# **Example 5.4.1a: Theis well pumping rate** time = 750 min Vol = 13500 ft^3 Q = Vol/time; ft^3/day #Fetter: 25,900 ft^3/day # **Example 5.4.1b: Theis well pumping rate** Q = 125 gal/min Q;ft^3/day #Fetter: 24,100 ft^3/day # **Example 5.4.1c: Theis well drawdown** Ko = 14.9 m/day S = 0.0051 bo = 20.1 m Q = 2725 m^3/day r = 7 m t = 1 day To = Ko*bo u = r^2*S/(4*To*t) #Fetter: 0.00021 # W(u) is more commonly known as Exponential Integral E(x) Wu = ExpInt(u) #For comparison, from Fetter table, for u = 0.0002, W(u) = 7.94 drawdown = Q*Wu/(4*3.14*To) #Fetter: 5.7 m # **Example 5.4.2.2: Leaky aquifer** bo = 5.2 m Ko = 0.73 m/day Stor = 0.0035 Tran = 3.8 m^2/day bo_p = 1.1 m Ko_p = 5.5e-5 m/day Stor_p = 0.00061 bo_dp = 25 m Ko_dp = 35 m/day rw = 0.15 m Qwell = 28 m^3/day d1 = 1.5 m t = 1 day #--Part A: Test if bo_dp * Ko_dp > 100 * bo * Ko bo_dp*Ko_dp; m^2/day #Fetter 875 m^2/day 100*bo*Ko; m^2/day #Fetter 380 m^2/day #--Part B: Test if t = 1 day > 0.036 * bo_p * Stor_p / Ko_p 0.036 * bo_p * Stor_p / Ko_p; day #Fetter 0.44 day #--Part C.1: Test if t = 1 day > (30 * rw^2 *Stor/Tran) * (1-(10*rw/bo)^2) (30 * rw^2 * Stor / Tran) * (1 - (10*rw/bo)^2); day #Fetter 5.7e-4 day #--Part C.2: Test if rw/(Tran * bo_p /Ko_p)^(1/2) < 0.1 rw/(Tran * bo_p/Ko_p)^(1/2) #Fetter 5.5e-4 u = (rw^2 * Stor) / (4 * Tran * t) #Fetter incorrectly gives 5.17e-4 r_over_B = rw / (Tran *bo_p/Ko_p)^(1/2) #Fetter incorrectly gives 5.45e-3 #--Part D: From Appendix 3 W(u, r_over_B) = 7.0 drawdown = (Qwell/(4*3.14*Tran)) * 7.0 #Fetter 4.1 m # **Example 5.4.2.3: Leaky aquitard with elastic storage** b = 4.3 m Ko = 1.1 m/day S = 0.00053 Tr = 4.7 m^2/day b_p = 7.2 m Ko_p = 5.5e-6 m/day S_p = 0.00012 b_dp = 17 m Ko_dp = 87 m/day S_dp = 0.055 Qw = 15 m^3/day t = 1.76 days do = 22 m #--Part A: Test if b_dp * Ko_dp > 100 * b * Ko b_dp * Ko_dp; m^2/day #Fetter 1497 m^2/day 100 * b * Ko; m^2/day #Fetter 473 m^2/day #--Part B: Test if t > 0.036 * b_p * S_p / Ko_p t; day 0.036 * b_p * S_p / Ko_p; days #Fetter 5.66 days #--Part C: Test if t < b_p * S_p / (10* Ko_p) t; day b_p * S_p / (10* Ko_p); days #Fetter 15.7 days #--Part D: Find H(u, beta) u = do^2*S/(4*Tr*t) #Fetter 7.75e-3 B = (Tr*b_p/Ko_p)^(1/2) #Fetter 2.48e3 m - Fetter omits units beta = (do/(4*B))*(S_p/S)^(1/2) #Fetter 1.06e-3 #--Part E: From Appendix 4 H(u, beta) = 4.3 drawdown = (Qw/(4*3.14*Tr))*4.3 #Fetter 1.11 m # **Example 5.5.2.1: Theim Solution** Qw = 220 gal/min t = 270 min r1 = 26 ft h1 = 29.34 ft r2 = 73 ft h2 = 32.56 ft Tr = (Qw/(2*3.14*(h2 - h1))) * ln(r2/r1); ft^2/day #Fetter 2170 ft^2/day # **Example 5.5.3.1: Theis Solution** #--From match point Wu = 1.0 One_over_u = 1.0 deltaH = 2.4 ft t = 