Find: f(4[hr]) [cm/hr] saturation 0% 100%

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Find: f(4[hr]) [cm/hr] 0.25 0.50 0.75 1.00 saturation 0% 100% Green-Ampt sandy clay η θi=27.0% Δθ continuous ponding ho negligible θi 0.25 0.50 0.75 1.00 Find the infiltration rate of the soil after 4 hours, in centimeters per hour. [pause] In this problem, ---

Find: f(4[hr]) [cm/hr] 0.25 0.50 0.75 1.00 saturation 0% 100% Green-Ampt sandy clay η θi=27.0% Δθ continuous ponding ho negligible θi 0.25 0.50 0.75 1.00 a sandy clay soil, with an initial saturation of 27%, undergoes continuous ponding, with a negligible ponding depth.

Find: f(4[hr]) [cm/hr] +1 ψ * Δθ saturation 0% 100% Green-Ampt sandy clay η θi=27.0% Δθ continuous ponding ho negligible θi The Green-Ampt equation for infiltration rate, is --- ψ * Δθ F(t) f(t)=K * +1

Find: f(4[hr]) [cm/hr] +1 ψ * Δθ saturation 0% 100% Green-Ampt sandy clay η θi=27.0% Δθ continuous ponding ho negligible θi the infiltration rate, at time t, equals, --- ψ * Δθ f(t)=K +1 * F(t) infiltration rate

Find: f(4[hr]) [cm/hr] +1 ψ * Δθ saturation 0% 100% Green-Ampt sandy clay η θi=27.0% Δθ continuous ponding ho negligible θi hydraulic conductivity the hydraulic conductivity, times the quantity, --- ψ * Δθ f(t)=K +1 * F(t) infiltration rate

Find: f(4[hr]) [cm/hr] +1 ψ * Δθ saturation 0% 100% Green-Ampt sandy clay η θi=27.0% Δθ continuous ponding ho negligible θi hydraulic conductivity The soil suction head, psi, times, --- ψ * Δθ f(t)=K +1 * F(t) infiltration suction rate head

Find: f(4[hr]) [cm/hr] +1 ψ * Δθ saturation 0% 100% Green-Ampt sandy clay η θi=27.0% Δθ continuous ponding ho negligible θi hydraulic conductivity change in moisture the change in moisture content when the wetting front passes dry soil, delta theta, divided by, --- ψ * Δθ f(t)=K +1 * content F(t) infiltration suction rate head

Find: f(4[hr]) [cm/hr] +1 ψ * Δθ saturation 0% 100% Green-Ampt sandy clay η θi=27.0% Δθ continuous ponding ho negligible θi hydraulic conductivity change in moisture the cumulative infiltration, at time t, plus one. ψ * Δθ f(t)=K +1 * content F(t) cumulative infiltration suction infiltration rate head

Find: f(4[hr]) [cm/hr] +1 ψ * Δθ saturation 0% 100% Green-Ampt sandy clay η θi=27.0% Δθ continuous ponding ho negligible θi hydraulic conductivity change in moisture Infiltration parameters, for Green-Ampt equations, can be looked up, based on soil type. For a sandy clay, --- ψ * Δθ f(t)=K +1 * content F(t) cumulative infiltration suction infiltration rate head

Find: f(4[hr]) [cm/hr] +1 ψ * Δθ saturation 0% 100% Green-Ampt sandy clay η θi=27.0% Δθ continuous ponding ho negligible θi hydraulic conductivity change in cm K=0.060 moisture the hydraulic conductivity is 0.060 centimeters per hour, --- ψ * Δθ hr f(t)=K +1 * content F(t) cumulative infiltration suction infiltration rate head

Find: f(4[hr]) [cm/hr] +1 ψ * Δθ ψ =23.90 [cm] saturation 0% 100% Green-Ampt sandy clay η θi=27.0% Δθ continuous ponding ho negligible θi hydraulic conductivity change in cm K=0.060 moisture the suction head is 23.90 centimeters, --- ψ * Δθ hr f(t)=K +1 * content F(t) ψ =23.90 [cm] cumulative infiltration suction infiltration rate head

