Find: c(x,t) [mg/L] of chloride continuous source of chloride contamination 10 25 50 100 i=0.002 Cl mg L co=725 x Find the concentration of chloride, in milligrams per liter. [pause] In this problem, --- x=15 [m] m s m2 s η=0.23 K=3*10-5 D=1*10-9 t=1 [yr] αx=0.83*(log(x))2.414 [m]
Find: c(x,t) [mg/L] of chloride continuous source of chloride contamination 10 25 50 100 i=0.002 Cl mg L co=725 x a continuous source of chloride contamination, with the initial concentration of 725 milligrams per liter, --- x=15 [m] m s m2 s η=0.23 K=3*10-5 D=1*10-9 t=1 [yr] αx=0.83*(log(x))2.414 [m]
Find: c(x,t) [mg/L] of chloride continuous source of chloride contamination 10 25 50 100 i=0.002 Cl mg L co=725 x travels in a one-dimensional flow field, in the x direction. [pause] The problem asks to find the concentration, --- x=15 [m] m s m2 s η=0.23 K=3*10-5 D=1*10-9 t=1 [yr] αx=0.83*(log(x))2.414 [m]
Find: c(x,t) [mg/L] of chloride continuous source of chloride contamination 10 25 50 100 i=0.002 Cl mg L co=725 x c sub x, t, at a distance x, 15 meters away from the source, and at a time t, 1 year after contamination begins. x=15 [m] m s m2 s η=0.23 K=3*10-5 D=1*10-9 t=1 [yr] αx=0.83*(log(x))2.414 [m]
Find: c(x,t) [mg/L] of chloride continuous source of chloride contamination 10 25 50 100 i=0.002 Cl mg L co=725 x The problem also provides other parameters shown here. For a continuous source of contamination, --- x=15 [m] m s m2 s η=0.23 K=3*10-5 D=1*10-9 t=1 [yr] αx=0.83*(log(x))2.414 [m]
Find: c(x,t) [mg/L] of chloride x=15 [m] co=725 mg L Cl i=0.002 m s the concentration at a downgrade distance, x, --- K=3*10-5 t=1 [yr] m2 s η=0.23 D=1*10-9 x αx=0.83*(log(x))2.414 [m]
Find: c(x,t) [mg/L] of chloride co x-vx*t c(x,t)= erfc * 2 2 Dx*t x+vx*t vx*x +exp * erfc Dx 2 Dx*t x=15 [m] co=725 mg L Cl i=0.002 m s after a time duration, t, equals one half the initial concentration times times an expression which involves the --- K=3*10-5 t=1 [yr] m2 s η=0.23 D=1*10-9 x αx=0.83*(log(x))2.414 [m]
Find: c(x,t) [mg/L] of chloride complimentary co x-vx*t error function c(x,t)= erfc * 2 2 Dx*t x+vx*t vx*x +exp * erfc Dx 2 Dx*t x=15 [m] co=725 mg L Cl i=0.002 m s complimentary error function, K=3*10-5 t=1 [yr] m2 s η=0.23 D=1*10-9 x αx=0.83*(log(x))2.414 [m]
Find: c(x,t) [mg/L] of chloride complimentary co x-vx*t error function c(x,t)= erfc * 2 2 Dx*t distance x+vx*t vx*x +exp * erfc Dx 2 Dx*t x=15 [m] co=725 mg L Cl i=0.002 m s the distance, --- K=3*10-5 t=1 [yr] m2 s η=0.23 D=1*10-9 x αx=0.83*(log(x))2.414 [m]
Find: c(x,t) [mg/L] of chloride complimentary co x-vx*t error function c(x,t)= erfc * 2 2 Dx*t distance x+vx*t vx*x +exp velocity * erfc Dx 2 Dx*t x=15 [m] co=725 mg L Cl i=0.002 m s the velocity, --- K=3*10-5 t=1 [yr] m2 s η=0.23 D=1*10-9 x αx=0.83*(log(x))2.414 [m]
