A water treatment plant treats 6000 m³ of water per day. As a part of the treatment process, discrete particles are required to be settled in a clarifier. A column test indicates that an overflow rate of 1.5 m per hour would produce the desired removal of particles through settling in the clarifier having a depth of 3.0 m. The volume of the required clarifier, (in m³, round off to 1 decimal place) would be
A water treatment plant treats 6000 m³ of water per day. As a part of the treatment process, discrete particles are required to be settled in a clarifier. A column test indicates that an overflow rate of 1.5 m per hour would produce the desired removal of particles through settling in the clarifier having a depth of 3.0 m. The volume of the required clarifier, (in m³, round off to 1 decimal place) would be
Principles of Geotechnical Engineering (MindTap Course List)
9th Edition
ISBN:9781305970939
Author:Braja M. Das, Khaled Sobhan
Publisher:Braja M. Das, Khaled Sobhan
Chapter8: Seepage
Section: Chapter Questions
Problem 8.1CTP
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![A water treatment plant treats 6000 m³ of water per day. As a part of the treatment
process, discrete particles are required to be settled in a clarifier. A column test indicates
that an overflow rate of 1.5 m per hour would produce the desired removal of particles
through settling in the clarifier having a depth of 3.0 m. The volume of the required clarifier,
(in m³, round off to 1 decimal place) would be](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0788f705-72f0-4f63-9631-3e863f4e0eef%2Ff8d327d8-284a-477e-8950-f536881ae333%2Fjqpehen_processed.jpeg&w=3840&q=75)
Transcribed Image Text:A water treatment plant treats 6000 m³ of water per day. As a part of the treatment
process, discrete particles are required to be settled in a clarifier. A column test indicates
that an overflow rate of 1.5 m per hour would produce the desired removal of particles
through settling in the clarifier having a depth of 3.0 m. The volume of the required clarifier,
(in m³, round off to 1 decimal place) would be
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