4.1 min Qw = 220 gal/min bo = 48 ft robs = 824 ft Tr = (Qw/(4*3.14*deltaH))*Wu; ft^2/day #Fetter 1400 ft^2/day Ko = Tr/bo; ft/day #Fetter 29 ft/day S = 4*Tr*t/(One_over_u * robs^2) #Fetter 2.4e-5 # **Example 5.5.3.2: Cooper-Jacob Solution** to = 5.2 min deltaH = 5.5 ft Qw = 42400 ft^3/day robs = 824 ft Tr = 2.3*Qw/(4*3.14*deltaH); ft^2/day #Fetter 1400 ft^2/day S = 2.25*Tr*to/robs^2 #Fetter 1.7e-5 # **Example 5.5.3.3: Jacob Straight Line Solution** Qw = 77000 ft^3/day deltaH = 8.8 ft ro = 460 ft t = 0.14 day Tr = 2.3*Qw/(2*3.14*deltaH); ft^2/day #Fetter 3200 ft^2/day S = 2.25*Tr*t/ro^2 #Fetter 0.0048 # **Example 5.5.3.4: Walton Graphical Solution** # Given Parameters b = 14 ft robs = 96 ft Qw = 25 gal/min # From match point Wu_rB = 1.0 u = 1.0/1.0 deltaH = 1.9 ft t = 33 min rB = 0.22 # Solution Tr = (Qw/(4*3.14*deltaH))*Wu_rB; ft^2/day #Fetter 200 ft^2/day S = 4 * Tr * u * t / robs^2 #Fetter incorrectly gives 0.00020 Kp = Tr*b*rB^2/robs^2; ft/day #Fetter 0.015 ft/day # **Example 5.5.3.5: Hantush Inflection-Point Solution** #Given Qw = 4800 ft^3/day b = 14 ft robs = 96 ft deltaHmax = 6.42 ft deltaH = 3.21 ft t = 32 min m_i = 3.1 ft f_rB = 2.3*deltaH/m_i # From table Ko_rB = 2.055 rB = 0.147 #Solution B = robs/rB; ft #Fetter 653 ft Tr = (Qw/(2*3.14*deltaHmax))*Ko_rB; ft^2/day #Fetter 240 ft^2/day S = 4*t*Tr/(2*robs*B) #Fetter 0.00017 Kp = Tr*b/B^2; ft/day #Fetter 0.0079 ft/day # **Example 5.5.5: Unconfined Aquifer** #Given Qw = 1000 gal/min b = 100 ft robs = 200 ft capR = 0.1 WuA_r = 0.1 uA = 1/1.0 deltaH = 0.041 ft t = 0.9 min #Solution Tr1 = (Qw/(4*3.14*deltaH))*WuA_r; ft^2/day #Fetter 3.7e4 ft^2/day S = 4*Tr1*uA*t/robs^2 #Fetter 0.0023 WuB_r = 0.1 uB = 1/10 deltH = 0.043 ft t = 128 min Tr2 = (Qw/(4*3.14*deltH))*WuB_r; ft^2/day #Fetter 3.5e4 ft^2/day Sy = 4*Tr2*uB*t/robs^2 #Fetter 0.031 Trave = (Tr1 + Tr2)/2; ft^2/day #Fetter 3.6e4 ft^2/day Kh = Trave/b; ft/day #Fetter 360 ft/day Kv = capR * b^2 * Kh/robs^2; ft/day #Fetter 9.0 ft/day # **Example 5.6.2: Confined Slug Test** #Given rc = 7.6 cm rs = 5.1 cm b = 5 m ho = 0.42 m # From plot Tt_r2 = 1.0 t = 13 s #Solution Tr = Tt_r2*rc^2/t; cm^2/s #Fetter 4.4 cm^2/s Ko = Tr/b; cm/s #Fetter 8.8e-3 cm/s mu = 1e-3 S = mu*rc^2/rs^2 #Fetter = 2.2e-3 # **Example 5.6.2.2: Hvorslev Slug Test** #Given rw = (2/2) in Le = 10 ft t37 = 1.8 s Ko = rw^2*ln(Le/rw)/(2*Le*t37); ft/day #Fetter 79 ft/day # **Example 5.6.3: Underdamped Response Slug Test** # The DimensionEngine calculator correctly reports that this problem # can not be solved as Eqn. 5.96 is dimensionally inconsistent rs= 0.5 ft S = 1.0 Le = 1.0 m a = 1.0 c = -a*ln(0.79*rs^2 * S * (grav()/Le)^(1/2)) # Homework Problems: TBD