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Mechanical Energy Kinetic Energy 1 2 Ek = 2 Mv Where Ek Is Energy (Kg-M2/S2) V Is Velocity (M/S) Gravitational Potential Energy

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Mechanical Energy Kinetic Energy 1 2 Ek = 2 Mv Where Ek Is Energy (Kg-M2/S2) V Is Velocity (M/S) Gravitational Potential Energy Mechanical Energy Kinetic Energy 1 2 Ek = 2 mv 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) 1 2 Etv = 2 r v + r gz + P Total Energy (per unit mass) or Bernoulli equation 1 P 2 Etm = 2 v + gz + r For a incompressible, frictionless fluid v2 P 2g + z + r g = 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 P h = z + r g Head in water of variable density hf = (r p/r f)hp where hp = pressure head hf = freshwater head Force Potential P F = gh = gz + r Darcy's Law dh KA dF Q = -KA dl = - g dl Applicability of Darcy's Law (Laminar Flow) r vd R = m < 1 where v = discharge velocity (m/s) d = diameter of passageway (m) R = Reynold's number Specific Discharge (Darcy Flux) and Average Linear Velocity Q dh v = A = -K dl Q K dh v = = - x neA ne dl 3 Equations of Ground-Water Flow Confined Aquifers Flux of water in and out of a Control Volume ¶ inMx = r wqxdydz; outMx = r wqxdydz + ¶x (r wqx) dx dydz; ¶ netMx = - ¶x (r wqx) dx dydz ¶ netMy = - ¶y (r wqy) dy dxdz ¶ netMz = - ¶z (r wqz) dz dxdy ¶ ¶ ¶ netMflux = - ( ¶x (r wqx) + ¶y (r wqy) + ¶z (r wqz) ) dxdydz From Darcy's Law ¶h ¶h ¶h qx = - K ¶x qy = - K ¶y qz = - K ¶z and assuming that w and K do not vary spatially ¶2h ¶2h ¶2h M = K + + r dxdydz net flux ( ¶x2 ¶y2 ¶z2 ) w Note hydraulic head and hydraulic conductivity are easier to measure than discharge. Mass of Water in the Control Volume ¶M ¶ M = r w n dxdydz => ¶t = ¶t (r w n dxdydz) if dxdy is constant 4 ¶M ¶t = [ r w n Error!,¶t) + r w dz Error!+ n dz Error!] dxdy compressibility of water dr w bdP = r w compressibility of control volume (vertical only) d(dz) a dP = dz Volume of Solids is constant dVs = 0 = d[(1 - n)dxdydz] => dzdn = (1 - n)d(dz) (1 - n)d(dz) dn = dz Fluid Pressure P = P0 + r wgh => dP = r wgdh dr w = r wb(r wgdh) d(dz) = dza (r wgdh) => dn = (1 - n)ar wgdh Substituting into equation for change of mass with time ¶M ¶h ¶t = (ar wg + nbr wg)r w dxdydz ¶t Conservation of Mass ¶M ¶t = netMflux ¶2h ¶2h ¶2h ¶h K + + = (ar g + nbr g) ( ¶x2 ¶y2 ¶z2 ) w w ¶t 5 In Two-Dimensions, this reduces to ¶2h ¶2h S ¶h + = ¶x2 ¶y2 T ¶t where Storativity = b(ar wg + nbr wg) Transmissivity = bK For Steady Flow, ¶h/¶t = 0 (Laplace equation) ¶2h ¶2h ¶2h + + = 0 ¶x2 ¶y2 ¶z2 For leakage into an aquifer from a confining layer ¶2h ¶2h e S ¶h + + = ¶x2 ¶y2 T T ¶t where e = leakage rate and from Darcy's Law (h0 - h) e = K' b' where K' = hydraulic conductivity of confining layer h0 = head at top of confining layer b' = thickness of confining layer Unconfined Aquifer (Boussinesq Equation) ¶ ¶h ¶ ¶h Sy ¶h ¶x ( ¶x ) + ¶y ( ¶y ) = K ¶t 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 6 If drawdown is small compared with saturated thickness then h can be replaced with average thickness ¶2h ¶2h Sy ¶h + = ¶x2 ¶y2 Kb ¶t Refraction of Flow Lines dh1 dh2 Q = K a and Q = K c 1 1 dl1 2 2 dl2 dh1 dh2 Q = Q => K a = K c 1 2 1 dl1 2 dl2 a c h = h => K = K 1 2 1 dl1 2 dl2 From geometry of right triangles a = b cos s 1 and dl1 = b sin s 1 c = b cos s 2 and dl2 = b sin s 2 cos s 1 cos s 2 K1 tan s 1 K = K => = 1 sin s 1 2 sin s 2 K2 tan s 2 Steady Flow in a Confined Aquifer dh q' = -Kb dl q' h = h1 - Kb x 7 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, dh q' = -Kh dx where h is saturated thickness of the aquifer. At x=0, h = h1 and at x=L, h = h2 L h2 ó ó õ q' dx = -K õ hdh 0 h1 Ingretrating 2 2 L h2 h2 h2 h1 q'x = -K 2 => q'L = -K( 2 - 2 ) |0 |h1 Rearranging yields Dupuit Equation: 2 2 K h1 - h2 q' = 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 ¶h q'x dy = -K ( h ¶x )x dy 8 Discharge through the right face q'x+dx is ¶h q'x+dx dy = -K ( h ¶x )x+dx dy h Note that ( h x ) has different values at each face. The change in flow rate in the x-direction between the two faces is ¶ ¶h (q'x+dx - q'x )dy = -K ¶x ( h ¶x ) dxdy Similarly, change in flow rate in the y-direction is ¶ ¶h (q'y+dy - q'y )dx = -K ¶y ( 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) ¶ ¶h ¶ ¶h - K ¶x ( h ¶x ) dxdy - K ¶y ( h ¶y ) dydx = wdxdy Note no significant flow in z-direction. Or simplifying ¶2h2 ¶2h2 - K + = 2w ( ¶x2 ¶y2 ) If w = 0, then equation reduces to a form of Laplace's equation: ¶2h2 ¶2h2 + = 0 ¶x2 ¶y2 If flow is in only one direction and we align x-axis parallel to the flow, then d2(h2) 2w = - dx2 K Integrating twice gives 9 wx2 2 h = - K + c1x + c2 Boundary conditions are: x = 0, h = h1 and x = L, h = h2, which requires 2 h1 = c2 2 h2 - h2 2 wL 2 2 1 wL h2 = - K + c1L + h1 => c1 = L + K Thus, 2 2 2 2 wx2 h - h wL h - h w 2 2 1 2 2 2 2 1 h = - K + L x + K x + h1 => h = h1 + L x + K(L - x) x or h2 - h2 2 2 1 w h = h1 + L x + K(L - x)x For w = 0 h2 - h2 2 2 1 h = h1 + L x It also follows from Darcy's Law, q'x = -Kh(dh/dx), that 2 2 K(h1 - h2) L q'x = 2L - w( 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 2 2 2 2 K(h1 - h2) L L K (h1 - h2) 0 = 2L - w( 2 - d) => d = 2 - w 2L Elevation of the water table divide is h2 - h2 2 2 1 w hmax = h1 + L d + K(L - d)d 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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