Find: f(4[hr]) [cm/hr] +1 ψ * Δθ ψ =23.90 [cm] θe =0.321 saturation 0% 100% Green-Ampt sandy clay η θi=27.0% Δθ continuous ponding ho negligible θi hydraulic conductivity change in cm K=0.060 moisture and the effective porosity equals 0.321. [pause] The change in moisture content can be determined by multiplying --- ψ * Δθ hr f(t)=K +1 * content F(t) ψ =23.90 [cm] cumulative θe =0.321 infiltration suction infiltration rate head

Find: f(4[hr]) [cm/hr] Δθ=(1-θi) * θe +1 ψ * Δθ ψ =23.90 [cm] saturation 0% 100% Green-Ampt sandy clay η θi=27.0% Δθ continuous ponding ho negligible θi change in cm K=0.060 moisture 1 minus the initial saturation, by the effective porosity. ψ * Δθ hr f(t)=K +1 * content F(t) ψ =23.90 [cm] Δθ=(1-θi) * θe θe =0.321

Find: f(4[hr]) [cm/hr] Δθ=(1-θi) * θe Δθ=(1-0.27) * 0.321 +1 ψ * Δθ saturation 0% 100% Green-Ampt sandy clay η θi=27.0% Δθ continuous ponding ho negligible θi change in cm K=0.060 moisture So 1 minus 0.27, times 0.321 equals a delta theta of --- ψ * Δθ hr f(t)=K +1 * content F(t) ψ =23.90 [cm] Δθ=(1-θi) * θe θe =0.321 Δθ=(1-0.27) * 0.321

Find: f(4[hr]) [cm/hr] Δθ=(1-θi) * θe Δθ=(1-0.27) * 0.321 Δθ=0.2343 +1 saturation 0% 100% Green-Ampt sandy clay η θi=27.0% Δθ continuous ponding ho negligible θi change in cm K=0.060 moisture 0.2343. [pause] So already, we know the hydraulic conductivity, --- ψ * Δθ hr f(t)=K +1 * content F(t) ψ =23.90 [cm] Δθ=(1-θi) * θe θe =0.321 Δθ=(1-0.27) * 0.321 Δθ=0.2343

Find: f(4[hr]) [cm/hr] +1 ψ * Δθ ψ =23.90 [cm] θe = 0.321 Δθ = 0.2343 saturation 0% 100% Green-Ampt sandy clay η θi=27.0% Δθ continuous ponding ho negligible θi hydraulic conductivity change in cm K=0.060 moisture suction head, and change in moisture content. The last unknown variable is --- ψ * Δθ hr f(t)=K +1 * content F(t) ψ =23.90 [cm] cumulative θe = 0.321 infiltration suction infiltration rate head Δθ = 0.2343

Find: f(4[hr]) [cm/hr] +1 ψ * Δθ ψ =23.90 [cm] θe = 0.321 Δθ = 0.2343 saturation 0% 100% Green-Ampt sandy clay η θi=27.0% Δθ continuous ponding ho negligible θi hydraulic conductivity change in cm K=0.060 moisture the cumulative infiltration, capital F sub t. [pause] The cumulative infiltrate, at a time t, equals --- ψ * Δθ hr f(t)=K +1 * content F(t) ψ =23.90 [cm] cumulative θe = 0.321 infiltration suction infiltration rate head Δθ = 0.2343

Find: f(4[hr]) [cm/hr] ψ * Δθ ψ =23.90 [cm] θe = 0.321 Δθ = 0.2343 F saturation 0% 100% Green-Ampt sandy clay η θi=27.0% Δθ continuous ponding cumulative ho negligible θi infiltration cm F (t) K=0.060 F(t) = K * t + ψ * Δθ * ln 1+ the hydraulic conductivity, times the time, plus --- hr ψ * Δθ ψ =23.90 [cm] time θe = 0.321 hydraulic conductivity Δθ = 0.2343