Find: c(x,t) [mg/L] of chloride complimentary co x-vx*t error function c(x,t)= erfc * 2 2 Dx*t distance x+vx*t vx*x +exp velocity * erfc Dx 2 Dx*t hydrodynamic dispersion x=15 [m] co=725 mg L Cl i=0.002 coefficient m s the hydrodynamic dispersion coefficient, --- K=3*10-5 t=1 [yr] m2 s η=0.23 D=1*10-9 x αx=0.83*(log(x))2.414 [m]
Find: c(x,t) [mg/L] of chloride complimentary co x-vx*t error function c(x,t)= erfc * 2 time 2 Dx*t distance x+vx*t vx*x +exp velocity * erfc Dx 2 Dx*t hydrodynamic dispersion x=15 [m] co=725 mg L Cl i=0.002 coefficient m s and the time. [pause] K=3*10-5 t=1 [yr] m2 s η=0.23 D=1*10-9 x αx=0.83*(log(x))2.414 [m]
Find: c(x,t) [mg/L] of chloride co x-vx*t c(x,t)= erfc * 2 2 Dx*t x+vx*t vx*x +exp * erfc Dx 2 Dx*t x=15 [m] co=725 mg L Cl i=0.002 m s We already know the initial concentration, c sub not, the distance, x, and the time, t, --- K=3*10-5 t=1 [yr] m2 s η=0.23 D=1*10-9 x αx=0.83*(log(x))2.414 [m]
Find: c(x,t) [mg/L] of chloride co x-vx*t c(x,t)= erfc * 2 2 Dx*t velocity x+vx*t vx*x +exp * erfc Dx 2 Dx*t hydrodynamic dispersion x=15 [m] co=725 mg L Cl i=0.002 coefficient m s but we still need to determine the velocity and the hydrodynamic dispersion coefficient. K=3*10-5 t=1 [yr] m2 s η=0.23 D=1*10-9 x αx=0.83*(log(x))2.414 [m]
Find: c(x,t) [mg/L] of chloride co x-vx*t c(x,t)= erfc * 2 2 Dx*t velocity x+vx*t vx*x +exp * erfc Dx 2 Dx*t hydrodynamic dispersion x=15 [m] co=725 mg L Cl i=0.002 coefficient m s Let’s first solve for the velocity. [pause] K=3*10-5 t=1 [yr] m2 s η=0.23 D=1*10-9 x αx=0.83*(log(x))2.414 [m]
Find: c(x,t) [mg/L] of chloride velocity dh 1 v=k * * η dL x=15 [m] co=725 mg L Cl i=0.002 m s The actual velocity of water through a porous medium --- K=3*10-5 t=1 [yr] m2 s η=0.23 D=1*10-9 x αx=0.83*(log(x))2.414 [m]
Find: c(x,t) [mg/L] of chloride velocity dh 1 v=k * * η dL Darcy velocity x=15 [m] co=725 mg L Cl i=0.002 m s equals the Darcy velocity, --- K=3*10-5 t=1 [yr] m2 s η=0.23 D=1*10-9 x αx=0.83*(log(x))2.414 [m]
Find: c(x,t) [mg/L] of chloride velocity dh 1 v=k * * η dL porosity Darcy velocity x=15 [m] co=725 mg L Cl i=0.002 m s divided by the porosity of the soil. K=3*10-5 t=1 [yr] m2 s η=0.23 D=1*10-9 x αx=0.83*(log(x))2.414 [m]
Find: c(x,t) [mg/L] of chloride velocity dh 1 v=k * * η dL porosity Darcy velocity x=15 [m] co=725 mg L Cl i=0.002 m Plugging the the appropriate values, the velocity in the x direction equals --- K=3*10-5 t=1 [yr] s m2 s η=0.23 D=1*10-9 x αx=0.83*(log(x))2.414 [m]
Find: c(x,t) [mg/L] of chloride velocity dh 1 v=k * * η dL porosity Darcy velocity m vx=2.61*10-7 s x=15 [m] co=725 mg L Cl i=0.002 m 2.61 times 10 to the –7 meters per second. [pause] With the velocity solved, ---- K=3*10-5 t=1 [yr] s m2 s η=0.23 D=1*10-9 x αx=0.83*(log(x))2.414 [m]
Find: c(x,t) [mg/L] of chloride hydrodynamic dispersion coefficient m s vx=2.61*10-7 x=15 [m] co=725 mg L Cl i=0.002 m we’ll next determine the hydrodynamic dispersion coefficient, --- K=3*10-5 s t=1 [yr] m2 s η=0.23 D=1*10-9 x αx=0.83*(log(x))2.414 [m]
