From the figure shown, the thickness of permeable layer is 1.1m making an angle of 14 degrees with the horizontal. K= 4.87x10^-12 cm/sec. If the vertical thickness depth of the aquifer at point of the first piezometer (left) is 1.5m and the other point at second piezometer is 1.1m, determine the seepage velocity. h=1.4 m Direction of seepage impervious layer 3 m 14 1.1 m impervious layer 14 -36 m-

From the figure shown, the thickness of permeable layer is 1.1m making an angle of 14 degrees with the horizontal. K= 4.87x10^-12 cm/sec. If the vertical thickness depth of the aquifer at point of the first piezometer (left) is 1.5m and the other point at second piezometer is 1.1m, determine the seepage velocity. h=1.4 m Direction of seepage impervious layer 3 m 14 1.1 m impervious layer 14 -36 m-

Principles of Foundation Engineering (MindTap Course List)

9th Edition

ISBN:9781337705028

Author:Braja M. Das, Nagaratnam Sivakugan

Publisher:Braja M. Das, Nagaratnam Sivakugan

Chapter2: Geotechnical Properties Of Soil

Section: Chapter Questions

Problem 2.8P

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From the figure shown, the thickness of permeable layer is 1.1m making an angle of 14 degrees with the horizontal. K= 4.87x10^-12 cm/sec. If the vertical thickness depth of the aquifer at point of the first piezometer (left) is 1.5m and the other point at second piezometer is 1.1m, determine the flow rate. h=1.4 m Direction impervious layer of seepage 3 m 14 1.1 m impervious layer L. 14
Question: A Blasius Exact Solution Equation For A Laminar Flat-Plate Boundary Layer Problem Which... A Blasius exact solution equation for a laminar flat-plate boundary layer problem which derived from Navier-Stoke equations, could be written as, ff"-2f"'=0 where f = f(n), =d'fm), dn? d'f(n). df (N) =- ,and f'=- dn' dn U is mainstream fluid velocity and u is fluid velocity in boundary layer. With suitable boundary conditions, the above equation had been solved by 4th-order Runge-Kutta numerical integration and the result is tabulated in Table Qla. TABLE Qla f'=w/U f" 7 = y, VX 0.000 0.000 0.332 0.323 0.267 1 0.166 0.330 0.630 0.846 0.956 0.992 0.998 0.650 1.397 2.310 3.283 0.161 0.064 3 4 0.016 0.002 6 7 4.280 0.001 5.279 6.280 0.999 1.000 0.000 Show that : CoRe , = 1.27 %3D
The figure shows a cross section through the strata underlying a site of which the permeability properties are of importance. Assume the coefficient of permeability in the horizontal and vertical directions to be the same in each stratum. (each layer is hydraulically isotropic). Estimate the ratio of equivalent hydraulic conductivity: kh(eq)/kv(eq) A: h=4.1 m, k=11x10^-4 mm/s B: h=1.49 m, k=69x10^-5 mm/s C: h=1.16 m, k=36x10^-7 mm/s D: h=2.91 m, k=51x10^-10 mm/s
A confined aquifer of 15 m constant thickness is sandwiched between two aquicludes as shown in the figure (not drawn to scale) Piezometer P 55.2 m Piezometric Surface Aquiclude Aquifier: Piezometer Q Aquiclude 34.1 m L = 2500m The heads indicated by two piezometers P and Q are 55.2 m and 34.1 m, respectively. The aquifer has a hydraulic conductivity of 80 m/day and its effective porosity is 0.25. If the distance between the piezometers is 2500 m, the time taken by the water to travel through the aquifer from piezometer location P to Q (in days, round off to 1 decimal place) is
Refer to Figure P3.3. Use Eqs. (3.10) and (3.11) to determine the variation of OCR and preconsolidation pressure c. FIGURE P3.3
2. A confined aquifer is recharged by a continuous supply of water from the certain source. The average thickness of the aquifer was determined to be 25 m and the average width is 4 km. The hydraulic conductivity of the aquifer was obtained at 40 m/day and its void ratio is 0.3333. The piezometer heads in two wells 1.325 km apart are 65 meters and 60 meters from a common reference datum. Determine the rate of flow in cubic meters per day. Determine: •compute the rate of flow through the aquifer in m3/day •compute the seepage velocity in m/day •compute the time of travel from the head of the aquifer to a point 8 km downstream in days
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(Situation 3) Determine the equivalent hydraulic conductivity of the soil strata given in the figure considering direction of flow of water parallel and perpendicular to the layers. k = 2.3 x 102 cm/s k, = 8 x 102 cm/s k, = 5.7 x 10 cm/s, k, = 25.5 x 104 cm/s 10 m 2 m! = 9.2 x 10 cm/s k, = 27 x 107 cm/s 10 m %3!
Find out the change in length in a rock mass. while the strain recorded is 3% and the original length of that mass is 4 cm. Q:5 b) How the property of permeability generates in soil and find out the hydraulic conductivity of a rock when change in height is 6 cm and the original height of the sample is 8 cm. the volume recorded in outlet is 20cm^3 in 20 min. the diameter of intact saturated rock mass is 5 cm. *
Problem 4: An inclined permeameter tube is filled with layers of soil of different permeability as shown below. Find the total head, elevation head and pore water pressure at points (A-B-C-D) with respect to the given datum, Given: 1.5k1=2k2=3k3=k4 k4=2x10-3 cm/sec 35 сm 9 cm K4 K3 K1 K2 15 cm 11 cm 7 cm 4 cm Datum شاشة 4cm
The upstream pressure on a rock sample is 0.113MPa and the downstream pressure is 0.1MPa. Under this pressure difference, a gas with the viscosity of 0.018 mPa·s flows through the rock sample. The diameter of this sample is 2.5cm, its length is 2.8cm. The volumetric flow rate at the outlet face is 0.514cm3/s. Calculate the gas permeability of this rock sample.
Question 4.9 A rock consists of six layers of different permeabilities. The geometry of the rock is given in the figure below. The permeability and dimensions of each layer are given in the table below. a) Find the average permeability of this rock for horizontal flow (k,). b) Find the average permeability of this rock for vertical flow (k). c) What is the k /k,? 1 2 4 6 Layer Permeability [mD] Thickness [in] Length [in] Width [in] 1 50 30 10 2 60 3.5 9. 10 3 30 1.7 21 10 4 40 1.8 21 10 20 2.5 18 10 10 2.5 12 10