Find: f(4[hr]) [cm/hr] ψ * Δθ ψ =23.90 [cm] θe = 0.321 Δθ = 0.2343 F saturation 0% 100% Green-Ampt sandy clay η θi=27.0% Δθ continuous ponding cumulative ho negligible θi infiltration cm F (t) K=0.060 F(t) = K * t + ψ * Δθ * ln 1+ the suction head times the change in moisture content times, --- hr ψ * Δθ ψ =23.90 [cm] time change in suction θe = 0.321 hydraulic moisture head conductivity content Δθ = 0.2343

Find: f(4[hr]) [cm/hr] ψ * Δθ ψ =23.90 [cm] θe = 0.321 Δθ = 0.2343 F saturation 0% 100% Green-Ampt sandy clay η θi=27.0% Δθ continuous ponding cumulative ho negligible θi infiltration cm F (t) K=0.060 F(t) = K * t + ψ * Δθ * ln 1+ the natural logarithm of the quantity 1 plus, --- hr ψ * Δθ ψ =23.90 [cm] time change in suction θe = 0.321 hydraulic moisture head conductivity content Δθ = 0.2343

Find: f(4[hr]) [cm/hr] ψ * Δθ ψ =23.90 [cm] θe = 0.321 Δθ = 0.2343 F saturation 0% 100% Green-Ampt sandy clay η θi=27.0% Δθ continuous ponding cumulative ho negligible θi infiltration cm F (t) K=0.060 F(t) = K * t + ψ * Δθ * ln 1+ the cumulative infiltration divided by --- hr ψ * Δθ ψ =23.90 [cm] time change in suction θe = 0.321 hydraulic moisture head conductivity content Δθ = 0.2343

Find: f(4[hr]) [cm/hr] ψ * Δθ ψ =23.90 [cm] θe = 0.321 Δθ = 0.2343 F saturation 0% 100% Green-Ampt sandy clay η θi=27.0% Δθ continuous ponding cumulative ho negligible θi infiltration cm F (t) K=0.060 F(t) = K * t + ψ * Δθ * ln 1+ the suction head times the change in moisture content. [pause] hr ψ * Δθ ψ =23.90 [cm] time change in suction θe = 0.321 hydraulic moisture head conductivity content Δθ = 0.2343

Find: f(4[hr]) [cm/hr] ψ * Δθ ψ =23.90 [cm] θe = 0.321 Δθ = 0.2343 F saturation 0% 100% Green-Ampt sandy clay η θi=27.0% Δθ continuous ponding 4 [hr] ho negligible θi cm F (4) K=0.060 F(4) = K * t + ψ * Δθ * ln 1+ Plugging in the known variables, the equation reduces down to --- hr ψ * Δθ ψ =23.90 [cm] θe = 0.321 Δθ = 0.2343

Find: f(4[hr]) [cm/hr] ψ * Δθ ψ =23.90 [cm] Δθ = 0.2343 F (4) saturation 0% 100% Green-Ampt sandy clay η θi=27.0% Δθ continuous ponding 4 [hr] ho negligible θi cm F (4) K=0.060 hr F(4) = K * t + ψ * Δθ * ln 1+ F sub t equals 0.24 centimeters, plus 5.6 centimeters times the natural logarithm of 1 plus F sub t divided by 5.6 centimeters. ψ * Δθ ψ =23.90 [cm] Δθ = 0.2343 F (4) F(4) = 0.24 [cm] + 5.6 [cm] * ln 1+ 5.6 [cm]