Find: c(x,t) [mg/L] of chloride hydrodynamic Dx = Dx + D dispersion coefficient m s vx=2.61*10-7 x=15 [m] co=725 mg L Cl i=0.002 m in the x direction, D sub x, which is the sum of the --- K=3*10-5 s t=1 [yr] m2 s η=0.23 D=1*10-9 x αx=0.83*(log(x))2.414 [m]
Find: c(x,t) [mg/L] of chloride diffusion hydrodynamic Dx = Dx + D dispersion coefficient coefficient mechanical dispersion coefficient m s vx=2.61*10-7 x=15 [m] co=725 mg L Cl i=0.002 m mechanical dispersion coefficient, and the diffusion coefficient. K=3*10-5 s t=1 [yr] m2 s η=0.23 D=1*10-9 x αx=0.83*(log(x))2.414 [m]
Find: c(x,t) [mg/L] of chloride diffusion hydrodynamic Dx = Dx + D dispersion coefficient coefficient mechanical dispersion coefficient m s vx=2.61*10-7 x=15 [m] co=725 mg L Cl i=0.002 m Since the problem statement provides the diffusion coefficient, K=3*10-5 s t=1 [yr] m2 s η=0.23 D=1*10-9 x αx=0.83*(log(x))2.414 [m]
Find: c(x,t) [mg/L] of chloride diffusion hydrodynamic Dx = Dx + D dispersion coefficient coefficient mechanical dispersion coefficient m s vx=2.61*10-7 x=15 [m] co=725 mg L Cl i=0.002 m we’ll solve for the the mechanical dispersion, which equals, --- K=3*10-5 s t=1 [yr] m2 s η=0.23 D=1*10-9 x αx=0.83*(log(x))2.414 [m]
Find: c(x,t) [mg/L] of chloride diffusion hydrodynamic Dx = Dx + D dispersion coefficient coefficient velocity Dx = αx * vx dynamic dispersivity m s vx=2.61*10-7 x=15 [m] co=725 mg L Cl i=0.002 m the dynamic dispersivity, times the velocity. K=3*10-5 s t=1 [yr] m2 s η=0.23 D=1*10-9 x αx=0.83*(log(x))2.414 [m]
Find: c(x,t) [mg/L] of chloride diffusion hydrodynamic Dx = Dx + D dispersion coefficient coefficient velocity Dx = αx * vx dynamic dispersivity m s vx=2.61*10-7 x=15 [m] co=725 mg L Cl i=0.002 m We already solved for the velocity, --- K=3*10-5 s t=1 [yr] m2 s η=0.23 D=1*10-9 x αx=0.83*(log(x))2.414 [m]
Find: c(x,t) [mg/L] of chloride diffusion hydrodynamic Dx = Dx + D dispersion coefficient coefficient velocity Dx = αx * vx dynamic dispersivity αx=0.83*(log(x))2.414 [m] m s vx=2.61*10-7 x=15 [m] co=725 mg L Cl i=0.002 m but we still need to figure out the dispersivity. When the distance of --- K=3*10-5 s t=1 [yr] m2 s η=0.23 D=1*10-9 x αx=0.83*(log(x))2.414 [m]
Find: c(x,t) [mg/L] of chloride diffusion hydrodynamic Dx = Dx + D dispersion coefficient coefficient velocity Dx = αx * vx dynamic dispersivity αx=0.83*(log(x))2.414 [m] m s vx=2.61*10-7 x=15 [m] co=725 mg L Cl i=0.002 m 15 meters is plugged into this equation, K=3*10-5 s t=1 [yr] m2 s η=0.23 D=1*10-9 x αx=0.83*(log(x))2.414 [m]
Find: c(x,t) [mg/L] of chloride diffusion hydrodynamic Dx = Dx + D dispersion coefficient coefficient velocity Dx = αx * vx dynamic dispersivity αx=0.83*(log(x))2.414 [m] αx=1.23 [m] m s vx=2.61*10-7 x=15 [m] co=725 mg L Cl i=0.002 m the dispersivity equals 1.23 meters. [pause] Multiplying this 1.23 meters by, --- K=3*10-5 s t=1 [yr] m2 s η=0.23 D=1*10-9 x αx=0.83*(log(x))2.414 [m]
