H2 = 0.3’ h = 1.25’ H1 = 0.2’ d = 1.5” H = 0.5’ D = 4” Ksand = 0.000075 ft/min Ksilt = 0.0038 ft/min Determine: Determine the total flow of water q. Find the equivalent permeability for the soil.

H2 = 0.3’ h = 1.25’ H1 = 0.2’ d = 1.5” H = 0.5’ D = 4” Ksand = 0.000075 ft/min Ksilt = 0.0038 ft/min Determine: Determine the total flow of water q. Find the equivalent permeability for the soil.

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Concept explainers

Field applications of hydraulic conductivity

Hydraulic conductivity is one of the hydraulic properties of soil defined as a measurement of the ability of a porous medium to permit water movement or fluid flow through it. It is denoted by the letter K. It is influenced by the density and viscosity of the fluid. It is also called permeability. Hydraulic conductivity is defined as the ratio of velocity to the hydraulic gradient. The different methods used to determine the hydraulic conductivity are laboratory methods (constant head permeability tests and falling head tests), in-situ field methods (pumping in and pumping out tests), small scale field tests (infiltration tests and slug tests), and indirect methods such as estimation from grain size, consolidation test data, and horizontal capillary test. The pedotransfer function (PTF) method is a specific empirical estimating approach employed in the soil sciences is also used to determine the hydraulic conductivity of the soil.

Question

H₂ = 0.3’ h = 1.25’

H₁ = 0.2’ d = 1.5”

H = 0.5’ D = 4”

K_sand = 0.000075 ft/min K_silt = 0.0038 ft/min

Determine:

Determine the total flow of water q.
Find the equivalent permeability for the soil.

h
sand
porous stone
Inflow
Id-
D
water
silt
H₂
H₁
H
water

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