Find: f(4[hr]) [cm/hr] ψ * Δθ ψ =23.90 [cm] Δθ = 0.2343 F (4) saturation 0% 100% Green-Ampt sandy clay η θi=27.0% Δθ continuous ponding 4 [hr] ho negligible θi cm F (4) K=0.060 hr F(4) = K * t + ψ * Δθ * ln 1+ We notice the the cumulative infiltration occurs twice in the equation, once in linear, and once in log. ψ * Δθ ψ =23.90 [cm] Δθ = 0.2343 cumulative infiltration F (4) F(4) = 0.24 [cm] + 5.6 [cm] * ln 1+ 5.6 [cm]

Find: f(4[hr]) [cm/hr] F (4) 1+ saturation 0% 100% η LHS RHS next F Δθ Therefore, we’ll have to solve the problem iteratively, unless we have access to a special calculator. cumulative infiltration F (4) F(4) = 0.24 [cm] + 5.6 [cm] * ln 1+ 5.6 [cm]

Find: f(4[hr]) [cm/hr] (t) F (4) 1+ saturation 0% 100% η LHS RHS next F (t) Δθ θi value If we had a graphing calculator, we could plot --- F (t) cumulative infiltration F (4) F(4) = 0.24 [cm] + 5.6 [cm] * ln 1+ 5.6 [cm]

Find: f(4[hr]) [cm/hr] (t) F (4) 1+ saturation 0% 100% η LHS RHS next F (t) Δθ θi LHS value the left hand side of the equation as a function of F sub t, and --- F (t) cumulative infiltration F (4) F(4) = 0.24 [cm] + 5.6 [cm] * ln 1+ 5.6 [cm]

Find: f(4[hr]) [cm/hr] (t) F (4) 1+ saturation 0% 100% η LHS RHS next F (t) Δθ θi LHS value RHS the right hand side of the equation as a function of F sub t, and --- F (t) cumulative infiltration F (4) F(4) = 0.24 [cm] + 5.6 [cm] * ln 1+ 5.6 [cm]

Find: f(4[hr]) [cm/hr] (t) F (4) 1+ saturation 0% 100% η LHS RHS next F (t) Δθ θi LHS value RHS the intersection point would be the correct value --- F (t) cumulative infiltration F (4) F(4) = 0.24 [cm] + 5.6 [cm] * ln 1+ 5.6 [cm]

Find: f(4[hr]) [cm/hr] (t) F (4) 1+ saturation 0% 100% η LHS RHS next F (t) Δθ θi LHS value RHS for the cumulative infiltration. [pause] To solve the problem by hand, we’ll begin by simply --- F (t) cumulative infiltration F (4) F(4) = 0.24 [cm] + 5.6 [cm] * ln 1+ 5.6 [cm]

Find: f(4[hr]) [cm/hr] (t) F (4) 1+ saturation 0% 100% F(4) = ? η LHS RHS next F (t) Δθ θi LHS value RHS guessing the cumulative infiltration in the soil, after 4 hours of continuous ponding. Since the soil is a sandy clay, we’ll guess --- F (t) cumulative infiltration F (4) F(4) = 0.24 [cm] + 5.6 [cm] * ln 1+ 5.6 [cm]

Find: f(4[hr]) [cm/hr] (t) F (4) 1+ initial guess saturation 0% 100% F(4) = 3 [cm] η LHS RHS next F (t) Δθ θi LHS value RHS 3 centimeters of infiltration. [pause] Then we’ll plug this value into --- F (t) cumulative infiltration F (4) F(4) = 0.24 [cm] + 5.6 [cm] * ln 1+ 5.6 [cm]

Find: f(4[hr]) [cm/hr] (t) F (4) 1+ initial guess saturation 0% 100% F(4) = 3 [cm] η LHS RHS next F (t) Δθ θi LHS value RHS our equation and solve the --- F (t) cumulative infiltration F (4) F(4) = 0.24 [cm] + 5.6 [cm] * ln 1+ 5.6 [cm]

