1. 500 x 100 m² of water has been stored in the reservoir of a dam at the beginning ofNovember 2008. In this month the river feeds the reservoir with average 20 m³/sinflow. Monthly evaporation from the dam lake surface is 9 x 106 m³ and noprecipitation has been observed during this month," 50 x 106 m³ of water has beenextracted from the reservoir for the nearby town consumption. The reservoir volumehas been measured as 480 x 106 m³ at the end of November..14-Check if there is any infiltration to the soil from the reservoir bottom, if there is, computethe volume of the infiltrated water.20 km².

Given, Volume of water in reservoir at the start of november month=500×106m3Volume of water in…

Answered: 1. 500 x 100 m³ of water has been…

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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1. 500 x 100 m² of water has been stored in the reservoir of a dam at the beginning of
November 2008. In this month the river feeds the reservoir with average 20 m³/s
inflow. Monthly evaporation from the dam lake surface is 9 x 106 m³ and no
precipitation has been observed during this month," 50 x 106 m³ of water has been
extracted from the reservoir for the nearby town consumption. The reservoir volume
has been measured as 480 x 106 m³ at the end of November..
14
-
Check if there is any infiltration to the soil from the reservoir bottom, if there is, compute
the volume of the infiltrated water.
20 km².

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Given,

$\begin{array}{rcl} V o l u m e o f w a t e r i n r e s e r v o i r a t t h e s t a r t o f n o v e m b e r m o n t h & = & 500 \times 10^{6} m^{3} \\ V o l u m e o f w a t e r i n r e s e r v o i r a t t h e e n d o f n o v e m b e r m o n t h & = & 480 \times 10^{6} m^{3} \\ A v e r a g e i n f l o w f r o m r i v e r & = & 20 m^{3} / s \\ M o n t h l y e v a p o r a t i o n f r o m d a m l a k e s u r f a c e & = & 9 \times 10^{6} m^{3} \\ P r e c e p i t a t i o n i n r e s e r v o i r & = & 0 m^{3} \\ W a t e r u s e d b y n e a r b y t o w n & = & 50 \times 10^{6} m^{3} \end{array}$

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