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

1 2 3 4 5 6 7 8 9 10 11 12 13 14

-140 -155 -115 -145 -125 -105 -135 -145 -150 -165 -185 -200 -225 -260

-170 -185 -150 -165 -170 -160 -110 -150 -155 -185 -200 -240 -250 -280

Suction head at 0.8 m -60 -75 -35 -65 -45 -25 -55 -65 -70 -85 -105 -120 -145 -180

Suction head at 1m -70 -85 -50 -65 -70 -60 -10 -50 -55 -85 -100 -140 -150 -180

hydraulic conductivity K 0.037384608 0.024122379 0.091630354 0.037384608 0.048421792 0.091630354 0.161386096 0.048421792 0.040610017 0.021225898 0.01429923 0.008660033 0.006634204 0.004358279

Head difference h1-h2 30 30 35 20 45 55 -25 5 5 20 15 40 25 20

Moisture flux q -0.05607691 -0.03618357 -0.16035312 -0.03738461 -0.10894903 -0.25198347 0.20173262 -0.01210545 -0.0101525 -0.0212259 -0.01072442 -0.01732007 -0.00829276 -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) = 𝑓𝑐 + (𝑓0 − 𝑓𝑐 )𝑒 −𝑘𝑡 𝐹(𝑡) = 𝑓𝑐 𝑡 +

𝑓0 − 𝑓𝑐 (1 − 𝑒 −𝑘𝑡 ) 𝑘

Time

f0

0 0.5 1 1.5 2

f 5 5 5 5 5

k 1 1 1 1 1

2 2 2 2 2

f(t)

F(t)

5 2.471518 1.541341 1.199148 1.073263

0 1.764241 2.729329 3.400426 3.963369

Cumulative Infiltration and Infiltration Rate vs. Time 6 5 4 3 2 1 0 0

0.5

1

1.5

Cumulative Infiltration

2

2.5

Infiltration Rate

Cumulative Infiltration vs. Infiltration Rate 6 5 4 3 2 1 0 0

1

2

3

4

5

Series1

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.

Time

S

K 5 5 5 5 5

0 0.5 1 1.5 2

F(t) f(t) 0.4 0 0 0.4 3.735534 3.935534 0.4 5.4 2.9 0.4 6.723724 2.441241 0.4 7.871068 2.167767

Cumulative Infiltration and Infiltration Rate vs. Time 6 4 2 0 -2 0

0.5

1

1.5

2

2.5

-4 -6 -8 -10 Cumulative Infiltration

Infiltration Rate

Cumulative Infiltration vs. Infiltration Rate 9 8 7 6 5 4 3 2 1 0 0

1

2

3

4

5

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):

Determine the parameters fo, fc, and k for Horton's equation. Assume that ponding occurs at t = 1.07 h. Cumulative Infiltration (cm) 0 0.54 0.75 1 1.2 1.4 1.6 2

t (h) 0 1.07 1.53 2.3 3.04 3.89 4.85 7.06

Infiltration Rate (cm/h) 0.5 0.5 0.37 0.29 0.25 0.22 0.2 0.17

In (fp-fc) -1.10866 -1.10866 -1.60944 -2.12026 -2.52573 -2.99573 -3.50656

Chart Title 0 -0.5

0

1

2

3

4

5

6

-1 -1.5 -2 -2.5 -3 y = -0.5424x - 0.8469 R² = 0.9654

-3.5 -4

𝑦 = 𝑎𝑥 + 𝑏 𝑎 = 𝐾 = 0.5424 ℎ−1 𝑓0 − 𝑓𝑐 = 𝑒 −0.8469 = 0.4287 𝑓𝑐 = 0.17 𝑐𝑚⁄ℎ 𝑓0 = (𝑓0 − 𝑓𝑐 ) + 𝑓𝑐 = 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.

Time

S

K 9 9 9 9 9 9 9

0 0.5 1 1.5 2 2.5 3

10 10 10 10 10 10 10

F(t)

f(t)

0 11.36396 19 26.0227 32.72792 39.23025 45.58846

0 16.36396 14.5 13.67423 13.18198 12.84605 12.59808

Cumulative Infiltration and Infiltration Rate vs. Time 50 40 30 20 10 0 0

0.5

1

1.5

2

2.5

3

3.5

Cumulative Infiltration vs. Infiltration Rate 18 16 14 12 10 8 6 4 2 0 0

10

20

30

40

50

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



Nilai F(t) di ruas kanan didapatkan melalui beberapa kali iterasi hingga tercapai konvergensi nilai F(t) Initial 0 20 40 60 80 100

Δθ 0.423 0.3384 0.2538 0.1692 0.0846 0

ΨΔθ 0.143143 0.114515 0.085886 0.057257 0.028629 0

Kt 0.05 0.05 0.05 0.05 0.05 0.05

F(t) 0.155054 0.142639 0.128608 0.112089 0.090919 0

Initial Effective Saturation vs. Cumulative Infiltration 0.2 0.15 0.1 0.05 0 0

20

40

60

80

100

120

Series1

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)

𝐹𝑝 = 𝑖. 𝑡𝑝 𝛹𝛥𝜃 + 𝐹 𝐹 − 𝐹𝑝 − 𝛹𝛥𝜃 𝑙𝑛 ( ) = 𝐾 (𝑡 − 𝑡𝑝 ) 𝛹𝛥𝜃 + 𝐹𝑝 i 1 3

initial 0.25 0.25

Δθ 0.23175 0.23175

ΨΔθ 4.83894 4.83894

tp 0.53766 0.05562

Fp F 0.53766 0.59439 0.16686 0.284533



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. 𝑡𝑝 =

𝐾 𝛹𝛥𝜃 𝑖(𝑖 − 𝐾)

𝐹𝑝 = 𝑖. 𝑡𝑝 𝛹𝛥𝜃 + 𝐹 𝐹 − 𝐹𝑝 − 𝛹𝛥𝜃 𝑙𝑛 ( ) = 𝐾 (𝑡 − 𝑡𝑝 ) 𝛹𝛥𝜃 + 𝐹𝑝 𝑓= 𝐾 (

𝛹𝛥𝜃 + 1) 𝐹

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 =

𝑡𝑝 =

1 𝑓0 − 𝑓𝑐 [𝑓0 − 𝑖 + 𝑓𝑐 ln ( )] 𝑖. 𝑘 𝑖 − 𝑓𝑐

1 10 − 4 [10 − 6 + 4 ln ( )] = 0.699 ℎ 6∗2 6− 4 𝐹 = 𝑓𝑐 𝑡 +

𝑓0 − 𝑓𝑐 (1 − 𝑒 −𝑘𝑡 ) 𝐾

𝐹 = (4 ∗ 1) +

10 − 4 (1 − 𝑒 −2∗1 ) 2

𝐹 = 5.0376 𝑐𝑚