Qr :An inverse conical tank with adiameter of (2 m) and height of (6 m)filled with water. A hole has a radius of(0.02 m) in the center of the base. After atime of 100 s the depth at the tankbecame 3.847 m. the depths through thetime t = o (25) 100 seconds are:O y1 = 2.112, y2= 2.012, y3= 2.001O y1 = 5.913 y2= 5.578 y3= 5.216O y1 = 4.587, y2= 4.324 , y3= 4.034y1 = 4.945, y2= 4.407, y3= 4.285

Given:- d=2m h=6m t=100s To find:- depth

Answered: center of the he depth at the m. the…

Engineering

Civil Engineering

center of the he depth at the m. the depths

Fundamentals of Geotechnical Engineering (MindTap Course List)

5th Edition

ISBN:9781305635180

Author:Braja M. Das, Nagaratnam Sivakugan

Publisher:Braja M. Das, Nagaratnam Sivakugan

Chapter8: Stresses In A Soil Mass

Section: Chapter Questions

Problem 8.14P

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Concept explainers

Open Channel Flow

Open channel flow is a branch of fluid mechanics. It deals with the flow of fluid in an open conduit or free surface. In an open channel flow, gravity induces the flow of water whose surface is open to the air. This flow is free-falling rather than confined in rigid body conduits. The open channel flow depends upon two parameters that are the cross-section of the conduit and the velocity profile. Practical examples of open channel flow are streams, rivers, canals, and other drainage systems.In open channel flow, the cross-sectional area of at least one of the sides is exposed to air, and the surface of water faces atmospheric pressure. This flow is also known as free gravity flow. The cross-sectional area of the conduit can be either man-made or a natural channel. For instance, the flow in rivers and streams are natural conduits, and the flow in a canal and sanitary systems are man-made or engineered channels.

Question

Qr :An inverse conical tank with a
diameter of (2 m) and height of (6 m)
filled with water. A hole has a radius of
(0.02 m) in the center of the base. After a
time of 100 s the depth at the tank
became 3.847 m. the depths through the
time t = o (25) 100 seconds are:
O y1 = 2.112, y2= 2.012, y3= 2.001
O y1 = 5.913 y2= 5.578 y3= 5.216
O y1 = 4.587, y2= 4.324 , y3= 4.034
y1 = 4.945, y2= 4.407, y3= 4.285

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