Find: f(4[hr]) [cm/hr] (t) LHS F (4) 1+ initial guess saturation 0% 100% F(4) = 3 [cm] η LHS RHS next F (t) Δθ θi LHS value RHS left hand side of the equation, which is just the cumulative infiltration itself, 3, --- F (t) LHS F (4) F(4) = 0.24 [cm] + 5.6 [cm] * ln 1+ 5.6 [cm]

Find: f(4[hr]) [cm/hr] (t) LHS F (4) 1+ initial guess saturation 0% 100% F(4) = 3 [cm] η LHS RHS next F (t) Δθ 3.00 θi LHS value RHS then we’ll solve the right hand side of the equation, F (t) LHS F (4) F(4) = 0.24 [cm] + 5.6 [cm] * ln 1+ 5.6 [cm]

Find: f(4[hr]) [cm/hr] (t) RHS LHS F (4) 1+ initial guess saturation 0% 100% F(4) = 3 [cm] η LHS RHS next F (t) Δθ 3.00 2.64 θi LHS value RHS which equals 2.64, when we guess--- F (t) RHS LHS F (4) F(4) = 0.24 [cm] + 5.6 [cm] * ln 1+ 5.6 [cm]

Find: f(4[hr]) [cm/hr] (t) RHS LHS F (4) 1+ initial guess saturation 0% 100% F(4) = 3 [cm] η LHS RHS next F (t) Δθ 3.00 2.64 θi LHS value RHS 3 centimeters for the cumulative infiltration. [pause] From this first iteration, we notice --- F (t) 3 [cm] RHS LHS F (4) F(4) = 0.24 [cm] + 5.6 [cm] * ln 1+ 5.6 [cm]

Find: f(4[hr]) [cm/hr] (t) RHS LHS F (4) 1+ initial guess saturation 0% 100% F(4) = 3 [cm] η LHS RHS next F (t) Δθ 3.00 2.64 θi LHS value RHS the left hand side of the equation is greater than the right hand side of the equation. F (t) 3 [cm] RHS LHS F (4) F(4) = 0.24 [cm] + 5.6 [cm] * ln 1+ 5.6 [cm]

Find: f(4[hr]) [cm/hr] (t) RHS LHS F (4) 1+ initial guess saturation 0% 100% F(4) = 3 [cm] η LHS RHS next F (t) Δθ 3.00 2.64 θi LHS value RHS Looking at the graph, this condition is true when the guess for the cumulative infiltration, F sub t, too high. F (t) RHS LHS F (4) F(4) = 0.24 [cm] + 5.6 [cm] * ln 1+ 5.6 [cm]

Find: f(4[hr]) [cm/hr] (t) RHS LHS F (4) 1+ initial guess saturation 0% 100% F(4) = 3 [cm] η LHS RHS next F (t) Δθ 3.00 2.64 θi LHS value RHS So our second guess will be less than 3 centimeters. [pause] F (t) RHS LHS F (4) F(4) = 0.24 [cm] + 5.6 [cm] * ln 1+ 5.6 [cm]

Find: f(4[hr]) [cm/hr] (t) RHS LHS F (4) 1+ second guess saturation 0% 100% F(4) = 1.5 [cm] η LHS RHS next F (t) Δθ 3.00 2.64 θi LHS value RHS Using a second guess of 1.5 centimeters, --- F (t) RHS LHS F (4) F(4) = 0.24 [cm] + 5.6 [cm] * ln 1+ 5.6 [cm]

Find: f(4[hr]) [cm/hr] (t) RHS LHS F (4) 1+ second guess saturation 0% 100% F(4) = 1.5 [cm] η LHS RHS next F (t) Δθ 3.00 2.64 θi LHS 1.50 value RHS the left hand side of the equation is 1.50 centimeters, --- F (t) RHS LHS F (4) F(4) = 0.24 [cm] + 5.6 [cm] * ln 1+ 5.6 [cm]

