Mechanical Energy Kinetic Energy E k = 1 2 mv2 where E k is energy (kg-m 2 /s 2 ) v is velocity (m/s) Gravitational Potential Energy E g = W = mgz where w is work (kg-m 2 /s 2 ) m is mass (kg) z is elevation above datum Pressure of surrounding fluid (potential energy per unit volume) Total Energy (per unit volume) E tv = 1 2 ρv2 + ρgz + P Total Energy (per unit mass) or Bernoulli equation E tm = 1 2 v2 + gz + P ρ For a incompressible, frictionless fluid v 2 2g + z + P ρg = 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
2 h = z + P ρg Head in water of variable density h f = (ρ p /ρ f )h p where h p = pressure head h f = freshwater head Force Potential Φ = gh = gz + P ρ Darcy's Law Q = -KA dh dl = - KA g dφ dl Applicability of Darcy's Law (Laminar Flow) R = ρvd µ < 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 = Q A = -K dh dl v x = Q n e A = - K dh n e dl
Equations of Ground-Water Flow Confined Aquifers 3 Flux of water in and out of a Control Volume inm x = ρ w q x dydz; outm x = ρ w q x dydz + x (ρ wq x ) dx dydz; netm x = - x (ρ wq x ) dx dydz netm y = - y (ρ wq y ) dy dxdz netm z = - z (ρ wq z ) dz dxdy netm flux = - ( x (ρ wq x ) + y (ρ wq y ) + z (ρ wq z ) ) dxdydz From Darcy's Law q x = - K h x q y = - K h y q z = - K h z and assuming that w and K do not vary spatially netm flux = K ( 2 h y 2 + 2 h z 2 ) ρ wdxdydz Note hydraulic head and hydraulic conductivity are easier to measure than discharge. Mass of Water in the Control Volume M = ρ w n dxdydz => Μ t = t (ρ w n dxdydz) if dxdy is constant
Μ t = [ ρ w n Error!, t) + ρ w dz Error!+ n dz Error!] dxdy 4 compressibility of water βdp = dρ w ρ w compressibility of control volume (vertical only) αdp = d(dz) dz Volume of Solids is constant dv s = 0 = d[(1 - n)dxdydz] => dzdn = (1 - n)d(dz) dn = (1 - n)d(dz) dz Fluid Pressure P = P 0 + ρ w gh => dp = ρ w gdh dρ w = ρ w β(ρ w gdh) d(dz) = dzα(ρ w gdh) => dn = (1 - n)αρ w gdh Substituting into equation for change of mass with time Μ t = (αρ wg + nβρ w g)ρ w dxdydz h t Conservation of Mass Μ t = netm flux K ( 2 h y 2 + 2 h ) z 2 = (αρ wg + nβρ w g) h t
5 In Two-Dimensions, this reduces to 2 h y 2 = S h T t where Storativity = b(αρ w g + nβρ w g) Transmissivity = bk For Steady Flow, h/ t = 0 (Laplace equation) 2 h y 2 + 2 h z 2 = 0 For leakage into an aquifer from a confining layer 2 h y 2 + e T = S h T t where e = leakage rate and from Darcy's Law e = K' (h 0 - h) b' where K' = hydraulic conductivity of confining layer h 0 = head at top of confining layer b' = thickness of confining layer Unconfined Aquifer (Boussinesq Equation) ( h ) x x + ( h ) y y = S y h K t where S y = 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
6 If drawdown is small compared with saturated thickness then h can be replaced with average thickness 2 h y 2 = S y h Kb t Refraction of Flow Lines Q 1 = K 1 a dh 1 dl 1 and Q 2 = K 2 c dh 2 dl 2 Q 1 = Q 2 => K 1 a dh 1 dl 1 = K 2 c dh 2 dl 2 h 1 = h 2 => K 1 a dl 1 = K 2 c dl 2 From geometry of right triangles a = b cos σ 1 and dl 1 = b sin σ 1 c = b cos σ 2 and dl 2 = b sin σ 2 K 1 cos σ 1 sin σ 1 = K 2 cos σ 2 sin σ 2 => K 1 K 2 = tan σ 1 tan σ 2 Steady Flow in a Confined Aquifer q' = -Kb dh dl h = h1 - q' Kb x
Steady Flow in an Unconfined Aquifer Dupuit Assumptions 7 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 dh dx where h is saturated thickness of the aquifer. At x=0, h = h 1 and at x=l, h = h 2 L 0 h 2 q' dx = -K h 1 hdh Ingretrating q'x L 0 h2 = -K 2 h 2 => q'l = -K( h 2 2 h 2 - h2 1 2 ) 1 Rearranging yields Dupuit Equation: q' = K 2 ( h 2 1 - h2 2 L ) 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 h x ) x dy
8 Discharge through the right face q' x+dx is q' x+dx dy = -K ( h h x ) x+dx dy Note that ( h h x ) 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 x ( h h x ) dxdy Similarly, change in flow rate in the y-direction is (q' y+dy - q' y )dx = -K y ( h h y ) 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 x ( h h x ) dxdy - K y ( h h y ) dydx = wdxdy Note no significant flow in z-direction. Or simplifying - K ( 2 h 2 2 y 2 ) = 2w If w = 0, then equation reduces to a form of Laplace's equation: 2 h 2 2 y 2 = 0 If flow is in only one direction and we align x-axis parallel to the flow, then d 2( h 2 ) dx 2 = - 2w K Integrating twice gives
9 h 2 = - wx2 K + c 1x + c 2 Boundary conditions are: x = 0, h = h 1 and x = L, h = h 2, which requires h 2 1 = c 2 h 2 2 = - wl2 K + c 1L + h 2 1 => c 1 = h2 2 - h2 1 L + wl K Thus, h 2 = - wx2 K + h2 2 - h2 1 L x + wl K x + h2 1 => h2 = h 2 1 + h2 2 - h2 1 L x + w K (L - x) x or h = h 2 1 + h2 2 - h2 1 L x + w K (L - x)x For w = 0 h = h 2 1 + h2 2 - h2 1 L x It also follows from Darcy's Law, q' x = -Kh(dh/dx), that q' x = K(h 2 1 - h2 2 ) 2L - w( L 2 - x) If the water table is subject to infiltration, there may be a water table divide where q' x = 0 at x = d
10 0 = K(h 2 1 - h2 2 ) 2L - w( L 2 - d) => d = L 2 - K (h2 1 - h2 2 ) w 2L Elevation of the water table divide is h max = h 2 1 + h2 2 - h2 1 L d + w K (L - d)d Example 1 Given: K = 0.002 cm/s, n e = 0.27, x=0, h 1 = 10 m, and x=175 m, h 2 = 7.5 m Determine: q', v x, 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, h 1 = 31 ft, and x=1500 m, h 2 = 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