HW4 Chapter 4

Nadia Karima Izzaty 1306369466 Dept. Teknik Sipil

4.1.1. Figure 4.1.5(b) shows the profiles through time of soil moisture head h, with vertical lines at weekly intervals. Calculate the soil moisture flux q between 0.8 m and 1.0 m at weekly intervals using the relationship K = 250(-Ψ)-2.11_{, where K is hydraulic} conductivity (cm/day) and if/ is soil suction head (cm).

Week Total head at 0.8 m Total head at 1 m Suction head at 0.8 m Suction head at 1 m hydraulic conductivity K Head difference h1-h2 Moisture flux q 1 -140 -170 -60 -70 0.037384608 30 -0.05607691 2 -155 -185 -75 -85 0.024122379 30 -0.03618357 3 -115 -150 -35 -50 0.091630354 35 -0.16035312 4 -145 -165 -65 -65 0.037384608 20 -0.03738461 5 -125 -170 -45 -70 0.048421792 45 -0.10894903 6 -105 -160 -25 -60 0.091630354 55 -0.25198347 7 -135 -110 -55 -10 0.161386096 -25 0.20173262 8 -145 -150 -65 -50 0.048421792 5 -0.01210545 9 -150 -155 -70 -55 0.040610017 5 -0.0101525 10 -165 -185 -85 -85 0.021225898 20 -0.0212259 11 -185 -200 -105 -100 0.01429923 15 -0.01072442 12 -200 -240 -120 -140 0.008660033 40 -0.01732007 13 -225 -250 -145 -150 0.006634204 25 -0.00829276 14 -260 -280 -180 -180 0.004358279 20 -0.00435828

4.2.3. For Horton's equation suppose f0 = 5 cm/h, f = 1 cm/h, and k = 2 h -1. Determine the cumulative infiltration after 0, 0.5, 1.0, 1.5, and 2.0 h. Plot the infiltration rate and cumulative infiltration as functions of time. Plot the infiltration rate as a function of the cumulative infiltration. Assume continuously ponded conditions.

𝑓(𝑡) = 𝑓_𝑐+ (𝑓₀ − 𝑓_𝑐)𝑒−𝑘𝑡

𝐹(𝑡) = 𝑓_𝑐𝑡 + 𝑓0− 𝑓𝑐

𝑘 (1 − 𝑒 −𝑘𝑡₎

Time f0 f k f(t) F(t) 0 5 1 2 5 0 0.5 5 1 2 2.471518 1.764241 1 5 1 2 1.541341 2.729329 1.5 5 1 2 1.199148 3.400426 2 5 1 2 1.073263 3.963369

4.2.6. Suppose the parameters for Philip's equation are sorptivity S = 5 cm-h~1/2 and K = 0.4 cm/h. Determine the cumulative infiltration after 0, 0.5, 1.0, 1.5, and 2.0 h. Plot the infiltration rate and the cumulative infiltration as functions of time. Plot the infiltration rate as a function of the cumulative infiltration. Assume continuously ponded conditions.

0 1 2 3 4 5 6 0 0.5 1 1.5 2 2.5

Cumulative Infiltration and Infiltration Rate

vs. Time

Cumulative Infiltration Infiltration Rate

0 1 2 3 4 5 6 0 1 2 3 4 5

Cumulative Infiltration vs. Infiltration Rate

Time S K F(t) f(t) 0 5 0.4 0 0 0.5 5 0.4 3.735534 3.935534 1 5 0.4 5.4 2.9 1.5 5 0.4 6.723724 2.441241 2 5 0.4 7.871068 2.167767

4.2.8. The infiltration into a YoIo light clay as a function of time for a steady rainfall rate of 0.5 cm/h is as follows (Skaggs, 1982):

-10 -8 -6 -4 -2 0 2 4 6 0 0.5 1 1.5 2 2.5

Cumulative Infiltration and Infiltration Rate vs. Time

Cumulative Infiltration Infiltration Rate

0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5

occurs at t = 1.07 h. t (h) Cumulative Infiltration (cm) Infiltration Rate (cm/h) In (fp-fc) 0 0 0.5 -1.10866 1.07 0.54 0.5 -1.10866 1.53 0.75 0.37 -1.60944 2.3 1 0.29 -2.12026 3.04 1.2 0.25 -2.52573 3.89 1.4 0.22 -2.99573 4.85 1.6 0.2 -3.50656 7.06 2 0.17 𝑦 = 𝑎𝑥 + 𝑏 𝑎 = 𝐾 = 0.5424 ℎ−1 𝑓₀− 𝑓_𝑐 = 𝑒−0.8469 = 0.4287 𝑓_{𝑐 = 0.17 𝑐𝑚 ℎ}⁄ 𝑓₀ = (𝑓₀ − 𝑓_𝑐) + 𝑓_{𝑐 = 0.5987 𝑐𝑚 ℎ}⁄

