8.5 A small reservoir has a spillway crest at elevation 200.00 m. Above this elevation, the storage and outflow from the reservoir can be expressed as Storage: S=36000 - 18000 y (m³) Q=10y (m³/s) Outflow: where y= height of the reservoir level above the spillway crest in m. Route an inflow flood hydrograph which can be approximated by a triangle as 1=0 at 1=0 h 1-30 m³/s at 1-6 h (peak flow) 1=0 at 1=26h (end of inflow). Assume the reservoir elevation as 200.00 m at r=0 h. Use a time step of 2 h.

8.5 A small reservoir has a spillway crest at elevation 200.00 m. Above this elevation, the storage and outflow from the reservoir can be expressed as Storage: S=36000 - 18000 y (m³) Q=10y (m³/s) Outflow: where y= height of the reservoir level above the spillway crest in m. Route an inflow flood hydrograph which can be approximated by a triangle as 1=0 at 1=0 h 1-30 m³/s at 1-6 h (peak flow) 1=0 at 1=26h (end of inflow). Assume the reservoir elevation as 200.00 m at r=0 h. Use a time step of 2 h.

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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Highway Drainage

Highway drainage is the process of extracting and regulating the excess surface water on the pavement by giving the desired slope. The construction of a highway drain helps to divert the water present on the surface, and sub-grade by the means of slope. Highway drainage systems are a necessary part of smooth functioning of traffic, since the highway drainage systems provide proper disposal of unwanted substances from the highways.

Question

8.5 A small reservoir has a spillway crest at elevation 200.00 m. Above this elevation, the
storage and outflow from the reservoir can be expressed as
Storage:
Outflow:
where
$=36000 - 18000 y (m²)
Q=10y (m³/s)
y = height of the reservoir level above the spillway crest in m.
Route an inflow flood hydrograph which can be approximated by a triangle as
I=0 at 1=0 h
I-30 m³/s at 1 6 h (peak flow)
I=0 at 1 26 h (end of inflow).
Assume the reservoir elevation as 200.00 m at t=0 h.
Use a time step of 2 h.

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