Chapter one introduction to hydrology
- 1. KABRI-DAHAR UNIVERSITY COLLEGE OF ENGINEERING AND TECHNOLOGY DEPARTMENT OF CIVIL COURSE :- ENGINEERING HYDROLOGY BY ENG. MAHAD
- 2. CHAPTER ONE INTRODUCTION CONTINENT • Introduction • Hydrologic Cycle &Its Component • Application In Engineering • Water Budget Equation
- 3. Introduction • Hydrology Means The Science Of Water. It Is The Science That Deals With The Occurrence, Circulation And Distribution Of Water Of The Earth And Earth's Atmosphere. • In A General Sense, Hydrology Is A Very Broad Subject Of An Inter-disciplinary Nature Drawing Support From Allied Sciences, Such As Meteorology, Geology, Statistics, Chemistry, Physics And Fluid Mechanics. • Hydrology Is Basically An Applied Science. To Further Emphasize The Degree Of Applicability, The Subject Is Sometimes Classified AS • Scientific Hydrology—the Study Which Is Concerned Chiefly With Academic Aspects. • Engineering Or Applied Hydrology—a Study Concerned With Engineering Applications.
- 4. Cont…. In A General Sense Engineering Hydrology Deals With I. Estimation Of Water Resources, II. The Study Of Processes Such As Precipitation ,Run Off, Evapotranspiration And Their Interaction And III.The Study Of Problems Such As Floods And Droughts, And Strategies To Combat Them.
- 5. HYDROLOGIC CYCLE Water Occurs On The Earth In All Its Three States, Viz. Liquid, Solid and Gaseous, and In Various Degrees Of Motion. Evaporation Of Water From Water Bodies Such As Oceans and Lakes, Formation and Movement Of Clouds, Rain and Snowfall, Stream Flow and Ground Water Movement are Some Examples Of The Dynamic Aspects Of Water. The Various Aspects Of Water Related To The Earth Can Be Explained In Terms of a Cycle Known as The Hydrologic Cycle.
- 7. WATER BUDGET EQUATION CAТCНМENТ AREA The Area Of Land Draining Into A Stream Or A Water Course At A Given Location Is Known As Catchment Area. It Is Also Called As Drainage Area Or Drainage Basin. In USA, it Is Known As Watershed. A Catchment Area Is Separated Form Its Neigh bouring Areas By A Ridge Called Divide In USA And Watershed In UK (Fig. 1.3). The Areal Extent Of The Catchment Is Obtained By Tracing The Ridge On A Topographic Map To Delineate The Catchment And Measuring The Area By A Planimeter. It Is Obvious That For A River While Mentioning The Catchment Area The Station To Which It Pertains (Fig.1.3) Must Also Be Mentioned. It Is Normal To Assume The Ground Water Divide To Coincide With The Surface Divide.
- 8. Schematic Sketch Of Catchment Of River AAt Station M
- 9. WAТER BUDGEТ EQUAТION For A Given Problem Area, Say A Catchment, In An Interval Of Time ∆T, The Continuity Equation For Water In Its Various Phases Is Written As Mass inflow – mass outflow = change in mass storage If the density of the inflow, outflow and storage volumes are the same Vi – V0= ∆S Where Vi = inflow volume of water into the problem area during the time period, V0= Outflow Volume of Water From The Problem Area During The Time Period, And ∆S = Change In The Storage of The Water Volume over And Under The Given Area During The Given period. While realizing that all the terms in a hydrological water budget may not be known To the same degree of accuracy ,an expression for the water budget of a catchment for a Time Interval ∆T Is Written As P–r–g –E–t =∆S In This P = Precipitation, R = Surface Runoff, G = Net Groundwater Flow Out Of The Catchment, E = Evaporation, T = Transpiration And ∆S = Change In Storage.
- 10. CONT…. The storage s consists of three components as S = Ss + Ssm + Sg Where Ss = surface water storage Ssm = water in storage as soil moisture and Sg = water in storage as ground water. Thus in Eq. (1.2-a) ∆s = ∆Ss + ∆ssm + ∆Sg All terms in eq.(1.2-a) have the dimensions of volume Note that all these Terms Can Be Expressed As Depth Over The Catchment Area (E.G. In Centimeters), And In Fact This Is A Very Common Unit. In terms of rainfall – run off relationship, eq.(1.2-a) can be represented as R = p – l (1.2-b) Where l = losses = water not available to run off due to infiltration (causing addition To soil moisture and ground water storage), evaporation, transpiration and surface storage. Details Of Various Components Of The Water Budget Equation Are Discussed In Subsequent Chapters. Note That In Eqs (1.2-a And B) The Net Import Of Water Into The Catchment, from sources outside the catchment, by action of man is assumed to be zero.
