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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