Find: f(4[hr]) [cm/hr] (t) RHS LHS F (4) 1+ second guess saturation 0% 100% F(4) = 1.5 [cm] η LHS RHS next F (t) Δθ 3.00 2.64 θi LHS 1.50 1.57 value RHS and the right hand side of the equation computes to 1.57 centimeters. [pause] Now we have a situation where, --- F (t) RHS LHS F (4) F(4) = 0.24 [cm] + 5.6 [cm] * ln 1+ 5.6 [cm]

Find: f(4[hr]) [cm/hr] (t) RHS LHS F (4) 1+ second guess saturation 0% 100% F(4) = 1.5 [cm] η LHS RHS next F (t) Δθ 3.00 2.64 θi LHS 1.50 1.57 value RHS the left hand side of the equation is less than the right hand side of the equation, --- F (t) RHS LHS F (4) F(4) = 0.24 [cm] + 5.6 [cm] * ln 1+ 5.6 [cm]

Find: f(4[hr]) [cm/hr] (t) RHS LHS F (4) 1+ second guess saturation 0% 100% F(4) = 1.5 [cm] η LHS RHS next F (t) Δθ 3.00 2.64 θi LHS 1.50 1.57 value RHS as would occur when the guessed value for cumulative infiltration is too low, and should be higher. F (t) RHS LHS F (4) F(4) = 0.24 [cm] + 5.6 [cm] * ln 1+ 5.6 [cm]

Find: f(4[hr]) [cm/hr] (t) RHS LHS F (4) 1+ third guess saturation 0% 100% F(4) = 2 [cm] η LHS RHS next F (t) Δθ 3.00 2.64 θi LHS 1.50 1.57 value RHS 2.00 2 centimeters is tried on the third iteration, the right hand side equals --- F (t) RHS LHS F (4) F(4) = 0.24 [cm] + 5.6 [cm] * ln 1+ 5.6 [cm]

Find: f(4[hr]) [cm/hr] (t) RHS LHS F (4) 1+ third guess saturation 0% 100% F(4) = 2 [cm] η LHS RHS next F (t) Δθ 3.00 2.64 θi LHS 1.50 1.57 value RHS 2.00 1.95 1.95 centimeters. [pause] Therefore, 2 centimeters is too high. [pause] F (t) RHS LHS F (4) F(4) = 0.24 [cm] + 5.6 [cm] * ln 1+ 5.6 [cm]

Find: f(4[hr]) [cm/hr] (t) RHS LHS F (4) 1+ fourth guess saturation 0% 100% F(4) = 1.75 [cm] η LHS RHS next F (t) Δθ 3.00 2.64 θi LHS 1.50 1.57 value RHS 2.00 1.95 For the fourth iteration we’ll try 1.75 centimeters, --- 1.75 F (t) RHS LHS F (4) F(4) = 0.24 [cm] + 5.6 [cm] * ln 1+ 5.6 [cm]

Find: f(4[hr]) [cm/hr] (t) RHS LHS F (4) 1+ fourth guess saturation 0% 100% F(4) = 1.75 [cm] η LHS RHS next F (t) Δθ 3.00 2.64 θi LHS 1.50 1.57 value RHS 2.00 1.95 which is slightly too low. [pause] Our fifth iteration using --- 1.75 1.76 F (t) RHS LHS F (4) F(4) = 0.24 [cm] + 5.6 [cm] * ln 1+ 5.6 [cm]

Find: f(4[hr]) [cm/hr] (t) LHS RHS F (4) 1+ fifth guess saturation 0% 100% F(4) = 1.80 [cm] η LHS RHS next F (t) Δθ 3.00 2.64 θi LHS 1.50 1.57 value RHS 2.00 1.95 1.80 centimeters, converges, so now we’ve found our cumulative infiltration. [pause] Returning to our equation --- 1.75 1.76 F (t) 1.80 1.80 - LHS 1.80 [cm] RHS F (4) F(4) = 0.24 [cm] + 5.6 [cm] * ln 1+ 5.6 [cm]

