. 1000 m)/ (0,25 m/day)
T = 3200 days = 8,7 years

11. Flow Net
To determine the solution of flow equations, we can use
analytical solutions or numerical solutions.
 Graphical analysis is one of considered method to traces
the path which a particle of groundwater would flows
through an aquifier by draw a flow lines. In steady
groundwater flow, we can use a flow net to construct a
graphical solution. The flow net consists of flow lines and
equipotential lines.
The equipotential lines show the distribution of potential
energy. It must fulfill the assumptions the material zones
flow through a confined aqifier are homogenous and
isotropic.

12. Flow Net Assumptions
Material zones are homogenous
K is isotropic
Fully saturated and steadyflow
Laminar, continous and irrotational flow
Fluid has a constant density
Darcy’s Law is valid

13. Step by step drawing a flow net
1.Make your 2D system picture
2.Determine boundary conditions
constant head (AB , AE)
impermeable boundary / no flow boundary (AE)
water table, where P = P atm (WL)
athmospheric P (preatic surface) (BC)
athmospheric P (seepage surface) (CD)
WL
WL
datum
B
A
C
D
E

14. Rules of drawing flow net
1.Flow lines is parallel with impermeable boundary/
no flow boundary
2.Equipotential lines is parallel with constant head
boundary
3.Flow lines is perpendicular wit equipotential lines
4.Equipotential lines is perpendicular with no flow
boundary

15. 3 flow tubes (Nf)
h1
h2 h3 h4 h5
l
w
4 stream lines
6 equipotential lines
5 head drops
(hd)– (Nd = 5)
Determine potential drops (hd) = Δh/N
d
Nd = head difference

16. Flow Net …(Example)
Boundary
Conditions

Lines kb (h1) and hl
(h9) are equipotentials
because they are at
constant elevation and
have constant water
depth above them.

Lines mn (q1), be (q5)
and he (q5) are flow
lines because they are
impermeable surfaces

17. Determine boundary heads
Head along lines kb and hl are equal to elevation of boundary above
the datum plus depth of water above the boundary:
h1 = 60 ft + 30 ft = 90 ft
h9 = 60 ft + 5 ft = 65 ft
Total Δh along flow lines (e.g., along q2)
Δh = (h1 – h9) = (90 ft – 65 ft) = 25 ft

18. How do anisotropic materials influence
flownets?
Flow lines will not meet equipotential lines at right angles, but they will if we
transform the domain into an equivalent isotropic section, draw the flow net, and
transform it back.
For the material above, we would either expand z dimensions or compress x
dimensions To do this we establish revised coordinate . We can use transformed K

19. Drawing flow net procedure for
anisotropic

21. Flow net with variation of K

24. Dupuit Equation (Unconfined Flow)
Assumptions :
1.The water table / free surface is only slightly inclined
2.Streamlines may be continued horizontal and
equipotential lines is vertical
3.Slope of the free surface and hydraulic gradient are
equal

25. Derivation of Dupuit Law
From Darcy’s Law
At steady state, rate of change of Q with distance, is 0

26. By subtitute eq (3) to eq (2) and (5), thus we have
If eq (6) setting x =L , so
Dupuit Equation
Dupuit Parabola

27. Application of Dupuit Equation
2 Rivers 1000 m apart , with K = 0,5 m/day
Average rainfall = 15 cm/years
Evaporation = 10 cm/ years
Water elevation of river 1 = 20 m ; and river 2 = 18 m
w= 5 cm /years = 1,369. 10
-4
m/day

Determine the daily discharge per-meter width into
each river

28. Water
table
Groundwater
divide
h0 =20 m
h
L =18 m
WL
WL
L
d
x

29. Discharge into river 1
The (-) sign indicates that flow is in the opposite direction
from x direction , so it means Q = 0,05 m
2
/day to river 1

30. Discharge into river 2, x = 1000 m
Flow is in the same direction from x direction , so it means
Q = 0,087 m
2
/day to river2

31. Find groundwater divide by setting Q = 0

32. “Make everything as simple as possible, but not simpler”
–Albert Einstein

Thank you 