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الكلية كلية الهندسة     القسم  الهندسة المدنية     المرحلة 7
أستاذ المادة كاظم نايف كاظم اليساري       10/07/2018 06:07:47
1- Basic Hydrologic cycle
The main processes in the hydrologic cycle are:
1- Precipitation (rainfall) P
2- Runoff (surface) R
3- Transpiration (from plants) T
4- Evaporation E
5- Infiltration F
6- Groundwater flow G
1-1 Water budget equation
For any hydrologic system, a water budget can be developed to account for various flow
pathways and storage components. The simplest system is an impervious inclined plan, confined
on all four sides with a single outlet. The hydrologic continuity equation for any system is:
dt
Q Qout dS in ? = ………. (1-1)
Where:
Qin inflow rate (volume / time)
Qout outflow rate (volume / time)
dt
dS Change in storage in (volume / time)
You can think of inflow in the form of (Precipitation),When dealing with the surface hydrology, infiltration “I” is considered as “loss”, but when
dealing with subsurface hydrology, infiltration “I” is considered as a gain to the ground water
surface, which is called “recharge”.
2 Prepared by Amr A. El Sayed, aelsayed@vt.edu
1-3 Watershed boundaries
Watershed
A watershed is defined as an area of land that drains to a single outlet and is separated
from other watersheds by a watershed divide.
Base Flow
The channel or the stream may contain a certain amount of “Base Flow” coming from
groundwater and soil contribution even if there is no rainfall, and that amount of water appeared
in the gage devices when estimating the runoff into channels or streams. Discharge from rainfall
excess, after losses have been subtracted, makes up the direct runoff hydrograph. The total runoff
hydrograph is direct runoff + Base flow.
3 Prepared by Amr A. El Sayed, aelsayed@vt.edu
4 Prepared by Amr A. El Sayed, aelsayed@vt.edu
Water balance unit conversion:
Each process in the hydrologic cycle may be presented in the same flow rate units
(Volume / time), and when this volume is spread over a certain area (like watershed), one can use
depth units. The depth units represent a volume of water when multiplied by the surface area of
the watershed.
Length units:
1.0 inch = 2.54 cm 1 ft = 12 inches 1 yard = 3 feet
1 mile = 1.6093 km 1 ft = 30.48 cm 1 mile = 5280 ft
Area units:
1 acre = 4047 square meters 1 acre = 43560 ft2 1 mile2 = 640 acre
1 hectare (ha) = 2.471 acre 1 ha = 10000 m2
1- Volume:
1 cfs-hr = 3600 ft3
1 cfs.day = 1 * 24 * 60 * 60 = 86400 ft3
2- Depth:
The volume is expressed as a depth of water over a certain area, thus:
Volume = Depth of water * Area.
1 cfs.hr = 3600 ft3, this volume is spread over an area of 1 acre.
depth =
ft
inches
ft
ft acre *12
43560
3600 * 1 2
3 = 0.9917 in ? 1.0 inch over 1.0 acre
1 cfs.hr = 1.0 acre-inch
1 cfs.day = 86400 ft3 (over an area of 1.0 acre = 43560 ft2)
depth = 1.9834 2.0
43560
86400 = ft ? ft over 1.0 acre
1 cfs.
And the outflow (Runoff, Groundwater flow, Evaporation, and Transpiration)
P – R – G – E – T = ? S ………. (1-2)
Note:
When

المادة المعروضة اعلاه هي مدخل الى المحاضرة المرفوعة بواسطة استاذ(ة) المادة . وقد تبدو لك غير متكاملة . حيث يضع استاذ المادة في بعض الاحيان فقط الجزء الاول من المحاضرة من اجل الاطلاع على ما ستقوم بتحميله لاحقا . في نظام التعليم الالكتروني نوفر هذه الخدمة لكي نبقيك على اطلاع حول محتوى الملف الذي ستقوم بتحميله .