Fundamentals of Geotechnical Engineering (MindTap Course List)
5th Edition
ISBN: 9781305635180
Author: Braja M. Das, Nagaratnam Sivakugan
Publisher: Cengage Learning
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Transcribed Image Text:Estimate the capillary soil suction using the Green-Ampt equation if it takes 3.5 hours
for an infiltration wetting front to travel 40 cm below the ground. The effective
porosity of the soil (n) is 0.2 while its hydraulic conductivity is 2.5 cm/hour.
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- 12. (Horton's equation for infiltration rate, ). One can use Horton's equation to estimate the infiltration rate of water into soil if the rainfall rate exceeds the infiltration rate. The Horton's equation states that: f = fe+ (fo-fc) x e-kt where f is the infiltration rate (cm h:¹), fe is the equilibrium infiltration rate (cm h 1), fo is the initial infiltration rate (cm h:¹), k is an empirical constant (h:¹), and t is time (h). Horton's equation can be integrated to yield an equation that represents the total volume of water that would infiltrate over a given period: V=A₁ [Fxα= 4, 16+ V = A₁ [fx dt = A₁ [lfc + (fo-fc) × e-k¹] x dt = A₂ × [fet + fo=fc (1 - e-*)] 0 where As is the unit area (1 m²) and V is the volume (m³). Sothan loamy sand has the following characteristics: fe= 6.68 cm h¹, fo = 8.81 cm h¹, k = 1.40 h ¹. Assuming the rate of precipitation exceeds the rate of infiltration throughout the storm event. 12.1. Find the infiltration rate at times of 12, 30, 60, and 120 min.…arrow_forwardA falling-head permeability test was carried out on a clay soil of diameter 10 cm and length 15 cm. In 1 hour the head in the standpipe of diameter 5 mm dropped from 68 cm to 50.2 cm. Calculate the hydraulic conductivity of this clay.arrow_forwardThe initial infiltration capacity of the soil is 0.0386cm per min. the time constant as 10.18 per min and the final capacity of soil for infiltration is 0.0683 cm per hour. Use Hortons equation to solve for ft in mm per hour at 3 minarrow_forward
- Problem 7.11arrow_forwardNeed only handwritten solution only (not typed one).arrow_forwardLength of soil sample for constant head permeability test is 30 cm, Area of specimen is 177cm, constant head difference is 50 cm, total volume of water collected in 5 min is 350cm'. Calculate the hydraulic conductivity of the soil. A) 2.85x10'cm/sec D) 1.09x10 cm/sec 8) 1.09x10 cm/sec 9 3.95x10 cm/sec Answer A Submitarrow_forward
- Calculate the equivalent hydraulic conductivity for the soil profile shown below:arrow_forwardA 15 cm vertical soil sample is taken and placed in constant head permeameter. The surface area of the soil sample is 20 cm2. h1 is measured at 25 cm and h2 is 10 cm. The water that flows through the soil sample has a constant discharge and 95 cm3 is collected in 15 minutes. Calculate the hydraulic conductivity of the soil sample. What kind of soil you think this is?arrow_forwardA falling-head permeability test was carried out on a clay soil of diameter 100 mm and length150 mm. In 4 minutes the head in the standpipe of diameter 5mm dropped from 680 to 350mm. Calculate the hydraulic conductivity of this clay.arrow_forward
- 3. The Horton infiltration equation f(t) describes how the infiltration capacity (mm/hr) for the soil is changing with time as the soil becomes wetter. The accumulated infiltrated water (in mm) can be calculated using the integral of f(t), i.e., F(t). where, fo = initial infiltration capacity of dry soil (mm/hr) fc = final infiltration capacity when the soil has become saturated soil (mm/hr) k = time constant (hr4) Note that the Horton equation is only valid for ponding conditions. In a specific soil, the following parameters are known: fo = 35 (mm/hr) fc = 6 (mm/hr) k = 2 (hr 1) %3D What is the infiltration capacity after 15,30, and 60 min? How much water has infiltrated after 15, 30, and 60 min?arrow_forwardQ. Solve this seepage problem.arrow_forward← Darcy' Law A sand sample of 35 m² cross sectional area and 20 cm long was tested in a constant head permeameter. Under a head of 60 km, the discharge was 120 ml in 6 min. The dry unit weight of sand used for the test was 1120 g. and GS = 2.68 Take Sw = 1 g/ cm³ The hydraulic conductivity (in cm/sec) is × 10-³ Xarrow_forward
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