Eqns. Chap. 5a (Fetter)



Mechanical Energy

Kinetic Energy

Ek = mv2

where Ek is energy (kg-m2/s2)

v is velocity (m/s)

Gravitational Potential Energy

Eg = W = mgz

where w is work (kg-m2/s2)

m is mass (kg)

z is elevation above datum

Pressure of surrounding fluid (potential energy per unit volume)

Total Energy (per unit volume)

Etv = ρv2 + ρgz + P

Total Energy (per unit mass) or Bernoulli equation

Etm = v2 + gz +

For a incompressible, frictionless fluid

+ z + = constant

This is the total mechanical energy per unit weight or hydraulic head (m)

Groundwater velocity is very low, m/yr, so kinetic energy can be dropped

h = z +

Head in water of variable density

hf = (ρp/ρf)hp

where hp = pressure head

hf = freshwater head

Force Potential

Φ = gh = gz +

Darcy's Law

Q = -KA = -

Applicability of Darcy's Law (Laminar Flow)

R = < 1

where v = discharge velocity (m/s)

d = diameter of passageway (m)

R = Reynold's number

Specific Discharge (Darcy Flux) and Average Linear Velocity

v = = -K

vx = = -

Equations of Ground-Water Flow

Confined Aquifers

Flux of water in and out of a Control Volume

inMx = ρwqxdydz; outMx = ρwqxdydz + dx dydz;

netMx = - dx dydz

netMy = - dy dxdz

netMz = - dz dxdy

netMflux = - ( + + ) dxdydz

From Darcy's Law

qx = - K qy = - K qz = - K

and assuming that ρw and K do not vary spatially

netMflux = K ( + + ) ρwdxdydz

Note hydraulic head and hydraulic conductivity are easier to measure than discharge.

Mass of Water in the Control Volume

M = ρw n dxdydz => =

if dxdy is constant

= [ ρw n ,∂t) + ρw dz + n dz ] dxdy

compressibility of water

βdP =

compressibility of control volume (vertical only)

αdP =

Volume of Solids is constant

dVs = 0 = d[(1 - n)dxdydz] => dzdn = (1 - n)d(dz)

dn =

Fluid Pressure

P = P0 + ρwgh => dP = ρwgdh

dρw = ρwβ(ρwgdh)

d(dz) = dzα(ρwgdh) => dn = (1 - n)αρwgdh

Substituting into equation for change of mass with time

= (αρwg + nβρwg)ρw dxdydz

Conservation of Mass

= netMflux

K ( + + ) = (αρwg + nβρwg)

In Two-Dimensions, this reduces to

+ =

where Storativity = b(αρwg + nβρwg)

Transmissivity = bK

For Steady Flow, ∂h/∂t = 0 (Laplace equation)

+ + = 0

For leakage into an aquifer from a confining layer

+ + =

where e = leakage rate and from Darcy's Law

e = K'

where K' = hydraulic conductivity of confining layer

h0 = head at top of confining layer

b' = thickness of confining layer

Unconfined Aquifer (Boussinesq Equation)

+ =

where Sy = specific yield

h = saturated thickness of aquifer (it is now a proxy for both hydraulic head and thickness of aquifer)

This is a nonlinear equation

If drawdown is small compared with saturated thickness then h can be replaced with average thickness

+ =

Refraction of Flow Lines

Q1 = K1a and Q2 = K2c

Q1 = Q2 => K1a = K2c

h1 = h2 => K1 = K2

From geometry of right triangles

a = b cos σ1 and dl1 = b sin σ1

c = b cos σ2 and dl2 = b sin σ2

K1 = K2 => =

Steady Flow in a Confined Aquifer

q' = -Kb

h = h1 - x

Steady Flow in an Unconfined Aquifer

Dupuit Assumptions

1) Hydraulic gradient is equal to the slope of the water table;

2) for small water table gradients, the streamlines are horizontal and the equipotential lines are vertical

From Darcy's law,

q' = -Kh

where h is saturated thickness of the aquifer. At x=0, h = h1

and at x=L, h = h2

q' dx = -K hdh

Ingretrating

q'x|= -K | => q'L = -K( - )

Rearranging yields Dupuit Equation:

q' =

Consider a small prism of the unconfined aquifer

From Darcy's Law total flow in the x-direction through the left face of the prism is

q'x dy = -K ( h )x dy

Discharge through the right face q'x+dx is

q'x+dx dy = -K ( h )x+dx dy

Note that ( h ) has different values at each face.

The change in flow rate in the x-direction between the two faces is

(q'x+dx - q'x )dy = -K dxdy

Similarly, change in flow rate in the y-direction is

(q'y+dy - q'y )dx = -K dydx

For steady flow, any change in flow through the prism must be equal to a gain or loss of water across the water table, w (Conservation of Mass)

- K dxdy - K dydx = wdxdy

Note no significant flow in z-direction. Or simplifying

- K ( + ) = 2w

If w = 0, then equation reduces to a form of Laplace's equation:

+ = 0

If flow is in only one direction and we align x-axis parallel to the flow, then

= -

Integrating twice gives

h2 = - + c1x + c2

Boundary conditions are: x = 0, h = h1 and x = L, h = h2, which requires

h= c2

h= - + c1L + h => c1 = +

Thus,

h2 = - + x + x + h=> h2 = h+ x + x

or

h =

For w = 0

h =

It also follows from Darcy's Law, q'x = -Kh(dh/dx), that

q'x = - w( - x)

If the water table is subject to infiltration, there may be a water table divide where q'x = 0 at x = d

0 = - w( - d) => d = -

Elevation of the water table divide is

hmax =

Example 1

Given: K = 0.002 cm/s, ne = 0.27, x=0, h1 = 10 m, and x=175 m, h2 = 7.5 m

Determine: q', vx, and h at 87.5 m

Example 2

Given: K = 1.2 ft/day, w = 0.5 ft/y or 0.0014 ft/day,

x=0, h1 = 31 ft, and x=1500 m, h2 = 27 ft

Determine: location of water divide and maximum water table elevation,

daily discharge per 1000 ft at x = 0

daily discharge per 1000 ft at x = 1500 ft

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