By : Putika Ashfar .K May 2016

Principles of Groundwater Flow

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.

Governing equation of groundwater flow

Darcy’s Law Darcy’s law is the most fundamental equation, govern

groundwater movement

K = hydraulic conductivity (m/d) Q = water discharge (m3/d) dh/dl = hydraulic gradient

Equation of groundwater flow Confined Aquifer

Equation of groundwater flow Unconfined Aquifer

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

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

Direction of flow

55 m

50 m 1000 m

5 m

30 m Bed rocks

Confining unit

Cross-sectional area of aquifer (A) b= 30 m w = 5 km x 1000 = 5000 m A = b . w = 150.000 m2

Hydraulic gradient (HL) HL=Δh/L = (55 m – 50 m)/ 1000 m = 0,005 Rate of flow through aquifer (Q) Q = K.A. HL Q = 50 m/ day. 150.000 m2. 0,005 Q= 37.500 m3/day

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

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.

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

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

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

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/Nd

Nd = head difference

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

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

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

Drawing flow net procedure for anisotropic

Flow net with variation of K

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

Derivation of Dupuit Law From Darcy’s Law

At steady state, rate of change of Q with distance, is 0

By subtitute eq (3) to eq (2) and (5), thus we have

If eq (6) setting x =L , so

Dupuit Equation Dupuit Parabola

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

Water table

Groundwater divide

h0 =20 m

hL =18 m

WL

WL

L d

x

Discharge into river 1

The (-) sign indicates that flow is in the opposite direction from x direction , so it means Q = 0,05 m2/day to river 1

Discharge into river 2, x = 1000 m

Flow is in the same direction from x direction , so it means Q = 0,087 m2/day to river2

Find groundwater divide by setting Q = 0

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

Thank you