Find: c(x,t) [mg/L] of chloride diffusion hydrodynamic Dx = Dx + D dispersion coefficient coefficient Dx = αx * vx αx=1.23 [m] m vx=2.61*10-7 x=15 [m] co=725 mg L Cl i=0.002 s m the velocity of 2.61 times 10 to the –7 meters per second, --- K=3*10-5 s t=1 [yr] m2 s η=0.23 D=1*10-9 x αx=0.83*(log(x))2.414 [m]
Find: c(x,t) [mg/L] of chloride diffusion hydrodynamic Dx = Dx + D dispersion coefficient coefficient Dx = αx * vx αx=1.23 [m] m2 s Dx=3.21*10-7 m vx=2.61*10-7 x=15 [m] co=725 mg L Cl i=0.002 s m the mechanical dispersion coefficient equals 3.21 times 10 to the –7, meters squared per second. K=3*10-5 s t=1 [yr] m2 s η=0.23 D=1*10-9 x αx=0.83*(log(x))2.414 [m]
Find: c(x,t) [mg/L] of chloride diffusion hydrodynamic Dx = Dx + D dispersion coefficient coefficient m2 s Dx=3.21*10-7 m vx=2.61*10-7 x=15 [m] co=725 mg L Cl i=0.002 s m With a little addition, we can solve for the hydrodynamic dispersion coefficient, which equals, --- K=3*10-5 s t=1 [yr] m2 η=0.23 D=1*10-9 s x αx=0.83*(log(x))2.414 [m]
Find: c(x,t) [mg/L] of chloride diffusion hydrodynamic Dx = Dx + D dispersion coefficient coefficient m2 s Dx=3.21*10-7 m2 s Dx = 3.22*10-7 m vx=2.61*10-7 x=15 [m] co=725 mg L Cl i=0.002 s m 3.22 times 10 to the –7, meters squared per second. [pause] Having solved for the --- K=3*10-5 s t=1 [yr] m2 s η=0.23 D=1*10-9 x αx=0.83*(log(x))2.414 [m]
Find: c(x,t) [mg/L] of chloride velocity hydrodynamic dispersion m vx=2.61*10-7 coefficient s m2 Dx = 3.22*10-7 x=15 [m] co=725 mg L Cl i=0.002 s m s velocity and dispersion coefficient, we can now plug these values into --- K=3*10-5 t=1 [yr] m2 s η=0.23 D=1*10-9 x αx=0.83*(log(x))2.414 [m]
Find: c(x,t) [mg/L] of chloride co x-vx*t c(x,t)= erfc * 2 2 Dx*t x+vx*t vx*x +exp * erfc m vx=2.61*10-7 Dx 2 Dx*t s m2 Dx = 3.22*10-7 x=15 [m] co=725 mg L Cl i=0.002 s m s the original equation for the concentration, as well as the variables for, --- K=3*10-5 t=1 [yr] m2 s η=0.23 D=1*10-9 x αx=0.83*(log(x))2.414 [m]
Find: c(x,t) [mg/L] of chloride co x-vx*t c(x,t)= erfc * 2 2 Dx*t x+vx*t vx*x +exp * erfc m vx=2.61*10-7 Dx 2 Dx*t s m2 Dx = 3.22*10-7 s i=0.002 m s the initial concentration, the distance, and the time. The equation simplifies to --- K=3*10-5 t=1 [yr] Cl mg m2 s η=0.23 co=725 D=1*10-9 L x αx=0.83*(log(x))2.414 [m] x=15 [m]
Find: c(x,t) [mg/L] of chloride erfc(1.06) * L +exp(12.16) * erfc(3.64) m vx=2.61*10-7 s m2 Dx = 3.22*10-7 s i=0.002 m s 362.5 milligrams per liter, times the quantity, the complimentary error function of 1.06, plus e raised to 12.16 times the complimentary error of 3.64. [pause] K=3*10-5 t=1 [yr] Cl mg m2 s η=0.23 co=725 D=1*10-9 L x αx=0.83*(log(x))2.414 [m] x=15 [m]
Find: c(x,t) [mg/L] of chloride erfc(1.06) * L +exp(12.16) * erfc(3.64) m vx=2.61*10-7 s m2 Dx = 3.22*10-7 s i=0.002 m s The complimentary error function, of a variable, beta, is equal to --- K=3*10-5 t=1 [yr] Cl mg m2 s η=0.23 co=725 D=1*10-9 L x αx=0.83*(log(x))2.414 [m] x=15 [m]