- 11. CONT…. EXAMPLE I.I A lake had a water surface elevation of 103.200 m above datum at the beginning of a certain month. in that month the lake received an average inflow of 6.0 m3 /s from surface runoff sources. in the same period the outflow from the lake had an average value of 6.5m3 /s. further, in that month ,the lake received a rainfall of 145 mm and the evaporation from the lake surface was estimated as 6.10 cm. write the water budget equation for the lake and calculate the water surface elevation of the lake at the end of the month. the average lake surface area can be taken as 5000 ha. assume that there is no contribution to or from the groundwater storage. SOLUTION: in a time interval ∆t the water budget for the lake can be written as input volume – output volume = change in storage of the lake (I∆T +PA)– (Q∆T +EA)=∆S Where I =average rate of inflow of water in to the lake, q=average rate of outflow from the lake ,P=precipitation, E=evaporation, A=average surface area of the lake and ∆s = change in storage volume of the lake. Here ∆t =1month=30×24×60 ×60=2.592 ×106 s=2.592ms in one month: inflow volume = I ∆t = 6.0 × 2.592 = 15.552 mm3 outflow volume = Q∆t= 6.5×2.592 = 16.848 mm3
- 12. CONT… input due to precipitation= Output due to evaporation = Hance ∆S = 15.552+7.25–16.848 –3.05 = 2.904 Mm3 change in elevation = new water surface elevation at the end of the month =103.200+0.058 = 103.258 m above the datum.
- 13. Cont.. Example 1.2 a small catchment of area 150 ha received a rainfall of 10.5 cm in 90 minutes due to astorm.at the outlet of the catchment, the stream draining the catchment was dry before the storm and experienced a runoff lasting for 10 hours with an average discharge of 1.5 m3 /s. the stream was again dry after the runoff event. (a) what is the amount of water which was not available to runoff due to combined effect of infiltration, evaporation and transpiration? what is the ratio of runoff to precipitation? SOLUTIOH: the water budget equation for the catchment in a time ∆t is R= P– L (1.2-b) where L = losses = water not available to runoff due to infiltration (causing addition to soil moisture and groundwater storage), evaporation, transpiration and surface storage. in the present case ∆t = duration of the runoff = 10 hours. Note that the rainfall occurred In the first 90 minutes and the rest 8.5hours the precipitation was zero.
- 14. Cont… a) P = input due to precipitation in 10 hours =150×100×100×(10.5/100)=157,500m3 R = runoff volume = outflow volume at the catchment outlet in10 hours = 1.5 ×10×60 ×60= 54,000 m3 Hence Losses L=157,500 –54,000 =103,500m3 b) runoff/rainfall = 54,000/157,500 = 0.343 (this ratio is known as runoff coefficient
- 15. APPLICATIONS INENGINEERING • Hydrology finds its greatest application in the design and operation of water-resources engineering projects, such as those for (i)irrigation, (ii) water supply, (iii) flood control, (iv)water power, and (v) navigation. In all these projects hydrological investigations for the proper assessment of the following factors are necessary: 1) The capacity of storage structures such as reservoirs. 2) The magnitude of flood flows to enable safe disposal of the excess flow. 3) The minimum flow and quantity of flow available at various seasons. 4) The interaction of the flood wave and hydraulic structures, such as levees, reservoirs, barrages and bridges.
- 16. Cont… The hydrological study of a project should necessarily precede structural and other detailed design studies. It involves the collection of relevant data and analysis of the data by applying the principles and theories of hydrology to seek solutions to practical problems. Many important projects in the past have failed due to improper assessment of the hydrological factors. Some typical failures of hydraulic structures are: (i) Overtopping and consequent failure of an earthen dam due to an inadequate spillway capacity, (ii) failure of bridges and culverts due to excess flood flow and (iii) Inability of a large reservoir to fill up with water due to overestimation of the stream flow. Such failure, often called hydrologic failures under score the uncertainty aspect inherent in hydrological studies.
- 17. SOURCESOFDATA Depending up on the problem at hand, a hydrologist would require data relating to the various relevant phases of the hydrological cycle playing on the problem catchment. the data normally required in the studies are: Weather rerecords—temperature, humidity and wind velocity Precipitation data Stream flow records Evaporation and evapotranspiration data Infiltration characteristics of the study area Soils of the area Land use and land cover Ground water characteristics Physical and geological characteristics of the area Water quality data