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K VALUES VARY WITHIN SPACE AND WITH DIRECTION HETEROGENEITY - describes spatial variation ANISOTROPY - describes directional variation HOMOGENEOUS - uniform throughout (K independent of position) ISOTROPIC - properties do not vary with direction LAYERED HETEROGENEITY Individual layers are homogeneous K1 K2 K3 1 DISCONTINUOUS HETEROGENEITY e.g. across a fault K1 K3 K2 K4 K3 K5 K6 TRENDING HETEROGENEITY variation in sedimentation patterns K1 K2 K3 2 RANDOM HETEROGENEITY small scale variation with a structure that isn't easily tied to geologic process, although we know it is its basis we can describe with geostatistics (spatial statistics) AVERAGING OPTIONS Weighted Arithmetic ∑K d ∑d i i i Geometric N ⎛1 ⎞ ⎜ (logK1 +logK2 +...+logKN ) ⎟ ⎝N ⎠ K1K2 ...KN or 10 ⎛1 ⎞ ⎜ log(K1K2 ...KN ) ⎟ ⎝N ⎠ or 10 Above assumes each K represents an equal volume, each could be weighted by the volume represented Weighted Harmonic ∑d d ∑K i i i 3 Random distribution of K * * * * * * * * * * * * 6 1x10-3cm/s 4 1x10-5cm/s 2 1x10-6cm/s * sample locations assumed to be representative of the proportions Realize you would only know the values shown * * * * * * * * * * * * You might have clues based on lithology * * * * * * * * * * * * 4 ANISOTROPY CONSIDER K IN THE PRINCIPLE DIRECTIONS X, Y, Z ISOTROPIC - KX = KY = KZ TRANSVERSELY ISOTROPIC - KX = KY not = KZ CONSIDER THE RELATIONS OF K IN A VERTICAL CROSS SECTION HOMOGENEOUS ISOTROPIC HOMOGENEOUS ANISOTROPIC HETEROGENEOUS ANISOTROPIC HETEROGENEOUS ISOTROPIC @ each point z THERE IS A RELATION BETWEEN LAYERED HETEROGENEITY & ANISOTROPY an equivalent K can be calculated to simplify complex systems thus making it possible to apply Darcy's Law dT d1 K1 Kx1 =Kz1 d2 K2 Kx2 =Kz2 d3 K3 Kx3 =Kz3 d4 Kx4 =Kz4 K4 x 5 Consider Flow Perpendicular to Layering: Say that K4= K1 < K3 << K2 V = Q/A Qin = Qout therefore Vin = Vout Δhi = head drop across layer i Δh = Δh1 + Δh2 + Δh3 + Δh4=htop-hbot htop Δh1 K1 Δh2 K2 d3 Δh3 K3 d4 Δh4 d1 d2 dT Δh Kx1 =Kz1 Δh=htop-hbot Kx2 =Kz2 Kx3 =Kz3 Kx4 =Kz4 K4 hbot By Darcy’s Law: Rearrange: V= K eq Δh K 1 Δ h 1 K 2 Δh 2 K Δh = = ... n n = d1 d2 dn dT Expand head difference term: Substitute Darcy’s Law for head differences: Vd T Vd T Vd T = == Vd1 Vd 2 Vd n Δh Δh1 + Δh 2 + ... + Δh n + + ... + K1 K2 Kn dT = Equivalent K for flow di perpendicular to layers ∑K i K eq = K eq d1 d2 dT Δh Δh1 Δh2 d3 Δh3 d4 Δh4 K1 K2 K3 K4 6 Consider Flow Parallel to Layering: Say for example K4= K1 < K3 << K2 V = Q/A Qin = Qout therefore Vin = Vout dT d1 K1 d2 K2 d3 K3 d4 hright hleft Δh = hleft-hright K4 By Darcy’s Law: Δh d 2 (1) L Δh Q4 = K 4 d 4 (1) L Δh d1 (1) L Δh Q3 = K 3 d 3 (1) L Q1 = K 1 Q2 = K 2 1 for unit width into the paper Total flow is their sum: dT Δh Δh Δh Δh + K 2d 2 + K 3d 3 + K 4d 4 L L L L d1 K1 d2 K2 d3 K3 d4 hright hleft Q T = K 1d 1 K4 7 V= Q Q = A d T (1) 1 for unit width into the