i) A pipe of 1.2 m diameter was provided in a reservoir to act as an outlet. Due to disuse, it was buried and completely clogged up for some length by sediment. Measurements indicated the presence of fine sand (coefficient of permeability, K, = 10 m/day) deposit for a length of 100 m, at the upstream end and of coarse sand (coefficient of permeability, K₂ = 50 m/day) at the downstream end for a length of 50 m. In between these two layers the presence of silty sand (coefficient of permeability, K3 = 0.10 m/day) for some length is identified. For a head difference of 20 m on either side of the clogged length, the seepage discharge is found to be 0.8 m³/day. Estimate the length of the pipe filled up by silty sand.

i) A pipe of 1.2 m diameter was provided in a reservoir to act as an outlet. Due to disuse, it was buried and completely clogged up for some length by sediment. Measurements indicated the presence of fine sand (coefficient of permeability, K, = 10 m/day) deposit for a length of 100 m, at the upstream end and of coarse sand (coefficient of permeability, K₂ = 50 m/day) at the downstream end for a length of 50 m. In between these two layers the presence of silty sand (coefficient of permeability, K3 = 0.10 m/day) for some length is identified. For a head difference of 20 m on either side of the clogged length, the seepage discharge is found to be 0.8 m³/day. Estimate the length of the pipe filled up by silty sand.

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

See similar textbooks

Similar questions

glass container with pervious bottom containing fine sand in loose state (void ratio 0.8) is subjected to hydrostatic pressure from underneath until quick condition occurs in the sand. If the specific gravity of sand particles -2.65, area of cross-section of sand sample 10 cm² and height of sample = 10 cm, compute the head of water required to cause quicksand condition and also the seepage force acting from below. Seepag
For a constant head laboratory permeability test on a fine sand, the following are given:Length of specimen = 250 mmDia. of specimen = 62.5 mmHead difference = 450 mmWater collected in 2 min. = 484 cu.mmDetermine the discharge velocity in mm/sec.
Value of hydraulic conductivity of a sand is 0.010 cm/sec at a porosity of 0.45. Determine the hydraulic conductivity of sand at a porosity of 0.50?
The result of the liquid limit tests where N is the no of blows and w is the water content in % are given as follows: Trial 1 N=11 w=53.9; Trial 2 N=15 w=50.6; Trial 3 N=23 w=48.1; Trial 4 N= 30 w= 46; Trial 5 N= 46 w=43.3; Trial 6 N= 53 w=41. The PL= 30 and the in situ moisture content is 40%. Determine the liquidity index. a.0.56 b.0.59 c.0.91 d.1.25
Show solution 4
A vertical sand column of length 100 cm and cross section area of 150 cm² is allowing the water to flow downward. The head difference was found to be 110 cm. The soil has a hydraulic conductivity of 20 mld and void ratio was taken as 0.5. Calculate the (a) rate of flow, (b) the Darcy velocity and (c) the average velocity.
Geotechnical engineering
solutionn.
The results of a constant head permeability test for a fine sand are as follows: Diameter of the sample = 35 cm Length of sample = 83 cm Constant head difference = 59 cm Time of collection = 310 secs Weight of water collected = 372 grams Find the seepage velocity in cm/min. if the void ratio is 0.48. Round off to four decimal places.
8. The lower sand layer in the soil profile is in artesian condition. Calculate the maximum depth z of the proposed excavation. Standpipe 3.0 m 0.0 m 9 WT Sand Pary 1.70 Mg/m Paar-2.10 Mg/m WT 2.0 m Sand 4.0 m Clay Peat 2.10 Mg/m Clay 8.0 m Sand Sand (b) (a) Fig P-63D B. 2. 43 m C. 2.58m D. 2.76m A. 2.1 m
The result of the liquid limit tests where N is the no of blows and w is the water content in % are given as follows: Trial 1 N=11 w=53.9; Trial 2 N=15 w=50.6; Trial 3 N=23 w=48.1; Trial 4 N= 30 w= 46; Trial 5 N= 46 w=43.3; Trial 6 N= 53 w=41. The PL= 20 and the in situ moisture content is 45%. Determine the plasticity index. a.22 b.17 c.27 d.12
Problem attached