Find: c(x,t) [mg/L] of chloride erfc(1.06) * L +exp(12.16) * erfc(3.64) erfc(β)=1-erf(β) i=0.002 m s 1 minus the error function, of beta, where the error function of beta equals, --- K=3*10-5 t=1 [yr] Cl mg m2 s η=0.23 co=725 D=1*10-9 L x αx=0.83*(log(x))2.414 [m] x=15 [m]
Find: c(x,t) [mg/L] of chloride erfc(1.06) * L +exp(12.16) * erfc(3.64) erfc(β)=1-erf(β) β 2 2 erf(β)= e-u du π i=0.002 m s 2 over root PI time the integral of e to the negative u squared du evaluated from 0 to beta. [pause] Error function values are typically looked up in tables. K=3*10-5 t=1 [yr] Cl mg m2 s η=0.23 co=725 D=1*10-9 L x αx=0.83*(log(x))2.414 [m] x=15 [m]
Find: c(x,t) [mg/L] of chloride erfc(1.06) * L +exp(12.16) * erfc(3.64) erfc(1.0) = 0.1573 erfc(1.1) = 0.1198 i=0.002 m s For example, complimentary error function values are looked up for 1.0 and 1.1, and the value corresponding to 1.06 --- K=3*10-5 t=1 [yr] Cl mg m2 s η=0.23 co=725 D=1*10-9 L x αx=0.83*(log(x))2.414 [m] x=15 [m]
Find: c(x,t) [mg/L] of chloride 0.1348 mg c(x,t)=362.5 erfc(1.06) * L +exp(12.16) * erfc(3.64) erfc(1.0) = 0.1573 interpolate erfc(1.1) = 0.1198 erfc(1.06) = 0.1348 i=0.002 m s is determined by interpolation, and equals 0.1348. The error function of 3.64, K=3*10-5 t=1 [yr] Cl mg m2 s η=0.23 co=725 D=1*10-9 L x αx=0.83*(log(x))2.414 [m] x=15 [m]
Find: c(x,t) [mg/L] of chloride 0.1348 mg c(x,t)=362.5 erfc(1.06) * L +exp(12.16) * erfc(3.64) erf(3.64) ~1.0 ~ i=0.002 m s is practically 1, therefore, its compliment is considered to be zero, --- K=3*10-5 t=1 [yr] Cl mg m2 s η=0.23 co=725 D=1*10-9 L x αx=0.83*(log(x))2.414 [m] x=15 [m]
Find: c(x,t) [mg/L] of chloride 0.1348 mg c(x,t)=362.5 erfc(1.06) * L +exp(12.16) * erfc(3.64) erf(3.64) ~1.0 ~ erfc(3.64) ~ 0.0 ~ i=0.002 m s and this last term cancels out of the equation. [pause] Solving for the concentration then becomes --- K=3*10-5 t=1 [yr] Cl mg m2 s η=0.23 co=725 D=1*10-9 L x αx=0.83*(log(x))2.414 [m] x=15 [m]
Find: c(x,t) [mg/L] of chloride 0.1348 * L i=0.002 m s 362.5 milligrams per liter, times 0.1348, which equals, ---- K=3*10-5 t=1 [yr] Cl mg m2 s η=0.23 co=725 D=1*10-9 L x αx=0.83*(log(x))2.414 [m] x=15 [m]
Find: c(x,t) [mg/L] of chloride 0.1348 * L mg c(x,t)=48.9 L i=0.002 m s 48.9 milligrams per liter. K=3*10-5 t=1 [yr] Cl mg m2 s η=0.23 co=725 D=1*10-9 L x αx=0.83*(log(x))2.414 [m] x=15 [m]
Find: c(x,t) [mg/L] of chloride 10 25 50 100 mg c(x,t)=362.5 0.1348 * L mg c(x,t)=48.9 L i=0.002 m s Looking over the possible solutions, --- K=3*10-5 t=1 [yr] Cl mg m2 s η=0.23 co=725 D=1*10-9 L x αx=0.83*(log(x))2.414 [m] x=15 [m]
Find: c(x,t) [mg/L] of chloride 10 25 50 100 mg c(x,t)=362.5 0.1348 * L mg c(x,t)=48.9 L AnswerC i=0.002 m s the answer is C. K=3*10-5 t=1 [yr] Cl mg m2 s η=0.23 co=725 D=1*10-9 L x αx=0.83*(log(x))2.414 [m] x=15 [m]
? Index σ’v = Σ γ d γT=100 [lb/ft3] +γclay dclay 1 Find: σ’v at d = 30 feet (1+wc)*γw wc+(1/SG) σ’v = Σ γ d d Sand 10 ft γT=100 [lb/ft3] 100 [lb/ft3] 10 [ft] 20 ft Clay = γsand dsand +γclay dclay A W S V [ft3] W [lb] 40 ft text wc = 37% ? Δh 20 [ft] (5 [cm])2 * π/4