paper Substitute Eqtn for Q from previous slide Simplify with a summation Δh Δh Δh Δh + K 2d2 + K 3d 3 + K 4d4 L L L L == V= dT We can divide V by K1d1 hleft K eq V = i dT V = Δh L i i )Δh dT L Cancel the gradients: ∑ = n i =1 K id i dT d1 K1 d2 K2 d3 K3 d4 hright gradient to get K: ∑ (K d K4 Calculate Flow and Heads between boundaries L1 H1 H3 D K1 K2 H4 L2 H2 H1 = 20cm H2 = 10cm K1 = 1cm/sec K2 = 0.2cm/sec L1 = 30cm L2 = 30cm D = 2cm Q @ H2 = ?? H3 = ??, H4 = ?? 8 Calculate Flow and Heads between boundaries L1 H1 K1 H3 D L2 K2 H2 H1 = 20cm H2 = 10cm K1 = 1cm/sec K2 = 0.2cm/sec L1 = 30cm L2 = 30cm D = 2cm Q @ H2 = ?? H3 = ?? Calculate Equivalent K's of the "Ant Farms" 9 Looking Back Recall the options and when each is appropriate AVERAGING OPTIONS Weighted Arithmetic ∑K d ∑d i i i Geometric N ⎛1 ⎞ ⎜ (logK1 +logK2 +...+logKN ) ⎟ ⎝N ⎠ K1K2 ...KN or 10 ⎛1 ⎞ ⎜ log(K1K2 ...KN ) ⎟ ⎝N ⎠ or 10 Above assumes each K represents an equal volume, each could be weighted by the volume represented e.g. each K (or for the middle option log(K)) would be multiplied by di/dmin & then N= sum of di/dmin Weighted Harmonic ∑d d ∑K i i i ⎛h −h ⎞ VDarcy = − K ⎜ 2 1 ⎟ ⎝ l ⎠ K is equivalent K over the distance l l is the flow path distance between h2 and h1 gradient is in the direction of flow when using velocity to calculate volumetric discharge, area is perpendicular to the gradient 10 Consider a Diagonal Gradient Given a Keqx and a Keqz H top left = htl Δhx = htl -htr = hbl -hbr iz = Δhz /Lz 18 10 ix = Δhx /Lx Vx = Keqx ix Vz = Keqz iz Δhz = htl -hbl = htr -hbr 20 Lz H top right = htr Lx H bot left = htl- Δhz Resultant V H bot right =htr - Δhz 8 Calculate a resultant velocity at the center now the gradient is different at top and bottom and from side to side Kx=1x10-5 m/s Ky=1x10-6m/s 42m 32m 26m 26 21m 29 33m 29m 33 32 11 Consider a Gradient Askew to Layering: The overall velocity is actually composed of individual velocity vectors in each layer H top left = htl Δhx = htl -htr = hbl -hbr H top right = htr Lx dT d1 ix = Δhx /Lx K1 iz1 = Δhz1 /dz1 Kx1 =Kz1 d2 ix = Δhx /Lx K2 iz2 = Δhz2 /dz2 Kx2 =Kz2 d3 ix = Δhx /Lx K3 iz3 = Δhz3 /dz3 Kx3 =Kz3 d4 ix = Δhx /Lx iz4 = Δhz4 /dz4 Kx4 =Kz4 K4 H bot left = htl- Δhz H bot right =htr - Δhz dT Δh di ∑K Δh K 1 Δh 1 K 2 Δh 2 K n Δh n K eq Δh i = = V= = = ... = d dT d1 d2 dn dT ∑ Ki i K i Δh i Δh so = d di ∑ Ki i To calculate Δhzi: Δh i = Δhd i d Ki ∑ i Ki d1 d2 dT Δh Δh1 Δh2 d3 Δh3 d4 Δh4 K1 K2 K3 K4 12 Consider a Gradient Askew to Layering: The overall velocity is actually composed of individual velocity vectors in each layer K4= K1 < K3 << K2 H top left = htl H top right = htr Lx d2 d3 d4 K1 Kx1 =Kz1 K2 Kx2 =Kz2 Kx3 =Kz3 K3 Vx = Keqx ix Vz = Keqz iz dT ix = Δhx /Lx iz = Δhz /Lz d1 Kx4 =Kz4 K4 Resultant V H bot left = htl- Δhz H bot right =htr - Δhz Refraction at Layer Interfaces K 1 tan θ1 = K 2 tan θ 2 θ1 K4= K1 < K3 << K2 K1 θ2 K1 K2 K2 K2 K3 K3 θ2 θ3 K4 13