Find: f(4[hr]) [cm/hr] +1 ψ =23.90 [cm] ψ * Δθ θe = 0.321 Δθ = 0.2343 fifth guess saturation 0% 100% F(4) = 1.80 [cm] η LHS RHS next F Δθ 1.80 1.80 - θi hydraulic cm K=0.060 conductivity hr change in ψ =23.90 [cm] for the infiltration rate, we now know --- ψ * Δθ moisture f(4)=K +1 * content θe = 0.321 F(4) Δθ = 0.2343 cumulative infiltration F(4) =1.80 [cm] suction infiltration rate head

Find: f(4[hr]) [cm/hr] +1 ψ =23.90 [cm] ψ * Δθ θe = 0.321 Δθ = 0.2343 fifth guess saturation 0% 100% F(4) = 1.80 [cm] η LHS RHS next F Δθ 1.80 1.80 - θi hydraulic cm K=0.060 conductivity hr change in ψ =23.90 [cm] the cumulative infiltration, after 4 hours, is 1.80 centimeters. [pause] And from earlier, we’ve already determined --- ψ * Δθ moisture f(4)=K +1 * content θe = 0.321 F(4) Δθ = 0.2343 cumulative infiltration F(4) =1.80 [cm] suction infiltration rate head

Find: f(4[hr]) [cm/hr] +1 ψ =23.90 [cm] ψ * Δθ θe = 0.321 Δθ = 0.2343 fifth guess saturation 0% 100% F(4) = 1.80 [cm] η LHS RHS next F Δθ 1.80 1.80 - θi hydraulic cm K=0.060 conductivity hr change in ψ =23.90 [cm] the values for hydraulic conductivity, suction head and change in moisture content. After making these substitutions, the infiltration rate equals, --- ψ * Δθ moisture f(4)=K +1 * content θe = 0.321 F(4) Δθ = 0.2343 cumulative infiltration F(4) =1.80 [cm] suction infiltration rate head

Find: f(4[hr]) [cm/hr] +1 ψ =23.90 [cm] ψ * Δθ θe = 0.321 Δθ = 0.2343 fifth guess saturation 0% 100% F(4) = 1.80 [cm] η LHS RHS next F Δθ 1.80 1.80 - θi hydraulic cm K=0.060 conductivity hr change in ψ =23.90 [cm] 0.247 centimeters per hour. [pause] ψ * Δθ moisture f(4)=K +1 * content θe = 0.321 F(4) cm Δθ = 0.2343 f(4)=0.247 hr F(4) =1.80 [cm]

Find: f(4[hr]) [cm/hr] +1 ψ =23.90 [cm] ψ * Δθ θe = 0.321 Δθ = 0.2343 fifth guess saturation 0% 100% F(4) = 1.80 [cm] η LHS RHS next F Δθ 1.80 1.80 - θi hydraulic cm K=0.060 conductivity hr 0.25 0.50 0.75 1.00 ψ =23.90 [cm] When reviewing the possible solutions, --- ψ * Δθ f(4)=K +1 * θe = 0.321 F(4) cm Δθ = 0.2343 f(4)=0.247 hr F(4) =1.80 [cm]

Find: f(4[hr]) [cm/hr] +1 ψ =23.90 [cm] ψ * Δθ θe = 0.321 Δθ = 0.2343 fifth guess saturation 0% 100% F(4) = 1.80 [cm] η LHS RHS next F Δθ 1.80 1.80 - θi hydraulic cm K=0.060 conductivity hr 0.25 0.50 0.75 1.00 ψ =23.90 [cm] the answer is A. ψ * Δθ f(4)=K +1 * θe = 0.321 F(4) cm Δθ = 0.2343 f(4)=0.247 hr F(4) =1.80 [cm] AnswerA

? 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