4.2.11. Solve Prob. 4.2.10 for a sandy soil with parameters S = 9.0 cm-h~m and K = 10 cm/h. Assume continuously ponded conditions.

y = -0.5424x - 0.8469 R² = 0.9654 -4 -3.5 -3 -2.5 -2 -1.5 -1 -0.5 0 0 1 2 3 4 5 6

Chart Title

Time S K F(t) f(t) 0 9 10 0 0 0.5 9 10 11.36396 16.36396 1 9 10 19 14.5 1.5 9 10 26.0227 13.67423 2 9 10 32.72792 13.18198 2.5 9 10 39.23025 12.84605 3 9 10 45.58846 12.59808

4.3.4. For the soil of Prob. 4.3.3, compute the cumulative infiltration after one hour for initial effective saturations of 0, 20, 40, 60, 80, and 100 percent. Draw a graph of cumulative infiltration vs. initial effective saturation.

𝐹(𝑡) = 𝐾𝑡 + 𝛹𝛥𝜃 𝑙𝑛 (1 + 𝐹(𝑡) 𝛹𝛥𝜃) 0 10 20 30 40 50 0 0.5 1 1.5 2 2.5 3 3.5

Cumulative Infiltration and Infiltration Rate vs.

Time

konvergensi nilai F(t) Initial Δθ ΨΔθ Kt F(t) 0 0.423 0.143143 0.05 0.155054 20 0.3384 0.114515 0.05 0.142639 40 0.2538 0.085886 0.05 0.128608 60 0.1692 0.057257 0.05 0.112089 80 0.0846 0.028629 0.05 0.090919 100 0 0 0.05 0

4.4.3. Compute the ponding time and cumulative infiltration at ponding for a clay loam soil with a 25 percent initial effective saturation subject to a rainfall intensity of (a) 1 cm/h (b) 3 cm/h. 𝑡𝑝 = 𝐾 𝛹𝛥𝜃 𝑖(𝑖 − 𝐾) 𝐹_𝑝 = 𝑖. 𝑡𝑝 𝐹 − 𝐹_𝑝 − 𝛹𝛥𝜃 𝑙𝑛 (𝛹𝛥𝜃 + 𝐹 𝛹𝛥𝜃 + 𝐹_𝑝) = 𝐾 (𝑡 − 𝑡𝑝) i initial Δθ ΨΔθ tp Fp F 1 0.25 0.23175 4.83894 0.53766 0.53766 0.59439 3 0.25 0.23175 4.83894 0.05562 0.16686 0.284533 0 0.05 0.1 0.15 0.2 0 20 40 60 80 100 120

Initial Effective Saturation vs. Cumulative

Infiltration

 Nilai F(t) di ruas kanan didapatkan melalui beberapa kali iterasi hingga tercapai konvergensi nilai F(t)

4.4.4. Calculate the cumulative infiltration and the infiltration rate after one hour of rainfall at 3 cm/h on a clay loam with a 25 percent initial effective saturation.

𝑡_𝑝 = 𝐾 𝛹𝛥𝜃 𝑖(𝑖 − 𝐾) 𝐹𝑝 = 𝑖. 𝑡𝑝 𝐹 − 𝐹_𝑝 − 𝛹𝛥𝜃 𝑙𝑛 (𝛹𝛥𝜃 + 𝐹 𝛹𝛥𝜃 + 𝐹_𝑝) = 𝐾 (𝑡 − 𝑡𝑝) 𝑓 = 𝐾 ( 𝛹𝛥𝜃 𝐹 + 1) 𝐹 = 0.284533 𝑐𝑚 (𝑑𝑎𝑟𝑖 𝑠𝑜𝑎𝑙 4.4.3) 𝑓 = 1.8 𝑐𝑚/ℎ

4.4.10. A soil has Horton's equation parameters fo = 10 cm/h, f = 4 cm/h and k = 2 h-1. Calculate the ponding time and cumulative infiltration at ponding under a rainfall of 6 cm/h. 𝑡𝑝 = 1 𝑖. 𝑘 [𝑓0− 𝑖 + 𝑓𝑐ln ( 𝑓₀− 𝑓_𝑐 𝑖 − 𝑓𝑐 )] 𝑡𝑝 = 1 6 ∗ 2 [10 − 6 + 4 ln ( 10 − 4 6 − 4)] = 0.699 ℎ 𝐹 = 𝑓_𝑐𝑡 + 𝑓0− 𝑓𝑐 𝐾 (1 − 𝑒 −𝑘𝑡₎ 𝐹 = (4 ∗ 1) + 10 − 4 2 (1 − 𝑒 −2∗1₎ 𝐹 = 5.0376 𝑐𝑚